Differential Geometry, Wang Zuoqin USTC 2021

Prévia do material em texto

Don’t just read it; fight it!
Ask your own questions,
look for your own examples,
discover your own proofs.
Is the hypothesis necessary?
Is the converse true?
What happens in the classical special case?
What about the degenerate cases?
Where does the proof use the hypothesis?
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Course Information
♣ About the course:
• Instructor: Zuoqin Wang
– Email: wangzuoq@ustc.edu.cn
– Office: 1601
• TAs:
– HONG Fang, hongf@mail.ustc.edu.cn
– HUANG Shuping, hsp@mail.ustc.edu.cn
– CHEN Hengyu, hichychen@mail.ustc.edu.cn
Question Answering: TBA
• Lecture time/room: Tu 7:50-9:25 am & Th 15:55-17:30 pm @ 1102
• Webpage: http://staff.ustc.edu.cn/∼wangzuoq/Courses/21F-Manifolds/
• PSets: Will be posted every week on the course webpage, and will be collected
every two weeks (on Tuesdays before class).
• Exams: There will be one midterm and one final exam.
• Course grades:
– Regular: HWs (30%), Midterm (30%), Final (40%)
– To get A+: An additional term paper is required. (More details after
midterm)
♦ Notes and Reference books:
Course Notes will be uploaded to the course webpage after each lecture.
We will not follow any single book. The following are some nice reference books:
• Introduction to Smooth Manifolds, 2nd ed, by John Lee
• An Introduction to Manifolds, 2nd ed, by Loring W. Tu
• Differential Topology, by Victor Guillemin and Alan Pollack
♥ Prerequisites:
• Basic Analysis: Ck maps, multiple integrals, the inverse and implicit function
theorems, existence and uniqueness theory for ODEs
——c.f. Appendix C and Appendix D in Lee’s book
• Basic Topology: Topological spaces, quotient spaces, connectedness, compact-
ness, Hausdorff, second countable, continuity, proper
——c.f. Appendix A in Lee’s book
• Basic Algebra: Linear spaces, direct sums, inner products, linear transforma-
tions, matrices, groups, quotient groups
——c.f. Appendix B in Lee’s book
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♠ Contents:
Smooth manifolds are nice geometric objects on which one can do analysis: they
are higher dimensional generalizations of smooth curves and smooth surfaces; they
appear as the solution sets of systems of equations, as the phase spaces of many physics
systems, etc. They are among the most important objects in modern mathematics and
physics.
In this course we plan to cover
• Basic theory: definitions, examples, structural theorems etc.
– Smooth manifolds and submanifolds
– Smooth maps and differentials
– Vector fields and flows
– Basic structure theory (differential topology)
– Manifolds with special structures: Lie groups, vector/fiber bundles
• Geometry of differential forms
– Tensors, differential forms
– Exterior derivatives and integrations of differential forms
– de Rham cohomology and applications
– (If time permits) Geometric structures related to differential forms: Rie-
mannian/symplectic structures, Chern-Weil
LECTURE 1: TOPOLOGICAL MANIFOLDS
1. Review of Topology
¶ Topology.
Recall that a topology on a set X is a collection T of subsets of X whose elements
are called open sets, such that
(1) The set X itself and the empty set ∅ are open sets.
(2) Any union of open sets is an open set.
(3) Any finite intersection of open sets is an open set.
As usual, the complement of an open set is called a closed set. There are many natural
ways to construct new topological spaces from old ones. For example,
• If T is a topology on X, and Y ⊂ X, then
TY = {U ∩ Y | U ∈ T }
is a topology on Y , which is known as the induced subspace topology.
• If TX and TY are topologies on X and Y respectively, then
TX×Y = {W ⊂ X × Y | ∀(x, y) ∈ W,∃U ∈ TX , V ∈ TY s.t. (x, y) ∈ U × V ⊂ W.}
is a topology on X × Y , which is known as the product topology
• If (X,TX) is a topological space, and ∼ is an equivalence relation on X, then
TX/∼ = {V ⊂ X/∼ | π−1(V ) ∈ TX}
is a topology on the quotient X/∼, which is known as the quotient topology.
¶ Topological properties.
We have studied many topological properties in undergraduate topology course.
Here we only mention the most important ones. Recall that a topological space X is
• Hausdorff (i.e. T2) if for any x 6= y ∈ X, there exist open sets U 3 x and V 3 y
so that U ∩ V = ∅.
• second-countable (i.e. A2) if there exists a countable sub-collection T0 of T so
that any open set is a union of (not necessarily finite) open sets in T0.
• compact if for any open covering {Uα} of X, there exists a finite sub-collection
{Uα1 , · · · , Uαk} so that X = Uα1 ∪ · · · ∪ Uαk .
4
LECTURE 1: TOPOLOGICAL MANIFOLDS 5
• disconnected if there exist two non-empty open sets U1 and U2 in X so that
U1 ∪ U2 = X and U1 ∩ U2 = ∅.
X isconnected if it is not disconnected. (If X is disconnected, then any maximal
connected subset of X is called a connected component of X.)
• path-connected if for any p, q ∈ X, there is a continuous map f : [0, 1]→ X so
that f(0) = p and f(1) = q. Such a map is called a path from p to q.
Unless otherwise stated, all spaces we are going to study in this course will be Haus-
dorff and second countable. Note that if X is Hausdorff or second-countable, then
any subspace of X is also Hausdorff or second-countable with respect to the induced
subspace topology. Similarly the products of Hausdorff or second-countable spaces will
remain to be Hausdorff or second-countable. However, one should be aware that in
general, the quotient of a Hausdorff or second-countable space need not be Hausdorff
or second-countable.
The compactness and connectedness are also very important and will be frequently
used in this course. In most cases we may assume our objects to be connected, since for
disconnected ones we may simply study their connected components. But we do study
non-compact objects, although as we will see, usually the compactness assumption will
make our life much easier.
¶ Topological invariants.
Topological spaces are spaces on which one can define continuous maps. Recall
that a map f : X → Y between topological spaces is called continuous if for any open
set V in Y , the pre-image f−1(V ) is an open set in X. Two topological spaces X and
Y are homeomorphic, if there is a continuous map f : X → Y which is one-to-one and
onto, so that f−1 is also continuous. Such a map f is called a homeomorphism. Note
that (by default) any topological property is preserved under homeomorphism.
A quantity (or “mathematical objects” like groups) associated to topological s-
paces that is invariant under homeomorphisms is usually called a topological invariant.
The simplest topological invariants include the number of connected/path-connected
components, the fundamental group and the homology/homotopy groups that you will
learn in algebraic topology course. The following elementary theorem tells us that the
dimension of Euclidian spaces is a topological invariant:
Theorem 1.1 (Topological Invariance of Dimension). If U is an open set in Rn and
V is an open set in Rm, and f : U → V is a homeomorphism, then m = n.
The theorem can be easily proved for n = 1 via connectedness, and for n = 2
via the fundamental group. We will prove the theorem for all n later in this course,
using more sophisticated topological invariants: the de Rham cohomology groups. For
an elementary proof (through Brouwer’s invariance of domain theorem) without using
(co)homology or homotopy theory, see my Topology(H) notes.
6 LECTURE 1: TOPOLOGICAL MANIFOLDS
2. Topological manifolds
¶ Locally Euclidean space.
We will only study “nice” topological spaces so that we can do analysis on them.
We know how to do analysis (namely the theory of differentiation, integration etc, not
just the theory of limits which we already have on any topological space) on Euclidean
spaces which are the simplest topological spaces. Since analysis is mainly built up
locally, it is natural to study spaces that are locally Euclidean:
Definition 2.1. A topological space M is locally Euclidean (of dimension n) if for
every x ∈ M , there exists a triple {ϕ,U, V }, called a chart (around p), such that U
is an open neighborhood of x in M , V is an open subsetof Rn, and ϕ : U → V is a
homeomorphism.
Remark. There are many different charts near any given point x. In fact, suppose
(ϕ,U, V ) is a chart near x. Then
• for any open neighborhood U1 ⊂ U of x, if we denote ϕ1 = ϕ|U1 and let
V1 = ϕ(U1), then (ϕ1, U1, V1) is also a chart near x.
• for any v ∈ Rn, if we let Vv = V +{v} and let ϕv(x) = ϕ(x)+v, then (ϕv, U, Vv)
is a chart near p.
More generally, one may have overlapping charts looks like the following:
Unfortunately locally Euclidean spaces could be bad either semi-locally or globally:
• The line with two origins is locally Euclidean, but it is bad “semi-locally” in
the sense that it is not Hausdorff.
• The long line is locally Euclidean, but it is bad “globally” in the sense that it
is not second countable.
If you are interested in these two weird lines but have never seen them before, you
may consult Munkres’ book Topology for their definitions and basic properties. I don’t
want to spend too much time on bad examples in this nice course. (In fact the first
one is quite simple, while the second one is quite complicated.)
LECTURE 1: TOPOLOGICAL MANIFOLDS 7
¶ Topological manifolds.
Now we define topological manifolds. Roughly speaking, topological manifolds are
nice topological spaces that locally Euclidean and are not bad semi-locally/globally:
Definition 2.2. An n-dim topological manifold M is a topological space so that
(1) M is Hausdorff.
(2) M is second-countable.
(3) M is locally Euclidean space of dimension n.
Remark. Both the Hausdorff and the second-countable conditions are important in
defining reasonably nice geometric objects. For example, by the Hausdorff property,
the limit of a convergent sequence is unique (try to prove this!). We will prove later
that the Hausdorff property together with the second countability property imply the
existence of a partition of unity, which is a fundamental tool in studying manifolds.
Remark. The history of the notion of manifold:
• The name manifold was first introduced by B. Riemann in his inaugural lecture
“On the hypotheses which lie at the bases of geometry” at Göttingen university
on July 10, 1854. He described manifolds intuitively by realizing n-manifold as
a continuous stack of (n− 1)-dimensional manifold. This is a breakthrough in
the history of human thought: Euclidean space is no longer the unique space
(but still keeps its special role as the local model). Riemann also introduced the
concept of Riemann surface (=1 dimensional complex manifold ≈ 2 dimensional
real manifold) as a geometric foundation in the study of multi-valued functions
in complex analysis.
• In 1895 H. Poincaré published his famous paper “Analysis Situs” (together
with five supplements between 1899 and 1904) in which he studied three and
higher-dimensional manifolds. He first defined a manifold as the level set of a
continuously differentiable function (satisfying nondegeneracy hypothesis) be-
tween Euclidean spaces, and then gave a more general definition based on a
chain of manifolds which is a precursor to the modern notion of atlas.
• The current intrinsic definition of manifold was given by H. Weyl in his lecture
course on Riemann surfaces in 1911-1912, in which he utilized point set topology
to make the theory more rigorous.
• Finally during the 1930s H. Whitney showed that the intrinsic definition in
terms of charts was equivalent to Poincaré’s definition in terms of subsets of
Euclidean space.
8 LECTURE 1: TOPOLOGICAL MANIFOLDS
Remark. Also note the difference between English and French:
• In English, “manifold” refers to the space with topological (or differentiable)
structure above, while “variety” refers to spaces with an algebraic structure.
• In French, “manifold” is translated as “variety”, so that spaces with topologi-
cal/differentiable/algebraic structure are translated as topological/analytic/alge-
braic varieties respectively.
Remark. Finally a couple words on the Chinese translation: Jiang Zehan(ôLº)(also
known as Kiang Tsai-han), the founder of China’s topology, gave a very beautiful
translation of the word “manifold”, namely “6/”. In traditional Chinese literature,
the word 6/ appeared in a famous poem in Song dynasty
U/k�í§,,D6/" —©U�5�íy6
and also in a much earlier classics that dates back to ancient time,
�1��§¬Ô6/" —5´²#_#ßD6
¶ Simplest examples of topological manifolds.
The simplest examples of topological manifolds include
(1) Any set of countable many points is a topological manifold of dimension 0.
(2) The Euclidean space Rn is an n-dim topological manifold.
(3) Any open subset of Rn is a topological manifold, with the chart map ϕ the
identity map. More generally, any open subset of a topological manifold, with
the induced topology, is automatically a topological manifold. [Here we already
used the fact that any subset of a Hausdorff/second countable topological space remains to
be Hausdorff/second countable.]
• (The General Linear Group). Let M(n,R) be the set of all n× n real
matrices. Then M(n,R) is a linear space that is isomorphic to Rn2 . So
M(n,R) is a topological manifold in the natural way. A more interesting
example is the general linear group
GL(n,R) = {A ∈M(n,R) | det(A) 6= 0}.
It is an open subset in M(n,R), and thus is a topological manifold of
dimension n2. Later we will construct many other manifolds consists of
matrices. They are important examples of Lie groups.
(4) If M1 and M2 are topological manifolds of dimension n1 and n2 respectively,
then the product M1 ×M2, endowed with product topology, is a topological
manifold of dimension n1+n2. In fact, if {ϕ1, U1, V1} and {ϕ2, U2, V2} are charts
on M1 and M2 around p and q respectively, then {ϕ1 × ϕ2, U1 ×U2, V1 × V2} is
a chart around (p, q) in M1 ×M2. [Here we also used the fact that the product of two
Hausdorff/second countable topological spaces remains to be Hausdorff/second countable.]
We will give more sophisticated examples next time, including graphs of functions,
spheres, real projective spaces etc.
LECTURE 2: SMOOTH MANIFOLDS
1. Smooth Manifolds: The definition
¶ Smooth functions and smooth maps.
Let U be an open set in Rn, and f : U → R a continuous function. Recall that f
is said to be a Ck-function, if all its partial derivatives of order at most k,
∂αf :=
∂|α|f
∂xα
:=
∂|α|f
(∂x1)α1 · · · (∂xn)αn
, |α| = α1 + · · ·+ αn ≤ k
exist and are continuous on U . We say that f is a C∞ function, or a smooth function,
if it is of class Ck for all positive integers k. A function f is an analytic function (or
a Cω function) if it is smooth and agrees with its Taylor series in a neighborhood of
every point. Note that not all smooth functions are analytic.
Now let U be an open set in Rn and V an open set in Rm. Let
f = (f1, · · · , fm) : U → V
be a continuous map. We say f is C∞ (or Ck, or Cω) if each component fi, 1 ≤ i ≤ m,
is a C∞ (or Ck, or Cω) function. 1
Definition 1.1. A smooth map f : U → V is a diffeomorphism if f is one-to-one and
onto, and f−1 : V → U is also smooth.
Obviously
• If f : U → V is a diffeomorphism, so is f−1.
• If f : U → V and g : V → W are diffeomorphisms, so is g ◦ f : U → W .
¶ Definition of smooth manifolds.
We would like to define smooth structures on topological manifolds so that one
can do calculus on it. In particular, we should be able to talk about smoothness of
continuous functions on a given smooth manifold M . Since near each point in M , one
has a chart {ϕ,U, V } which identify the open set U in M with the open set V in Rn,
it is natural identify any function f on U with the function f ◦ ϕ−1 on V , and use
the smoothness of f ◦ ϕ−1 to define the smoothness of f itself. This idea is of course
correct. The only issue is that a point on M could sit in many different open charts,
and the smoothness of a function at this point should be independent of the choice
1In this course we will mainly consider C∞ functions/maps. However, most definitions/theorems
can be easily extended to the Ck setting. On the other hand, Cω theory will be quitedifferent.
1
2 LECTURE 2: SMOOTH MANIFOLDS
of charts. In other words, if both ϕ and ψ are chart maps near a point, we want the
maps f ◦ ϕ−1 and f ◦ ψ−1 to be simultaneously smooth or non-smooth. This amounts
to require ϕ ◦ ψ−1 to be smooth. (Note: even though f ◦ ϕ−1 and f ◦ ψ−1 are both smooth, we
still want the map ϕ ◦ ψ−1 to be smooth so that the differentials of f ◦ ϕ−1 and f ◦ ψ−1 are nicely
related by the chain rule.) With this requirement at hand, we define
Definition 1.2. Let M be a topological manifold of dimension n. We say two charts
{ϕα, Uα, Vα} and {ϕβ, Uβ, Vβ} of M are compatible if the transition map
ϕαβ = ϕβ ◦ ϕ−1α : ϕα(Uα ∩ Uβ)→ ϕβ(Uα ∩ Uβ)
is a diffeomorphism. [Note that both ϕα(Uα ∩Uβ) and ϕβ(Uα ∩Uβ) are open in Rn, so
the smoothness of ϕαβ is well understood.]
Definition 1.3. (1) An atlas A on M is a collection of charts {ϕα, Uα, Vα} with⋃
α Uα = M , such that all charts in A are compatible to each other.
(2) Two atlas on M are said to be equivalent if their union is still an atlas on M .
Example. We can define three atlas on R by Ai = {ϕi,R,R} (1 ≤ i ≤ 3), where
ϕ1(x) = x, ϕ2(x) = 2x and ϕ2(x) = x
3. Then A1,A2 are equivalent, but A1,A3 are
non-equivalent since
ϕ31(x) = ϕ1 ◦ ϕ−13 (x) = x1/3
is not smooth on R.
Definition 1.4. An n-dimensional smooth manifold is an n-dimensional topological
manifold M equipped with an equivalence class of atlas. This equivalence class is called
its smooth structure.
So a smooth manifold is a pair (M,A). In the future we will always omit A if there
is no confusion of the smooth structure.
Remark. Similarly one can define Ck manifolds, real analytic (=Cω) manifolds and
complex manifolds. For example, a complex manifold is a Hausdorff and second count-
able topological space that locally looks like Cn, so that the transition maps are all
holomorphic maps.
LECTURE 2: SMOOTH MANIFOLDS 3
Remark. One can also define and study infinitely dimensional manifolds. There are
many different theories on infinitely dimensional manifolds, depending on whether the
manifold is modelled locally on a Banach space, a Hilbert space, a Fréchet space etc,
and one can define smooth structures on such manifolds.
Remark. Some deep results from differential topology:
• There exist topological manifolds that do not admit smooth structure. The
first example is a compact 10-dimensional manifold found by M. Kervaire.
• If a topological manifold admits a C1 structure, it also admits a C∞ structure.
• Any manifold M admits a finite atlas consisting of dimM + 1 charts (not
necessarily connected).
2. Examples of Smooth Manifolds
¶ First examples.
Note that by definition, we immediately have
Proposition 2.1. If a topological manifold M can be covered by a single chart, then
such a chart automatically determines a smooth structure on M .
As a consequence,
• Rn and any open subset of Rn is a smooth manifold.
• The general linear group GL(n,R) that we studied last time is a smooth man-
ifold.
Example. (Graphs). For any open set U ⊂ Rm and any continuous map f : U → Rn,
the graph of f is the subset in Rm+n = Rm × Rn defined by
Γ(f) = {(x, y) | x ∈ U, y = f(x)} ⊂ Rm+n.
With the subspace topology inherited from Rm+n, Γ(f) is Hausdorff and second-
countable. It is locally Euclidean since it has a global chart {ϕ,Γ(f), U}, where
ϕ : Γ(f)→ U, ϕ(x, y) = x
is the projection onto the first factor map. [To see why ϕ is a homeomorphism: obviously ϕ
is continuous, invertible, and its inverse
ϕ−1 : U → Γ(f), ϕ−1(x) = (x, f(x))
is continuous.] So Γ(f) is a topological manifold of dimension m. Since it can be covered
by one chart, we conclude
The graph Γ(f) of any continuous function f : U ⊂ Rn → R admits
an intrinsic structure of a smooth manifold. [However, it is possible that
Γ(f) is not a smooth submanifold of Rn+1.]
4 LECTURE 2: SMOOTH MANIFOLDS
¶ The spheres as smooth manifolds.
Example. (Spheres). For each n ≥ 0, the unit n-sphere
Sn = {(x1, · · · , xn, xn+1) | (x1)2 + · · ·+ (xn)2 + (xn+1)2 = 1} ⊂ Rn+1
with the subspace topology is Hausdorff and second-countable. To show that it is
locally Euclidean, we can cover Sn by two open subsets
U+ = S
n \ {(0, · · · , 0,−1)}, U− = Sn \ {(0, · · · , 0, 1)}
and define two charts {ϕ+, U+,Rn} and {ϕ−, U−,Rn} by the stereographic projections
ϕ±(x
1, · · · , xn, xn+1) = 1
1± xn+1
(x1, · · · , xn).
It is easy to check that ϕ± are continuous, invertible, and the inverse
ϕ−1± (y
1, · · · , yn) = 1
1 + (y1)2 + · · ·+ (yn)2
(
2y1, · · · , 2yn,±(1− (y1)2 − · · · − (yn)2)
)
is also continuous.
It follows that on ϕ−(U+ ∩ U−) = Rn \ {0},
ϕ−+(y
1, · · · , yn) = ϕ+ ◦ ϕ−1− (y1, · · · , yn)
= ϕ+
(
1
1 + |y|2
(
2y1, · · · , 2yn,−1 + |y|2
))
=
1
|y|2
(y1, · · · , yn),
which is a diffeomorphism from Rn \ {0} to itself. So these two charts are compatible.
Remark. We can also cover Sn by 2n+ 2 charts using hemispheres. More precisely, for
any 1 ≤ i ≤ n+ 1, we let
U+i = {(x1, · · · , xn+1) ∈ Sn : xi > 0}
LECTURE 2: SMOOTH MANIFOLDS 5
be the “upper hemisphere” in the ith direction and define ϕ+i : U
+
i → Bn(1) be the
projection map
ϕ+i (x
1, · · · , xn+1) = (x1, · · · , xi−1, xi+1, xn+1),
where Bn(1) is the unit ball in Rn. Then one can check that (ϕ+i , U
+
i , B
n(1)) are chart-
s. Similarly one can construct charts (ϕ−i , U
−
i , B
n(1)) on each “lower hemispheres”.
[Check: charts of Sn defined via hemispheres are compatible with these two charts.]
¶ The real projective spaces as smooth manifolds.
Example. (The Real Projective Spaces).
The n dimensional real projective space RPn is by definition the set of lines passing
the origin in Rn+1. To give RPn a topology, we will regard it as the quotient space
RPn = Rn+1 − {(0, · · · , 0)}
/
∼,
where the equivalent relation ∼ is given by
(x1, · · · , xn+1) ∼ (tx1, · · · , txn+1), ∀t 6= 0.
One can also regard RPn as the quotient of Sn by gluing the antipodal points
RPn = Sn/ ∼ .
Form these descriptions it is easy to see that RPn is Hausdorff and second-countable,
and in fact is compact.[Prove it.]
Usually people will denote the element (=the equivalence class or the “line”) in
RPn containing the point (x1, · · · , xn+1) by [x1 : · · · : xn+1].
Now we construct local charts on RPn. Consider the open sets
Ui = {[x1 : · · · : xn+1] | xi 6= 0}, 1 ≤ i ≤ n+ 1.
For each i, define ϕi : Ui → Rn to be
6 LECTURE 2: SMOOTH MANIFOLDS
ϕi([x
1 : · · · : xn+1]) =
(
x1
xi
, · · · , x
i−1
xi
,
xi+1
xi
, · · · , x
n+1
xi
)
.
It is not hard to check that each ϕi is well-defined, is continuous, and the inverse
ϕ−1i (y
1, · · · , yn) = [y1 : · · · : yi−1 : 1 : yi : · · · : yn].
is continuous. So each (ϕi, Ui,Rn) is a chart and RPn is a topological manifold.
Finally we will show that RPn is in fact a smooth manifold. Without loss of
generality, let’s verify that ϕ1,n+1 is a diffeomorphism between
ϕ1(U1 ∩ Un+1) = {(y1, · · · , yn) | yn 6= 0} =: Vn
and
ϕn+1(U1 ∩ Un+1) = {(y1, · · · , yn) | y1 6= 0} =: V1.
In fact, by definition
ϕ1,n+1(y
1, · · · , yn) = ϕn+1 ◦ ϕ−11 (y1, · · · , yn)
= ϕn+1([1 : y
1 : · · · : yn])
=
(
1
yn
,
y1
yn
, · · · , y
n−1
yn
)
which is obviously a diffeomorphism from Vn to V1. Similarly one can show that all
other transition maps ϕij are diffeomorphisms.
Remark. By a similar way one can define the n dimensional complex projective space
CPn as the space of “complex lines” in Cn and verify that it is a smooth manifold. More
generally, one can show that Grassmannian 2 Gr(k, V ), the space of all k-dimensional
linear subspaces of an n-dimensional (real or complex) vector space V , is a smooth
manifold. For details, c.f. John Lee, page 22-24.
Example. (The set of all straight lines in R2). The set of all straight lines in R2 is
a manifold. To see this, we just notice that any straight line is of the form
ax+ by + c = 0
for some a, b, c ∈ R, with two triples (a, b, c) and (a′, b′, c′) defines the same line if and
only if [a : b : c] = [a′ : b′ : c′], that is, if and only if they give the same point in RP2.
Also notice that[0 : 0 : 1] will not give us a line, while each other element in RP2 gives
us a line in R2. Thus we get a well-defined bijective map [the image is a Möbius band!]
φ : {the set of lines in R2} → RP2 \ {[0 : 0 : 1]},
ax+ by + c = 0 7→ [a : b : c].
Now we just “move” all structures on RP2 to the set of all straight lines in R2 and thus
define a smooth manifold structure.[Try to construct manifold structure by local parametriza-
tions directly.]
2Named after H. Grassmann, the founder of linear algebra!
LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY
1. Smooth Functions
¶ Smooth functions on manifolds.
Definition 1.1. Let (M,A) be a smooth manifold, and f : M → R a function.
(1) We say f is smooth at p ∈ M if there exists a chart (ϕα, Uα, Vα) ∈ A with
p ∈ Uα, such that the function f ◦ ϕ−1α : Vα → R is smooth at ϕα(p).
(2) We say f is a smooth function on M if it is smooth at every x ∈M .
Remark. Suppose f ◦ ϕ−1α is smooth at ϕ(p). Let (ϕβ, Uβ, Vβ) be another chart in A
with p ∈ Uβ. Then by the compatibility of charts, the function
f ◦ ϕ−1β = (f ◦ ϕ
−1
α ) ◦ (ϕα ◦ ϕ−1β )
must be smooth at ϕβ(p). So the smoothness of a function is independent of the choice
of charts in the given atlas.
Remark. According to the chain rule, it is easy to see that if f : M → R is smooth at
p ∈M , and h : R→ R is smooth at f(p), then h ◦ f is smooth at p.
Example. Each coordinate function fi(x
1, · · · , xn+1) = xi is smooth on Sn since
fi ◦ ϕ−1± (y1, · · · , yn) =
{
2yi
1+|y|2 , 1 ≤ i ≤ n
±1−|y|
2
1+|y|2 , i = n+ 1
are smooth functions on Rn. Similarly the latitude is a smooth function on S2, since it
can be written as the composition of the “height function” x3 with a smooth function.
However, the longitude is not even a well-defined real-valued function on S2.
We will denote the set of all smooth functions on M by C∞(M). Note that this is a
(commutative) algebra, i.e. it is a vector space equipped with a (commutative) bilinear
“multiplication operation”: If f, g are smooth, so are af + bg and fg; moreover, the
multiplication is commutative, associative and satisfies the usual distributive laws.
Now suppose f ∈ C∞(M). As usual, the support of f is by definition the set
supp(f) = {p ∈M | f(p) 6= 0}.
We say that f is compactly supported, denoted by f ∈ C∞0 (M), if the support of f is a
compact subset in M . Obviously
• if f, g ∈ C∞0 (M), then af + bg ∈ C∞0 (M).
• if f ∈ C∞0 (M) and g ∈ C∞(M), then fg ∈ C∞0 (M).
So C∞0 (M) is an ideal of of the algebra C
∞(M). Note that if M is compact, then any
smooth function is compactly supported.
1
2 LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY
¶ Bump functions.
A bump function (sometimes also called a test function) is a compactly supported
smooth function, which is usually supposed to be non-negative, no more than 1, and
equals to 1 on a given compact set (or has total integral 1 on a given set).
Example (A bump function on Rn). In what follows we first define two auxiliary func-
tions f1 and f2 on R. Then we define a bump function f3 on Rn. We list the definition of
fk in the left, and list the properties of fk in the right. The smoothness and properties
of fk follows from those of fk−1 (so you should check f1 is a smooth function):
f1(x) =
{
e−1/x, x > 0
0, x ≤ 0 =⇒ f1(x) =
{
∈ (0, 1), x > 0,
0, x ≤ 0,
f2(x) =
f1(x)
f1(x) + f1(1− x)
=⇒ f2(x) =
 0, x ≤ 0,∈ (0, 1), 0 < x < 1,1, x ≥ 1
f3(x) = f2(2− |x|) =⇒ f3(x) =
 0, |x| ≥ 2,∈ (0, 1), 1 < |x| < 2,1, |x| ≤ 1.
The graphs of f1, f2 and f3 (with n = 1) are shown below:
With the help of this Euclidean bump function, we can construct bump functions
on any smooth manifold with prescribed support and prescribed “equal to one region”:
Theorem 1.2. Let M be a smooth manifold, A ⊂M is a compact subset, and U ⊂M
an open subset that contains A. Then there is a bump function ϕ ∈ C∞0 (M) so that
0 ≤ ϕ ≤ 1, ϕ ≡ 1 on A and supp(ϕ) ⊂ U .
Proof. [The idea of the proof: Cover the compact set A by finitely many small pieces,
where each piece is contained in one (carefully chosen) chart, so that one can copy the
“Euclidean bump function” that we constructed above to such pieces.]
For each q ∈ A, there is a chart (ϕq, Uq, Vq) near q so that Uq ⊂ U and Vq contains
the open ball B3(0) of radius 3 centered at 0 in Rn. (Question: Why one can find such
a chart which is compatible with given charts?) Let Ũq = ϕ
−1
q (B1(0)), and let
fq(p) =
{
f3(ϕq(p)), p ∈ Uq,
0, p /∈ Uq.
Then fq ∈ C∞0 (M), supp(fq) ⊂ Up and fq ≡ 1 on Ũq. (Question: which assumption on
manifold do we need here?)
LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY 3
Now the family of open sets {Ũq}q∈A is an open cover of A. Since A is compact,
there is a finite sub-cover {Ũq1 , · · · , ŨqN}. Let ψ =
∑N
i=1 fqi . Then ψ is a compactly
supported smooth function on M so that ψ ≥ 1 on A and supp(ψ) ⊂ U . It follows
that the function ϕ(p) = f2(ψ(p)) satisfies all the conditions we want. �
Here is what such a bump function will look like:
As a simple consequence, we see that the vector space C∞0 (M) (and thus C
∞(M))
is infinitely dimensional (assuming dimM > 0).
2. Partition of unity
¶ Partition of unity.
So for any compact subset K ⊂ M , one can always cover it by finitely many nice
neighborhoods on which we can construct nice “local” functions. By adding these
(finitely many) local functions, we can get nice global functions on M that behaves
nicely on K. It turns out that the same idea applies to the whole manifold M : we can
generate an infinite collection of smooth functions on M , and add them to get a global
smooth function, provided that near each point, there are only finitely many nonzero
functions in our collection. More importantly, we can use such a collection of functions
to “glue” geometric/analytic objects that can be defined locally using charts.
Definition 2.1. Let M be a smooth manifold, and {Uα} an open cover of M . A (s-
mooth) partition of unity (P.O.U. in brief)1 subordinate to the cover {Uα} is a collection
of smooth functions {ρα} defined on the whole manifold M so that
(1) 0 ≤ ρα ≤ 1 for all α.
(2) supp(ρα) ⊂ Uα for all α.
(3) each p ∈M has a neighborhood which intersects only finitely many supp(ρα)’s.
(4)
∑
α ρα(p) = 1 for all p ∈M .
Remark. Two consequences of the local finiteness condition (3): Let’s denote by Wp a
neighborhood of p which intersect only finitely many supp(ρα)’s. Then
1In this course, when we talk about P.O.U., we always mean smooth partition of unity. See my
general topology course notes for a theory of continuous partition of unity for paracompact spaces.
Recall that a topological space X is paracompact if every open covering admits a locally finite open
refinement. It is not hard to show each topological manifold is paracompact.
4 LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY
• Since {Wp}p∈M is an open cover of M , and since M is second countable, one
can find countably many Wpi ’s which also cover M . Since each Wpi intersect
only finitely many supp(ρα)’s, we conclude that there are only countable many
ρα’s whose support are non-empty. So even if we may start with uncountably
many open sets, the P.O.U. automatically “delete” most of them so that only
countably many of them are left (which still form an open cover of M).
• For each p, on the open set Wp, a sum like (4) [which looks like an uncountable
sum, or maybe a countable infinite sum in view of the previous paragraph] is
in fact a finite sum. This fact is CRUCIAL in applications.
The main result in this section is to prove
Theorem 2.2 (The existence of P.O.U.). Let M be a smooth manifold, and {Uα} an
open cover of M . Then there exists a P.O.U. subordinate to {Uα}.
Locally each manifold looks like Rn, so that one have rich mathematics on it.
P.O.U. is the tool that can “glue” local smooth objects into a global smooth object.
We will see many such examples in the future. For example, we will
• approximate continuous functions/maps via smooth functions/maps.
• define integrals of differential forms in local charts,and use P.O.U. to define
the integral of a differential form on the whole manifold.
• (in future course) construct Riemannian metric, linear connection etc.
As an immediate application of P.O.U., we generalize Theorem 1.2 to closed subsets:
Corollary 2.3. Let M be a smooth manifold, A ⊂ M is a closed subset, and U ⊂ M
an open subset that contains A. Then there is a “bump” function ϕ ∈ C∞(M) so that
0 ≤ ϕ ≤ 1, ϕ ≡ 1 on A and supp(ϕ) ⊂ U .
Proof. Obviously {U,M \ A} is an open covering of M . Let {ρ1, ρ2} be a P.O.U.
subordinate to this open covering. Then the function ϕ = ρ1 is what we need: ρ1 is
smooth, 0 ≤ ρ1 ≤ 1, supp(ρ1) ⊂ U , and finally ρ1 = 1 on A since ρ2 = 0 on A. �
Note that this implies a smooth version of Urysohn’s lemma [See today’s PSet].
¶ Existence of P.O.U.: The proof.
The proof relies on the following technical lemma from general topology.
Lemma 2.4. For any open covering U = {Uα} of a topological manifold M , one can
find two countable family of open covers V = {Vj} and W = {Wj} of M so that
• For each j, V j is compact and V j ⊂ Wj.
• W is a refinement of U : For each j, there is an α = α(j) so that Wj ⊂ Uα.
• W is locally finite: Any p ∈ M has a neighborhood W such that W ∩Wj 6= ∅
for only finitely many Wj’s.
We will first prove Theorem 2.2 and postpone the proof Lemma 2.4 as an appendix.
LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY 5
Proof of Theorem 2.2. [Please compare the first paragraph of this proof with the proof
of Theorem 1.2.] Since V j is compact and V j ⊂ Wj, according to Theorem 1.2 we can
find nonnegative functions ϕj ∈ C∞0 (M) so that
0 ≤ ϕj ≤ 1, ϕj ≡ 1 on V j, supp(ϕj) ⊂ Wj.
Since W is a locally finite covering, the function
ϕ =
∑
j
ϕj
is a well-defined smooth function on M . Since each ϕj is nonnegative, and V is a
covering of M , ϕ is strictly positive on M . It follows that the functions
ψj =
ϕj
ϕ
are smooth and satisfy 0 ≤ ψj ≤ 1 and
∑
j ψj = 1.
Next let’s re-index the family {ψj} to get the demanded P.O.U. subordinate to
{Uα}. For each j, we fix an index α(j) so that Wj ⊂ Uα(j), and define
ρα =
∑
α(j)=α
ψj.
Note that the right hand side is a finite sum near each point, so it does define a smooth
function. By local finiteness of W ,
suppρα =
⋃
α(j)=α
suppψj =
⋃
α(j)=α
suppψj =
⋃
α(j)=α
suppψj ⊂ Uα.
Clearly the family {ρα} is a P.O.U. subordinate to {Uα}. �
¶ Appendix: The proof of Lemma 2.4.
It remains to prove Lemma 2.4. In particular, we want to prove the existence of
locally finite open refinement. The proof is quite geometric. First we prove
Lemma 2.5. For any topological manifold M , there exists a countable collection of
open sets {Xi} so that [Such a collection of subsets is called an exhaustion of M ]
(1) For each j, the closure Xj is compact.
(2) For each j, Xj ⊂ Xj+1.
(3) M = ∪jXj.
Proof. Since M is second countable, there is a countable basis of the topology of M .
Out of this countable collection of open sets, we pick those that have compact closures,
and denote them by Y1, Y2, · · · . Since M is locally Euclidean, it is easy to see that
Y = {Yj} is an open cover of M .
6 LECTURE 3: SMOOTH FUNCTIONS; PARTITION OF UNITY
We let X1 = Y1. Since Y is an open cover of X1 which is compact, there exist
finitely many open sets Yi1 , · · · , Yik so that
X1 ⊂ Yi1 ∪ · · · ∪ Yik .
Let
X2 = Y2 ∪ Yi1 ∪ · · · ∪ Yik .
Obviously X2 is compact. By repeating this procedure we get a sequence of open sets
X1, X2, X3, · · · which satisfies (1) and (2). It satisfies (3) since Xk ⊃ ∪kj=1Yj �
Proof of Lemma 2.4. For each p ∈M , there is an j and an α(p) so that p ∈ Xj+1 \Xj
and p ∈ Uα(p). Since M is locally Euclidean, one can choose open neighborhoods Vp,Wp
of p so that V p is compact and
V p ⊂ Wp ⊂ Uα(p) ∩ (Xj+2 \Xj−1).
Now for each j, since the “stripe” Xj+1 \ Xj is compact, one can choose finitely
many points pj1, · · · , p
j
kj
so that Vpj1
, · · · , Vpjkj
is an open cover of Xj+1 \Xj. Denote all
these Vpjk
’s by V1, V2, · · · , and the corresponding Wpjk ’s by W1,W2, · · · . Then V = {Vk}
and W = {Wk} are open covers of M that satisfies all the conditions in Lemma 2.4.
For example, the local finiteness property ofW follows from the fact that there are only
finitely many Wk’s (that correspond to j and j − 1 above) intersect Xj+1 \Xj−1. �
We end this section with the following question:
Where did we use the Hausdorff condition in proving P.O.U.?
LECTURE 4: SMOOTH MAPS
1. Smooth Maps
¶ Smooth maps between manifolds.
Since manifolds are locally Euclidean, we are able to transplant conceptions defined
on Euclidean spaces to manifolds. Recall that a smooth function on a smooth manifold
M is a function f : M → R so that for any chart1 (ϕα, Uα, Vα) of M , the function f ◦ϕ−1α
is a smooth function on Vα. More generally, we can define smooth maps between smooth
manifolds:
Definition 1.1. Let M,N be smooth manifolds. We say a continuous map f : M → N
is smooth if for any chart (ϕα, Uα, Vα) of M and chart (ψβ, Xβ, Yβ) of N , the map
ψβ ◦ f ◦ ϕ−1α : ϕα(Uα ∩ f−1(Xβ))→ ψβ(Xβ)
is smooth. [Again: Both ϕα(Uα ∩ f−1(Xβ)) and ψβ(Xβ) are Euclidian open sets.]
Similarly one can define Ck or Cω maps between Ck or Cω manifolds.
Remark. In the definition we require the map f to be continuous. This is to guarantee
that the map ψβ ◦ f ◦ ϕ−1α is defined on a neighborhood of ϕα(p). In general the
smoothness of all ψβ ◦ f ◦ ϕ−1α ’s does not imply the continuity of f .2 (See Problem Set)
The following proposition (which claims that the conception of smoothness is independent
of the choice of equivalent smooth structures) can be easily checked:
Proposition 1.2. If f : (M,A)→ (N,B) is smooth, A1 and B1 are atlas on M and N
that are compatible with A and B respectively, then f : (M,A1)→ (N,B1) is smooth.
1In this course, when we say “any chart of a smooth manifold M”, we always mean “any chart in
a given atlas A that defines the smooth structure of M”.
2Alternatively, it is also enough to assume that each point p has a coordinate neighborhood U so
that f(U) is contained in a coordinate domain on N . c.f. Hirsch, Differential Topology, GTM 33.
1
2 LECTURE 4: SMOOTH MAPS
The set of all smooth maps from M to N is denoted by C∞(M,N). We will
leave it as an exercise to prove that if f ∈ C∞(M,N) and g ∈ C∞(N,P ), then
g ◦ f ∈ C∞(M,P ). As a consequence, any smooth map f : M → N induces a “pull-
back” map
f ∗ : C∞(N)→ C∞(M), g 7→ g ◦ f.
The pull-back will play an important role in the future.
¶ Examples of smooth maps.
Example. If we equip R with the standard smooth structure (i.e., the smooth structure
{(ϕ1(x) = x,R,R)}, which we will always assume), a map f : M → R is a smooth
map in the definition above if and only if it is a smooth function in the sense that we
learned last time. More generally, a map
f = (f1, · · · , fk) : M → Rk
is a smooth map if and only if each fi ∈ C∞(M).
Example. Recall that the general linear group GL(n,R) is an open subset of Rn2 and
thus is a smooth manifold. It is then easy to check (by using the standard atlas with
only one chart) that (why?)
• the determinant
det : GL(n,R)→ R, A 7→ detA
is a smooth function,
• the matrix product map
m : GL(n,R)×GL(n,R)→ GL(n,R), (A,B) 7→ AB
is a smooth map,
• the matrix inversion map
i : GL(n,R)→ GL(n,R), A 7→ A−1
is a smooth map.
As we will see, these facts imply that GL(n,R) is a Lie group.
Example. The inclusion map ι : Sn ↪→ Rn+1 is smooth, since
ι ◦ ϕ−1± (y1, · · · , yn) =
1
1 + |y|2
(
2y1, · · · , 2yn,±(1− |y|2)
)
are smooth maps from Rn to Rn+1. Note that by definition, if g is any smooth function
on Rn+1, the pull-back function ι∗g is just the restriction of g to Sn:
ι∗g = g|Sn .
So the restriction of any smooth function on Rn+1 to Sn is a smooth function on Sn.
LECTURE 4: SMOOTH MAPS 3
Example. The projection map π : Rn+1 \ {0} → RPn is smooth, since
ϕi ◦ π(x1, · · · , xn+1) =
(
x1
xi
, · · · , x
i−1
xi
,
xi+1
xi
, · · · , x
n+1
xi
)
is smooth on π−1(Ui) = {(x1, · · · , xn+1) : xi 6= 0} for each i.
¶ Diffeomorphisms.
As inthe Euclidean case, we can define diffeomorphisms between smooth manifolds.
Definition 1.3. Let M,N be smooth manifolds. A map f : M → N is a diffeomor-
phism if it is smooth, bijective, and the inverse f−1 is smooth.
If there exists a diffeomorphism f : M → N , then we say M and N are diffeo-
morphic. Sometimes we will denote M ' N . The following properties can be easily
deduced from the corresponding properties in the Euclidean case:
• The identity map Id : M →M is a diffeomorphism.
• If f : M → N and g : N → P are diffeomorphisms, so is g ◦ f .
• If f : M → N is a diffeomorphism, so is f−1. Moreover, dimM = dimN .
So “diffeomorphism” defines an equivalence relation on the set of all smooth manifolds.
We will regard diffeomorphic smooth manifolds as the same.
Remark. In particular, for any smooth manifold M ,
Diff(M) = {f : M →M | f is a diffeomorphism}
is a group, which is known as the diffeomorphism group of M .
Example. If M is a smooth manifold, then any chart (ϕ,U, V ) gives a diffeomorphism
ϕ : U → V from U ⊂M (where we regard U as a smooth manifold) to V ⊂ Rn. More
generally, one can check: two atlas A = {(φα, Uα, Vα)}α and B = {(ϕβ, Uβ, Vβ)}β of M
are equivalent if and only if the identity map Id : (M,A)→ (M,B) is a diffeomorphism.
Example. We have seen that on M = R, the two atlas A = {(ϕ1(x) = x,R,R)} and
B = {(ϕ2(x) = x3,R,R)} define non-equivalent smooth structures. However, the map
f : (R,A)→ (R,B), f(x) = x1/3
is a diffeomorphism. So we still think these two smooth structures are the same,
although they are not equivalent.
Remark. Here are some deep results on different (=non-diffeomorphic) smooth struc-
tures on a given topological manifold:
• (T. Rado, E. Moise) There is only one smooth structure for any topological
manifold of dimension smaller than 4.
• (J. Milnor and M. Kervaire) The topological 7-sphere admits exactly 28 different
smooth structures. More generally, it is known that for spheres, the number of
different smooth structures are (from Wikipedia)
4 LECTURE 4: SMOOTH MAPS
The very special case is S4: we don’t know whether the smooth structure on
S4 is unique. This is known as the smooth Poincaré conjecture.
• (S. Donaldson and M. Freedman) For any n 6= 4, Rn has a unique smooth struc-
ture up to diffeomorphism; but on R4 there exist uncountable many distinct
pairwise non-diffeomorphic smooth structures.
On the other hand, it is also known that the obstruction of the existence/uniqueness
of smooth structure lies only in topological manifold:
• (Whitney) Any C1 manifold admits a unique smooth structure. (c.f. Theorem
2.9 in Hirsch, Differential Topology, GTM 33)
2. The differential of a smooth map: the Euclidean case
¶ The differentials of Euclidean smooth maps.
Let U, V be Euclidean open sets, and f : U → V be a smooth map. The differential
of f , df , assigns to each point a ∈ U a linear map dfa : Rn → Rm whose matrix (with
respect to the canonical basis) is the Jacobian matrix of f at a,
dfa =

∂f1
∂x1
(a) · · · ∂f1
∂xn
(a)
...
. . .
...
∂fm
∂x1
(a) · · · ∂fm
∂xn
(a)
 .
As we have seen in multi-variable calculus, the differential dfx plays a crucial role in
studying the map f , since it is the “linearization” of the map f near the point x:
lim
x→a
‖f(x)− f(a)− dfa(x− a)‖
‖x− a‖
= 0.
A very useful fact for the differential is the well-known chain rule: if f : U → V and
g : V → W are smooth maps, so is the composition map g ◦ f : U → W , and
d(g ◦ f)x = dgf(x) ◦ dfx.
One should regard d as a “functor” from the category of whose objects are “Euclidean
open sets” and morphisms are “smooth maps to the category with objects “vector
spaces” and morphisms “linear maps”. Then the chain rule is merely part of the
functor proprety. As a consequence of the use of the linear approximation, we can
prove the following smooth “invariance of domain” theorem:
Theorem 2.1 (Invariance of Dimension, smooth version). If f : U → V is a diffeomor-
phism, then for each x ∈ U , the differential dfx is a linear isomorphism. In particular,
dimU = dimV .
Proof. Applying the chain rule to f−1 ◦ f = IdU , and notice that the differential of the
identity map IdU : U → U is the identity transformation IdRn : Rn → Rn, we get
(df−1)f(x) ◦ dfx = IdRn .
The same argument applies to f ◦ f−1, which yields dfx ◦ (df−1)f(x) = IdRm . By basic
linear algebra, we conclude that m = n and that dfx is an isomorphism. �
LECTURE 4: SMOOTH MAPS 5
¶ Reading material: The inverse/implicit function theorems.
Conversely, one would like to ask: if the linearization dfx is an linear isomorphism,
what can we say about the map f itself? In general f is no longer a diffeomorphism:
Example. Consider the map
f : C \ {0} → C \ {0}, z 7→ f(z) = z2.
Then f(z) = f(−z). So f is not a diffeomorphism since it is not invertible. However,
at each point z = (x, y) ∈ R2 \ {0},
dfz =
(
2x −2y
2y 2x
)
,
which is an isomorphism for each z = (x, y) 6= (0, 0). Fortunately, the situation is not
that bad: f is not very far away from being a diffeomorphism: for any x ∈ C \ {0},
one can find a small neighborhood Ux of x, such that the restriction f |Ux : Ux → f(Ux)
is a diffeomorphism. This motivates the following definition:
Definition 2.2. Let f : U → V be a smooth map. We say f is a local diffeomorphism
near x ∈ U if there exists a neighborhood Ux containing x and a neighborhood Vf(x)
containing f(x) such that
f |Ux : Ux → Vf(x)
is a diffeomorphism
Now we can state the inverse function theorem, which can be viewed as a partial
inverse of the “smooth invariance of domain” theorem.
Theorem 2.3 (The inverse function theorem). If f : U → V is a smooth map, and
dfx is an isomorphism, then f is a local diffeomorphism near x.
In other words, an isomorphism in the linear category implies a local diffeomor-
phism in the smooth category. The inverse function theorem is a special case (why) of
the following implicit function theorem in multivariable analysis course:
Theorem 2.4 (The implicit function theorem). Let W be an open set in Rnx × Rmy ,
and F = (F1, · · · , Fm) : W → Rm a smooth map. Let (x0, y0) be a point in W so that
the m×m matrix 
∂F1
∂y1
(x0, y0) · · · ∂F1∂ym (x0, y0)
...
...
...
∂Fm
∂ym
(x0, y0) · · · ∂Fm∂ym (x0, y0)
 .
is nonsingular, then there exists a neighborhood U0× V0 of (x0, y0) in W and a smooth
map f : U0 → V0 so that the graph of f is the level set of F near (x0, y0), i.e.
• f(x0) = y0,
• If we denote c = F (x0, y0), then F−1(c) ∩ (U0 × V0) is the graph of f , i.e.
F (x, f(x)) = c for all x ∈ U0.
LECTURE 5: THE DIFFERENTIAL OF A SMOOTH MAP
1. The tangent space
¶ The idea behind the definition.
Now suppose M,N are smooth manifolds, and f : M → N smooth. As in the
Euclidean case we would like to define its differential dfp at p to be a linear map
between corresponding tangent spaces, which serves as a linearization of f near p. But
the first question is: what is the tangent space of a smooth manifold at a point?
If M is a concrete manifold that sits in some RN (which is always true as we will
see in the future), then we may choose a coordinate chart (ϕ,U, V ) near p, so that
ϕ−1 : V → U is a diffeomorphism. If we denote the embedding of M into RN to be
ι : M ↪→ RN , then we get a smooth map ι ◦ ϕ−1 : V → RN between Euclidean open
sets, and we may define the tangent space TpM to be the image d(ι ◦ ϕ−1)ϕ(p)(Rn) of
the differential (which is a linear map)
d(ι ◦ ϕ−1)ϕ(p) : Rn → RN .
Of course one has to check that the space TpM defined by this way is independent of
the choice of coordinate charts, AND, one need to identify different TpM ’s obtained by
this way when we use two different embeddings of M into Euclidean spaces.
In what follows we will define the tangent space TpM using data on M itself only,
since we don’t know a priori whether M can be embedded into an Euclidean space, and
Since M is an abstract manifold, we don’t have a simple nice “geometric picture”. To
get some idea on how to define the tangent space at each point on a smoothmanifold,
we shall take a closer look at the Euclidean case. The idea is:
• ∀~v ∈ Rn at a given point a can be identified with a directional derivative at a,
• there is a pure algebraic characterization of the directional derivative, which
can be generalized to manifolds.
That vector space is the abstract tangent space we are looking for!
¶ The Euclidean directional derivative: an algebraic characterization.
We recall: for any ~v ∈ Rn, the directional derivative of f at x in the direction v is
Da~vf := dfx(~v) = lim
h→0
f(x+ h~v)− f(x)
h
=
d
dt
∣∣∣∣
t=0
f(a+ t~v).
So ~v gives us an operator Da~v : C
∞(Rn)→ R. In coordinates, if ~v = 〈v1, · · · , vn〉T , then
Da~vf =
∑
vi
∂f
∂xi
,
1
2 LECTURE 5: THE DIFFERENTIAL OF A SMOOTH MAP
in other words, as an operator on C∞(Rn),
Da~v =
∑
i
vi
∂
∂xi
.
Of course we know that Da~v : C
∞(Rn)→ R is a very special operator: it is a linear
Da~v(αf + βg) = αD
a
~vf + βD
a
~vg, ∀α, β ∈ R
and it satisfies the Leibnitz law at a:
Da~v(fg) = f(a)D
a
~vg + g(a)D
a
~vf.
Conversely,
Proposition 1.1. If D : C∞(Rn)→ R is linear and satisfies the Leibnitz law a, i.e.
D(fg) = f(a)D(g) + g(a)D(f),
then D = Da~v for some vector ~v at a.
Proof. For any f ∈ C∞(Rn), we have
f(x) = f(a) +
∫ 1
0
d
dt
f(a+ t(x− a))dt = f(a) +
n∑
i=1
(xi − ai)hi(x),
where
hi(x) =
∫ 1
0
∂f
∂xi
(a+ t(x− a))dt.
Note that by the Leibnitz law, when we apply D to the constant function 1, we must
have D(1) = 0 since
D(1) = D(1 · 1) = 2D(1).
By linearity, D(c) = 0 for any constant c. So
D(f) = 0 +
n∑
i=0
D(xi)hi(a) +
n∑
i=0
(ai − ai)D(hi) =
n∑
i=0
D(xi)
∂f
∂xi
(a).
It follows that as an operator on C∞(Rn),
D =
n∑
i=1
D(xi)
∂
∂xi
∣∣∣∣
x=a
.
In other words, if we let ~v = 〈D(x1), · · · , D(xn)〉, then D = Da~v . �
This motivates the following “algebraic” definition of derivatives:
Definition 1.2. Any linear operator Da : C∞(Rn)→ R satisfying the Leibnitz law at
a is called a derivative at a.
It is obvious that the set D of all derivatives at a is a vector space. Now consider
the correspondence
~v Da~v .
We have
LECTURE 5: THE DIFFERENTIAL OF A SMOOTH MAP 3
• It is a linear map from the vector space Rn (of all tangent vectors at a) to the
vector space D of all derivatives at a:
Daα~v+β ~w = αD
a
~v + βD
a
~w.
• It is injective: if ~v1 6= ~v2, then Da~v1 6= D
a
~v2
(try to prove it).
• It is surjective: This follows from Proposition 1.1.
So the space of tangent vectors ~v at a is linearly isomorphic to the space of derivatives
at a, i.e. we can identify the set (vector space) of all tangent vectors at a with the set
(vector space) of all derivatives at a!
¶ Tangent vector on manifolds.
Now back to the study of manifolds. Although we don’t have the geometric vectors
in the abstract setting, we do have the space C∞(M) of all smooth functions. So as in
the Euclidean case we may define derivatives at a point and simply call them tangent
vectors at that point:
Definition 1.3. Let M be an n-dimensional smooth manifold. A tangent vector at a
point p ∈M is a R-linear map Xp : C∞(M)→ R satisfying the Leibnitz law
(1) Xp(fg) = f(p)Xp(g) +Xp(f)g(p)
for any f, g ∈ C∞(M).
It is easy to see that the set of all tangent vectors of M at p is a linear space. We
will denote this set by TpM , and call it the tangent space TpM to M at p..
As argued above, we can easily prove: if f ≡ c is a constant function, then Xp(f) =
0. More generally,
Lemma 1.4. If f = c in a neighborhood of p, then Xp(f) = 0.
Proof. Let ϕ be a smooth function on M that equals 1 near p, and equals 0 at points
where f 6= c. (How to construct such an f?) Then
(f − c)ϕ ≡ 0.
So
0 = Xp((f − c)ϕ) = (f(p)− c)Xp(ϕ) +Xp(f)ϕ(p) = Xp(f). �
As a consequence, we see that if f = g in a neighborhood of p1, then Xp(f) = Xp(g).
In other words, Xp(f) is determined by the values of f near p. So one can replace
C∞(M) in Definition 1.3 by C∞(U), where U is any open set that contains p. In other
words, as linear spaces,
Proposition 1.5. If M is a smooth manifold, p ∈ U ⊂M , where U is open, then
TpM ' TpU.
1There is a notion “germ”: we say f and g define the same germ at p if f = g in a neighborhood
of p. It is easy to see that “germ at p” gives an equivalence relation on C∞(M) (or more generally
on C∞(M,N)). When studying local properties, it is always enough to work on germs.
4 LECTURE 5: THE DIFFERENTIAL OF A SMOOTH MAP
2. The Differentials
¶ The differential of smooth maps between smooth manifolds.
Finally we try to define the differential of a smooth map between smooth manifolds.
Recall that the differential of a smooth map f : U → V between open sets in Euclidean
spaces at a ∈ U is a linear map dfa : TaU = Rnx → Tf(a)V = Rmy whose matrix is the
Jacobian matrix ( ∂fi
∂xj
) of f at a. To transplant this conception to smooth maps between
smooth manifolds, we need to take a closer look at the two interpretation of TaU : We
have seen that we can identify the (geometric) vector ~v ∈ Rn at a with the (algebraic)
derivative Da~v =
∑
vi ∂
∂xi
. Note that geometrically,
dfa(~v) =
(
∂fi
∂xj
)
~v =
〈∑
j
∂f1
∂xj
vj, · · · ,
∑
j
∂fn
∂xj
vj
〉T
.
The vector in the right hand side is a vector in Rmy . When interpreted as a derivative
on V at f(a), it is a map that maps g ∈ C∞(Rmy ) to∑
i
∑
j
vj
∂fi
∂xj
∂g
∂yi
=
∑
j
vj
∂
∂xj
(g ◦ f) = Da~v(g ◦ f).
In other words, the derivative that corresponds to the vector dfa(~v) is the derivative at
f(a) that maps g ∈ C∞(Rm) to Da~v(g ◦ f).
Motivated by these computations, we define
Definition 2.1. Let f : M → N be a smooth map. Then for each p ∈ M , the
differential of f is the linear map dfp : TpM → Tf(p)N defined by
dfp(Xp)(g) = Xp(g ◦ f)
for all Xp ∈ TpM and g ∈ C∞(N).
¶ Properties of the differential.
The chain rule still holds for composition of smooth maps between smooth mani-
folds:
Theorem 2.2 (Chain rule). Suppose f : M → N and g : N → P are smooth maps,
then d(g ◦ f)p = dgf(p) ◦ dfp.
Proof. For any Xp ∈ TpM and h ∈ C∞(P ),
d(g ◦ f)p(Xp)(h) = Xp(h ◦ g ◦ f) = dfp(Xp)(h ◦ g) = dgf(p)(dfp(Xp))(h). �
Obviously the differential of the identity map is the identity map between tangent
spaces. So again the differential d is a functor from the category of smooth manifolds
(with morphisms smooth maps) to the category of vector spaces (with morphisms linear
maps). By the standard functorality argument (c.f. the proof of Theorem 2.1 in Lecture 4),
we get
Corollary 2.3. If f : M → N is a diffeomorphism, then dfp : TpM → Tf(p)N is a
linear isomorphism.
LECTURE 5: THE DIFFERENTIAL OF A SMOOTH MAP 5
In particular, we have
Corollary 2.4. If dimM = n, then TpM is an n-dimensional linear space.
Proof. Let (ϕ,U, V ) be a chart near p. Then ϕ : U → V is a diffeomorphism. It follows
that dimTpM = dimTpU = dimTf(p)V = n. �
Note that in the proof we really showed: For any local chart (ϕ,U, V ) around p,
we have
TpM = span{∂1, · · · , ∂n},
where ∂i := dϕ
−1( ∂
∂xi
). We will abuse the notation and think of xi as a function on U
(which really should be xi ◦ ϕ). In other words, we will denote ϕ = (x1, · · · , xn) with
each xk the kth coordinate function on U , and denote the chart by {U ;x1, · · · , xn}. In
such coordinate charts, one has the following explicit formula for ∂i:
∂i : C
∞(U)→ R, ∂i(f) =
∂(f ◦ ϕ−1)
∂xi
(ϕ(p)).
LECTURE 6: LOCAL BEHAVIOR VIA THE DIFFERENTIAL
1. The Inverse function theorem
¶ The inverse function theorem.
Last time we showed that if f : M → N is a diffeomorphism, then the differential
dfp : TpM → Tf(p)N is a linear isomorphism. As in the Euclidean case (see Lecture
4), the converse is not true in general (i.e., “f : M → N is smooth and dfp is a linear
isomorphism for every p ∈ M” does not imply “f is a diffeomorphism”) but we still
have the following partial converse:
Theorem 1.1 (The Inverse Function Theorem). Let f : M → N be a smooth map such
that dfp : TpM → Tf(p)N is a linear isomorphism, then there exists a neighborhood U1
of p and a neighborhood X1 of q = f(p) such that f |U1 : U1 → X1 is a diffeomorphism.Proof. Take a chart (ϕ,U, V ) near p and a chart (ψ,X, Y ) near f(p) so that f(U) ⊂ X
(which is always possible after shrinking U and V ). Since ϕ : U → V and ψ : X → Y
are diffeomorphisms,
d(ψ ◦ f ◦ ϕ−1)ϕ(p) = dψq ◦ dfp ◦ dϕ−1ϕ(p) : Tϕ(p)V = R
n → Tψ(q)Y = Rn
is a linear isomorphism. It follows from the inverse function theorem (c.f. Lecture
4) that there exist neighborhoods V1 of ϕ(p) and Y1 of ψ(q) so that ψ ◦ f ◦ ϕ−1 is a
diffeomorphism from V1 to Y1. Take U1 = ϕ
−1(V1) and X1 = ψ
−1(Y1). Then
f = ψ−1 ◦ (ψ ◦ f ◦ ϕ−1) ◦ ϕ
is a diffeomorphism from U1 to X1. �
¶ Local diffeomorphism v.s. global diffeomorphism.
Definition 1.2. We say a smooth map f : M → N is a local diffeomorphism near p, if
it maps an open neighborhood of p diffeomorphically to an open neighborhood of f(p).
Note that it is possible that a map is a local diffeomorphism everywhere, but still
fails to be diffeomorphism.
Example. Let f : S1 → S1 be given by f(eiθ) = e2iθ. Then it is a local diffeomorphism
everywhere, but it is not a global diffeomorphism since it is not invertible. [Please
compare this example with the example on page 5 of Lecture 4.]
It turns out that the invertibility is the only obstruction for an “everywhere local
diffeomorphism” to be a global diffeomorphism:
Proposition 1.3. Suppose f : M → N is a local diffeomorphism near every p ∈ M .
If f is invertible, then f is a global diffeomorphism.
1
2 LECTURE 6: LOCAL BEHAVIOR VIA THE DIFFERENTIAL
Proof. We only need to show f−1 is smooth. Fix any q = f(p). The smoothness of f−1
at q depends only on the behaviour of f−1 near q. Since f is a diffeomorphism from a
neighborhood of p onto a neighborhood of q, f−1 is smooth at q. �
2. The constant rank theorem
¶ Submersion and immersion.
What if dfp is not a linear isomorphism? Note that a linear isomorphism is both
surjective and injective. It is natural to study the those smooth maps whose differential
is either surjective or injective:
Definition 2.1. Let f : M → N be a smooth map.
(1) f is a submersion at p if dfp : TpM → Tf(p)N is surjective.
(2) f is an immersion at p if dfp : TpM → Tf(p)N is injective.
We say f is a submersion/immersion if it is a submersion/immersion at each point.
Obviously
• If f is a submersion, then dimM ≥ dimN .
• If f is an immersion, then dimM ≤ dimN .
Example. In PSet 2-1-6, we see: the natural projection π : TM →M is a submersion.
Similarly, the “zero section” embedding ι : M → TM, p 7→ (p, 0) is an immersion.
Example. A local diffeomorphism is both a submersion and an immersion.
Example (Canonical submersion). If m ≥ n, then the projection map
π : Rm → Rn, (x1, · · · , xm) 7→ (x1, · · · , xn)
is a submersion.
Example (Canonical immersion). If m ≤ n, then the inclusion map
ι : Rm ↪→ Rn, (x1, · · · , xm) 7→ (x1, · · · , xm, 0, · · · , 0)
is an immersion.
It turns out that any submersion/immersion locally looks like these two canonical
ones.
Theorem 2.2 (Canonical Submersion Theorem). Let f : M → N be a submersion at
p ∈ M , then m = dimM ≥ n = dimN , and there exist charts (ϕ1, U1, V1) around p
and (ψ1, X1, Y1) around q = f(p) such that
ψ1 ◦ f ◦ ϕ−11 = π|V1 .
Theorem 2.3 (Canonical Immersion Theorem). Let f : M → N be an immersion at
p ∈ M , then m = dimM ≤ n = dimN , and there exist charts (ϕ1, U1, V1) around p
and (ψ1, X1, Y1) around q = f(p) such that
ψ1 ◦ f ◦ ϕ−11 = ι|V1 .
LECTURE 6: LOCAL BEHAVIOR VIA THE DIFFERENTIAL 3
¶ The constant rank theorem.
We will not prove the canonical submersion/immersion theorems above. Instead,
we will prove a more general theorem which has the canonical submersion/immersion
theorems as special cases. For this purpose, we define
Definition 2.4. We say a smooth map f : M → N is a constant rank map near p ∈M
if there is a neighborhood U of p so that dfq has constant rank (i.e. there exists r ∈ N
so that rank(df)q ≡ r) for all q ∈ U .
Example. If f is a submersion/immersion at p, then it is a submersion/immersion near
p (why?), and thus is a constant rank map near p.
Example (“Canonical” constant rank map). More generally, by composing suitable
canonical submersion and canonical immersion, we get a constant rank map
Rm = Rr+m−r π−→ Rr ι−→ Rr+n−r = Rn
which sends (x1, · · · , xr, xr+1, · · · , xm) ∈ Rm to (x1, · · · , xr, 0, · · · , 0) ∈ Rn.
We shall prove:
Theorem 2.5 (The Constant Rank Theorem). Let f : M → N be a smooth map so
that rank(df) ≡ r near p. Then there exists charts (ϕ1, U1, V1) around p and (ψ1, X1, Y1)
near f(p) such that that
ψ1 ◦ f ◦ ϕ−11 (x1, · · ·xm) = (x1, · · · , xr, 0, · · · , 0).
Proof. As usual we will convert the general case to the Euclidian case.
Step 1: The Euclidean case.
We first assume U ⊂ Rm is open, and f : U → Rn is a smooth map so that dfx
has constant rank r for all x ∈ U . By translation (in both Rm and Rn, which amounts
to composing f with suitable “translation diffeomorphisms” in both sides) we may
assume 0 ∈ U and f(0) = 0. Since rank(df)0 = r, by switching coordinates (again in
both Rm and Rn, which amounts to composing f with suitable “switching coordinates
diffeomorphisms” in both sides) we may assume that the upper-left r × r submatrix,(
∂fi
∂xj
)
1≤i,j≤r
,
of the Jacobian df = ( ∂fi
∂xj
)1≤i≤n,1≤j≤m is nonsingular at x = 0 (and thus is nonsingular
near x = 0).
[The idea: Since rank
(
∂fi
∂xj
)
1≤i,j≤r = rank(
∂fi
∂xj
)1≤i≤n,1≤j≤m, we may try to take
f1, · · · , fr as part of our coordinates, so that with respect to these new coordinates, f
will keep the first r coordinates unchanged.] Now define ϕ : U → Rm by
ϕ(x) = (f1(x), · · · , fr(x), xr+1, · · · , xm).
Then ϕ(0) = 0, and the differential
dϕ =
(( ∂fi
∂xj
)
1≤i,j≤r ∗
0 Idn−r
)
4 LECTURE 6: LOCAL BEHAVIOR VIA THE DIFFERENTIAL
is nonsingular at x = 0. By the inverse function theorem, ϕ is a local diffeomorphism
near 0, i.e., there exists neighborhood U1 of 0 in Rm and V1 of 0 in Rm such that
ϕ : U1 → V1 is a diffeomorphism. Note that by definition,
f ◦ ϕ−1(f1(x), · · · , fr(x), xr+1, · · · , xm) = f ◦ ϕ−1(ϕ(x)) = f(x) = (f1(x), · · · , fn(x))
i.e., locally near 0 we have
f ◦ ϕ−1(x) = (x1, · · · , xr, gr+1(x), · · · , gn(x))
for some smooth functions gr+1, · · · , gn (with gi(0) = 0). Moreover, by chain rule,
dfϕ−1(x) ◦ (dϕ−1)x =
(
Idr 0
∗
(
∂gi
∂xj
)
r+1≤i≤n,r+1≤j≤m
)
.
Crucial observation: Since (dϕ−1)x is a linear isomorphism, “rank(dfx) = r near 0”
implies “rank(dfϕ−1(x) ◦ (dϕ−1)x) = r near 0”, and thus implies
∂gi
∂xj
= 0, ∀r + 1 ≤ i ≤ n, r + 1 ≤ j ≤ m
near 0. It follows that in a small neighborhood of 0, we have
gi(x) = gi(x
1, · · · , xr), ∀r + 1 ≤ i ≤ n.
In other words, near 0 we have
f ◦ ϕ−1(x) = (x1, · · · , xr, gr+1(x1, · · · , xr), · · · , gn(x1, · · · , xr)).
It remains to kill these gi’s. So we define
ψ(y) =
(
y1, · · · , yr, yr+1 − gr+1(y1, · · · , yr), · · · , yn − gn(y1, · · · , yr)
)
.
in a small neighborhood of 0, and get
ψ ◦ f ◦ ϕ−1(x1, · · · , xr, xr+1, · · · , xn) = (x1, · · · , xr, 0, · · · , 0).
It remains to check that ψ is a local diffeomorphism. Again this follows from the inverse
function theorem and a simple computation dψ0 =
(
Idr 0
∗ Idn−r
)
.
Step 2: The general case.
The general case follows easily (by the standard trick): Take a coordinate neigh-
borhood (ϕ,U, V ) near p and (ψ,X, Y ) near f(p), so that f(U) ⊂ X, and dfq has
constant rank r on U . Then ψ ◦ f ◦ ϕ−1 : V → Y has constant rank r since
d(ψ ◦ f ◦ ϕ−1)x = dψf(ϕ−1(x)) ◦ dfϕ−1(x) ◦ (dϕ−1)x
and since (dϕ−1)x, dψf(ϕ−1(x)) are linear isomorphisms. Now the desired conclusion
follows from the Euclidean case. �
As a consequence, we see a map is a constant rank map if and only if it can be
written, locally, as a composition j ◦s, where s is a submersion while j is an immersion.
In particular,
• If a constant rank map is surjective, then it is a submersion.
• If a constant rank map is injective, then it is an immersion.
LECTURE 7: SARD’S THEOREM
1. Critical points and critical values
¶ Critical points and critical values: the definition.
Recall that in calculus,
• a number a ∈R is called a critical point of a smooth function f : R → R if
f ′(a) = 0.
• a point a ∈ Rm is called a critical point of a smooth function f : Rm → R if
∂f
∂xi
= 0 for all i (i.e., the differential dfa : Rm → R is not surjective).
• a point a ∈ Rm is a critical point of a smooth map f : Rm → Rn between
Euclidean open sets if dfa : Rm → Rn is not surjective (i.e., if f is not a
submersion at a).
This conception can be easily extended to smooth maps between smooth manifolds:
Definition 1.1. Let M,N be smooth manifolds and f : M → N a smooth map.
(1) We say p ∈ M is a critical point of f if dfp : TpM → Tf(p)N is not surjective1.
We say p ∈M is a regular point of f if it is not a critical point.
(2) We say q ∈ N is a regular value of f if any p ∈ f−1(q) is a regular point. We
say q ∈ N is a critical value of f if it is not a regular value.
We will denote the set of all critical points of f by Crit(f).
Remark. Let f : M → N be a smooth map.
• By definition, any q ∈ N\Im(f) is automatically a regular value. [So a regular
value of a smooth map need not be a value of the map!]
• Critical values are exactly the image of critical points, but the pre-image of
critical values may contain regular points. [So the image of a regular point
could be a critical value!]
• By definition, the set of regular points is open in M :
⇐⇒ p is a regular point of f
⇐⇒ f is a submersion at p
⇐⇒ f is a submersion near p, i.e. points “near” p are regular points.
As a consequence, the set of critical points is closed in M . However, the set of
critical values need not be closed in N , and the set of regular values need not
be open in N .
Critical/regular values are important in the study of smooth maps/manifolds. For
example, we will show that if q is a regular value of a smooth map f : M → N , then
f−1(q) is a smooth submanifold (the conception will be defined later) of M .
1We define critical points to be those points such that rank(dfp) < dimN . In some books critical
points are defined to be those points such that rank(dfp) < min(dimM,dimN).
1
2 LECTURE 7: SARD’S THEOREM
¶ Examples.
In calculus we have seen that any max/min value points are critical. This is still
true for smooth functions defined on smooth manifolds:
Proposition 1.2. Let f ∈ C∞(M) be a smooth function and p ∈ M be a maximal
or minimal value point of f . Then p is a critical point of f . [As a consequence, any
smooth function defined on a compact manifold admits at least two critical points.]
Proof. For any Xp =
∑
ai∂i|p ∈ TpM , we have
Xp(f) =
∑
ai∂i|p(f) =
∑
ai
∂(f ◦ ϕ−1)
∂xi
(ϕ(p)) = 0.
So when applied to the function g(t) = t (viewed as an element in C∞(R)), we get
dfp(Xp)(g) = Xp(g ◦ f) = Xp(f) = 0.
This implies dfp(Xp) = 0 (since if we write dfp(Xp) = c
d
dt
, then c = c d
dt
g). So p is a
critical point of f . �
Note that in the argument above, we really proved the following useful lemma:
Lemma 1.3. If we identify TtR with R [by identifying c ddt with c], then
dfp(Xp) = Xp(f), ∀f ∈ C∞(M), ∀Xp ∈ TpM,
Example. As a consequence of Proposition 1.2, the “height function”
f : Sn → R, f(x1, · · · , xn+1) = xn+1.
has the north pole (0, · · · , 0, 1) and the south pole (0, · · · , 0,−1) as its critical points
f . It is not hard to check that all other points in Sn are regular points of f (check!).
Thus the only critical values of f are 1 and −1.
Example. For the following two extremal cases,
• f : M → N is a constant map, i.e. f(p) ≡ q0 ∈ N ,
• f : M → N is any smooth map, but dimM < dimN ,
any point in M is a critical point, and thus any point in f(M) is a critical value of f .
Example. Here is a function f ∈ C∞(R) whose regular values are dense in R: we can
list all rational numbers as Q = {r1, r2, · · · }. Then we take a smooth bump function
f0 defined on R such that
supp(f0) ⊂ (−1/3, 1/3) and f0 ≡ 1 on (−1/4, 1/4).
Let
f(x) =
∞∑
k=1
rkf0(x− k).
Then each k ∈ N is a critical point of f , and thus the set of critical values of f
contains f(N) = Q. (Question: Can you find an example whose critical values form an
uncountable set?)
LECTURE 7: SARD’S THEOREM 3
2. Sard’s Theorem
¶ Measure zero sets on smooth manifolds.
The main theorem we will prove today is: For any smooth map f : M → N ,
“most” points in N (the complement of a negligible set in the sense of measure) are
regular values of f .
Let’s first explain the phrase “negligible in the sense of measure”, namely “measure
zero sets” on smooth manifolds. Note that we have not introduce any measure on M or
N yet. However, it turns out that we don’t need to define a measure when we are only
talking about “measure zero sets” on smooth manifolds! The idea is: use the Lebesgue
measure on Euclidean space.
We should be careful when we are trying to define something on manifolds via the
corresponding Euclidean ones: We may want to “transplant” the Lebesgue measure on
the Euclidean space to manifolds by using local charts. However, this does not give
us a well-defined measure on manifolds since it depends on the choice of local charts.
In fact, a measure structure is an extra structure on manifold. With only a smooth
structure at hand, we don’t have a canonical choice of measure structure. (We will see
that for an orientable manifold, each volume form gives rise to a measure).
However, “whether a set is of measure zero or not” make sense without introducing
a measure: again we use the Lebesgue measure on Euclidian space, but now it is
independent of the choices of local charts. Recall that a subset A ⊂ Rn is of measure
zero if for any ε > 0, there exists a countable union of open boxes Ui ∈ Rn so that
A ⊂
⋃
i
Ui and
∑
i
volume(Ui) < ε.
For Lebesgue measure zero sets in Rn, we have the following properties:
(i) A countable union of measure zero sets is a measure zero set.
(ii) If A ⊂ Rn is a measure zero set, and f : Rn → Rn is smooth, then f(A) is a
measure zero set in Rn. 2
Since any manifold M can by covered by countable many charts, and each chart
identify an open set in M with an open set in Rn, the following definition is independent
of coordinate charts (and thus is reasonable):
Definition 2.1. We say A ⊂M is a measure zero set if for any p ∈ A, one can find a
chart (ϕ,U, V ) of M near p so that ϕ(A ∩ U) is a measure zero set in V .
It is also obvious that a countable union of measure zero sets in a smooth manifold
is still a measure zero set. We will need the following special case of Fubini’s theorem:
Theorem 2.2 (Fubini’s Theorem — a special case). Let A be a measurable subset of
Rn such that the “slice” A ∩ ({c} × Rn−r) has Lebesgue measure zero in Rn−r for all
c ∈ Rr. Then A has Lebesgue measure zero in Rn.
2In real analysis we know that a continuous function could map a measure zero set in Rn to a set
with positive measure in Rn. However, a (local) Lipschitz map will always map a measure zero set in
Rn to a measure zero set in Rn.
4 LECTURE 7: SARD’S THEOREM
¶ Sard’s theorem.
Now we are ready to state and prove the following remarkable theorem in differen-
tial topology, which claim that the set of critical values is negligible in N . [However,
the theorem does not claim that the set of critical points is a measure zero subset in
M . In fact as we have seen, it could have that all points in M are critical points. ]
Theorem 2.3 (Sard’s Theorem3). For any smooth map f : M → N , the set of all
critical values of f is of measure zero in N .
Note that if n = dimN = 0, then there is no critical points (and thus no critical
values) at all. So in what follows we may assume n > 0.
Since we may cover M by at most countable many coordinate charts Ui, with
each f(Ui) ⊂ Xi for some coordinate charts Xi of N , and since a countable union of
measure zero sets is still of measure zero, it suffices to prove the theorem for smooth
maps between Euclidean open sets, i.e.
Theorem 2.4. If U ⊂ Rm and X ⊂ Rn are Euclidean open subsets, and f : U → V is
smooth, then the set of all critical values is of measure zero in Rn..
Proof. First observe that if m <n, then the theorem holds trivially. In fact, in this
case we can prove that the whole image f(U) is of measure zero in Rn. To see this,
we identify U with the subset U × {0} inside U ×Rn−m ⊂ Rn. Obviously U × {0} has
measure zero in Rn. Now define a map f̃ : U × Rn−m → Rn via f̃(x, y) = f(x). Then
f(U) = f̃(U × {0}). (f̃ is smooth since it is the composition of two smooth maps: the
projection map π : U × Rn−m → U and f : U → Rn.) Since f̃ is a smooth between
Euclidean spaces of the same dimension, and U × {0} is of measure zero in Rn, we
conclude that the image f̃(U × {0}) is of measure zero in Rn.
In what follows we will proceed by induction. The theorem is certainly true for
m = 0, since any countable set has measure zero. We will proceed to prove that the
theorem is true for m assuming that it is true for m− 1. Let C be the set of all critical
points of f , then we need to show that f(C) is of measure zero in N . Denote
Cj = {x ∈ U | ∂αf(x) = 0 for all |α| ≤ j}.
Obviously for any positive integer k,
f(C) = f(C\C1) ∪ f(C1\C2) ∪ · · · ∪ f(Ck−1\Ck) ∪ f(Ck).
Following J. Milnor, we will divide the proof into three steps:
Step 1: f(C\C1) has measure zero.
Step 2: f(Ci\Ci+1) has measure zero for each i.
Step 3: f(Ck) has measure zero for large k, say for k ≥ mn .
[Question: Why do we need Step 3? Are Steps 1 and 2 enough?]
3The theorem was first proven by Morse in 1939 for m = 1, and was extended to the general case
by Sard in 1942. A version for infinite-dimensional Banach manifolds was proven by Stephen Smale
in 1965.
LECTURE 7: SARD’S THEOREM 5
Proof of step 1. For each x ∈ C \C1, we will find an open set Ux 3 x such that
f(Ux∩C) has measure zero. Since C\C1 can be covered by countably many such open
sets (by second-countability), this implies f(C\C1) is of measure zero.
Note that if n = 1, then a point x is a critical point of f if and only if ∂f
∂xi
(x) = 0
for all i. It follows C\C1 = ∅, so f(C\C1) must be of measure zero. So we may assume
m ≥ n > 1. Since x 6∈ C1, there is some partial derivative, say ∂f1∂x1 , is not zero at x.
Consider
h : U → Rm, h(x) = (f1(x), x2, · · · , xm).
Then dhx is non-singular. According to the inverse function theorem, h maps a neigh-
borhood Ux of x diffeomorphically onto an open set V in Rm. The composition
g = f ◦ h−1 will then map V into Rn. Moreover, since h−1 is a diffeomorphism on
V , dh−1 is a linear isomorphism everywhere in V . So the set of critical values of g is
exactly f(Ux ∩ C).
Note that the map g we constructed is of the form (why?)
g(t, x2, · · · , xm) = (t, g2, · · · , gn).
So for each t, g induces a smooth map gt : ({t}×Rm−1)∩ V → {t}×Rn−1. Moreover,
dg =
(
1 0
∗
(
∂(gt)i
∂xj
)
i,j≥2
)
.
It follows that a point in ({t} × Rm−1) ∩ V is critical for gt if and only if it is critical
for g. But by the induction hypothesis, Sard’s theorem is true for m− 1, i.e. holds for
each gt. So the set of critical values of gt has measure zero in {t} × Rn−1. Finally by
applying Fubini’s theorem, we see that the set of critical values of g is of measure zero.
Proof of step 2. For each x ∈ Ci\Ci+1, one can find some multi-index α with |α| = i,
so that
• the partial derivative w := ∂αf vanishes on Ci,
• at least one first order partial derivative of w, say ∂w
∂x1
, does not vanish at x.
Again by applying the inverse function theorem, we conclude that
h : U → Rm, h(x) =
(
w(x), x2, · · · , xm
)
maps a neighborhood Ux of x diffeomorphically onto an open set V in Rm. (To get a
better understand and/or to avoid possible mistakes, you may want to think about the
meaning of this map for m = 1.) By construction, h carries Ci∩Ux into the hyperplane
{0} × Rm−1. Again we consider the map g = f ◦ h−1. Then the critical points of g of
type Ci are all in the hyperplane {0} × Rm−1. Let
ḡ : ({0} × Rm−1) ∩ V → Rn
be the restriction of g. Then the set of critical points of g of type Ci coincides with the
set of critical points of ḡ. By induction, the set of critical values of ḡ is of measure zero
in Rn. It follows that the image of the critical points of g of type Ci is of measure zero.
Therefore, f(Ci ∩ Ux) is of measure zero. Since Ci\Ci+1 can be covered by countable
many such sets Ux, f(Ci\Ci+1) is of measure zero.
6 LECTURE 7: SARD’S THEOREM
Proof of step 3. Let Q ⊂ U be a cube whose sides are of length δ. We will prove
that for k > m
n
−1, f(Ck ∩Q) has measure zero. Since Ck can be covered by countably
many such cubes, this implies that f(Ck) has measure zero.
From Taylor’s theorem, the compactness of Q and the definition of Ck, we see that
f(x+ h) = f(x) +R(x, h),
where |R(x, h)| < a|h|k+1 for x ∈ Ck ∩Q, x+ h ∈ Q, and the constant a depends only
on f and Q. Now we subdivide Q into rm cubes whose sides are of length δ
r
. Let Q1
be a cube of subdivision that contains a point x ∈ Ck. Then any point of Q1 can be
written as x + h with |h| <
√
m δ
r
. It follows that f(Q1) lies in a cube with sides of
length b
rk+1
centered about f(x), where b = 2a(
√
mδ)k+1 is a constant. So f(Ck ∩ Q)
is contained in the union of at most rm cubes having total volume
Vol ≤ rm( b
rk+1
)n = bnrm−(k+1)n.
Since k > m
n
− 1, we see Vol → 0 as r → ∞. It follows that f(Ck ∩ Q) is of measure
zero. �
Remark. Sard’s theorem holds for Cr maps between Cr manifolds, provided r ≥ 1 +
max(m − n, 0). Moreover, this is sharp. However, the proof above does not work in
this more general setting. [Can you see why?]
LECTURE 8: SMOOTH SUBMANIFOLDS
1. Smooth submanifolds
¶ Smooth submanifolds: The definition.
Let M be a smooth manifold of dimension n. What object can be called a “smooth
submanifold” of M? [Recall: what is a vector subspace W of a vector space V ? W
should satisfy three conditions:
• W is a subset of V ;
• W is a vector space by itself;
• the vector space structure on W should be the restriction of the vector space
structure on V .
Similarly, what is a subgroup of a group? What is a topological subspace of a topolog-
ical space? In each case you can always write down the three conditions: the inclusion
relation, the structure itself, and the compatibility.] A smooth submanifold S of M
should be something that satisfies three the following conditions:
• S should be a subset of M ;
• S itself should be a smooth manifold of dimension k ≤ n;
• the smooth structures on S and on M should be compatible.
The last condition, i.e. the compatibility, can be stated more precisely: the smooth
structure (=a set of coordinate charts) on S should be the “restriction” of the smooth
structure (=a set of coordinate charts) on M .
Definition 1.1. A subset S ⊂ M is a k-dimensional smooth submanifold of M if for
every p ∈ S, there is a chart (ϕ,U, V ) around p of M such that
ϕ(U ∩ S) = V ∩ (Rk × {0}) = {x ∈ ϕ(U) | xk+1 = · · · = xn = 0}.
We will call codim(S) = n− k the codimension of S.
Remark. Roughly speaking, smooth submanifolds are objects that are defined locally
(on a coordinate chart) by equations
ϕk+1 = · · · = ϕn = 0.
Note that ϕk+1, · · · , ϕn are smooth functions on U , since ϕ is a diffeomorphism.
1
2 LECTURE 8: SMOOTH SUBMANIFOLDS
¶ Smooth submanifolds: Examples.
Example. Let M,N be smooth manifolds, and f : M → N be smooth. Then the graph
Γf = {(p, q) | q = f(p)} ⊂M ×N
is a smooth submanifold of M ×N . To see this, we take a chart (ϕ,U, V ) of M near p
and a chart (ψ,X, Y ) of N near q = f(p). Then (c.f. PSet 1-1-6) (ϕ×ψ,U×X, V ×Y ) is
a chart of M×N near (p, q). But this chart is not good for our purpose. To get a chart
that is suitable for our purpose, we write the equation q = f(p) as ψ−1(y) = f(ϕ−1(x)),
i.e. y = ψ(f(ϕ−1(x))). Now we define a smooth map
Ψ : V × Y → Rm × Rn, (a, b) 7→ (a, b− ψ ◦ f ◦ ϕ−1(a)).
It is easy to see that Ψ is one-to-one and is a local diffeomorphism everywhere, thus is
a global diffeomorphism from V × Y onto its image Ψ(V × Y ), which is an Euclidian
open set. Thus (Ψ ◦ (ϕ × ψ), U × X,Ψ(V × Y )) is also a local chart of M × N near
(p, q). Moreover, with respect to this local chart,
(p, q) ∈ Γf ∩ (U ×X)=⇒ ψ(q) = ψ(f(ϕ−1(ϕ(p)))) =⇒ Ψ(ϕ(p), ψ(q)) = (ϕ(p), 0),
so the conclusion follows.
Remark. In Lecture 2 we mentioned that for any continuous function f : U ⊂ Rn → R,
there is a smooth structure on the graph Γf to make it a smooth manifold of dimension
n. However, in general Γf is not a smooth submanifold of Rn+1 if f is not smooth.
• If you repeat the arguments in the above example, which step does not work?
• However, it is possible f is not a smooth function, while Γf is still a smooth
submanifold: Consider the function f(x) = x1/3. Then the graph of y = f(x)
is the same as the graph of the function x = y3, which is of course a smooth
submanifold of R2!
Example. The sphere Sn is a smooth submanifold of Rn+1. Can you construct a local
chart of Rn+1 near every point of Sn which satisfies the condition in the definition 1.1?
¶ The induced smooth structures on smooth submanifolds.
Note that in the definition of a smooth submanifold above, we did not spell out
the smooth structure on S. To construct natural charts on S, we denote
π : Rn → Rk, (x1, · · · , xn) 7→ (x1, · · · , xk)
 : Rk ↪→ Rn, (x1, · · · , xk) 7→ (x1, · · · , xk, 0, · · · , 0).
Then we have
Proposition 1.2. Let (ϕ,U, V ) be a chart on M that satisfies Definition 1.1. Let
X = U ∩ S, Y = π ◦ ϕ(X) and ψ = π ◦ ϕ|X . Then (ψ,X, Y ) is a smooth chart on S
and charts of this form are compatible, so that S is a smooth manifold. Moreover, the
inclusion map ι : S ↪→ X is a smooth immersion.
Proof. By definition, ψ is invertible and the inverse ψ−1 = ϕ−1 ◦ . So (ψ,X, Y ) is a
chart on S. It remains to check that charts of this type are compatible. In fact, the
transition maps are
LECTURE 8: SMOOTH SUBMANIFOLDS 3
ψβ ◦ ψ−1α = π ◦ ϕβ ◦ ϕ−1α ◦  = π ◦ ϕα,β ◦ ,
which are smooth. Moreover, with respect to these smooth structures, the inclusion
map ι : S ↪→M is a smooth immersion since by definition,
ϕ ◦ ι ◦ ψ−1 = .
�
¶ The tangent space of a smooth submanifold (as a vector subspace).
Now let S ⊂ M be a submanifold, and p ∈ S. Since ι : S ↪→ M is an embedding,
dιp : TpS → TpM is injective. We might identify TpS with the vector subspace dιp(TpS)
of TpM for every p ∈ S. In other words, we can identify any vector Xp ∈ TpS with the
vector X̃p = dιp(Xp) in TpM so that for any f ∈ C∞(M),
X̃p(f) = (dιp(Xp))f = Xp(f ◦ ι) = Xp(f |S).
A natural question is: which vectors in TpM can be regarded as vectors in TpS?
Theorem 1.3. Suppose S ⊂M is a submanifold, and p ∈ S. Then
TpS = {Xp ∈ TpM | Xp(f) = 0 for all f ∈ C∞(M) with f |S = 0}.
Proof. From the description above, we see if Xp ∈ TpS, then for f ∈ C∞(M) with
f |S = 0, X̃p(f) = Xp(f |S) = 0.
Conversely, if Xp ∈ TpM satisfies Xp(f) = 0 for all f that vanishes on S, we need
to show Xp ∈ TpS. Take a coordinate chart (ϕ,U, V ) on M such that near p, S is
given by xk+1 = · · · = xn = 0. Then TpM is the span of ∂1, · · · , ∂n, while TpS is the
subspace spanned by ∂1, · · · , ∂k. In other words, a vector Xp =
∑
X i∂i lies in TpS if
and only if X i = 0 for i > k.
Now let h be a smooth bump function supported in U that equals 1 in a neighbor-
hood of p. For any j > k, consider the function fj(x) = h(x)x
j(ϕ(x)), extended to be
zero on M \ U . Then fj|S = 0. So
0 = Xp(fj) =
∑
X i
∂(h(ϕ−1(x))xj)
∂xi
(ϕ(p)) = Xj
for any j > k. It follows that Xp ∈ TpS. �
2. Smooth submanifolds as pre-images and images
¶ Smooth submanifolds as level sets.
Now we prove the result that we mentioned last time: the pre-image of a regular
value of any smooth map is a smooth submanifold. [Thus, as a consequence of Sard’s
theorem, for any smooth map f : M → N , if f(M) is not negligible in N , then for
most q ∈ f(M), the pre-image f−1(q) is a smooth submanifold of M .]
Theorem 2.1 (Regular Level Set Theorem). Let f : M → N be a smooth map and
q ∈ N a regular value of f . Then f−1(q) is a smooth submanifold of M of dimension
dimM − dimN . Moreover, for every p ∈ S, TpS = ker(dfp : TpM → TqN).
4 LECTURE 8: SMOOTH SUBMANIFOLDS
Note that if q is a regular value, then f is a full rank map near any p ∈ f−1(q). So
the regular level set theorem is a consequence of
Theorem 2.2 (Constant rank level set theorem). Let M,N be smooth manifold, and
f : M → N be a smooth map with constant rank r. Then each level set of f is a closed
submanifold of codimension r in M . Moreover, for every p ∈ S, TpS (viewed as a
vector subspace of TpM) is the kernel of the map dfp : TpM → TqN .
Proof. Let p ∈ S := f−1(q). Then by the constant rank theorem that we proved in
Lecture 6, there are charts (ϕ1, U1, V1) centered at p and (ψ1, X1, Y1) centered at q such
that f(U1) ⊂ X1, and
ψ1 ◦ f ◦ ϕ−11 (x1, · · · , xm) = (x1, · · · , xr, 0, · · · , 0).
It follows that ϕ1 maps U1 ∩ f−1(q) onto V1 ∩ {(0, · · · , 0, xr+1, · · · , xm)}. So f−1(q) is
a submanifold of M of codimension r.
Next denote the inclusion by ι : S → M . Then for any p ∈ S, f ◦ ι(p) = q. In
other words, f ◦ ι is a constant map on S. So dfp ◦ dιp = 0, i.e. dfp = 0 on the image of
dιp : TpS ↪→ TpM , or in other words, TpS ⊂ ker(dfp). But dimTpS = dimS = m − r,
and
dim ker(dfp) = dim ker((dψ1)f(p) ◦ dfp ◦ (dϕ−11 )ϕ(p)) = m− r.
By dimension counting we conclude that TpS coincides with the kernel of dfp. �
The regular level set theorem and the constant rank theorem are very powerful
tools in applications. For example, we immediately see
• Sn is a smooth submanifold of Rn+1.
• According to PSet2-2-4(a), SL(n,R) is a smooth submanifold of GL(n,R).
• According to PSet2-2-4(b), O(n,R) is a smooth submanifold of GL(n,R).
Note that the level set of a critical value may fail to be a smooth manifold. For example,
consider the function
f : R2 → R, f(x, y) = x2 − y2.
Then df(x,y) = (2x, 2y) and thus the only critical point is (0, 0). For the only critical
value 0, the level set f−1(0) is not a manifold.
¶ Smooth submanifolds as images: immersions are not enough.
As we have seen, any smooth submanifold S of M is the image of a smooth im-
mersion ι : S ↪→M . It is natural to ask: whether the image of any smooth immersion
is a smooth submanifold? First by the canonical immersion theorem, if f : N →M is
an immersion, then for each p ∈ N , there exists a coordinate neighborhood U of p so
that f(U) is a smooth submanifold of M (why? write down a proof). Unfortunately
in general f(N) is need not be a smooth submanifold of M :
Example. The following two graphs are the images of two immersions of R into R2.
For the first one, the immersion is not injective. For the second one, the immersion is
injective, while the image still have different topology than R.
LECTURE 8: SMOOTH SUBMANIFOLDS 5
Example. There is a more complicated/interesting example: consider the map
f : R→ T2 = S1 × S1, f(t) = (eit, ei
√
2t).
Then f is an immersion (why?), and the image f(R) is a “dense curve”1 in T2.
Remark. For the three immersions above whose image are not submanifolds, the first
one is “worst” since the image is not a manifold in any sense: at the crossing point,
the image is not a manifold, no matter what topology you give to the image. On the
other hand, for the second one and the third one, we can easily see that
• if we use the “subspace topology” inherited from R2 or T2, then the images are
not manifolds;
• if we endow the images with the topology that “borrowed” from R, then the
images are smooth manifolds by themselves!
In general, the image of any injective immersion is a manifold, where the manifold
structure is “borrowed” from the source manifold. So people call the images of injec-
tive immersions immersed submanifolds. To distinguish immersed submanifolds with
smooth submanifolds defined in Definition 1.1, sometimes people call smooth subman-
ifolds embedded submanifolds or regular submanifolds.
¶ Smooth submanifolds as images: embeddings.
What is the difference between smooth submanifolds and immersed submanifolds?
As we just described, the topology that makes an immersed submanifold a manifold is
the topology from the source manifold, not the “subspace topology” from the targetmanifold. On the other hand, if S is a smooth submanifold of M , then the underlying
topology of the smooth manifold S is the topology generated by charts (ψ,X, Y ) in
Proposition 1.2. By definition 1.1 it is easy to see
1This is a consequence of the fact that {{n
√
2} | n ∈ Z} is dense in [0, 1].
6 LECTURE 8: SMOOTH SUBMANIFOLDS
Proposition 2.3. Let S be a smooth submanifold of M . Then ι : S ↪→ M is a
homeomorphism from S to ι(S) (endowed with the subspace topology from M).
Remark. If S is a smooth submanifold of M , then there is a unique topology/smooth
structure on S so that the inclusion map ι : S ↪→M is a smooth immersion which is a
homeomorphism onto its image. (See Theorem 5.31 in Lee’s book.)
So for any smooth submanifold S, the inclusion map ι : S ↪→ M is a special
immersion which is a homeomorphism onto its image.
Definition 2.4. Let M,N be smooth manifolds, and f : N → M an immersion. f is
called an embedding if it is a homeomorphism onto its image f(N), where the topology
on f(N) is the subspace topology as a subset of M .
By definition, the inclusion map ι : S ↪→M is an embedding. So
each smooth submanifold is the image of an embedding.
Conversely,
Theorem 2.5. Let f : N → M be an embedding. Then the image f(N) is a smooth
submanifold of M .
Proof. Let p ∈ N and q = f(p). Since f is an immersion, the canonical immersion
theorem implies that there exists charts (ϕ1, U1, V1) near p and (ψ1, X1, Y1) near q such
that on V1, ψ1 ◦ f ◦ ϕ−11 is the canonical embedding  : Rm → Rn restricted to V1, i.e.
ψ1 ◦ f =  ◦ ϕ1
on U1. Since f is a homeomorphism onto its image, f(U1) is relative open in the
subspace f(N) ⊂ M . In other words, there exists an open set X ⊂ M such that
f(U1) = f(N) ∩X. Replace X1 by X1 ∩X, and Y1 by ψ1(X1 ∩X). Then for this new
chart (ψ1, X1, Y1),
ψ1(X1 ∩ f(N)) = Y1 ∩ ψ1(f(U1)) = Y1 ∩ (ϕ1(U1)) = Y1 ∩ (Rm × {0}).
[Please repeat the above argument for the “dense curve in T2” example to see what’s
wrong there.] �
Let’s summarize the main difference between an immersion and an embedding:
• If f : N → M is an immersion, then by the canonical immersion theorem, any
point p ∈ N has a neighborhood in N whose image is “nice” in M .
• If f : N → M is an embedding, then by Theorem 2.5, any point q ∈ f(N) has
a neighborhood in f(N) that is “nice” in M .
LECTURE 9: THE WHITNEY EMBEDDING THEOREM
As we mentioned in Lecture 1, historically there are two definitions of manifolds:
the extrinsic/concrete definition used by Poincaré (as the set of possible values of
some variables with certain constraints in Euclidean space), and the intrinsic/abstract
definition used by Weyl (as local charts patched together). During 1930s Whitney
showed that the two definitions are in fact the same. In fact, Whitney’s result is much
more stronger than this. He showed that not only one can embed any smooth manifold
into some Euclidian space, but that the dimension of the Euclidian space can be chosen
to be (as low as) twice the dimension of the manifold itself!
Theorem 0.1 (The Whitney embedding theorem). Any m dim smooth manifold M
can be embedded into R2m+1 (and immersed into R2m).
Remark. In 1944, by using completely different techniques (now known as the “Whitney
trick”), Whitney was able to prove
Theorem 0.2 (The Strong Whitney Embedding Theorem). Any m(≥ 2) dim
smooth manifold M can be embedded into R2m (and immersed into R2m−1).
We will not prove this stronger version in this course, but just mention that the Whitney
trick was further developed in h-cobordism theory by Smale, using which he proved the
Poincaré conjecture in dimension ≥ 5 in 1961!
Remark. Many shaper results were proved during the second half of 20th century:
• Any smooth compact orientable m-manifold can be embedded into R2m−1.
• For m 6= 2k, any smooth m-manifold can be embedded into R2m−1. (But if
m = 2k, RPm cannot be embedded into R2m−1).
• Any smooth m-manifold can be immersed into R2m−a(m), where a(m) is the
number of 1’s that appear in the binary expansion of m.
• Any compact hypersurface [i.e. has codimension 1] in Rn is orientable. In par-
ticular, non-orientable surfaces like RP2, the Klein bottle etc can’t be embedded
into R3. [But they can be immersed into R3. Note: “immersed into” 6= “the
image is an immersed submanifold”, since the immersion need not be injective.]
In what follows we prove Theorem 0.1. It is easier if M is compact, while the
non-compact case is a bit harder1. In both cases the proof is divided into three steps:
• Step 1: Injectively immerse M into RK for a large number K
• Step 2: For K > 2m+ 1, one can project RK to some RK−1.
• Step 3: Under certain compactness conditions, injective immersions must be
embeddings.
The philosophy behind each step:
1Since compact and non-compact manifolds are very different, they get special names: closed man-
ifold=compact manifold without boundary, open manifold=non-compact manifold without boundary.
1
2 LECTURE 9: THE WHITNEY EMBEDDING THEOREM
• Behind Step 1: Each small piece of the manifold can be embedded as a small
piece of Rm inside a larger ambient Euclidian space; to embed two “consec-
utive” pieces, they may “intersect”, but it should be possible to avoid the
“intersection” if the ambient space (in which the two pieces were embedded)
has dimension greater than the sum of the dimensions of two pieces!
• Behind Step 2: If the ambient space has a very high dimension, you can “see”
the whole immersed manifold in some directions. Project along that direction
to product an immersion into an ambient space that has lower dimension.
• Behind Step 3: Injective immersions are locally nice, compactness means there
are only finitely many local pieces.
1. The Whitney embedding theorem: Compact Case
¶ Injectively immerse a compact manifold into RN .
To prove the Whitney embedding theorem for compact manifolds, we start with
Theorem 1.1. Any compact smooth manifold M admits an injective immersion into
RK for sufficiently large K.
Proof. Let {(ϕi, Ui, Vi) | 1 ≤ i ≤ k} be a finite set of coordinate charts on M that
covers M . Let {ρi | 1 ≤ i ≤ k} be a P.O.U. subordinate to {Ui | 1 ≤ i ≤ k}. Define
Φ : M → Rk(m+1), p 7→ (ρ1(p)ϕ1(p), · · · , ρk(p)ϕk(p), ρ1(p), · · · , ρk(p)).
We shall prove
• Φ is an injective map. Suppose Φ(p1) = Φ(p2). Take an index i so that
ρi(p1) = ρi(p2) 6= 0. Then p1, p2 ∈ supp(ρi) ⊂ Ui. It follows that ϕi(p1) =
ϕi(p2). So we must have p1 = p2 since ϕi is bijective.
• Φ is an immersion. For any Xp ∈ TpM , by Leibnitz law,
dΦp(Xp) = (Xp(ρ1)ϕ1(p) + ρ1(p)(dϕ1)p(Xp), · · · ,
Xp(ρk)ϕk(p) + ρk(p)(dϕk)p(Xp), Xp(ρ1), · · · , Xp(ρk)).
It follows that if dΦp(Xp) = 0, thenXp(ρi) = 0 for all i, and thus ρi(p)(dϕi)p(Xp) =
0 for all i. Pick an index i so that ρi(p) 6= 0. We see (dϕi)p(Xp) = 0. Since ϕi
is a diffeomorphism, we conclude that Xp = 0. So dΦp is injective. �
Remark. In fact we proved a stronger result:
Theorem 1.2. If M can be covered by finitely many coordinate charts, then
there exists an injective immersion from M into some Euclidian space.
¶ Lowering the dimension.
Next we apply Sard’s theorem to prove (note: we don’t assume compactness here)
Theorem 1.3. If a smooth manifold M of dimension m admits an injective immersion
into RK for some K > 2m+ 1, then it admits an injective immersion into RK−1.
LECTURE 9: THE WHITNEY EMBEDDING THEOREM 3
Proof. Suppose we already have an injective immersion Φ : M → RK with K > 2m+1.
We want to produce an injective immersion of M into RK−1. To do so, we consider
the compositions of Φ with projections from RK to all possible K − 1 dimensional
vector subspaces in RK , and we will show that for “almost all” projections, we still get
injective immersions.
Note that any K−1 dimensional vector subspace in RK is uniquely determined by
its normal direction, which is a 1-dimensional line passing the origin, and the set of all
1-dimensional lines passing the origin in RK is the real projective space RPK−1, which
is a smooth manifold of dimension K − 1. For any [v]∈ RPK−1, we let
P[v] = {u ∈ RN | u · v = 0} ' RK−1
be the orthogonal complement space of [v] in RK . Let
π[v] : RK → P[v]
be the orthogonal projection to this hyperplane. We claim that the set{
[v] | Φ[v] = π[v] ◦ Φ is not an injective immersion
}
has measure zero in RPK−1. Hence for most [v] ∈ RPK−1, the map Φ[v] is an injective
immersion from M into some RK−1.
First let’s consider [v]’s so that Φ[v] is not injective. Then one can find p1 6= p2 so
that Φ[v](p1) = Φ[v](p2), i.e. 0 6= Φ(p1)− Φ(p2) lies in the line [v]. In other words,
[v] = [Φ(p1)− Φ(p2)].
So [v] must lie in the image of the smooth map
α : (M ×M) \∆M → RPK−1, (p1, p2) 7→ [Φ(p1)− Φ(p2)],
where ∆M = {(p, p) | p ∈ M} is the “diagonal” in M ×M . Since (M ×M) \ ∆M is
a smooth manifold of dimension 2m < K − 1, Sard’s theorem implies that the image
of α is of measure zero in RPK−1. So the set of [v]’s so that Φ[v] is not injective is of
measure zero.
Next let’s consider [v]’s so that Φ[v] is not an immersion. Then there exists some
p ∈M and some 0 6= Xp ∈ TpM so that (dΦ[v])p(Xp) = 0, i.e.
(dπ[v])Φ(p)(dΦ)p(Xp) = 0.
Since π[v] is linear, dπ[v] = π[v]. [A conceptional proof: What is the differential of a
smooth map? It’s the linear approximation of the smooth map. What if the smooth
map is already linear then?] It follows that 0 6= (dΦ)p(Xp) is in [v], i.e.
[v] = [(dΦ)p(Xp)].
In other words, [v] lies in the image of
β : TM \ {0} → RPK−1, (p,Xp) 7→ [(dΦ)p(Xp)],
where TM \ {0} = {(p,Xp) | Xp 6= 0} is an open submanifold of the tangent bundle
TM , which is a smooth manifold of dimension 2m. Again since TM has dimension
2m < K − 1, by Sard’s theorem, the image of β is of measure zero. So the set of [v]’s
so that Φ[v] is not an immersion is of measure zero. �
4 LECTURE 9: THE WHITNEY EMBEDDING THEOREM
In view of the fact
(dΦ[v])p(Xp) = 0⇐⇒ (dΦ[v])p(
Xp
|Xp|
) = 0,
one can modify the last step and prove (note: we don’t assume compactness here)
Theorem 1.4. If a smooth manifold M of dimension m can be embedded into R2m+1,
then it can be immersed into R2m. [We don’t require the immersion to be injective.]
Sketch of proof. We first embed M into R2m+1, then repeat the last step in the proof of
Theorem 1.3, with the modification that we choose Xp ∈ TpM so that |Xp| = 1 (here
the length of a vector Xp ∈ TpM (as a subset of TpR2m+1) is the Euclidian length). In
other words, the map β in the proof above can be replaced by the map
β̃ : SM → RP2m, (p,Xp) 7→ [(dΦ)p(Xp)],
where SM = {(p,Xp) | |Xp| = 1} is the “sphere bundle” of M , which is a smooth
manifold of dimension 2m− 1. (Please write down the details.) �
Remark. Here you already see the advantage of the existence of an embedding into
Euclidian spaces: you can now talk about the “length” of tangent vectors (which is an
“extrinsic” conception here). (One can also define length of tangent vectors on smooth
manifolds intrinsically, which leads to the subject Riemannian Geometry.)
¶ Proof of Whitney’s theorems for compact manifolds.
Let f : N →M be an injective immersion. As we mentioned at the end of Lecture
8, any p ∈ N has a neighborhood which is embedded into M via f . In general we can’t
conclude that f is an embedding. However, if N is compact, we can cover N by finitely
many such neighborhoods, and thus it is clear that f should be a global embedding:
Theorem 1.5. If f : N →M is an injective immersion, and N is compact, then f is
an embedding.
Proof. Since f is injective, f : N → f(N) is invertible. Since f : N →M is continuous,
f : N → f(N) is also continuous. It remains to show that f : N → f(N) is closed.
So we take an arbitrary closed set A in N . Since N is compact, A has to be compact.
So f(A) is compact. Because f(N) is Hausdorff, f(A) is a closed subset in f(N). So
f : N → f(N) is a closed map, and thus f−1 : f(N)→ N is continuous. �
As an immediate consequence, we see
Theorem 1.6 (The Whitney Embedding Theorem, Compact Case). Any smooth com-
pact manifold M of dimension m can be embedded into R2m+1.
In view of Theorem 1.4, we get
Theorem 1.7 (The Whitney Immersion Theorem, Compact Case). Any smooth com-
pact manifold M of dimension m can be immersed into R2m.
LECTURE 9: THE WHITNEY EMBEDDING THEOREM 5
2. The Whitney embedding theorem: Non-Compact Case
Of course the above arguments fail when the manifold M is non-compact, since we
can no longer cover M by finitely many coordinate charts, and an injective immersion
from a non-compact manifold could fail to be an embedding. In general non-compact
objects could behave quite differently from compact objects. Fortunately for the Whit-
ney embedding theorem, the non-compact version holds as well.
¶ Injectively immerse a non-compact manifold into RN .
We start with a non-compact analogue of Theorem 1.1:
Theorem 2.1. Any non-compact smooth manifold M admits an injective immersion
into RK for sufficiently large K.
Proof. [The idea: a non-compact manifold is the union of countably many compact
subsets!] According to PSet 1-2-4, one can find a positive smooth exhaustion function
f on M . For each i ∈ N, we define
Mi = f
−1([i, i+ 1]).
We cover the compact set Mi by finitely many coordinate charts U1, · · · , Uk, and let
Ni = (U1 ∪ · · · ∪ Uk) ∩ f−1
(
(i− 0.1, i+ 1.1)
)
.
Then each Ni is an open submanifold of M so that Mi ⊂ Ni. Moreover, Ni ∩ Nj = ∅
if |i − j| ≥ 2. By construction, each Ni can be covered by finitely many coordinate
charts. So according to Theorem 1.2, there is an injective immersion from Ni into some
RK for K large. Since Ni is a smooth manifold (without boundary) of dimension m,
Theorem 1.3 implies that we can find injective immersions ϕi from Ni into R2m+1.
Now pick smooth bump function ρi so that ρi = 1 in an open neighborhood of Mi
and suppρi ⊂ Ni (here we used smooth Urysohn’s lemma, i.e. PSet1-2-3). Define
Φ : M → R4m+3, p 7→
(∑
i odd
ρi(p)ϕi(p),
∑
i even
ρi(p)ϕi(p), f(p)
)
.
Note that near each point p ∈ M , at most one term in each summations above is
nonzero. So Φ is a smooth map. It remains to show that Φ is an injective immersion:
• Φ is injective If Φ(p1) = Φ(p2), then ∃i ∈ N so that f(p1) = f(p2) ∈ [i, i+ 1].
So p1, p2 ∈Mi ⊂ Ni and ϕi(p1) = ϕi(p2). Since ϕi is injective, we get p1 = p2.
• Φ is an immersion Suppose p ∈ Mi. Without loss of generality, we assume i
is odd. Then for any 0 6= Xp ∈ TpM ,
dΦp(Xp) = ((dϕi)p(Xp), ∗, ∗).
Since ϕi is an immersion on Ui 3 p, we get (dϕi)p(Xp) 6= 0. Thus dΦp(Xp) 6= 0.
This completes the proof. �
6 LECTURE 9: THE WHITNEY EMBEDDING THEOREM
¶ Proof of Whitney’s theorems for non-compact manifolds.
According to Theorem 2.1 and Theorem 1.3, any smooth manifold of dimension
m admits an injective immersion into R2m+1. But for non-compact manifolds, such
injective immersion need not to be an embedding. So we need a non-compact analogue
of Theorem 1.5. What is a good generalization of a continuous map whose domain is
compact? A proper map! (In some sense, properness is a substitution of compactness.)
In PSet3-1-5 we proved
Proposition 2.2. Any proper injective immersion f : N →M is an embedding.
Now we are ready to prove
Theorem 2.3 (The Whitney Embedding Theorem, Non-Compact Case). Any smooth
non-compact manifold M of dimension m can be embedded into R2m+1.
Proof. By Theorem 2.1 and Theorem 1.3, there exists an injective immersion Φ : M →
R2m+1. Composing Φ with the diffeomorphism
R2m+1 → B2m+1(1) = {x ∈ R2m+1 | |x| ≤ 1}, x 7→ x
1 + |x|2
if necessary, we may assume that |Φ(p)| ≤ 1 for all p ∈M .
Take any positive smooth exhaustion function f on M , and define
Φ̃ = (Φ, f) : M → R2m+2, p 7→ (Φ(p), f(p)).
Repeating the proof of Theorem 1.3, we get another injective immersion
Ψ = π[v] ◦ Φ̃ : M → R2m+1,
where π[v] is some projection π : R2m+2 → P[v] ' R2m+1. Moreover, since almost all
[v] ∈ RP2m+1 work for our purpose, we can always choose [v] so that [v] 6= [0 : · · · : 0 : 1].
We claim that Ψ is proper. So according to Proposition 2.2, Ψ is an embedding.
It remains to prove that Ψ is proper. WLOG, we may assume v ∈ S2m+1. Denotev = (v′, v2m+1). Then the condition [v] 6= [0 : · · · : 0 : 1] is equivalent to |v2m+1| < 1.
Since |v| = 1, we have π[v](x) = x− (x · v)v. So we get
Ψ(p) = (Φ(p), f(p))−
[
(Φ(p), f(p)) · (v′, v2m+1)
]
(v′, v2m+1)
=
(
∗, f(p)[1− (v2m+1)2]− (Φ(p) · v′)v2m+1
)
.
Now we prove properness. For any compact set K ⊂ R2m+1, ∃A > 0 so that
K ⊂ {x | |x2m+1| < A}.
Since |Φ(p)| ≤ 1, |v2m+1| ≤ 1 and |v′| ≤ 1, we have∣∣f(p)[1− (v2m+1)2]− (Φ(p) · v′)v2m+1∣∣ < A
=⇒
∣∣f(p)[1− (v2m+1)2]∣∣ ≤ A+ ∣∣(Φ(p) · v′)v2m+1∣∣ ≤ A+ 1.
It follows that Ψ−1(K) ⊂ f−1([− A+1
1−|v2m+1|2 ,
A+1
1−|v2m+1|2 ]). But Ψ
−1(K) is closed in M
since Ψ is continuous, and f−1([− A+1
1−|v2m+1|2 ,
A+1
1−|v2m+1|2 ]) is compact in M since f is
proper. So Ψ−1(K) is compact. So Ψ is proper and we are done. �
LECTURE 9: THE WHITNEY EMBEDDING THEOREM 7
In view of Theorem 1.4, we get
Theorem 2.4 (The Whitney Immersion Theorem, Non-Compact Case). Any smooth
non-compact manifold M of dimension m can be immersed into R2m.
Finally we remark that the Whitney embedding/immersion theorems also holds
for smooth manifolds with boundary. For details, c.f. Hirsch, Differential Topology,
Chapter 1, Theorem 4.3.
LECTURE 10: TUBULAR NEIGHBORHOOD THEOREM
1. Generalized Inverse Function Theorem
¶ Generalized Inverse Function Theorem.
In Lecture 6 we proved the inverse function theorem which claims that if dfp is a
linear isomorphism, then f is a local diffeomorphism near p, i.e. f is maps a neighbor-
hood of p diffeomorphically to a neighborhood of f(p). In applications, one may need
to map a neighborhood of a subset/submanifold X diffeomorphically to a neighborhood
of f(X). It turns out that under suitable assumptions, the inverse function theorem
can be extended to this setting to produce such a diffeomorphism on a neighborhood
a smooth submanifold:
Theorem 1.1 (Generalized Inverse Function Theorem).
Let f : M → N be a smooth map, and X ⊂ M a submanifold. Suppose dfx : TxM →
Tf(x)N is a linear isomorphism for each x ∈ X. Moreover,
Case 1: If X is compact, we simply assume f is injective on X.
Case 2: If X is non-compact, we assume f(X) is a submanifold of N
and f maps X diffeomorphically onto f(X).
Then f maps a neighborhood U of X in M diffeomorphically onto a neighborhood V of
f(X) in N .
We give a couple remarks on the conditions:
Remark. In fact, the simple condition “f is injective on X” is equivalent to the com-
plicated condition “X diffeomorphically onto f(X)” when X is compact: By assump-
tion, f |X : X → N is an immersion. So if f |X is injective and X is compact, then
f |X : X → N is an embedding, and thus f |X : X → f(X) is a diffeomorphism.
Remark. When X is non-compact, the theorem may fail if we only assume X is a
smooth submanifold and f |X is injective. Here is a simple example:
Consider the covering map
f : R2 → T2, (t, s) 7→ (eit, eis)
which is a local diffeomorphism near any (t, s) ∈ R2. We take X to be
the “irrational-slope line”
X = {(t,
√
2t) | t ∈ R}
in R2 (which is a perfectly nice smooth submanifold), then f |X is injec-
tive but there is no hope to find neighborhoods U of X in R2 and V of
f(X) in T2 so that f maps U diffeomorphically onto V .(why?)
1
2 LECTURE 10: TUBULAR NEIGHBORHOOD THEOREM
¶ Generalized Inverse Function Theorem: The compact case.
Proof of Theorem 1.1, Case 1 (compact version).
By the inverse function theorem, f is a local diffeomorphism near each point in X.
According to Proposition 1.3 in Lecture 6, it is enough to prove that f is injective
on a neighborhood of X. For this purpose we embed M into RK , and consider the
“ε-neighborhood” of X in M :
Xε = {x ∈M | d(x,X) < ε},
where d(·, ·) is the Euclidean distance between a point and a set, namely
d(x,X) = inf{d(x, y) | y ∈ X}.
Note that Xε is a open subset in RK , and it is bounded since X is compact. Moreover
we have X =
⋂
k>0X
1/k since X is closed. (Why?)
Now we proceed by contradiction. If f is not one-to-one on each X1/k, then one
can find ak 6= bk ∈ X1/k such that f(ak) = f(bk). Since all ak’s lie in a bounded closed
set in RK , which is compact, one can find a subsequence such that aki → a∞ ∈ X.
Similarly there is a subsequence bkij → b∞ ∈ X. Since f(a∞) = f(b∞), one must
have a∞ = b∞, since f is one-to-one on X. So in any neighborhood of a∞, f is not
one-to-one. But dfa∞ is linear isomorphism implies that f is a local diffeomorphism
near a∞, which is a contradiction. �
By staring at the proof, one can see that in the proof we don’t need the submanifold
structure of X, the only requirement is that X is a compact subset in M . So what we
really proved is a result with weaker assumption:
Theorem 1.2 (Generalized Inverse Function Theorem, compact subset version).
Let f : M → N be a smooth map that is injective on a compact subset X ⊂ M , and
suppose dfx : TxM → Tf(x)N is a linear diffeomorphism for each x ∈ X. Then f maps
a neighborhood U of X in M diffeomorphically onto a neighborhood V of f(X) in N .
¶ Generalized Inverse Function Theorem: the noncompact case.
Proof of Theorem 1.1, Case 2 (non-compact version).
[The idea: Any non-compact manifold can be written as a union of countably many
“compact stripes”] By using any positive smooth exhaustion function g on f(X), we
may decompose f(X) =
⋃∞
k=1Kk, where Kk = g
−1([k, k + 1]). Since f |X : X → f(X)
is a diffeomorphism, each Jk := f |−1X (Kk) is compact in X and give us a decomposition
X =
⋃∞
k=1 Jk. In particular, by Theorem 1.2, there is an open neighborhood Ũk of
Jk−1 ∪ Jk ∪ Jk+1 in M , such that f is a diffeomorphism on Ũk. Since f(X) is a smooth
submanifold in N , by embedding N into RK and using the induced distance function,
one can easily prove
dk := dist(Kk,
⋃
j>k+1
Kj) > 0.
(Here we need to use the compactness of Kk and use the fact that f(X) is a smooth
submanifold. Figure out the detail.)
LECTURE 10: TUBULAR NEIGHBORHOOD THEOREM 3
Now we choose a decreasing sequence of positive numbers ε1 > ε2 > · · · > 0 such
that εk < dk/2 for each k. Note that this implies K
εl
l ∩K
εk
k = ∅ if |l − k| > 1. Define
Uk = Ũk−1 ∩ Ũk ∩ Ũk+1 ∩ f−1(Kεkk ).
Then Uk is an open neighborhood of Jk, Vk = f(Uk) is an open neighborhood of
Kk, and f maps Uk diffeomorphically onto Vk. Moreover, the definition of Uk implies
Uk−1 ∪ Uk ∪ Uk+1 ⊂ Ũk.
We define U =
⋃
k≥1 Uk and V =
⋃
k≥1 Vk. Then U is an open neighborhood
of X in M , V is an open neighborhood of f(X) in N , and f : U → V is a local
diffeomorphism everywhere. It remains to show f is injective on U . Suppose x, y ∈ U
and f(x) = f(y). Then there exists k such that f(x) = f(y) ∈ Vk ⊂ Kεkk . It follows
x, y ∈ Uk−1∪Uk∪Uk+1 ⊂ Ũk. Since f is a diffeomorphism on Ũk, we conclude x = y. �
2. Tubular Neighborhood Theorem
Now let X ⊂M be a smooth submanifold. We want to study the followong prob-
lem: What does M looks like “near” X? The famous tubular neighborhood theorem
claims that X always admits a “tubular” neighborhood inside M . Moreover, the tubu-
lar neighborhood looks like a neighborhood of X inside its “normal bundle”. This gives
some kind of “canonical form” of a neighborhood of any smooth submanifold.
¶ Tubular neighborhood when embedded in RK.
We first prove the following “extrinsic version” of the tubular neighborhood theo-
rem, which gives us a nice neighborhood of any submanifold in RK :
Theorem 2.1 (ε-Neighborhood Theorem). Let ι : X ↪→ RK be a smooth submanifold.
Then there exists a continuous positive-valued function ε : X → R+, such that if we let
Xε be the ε-neighborhood of X (which is open in RK),
Xε := {y ∈ RK | |y − x| < ε(x) for some x ∈ X},
then
(1) each y ∈ Xε possesses a unique closest point πε(y) in X;
(2) the map π� : X
ε → X is a submersion.
(Note: If X is compact, then the function ε can be taken to be a constant.)
The ε-neighborhood described in the above theorem looks like
4 LECTURE 10: TUBULAR NEIGHBORHOOD THEOREM
We will prove the theorem by constructing a diffeomorphism between a neighbor-
hood of X inside its normal bundle (which can be think of as a nice “non-intersecting”way to put the green lines above together to form a smooth manifold) and a neigh-
borhood of X inside RK . So let’s start with the conception of the normal bundle of a
manifold embedded in RK , with details left as an exercise:
Let ι : X ↪→ RK be a smooth submanifold of dimension r. For each x ∈ X, we can
identify TxX with an r-dimensional vector subspace in RK via
TxX ' dιx(TxX) ⊂ TxRK ' RK .
Let Nx(X,RK) be the orthogonal complement of TxX in RK ,
Nx(X,RK) := {v ∈ TxRK ' RK | v ⊥ TxX},
which is a (K − r)-dimensional vector subspace of RK . Define
N(X,RK) = {(x, v) ∈ RK × RK | x ∈ X, v ∈ Nx(X,RK)} ⊂ TRK .
We will leave it as an exercise for you to prove that N(X,RK) is a K-dimensional
smooth submanifold in TRK , and the canonical projection map
π : N(X,RK)→ X, (x, v) 7→ x
is a submersion.
Definition 2.2. Let ι : X ↪→ RK be a smooth submanifold. We call N(X,RK)
described above normal bundle of X in RK .
Remark. The conception of normal bundle N(X,RK) is extrinsic: it depends on the
ambient space RK and also depends on the way of embedding ι : X ↪→ RK .
Proof of the ε-Neighborhood Theorem.
Define a map
h : N(X,RK)→ RK , (x, v) 7→ x+ v.
Then at any point (x, 0) ∈ N(X,RK), dh is non-singular, because dh(x,0) maps T(x,0)(X×
{0}) ⊂ T(x,0)N(X,RK) bijectively onto TxX ⊂ TxRK , and maps the tangent space
T(x,0)({x} ×Nx(X,RK)) bijectively onto Nx(X,RK) ⊂ TxRK .
Also by definition, h maps X × {0} ⊂ N(X,RK) diffeomorphically onto X ⊂ RK .
According to the generalized inverse function theorem (GIFT) we just proved, h maps
a neighborhood U of X × {0} in N(X,RK) diffeomorphically onto a neighborhood V
of X in RK .
Now for each x ∈ X, we define
ε(x) = sup{r ≤ 1 | Br(x) ⊂ V }.
One can check that ϕ is a positive continuous function on X.(Check this!) Note that
by definition, Xε ⊂ V is an open submanifold. Consider the map
πε : X
ε → X, y 7→ πε(y) = π ◦ h−1(y).
It is a submersion since π is a submersion and h−1 is a diffeomorphism on V . It remains
to prove that πε(y) is the unique closest point to y in X. In fact, let z ∈ X be a point in
X that is closest to y. Then the sphere centered at y with radius |y−z| is tangent to X
LECTURE 10: TUBULAR NEIGHBORHOOD THEOREM 5
at z. It follows that the vector y−z is perpendicular to X at z, i.e. y−z ∈ Nz(X,RK).
So we have
y = z + (y − z) = h(z, y − z),
i.e. πε(y) = z. So the point z is unique, and πε(y) is the unique point in X that is
closest to y. This completes the proof. �
¶ The tubular neighborhood theorem.
In general, suppose X ⊂M is a smooth submanifold. Then one can still define the
normal bundle N(X,M) (with respect to the ambient space M) as the set of points of
the form (x, v), where for any x ∈ X, v is any vector in the quotient space
Nx(X,M) = TxM/TxX.
One can show that N(X,M) is a smooth manifold whose dimension equals dimM . .
To get a more extrinsic (and geometric) description of N(X,M), we may embed M
into RK . Then we will get an inclusion TxX ⊂ TxM ⊂ TxRK , and the quotient space
TxM/TxX can be identified as the space of vectors in TxM which are perpendicular to
TxX. It follows
N(X,M) ' {(x, v) | x ∈ X, v ∈ TxM and v ⊥ TxX}.
Note that by using this identification, we have
T(x,0)N(X,M) ' TxX ⊕ T⊥x X,
where T⊥x X is the orthogonal complement of TxX inside TxM .
Now we prove
Theorem 2.3 (Tubular Neighborhood Theorem). Let X ⊂ M be a smooth submani-
fold. Then there exists a diffeomorphism from an open neighborhood of X in N(X,M)
onto an open neighborhood of X in M .
Proof. Embed M into RK . Let πε : M ε → M be as in the ε-neighborhood theorem
(for the embedding ι : M ↪→ RK). Again consider the map
h : N(X,M)→ RK , h(x, v)→ x+ v.
Then W := h−1(M ε) is an open neighborhood of X in N(X,M). Now consider the
composition
hε = πε ◦ h : W −→M.
Then hε is smooth, and is the identity map on X ⊂ N(X,M). Moreover, according to
the decomposition of T(x,0)N(X,M) above, (dhε)(x,0) maps T(x,0)N(X,M) bijectively
onto TxM . So the theorem follows from GIFT. �
LECTURE 11: APPROXIMATIONS AND DEFORMATIONS
1. Whitney Approximation Theorems
¶ Approximation of functions.
We begin with the following application of P.O.U.: [In the special case when M is
compact, this also follows from the Stone-Weierstrass approximation theorem.]
Theorem 1.1 (Whitney approximation theorem for continuous functions).
Let M be a smooth manifold. Then for any continuous function g : M → R and any
positive continuous function δ : M → R>0, there exists a smooth function f : M → R
so that |f(p)− g(p)| < δ(p) holds for all p ∈M .
We will prove a stronger version of this theorem. We need a definition:
Definition 1.2. We say a function g : M → R is smooth on a subset A ⊂ M if there
exists an open set U ⊃ A and a smooth function g0 defined on U s.t. g0 = g on A.
As a consequence, any function g is smooth on any single point set {p}, although
it may fail to be smooth at p. Now we state the relative version of Theorem 1.1:
Theorem 1.3 (Whitney approximation theorem for functions, relative version).
Let M be a smooth manifold, and A ⊂ M a closed subset. Then for any continuous
function g : M → R which is smooth on A and any positive continuous function
δ : M → R>0, there exists a smooth function f : M → R with f = g on A, so that
|f(p)− g(p)| < δ(p) holds for all p ∈M .
Proof. [The idea of the proof: For each p one can find a tiny small open set Up con-
taining p so that g is “almost constant” on Up. Then on Up one can approximate g by
the constant function f(·) ≡ g(p) (on Up). Then “glue” all these constant functions
together via a P.O.U. ρp subordinate to the open cover {Up}.] By definition, there
exists an open set U ⊃ A and a smooth function g0 defined on U so that g0 = g on A.
Let U0 = {p ∈ U : |g0(p)− g(p)| < δ(p)}. Then U0 is open in M and U0 ⊃ A.
Next we construct a (nice) open cover of M \A. For any q ∈M \A, we let
Uq = {p ∈M \A : |g(p)− g(q)| < δ(p)}.
Then {Uq | q ∈M\A} is an open covering of M\A. Since the topology on M is second
countable, one can find countable many such Uqi , i = 1, 2, · · · , which cover M \A.
Now let {ρ0, ρi} be P.O.U. subordinate to the open cover {U0, Uqi : i = 1, 2, · · · } of
M , and define a smooth function on M via
f(p) = ρ0(p)g0(p) +
∑
i≥1
ρi(p)g(qi).
1
2 LECTURE 11: APPROXIMATIONS AND DEFORMATIONS
Since the summation is locally finite, f is smooth. Also by definition, f = g0 = g on
A. Moreover, for any q ∈M one has
|f(p)− g(p)| =
∣∣∣∣ρ0(p)g0(p) +∑
i≥1
ρi(p)g(qi)−
∑
i≥0
ρi(p)g(p)
∣∣∣∣
≤ ρ0(p)|g0(p)− g(p)|+
∑
i≥1
ρi(p)|g(qi)− g(p)|
< ρ0(p)δ(p) +
∑
i≥1
ρi(p)δ(p)
= δ(p),
where in the last inequality, the fact ρ0(p)|g0(p) − g(p)| < ρ0(p)δ(p) follows from the
facts that if p ∈ U0, then by definition |g0(p) − g(p)| < δ(p), while if p 6∈ U0, then
ρ0(p)=0; the fact ρi(p)|gi(q) − g(p)| < ρi(p)δ(p) follows from a similar argument on
whether p is in Uqi or not (and note that at least one of the inequality holds). �
By working on each component, it is obvious that the same conclusion holds for
any RK-valued maps:
Theorem 1.4 (Whitney approximation theorem for RK-valued maps, relative version).
Let M be a smooth manifold, and A ⊂M a closed subset. Then for any continuous map
g : M → RK which is smooth on A and any positive continuous function δ : M → R>0,
there exists a smooth map f : M → RK with f = g on A, so that |f(p)− g(p)| < δ(p)
holds for all p ∈M .
Remark. (1) By taking A = ∅ we see that Theorem 1.3 implies Theorem 1.1.
(2) Theorem 1.3 also implies a smooth version of Tietze extension theorem:
[Smooth Extension] Any smooth function on a closed subset A can
be extended to a smooth function on the whole manifold M .
Note that our definition “smooth on a closed subset” is quite strong. There is an
extension theorem under much weaker assumption, also due to Whitney:
Theorem 1.5 (Whitney extension theorem). Let A ⊂ Rn be a closes
subset, and {fα}|α|≤m is a collection of functions defined on A which are
compatible in the following sense:
fα(x) =
∑
|β|≤m−|α|
fα+β(y)
β!
(x− y)β + o(|x− y|m−|α|).
Then there existsf ∈ Cm(Rn) so that
(1) For all |α| ≤ m, ∂αf = fα on A
(2) f is real-analytic on Ac = Rn \ A.
¶ Approximate continuous maps by smooth maps [in homotopy class].
Now let M,N be smooth manifolds and let g : M → N be a continuous map. A
natural question to ask is: Can we “approximate” g by a smooth map f : M → N?
Of course there is one issue in proposing the above question:
LECTURE 11: APPROXIMATIONS AND DEFORMATIONS 3
What do we mean by “approximate g ∈ C0(M,N) by f ∈ C∞(M,N)”?
Usually there are two different meanings:
• Endow with N a metric d [e.g. by embedding N into Euclidean space so that it
admits an induced metric], so that C0(M,N) is a metric space (endowed with
the uniform metric du)
• f can be obtained by “deforming” the map g.
In what follows we will mainly work on the second meaning (however, we will see the
use of the first meaning in the proof). We will not work directly on the first meaning,
since the approximation depends heavily on the extra structure: the metric or the
embedding. (For example, for any ε > 0, we can embed Nn into an ε-ball in R2n+1,
so that any smooth map f ∈ C∞(M,N) is a ε-approximation of any continuous map
g ∈ C0(M,N), which is not what we want.)
Recall from topology that “continuous deformations” are equivalent to “homo-
topies” [For details, see my topology (H) notes. Note that manifolds are always locally
compact Hausdorff]. Recall that two maps f0, f1 ∈ C0(X, Y ) are homotopic if there
exists F ∈ C0(X × [0, 1], Y ) so that
F (·, 0) = f0(·) and F (·, 1) = f1(·).
Now we prove the Whitney Approximation Theorem for continuous maps, which claims
that any continuous map between smooth manifolds can be continuously deformed to
a smooth map:
Theorem 1.6 (Whitney Approximation Theorem for Continuous Maps).
Given any continuous mapping g ∈ C0(M,N), one can find a smooth mapping f ∈
C∞(M,N) which is homotopic to g. Moreover, if g is smooth on a closed subset A ⊂M ,
then one can choose f so that f = g on A.
Proof. We embed N into RK . By the ε-neighborhood theorem we proved last time,
there is a continuous function ε : N → R>0 so that each y ∈ N ε possesses a unique
closest point πε(y) ∈ N .
Think of g as a continuous function from M to RK and apply Theorem 1.4 to the
positive continuous function ε = ε ◦ g, we get a smooth map f̃ : M → RK which is
ε-close to g, i.e.
|f̃(x)− g(x)| < ε(g(x)), ∀x ∈M.
So f̃(x) ∈ B(g(x), ε(g(x))) ⊂ N ε. It follows that
(1− t)g(x) + tf̃(x) ∈ B(g(x), ε(g(x))) ⊂ N ε, ∀ 0 ≤ t ≤ 1.
Now define F : M × [0, 1]→ N by
F (x, t) = πε((1− t)g(x) + tf̃(x)).
Then F is a homotopy that connects the continuous map g to the smooth map
f = F (·, 1) = πε ◦ f̃ : M → N.
4 LECTURE 11: APPROXIMATIONS AND DEFORMATIONS
Finally note that if g is smooth on a closed subset A, then the smooth function
f̃ can be chosen so that f̃ = g on A. It follows that f = g = F (·, t) on A. (In other
words, the homotopy connecting g to f can be chosen to be relative to A.) �
Remark. If M,N are real analytic manifolds, then one can approximate continuous
maps by analytic maps.
As an immediate consequence, we prove
Corollary 1.7. The homotopy group πk(S
n) ' {0} if k < n.
Proof. Any continuous map f : Sk → Sn is homotopic to a smooth map f̃ : Sk → Sn.
Since k < n, by Sard’s theorem, f̃(Sk) is of measure zero in Sn. In particular, f̃ is not
surjective, and thus is null-homotopic (why?). �
2. Smooth deformation of smooth maps
¶ Smooth homotopy.
Since in this course, we are mainly interested in smooth objects (smooth manifolds,
smooth submanifolds, smooth functions, smooth maps, smooth vector fields, smooth
vector bundle, smooth differential form etc), we are interested in homotopies connecting
two smooth maps via “smooth path”, i.e.
Definition 2.1. We say f0, f1 ∈ C∞(M,N) are smoothly homotopic if there exists
F ∈ C∞(M × [0, 1], N) so that
F (·, 0) = f0 and F (·, 1) = f1.
Of course if f0 and f1 are smoothly homotopic, then they are homotopic. Con-
versely, we have
Theorem 2.2 (Homotopy ≡ Smooth homotopy).
Suppose f0, f1 ∈ C∞(M,N) are homotopic, then they are smoothly homotopic.
Proof. Let F : M × [0, 1] → N be a homotopy connecting f0 and f1. Continuously
extend F to a mapping F̃ : M × R→ N by defining
F̃ (x, t) = F (x, 0) if t ≤ 0, and F̃ (x, t) = F (x, 1) if t ≥ 1.
Then F̃ is a continuous map from M×R to N , and is smooth on closed subsets M×{0}
and M × {1}. (Note that by Definition 1.2, F̃ is smooth on M × {0} means there is a
smooth function G̃ defined on M × (−ε, ε) so that G̃(x, 0) = F̃ (x, 0). We don’t require
F̃ to be smooth in a neighborhood of M × {0}.) So by Theorem 1.6, there exists a
smooth map F : M × R → N (that is homotopic to F̃ , which we don’t need here),
such that F = F̃ on M × {0} and M × {1}, i.e.
F (·, 0) = f0 and F (·, 1) = f1.
It follows that F is the desired smooth homotopy connecting f0 and f1. �
LECTURE 11: APPROXIMATIONS AND DEFORMATIONS 5
Recall that homotopy is an equivalence relation on the space of continuous maps
from M to N . The equivalence classes are called homotopy classes of maps. The-
orem 2.2 implies that smooth homotopy is an equivalence relation on the space of
smooth maps from M to N . Moreover, combining Theorem 1.6 and Theorem 2.2,
we immediately see that homotopy classes of continuous maps coincides with the s-
mooth homotopy classes of smooth maps. But in general the smooth theory is easier
to compute, since we have a power weapon: the differentiation.
¶ Stable properties.
In many applications, it is important that certain properties of maps will remain
unchanged under a small deformation.
Definition 2.3. We say a property P concerning maps in C∞(M,N) is a stable property
if it is preserved under small deformation, namely, if f ∈ C∞(M,N) satisfies P and
F is a smooth homotopy with F (x, 0) = f , then there exists ε > 0 so that for each
0 < t < ε, the map ft(·) = F (·, t) satisfies the property P .
Theorem 2.4. Suppose M is compact. Then the following properties of maps in
C∞(M,N) are stable:
(1) immersion,
(2) submersion,
(3) embedding,
(4) local diffeomorphism,
(5) diffeomorphism.
Proof. (1) Let f : M → N be an immersion. Then for any p ∈ M , there is a m ×m
sub-matrix of dfp which is non-singular. By continuity, there exists an open set Up 3 p
and [0, εp) so that the corresponding m×m sub-matrix of (dft)q is non-singular for all
(q, t) ∈ Up × [0, εp). (Strictly speaking, here we are working inside a coordinate chart
U around p so that f(U) is contained in a coordinate chart around f(p).) Now the
set
⋃
p Up × [0, εp) is an open neighborhood of M × {0} inside M × [0, 1]. By the tube
lemma in general topology, there exists ε > 0 so that
M × [0, ε) ⊂
⋃
p
Up × [0, εp).
As a consequence, for any t < ε, (dft)p is injective for all p ∈M , i.e. ft : M → N is an
immersion.
(2) The proof is almost the same as (1).
(3) Recall that a map from a compact manifold is an embedding if and only if it
is an injective immersion. In view of (1), we only need to show: there exists ε > 0 so
that for each 0 < t < ε, the map ft is injective. (The proof is just a modification of the
proof to GIFT (compact case) that we did last time.) Suppose we can find ti → 0 and
xi, x
′
i with fti(xi) = fti(yi). By compactness, after passing to subsequences we may
assume xi → x0 and x′i → x′0. Then x0 = x′0 since
f(x0) = F (x0, 0) = lim
i→∞
F (xi, t0) = lim
i→∞
F (x′i, t0) = F (x
′
0, 0) = f(x
′
0).
6 LECTURE 11: APPROXIMATIONS AND DEFORMATIONS
This will give us the desired contradiction, since the (m+ 1)× (n+ 1) matrix
dG(x0,0) =
(
dfx0 ∗
0 1
)
has rank m+1, which implies that G is an immersion at (x0, 0), and thus by the canon-
ical immersion theorem, G is an immersion (and thus is injective) in a neighborhood
of (x0, 0).
(4) This is a consequence of (1): A local diffeomorphism is merely an immersion
between manifolds of the same dimension.
(5) Let f : M → N be a diffeomorphism. Then f maps each connected component
of M diffeomorphically to a connected component of N , and F maps a connected
component of Minto the corresponding connected component of N . So without loss
of generality, we may assume M is connected.
Since a diffeomorphism is automatically an embedding, a submersion and a local
diffeomorphism. By (2), (3) and (4), there exists ε > 0 os that for each 0 < t < ε,
the map ft is an embedding (and thus is injective), an submersion and is a local
diffeomorphism. Also recall that in PSet 2-2-1, we showed that any submersion from a
compact manifold to a connected manifold is surjective. So each ft is a diffeomorphism
for 0 < t < ε. �
Remark. However, being “constant rank” is not a stable property. (why?)
Remark. The properties listed in Theorem 2.4 fails to be stable if M is non-compact.
See PSet for a counterexample.
LECTURE 12: TRANSVERSALITY
1. Transversal intersection
¶ The pre-image of a submanifold.
In Lecture 8, we proved the regular level set theorem (or more generally, the con-
stant rank level set theorem), which claims that the level set S = f−1(q) of a smooth
map f : M → N is a smooth submanifold of M if q is a regular value of f (or if f is a
constant rank map in a neighborhood of S), and the tangent space TpS for p ∈ S can
be identified with the subspace ker(dfp : TpM → TqN).
Now let f : M → N be smooth and let X ⊂ N be a smooth submanifold. We
consider the following question:
When will f−1(X) be a smooth submanifold in M?
Note that the problem “whether f−1(X) is a smooth submanifold in M or not” is
a local problem: it is enough to examine, for each p ∈ f−1(X), whether there is a
neighborhood U of p in M so that U ∩f−1(X) is a smooth submanifold of M . Since X
is a smooth submanifold of N , by definition locally in a neighborhood V of q = f(p)
in N , one can find a smooth map g : V → Rl, where l = dimN − dimX is the
codimension of X in N , such that g−1(0) = X. (In the language of Lec. 8, g can be
taking to be the map g = (ϕk+1, · · · , ϕn).) (Note that 0 is a regular value of the map
g, and thus TqX = ker(dgq).) As a consequence, we can write f
−1(X) as the level set
of the composition g ◦ f :
f−1(X) = f−1 ◦ g−1(0) = (g ◦ f)−1(0).
So according to the regular level set theorem, for f−1(X) to be a smooth submanifold
of M , it is enough to check whether 0 is a regular value of g ◦ f .
Now we would like to find conditions on f and S so that 0 is a regular value of
g ◦ f . Moreover, since given f and X, the map g is not unique, we would like to find
conditions that is independent of g. In other words, we need to figure out conditions
so that for any p ∈ (g ◦ f)−1(0) = f−1(X), the differential d(g ◦ f)p is surjective. By
chain rule,
d(g ◦ f)p = dgq ◦ dfp.
Note that by definition, dgq is already surjective, with kernel ker(dgq) = TqX. So for
the differential d(g ◦ f)p : TpU → Tg(q)Rl = Rl to be surjective, it is enough to assume
that
Im(dfp) contains some “complementary subspace” of TqX in TqN .
1
2 LECTURE 12: TRANSVERSALITY
¶ Transversal intersection between a map and a submanifold.
To be more precise, we will need the following linear algebra lemma:
Lemma 1.1. Let L : V → W be any surjective linear map, and V1 ⊂ V be any vector
subspace. Then L(V1) = W if and only if V1 + ker(L) = V .
Proof. If V1 + ker(L) = V , then L(V1) = L(V1 + ker(L)) = L(V ) = W .
If V1 + ker(L) 6= V , we pick any v ∈ V but v 6∈ V1 + ker(L). Then L(v) 6∈ L(V1),
since if there is v1 ∈ V1 such that L(v1) = L(v), then v − v1 ∈ ker(L), i.e.
v = v1 + (v − v1) ∈ V1 + ker(L),
a contradiction. It follows that L(V1) 6= W . �
According to this lemma, 0 is a regular value of g ◦ f if and only if
(1) Im(dfp) + Tf(p)X = Tf(p)N, ∀p ∈ f−1(X).
Note that this condition depends only on f and X, and does not depend on g.
Definition 1.2. Let f : M → N be a smooth map, and X ⊂ N be a smooth subman-
ifold. We say f intersect X transversally, and denote by f −t X, if (1) holds.
Remark. In the PSet you will be asked to prove that given submanifold X ⊂ N , the
property
f ∈ C∞(M,N) intersects X ⊂ N transversally
is a stable property. In other words, if f −t X and you deform f a bit to fε, then we
still have fε
−t X.
Remark. There are two extremal cases:
• If f(M) ∩ X = ∅, i.e. if f−1(X) = ∅, then f intersects X transversally, since
there is no condition to check at all.
• If X ⊂ N is a smooth submanifold so that any q ∈ X is a regular value
of f : M → N , then f is a submersion at any p ∈ f−1(X), and thus the
transversality condition (1) holds trivially. In particular,
Proposition 1.3. If f : M → N is a submersion, then f intersects with any
smooth submanifold X in N transversally.
Now we can answer the question proposed above.
Theorem 1.4. Let f : M → N be a smooth map, and let X ⊂ N be a smooth subman-
ifold so that f −t X. Then f−1(X) is a smooth submanifold in M whose codimension
equals to the codimension of X (in N), and
Tp(f
−1(X)) = df−1p (Tf(p)X), ∀p ∈ f−1(X).
Proof. The argument above shows that if f −t X, then 0 is a regular value of g ◦ f .
So f−1(X) = (g ◦ f)−1(0) is a smooth submanifold in M . The dimension of f−1(X) is
dimM − l, where l = dimN − dimX. So dimM − dim f−1(X) = dimN − dimX, i.e.
codimf−1(X) = codimX.
LECTURE 12: TRANSVERSALITY 3
Finally, also by the regular level set theorem, the tangent space of f−1(X) at p is
Tp(f
−1(X)) = ker(d(g ◦ f)p) = (dgf(p) ◦ dfp)−1(0) = df−1p (dg−1f(p)(0)) = df
−1
p (Tf(p)X).
This completes the proof. �
¶ Transversal intersection between two submanifolds/two maps.
In literature there are two other types of transversal intersections, both can be
interpreted as special case of f −t X.
Transversal intersection of two submanifolds: Let X1, X2 be two smooth submanifolds
of M and let ι : X1 ↪→ M be the canonical embedding. Then ι−1(X2) = X1 ∩ X2.
Moreover, since we have the identification TpX1 = dιp(TpX1), the condition ι
−t X2 is
equivalent to say TpX1 + TpX2 = TpM for any p ∈ X1 ∩X2.
Definition 1.5. We say two smooth submanifolds X1 and X2 in M intersect transver-
sally if for any p ∈ X1 ∩X2,
TpX1 + TpX2 = TpM.
In this case we write X1
−t X2.
Remark. Note: By definition, if X1 ∩X2 = ∅, then X1 −t X2.
So if X1
−t X2, then ι −t X2. So X1 ∩ X2 = ι−1(X2) is a smooth submanifold of
M .Moreover, the dimension of X1 ∩X2 is given by
dim(X1 ∩X2) = dimX1 − (dimM − dimX2) = dimX1 + dimX2 − dimM.
Moreover, the tangent space of X1 ∩ X2 at p equals dι−1p (TpX2) = TpX1 ∩ TpX2. We
conclude
Corollary 1.6. Let X1 and X2 be two smooth submanifolds in M which intersect
transversally, then X1 ∩X2 is a smooth submanifold whose dimension equals dimX1 +
dimX2 − dimM , and for any p ∈ X1 ∩X2,
Tp(X1 ∩X2) = TpX1 ∩ TpX2.
Remark. Note that if X1
−t X2, then the codimensions of X1, X2 and X1 ∩ X2 in M
satisfy the simple relation
codimX1 ∩X2 = codimX1 + codimX2.
Transversal intersection of two maps: More generally, one can define
Definition 1.7. Let f1 : M1 → N and f2 : M2 → N be smooth maps. We say f1 and
f2 intersect transversally, and denote by f1
−t f2, if the product map
f1 × f2 : M1 ×M2 → N ×N
intersect the “diagonal submanifold”
∆N = {(q, q) | q ∈ N} ⊂ N ×N
transversally.
In this setting, Theorem 1.4 becomes (CHECK)
4 LECTURE 12: TRANSVERSALITY
Corollary 1.8. If f1
−t f2, then the fiber product
F = (f1 × f2)−1(∆N)
is a submanifold of M1 ×M2 whose tangent space at (p1, p2) ∈ F is
T(p1,p2)F = {(X1, X2) | Xi ∈ TpiMi, (df1)p1(X1) = (df2)p2(X2).}.
Note that in the special case where f2 is ι2 : X ↪→ N , then f1 −t f2 is reduced to
f1
−t X.
Remark. There is a weaker assumption on intersections, called clean intersection, which
assumes that the intersections are submanifolds with “correct tangent spaces” without
assuming the transversality.
¶ Transversality is generic.
Finally we prove the following transversality theorem which is useful in finding
transversal maps: (In general M could be a manifold with boundary, but we will not
discuss that here.)
Theorem 1.9 (The Transversality Theorem). Let F : S ×M → N be a smooth map,
X ⊂ N be a smooth submanifold. For each s ∈ S, we let fs : M → N be the map
fs(p) = F (s, p).
Suppose F −t X. Thenfor every regular value s ∈ S of the projection map 1
π : F−1(X) ⊂ S ×M → S, π(s, p) = s,
one has fs
−t X. (So by Sard’s theorem, for most s ∈ S, we have fs −t X.)
Proof. Let s be any regular value of π. For any p ∈ f−1s (X), we need to show
Im(dfs)p + TqX = TqN,
where q = fs(p). Since F
−t X, for any Yq ∈ TqN , there exists (Zs, Zp) ∈ T(s,p)(S ×M)
and Zq ∈ TqX such that
Yq = (dF )(s,p)(Zs, Zp) + Zq.
But since s is a regular value of π, for Zs ∈ TsS, there exists Z ′p ∈ TpM so that
(Zs, Z
′
p) ∈ T(s,p)F−1(X). So we can write
Yq = (dF )(s,p)(0, Zp − Z ′p) + (dF )(s,p)(Zs, Z ′p) + Zq.
The conclusion follows since
(dF )(s,p)(0, Zp − Z ′p) = (dfs)p(Zp − Z ′p) ∈ Im(dfs)p,
and
(dF )(s,p)(Zs, Z
′
p) ∈ dF(s,p)
(
T(s,p)F
−1(X)
)
⊂ TqX.
�
As a corollary, we can show that transversal maps are generic:
1Note that since F −t X, the pre-image F−1(X) is a smooth submanifold in S×M . So the projection
map π is a smooth map.
LECTURE 12: TRANSVERSALITY 5
Corollary 1.10. Let f : M → RK be any smooth map and X ⊂ RK be any smooth
submanifold. Then for a.e. v ∈ RK, the “v-shifted” map
fv : M → RK , p 7→ fv(p) = f(p) + v
is transversal to X.
Proof. Define a smooth map F by
F : M × RK → RK , (p, v) 7→ f(p) + v.
Then for any fixed p ∈M , F (p, ·) is a submersion (a diffeomorphism in fact). It follows
that F is a submersion from M × RK onto RK . So by Proposition 1.3, F intersects
with any smooth submanifold X in RK transversally. Now the conclusion follows from
the transversality theorem above. �
In particular, if we take f to be the inclusion map, we have
Corollary 1.11 (General position lemma). Let M,N be smooth submanifolds of RK.
Then for a.e. a ∈ RK, M + a intersects N transversally.
(To get a better understanding of the previous two corollaries, think of the following
problem: Let M and N be two spheres in RK . When will M + a−t N? )
By using tubular neighborhoods, one can easily prove
Theorem 1.12 (Thom’s Homotopy Transversality Theorem). Let f : M → N be
any smooth map, and Y ⊂ N be any smooth submanifold. Then f is homotopic to a
smooth map g : M → N which intersects Y transversally. Moreover, if X ⊂ M is a
submanifold and f |X is transverse to Y , then g can be chosen so that g|X = f |X .
Proof. Exercise. [Hint: Embed N into RK . Let πε : N ε → N be the ε-neighborhood
of N . Now define F : B ×M → N as F (s, p) = πε(f(p) + ε(f(p))s), where B is the
open unit ball in RK .] �
As a consequence, we prove
Corollary 1.13. Let M ⊂ Rn be a smooth submanifold of dimension m ≤ n−2. Then
the complement Rn \M is connected.
Proof. Let γ be any path joining x and y in Rn. Then γ is homotopic to a smooth
curve γ′ joining x and y. By corollary above, γ′ is homotopic to a smooth map γ′′
joining x and y that is transversal to M . By dimension counting, we must have
Image(γ′′) ∩M = ∅. �
LECTURE 13: SMOOTH VECTOR FIELDS
1. Smooth vector fields
¶ Tangent vectors: a review.
Recall that a tangent vector Xp at p ∈M is a linear map Xp : C∞(M)→ R s.t.
Xp(fg) = Xp(f)g(p) + f(p)Xg(p).
We have known that in the definition above, we can replace C∞(M) by C∞(U), where
U is any open neighborhood of p. In particular, we can choose (ϕ,U, V ) to be a
coordinate chart, then at each p ∈ U , the tangent vectors {(∂i)p | 1 ≤ i ≤ m} form a
basis of the tangent space TpM , where
(∂i)p =
(
dϕ−1
)
ϕ(p)
(
∂
∂xi
∣∣∣∣
ϕ(p)
) : C∞(U)→ R,
f 7→ (∂if)p =
∂(f ◦ ϕ−1)
∂xi
(ϕ(p)),
where { ∂
∂xi
∣∣
ϕ(p)
| 1 ≤ i ≤ n} is the standard basis1 of Tϕ(p)Rn. By this way, for any
vector Xp ∈ TpM , we can write
Xp =
n∑
i=1
ai(∂i)p.
Here is a more geometric way to understand these vectors: if we embed M into
some RK , then one can visualize (∂i)p as the tangent vector (in the sense that we
learned in calculus) of the curve (which is a curve in RK)
γi(t) = ϕ
−1((0, · · · , 0, t, 0, · · · , 0)), t ∈ (−ε, ε)
at p, where t is located at the ith-position, ε is chosen so that (0, · · · , 0, t, 0, · · · , 0) ∈ V
for all −ε < t < ε, and we have chosen the coordinates so that ϕ(p) = 0 ∈ V ⊂ Rn.
If you don’t want such an “extrinsic” definition of the tangent vector of a curve,
we may give an “intrinsic” definition of tangent vector of a curve on a surface:
Definition 1.1. Let γ : R→M be a smooth curve on M with γ(0) = p,
and let d
dt
be the unit tangent vector on R. Then we call the vector
γ̇(0) := dγ0(
d
dt
)
the tangent vector of the curve γ at p = γ(0).
Note that by definition of the differential, for any f ∈ C∞(M),
γ̇(0)(f) = dγ0(
d
dt
)(f) =
d
dt
∣∣∣∣
0
(f ◦ γ).
1In other words, the identification Tϕ(p)Rn ' Rn is given by ∂∂xi
∣∣
ϕ(p)
! ei, where {e1, · · · , en} is
the standard basis of Rn.
1
2 LECTURE 13: SMOOTH VECTOR FIELDS
One may check [Check it!] that the curve γi described above is a curve on M whose
(intrinsic) tangent vector at p is (∂i)p.
¶ Smooth vectors fields on M .
In multi-variable calculus and in ODE, we have learned the conception of (smooth)
vector fields, that is, a “smooth assignment”2 of vectors to each point in the domain.
In other words, a vector field X on V ⊂ Rn assigns to each x ∈ V a vector Xx, and
the smoothness assumption requires that if we write
Xx =
n∑
i=1
ai(x)
∂
∂xi
∣∣∣∣
x
,
then each coefficient function ai(x) depends smoothly on x ∈ V . Here is the advantage
of writing vector fields using “partial derivatives” instead of using Euclidian vector ei’s:
X is not merely a collection of “geometric arrows”, but in fact a first order differential
operator acting on C∞(V ). So vector field is an important object in analysis.
Now we define smooth vector fields on smooth manifolds. An obvious one is
Definition 1.2. A smooth vector field X on a smooth manifold M is a smooth assign-
ment that assigns to each p ∈M a tangent vector Xp ∈ TpM .
Of course one should explain the meaning of the words “smooth assignment”. As
usual this can be understood via local charts (ϕ,U, V ). In any such local chart, we note
that ∂
∂xi
are Euclidian smooth (“constant”) vector fields that are defined everywhere
in V , and thus ∂i should be smooth vector fields that are defined everywhere in U . For
the vector field X under consideration, we can write
X = X1∂1 + · · ·+Xn∂n =
∑
X i∂i,
where X i’s are functions on U . Then we define
Definition 1.2 (continued):
We say X is smooth on U , if all coefficients X i’s are smooth functions on U .
Now let X be any vector field (not necessary smooth) on M . Take any f ∈ C∞(M),
then for any p ∈M we get a number Xp(f) ∈ R. So X maps f to a function
Xf : M → R, p 7→ Xf(p) := Xp(f).
Proposition 1.3. X is smooth if and only if
f ∈ C∞(M) =⇒ Xf ∈ C∞(M).
Proof. Exercise. [Hint: How do you get the coefficient X i of a vector field X in a local
chart? We did that in the proof of Theorem 1.3 in Lecture 8.] �
2Of course one can also define continuous vector fields or even discontinuous vector fields. But
unless otherwise stated, in this course we are only interested in smooth objects.
LECTURE 13: SMOOTH VECTOR FIELDS 3
In what follows when we say vector fields, we always mean smooth vector fields.
Notation: We will denote the set of all smooth vector fields on M by Γ∞(TM).
Note that Γ∞(TM) has a very nice algebraic structure:
X1, X2 ∈ Γ∞(TM), f1, f2 ∈ C∞(M) =⇒ f1X1 + f2X2 ∈ Γ∞(TM).
This implies that Γ∞(TM) is not only a vector space (whose dimension is infinity
(why)), but in fact a module over the algebra C∞(M).
2. Various faces of vector fields
¶ Smooth vectors fields as smooth sections.
One may worry that whether the above description of smoothness depends on the
choice of local charts. The following proposition gives us a global characterization of
smooth vector fields, and can be used as an alternative (and more precise) definition.
Note that another way to say
“sign to each p ∈M a vector Xp ∈ TpM”
is
“give a map X : M → TM so that π ◦X = Id”,
where π : TM → M is the canonical projection π(Xp) = p, which (as we have seen in
PSet 2-1-6) is a submersion from the smooth manifold TM (=the tangent bundle of
M) to M . In general, a maps X :M → TM satisfying π ◦X = Id is called a section
of the tangent bundle.
Proposition 2.1 (Smooth Vector Field=Smooth Sections of the tangent bundle).
A smooth vector field X on M is a smooth map X : M → TM such that π ◦X = Id.
Proof. We use local chart (ϕ,U, V ) for M , and local chart (Tϕ, π−1(U), V × Rn) for
TM , where Tϕ is given by (c.f. PSet 2-1-6)
Tϕ(p,Xp) = (ϕ(p), dϕp(Xp)),
where when viewed as a vector in Rn, dϕp(Xp) should be explained as the coefficients
of dϕp(Xp) in the basis
∂
∂xi
∣∣
ϕ(p)
. Note that by definition dϕ(∂i) =
∂
∂xi
, which implies
(by linearity)
Xp =
∑
X i(p)(∂i)p =⇒ dϕp(Xp) =
∑
X i(ϕ−1(x))
∂
∂xi
∣∣∣∣
ϕ(p)
.
So in these local charts, the map X becomes
Tϕ ◦X ◦ ϕ−1(x1, · · · , xn) =
(
x1, · · · , xn, X1(ϕ−1(x)), · · · , Xn(ϕ−1(x))
)
.
By definition, the map X : M → TM is a smooth map if and only if all these X i’s are
smooth. �
4 LECTURE 13: SMOOTH VECTOR FIELDS
Remark. It is very important that although the vector field X is globally defined (=is
defined everywhere AND is independent of local charts), the set of vector fields ∂i’s
and the coefficients X i’s are only locally defined (=are defined only on U AND depend
on the choice of the chart map). Although at each point, or even in each coordinate
chart, one can write vector fields as linear combinations of n special vector fields, in
general there do not exist exactly n global vector fields defined on M which “generate”
all vector fields on M . (→ a strong topological restriction for M)
¶ Vector fields as derivations.
Let X be a smooth vector field on M . According to the Leibnitz rule for the
tangent vector Xp : C
∞(M) → R, we immediately get the Leibnitz rule for vector
fields:
X(fg) = fX(g) +X(f)g, ∀f, g ∈ C∞(M)
Definition 2.2. A derivation of the algebra C∞(M) is a map D : C∞(M)→ C∞(M)
that satisfies the Leibnitz rule
D(fg) = fD(g) +D(f)g, ∀f, g ∈ C∞(M).
So any smooth vector field X on M is a derivation. Conversely we have
Proposition 2.3. Derivations are smooth vector fields: For any derivation D on
C∞(M), there is a smooth vector field X, such that Df = Xf .
Proof. Let D be a derivation. For any p ∈M , define Xp : C∞(M)→ R by
Xp(f) := Df(p).
Then one can check that Xp is a derivative at p, in the sense of Lecture 5. As in Lecture
5, Xp must be a tangent vector at p. So starting from D we get a vector field X on
M , with Df = Xf . The smoothness of X follows from Proposition 1.3. �
This can be thought of as a third (equivalent) definition of smooth vector field.
¶ Differential operators on manifolds.
Recall that a differential operator of order (at most) n on U ⊂ Rm is an operator
of the form
P =
∑
|j|≤n
aj(x
1, · · · , xm)
(
∂
∂x1
)j1
· · ·
(
∂
∂xm
)jm
,
where for simplicity we will assume that the coefficients aj’s are smooth. Similarly we
may define
Definition 2.4. A differential operator of order (at most) n on a smooth manifold M
is a linear map P : C∞(M) → C∞(M) that locally has the form above, i.e., in any
local charts (ϕ,U, V ), P can be written as
P =
∑
|j|≤n
aj(ϕ(p)) (∂1)
j1 · · · (∂m)jm ,
where again for simplicity we assume aj ∈ C∞(U).
LECTURE 13: SMOOTH VECTOR FIELDS 5
So by definition, a smooth vector field X on M is a differential operator of order
1. More generally, if X1, · · · , Xn are smooth vector fields on M , then the composition
X1◦· · ·◦Xn is a differential operator of order n onM . Conversely, locally any differential
operator of order n is a linear combination of the compositions of at most n vector fields
(here we regard “multiplying by a function” as “acting by 0 vector fields”)(Prove this
claim. Hint: use P.O.U. ).
Remark. Note that by definition, any differential operator is local:
supp(Pf) ⊂ supp(f), ∀f ∈ C∞(M).
Peetre proved that the converse is also true, so that we have a very nice intrinsic
characterization of differential operators on manifolds (and in fact more generally, for
differential operators acting on sections of vector bundles): A differential operator
on a smooth manifold M is a linear map P : C∞(M) → C∞(M) that is local, i.e.
supp(Pf) ⊂ supp(f) for all f ∈ C∞(M).
LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS
1. Integral Curves
¶ Integral Curves.
Suppose we have a smooth vector field defined on an Euclidian region. In calculus
and in ODE, we learned the conception of integral curves of such a vector field: an
integral curve is a parametric curve that represents a specific solution to the ordinary
differential equation represented by the vector field. Geometrically, they are curves so
that the given vector field is the tangent vector to the curves everywhere.
Here is an example of vector fields with many integral curves drawn:
The conception of integral curves above can be generalized to smooth manifolds
easily. Recall that a smooth curve in a smooth manifold M is a smooth map γ : I →M ,
where I is an interval in R. For any a ∈ I, the tangent vector of γ at the point γ(a) is
γ̇(a) =
dγ
dt
(a) := dγa(
d
dt
),
where d
dt
is the standard coordinate tangent vector of R.
Definition 1.1. Let X ∈ Γ∞(TM) be a smooth vector field on M . A smooth curve
γ : I →M is called an integral curve of X if for any t ∈ I,
γ̇(t) = Xγ(t).
Remark. By a curve we really mean a “parametrized curve”. The parametrization is
a part of the definition. Different parametrizations of the “same geometric picture”
represent different curves. In general, a re-parametrization of an integral curve is no
longer an integral curve.
1
2 LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS
¶ Examples of integral Curves.
However, it is easy to see that linear re-parametrizations produce integral curves:
Lemma 1.2. If γ : I →M is an integral curve of a vector field X, then
(1) Let Ia = {t | t+ a ∈ I}, then
γa : Ia →M, γa(t) := γ(t+ a)
is an integral curve of X.
(2) Let Ia = {t | at ∈ I} (a 6= 0), then
γa : Ia →M, γa(t) := γ(at)
is an integral curve for Xa = aX.
The proofs are simple and thus are omitted.
Example. Consider the coordinate vector field X = ∂
∂x1
on Rn. Then the integral curves
of X are the straight lines parallel to the x1-axis, parametrized as
γ(t) = (c1 + t, c2, · · · , cn).
To check this, we note that for any smooth function f on Rn,
dγ(
d
dt
)f =
d
dt
(f ◦ γ) = ∇f · dγ
dt
=
∂f
∂x1
.
Remark. Note that although the curve
γ̃(t) = (c1 + 2t, c2, · · · , cn)
has the same picture (i.e. the same “horizontal line” passing the point (c1, · · · , cn)) as
γ, it is not an integral curve of X, but an integral curve of 2X, since ˙̃γ(t) = 2 ∂
∂x1
.
Example. Consider the vector field X = x ∂
∂y
− y ∂
∂x
on R2. Then if γ(t) = (x(t), y(t)) is
an integral curve of X, we must have, for any f ∈ C∞(R2),
x′(t)
∂f
∂x
+ y′(t)
∂f
∂y
= ∇f · dγ
dt
= Xγ(t)f = x(t)
∂f
∂y
− y(t)∂f
∂x
,
which is equivalent to the system
x′(t) = −y(t), y′(t) = x(t).
The solution to this system is
x(t) = a cos t− b sin t, y(t) = a sin t+ b cos t.
These are circles centered at the origin in the plane parametrized by the angle (with
counterclockwise orientation).
LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS 3
¶ ODE in local charts: Existence, Uniqueness and Smoothness.
To study further properties of integral curves, we need to convert the equation
γ̇(t) = Xγ(t) (which is an equation relating tangent vectors on manifolds) into ODEs
on functions defined on Euclidian region. To do so we first notice the following nice
local formula for a vector field, whose proof is left as an exercise:
Lemma 1.3. Let X be a smooth vector field on M . Then in local chart (ϕ,U, V ) we
have X =
∑
X(xi)∂i, where x
i : U → R is the ith coordinate function defined by ϕ.
So we let γ : I →M be an integral curve of X. To study the equation γ̇(t) = Xγ(t)
at a given point γ(t), WLOG we may assume γ(t) ∈ U , and (ϕ,U, V ) is a coordinate
chart. By using the local chart map ϕ, one can convert the point γ(t) ∈ U to
ϕ(γ(t)) = (x1(γ(t)), · · · , xn(γ(t))) ∈ Rn.
If we denote yi = xi ◦ γ : I → R, then we can convert the (vector!) equation defining
integralcurves into equations on these one-variable functions yi’s. More precisely,
according to the previous lemma, we have
γ̇(t) = dγt(
d
dt
) =
∑
i
dγt(
d
dt
)(xi)∂i =
∑
i
(xi ◦ γ)′(t)∂i =
∑
i
(yi)′(t)∂i.
So the integral curve equation γ̇(t) = Xγ(t) becomes∑
i
(yi)′(t)∂i =
∑
i
X i(γ(t))∂i =
∑
i
X i ◦ ϕ−1(y1(t), · · · , yn(t))∂i.
for all t ∈ I. In conclusion, we convert the integral curve equation into the following
system of ODEs on the one-variable functions yi’s:
(yi)′(t) = X i ◦ ϕ−1(y1, · · · , yn), ∀t ∈ I, ∀1 ≤ i ≤ n.
This is a system of first order ODEs on the (one-variable) functions yi = xi◦γ : R→ R.
Conversely, any solution to this system of ODEs defines an integral curve of the vector
field X inside the open set U .
Recall:
Theorem 1.4 (The Fundamental Theorem for Systems of First Order ODEs). Suppose
V ⊂ Rn is open, and F = (F 1, · · · , F n) : V → Rn a smooth vector-valued function.
Consider the initial value problem
(1)
{
ẏi(t) = F i(y1(t), · · · , yn(t)), i = 1, · · · , n
yi(t0) = c
i, i = 1, · · · , n
Then
(1) Existence : For arbitrary t0 ∈ R and c0 ∈ V , there exist an open interval
I0 3 t0 and an open set V0 3 c0 so that for any c = (c1, · · · , cn) ∈ V0, the
system (1) has a smooth solution yc(t) = (y
1(t), · · · , yn(t)) ∈ V for t ∈ I0.
(2) Uniqueness : If y1 is a solution to the system (1) for t ∈ I0 and y2 is a solution
to the system (1) for t ∈ J0, then y1 = y2 for t ∈ I0 ∩ J0.
(3) Smoothness : The solution function Y (c, t) := yc(t) in part (1) is smooth on
(c, t) ∈ V0 × I0.
4 LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS
We will refer to Lee’s book, Appendix D (Page 663-671) for a proof. According to
the fundamental theorem of systems of ODEs, we conclude
Theorem 1.5 (Local Existence, Uniqueness and Smoothness). Suppose X is a smooth
vector field on M . Then for any point p ∈ M , there exists a neighborhood Up of p, an
εp > 0 and a smooth map
Γ : (−εp, εp)× Up →M
so that for any q ∈ U , the curve γq : (−ε, ε)→M defined by
γq(t) := Γ(t, q)
is an integral curve of X with γ(0) = q. Moreover, this integral curve is unique in the
sense that if σ : I →M is another integral curve of X with σ(0) = q, then σ(t) = γq(t)
for t ∈ I ∩ (−εp, εp).
2. Complete vector fields
¶ Complete/Noncomplete vector fields.
As a consequence of the uniqueness, any integral curve has a maximal defining
interval. We are interested in those vector fields whose maximal defining interval is R.
Definition 2.1. A vector field X on M is complete if for any p ∈ M , there is an
integral curve γ : R→M such that γ(0) = p.
Note every smooth vector field is complete.
Example. Consider the vector field X = t2 d
dt
on R. Let γ(t) = (x(t)) be its integral
curve. Then according to the integral curve equation,
x′(t)
d
dt
= Xγ(t) = x(t)
2 d
dt
=⇒ x′(t) = x(t)2.
The solution to this ODE is with initial condition x(0) = c is
xc(t) =
1
−t+ 1/c
for c 6= 0
and
x0(t) = 0 for c = 0.
Note that the maximal interval of xc(t) is
Ic = (−∞, 1/c) for c > 0
and
Ic = (1/c,+∞) for c < 0.
Since the integral curves starting at any c 6= 0 is not defined for all t ∈ R, we conclude
that X is not complete.
LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS 5
¶ Compactly supported vector fields are complete.
We will use complete vector fields to construct global flows next time. We will end
this lecture with a sufficient condition for a vector field to be complete. As in the case
of functions, we can define the support of a vector field by
supp(X) = {p ∈M | X(p) 6= 0}.
Our criteria is
Theorem 2.2. If X is a compactly supported vector field on M , then it is complete.
Proof. Let C = supp(X). First suppose q ∈M \C, i.e. Xq = 0. We define a “constant
curve” γq on M by letting γq(t) = q for all t ∈ R, then we see
γ̇q(t) = 0 = Xq = Xγq(t).
In other words, the constant curve γq (whose domain is R) is the unique integral curve
of X passing q.
Now suppose p ∈ C. (The idea: use compactness to find a uniform constant ε0 so
that any integral curve starting at a point in C is defined on an interval of length 2ε0.
Then if we have an integral curve, we can always extend the domain by ε0.) Since any
integral curve starting at q ∈M \C stays at q, we see that every integral curve starting
at p ∈ C stays in C. By Theorem 1.5, for any q ∈ C, there is an interval Iq = (−εq, εq),
a neighborhood Uq of q and a smooth map
Γ : Iq × Uq → C
such that for all p ∈ Uq,
γp(t) = Γ(t, p)
is an integral curve of X with γp(0) = p. Since ∪qUq ⊃ C, and C is compact, one
can find a finite many points q1, · · · qN in C so that {Uq1 , · · · , UqN} cover C. Let
I = ∩kIqk = (−ε0, ε0), then for any q ∈ C, there is an integral curve γq : I → C. Now
suppose the maximal defining interval for p ∈ C is Ip. We need to prove Ip = R. In
fact, if Ip 6= R, WLOG, we may assume that sup Ip = c <∞. Then starting from the
point q = γp(c− ε02 ), there is an integral curve
γq : (−ε0, ε0)→M
of the vector field X. By uniqueness, γq(t) = γp(t+ c− ε02 ). It follows that the defining
interval of γp extends to c+
ε0
2
, which is a contradiction. �
In particular, if M itself is compact, then the set Supp(X), as a closed set in the
compact manifold, is always compact. So we get
Corollary 2.3. Any smooth vector field on a compact manifold is complete.
6 LECTURE 14: INTEGRAL CURVES OF SMOOTH VECTOR FIELDS
¶ The flow generated by a complete vector field.
Now suppose M is a smooth manifold and X is a complete vector field on M . By
definition, for any p ∈ M , there is a unique integral curve γp : R → M such that
γp(0) = p. From this one can, for any t ∈ R, define a map
φt : M →M, p 7→ γp(t).
By definition, φt(p) is smooth in t for any fixed p. In fact, the family of maps {φt | t ∈ R}
satisfies the following very important group law :
Proposition 2.4. For any t, s ∈ R, we have φt ◦ φs = φt+s.
Proof. Notice that for any p ∈M and any fixed s ∈ R, the two curves
γ1(t) = φt ◦ φs(p) and γ2(t) = φt+s(p)
are both integral curves for X starting at the same point
γ1(0) = φs(p) = γ2(0).
By uniqueness of integral curves, we have
φt ◦ φs = φt+s.
�
Since φ0 = Id, we conclude that
Corollary 2.5. φt : M →M is bijective, and φ−1t = φ−t.
Next time we will prove that each φt, as a map from M to M , is smooth, and thus
is a diffeomorphism. As a consequence, the map
t 7→ φt
is a group homomorphism from R to Diff(M). We will call the family {φt | t ∈ R} a
one-parameter subgroup of diffeomorphisms.
In fact, the result we are going to prove below is a bit stronger: if we let Φ be the
collection of all φt’s,
Φ : R×M →M, (t, p) 7→ φt(p).
We will show that the map Φ is smooth as a map on joint variables (t, p).
LECTURE 15: THE DYNAMICAL SYSTEM ASSOCIATED WITH A
VECTOR FIELD
1. The flow generated by a vector field
¶ The local flow generated by a vector field.
Let X be a smooth vector field (which need not to be complete) on a smooth
manifold M . Last time we have seen that for each point p, there is a positive number
tp > 0, so that there exists a smooth map
Φ :M = {(t, p) | − tp < t < tp}⊂ R×M −→M
(t, p) 7→ Φ(t, p) = γp(t),
where γp is the unique integral curve of X passing p, such that if we write
φt(p) := Φ(t, p), t ∈ (−tp, tp),
then (by modifying the proof we did last time a bit)
(1) φ0 = IdM,
(2) φt+s(p) = φt ◦ φs(p) if t, s, t+ s ∈ (−tp, tp).
As a consequence, if there exists t0 > 0 so that t0 < tp for all p, i.e. {t0} ×M ⊂ M,
then φt0 (and thus φt for each 0 < t < t0) is a diffeomorphism. In this case, the map
Φ|[0,t0]×M : [0, t0]×M →M
is an isotopy (c.f. PSet3-2-3) between IdM and φt0 .
Definition 1.1. We call the map Φ :M→M the local flow generated by X.
¶ The flow generated by a complete vector field.
For the case where X is complete, as we showed last time, the map Φ is defined
over the whole R×M . So we define
Definition 1.2. If X is a complete vector field on M , then we call the map
Φ : R×M →M, (t, p) 7→ Φ(t, p) = γp(t) = φt(p)
the flow generated by X.
Example. The flow generated by the vector field X = ∂
∂x1
on Rnis the translation
Φ : R× Rn → Rn, (t, x1, x2, · · · , xn) 7→ (t+ x1, x2, · · · , xn).
More generally, the flow generated by a constant vector field X =
∑
ci ∂
∂xi
on Rn is
Φ : R× Rn → Rn, (t, x1, x2, · · · , xn) 7→ (c1t+ x1, · · · , cnt+ xn).
1
2 LECTURE 15: THE DYNAMICAL SYSTEM ASSOCIATED WITH A VECTOR FIELD
Example. If we identify R2 with C, then the flow generated by the vector field
X = x
∂
∂y
− y ∂
∂x
is the counterclockwise rotation
Φ : R× C→ C, (t, z) 7→ eitz.
Note that this vector field is tangent to circles centered at the origin. We will denote
the induced vector fields on such circles by d
dθ
.
Now we prove that the flow Φ is smooth:
Theorem 1.3. For any complete vector field X, the flow Φ is smooth.
Proof. According to the fundamental theorem of ODE, for any p ∈ M , there is a
neighborhood Up of p and an interval Ip = (−εp, εp) 3 0 such that
Φ|Ip×Up : Ip × Up →M
is smooth. We need to prove that Φ is smooth near a point (t0, p) ∈ R×M for any t0.
[Idea: By group law we know γp(t0 + t) = γγp(t0)(t). Since “joint smoothness”
holds for small t, we can try to connect p to γt0(p) by a sequence of points, so that
each point can be joint to the next point by a “short” integral curve, so short so
that the smoothness holds uniformly. To get such a finite sequence of points, we
apply our favorite compactness argument.] Since γp : R → M is smooth, the set
K = γp([−εp, t0 + εp]) is compact [Note that X need not be compactly supported][Try
to write down a simpler proof if X is compactly supported.]. So one can find p1, · · · , pm
in K so that the open sets Up1 , · · · , Upm cover K. We denote
U := Up1 ∪ · · · ∪ Upm and I = (−ε0, ε0) := Ip1 ∩ · · · ∩ Ipm .
Then U is an open neighborhood of K, I is an open interval containing 0, and the map
Φ|I×U : I × U →M
is smooth. In particular, for any s0 ∈ I, the map φs0 = Φ(s0, ·) : U →M is smooth.
Now take N large so that s0 := t0/N ∈ I. We set U1 = U and define iteratively
Uk+1 := φ
−1
s0
(Uk) ∩ U, k = 1, · · · , N.
Then for any k = 1, · · · , N + 1, Uk is open, and φs0 : Uk → Uk−1 is smooth. Finally we
consider the open set Φ|−1I×U(UN+1). Since
Φ(s0,Φ(s0, · · · ,Φ(s0,Φ(0, p)) · · · )) = Φ(t0, p) = γp(t0) ∈ K ⊂ U,
we see (0, p) ∈ Φ|−1I×U(UN+1). So there exists a neighborhood I0 × U0 of (0, p) so that
I0 × U0 ⊂ Φ|−1I×U(UN+1), i.e. Φ|I0×U0 : I0 × U0 → UN+1 is smooth. It follows that Φ is
smooth on (t0 + I0)× U0, since it can be written as a composition
(t0 + I0)× U0 −→ I0 × U0
Φ−→ UN+1
φs0−→ UN
φs0−→ · · ·
φs0−→ U1
(t, p) 7→ (t− t0, p) 7→ Φ(t− t0, p) 7→ · · · 7→ Φ(t, p)
of smooth maps. This completes the proof. �
LECTURE 15: THE DYNAMICAL SYSTEM ASSOCIATED WITH A VECTOR FIELD 3
¶ The flow as a one-parameter group of diffeomorphisms.
As we have shown last time, the family of maps φt : M →M form a one-parameter
group, namely
(1) φ0 = IdM,
(2) φt+s = φt ◦ φs,∀t, s.
Since each φt is smooth, invertible, and the inverse φ
−1
t = φ−t is also smooth, they are
diffeomorphisms. As a consequence, the map
t 7→ φt
is a group homomorphism from R to Diff(M). We will call the family {φt | t ∈ R} a
one-parameter subgroup of diffeomorphisms.
So given us a complete vector field X, we get a one-parameter subgroup of diffeo-
morphisms. In fact Theorem 1.3 tells us more: {φt} is a family of diffeomorphisms that
depends on t smoothly (and thus Φ is an isotopy). Conversely, given a smooth map
Φ : R ×M → M so that φt := Φ(t, ·) is one-parameter subgroup of diffeomorphisms,
we may define a vector field X on M by
Xp :=
d
dt
∣∣∣∣
t=0
φt(p)= γ̇p(0).
Then (check the details)
• the smoothness of Φ implies the smoothness of the vector field X,
• The vector field X is complete since γp(t) := Φ(t, p) is an integral curve of X,
• The map Φ is the flow generated by X.
We will call X the infinitesimal generator of the flow Φ.
Remark. More generally, we may consider a time-dependent vector field Xt, i.e. a
vector field that depends smoothly on the parameter t, in the sense that locally Xt can
be written as
Xt =
∑
X i(t, x)∂i
with coefficients X i(t, x) that also depends smoothly on t. For simplicity let’s assume
M is compact. Then Xt also generates a smooth family of diffeomorphisms
1. In
general this family of diffeomorphisms will no longer be a one-parameter subgroup of
diffeomorphisms (why?). It is called the time-dependent flow generated by the time-
dependent vector field Xt.
¶ An application to Morse theory.
As an application, we prove the following fundamental result in Morse theory2:
1One way to see this: A time dependent flow Xt on M determines a vector field X̃ = (∂t, Xt(p)) on
I ×M . By this way we can treat the time-dependent vector fields and its flow on M via the theory
of ordinary vector fields/flows on I ×M .
2Morse theory is a sub-branch of differential topology that enables one to analyze the topology of
a manifold by studying differentiable functions on that manifold. It allows one to find CW structures
and handle decompositions on manifolds and to obtain substantial information about their homology.
4 LECTURE 15: THE DYNAMICAL SYSTEM ASSOCIATED WITH A VECTOR FIELD
Theorem 1.4. Let M be a compact manifold and f a smooth real-valued function on
M . For any a ∈ R, denote
Ma = f−1
(
(−∞, a)
)
.
Suppose a < b and each c ∈ [a, b] is a regular value of f . Then there is a diffeomorphism
ϕ : M →M so that ϕ(Ma) = M b.
Proof. We endow M with a Riemannian metric 〈·, ·〉, and define a vector field ∇f on
M by requiring
〈∇f,Xp〉 = dfp(Xp)
Since the set of critical points of f is closed, one can find an open set U without
critical point such that f−1([a, b]) ⊂ U . Now let h be a smooth bump function with
supp(h) ⊂ U and h = 1 on f−1([a, b]). Since df 6= 0 in U , we get ∇f 6= 0 in U , and
thus
X :=
h
〈∇f,∇f〉
∇f
is a well-defined smooth vector field on M . Let ϕt be the flow of X. Then
d
dt
ϕ∗tf(p) =
d
dt
f(ϕt(p)) = dfϕt(p)(Xϕt(p)) = g(p).
If follows that
d
dt
f(ϕt(p) = 1 for p ∈ f−1([a, b]).
This implies that if p ∈Ma, then ϕb−a(Ma) = M b. �
As a consequence, one can prove (left as an exercise)
Theorem 1.5 (Reeb’s theorem). Let M be a compact manifold of dimension n, and
suppose there exists a smooth real valued function f ∈ C∞(M) so that f has exactly
two critical points, both are non-degenerate. Then M is homeomorphic to Sn.
2. Reading Materials: The Lie Derivatives
¶ The Lie derivative of a function along a vector field.
A smooth manifold M together with a flow Φ is called a dynamical system. In
what follows we study the dynamical system generated by the flow of a vector field X.
For simplicity we assumeX is complete, so that we have a family of diffeomorphisms
φt : M →M . Now let f ∈ C∞(M) be any smooth function. (In physics, M represents
the phase space, and f represents an observable.) We would like to calculate “the rate
of change of f” along the flow. As in calculus, we define
Definition 2.1. The Lie derivative of a f ∈ C∞(M) with respect to X ∈ Γ∞(TM) is
LX(f) :=
d
dt
∣∣∣∣
t=0
φ∗tf
(
= lim
t→0
φ∗tf − f
t
)
.
LECTURE 15: THE DYNAMICAL SYSTEM ASSOCIATED WITH A VECTOR FIELD 5
¶ The Lie derivative of a vector field along another vector field.
So the Lie derivative LXf measures “the rate of change of a smooth function f
along the direction of X”. Now let Y ∈ Γ∞(TM) be another smooth vector field. We
would like to measure “the rate of change of a Y along the direction of X”. One way
to do so is to embed M into some Euclidian space and measure the rate of change of
“coordinates components” of Y . In what follows we will take an intrinsic way. Naively
one would like to compute the limit “limt→0
Yφt(p)−Yp
t
”, where φt is the flow generated by
X. Unfortunately Yφ(p) is not a tangent vector at p and thus the expression “Yφt(p)−Yp”
makes no sense at all. To fix the problem, we need to “push-forward” the tangent vector
Yφt(p) at φt(p) to a tangent vector at p. This can be done via the so-called “push-
forward” operation for vector fields, which is an analogy of the “pull-back” operation
on smooth functions.In general, given a smooth vector field X ∈ Γ∞(TM), one cannot “push-forward”
X to get a smooth vector field on N : for any p ∈ M , we know dϕp(Xp) ∈ Tϕ(p)N .
However, “Yϕ(p) := dϕp(Xp)” does not define a smooth vector field on N , since
• There may exists q ∈ N which is not in the image of ϕ.
• There may exists p1, p2 ∈M such that ϕ(p1) = ϕ(p2) but dϕp1(Xp1) 6= dϕp2(Xp2).
However, if ϕ : M → N is a diffeomorphism, then the two issues above do not exist
and thus the formula Yϕ(p) := dϕp(Xp) does define a smooth vector field on N .
Definition 2.2. If ϕ : M → N is a diffeomorphism, and X ∈ Γ∞(TM), then the
push-forward of X is the smooth vector field ϕ∗X on N defined by
(ϕ∗X)ϕ(p) = dϕp(Xp), ∀p ∈M.
Now we define the “rate of change of Y ∈ Γ∞(TM) with respect to X ∈ Γ∞(TM).
Definition 2.3. The Lie derivative of Y ∈ Γ∞(TM) with respect to X ∈ Γ∞(TM) is
LX(Y ) :=
d
dt
∣∣∣∣
t=0
(φ−t)∗Y
(
= lim
t→0
(φ−t)∗Yφt(p) − Yp
t
)
.
Remark. In both definitions we don’t really need to assume the completeness of X.
To see this, one just need to notice that the differentiations above are local, and that
for any p ∈ M , there is a neighborhood U of p and a uniform t0 so that φt(q) = γq(t)
makes sense for all q ∈ U and all |t| < t0. Then repeat the proof above.
In Today’s PSet, you will prove
Proposition 2.4. Let X, Y ∈ Γ∞(TM) and f ∈ C∞(M). Then
(1) LXf = X(f).
(2) LX(Y ) = [X, Y ].
So roughly speaking, Xf can be think of as “the derivative of f along the direction
of X”, and [X, Y ] can be think of as “the derivative of Y along the direction of X”.
Later on we will define Lie derivatives for differential forms.
LECTURE 16: DISTRIBUTIONS AND FOLIATIONS
1. Distributions
¶ Distributions.
Suppose M is an n-dimensional smooth manifold. We have seen that any smooth vector
field X on M can be integrated locally near any point p to an integral curve γp. Moreover,
• If Xp = 0, then γp is the constant curve γp(t) ≡ p. (0-dimensional points)
• If Xp 6= 0, then γp is a 1-dimensional curve passing p. (1-dimensional curves)
In what follows we will develop a higher dimensional analogue to the “vector field! integral
curve” correspondence. We first generalize the conception of vector field.
Definition 1.1. Let M be a smooth manifold.
(1) A k-dimensional distribution V on M is a map which assigns to every point p ∈ M
a k-dimensional vector subspace Vp of TpM .
(2) A distribution V is called smooth if for every p ∈ M , there is a neighborhood
U of p and k smooth vector fields X1, · · · , Xk on U such that for every q ∈ U ,
{X1(q), · · · , Xk(q)} form a basis of Vq. (In particular, Xi(q) 6= 0 for all 1 ≤ i ≤ k.)
(3) We say a vector field X belongs to a distribution V if Xp ∈ Vp for all p ∈M .
In what follows, all distributions will be smooth.
We also generalize the conception of integral curves:
Definition 1.2. Suppose V is a k-dimensional smooth distribution on M .
(1) An immersed submanifold ι : N ↪→M is called an integral manifold for V if for every
p ∈ N , the image of dιN : TpN → TpM is Vp.
(2) We say the distribution V is integrable if through each point of M there exists an
integral manifold of V .
1
2 LECTURE 16: DISTRIBUTIONS AND FOLIATIONS
¶ Examples of integrable and non-integrable distributions.
Example. The vector fields ∂
∂x1
, · · · , ∂
∂xk
span a k-dimensional distribution V in Rn. The
integral manifolds of V are “hyperplanes” defined by the system of equations
xi = ci (k + 1 ≤ i ≤ n).
We will see below (Theorem 2.2, the local Frobenius theorem) that locally any integrable
distribution can be written as this form.
Example. An integral manifold need not to be an embedded submanifold of M . For example,
consider M = S1 × S1 ⊂ R2x × R2y. Fix any irrational number a, the integral manifolds of
the non-vanishing vector field
Xa = (x2
∂
∂x1
− x1 ∂
∂x2
) + a(y2
∂
∂y1
− y1 ∂
∂y2
)
are “dense curves” in M . (However, they are immersed submanifolds.)
Example. Any non-vanishing vector field X is a 1-dimensional distribution. It is always
integrable: the image of any integral curve of X is an integral manifold.
It turns out that a higher dimensional distribution may fail to be integrable.
Example. Consider the smooth distribution V on R3 spanned by two vector fields
X1 =
∂
∂x1
+ x2
∂
∂x3
, X2 =
∂
∂x2
.
I claim that there is no integral manifold through the origin. In fact, if V is integrable,
then the integrable manifold N of V containing the origin must also contain the integrable
curve of X1 passing the origin (for a reason, see Corollary 2.3 below), which is a piece of the
x1-axis, i.e.
N ⊃ {(t, 0, 0) | |t| < ε}.
Similarly N must also contain the integral curves of the vector field X2 passing all these
points (t, 0, 0). It follows that for each |t| < ε, N contains a small piece of line segment
parallel to the x2-axis, i.e.
N ⊃ {(t, s, 0) | |t| < ε, |s| < δt.}
In other words, N contains a piece of the x1-x2 plane that contains the origin. This is a
contradiction, because the vector ∂
∂x1
is a tangent vectors of this piece of plane but is not in
Vp for any p 6= (t, 0, 0).
¶ Frobenius condition and involutive distributions.
Since not all distributions are integrable, we are interested in the conditions to make a
distribution integrable. A necessary condition is easy to find. Intuitively, if two vector fields
of M are tangent to a submanifold N of M , then their Lie bracket should also tangent to
N . Motivated by this fact, we have
LECTURE 16: DISTRIBUTIONS AND FOLIATIONS 3
Theorem 1.3. If a distribution V is integrable, then for any two vector fields X and Y
belonging to V, their Lie bracket [X, Y ] belongs to V also.
Proof. Fix any p ∈M , suppose ι : N ↪→M is an integrable manifold of V passing p. Since N
is an immersed submanifold of M , one can “shrink” N so that ι(N) is in fact an embedded
submanifold of M . Now suppose X, Y are vector fields belonging to V , then the restrictions
X|N , Y |N of X, Y to N are vector fields that are tangent to the submanifold N . More
precisely, there exists X̃, Ỹ ∈ Γ∞(TN) such that
Xp = dιp(X̃p) and Yp = dιp(Ỹp), ∀p ∈ N = ι(N).
It follows from today’s PSet (PSet 5-1-3-a-ii)) that for any p ∈ N ,
[X, Y ]Mp = dιp([X̃, Ỹ ]
N
p ) ∈ Vp.
So [X, Y ] belongs to V also. �
Definition 1.4. A distribution V is involutive if it satisfies the following Frobenius condition:
Frobenius condition: If X, Y ∈ Γ∞(TM) belong to V , so is [X, Y ].
Example. Any 1 dimensional distribution is involutive since
[fX, gX] = (fX(g)− gX(f))X
is a multiple of X.
Note that by definition, to check a distribution V is involutive, one need to check [X, Y ] ∈
V for all X, Y ∈ V . It turns out that it is enough to check this for a set of local smooth basis
of V :
Lemma 1.5. Let V be a k dimensional distribution on M . Suppose for each p ∈ M ,
there exist a neighborhood U of p and k pointwise linearly independent smooth vector fields
X1, · · · , Xk on U so that [Xi, Xj] belong to V for all 1 ≤ i, j ≤ k. Then V is involutive.
Proof. Left as an exercise. �
Example. The distribution spanned by ∂
∂x1
, · · · , ∂
∂xk
in Rn is involutive, since
[
∂
∂xi
,
∂
∂xj
] = 0, ∀1 ≤ i, j ≤ k.
Example. The distribution V spanned by
X1 =
∂
∂x1
+ x2
∂
∂x3
, X2 =
∂
∂x2
is not involutive, since
[X1, X2] = −
∂
∂x3
is not in V .
4 LECTURE 16: DISTRIBUTIONS AND FOLIATIONS
¶ Frobenius Theorem.
Theorem 1.3 tells us that any integrable distribution is involutive. It turns out that the
converse is also true, namely, any involutive distribution is integrable:
Theorem 1.6 (Global Frobenius Theorem). Let V be an involutive k-dimensional distribu-
tion. Then through every point p ∈M , there is a unique maximal connected integral manifold
of V (in particular, V is integrable).
Example. Let f : M → N be a submersion. For any p ∈M , let Vp = ker(dfp). Then
• V is a distribution since dimVp = dimM − dimN is constant for all p ∈M .
• V is involutive: if X, Y are vector fields belonging to V , then dfp(Xp) = dfp(Yp) = 0
for all p. It follows that for any g ∈ C∞(N),X(g ◦ f)(p) = dfp(Xp)(g) = 0 for all p,
i.e. X(g ◦ f) = 0. Similarly Y (g ◦ f) = 0. It follow that
dfp([X, Y ]p)(g) = [X, Y ]p(g ◦ f) = Xp(Y (g ◦ f))− Yp(X(g ◦ f)) = 0.
• V is integrable: the integrable manifold passing p ∈M is the submanifold f−1(f(p)).
Example. Consider the distribution V on R3 spanned by
X1 = x
1 ∂
∂x2
− x2 ∂
∂x1
, X2 =
∂
∂x3
on M = R3 − {x1 = x2 = 0}. Since [X1, X2] = 0, V is involutive. What are its integral
manifolds? Well, let’s first compute the integral curves of X1 and X2. Through any point
(x1, x2, x3), the integral curves of X1 are circles in the x
3-plane with origin the center,
and the integral curves of X2 are the lines that are parallel to the x
3-axis. Note that the
integral manifold passing (x1, x2, x3) of the distribution should contain all points of the form
ϕX1t (ϕ
X2
s (x
1, x2, x3)) for all t, s. In our case, these are the cylinders centered at the x3-axis.
¶ Foliation.
As a consequence, given any k-dimensional involutive distribution on M , one can “de-
compose” M into a disjoint union of k-dimensional connected immersed submanifolds. Such
a decomposition is called a foliation structure on M .
Definition 1.7. A k-dimensional foliation F of an m-dimensional manifold M is a decom-
position of M into a union of disjoint connected immersed submanifolds {Lα}α∈A, called the
leaves of the foliation, with the following property:
Every point in M has a neighborhood U and a system of local coordinates
x = (x1, · · · , xm) on U such that for each leaf Lα, the connected components
of U ∩ Lα are described by the equations xk+1 = ck+1, · · · , xm = cm.
A foliation is called regular if each leaf is an embedded submanifold.
Using the language of foliation, we can rewrite the Frobenius theorem as: Let V be an
involutive k-dimensional distribution. Then the collection of all maximal connected integral
manifold of V form a k-dimensional foliation F of M . Conversely, given any k-dimensional
foliation F , by definition the collection of all tangent spaces of all leaves form an integrable
k-dimensional distribution.
LECTURE 16: DISTRIBUTIONS AND FOLIATIONS 5
2. The proof of Frobenius theorem
¶ Flatten one vector field locally.
We first prove the following useful lemma:
Lemma 2.1. Let X be any smooth vector field on M with Xp 6= 0. Then there is a local
chart (ϕ,U, V ) near p so that X = ∂1 on U .
Proof. Choose a local chart (Ũ , y1, · · · , yn) around p such that Xp = ∂∂y1
∣∣∣
p
. Denote X =∑n
i=1 ξi
∂
∂yi
on Ũ , where ξi are smooth functions on Ũ . Shrinking Ũ if necessary, we may
assume ξ1 6= 0 on Ũ . Consider the system of ODEs
dyi
dy1
=
ξi(y
1; y2, · · · , yn)
ξ1(y1; y2, · · · , yn)
, 2 ≤ i ≤ n.
By basic theory of ODE, locally for any given initial data (z2, · · · , zn), with |z| < ε, the
system above has a unique solution
yi = yi(y1; z2, · · · , zn), |y1| < ε
with initial condition
yi(0; z2, · · · , zn) = zk, 2 ≤ i ≤ n
and the functions yi depends smoothly on y1 and on zj’s. Consider
y1 = z1,
yi = yi(z1; z2, · · · , zn), 2 ≤ i ≤ n.
Since the Jacobian
∂(y1, · · · , yn)
∂(z1, · · · , zn)
∣∣∣∣
z1=0
= 1,
we can make the change of variables from (y1, · · · , yn) to (z1, · · · , zn), i.e. there exists a
neighborhood U ⊂ Ũ of p, with (z1, · · · , zn) as local coordinate functions. We have in this
new chart
X =
∑
ξi
∂
∂yi
= ξ1
∑ ∂yi
∂z1
∂
∂yi
= ξ1
∂
∂z1
.
Finally if we let x1(z1, · · · , zn) =
∫ z1
0
dt
ξ1(t,z2,··· ,zn) and xj = zj for j ≥ 2, then {x
1, · · · , xn}
are local coordinate functions on U such that X = ∂
∂x1
on U . �
Remark. More generally, if [Xi, Xj] = 0 and if Xi’s are linearly independent, then there
exists a coordinate chart so that Xi = ∂i. The proof is left as an exercise. Note that the
condition [Xi, Xj] = 0 is necessary since in any local charts, we have [∂i, ∂j] = 0.
6 LECTURE 16: DISTRIBUTIONS AND FOLIATIONS
¶ Local Frobenius Theorem: “Flatten” an involutive distribution.
Before proving the global Frobenius theorem, we first prove the following local version:
locally any involutive distribution is “flat”, and thus is integrable.
Theorem 2.2 (Local Frobenius Theorem). Let V be an involutive k-dimensional distribu-
tion. Then for every p ∈ M , there exists a coordinate patch (U, x1, · · · , xn) centered at p
such that for all q ∈ U , Vq = span{∂1(q), · · · , ∂k(q)}.
Proof. By Lemma 2.1, the theorem holds for k = 1. Now suppose the theorem holds for
k − 1 dimensional involutive distributions. Let V be an k dimensional distribution spanned
locally by X1, X2, · · · , Xk near p. Suppose V is involutive, i.e.
[Xi, Xj] ≡ 0 mod (X1, · · · , Xk), 1 ≤ i, j ≤ k.
According to Lemma 2.1, there exits a local chart (U ; y1, · · · , yn) near p such that Xk = ∂yk .
For 1 ≤ i ≤ k − 1 let
X ′i = Xi −Xi(yk)Xk,
then X ′i(y
k) = 0 for 1 ≤ i ≤ k − 1, and Xk(yk) = 1. Note that the vector fields
X ′1, · · · , X ′k−1, Xk still span V . Moreover, if we denote
[X ′i, X
′
j] = aijXk mod (X
′
1, · · · , X ′k−1), 1 ≤ i, j ≤ k − 1,
then after applying both sides to the function yk, we see aij = 0 for all 1 ≤ i, j ≤ k − 1.
According to Lemma 1.5, the k − 1 dimensional distribution
V ′ = span{X ′1, · · · , X ′k−1}
is involutive. By induction hypothesis, there is a local chart (U, z1, · · · , zn) near p such that
V ′ is spanned by {∂z1 , · · · , ∂zk−1}. Since each ∂zi(1 ≤ i ≤ j) is a linear combination of X ′j
for 1 ≤ j ≤ k − 1, we conclude ∂zi(yk) = 0.
Now denote
[∂zi , Xk] = biXk mod (∂z1 , · · · , ∂zm−1).
Apply both sides to the function yk, we see bi = 0 for all i. So we can write
[∂zi , Xk] =
k−1∑
j=1
Cij∂zj .
Suppose Xk =
∑n
j=1 ξj∂zj . Insert this into the previous formula, we see
∂ziξj = 0, 1 ≤ i ≤ k − 1, k ≤ j ≤ n.
In other words, for j ≥ k, ξj = ξj(zk, · · · , zn). Let
X ′k =
n∑
j=k
ξj∂zj .
Then {∂z1 , · · · , ∂zk−1 , X ′k} still span V . Finally according to Lemma 2.1 again, there is a
local coordinate change from (z1, · · · , zk, · · · , zn) to (x1, · · · , xk, · · · , xn) with xi = zi for
1 ≤ i ≤ k − 1, such that X ′k = ∂xk . This completes the proof. �
LECTURE 16: DISTRIBUTIONS AND FOLIATIONS 7
As a consequence, we get the following result that we used in previous examples:
Corollary 2.3. Let X be a smooth vector field belongs to V. Then for any p ∈ M , the
integral curve of X passing the point p lies in the integral manifold of V passing p.
¶ Sketch of proof of the Global Frobenius theorem:
Sketch of proof of the Global Frobenius theorem: For any p ∈M , let
Np = {q ∈M | ∃ a piecewise smooth integral curve of V jointing p to q}1.
We claim that Np is the maximal connected integral manifold of V containing p.
The manifold structure is defined as follows: for any q ∈ Np, there is a coordinate patch
(φ, U, V ) (with φ(w) = (x1(w), · · · , xn(w))) centered at q such that V = span{∂1, · · · , ∂k} in
U . For each small ε, let
Wε = {w ∈ U | (x1)2(w) + · · ·+ (xk)2(w) < ε, xk+1(w) = · · · = xn(w) = 0}.
Then any point w ∈ Wε can be joint to q by the integral curve
γ(t) = φ−1(tx1(w), · · · , txk(w), 0, · · · , 0).
So Wε ⊂ Np. Let
ϕ : Wε → Bk(ε) ⊂ Rk, w 7→ (x1(w), · · · , xk(w)).
Now we define the topology on Np by giving it the weakest topology such that all these
ϕ’s are homeomorphisms. The atlas on Np is defined to be the set of charts (ϕ,W,B
k(ε)).
One can check that Np is a manifold with this given atlas. For more details, c.f. Warner,
pg.48-49. �
1We say γ is a piecewise smooth integral curve of V if γ is of the form “γ1 connected to γ2 connected · · ·
connected to γm”, where each γi is a smooth integral curve of a smooth vector field in V.
LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS
1. Lie groups
¶ Lie groups: Definition and examples.
A Lie group is a special smooth manifold on which there is a group structure, and
moreover, the manifold structure (which is already a mixture of an topological structure
with an “analysis structure”, namely, the smooth structure) and the group structure
are compatible. In some sense, Lie groups are the best objects in mathematics: they
stand atthe crossing point of algebra, analysis and geometry. Lie groups are not just
special examples of smooth manifolds with nice properties. They represent smooth
families of symmetries1 and play very important roles in various places in geometry
and physics.
Definition 1.1. A Lie group G is a smooth manifold equipped with a group structure
so that the group multiplication
µ : G×G→ G, (g1, g2) 7→ g1 · g2
is a smooth map.
Example. Here are some basic examples:
• Rn, considered as a group under addition.
• R∗ = R− {0}, considered as a group under multiplication.
• S1, considered as a group under multiplication.
• Linear Lie groups with matrix multiplications
GL(n,R) = {X ∈M(n,R) | detX 6= 0}, (the general linear group)
SL(n,R) = {X ∈M(n,R) | detX = 1}, (the special linear group)
O(n) = {X ∈M(n,R) | XXT = In}, (the orthogonal group).
(GL(n,R) is a smooth manifold since it is an open set in M(n,R) ' Rn2 . By
PSet 2-2-4, SL(n,R) and O(n) are smooth submanifolds of GL(n,R), and thus
are smooth manifolds.)
• If G1 and G2 are Lie groups, so is their product G1×G2. (So Tn = S1×· · ·×S1
is a Lie group.)
1Wiki: Lie groups are named after Norwegian mathematician Sophus Lie (1842-1899), who laid
the foundations of the theory of continuous transformation groups. Lie’s original motivation for
introducing Lie groups was to model the continuous symmetries of differential equations, in much the
same way that finite groups are used in Galois theory to model the discrete symmetries of algebraic
equations.
1
2 LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS
Remarks. (1) (Hilbert’s 5th problem, [Gleason and Montgomery-Zippin, 1950’s]): Let
G be any topological group whose underlying space is a topological manifold, then G
admits a smooth structure to make it a Lie group.
(2) Not every smooth manifold admits a Lie group structure. For example, the only
spheres that admit a Lie group structure are S0, S1 and S3 (Can you write down a Lie
group structure on S3?). Here are some simple topological restrictions for a smooth
manifold to admits a Lie group structures:
• The underlying space of a Lie group must be orientable.
• For any Lie group G (and in fact for any topological group G), the fundamental
group π1(G) is always abelian. (Left as an exercise)
• The tangent bundle TG of any Lie group is trivial: TG ' G×Rn. (We will see
TS2 6' S2 × R2. As a consequence, there is no Lie group structure on S2.)
As a result, among all the 2 dimensional closed manifolds (i.e. closed surfaces), the
only one that admits a Lie group structure is T2 = S1 × S1.
¶ Left/right multiplications.
Now suppose G is a Lie group. For any elements a, b ∈ G, there are two natural
maps, the left multiplication
La : G→ G, g 7→ a · g
and the right multiplication
Rb : G→ G, g 7→ g · b.
La is smooth since it can be viewed as a composition of smooth maps,
La : G
ja
↪−→ G×G µ−→ G,
g 7−→ (a, g) 7−→ a · g.
Similarly Rb = µ ◦ ib is smooth, where ib : G ↪→ G × G is the smooth inclusion map
ib(g) = (g, b). It is obviously that L
−1
a = La−1 and R
−1
b = Rb−1 . So both La and Rb are
diffeomorphisms. Moreover, La and Rb commutes with each other: LaRb = RbLa.
As an application of the left multiplication we prove
Proposition 1.2. For any Lie group G, we have TG ' G× Rn.
Proof. We identify TeG = Rn, and define
φ : G× TeG→ TG, φ(a, ξ) = (a, dLa(ξ))
This is obviously a bijective map, with inverse φ−1(a, ξ) = (x, dL−a(ξ)). Note that
when we fix x, both φ and φ−1 are linear isomorphisms between {a} × TeG and TaG.
This implies that both φ and φ−1 are smooth, and thus are diffeomorphisms. �
¶ The smoothness of the inverse.
We can use left/right multiplication to compute the differential of µ:
LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS 3
Lemma 1.3. The differential of the multiplication map µ : G×G→ G is given by
dµa,b(Xa, Yb) = (dRb)a(Xa) + (dLa)b(Yb)
for any (Xa, Yb) ∈ TaG× TbG ' T(a,b)(G×G).
Proof. For any function f ∈ C∞(G), we have
(dµa,b(Xa, Yb))(f) = (Xa, Yb)(f ◦ µ) = Xa(f ◦ µ ◦ ib) + Yb(f ◦ µ ◦ ja)
= Xa(f ◦Rb) + Yb(f ◦ La)
= (dRb)a(Xa)(f) + (dLa)b(Yb)(f). �
As an application, we can prove
Proposition 1.4. For any Lie group G, the group inversion map
i : G→ G, g 7→ g−1
is smooth, and
dιa(Xa) = −(dLa−1)e(dRa−1)a(Xa), ∀Xa ∈ TaG.
Proof. [Idea: write i as a composition of smooth maps...] Consider the smooth map
f : G×G→ G×G, (a, b) 7→ (a, ab).
Obviously it is bijective. According to the lemma above, the derivative of f is
df(a,b) : TaG× TbG→ TaG× TabG, (Xa, Yb) 7→ (Xa, (dRb)a(Xa) + (dLa)b(Yb)).
This is a bijective linear map since dRb, dLa are. (Can you write down the inverse of
df(a,b)?) It follows by the inverse function theorem that f is locally a diffeomorphism
near each pair (a, b). However, since f is globally bijective, it must be a globally
diffeomorphism. We conclude that its inverse,
f−1 : G×G→ G×G, (a, c) 7→ (a, a−1c)
is a diffeomorphism. Thus the inversion map i, as the composition
G ↪−→ G×G f
−1
−→ G×G π2−→ G
a 7−→ (a, e) 7−→ (a, a−1) 7−→ a−1
is smooth. This also implies
dιa(Xa) = dπ2 ◦ (df−1)(a,e)(Xa, 0) = dπ2
(
Xa,−(dLa−1)e(dRa−1)a(Xa)
)
= −(dLa−1)e(dRa−1)a(Xa),
where in the second last step, we used the fact
(df)(a,a−1)(Xa, Ya−1) = (Xa, 0) =⇒ Ya−1 = −(dLa)a−1(dRa−1)a(Xa)
= −(dLa−1)e(dRa−1)a(Xa). �
Since dLe = dRe = Id, we get
(dι)e(Xe) = −Xe, ∀Xe ∈ TeG.
4 LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS
2. Lie algebras associated to Lie groups
¶ Abstract Lie algebras.
We start with the abstract definition.
Definition 2.1. A Lie algebra is a real vector space V together with a binary bracket
operation
[·, ·] : V × V → V,
such that for any X, Y, Z ∈ V ,
(1) (Linearity) [aX1 + bX2, Y ] = a[X1, Y ] + b[X2, Y ],
(2) (Skew-symmetry) [X, Y ] = −[Y,X],
(3) (Jacobi identity) [X, [Y, Z]] + [Y, [Z,X]] + [Z, [X, Y ]] = 0.
The bracket [·, ·] is called the Lie bracket.
Example. Any vector space admits a trivial Lie algebra structure: [X, Y ] ≡ 0.
Example. By PSet 4-2-3, the set of all smooth vector fields Γ∞(TM) on any smooth
manifold form a Lie algebra.
Example. The set of all n× n real matrices, M(n,R), is a Lie algebra if we define the
Lie bracket to be the commutator of matrices:
[A,B] := AB −BA.
(More generally, the commutator [X, Y ] = XY −Y X define a Lie algebra structure on
any associative algebra.
¶ Left invariant vector fields on Lie groups.
Now we will associate to any Lie group G a God-given Lie algebra. Suppose G is
a Lie group. From the left translation La one can, for any vector Xe ∈ TeG, define a
vector field X on G by
Xa = (dLa)e(Xe).
[Can you prove the smoothness of X?]
It is not surprising that the vector field X is “invariant” under any left translation:
(dLa)b(Xb) = (dLa)b ◦ dLb(Xe) = dLab(Xe) = Xab.
Definition 2.2. A left invariant vector field on a Lie group G is a smooth vector field
X on G which satisfies
(dLa)b(Xb) = Xab, ∀a, b ∈ G.
So any tangent vector Xe ∈ TeG determines a left invariant vector field on G.
Conversely, any left invariant vector field X is uniquely determined by its “value” Xe
at e ∈ G, since for any a ∈ G, X(a) = (dLa)Xe.
We will denote the set of all left invariant vector fields on Lie group G by g, i.e.
g = {X ∈ Γ∞(TG) | X is left invariant}.
Then it is not hard to see that
LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS 5
• g is a vector subspace of Γ∞(TG).
• the correspondence Xe ∈ TeG! X ∈ g preserves the linear structures.
So we have a linear isomorphism
g ' TeG
as vector spaces. In particular, dim g = dimG.
Next we show that g is a Lie algebra, the Lie algebra structure being the one
induced from the Lie algebra structure on Γ∞(TG):
Proposition 2.3. If X, Y ∈ g, so is their Lie bracket [X, Y ].
Proof. We want to show that [X, Y ] is left-invariant if X and Y are. First notice
Y (f ◦ La)(b) = Yb(f ◦ La) = (dLa)b(Yb)f = Yabf = (Y f)(Lab) = (Y f) ◦ La(b)
for any smooth function f ∈ C∞(G). Thus
Xab(Y f) = (dLa)b(Xb)(Y f) = Xb((Y f) ◦ La) = XbY (f ◦ La).
Similarly YabXf = YbX(f ◦ La). Thus
dLa([X, Y ]b)f = XbY (f ◦ La)− YbX(f ◦ La)= Xab(Y f)− Yab(Xf) = [X, Y ]ab(f).
�
¶ The Lie algebras associated with Lie groups.
It follows that the space g of all left invariant vector fields on G together with the
Lie bracket operation [·, ·] is an n-dimensional Lie subalgebra of the Lie algebra of all
smooth vector fields Γ∞(TG).
Definition 2.4. The Lie algebra g of is called the Lie algebra of G.
We give a couple examples:
Example. (The Euclidean space Rn.) This is obviously a Lie group, since the group
operation
µ((x1, · · · , xn), (y1, · · · , yn)) := (x1 + y1, · · · , xn + yn).
is smooth.
Moreover, for any a ∈ Rn, the left translation La is just the usual translation map
on Rn. So dLa is the “identity map”, as long as we identify TaRn with Rn in the usual
way. It follows that any left invariant vector field is in fact a constant vector field, i.e.
Xv = v1
∂
∂x1
+ · · ·+ vn
∂
∂xn
for ~v = (v1, · · · , vn) ∈ T0Rn. Since ∂∂xi commutes with
∂
∂xj
for any pair (i, j), we
conclude that the Lie bracket of any two left invariant vector fields vanishes. In other
words, the Lie algebra of G = Rn is g = Rn with vanishing Lie bracket.
6 LECTURE 17: LIE GROUPS AND THEIR LIE ALGEBRAS
Example. (The affine group R∗ nR1)
One can equip with R∗×R the usual manifold structure and with the group oper-
ation
µ((x1, y1), (x2, y2)) = (x1x2, x1y2 + y1).
This is called the affine group of R1. It is obviously a non-abelian Lie group.
Taking derivative from the previous expression of L(x1,y1)(x2, y2), we get
dL(x1,y1)(v1
∂
∂x
+ v2
∂
∂y
) = x1v1
∂
∂x
+ x1v2
∂
∂y
.
Thus if a(x, y) ∂
∂x
+ b(x, y) ∂
∂y
is a left invariant vector field, we must have
a(x1x, x1y + y1) = a(x, y)x1 and b(x1x, x1y + y1) = b(x, y)x1
for all x, y, x1, y1 ∈ R. In particular, X = x ∂∂x and Y = x
∂
∂y
are two left invariant
vector fields. Since the Lie group is 2-dimensional, X, Y form a basis of its Lie algebra.
The Lie brackets are given by [X,X] = [Y, Y ] = 0 and
[X, Y ] = [x
∂
∂x
, x
∂
∂y
] = x
∂
∂y
= Y.
Example. (The general linear group)
GL(n,R) = {X ∈M(n,R) | detX 6= 0}.
Obviously GL(n,R) is n2-dimensional noncompact Lie group. Moreover, it is not con-
nected, but consists of exactly two connected components (why?),
GL+(n,R) = {X ∈M(n,R) | detX > 0}
and
GL−(n,R) = {X ∈M(n,R) | detX < 0}.
The fact that GL(n,R) is an open subset of M(n,R) ' Rn2 implies that the Lie algebra
of GL(n,R), as the tangent space at e = In, is the set M(n,R) itself, i.e.
gl(n,R) = {A | A is an n× n matrix}.
To figure out the Lie bracket operation, we take a matrix A = (Aij)n×n ∈ gl(n,R), and
denote the global coordinate system on GL(n,R) by (X ij). Then the corresponding
tangent vector at TInGL(n,R) is
∑
Aij
∂
∂Xij
, and the corresponding left-invariant vector
on G at the matrix X = (X ij) is
∑
X ikAkj
∂
∂Xij
. It follows that the Lie bracket [A,B]
between matrices A,B ∈ g is the matrix corresponding to[∑
X ikAkj
∂
∂X ij
,
∑
XpqBqr
∂
∂Xpr
]
=
∑
X ikAkjBjr
∂
∂X ir
−
∑
XpqBqrArj
∂
∂Xpj
=
∑
X ik (AkrBrj −BkrArj)
∂
∂X ij
.
In other words, the Lie bracket operation on g is the matrix commutator
[A,B] = AB −BA.
LECTURE 18: THE EXPONENTIAL MAP
1. Lie Homomorphisms
¶ Lie group/Lie algebra homomorphisms.
It is natural to study morphisms between Lie groups and between Lie algebras.
As usual, they are maps between corresponding objects that preserves corresponding
definition properties.
Definition 1.1. Let G,H be Lie groups.
(1) A map φ : G→ H is called a Lie group homomorphism if it is smooth and is a
group homomorphism, i.e.
φ(g1 · g2) = φ(g1) · φ(g2), ∀g1, g2 ∈ G.
(2) A Lie group homomorphism φ : G → H is called an Lie group isomorphism if
it is invertible and the inverse φ−1 : H → G is also a Lie group homomorphism.
Note that by definition, if two Lie groups are isomorphic, then they are diffeomor-
phic as manifolds, and are isomorphic as groups.
Example. For any Lie group G and any a ∈ G, the conjugation map
c(a) = La ◦Ra−1 : G→ G, g 7→ aga−1
is a Lie group isomorphism.
Remark. In PSet 1-1-1 we have seen that for any connected topological group G, there
is an open neighborhood U of e ∈ G such that G = ∪∞n=1Un. As a consequence, any Lie
group homomorphism φ : G→ H is determined by its restriction to a neighborhood of
e ∈ G, provided G is connected.
Similarly one can define Lie algebra homomorphisms to be
Definition 1.2. Let g, h be Lie algebras.
(1) A linear map L : g→ h is called a Lie algebra homomorphism if
L([X1, X2]) = [L(X1), L(X2)], ∀X1, X2 ∈ g.
(2) A Lie algebra homomorphism L : g → h is called an Lie algebra isomorphism
if it is invertible.
Note that if a Lie algebra homomorphism L : g → h is invertible, the inverse
L−1 : h → g is automatically a Lie algebra homomorphism. So again we see the
phenomena that “linear objects are much easier to handle”.
1
2 LECTURE 18: THE EXPONENTIAL MAP
Example. For any X ∈ GL(n,R), the adjoint map
AdX : gl(n,R)→ gl(n,R), A 7→ XAX−1
is a Lie algebra isomorphism. (Check this. What is (AdX)
−1?)
¶ From Lie group homomorphisms to Lie algebra homomorphisms.
Now suppose φ : G → H is a Lie group homomorphism, then its differential at e
gives a linear map dφe : TeG→ TeH. Under the identification
TeG ' g and TeH ' h,
we get an induced map, which we will denote by dφ, from g to h:
dφ : g→ h.
In other words, when viewed as left invariant vector fields, for any X ∈ g, the image
dφ(X) is the left invariant vector field on H whose value at e ∈ H is dφe(Xe), i.e.
(dφ(X))h = dL~(dφe(Xe)).
Example. Start with the conjugation map c(X) : GL(n,R) → GL(n,R) on GL(n,R).
Taking differential at the identity matrix In, we get
(dc(X))In(A) =
d
dt
∣∣∣∣
t=0
c(X)(I + tA) ==
d
dt
∣∣∣∣
t=0
X(I + tA)X−1 = XAX−1.
In other words, the induce map is the Lie algebra homomorphism
dc(X) = AdX : gl(n,R)→ gl(n,R).
In what follows we show that the induced map dφ is always a Lie algebra homo-
morphism. We need
Lemma 1.3. Any X ∈ g is φ-related to dφ(X) ∈ h.
Proof. Take any X ∈ g. Write h = φ(g). Since φ is a group homomorphism, we have
φ ◦ Lg = Lh ◦ φ.
It follows
dφg(Xg) = dφg ◦ (dLg)e(Xe) = dLh ◦ dφe(Xe) = (dφ(X))h.
This completes the proof. �
As a consequence, we can prove
Theorem 1.4. If φ : G → H is a Lie group homomorphism, then the induced map
dφ : g→ h is a Lie algebra homomorphism.
Proof. Let X, Y ∈ g. According to the lemma above,
• X is φ-related to dφ(X), Y is φ-related to dφ(Y ).
 So by PSet 5-1-3, [X, Y ] is φ-related to [dφ(X), dφ(Y )].
• [X, Y ] is φ-related to dφ([X, Y ]).
LECTURE 18: THE EXPONENTIAL MAP 3
It follows that
[dφ(X), dφ(Y )]e = dφe([X, Y ]e) = (dφ([X, Y ]))e.
Since both dφ([X, Y ]) and [dφ(X), dφ(Y )] are left-invariant vector fields on H, we
conclude dφ([X, Y ]) = [dφ(X), dφ(Y )]. �
Example. Last time we have seen that GL(n,R) (with matrix multiplication) is a Lie
group, and the corresponding Lie algebra is gl(n,R) (with matrix commutator). It is
also easy to see that R∗ (with real number multiplication) is a Lie group (= GL(1,R)),
and the corresponding Lie algebra is the trivial Lie algebra R (with real number addi-
tion) (= gl(1,R)). It is easy to see that the determinant det : GL(n,R)→ R∗ is a Lie
group homomorphism, since
det(XY ) = detX detY, ∀X, Y ∈ GL(n,R).
According to PSet 2-2-4, we have
d detX(A) = (detX)tr(X
−1A), ∀X ∈ GL(n,R), A ∈ gl(n,R).
By taking X = In, we conclude that the induced Lie algebra homomorphism for det is
d det = tr : gl(n,R)→ R, A 7→ tr(A).
Since R is the trivial Lie algebra, we thus get a conceptional proof of the fact
tr(AB) = tr(BA), ∀A,B ∈ gl(n,R).
Remark. In summary, we have two categories, the category LIEGROUP
• Objects are Lie groups,
• Morphisms are Lie group homomorphisms
and the category LIEALGEBRA
• Objects are Lie algebras,
• Morphisms are Lie algebra homomorphisms.
Moreover, we have a functor LIE that
• associates to each Lie group G its Lie algebra g,
• associates to Lie group homomorphism φ : G → H the induced Lie algebra
homomorphism dφ : g→ h.1
Of course it is easy to find different Lie groups whose Liealgebras are the same.
However, it turns out that the functor LIE is “invertible” when we are restricted to the
subcategory of LIEGROUP whose objects are simply connected Lie groups: According
to Lie’s third theorem, any finitely dimensional Lie algebra is the Lie algebra of some
simply connected Lie group; moreover, if G is simply connected, then any Lie algebra
homomorphism L : g → h can be “lifted” to a Lie group homomorphism φ : G → H
so that L = dφ.
1One need to check d(IdG) = Idg and d(φ1 ◦ φ2) = dφ1 ◦ dφ2, both follows from definitions easily.
4 LECTURE 18: THE EXPONENTIAL MAP
2. The Exponential Map
¶ The exponential map.
Let G be any Lie group, g its Lie algebra. So by definition, any X ∈ g is a
left-invariant vector field. In general G need not be compact, and thus X need not
be compactly-supported. However, thanks to the left translation maps, we have a
“universal way” to control the vectors at different points of a left invariant vector field.
As a result, one can prove(left as an exercise)
Lemma 2.1. Any left invariant vector field X ∈ g on a Lie group G is complete.
Let φXt : G → G be the flow generated by X ∈ g. The completeness guarantees
that φXt is well-defined for all t. In particular, it is defined for t = 1.
Definition 2.2. The exponential map2 of G is the map
exp : g→ G, X 7→ φX1 (e).
Remark. It is not hard to see φXts = φ
tX
s . So
exp(tX) = φtX1 (e) = φ
X
t (e).
Moreover, one can easily show that {exp(tX)|t ∈ R} is a one-parameter subgroup of G:
exp(tX) · exp(sX) = exp((t+ s)X).
Note that in general
exp(tX) exp(tY ) 6= exp(t(X + Y )).
Example. For G = R∗, we can identify T1G = R. For any x ∈ R = T1G, the left
invariant vector field corresponding to x = x d
dt
∈ T1G, when evaluated at a ∈ G, is
Xa = ax
d
dt
.
By solving corresponding ODE, one can find the integral curve of X starting at e = 1,
γXe (t) = e
tx. (γ̇Xe (t) = xe
tx d
dt
= XγXe (t))
So we get
exp(x) = φX1 (e) = γ
X
e (1) = e
x.
Example. Similarly one can show (exercise)
(1) for G = (S1, ·),
exp : iR = TeS1 → S1, exp(ix) = eix,
(2) for G = (Rn,+),
exp : Rn = T0Rn → Rn, exp(x) = x,
2There is a conception of exponential map in Riemannian geometry. It turns out that if G is a
compact Lie group endowed with the bi-invariant metric, then the Riemannian geometry exponential
map coincide with the Lie theory exponential map.
LECTURE 18: THE EXPONENTIAL MAP 5
(3) for G = GL(n,R),
exp : gl(n,R)→ GL(n,R), exp(A) = eA = I + A+ A
2
2!
+
A3
3!
+ · · · .
¶ The differential of the exponential map.
One of the most useful properties for the exponential map is
Lemma 2.3. The exponential map exp : g → G is a smooth map. Moreover if we
identify T0g with g, then
d exp0 = Id.
Proof. Consider the map
Φ̃ : R×G× g→ G× g, (t, g,X) 7→ (g · exp(tX), X).
One can check that this is the flow on G× g corresponding to the left invariant vector
field (X, 0) on G × g, thus it is smooth. It follows that exp is smooth, since it is the
composition
g ↪−→ R×G× g Φ̃−→ G× g π1−→ G,
X 7−→ (1, e,X) 7−→ (exp(tX), X) 7→ exp(tX).
Since exp(tX) = φXt (e) = γ
X
e (t), we get
d
dt
∣∣∣∣
t=0
exp(tX) = X.
On the other hand,
d
dt
∣∣∣∣
t=0
exp ◦tX = (d exp)0
d(Xt)
dt
= (d exp)0X.
We conclude that (d exp)0 equals to the identity map. �
Since (d exp)0 is bijective, we have
Corollary 2.4. exp is a local diffeomorphism near 0, i.e. it is a diffeomorphism from
a neighborhood of 0 ∈ TeG to a neighborhood of e ∈ G.
Remark. According to the examples above, we see that in general exp is not a global
diffeomorphism. However, it is still reasonable to ask:
Is the exponential map exp : g→ G surjective?
Of course for exp to be surjective, a necessary condition is that G should be connected.
It turns out that for any compact connected Lie group G, the exponential map is always
surjective. However, the exponential map need not be surjective for non-compact
connected Lie groups (e.g. SL(2,R)).
6 LECTURE 18: THE EXPONENTIAL MAP
¶ The Baker-Campbell-Hausdorff formula.
As another consequence, we can prove
Proposition 2.5. For any X, Y ∈ g, there is a smooth map Z : (−ε, ε) → g so that
for all t ∈ (−ε, ε),
exp(tX) exp(tY ) = exp(t(X + Y ) + t2Z(t)).
Proof. Since exp is a diffeomorphism near 0 ∈ g, there is an ε > 0 so that the map
ϕ : (−ε, ε)→ g, t 7→ ϕ(t) = exp−1(exp(tX) exp(tY ))
is smooth. Note that we can write ϕ as the composition
R γ
X
e ×γYe−→ G×G µ−→ G exp
−1
−→ R.
According to Lemma 1.3 in Lecture 17, dµe,e(X, Y ) = X + Y . It follows
ϕ′(0) = (d exp0)
−1(γ̇Xe (0) + γ̇
Y
e (0)) = X + Y.
Since ϕ(0) = 0, by Taylor’s theorem,
ϕ(t) = t(X + Y ) + t2Z(t)
for some smooth function Z. (Why? Can you write an explicit expression of Z involving
integrals and derivatives of ϕ to show the smoothness?) �
Remark. The mysterious function Z is explicitly given by the Baker-Campbell-Hausdorff
formula:
Z(t) =
1
2
[X, Y ] +
t
12
([X, [X, Y ]]− [Y, [X, Y ]]) + t
2
24
[X, [Y, [X, Y ]]] + · · ·
For a proof, c.f. my Lie group notes.
¶ The naturality of exp.
Next we relates the exponential maps with the Lie group/Lie algebra homomor-
phisms. It turns out that exp is natural in the following sense:
Proposition 2.6 (exp is Natural). Given any Lie group homomorphism ϕ : G → H,
the diagram
g
dϕ−−−→ hyexpG yexpH
G
ϕ−−−→ H
commutes, i.e. ϕ ◦ expG = expH ◦(dϕ).
Proof. Left as an exercise. �
As an application, we have
Corollary 2.7. If G is connected, any Lie group homomorphism ϕ : G → H is
determined by the induced Lie algebra homomorphism dϕ : g→ h.
Proof. Left as an exercise. �
LECTURE 18: THE EXPONENTIAL MAP 7
¶ [Reading Material] Define Lie bracket on TeG directly.
As we have seen, each element g ∈ G gives rise to an automorphism
c(g) : G→ G, x 7→ gxg−1.
Notice that c(g) maps e to e, its differential at e gives us a linear map
Adg = (dc(g))e : TeG→ TeG.
In other words, we get a map (the adjoint representation of the Lie group G)
Ad : G→ End(TeG), g 7→ Adg.
Note that Ad(e) is the identity map in End(TeG). Moreover, since End(TeG) is a linear
space, its tangent space at Id can be identified with End(TeG) itself in a natural way.
Taking derivative again at e, we get (the adjoint representation of the Lie algebra g)
ad : TeG→ End(TeG).
Applying the naturality of exp to the Lie group homomorphism Ad : G→ End(g) and
to the conjugation map c(g) : G→ G, we have
Proposition 2.8. (1) Ad(exp(tX)) = exp(tad(X)).
(2) g(exp tX)g−1 = exp(tAdgX).
Now we show that ad is nothing else but the Lie bracket operation:
Theorem 2.9. We have ad(X)(Y ) = [X, Y ].
Proof. Since Adg is the differential of c(g), we have
Ad(exp tX)Y =
d
ds
∣∣∣∣
s=0
c(exp tX) exp sY =
d
ds
∣∣∣∣
s=0
exp(tX) exp(sY ) exp(−tX).
So for any f ∈ C∞(G), according to Proposition 2.8,
(ad(X)Y )f = (
d
dt
∣∣∣∣
t=0
(Ad(exp tX)Y ))f
=
∂2
∂s∂t
∣∣∣∣
s=t=0
f(exp(tX) exp(sY ) exp(−tX))
=
∂2
∂s∂t
∣∣∣∣
s=t=0
f(exp(tX) exp(sY )) +
∂2
∂s∂t
∣∣∣∣
s=t=0
f(exp(sY ) exp(−tX))
= XY f(e)− Y Xf(e) = [X, Y ]e(f). �
LECTURE 19: LIE SUBGROUPS
1. Lie subgroups v.s. Lie subalgebras
¶ Lie subalgebras.
We have defined Lie groups and Lie algebras, and studied the correspondence
between them (in a primitive level). Now we would like to define sub-objects for Lie
groups and Lie algebras, and in particular we would like to set up a nice correspondence
between Lie subgroups and Lie subalgebras.
Of course a Lie subgroup should be both a submanifold and a subgroup, while a Lie
subalgebra should be both a vector subspace and a Lie algebra. There is one ambiguity
in defining Lie subgroups: should we require a Lie subgroup to be a smooth subman-
ifold (which, by our definition, is an embedded submanifold), or just an immersed
submanifold?
Since we want a nice correspondence between Lie subgroups of a given Lie group G
and Lie subalgebras of the corresponding Lie algebra g, let’s first study Lie subalgebras
(whose definition is clear), and then look at the corresponding Lie subgroups.
Definition 1.1. A Lie subalgebra h of a Lie algebra g is a vector subspaceso that
[X, Y ]g ∈ h holds for all X, Y ∈ h.
Example. Consider the Lie algebra g = gl(2,R). Inside g we consider
h1 =
{(
a 0
0 0
)
: a ∈ R
}
, h2 =
{(
a 0
0 b
)
: a, b ∈ R
}
, h3 =
{(
a b
0 0
)
: a, b ∈ R
}
It is easy to check that they are all Lie subalgebras of g = gl(2,R). Note that h1 is an
abelian Lie subalgebra of dimension 1, h2 is an abelian Lie subalgebra of dimension 2,
while h3 is a non-abelian Lie subalgebra of dimension 2. [Can you find Lie subgroups
of GL(2,R) whose Lie algebras are h1, h2 and h3? Note: They are not unique.] [Can
you construct a vector subspace of g = gl(2,R) that is not a Lie subalgebra?]
¶ Lie subgroups.
To get a clear idea on the Lie subgroup-Lie subalgebra correspondence, let’s explore
an even simpler example:
Example. Consider the abelian Lie group G = T2 whose Lie algebra is the 2-dimensional
Lie algebra g = R2, endowed with trivial Lie bracket [·, ·] ≡ 0. Then any vector
subspace of R2 is a Lie subalgebra of g. We take an arbitrary Lie subalgebra of g,
which is a one-dimensional vector subspace in g, (we omit the 0-dimensional case)
hα = the line passing the origin in R2 whose slope equals α.
To get the corresponding Lie subgroups of G whose Lie algebra is ha, we have to
distinguish lines with rational slopes with lines with irrational slopes: If α = p/q,
1
2 LECTURE 19: LIE SUBGROUPS
where p, q are co-prime integers, then the Lie subgroup of G whose Lie algebra equals
hα is
Hα = Hp,q := {(eipt, eiqt) | t ∈ R}.
These subgroups are all isomorphic to S1 and are all embedded submanifolds in T2.
However, if α is irrational, then the corresponding Lie subgroups are “dense curves”
(as we have seen in Lecture 8) of the form
Hα := {(eit, eiαt) | t ∈ R},
which are immersed submanifolds in T2 and are isomorphic to R. Since Hα = T2, they
are not embedded submanifolds.
In conclusion, Lie subgroups “should” be defined as immersed submanifolds:
Definition 1.2. A subgroup H of a Lie group G is called a Lie subgroup if it is an
immersed submanifold, and the group multiplication µH = µG|H×H is smooth.
So if H is a Lie subgroup of G, then the inclusion ιH : H ↪→ G is an injective Lie
group homomorphism:
ιH(µH(h1, h2)) = µG(ιH(h1), ιH(h2)).
Note that we don’t require the topology and the smooth structure of a Lie subgroup
H of G to be the ones inherited from G.
Remark. According the previous example, a Lie subgroup of a compact Lie group could
be non-compact.
¶ The Lie subalgebra of a Lie subgroup.
Now suppose ιH : H ↪→ G a Lie subgroup, and let h be the Lie algebra of H.
Then dιH : h → g is injective since by definition, dιH(X) is the left-invariant vector
field on G generated by the vector Xe ⊂ TeH ⊂ TeG. Last time we also showed that
dιH : h→ g Lie algebra homeomorphism, namely
dιH([X, Y ]
h) = [dιH(X), dιH(Y )]
g, ∀X, Y ∈ h.
So by identifying X ∈ h with dιH(X) ∈ g, we may regard h as a Lie subalgebra of g.
According to the naturality of the exponential map,
h
dιH−−−→ gyexpH yexpG
H
ιH−−−→ G
we conclude
Lemma 1.3. Let H be a Lie subgroup of G. Then the subgroup exponential map
expH : h→ H is the restriction of Lie exponential map expG : g→ G onto h.
By using the exponential map, we get the following nice characterization of the Lie
algebra of a Lie subgroup:
LECTURE 19: LIE SUBGROUPS 3
Theorem 1.4. Suppose H is a Lie subgroup of G. Then as a Lie subalgebra of g,
h = {X ∈ g | expG(tX) ∈ H for all t ∈ R}.
Proof. If X ∈ h, then for any t ∈ R,
expG(tX) = expH(tX) ∈ H.
Conversely we fix X 6∈ h and consider the map
ϕ : R× h→ G, (t, Y ) 7→ expG(tX) expG(Y ).
Using the facts d exp0 = Id and dµe,e(X, Y ) = X + Y , we get
dϕ0,0(τ, Ỹ ) = τX + Ỹ .
Since X 6∈ h, dϕ0,0 is injective. It follows that there exists a small ε > 0 and a
neighborhood U of 0 in h such that ϕ maps (−ε, ε)×U injectively into G. Shrinking U
if necessary, we may assume that expH maps U diffeomorphically onto a neighborhood
U of e in H. Choose a smaller neighborhood U0 of e in H such that U−10 U0 ⊂ U . (Such
an open neighborhood of e exists on any topological group, c.f. PSet 1-1-1(a)!) We
pick a countable collection {hj | j ∈ N} ⊂ H such that hjU0 cover H. (This is always
possible since any open covering of H admits a countable sub-covering.)
For each j denote Tj = {t ∈ R | expG(tX) ∈ hjU0}. We claim that Tj is a countable
set. In fact, if |t− s| < ε and t, s ∈ Tj, then
expG(t− s)X = expG(−sX) expG(tX) ∈ U .
So expG(t−s)X = expG(Y ) for a unique Y ∈ U . It follows ϕ(t−s, 0) = ϕ(0, Y ). Since
ϕ is injective on (−ε, ε)× U , we conclude Y = 0 and t = s.
Now each Tj is a countable set. So one can find t ∈ R such that t 6∈ Tj for all j. It
follows that expG(tX) 6∈ ∪jhjU0 = H. So
X 6∈ {X ∈ g | expG(tX) ∈ H for all t ∈ R}.
This completes the proof. �
Remark. However, X ∈ g and exp(X) ∈ H does not imply X ∈ h. Can you find an
example? (Consider T2.)
¶ Linear Lie groups/linear Lie algebras.
Definition 1.5. Any Lie subgroup of GL(n,R) is called a linear Lie group.
Remark. According to a deep theorem in Lie theory, any compact Lie group is isomor-
phic to a linear Lie group. On the other hand, there exist non-compact Lie groups
which are not linear Lie groups.
Since we are familiar with the general linear group GL(n,R), its Lie algebra (the
general linear Lie algebra) gl(n,R) , and the exponential map
exp : gl(n,R)→ GL(n,R), A 7→ eA = I + A+ A
2
2!
+ · · · ,
we are able to calculate the Lie algebras of linear Lie groups via Theorem 1.4.
4 LECTURE 19: LIE SUBGROUPS
Example. (The special linear group) The special linear group is defined as
SL(n,R) = {X ∈ GL(n,R) : detX = 1}.
It is a noncompact (n2 − 1)-dimensional Lie subgroup of GL(n,R). To determine its
Lie algebra sl(n,R), we notice that (Why? Use Jordan canonical form!)
det eA = eTr(A).
So for an n× n matrix A, eA ∈ SL(n,R) if and only if Tr(A) = 0. We conclude
sl(n,R) = {A ∈ gl(n,R) | Tr(A) = 0}.
Example. (The orthogonal group) Next let’s consider the orthogonal group
O(n) = {X ∈ GL(n,R) : XTX = In}.
This is a compact n(n+1)
2
dimensional Lie subgroup of GL(n,R). To figure out its Lie
algebra o(n), we note that (eA)T = eA
T
, so
(etA)T etA = In ⇐⇒ etA
T
= e−tA.
Since exp is locally bijective, we conclude that A ∈ o(n) if and only if AT = −A. So
o(n) = {A ∈ gl(n,R) | AT + A = 0},
which is the space of n× n skew-symmetric matrices.
Notice that O(n) is not connected. It consists of two connected components, and
the connected component of identity is the called the special orthogonal groups
SO(n) = {X ∈ GL(n,R) : XTX = In, detX = 1} = O(n) ∩ SL(n,R).
Its Lie algebra so(n) is the same as o(n).
¶ Lie subgroup/Lie subalgebra correspondence.
So from any Lie subgroup H of a Lie group G we get a Lie subalgebra h of g. Next
we will show that there is a one-to-one correspondence between Lie subalgebras of g
and connected Lie subgroups of G:
Theorem 1.6. Let G be a Lie group whose Lie algebra is g. If h is a Lie subalgebra
of g, then there is a unique connected Lie subgroup H of G with Lie algebra h.
Proof. (Idea: Our data is h, a set of vector fields. We want to construct a Lie subgroup,
which is a smooth manifold. How? Use the Frobenius theorem!)
Construct H: Let X1, · · · , Xk be a basis of h ⊂ g. Since X ′is are left invariant vector
fields on G, linearly independent at e, they are linearly independent at all g ∈ G. Let
Vg = span{X1(g), · · · , Xk(g)}.
Then
• V is a distribution of dimension k on G.
• The distribution is involutive, since [Xi, Xj] ∈ h for all i, j. (PSet 5-1-5)
LECTURE 19: LIE SUBGROUPS 5
According to the Frobenius theorem, there is a unique maximal connected integral
manifold of V through e. Denote this by H. It is an immersed submanifold of G.
Note that the distribution V is left invariant:
dLg1(Vg2) = Vg1g2 .
It follows that if N is an integral manifold of V , so is Lg(N) for any g.
(Idea: How to prove that an element is contained in the maximal connected integral
submanifold H? use: if H ′ is another integral manifold and H ′∩H 6= ∅, then H ′ ⊂ H.)
H is a subgroup:Take any h1, h2 ∈ H.
• Since h1 = Lh1e ∈ H ∩Lh1H 6= ∅, and since H is maximal, we have Lh1H ⊂ H.
So in particular h1h2 = Lh1h2 ∈ H.
• Since Lh−11 (h1) = e ∈ H∩Lh−11 H, and since H is maximal, we have Lh−11 H ⊂ H.
So in particular h−11 = Lh−11 e ∈ H.
It follows that H is a subgroup of G.
(Idea: The smoothness is an issue since the topology/smooth structure on H 6=
that of G. To overcome, use: locally any immersed submanifold is embedded.)
The smoothness of µH . First notice that the composition map
H ×H ↪→ G×G µG→ G
is smooth. Let h1, h2 ∈ H. Since H is an immersed submanifold of G, we can take
small open neighborhoods H1, H2, H3 of h1, h2 and h1h2 in H respectively, so that each
Hi, i = 1, 2, 3, is an embedded submanifold of G. Shrinking H1 and H2 if necessary, we
can assume H1 ·H2 ⊂ H3. (Can you write down the details here?) Since the smooth
structures on Hi’s are the ones inherited from G, we conclude that µH : H1×H2 → H3
is smooth. Since h1, h2 are arbitrary, µH is smooth.
(Idea: How to prove two connected Lie groups are the same? Use: each connected
Lie group is generated by any neighborhood of e.)
Uniqueness of H: Let K be another connected Lie subgroup of G with Lie algebra
h. Then K is also an integral manifold of V passing e. So we have K ⊂ H. Since
TeK = TeH, the inclusion has to be a local diffeomorphism near e. In other words,
there exist an open subset Ke of K near e and an open subset He of H near e so that
Ke = He. By connectedness of K and H, we get
K = ∪j≥1Kje = ∪j≥1Hje = H.
�
6 LECTURE 19: LIE SUBGROUPS
2. Closed Lie Subgroups (Reading Material)
¶ Closed subgroups.
We are most interested in those Lie subgroups H of G that are not just immersed
submanifolds, but in fact smooth submanifolds.
Proposition 2.1. Suppose G is a Lie group, H is a subgroup of G which is a subman-
ifold as well. Then H is a closed subset in G.
Proof. Since H is a submanifold of G, it is locally closed everywhere. In particular,
one can find an open neighborhood U of e in G such that U ∩ H = U ∩ H. Now
take any h ∈ H. Since hU is an open neighborhood of h in G, hU ∩ H 6= ∅. Let
h′ ∈ hU ∩H, then h−1h′ ∈ U . On the other hand, since h ∈ H, there is a sequence hn
in H converging to h. It follows that the sequence h−1n h
′ ∈ H converges to h−1h′. In
other words, h−1h′ ∈ U ∩H = U ∩H. So h ∈ H. This completes the proof. �
Definition 2.2. A subgroup H of a Lie group G is called a closed subgroup if it is a
closed subset in G. (Here we don’t require H to be a Lie subgroup!)
¶ Cartan’s closed subgroup theorem.
So by Proposition 2.1, if H is a Lie subgroup of G and is a smooth submanifold of
G, then H is a closed subgroup of G. In what follows we will prove the following very
powerful theorem, which claims that the converse is true:
Theorem 2.3 (E. Cartan’s closed subgroup theorem). Any closed subgroup H of a Lie
group G is a Lie subgroup (and thus a smooth submanifold) of G.
Remark. In this theorem we only assume that algebraically H is a subgroup, topo-
logically H is a closed subset, and we arrive at the strong conclusion that H is a Lie
subgroup (and in particular H itself is a Lie group). This is a very powerful tool to
prove various groups are Lie groups. For example, O(n), SL(n,R) are Lie groups, since
they are closed subgroups of GL(n,R).
As an immediate consequence of Cartan’s theorem, we get
Corollary 2.4. If ϕ : G → H is Lie group homomorphism, then ker(ϕ) is a Lie
subgroup of G.
Another very interesting consequence of Cartan’s theorem is
Corollary 2.5. Every continuous homomorphism between Lie groups is smooth.
Proof. Let φ : G→ H be a continuous homomorphism, then Γφ = {(g, φ(g)) | g ∈ G}
is a closed subgroup, and thus a Lie subgroup of G×H. It follows that the projection
p : Γφ
i
↪→ G×H π1−→ G
is bijective and smooth. Moreover, dp(eG,eH) is bijective. So p is local diffeomorphism
near (eG, eH). By left translation, p is a local diffeomorphism everywhere. Since p is
bijective, it has to be a global diffeomorphism. Thus φ = π2 ◦ p−1 is smooth. �
LECTURE 19: LIE SUBGROUPS 7
¶ Cartan’s closed subgroup theorem: The proof.
Now we proceed to prove Theorem 2.3. Suppose H is a closed subgroup of G. Let
h = {X ∈ g | exp(tX) ∈ H for all t ∈ R}.
We will need the following lemmas:
Lemma 2.6. h is a linear subspace of g.
Proof. Clearly h is closed under scalar multiplication. We have seen that there exists
smooth function Z : (−ε, ε)→ g so that
exp(tX) exp(tY ) = exp(t(X + Y ) + t2Z(t)).
It follows
exp(t(X + Y )) = lim
n→∞
(
exp(
tX
n
) exp(
tY
n
)
)n
.
So by closedness of H, h is also closed under addition. �
Lemma 2.7. Endow with g an inner product. Suppose X1, X2, · · · ∈ g satisfy
(1) Xi 6= 0 and Xi → 0 as i→∞.
(2) exp(Xi) ∈ H for all i.
(3) limi→∞
Xi
|Xi| = X ∈ g.
Then X ∈ h.
Proof. Fix any t 6= 0. Let ni =
[
t
|Xi|
]
be the integer part of t|Xi| . Then
|niXi − tX| ≤
∣∣∣∣[ t|Xi|
]
− t
|Xi|
∣∣∣∣ |Xi|+ t ∣∣∣∣ Xi|Xi| −X
∣∣∣∣→ 0.
So
exp(tX) = lim
i→∞
exp(niXi) = lim
i→∞
(expXi)
ni ∈ H.
�
Lemma 2.8. The exponential map exp : g → G maps a neighborhood of 0 in h bijec-
tively to a neighborhood of e in H.
Proof. Take a vector subspace h′ of g so that g = h⊕ h′. Define Φ : g = h⊕ h′ → G by
Φ(X + Y ) = exp(X) exp(Y ).
Then dΦ0(X̃ + Ỹ ) = X̃ + Ỹ . So near 0, Φ is a local diffeomorphism from g to G. Since
exp |h = Φ|h, to prove the lemma, it is enough to prove that Φ maps a neighborhood
of 0 in h bijectively to a neighborhood of e in H.
Suppose the lemma is false, then we can find Xi + Yi ∈ h⊕ h′ with Yi 6= 0 so that
Xi + Yi → 0 and Φ(Xi + Yi) ∈ H. Since exp(Xi) ∈ H, we must have exp(Yi) ∈ H for
all i. We let Y be a limit point of Yi|Yi|
′
s. Then by Lemma 2.7, Y ∈ h. Since Y ∈ h′, we
must have Y = 0, which is a contradiction since by construction, |Y | = 1. �
8 LECTURE 19: LIE SUBGROUPS
Proof of Cartan’s closed subgroup theorem. According to Lemma 2.8, one can find a
neighborhood U of e in G and a neighborhood V of 0 in g so that exp−1 : U → V is a
diffeomorphism, and so that exp−1(U ∩H) = V ∩ h. It follows that (exp−1, U, V ) is a
chart on G which makes H a submanifold near e. For any other point h ∈ H, we can
use left translation to get such a chart. This proves that H is a smooth submanifold
of G. The smoothness of µH is obvious. �
LECTURE 20: LIE GROUP ACTIONS
1. Smooth actions
¶ Smooth actions.
As we mentioned, Lie groups are “smooth families of symmetries”. In this lecture,
we will regard Lie groups as the symmetry groups acting on smooth manifolds, and
discuss basic facts from this point of view.
Let M be a smooth manifold. Then the “largest symmetry group” of M is Diff(M),
the group of diffeomorphisms on M , which is not a Lie group since it is “too large”:
its dimension is infinite. 1 Now let G be a Lie group. What do we mean when we say
G is a symmetry group of M? G must be a subgroup of Diff(M)! So the following
definition is natural:
Definition 1.1. Let G be a Lie group and M a smooth manifold.
(1) An action of a G on M is a homomorphism of groups τ : G → Diff(M). In
other words, for any g ∈ G, τ(g) is a diffeomorphism from M to M such that
τ(g1g2) = τ(g1) ◦ τ(g2).
(2) The action τ of G on M is smooth if the evaluation map
ev : G×M →M, (g,m) 7→ τ(g)(m)
is smooth. For simplicity we will denote τ(g)(m) by g ·m.
(3) A smooth manifold M endowed with a smooth G-action is called a G-manifold.
Remark. What we defined above is the left action. One can also define a right action
to be an anti -homomorphism τ̂ : G→ Diff(M), i.e. such that
τ̂(g1g2) = τ̂(g2) ◦ τ̂(g1).
Any left action τ can be converted to a right action τ̂ by letting τ̂(g)(m) = τ(g−1)(m).
¶ Examples of smooth actions.
Example. GL(n,R) (and thus any linear group) acts on Rn as linear transformations:
∀X ∈ GL(n,R) X : Rn → Rn, v 7→ Xv.
Can you write down a right action of GL(n,R) on Rn? This is why subgroups of
GL(n,R) are called linear Lie groups.
Note that this induces an action of O(n) on Sn−1.
1However, one can regard Diff(M) as an infinite dimensional Lie group.Can you guess what is the
Lie algebra associated to Diff(M)? It is Γ∞(TM), the space of smooth vector fields on M !
1
2 LECTURE 20: LIE GROUP ACTIONS
Example. We can define an S1-action on S2 by letting
∀θ ∈ S1 = R/2πZ rθ(x1, x2, x3) = (x1 cos θ − x2 sin θ, x1 sin θ + x2 cos θ, x3).
(This action can be viewed as the restriction of the linear action of GL(3,R) on R3
to a suitable subgroup (write down it!) that is diffeomorphic to S1. Since the action
preserves S2, it gives an smooth action on S2.) It is not hard to see that these rθ’s
are diffeomorphisms on S2. Geometrically they are “counterclockwise rotations” with
respect to the x3-axis.
Example. Any Lie group G acts on itself by many ways, e.g. by left multiplication, by
right multiplication and by conjugation. For example, the conjugation action of G on
G is given by
g ∈ G c(g) : G→ G, x 7→ gxg−1.
More generally, any Lie subgroup H of G can act on G by left multiplication, by right
multiplication and by conjugation.
Example. Any Lie group G acts on its Lie algebra g = TeG by the adjoint action:
g ∈ G Adg = (dc(g))e : g→ g.
For example, one can show that the adjoint action of GL(n,R) on gl(n,R) is given by
X ∈ GL(n,R) AdX : gl(n,R)→ gl(n,R), A 7→ XAX−1.
Example. Let X be a complete vector field on M . Then the flow
ρ : R→ Diff(M), t 7→ ρt = φXt
is a smooth action of R on M . (We have called the image of ρ a one-parameter subgroup
of Diff(M).)
Conversely, given any every smooth action of R on M , we can define a smooth
vector field X on M by letting X(m) = γ̇m(0), where γm(t) := ρt(m), so that the
R-action is exactly the flow of X.
¶ The induced vector field.
The last example above can be extended to more general smooth Lie group actions
easily: for any X ∈ g, there is a one-parameter subgroup exp(tX) in G.
Definition 1.2. Suppose Lie group G acts smoothly on M . For any X ∈ g, the induced
vector field XM on M associated to X is
XM(m) =
d
dt
∣∣∣∣
t=0
exp(tX) ·m.
Note that the integral curve of XM starting at m ∈ M is γm(t) = exp(tX) · m,
since by definition, γm(0) = m and
γ̇m(t) =
d
dt
(exp tX ·m) = d
ds
∣∣∣∣
s=0
(exp sX ◦ exp tX ·m) = XM(γm(t)).
In other words, the flow generated by XM is ρt = τ(exp(tX)).
LECTURE 20: LIE GROUP ACTIONS 3
Remark. So from any smooth Lie group action of G on M we get a map
dτ : g→ Γ∞(M), X 7→ XM .
This can be viewed as the differential of the map τ : G → Diff(M). One can prove
that dτ is a Lie algebra anti-homomorphism. It is called the infinitesimal action of g
on M .
2. Orbits and the quotient space
¶ Orbits and stabilizer.
Definition 2.1. Let τ : G→ Diff(M) be a smooth action.
(1) The orbit of G through m ∈M is
G ·m = {g ·m | g ∈ G} ⊂M.
(2) The stabilizer (also called the isotropic subgroup) of m ∈M is the subgroup
Gm = {g ∈ G | g ·m = m} < G.
Proposition 2.2. Let τ : G→ Diff(M) be a smooth action, m ∈M . Then
(1) The orbit G ·m is an immersed submanifold whose tangent space at m is
Tm(G ·m) = {XM(m) | X ∈ g}.
(2) The stabilizer Gm is a closed Lie subgroup of G whose Lie algebra is
gm = {X ∈ g | XM(m) = 0}.
Proof. (1) Let evm : G→M be the map evm(g) = g ·m. Then
evm ◦ Lg = τg ◦ evm.
Taking derivative at h ∈ G, we get
(devm)gh ◦ (dLg)h = (dτg)h·m ◦ (devm)h.
Since (dLg)h and (dτg)h·m are bijective, the rank of (devm)gh equals the rank of (devm)h
for any g and h. It follows that the map evm is of constant rank. By the constant rank
theorem in Lecture 6, its image, evm(G) = G ·m, is an immersed submanifold of M .
The tangent space of G ·m at m is the image under devm of TeG = g. But for any
X ∈ g and any f ∈ C∞(M), we have
(devm)e(X)f = Xe(f ◦ evm) =
d
dt
∣∣∣∣
t=0
f(exp(tX) ·m) = XM(m)(f).
So the conclusion follows.
(2) Gm is closed in G since Gm = ev
−1
m (m). It is a subgroup of G since τ is a group
homomorphism. By Cartan’s closed subgroup theorem, Gm is a closed Lie subgroup
of G. The Lie algebra of the Lie subgroup Gm is
gm = {X ∈ g | exp(tX) ∈ Gm,∀t ∈ R}.
4 LECTURE 20: LIE GROUP ACTIONS
It follows that exp(tX) ·m = m for X ∈ gm. Taking derivative at t = 0, we get
gm ⊂ {X ∈ g | XM(m) = 0}.
Conversely, if XM(m) = 0, then γ(t) ≡ m, t ∈ R, is an integral curve of the vector
field XM passing m. It follows that exp(tX) ·m = γ(t) = m, i.e. exp(tX) ∈ Gm for all
t ∈ R. So X ∈ gm. �
Definition 2.3. We say a Lie group action of G on M is
(1) transitive if there is only one orbit, i.e. M = G ·m.
(2) free if Gm = {e} for each m ∈M .
For example, for any subgroup H ⊂ G, the natural action of H on G by left/right
multiplications is free. We will see many examples of transitive actions below.
¶ The orbit space.
Suppose G acts on M smoothly. We will denote the set of orbits by M/G. For
example, if the action is transitive, then M/G contains only one element. We will
equip with M/G the quotient topology. This topology might be very bad in general,
e.g. non-Hausdorff.
Example. Consider the natural action of R>0 on R by multiplications, then there are
three orbits, {+, 0,−}. The open sets with respect to the quotient topology are
T = {{+}, {−}, {+.−}, {+, 0,−}, ∅}.
So the quotient is not Hausdorff.
One of the major theorems in the theory of group actions is
Theorem 2.4. Suppose a compact Lie group G acts on M smoothly, then
(1) (a) Each orbit G ·m is an embedded closed submanifold of M with
Tm(G ·m) = {XM(m) | X ∈ g}.
(b) The orbit space M/G is Hausdorff.
(2) If the action is free, then
(a) The orbit space M/G is a smooth manifold.
(b) The quotient map π : M →M/G is a submersion.
(3) If the action is transitive, then for each m ∈M , the map 2
F : G/Gm →M, gGm 7→ g ·m
is a diffeomorphism.
Remark. If G is not compact, the theorem still holds if we assume that the G-action
is proper. For more details of the definition of proper action as well as the proof of the
theorem, c.f. my Lie group course notes.
2Note: Gm acts on G freely. Here we use the right action.
LECTURE 20: LIE GROUP ACTIONS 5
¶ Homogeneous spaces.
In particular, we see that if the G-action on M is transitive, then for any m ∈M ,
M ' G/Gm.
Definition 2.5. A smooth manifold on which a Lie group acts transitively is called a
homogeneous space.
So by definition, homogeneous spaces are particularly nice smooth manifolds: they
have lots of nice symmetries, so that “geometrically (in contrast to the word “topolog-
ically”, which is true for all topological manifolds) all points are the same” in homoge-
neous spaces. Note that any homogeneous space is of the form G/H, where G is a Lie
group, and H is a closed Lie subgroup of G. Note that for a given homogeneous space
M , the Lie group G is not unique.
Example. The Euclidean group E(n) = RnnO(n) is the set Rn×O(n) equipped with
the multiplication
(b, A) · (b′, A′) = (b+ Ab′, AA′).
E(n) acts on Rn via
(b, A) · x = b+ Ax.
The action is obviously transitive. (This is known as rigid body motions in mechanics.)
So Rn is a homogeneous space. Moreover, the stabilizer of the origin is O(n). So
Rn ' E(n)/O(n).
Example. According to Gram-Schmidt, the natural action of O(n) on Sn−1 is transitive.
It follows that Sn−1 is a homogeneous space. Moreover, if we choose m to be the “north
pole” of Sn−1, then one can check that the stabilizer Gm is O(n− 1). It follows
Sn−1 ' O(n)/O(n− 1).
Example. The special linear group SL(2,R) acts on the upper half plane H = {z :
Im(z) > 0} by Möbius transformations,(
a b
c d
)
· z = az + b
cz + d
.
The action is transitive, and the stabilizer of i ∈ H is SO(2) (check). So
H ' SL(2,R)/SO(2).
Example. Let k < n. Consider
Grk(n) = {k dimensional linear subspaces of Rn}.
Then O(n) acts transitively on Grk(n), and the stabilizer of Rk × 0 ⊂ Rn is O(k) ×
O(n− k). It follows
Grk(n) ' O(n)/(O(k)×O(n− k))
The manifold Grk(n) is called a Grassmannian manifold. Note that Gr1(n) = RPn−1
is the real projective space.
LECTURE 21: TENSORS AND DIFFERENTIAL FORMS
1. Tensors as multi-linear maps
¶ Multi-linear maps.
Let V1, · · · , Vk be finite dimensional vector spaces.
Definition 1.1. A function T : V1×· · ·×Vk → R is calledmulti-linear if it is linear in each entry,
i.e. for each i and any fixed vectors v1 ∈ V1, · · · , vi−1 ∈ Vi−1, vi+1 ∈ Vi+1, · · · , vk ∈ Vk, the map
Ti : Vi → R, vi 7→ T (v1, · · · , vi, · · · , vk)
is linear.
Note that if T1, T2 are two multi-linear maps on V1 × · · · × Vk, so is their linear combinations.
Thus the set of all multi-linear maps on V1 × · · · × Vk is a vector space.
Example. For any f 1 ∈ V ∗1 , · · · , fk ∈ V ∗k (the dual spaces), we define
f 1 ⊗ · · · ⊗ fk : V1 × · · · × Vk → R, (v1, · · · , vk) 7→ f 1(v1) · · · fk(vk).
Obviously f 1 ⊗ · · · ⊗ fk is a multi-linear map. Note that by definition, for each 1 ≤ i ≤ k and
λ ∈ R, we have
f 1 ⊗ · · · ⊗ f i−1 ⊗ λf i ⊗ f i+1 ⊗ · · · ⊗ fk = λf 1 ⊗ · · · ⊗ f i−1 ⊗ f i ⊗ f i+1 ⊗ · · · ⊗ fk.
Not surprisingly, any multi-linear map is a linear combination of these special maps:
Theorem 1.2. Let {f 1i , · · · , f
n(i)
i } be a basis of V ∗i . Then the set of multi-linear maps
{f i11 ⊗ f i22 ⊗ · · · ⊗ f
ik
k | 1 ≤ ij ≤ n(j)}
form a basis of the vector space of multi-linear maps on V1×· · ·×Vk. In particular, dim⊗kV ∗ = nk.
Proof. We will denote by {ei1, · · · , ein(i)} the basis in V that is dual to the basis {f 1i , · · · , f
n(i)
i } of
V ∗i . For any multi-index I = (i1, · · · , ik), we will denote F I = f
i1
1 ⊗ f i22 ⊗ · · · ⊗ f
ik
k . Then the fact
F I(e1j1 , · · · , e
k
jk
) = δi1,··· ,ikj1,··· ,jk
implies that the multi-linear maps F I ’s are linearly independent.
Moreover, for any multi-linear map T on V1 × · · · × Vk, if we let TI = T (e1i1 , · · · , e
k
ik
), and
consider the multi-linear map
S = T −
∑
I
TIF
I ,
then S(e1j1 , · · · , e
k
jk
) = 0 for any multi-index J = (j1, · · · , jk). It follows from multi-linearity that
S ≡ 0. In other words, T =
∑
TIF
I is a linear combination of these F I ’s. �
1
2 LECTURE 21: TENSORS AND DIFFERENTIAL FORMS
Notation: We denote the vector space of multi-linear maps on V1 × · · · × Vk by V ∗1 ⊗ · · · ⊗ V ∗k .
Any element in this space is called a k-tensor.
Note that if T ∈ V ∗1 ⊗ · · · ⊗ V ∗k and S ∈ V ∗k+1 ⊗ · · · ⊗ V ∗k+l, then we can define the tensor
product T ⊗ S ∈ V ∗1 ⊗ · · · ⊗ V ∗k+l to be the tensor
(T ⊗ S)(v1, · · · , vk+l) = T (v1, · · · , vk)S(vk+1, · · · , vk+l).
By definition it is easy to see that ⊗ is a bilinear map and is associative:
(T ⊗ S)⊗R = T ⊗ (S ⊗R).
In fact, we may simply regard elements in V ∗i as 1-tensors on Vi. Then the k-tensor f
1 ⊗ · · · ⊗ fk
that we used above is the tensor product of the 1-tensors f 1, · · · , fk. [One may need to check the
consistency of the definition. What do I mean by consistency?]
Similarly we can define V1⊗· · ·⊗Vk as the vector space of multi-linear maps on V ∗1 ×· · ·×V ∗k .
By definition it has a basis {e1i1 ⊗ e
2
i2
⊗ · · · ⊗ ekik | 1 ≤ ij ≤ n(j)}.
Remark. For abstract vector spaces (which could be infinite dimensional), one can’t regard V as
(V ∗)∗ but one can still define the tensor products algebraically. More precisely, we can define V ⊗W
to be the quotient space V ⊗W = F (V ×W )/ ∼, where F (V ×W ) is the (infinite dimensional)
free vector space over V ×W , and ∼ is the equivalence relations generated by (v, w) ∼ (v, w),
(c1v1 + c2v2, w) ∼ c1(v1, w) + c2(v2, w) and (v, c1w1 + c2w2) ∼ c1(v, w1) + c2(v, w2). For any finite
dimensional vector spaces V and W , one has a natural linear isomorphism V ⊗W ∗ ' L(W,V ),
where L(W,V ) is the set of all linear maps from W to V . (Details will be left as an exercise.)
¶ Tensor powers of a vector space.
Now let V be an n-dimensional vector space, and V ∗ its dual space. We will call
⊗l,kV := (⊗lV )⊗ (⊗kV ∗)
the space of (l, k)-tensors on V . In other words, T ∈ ⊗l,kV if and only if
T = T (β1, · · · , βl, v1, · · · , vk)
is multi-linear with respect to each βi ∈ V ∗ and each vj ∈ V .
Remark. Note that ⊗1,0V = V and ⊗0,1V = V ∗. We will also abbreviate ⊗k,0V = ⊗kV . For the
case k = 0, we denote ⊗0V = R.
Next we will define a very useful operation on (l, k)-tensors.
Definition 1.3. For any 1 ≤ r ≤ l and 1 ≤ s ≤ k, we define the (r, s)-contraction to be the map
Crs : ⊗l,kV → ⊗l−1,k−1V given by
Crs (T )(β
1, · · · , βk−1, v1,· · ·, vl−1)=
∑
i
T (β1, · · · , βr−1, f i, βr, · · · , βk−1, v1,· · · , vs−1, ei, vs, · · · , vl−1)
where {e1, · · · , en} is a basis of V , and {f 1, · · · , fn} the dual basis.
LECTURE 21: TENSORS AND DIFFERENTIAL FORMS 3
One should check that this definition is independent of the choices of the basis {ei} of V .
Moreover, Crs (T ) is the (l − 1, k − 1)-tensor obtained from T by pairing the rth vector in T with
the sth co-vector in T :
Lemma 1.4. Let T be an (l, k)-tensor. For 1 ≤ r ≤ l, 1 ≤ s ≤ k, we have
(1) The definition of Crs is independent of the choices of the basis {ei} of V .
(2) For any v1, · · · , vl ∈ V and β1, · · · , βk ∈ V ∗,
Crs (v1 ⊗· · ·⊗ vl ⊗ β1 ⊗· · ·⊗ βk)=βs(vr)v1 ⊗· · ·⊗ v̂r ⊗· · ·⊗ vl ⊗ β1 ⊗· · ·⊗ β̂s ⊗· · ·⊗ βk,
where ·̂ means “remove the corresponding entry”.
Proof. Left as an exercise. �
Example. For example, if v, w ∈ V and α, β, γ ∈ V ∗, one has
C12(v ⊗ w ⊗ α⊗ β ⊗ γ) = β(v)w ⊗ α⊗ γ.
To see this, we compute by definition:
C12(v ⊗ w ⊗ α⊗ β ⊗ γ)(β1, v1, v2) =
∑
i
v ⊗ w ⊗ α⊗ β ⊗ γ(f i, β1, v1, ei, v2)
=
∑
i
f i(v)β1(w)α(v1)β(ei)γ(v2)
=
(∑
i
f i(v)β(ei)
)
β1(w)α(v1)γ(v2)
= β(v)β1(w)α(v1)γ(v2)
= β(v)w ⊗ α⊗ γ(β1, v1, v2).
2. Linear p-forms
¶ Symmetric and anti-symmetric tensors.
Now let’s fix a vector space V , and consider a k-tensor T on V , i.e. T ∈ ⊗kV ∗ = ⊗0,kV .
Definition 2.1. Let T ∈ ⊗kV ∗ be a k-tensor on V .
(1) We say T is symmetric if for any permutation σ of (1, 2, · · · , k),
T (v1, · · · , vk) = T (vσ(1), · · · , vσ(k)).
(2) We say T is alternating (or a linear k-form) if it is skew-symmetric, i.e.
T (v1, · · · , vi, · · · , vj, · · · , vk) = −T (v1, · · · , vj, · · · , vi, · · · , vk).
for all v1, · · · , vk ∈ V and any 1 ≤ i 6= j ≤ k
Example. • An inner product on V is a positive symmetric 2-tensor.
• det is a linear n-form on Rn.
4 LECTURE 21: TENSORS AND DIFFERENTIAL FORMS
We will denote the vector space of k-forms by ΛkV ∗. Note that ΛkV ∗ is not a brand new
space: it is a linear subspace of ⊗kV ∗. We will set Λ1V ∗ = ⊗1V ∗ = V ∗ and Λ0V ∗ = R.
Recall that a permutation σ ∈ Sk is called even or odd, depending on whether it is expressible
as a product of an even or odd number of simple transpositions. For any k-tensor T and any
σ ∈ Sk, we define another k-tensor T σ by
T σ(v1, · · · , vk) = T (vσ(1), · · · , vσ(k)).
Clearly
• For all k-tensor T , (T σ)π = T π◦σ for all σ, π ∈ Sk.
• A k-tensor T is symmetric if and only if T σ = T for all σ ∈ Sk.
• A k-tensor T is a k-form if and only if T σ = (−1)σT for all σ ∈ Sk, where (−1)σ = 1 if σ
is even, and (−1)σ = −1 if σ is odd.
¶ Anti-symmetrization.
For any k-tensor T on V , we consider the anti-symmetrization map
Alt(T ) =
1
k!
∑
π∈Sk
(−1)πT π.
Lemma 2.2. The map Alt is a projection from ⊗kV ∗ to ΛkV ∗, i.e. it is a linear map satisfying
(1) For any T ∈ ⊗kV ∗, Alt(T ) ∈ ΛkV ∗.
(2) For any T ∈ ΛkV ∗, Alt(T ) = T .
Proof. (1) For any T ∈ ⊗kV ∗ and any σ ∈ Sk,
[Alt(T )]σ =
1
k!
∑
π∈Sk
(−1)π(T π)σ = 1
k!
(−1)σ
∑
π∈Sk
(−1)σ◦πT σ◦π = (−1)σAlt(T ).
(2) If T ∈ ΛkV ∗, then each summand (−1)πT π equals T . So Alt(T ) = T since |Sk| = k!. �
We will need
Lemma 2.3. Let T, S,R be k-, l-, and m-forms respectively. Then
(1) Alt(T ⊗ S) = (−1)klAlt(S ⊗ T ).
(2) Alt(Alt(T ⊗ S)⊗R) = Alt(T ⊗ S ⊗R) = Alt(T ⊗ Alt(S ⊗R)).
Proof. Exercise. �
¶ The wedge product.
Now we can define a “product operation” for linear forms:
Definition 2.4. The wedge product of T ∈ ΛkV ∗ and S ∈ ΛlV ∗ is the (k + l)-form
T ∧ S = (k + l)!
k!l!
Alt(T ⊗ S).
The wedge product operation satisfies
LECTURE 21: TENSORS AND DIFFERENTIAL FORMS 5
Proposition 2.5. The wedge product operation ∧ : (ΛkV ∗)× (ΛlV ∗)→ Λk+lV ∗ is
(1) Bi-linear: (T, S) 7→ T ∧ S is linear in T and in S.
(2) Anti-commutative: T ∧ S = (−1)klS ∧ T .
(3) Associative: (T ∧ S) ∧R = T ∧ (S ∧R).
Proof. (1) follows from Definition 2.4. (2) follows from Lemma 2.3(1). (3)follows from Definition
2.4 and Lemma 2.3(2). �
So it makes sense to talk about wedge products of three or more linear forms. For example,
if T ∈ ΛkV ∗, S ∈ ΛlV ∗ and R ∈ ΛmV ∗, then we have
T ∧ S ∧R = (k + l +m)!
k!l!m!
Alt(T ⊗ S ⊗R).
One can easily extend this to wedge products of more than three linear forms. In particular, by
definition we have: if f 1, · · · , fk ∈ V ∗, then
f 1 ∧ · · · ∧ fk = k!Alt(f 1 ⊗ · · · ⊗ fk).
As a consequence,
Proposition 2.6. For any f 1, · · · , fk ∈ V ∗ and v1, · · · , vk ∈ V ,
(f 1 ∧ · · · ∧ fk)(v1, · · · , vk) = det(f i(vj)).
Proof. We have
(f 1 ∧ · · · ∧ fk)(v1, · · · , vk) = k! Alt(f 1 ⊗ · · · ⊗ fk)(v1, · · · , vk)
=
∑
σ∈Sk
(−1)σf 1(vσ(1)) · · · fk(vσ(k))
= det((f i(vj))).
�
¶ The vector space of linear k-forms.
Now we are ready to prove
Theorem 2.7. Let {f 1, · · · , fn} be a basis of V ∗. Then the set of k-forms
{f i1 ∧ f i2 ∧ · · · ∧ f ik | 1 ≤ i1 < i2 < · · · < ik ≤ n}
form a basis of ΛkV ∗. In particular, dim ΛkV ∗ =
(
n
k
)
.
Proof. Again we denote by {e1, · · · , en} the dual basis in V . For any multi-index I = (i1, · · · , ik)
with i1 < · · · < ik, we let ΩI = f i1 ∧ · · · ∧ f ik . Then for any multi-index J = (j1, · · · , jk) with
j1 < · · · < jk,
ΩI(ej1 , · · · , ejk) = det((f ir(ejs))1≤r,s≤k) = δ
i1,··· ,ik
j1,··· ,jk .
It follows that these ΩI ’s are linearly independent.
Moreover, since any T ∈ ΛkV ∗ is a k-tensor, we can write T =
∑
I TIF
I , where I = (i1, · · · , ik)
runs over all 1 ≤ i1, · · · , ik ≤ n, and F I is as in the proof of Theorem 1.2. Note that ΩI =
6 LECTURE 21: TENSORS AND DIFFERENTIAL FORMS
k!Alt(F I). Here, the indices I need not be increasing, but we have: ΩI = 0 if two indices in
I equal, and ΩI = ±ΩI′ if I contains no equal indices, where I ′ is the re-arrangement of I in
increasing order. So
T = Alt(T ) =
∑
all I
TIAlt(F
I) =
1
k!
∑
I increasing
( ∑
I′=I as sets
(±TI′)
)
ΩI
is a linear combination of ΩI with I’s being only increasing indices. �
Remark. As an immediate consequence, we see
• dim Λn(V ∗) = 1.
– So any n-form on an n-dimensional vector space V is a multiple of the non-trivial
n-form “det”.
• Moreover, for k > n, Λk(V ∗) = 0.
¶ The interior product and the pull-back.
Finally we define two more operators on linear k-forms.
Definition 2.8. The interior product of a vector v ∈ V with a linear k-form α ∈ Λk(V ∗) is the
(k − 1)-covector
ιvα(X1, · · · , Xk−1) := α(v,X1, · · · , Xk−1).
Definition 2.9. Let L : W → V be linear. The pullback L∗ : Λk(V ∗)→ Λk(W ∗) is defined to be
(L∗α)(X1, · · · , Xk) := α(L(X1), · · · , L(Xk))
The following proposition will be important in the rest of this semester. The proof is left as
an exercise.
Proposition 2.10. Let α be a linear k-form on V and β a linear l-form on V . Then
(1) For any v ∈ V , ιvιvα = 0.
(2) For any v ∈ V , ιv(α ∧ β) = (ιvα) ∧ β + (−1)kα ∧ ιvβ.
(3) For any linear map L : W → V , L∗(α ∧ β) = L∗α ∧ L∗β.
Proof. Left as an exercise. �
3. Reading: Tensors fields and differential forms on smooth manifolds
¶ Cotangent spaces.
Let M be a smooth manifold. We have associated to each p ∈ M a vector space TpM . If we
take any local chart (ϕ,U, V ) around p, then we can write down an explicit basis for TpM :
∂i|p : C∞(U)→ R, ∂i|p(f) =
∂(f ◦ ϕ−1)
∂xi
(ϕ(p)), (1 ≤ i ≤ n).
Note that not only the ∂i|p’s form a basis for the tangent space TpM , but in fact ∂i’s are smooth
vector fields on U , and for any q ∈ U , the ∂i|q’s form a basis for the tangent space TqM .
LECTURE 21: TENSORS AND DIFFERENTIAL FORMS 7
Now let’s study the dual space T ∗pM of TpM . We introduced T
∗
pM in PSet 3-1-6. It is called
the cotangent space of M at p, and elements in T ∗pM are called cotangent vectors at p. It is also
quit easy to write down an explicit basis of T ∗pM , (and in fact a basis of T
∗
qM , for each q ∈ U ,
varying smoothly in q), in any given local chart (ϕ,U, V ): We first note that for each 1 ≤ i ≤ n,
xi ◦ ϕ : U → R
is a smooth function on U . The differential of this function, which we will denote by dxi for
simplicity, is a linear map (when restricted to any q ∈ U)
dxi|q : TqM = TqU → Txi◦ϕ(q)R = R.
In other words, each dxi|q is an element in T ∗qM . More over, by definition,
dxi|q(∂j|q) = ∂j|q(xi ◦ ϕ) = δij.
So we conclude
Proposition 3.1. In any local chart (ϕ,U, V ), {dxi|q : 1 ≤ i ≤ n} is a basis of T ∗qM . Moreover,
this basis is the dual basis to the basis {∂i|q : 1 ≤ i ≤ n} of TqM .
In fact, for any f ∈ C∞(U), by the same way we get a linear map dfq : TqM → R. In other
words, we get a cotangent vector dfq ∈ T ∗qM . By definition, dfp(∂i|p) = ∂i|p(f). It follows
dfp = (∂1|pf)dx1|p + · · ·+ (∂n|pf)dxn|p,
and moreover, for any X ∈ Γ∞(TU),
df(X) = Xf,
where both sides are regarded as functions on U . We will call df a 1-form on U .
¶ Tensor fields on smooth manifolds.
Now we are ready to define (Compare: the definition of vector fields on manifolds)
Definition 3.2. An (l, k)-tensor field T on M is an assignment that assigns to each point p ∈M
an (l, k)-tensor Tp ∈ ⊗l,kTpM .
Remark. By definition, T is a tensor at if and only if it is point-wise linear in each entry. It follows
that T is tensor field on M if and only if it is “function-linear” in each entry. So it is more than a
multi-linear map, i.e., we also have (where ω’s are 1-forms on U , and X’s are vector fields on U)
T (f1ω
1, · · · , flωl, g1X1, , · · · , gkXk) = f1 · · · flg1 · · · gkT (ω1, · · · , ωl, X1, · · · , Xk).
If we fix any local chart (ϕ,U, V ) near p, then we can write
Tp =
∑
T i1···ilj1···jk ∂i1|p ⊗ · · · ⊗ ∂il |p ⊗ dx
j1|p ⊗ · · · ⊗ dxjk |p,
where T i1···ilj1···jk ’s are constants (which depends on p). In other words, in any coordinate chart U one
can write
T =
∑
T i1···ilj1···jk∂i1 ⊗ · · · ⊗ ∂il ⊗ dx
j1 ⊗ · · · ⊗ dxjk ,
where T i1···ilj1···jk ’s are functions on U .
8 LECTURE 21: TENSORS AND DIFFERENTIAL FORMS
Definition 3.3. We say a (l, k)-tensor field T on M is smooth if in any coordinate chart U , the
functions T i1···ilj1···jk ’s are smooth.
Note that when (l, k) = (1, 0), we will get a smooth vector field on M . The set of all smooth
(l, k)-tensors is denoted by Γ∞(⊗l,kTM). Again this is an infinite dimensional vector space.
Remark. The coefficient functions T i1···ilj1···jk ’s are only defined in local charts. If one uses another
chart U ′, one gets another set of coefficient functions (even if at the same point).
Example. A symmetric positive smooth (0, 2)-tensor field g on M is called a Riemannian metric
on M . Locally a Riemannian metric is of the form
g =
∑
gij(x)dx
i ⊗ dxj,
where (gij(x)) is a positive definite symmetric matrix depending smoothly on x. We have seen
the existence of Riemannian metric on any smooth manifold in PSet 3-2-5.
¶ Differential forms on smooth manifolds.
Similarly one can define smooth k-forms on a smooth manifold M :
Definition 3.4. A k-form ω on a smooth manifold M is an assignment that assigns to each point
p ∈M a linear k-form ωp ∈ ΛkT ∗pM . A k-form ω is smooth if locally one can write
ω =
∑
I
ωIdx
I =
∑
I
ωi1,··· ,ikdx
i1 ∧ · · · ∧ dxik ,
where the summation is over increasing k-tuples I={1≤ i1< · · ·< ik≤n}, and each ωI ∈C∞(U).
Since k-forms will be frequently used in the rest of this course, we will denote the set of all
smooth k-forms by Ωk(M) (instead of the lengthy expression Γ∞(ΛkT ∗M)). Note that any smooth
function on M can be viewed as a smooth 0-form. So
Ω0(M) = C∞(M).
Since there is no linear k-form on TpM for k > n = dimM , we get
Ωk(M) = 0, ∀k > n.
Note that if ω ∈ Ωk(M), and X1, · · · , Xk ∈ Γ∞(TM), then ω(X1, · · · , Xk) ∈ C∞(M).
¶ Operations on differential forms on smooth manifolds.
Of course the pointwise operations for linear k-forms that we learned last time still make sense
for differential forms on manifolds. So on differential forms we have the following operations:
• The wedge product ∧ : Ωk(U)× Ωl(U)→ Ωk+l(U).
– For example,
(dx1 + 2dx2) ∧ (dx1 ∧ dx2 − dx2 ∧ dx3 + 3dx1 ∧ dx3) = −7dx1 ∧ dx2 ∧ dx3.
• For any X ∈ Γ∞(TU), one has the interior productιX : Ωk(U)→ Ωk−1(U).
LECTURE 21: TENSORS AND DIFFERENTIAL FORMS 9
– For example,
ιX(dx
i1 ∧ · · · ∧ dxik) =
∑
r
(−1)r−1dxir(X)dxi1 ∧ · · · ∧ d̂xir ∧ · · · ∧ dxik .
• For any smooth map ϕ : U ′ → U , one has the pull-back ϕ∗ : Ωk(U)→ Ωk(U ′).
– This is defined pointwise via the linear map dϕp : TpU
′ → Tϕ(p)U . So if ω ∈ Ωk(U),
then
(ϕ∗ω)p(X1, · · · , Xk) = ωϕ(p)(dϕp(X1), · · · , dϕp(Xk)).
Note: if k = 0, then ϕ∗ is exactly the pull-back ϕ∗ : C∞(U)→ C∞(U ′) on functions.
These operations are all linear (where ∧ is bilinear). Here we list some important properties of
these operations.
Proposition 3.5. Suppose ω ∈ Ωk(U), η ∈ Ωl(U), X ∈ Γ∞(TU) and ϕ ∈ C∞(U ′, U) and ψ ∈
C∞(U, Ũ). Then
(1) ω ∧ η = (−1)klη ∧ ω.
(2) ϕ∗(ω ∧ η) = ϕ∗ω ∧ ϕ∗η.
(3) ιX(ω ∧ η) = (ιXω) ∧ η + (−1)kω ∧ ιXη.
(4) ιX ◦ ιX = 0.
(5) (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗.
Proof. All follows from definitions and corresponding results for linear differential forms. �
LECTURE 22: THE EXTERIOR DERIVATIVE
1. The exterior derivative
¶ The exterior derivative: A local definition.
Now we define the exterior derivative for differential forms. It generalizes the
conception of the differential on functions, and it is the most important operation for
the rest of this semester. Unlike the wedge product, the interior product and the
pull-back operations that we defined last time, the exterior derivative is no longer a
pointwise operation, but is a local operation (i.e. depends on the “nearby values”)
We start with f ∈ Ω0(U) = C∞(U). In this case we have already seen that
df ∈ Ω1(U). So we get a linear map
d : Ω0(U)→ Ω1(U), f 7→ df.
Locally on each coordinate chart we have
df =
∑
i
(∂if)dx
i.
We also have an “invariant definition” of df ∈ Ω1(U), via
df(X) = Xf, ∀X ∈ Γ∞(TU).
Now suppose ω is a k-form on M , so that locally
ω =
∑
I
ωi1,··· ,ikdx
i1 ∧ · · · ∧ dxik .
We want to define dω as a (k + 1)-form. It is natural to define
Definition 1.1. The exterior derivative of ω is the (k+1)-form dω given by the formula
(1) dω =
∑
I
dωi1,··· ,ik ∧ dxi1 ∧ · · · ∧ dxik =
∑
I,i
∂i(ωi1,··· ,ik)dx
i ∧ dxi1 ∧ · · · ∧ dxik .
¶ The exterior derivative: A coordinate-free definition.
Before we proceed, we need to clarify that dω defined above is well-defined. In
other words, the (k + 1)-form dω defined above should be independent of the choices
of coordinate patches.
Usually one has two ways to prove the well-definedness of a conception on mani-
folds. One is to check that the definition is unchanged if one use another coordinate
chart (for example, the definition of smoothness of a function), the other is to give
an equivalent but coordinate-free definition (usually called the invariant formulation).
We will take the second approach here since the coordinate-free expression of dω is also
very useful.
Let’s start with small k’s to find out the invariant formula of dω.
1
2 LECTURE 22: THE EXTERIOR DERIVATIVE
• For k = 0, i.e. ω = f ∈ C∞(U), we can regard df as a C∞(U)-linear map
df : Γ∞(TU)→ C∞(U)
such that
df(X) = Xf.
• For k = 1, i.e. ω ∈ Ω1(U), we want to regard dω as a C∞(U)-bilinear map
dω : Γ∞(TU)× Γ∞(TU)→ C∞(U).
We write ω =
∑
i ωidx
i, X =
∑
kX
k∂k and Y =
∑
l Y
l∂l. Then
dω(X, Y ) =
∑
i,j,k,l
(∂jωi)dx
j ∧ dxi(Xk∂k, Y l∂l)
=
∑
i,j
(
(∂jωi)X
jY i − (∂jωi)X iY j
)
=
∑
i,j
(
Xj∂j(ωiY
i)− ωiXj∂j(Y i)− Y j∂j(ωiX i) + ωiY j∂j(X i)
)
= X(ω(Y ))− Y (ω(X))− ω([X, Y ]).
So we arrive at
dω(X, Y ) = X(ω(Y ))− Y (ω(X))− ω([X, Y ]).
• For k = 2, i.e. ω ∈ Ω2(U), by tedious computations one gets: as a C∞(U)-
trilinear map
dω : Γ∞(TU)× Γ∞(TU)× Γ∞(TU)→ C∞(U),
one has
dω(X,Y,Z)=X(ω(Y,Z))−Y (ω(X,Z))+Z(ω(X,Y ))−ω([X,Y ],Z)+ω([X,Z],Y )−ω([Y,Z],X).
So we are naturally led to the following the invariant formula for dω:
Theorem 1.2. For any ω ∈ Ωk(U), the (k+1)-form dω, viewed as a C∞(U)-multilinear
map
dω : Γ∞(TU)× · · · × Γ∞(TU)→ C∞(U),
is given by the formula
(2)
dω(X1, · · · , Xk+1) :=
∑
i
(−1)i−1Xi(ω(X1, · · · , X̂i, · · · , Xk+1))
+
∑
i<j
(−1)i+jω([Xi, Xj], X1, · · · , X̂i, · · · , X̂j, · · · , Xk+1).
Sketch of proof. Define (k + 1)-form d̃ω via the formula (2). We need to show that
(1) d̃ω is anti-symmetric, i.e. for any r < s, a simple but messy computation yields
d̃ω(X1, · · · , Xr, · · · , Xs, · · · , Xk+1) = −d̃ω(X1, · · · , Xs, · · · , Xr, · · · , Xk+1).
LECTURE 22: THE EXTERIOR DERIVATIVE 3
(2) d̃ω is multi-linear at each point, i.e. d̃ω is C∞(U)-linear on U . Note that d̃ω is
obviously R-linear. So in view of (1), it is enough to prove for any f ∈ C∞(U),
d̃ω(fX1, X2, · · · , Xk+1) = fd̃ω(X1, · · · , Xk+1).
This can be checked by a direct computation:
d̃ω(fX1, X2, · · · , Xk+1) =fX1(ω(X2, · · · , Xk+1))
+
∑
i>1
(−1)i−1Xi(ω(fX1, · · · , X̂i, · · · , Xk+1))
+
∑
i>1
(−1)i+1ω([fX1, Xi], X2, · · · , X̂i, · · · , Xk+1)
+
∑
1<i<j
(−1)i+jω([Xi, Xj], fX1,· · ·, X̂i,· · ·, X̂j,· · ·, Xk+1)
=fd̃ω(X1, · · · , Xk+1)
+
∑
i>1
(−1)i−1(Xif)ω(X1, · · · , X̂i, · · · , Xk+1)
−
∑
i>1
(−1)i+1(Xif)ω(X1, · · · , X̂i, · · · , Xk+1)
=fd̃ω(X1, · · · , Xk+1).
(3) It remains to check that d̃ω has the local expression (1) as desired. Obviously
the map
d : Ωk(U)→ Ωk+1(U)
is linear. So without loss of generality, we may assume
ω = fdx1 ∧ · · · ∧ dxk
in a local chart U . Note that [∂i, ∂j] = 0. It follows that for any increasing
indices j1 < · · · < jk+1, the right hand side of
d̃ω(∂j1 , · · · , ∂jk+1) =
∑
i
(−1)i−1∂ji(ω(∂j1 , · · · , ∂̂ji , · · · , ∂jk+1))
vanishes except for the case j1 = 1, · · · , jk = k and i = k + 1 (and thus
ji ≥ k + 1). In other words, the only non-zero terms in all possible expressions
d̃ω(∂j1 , · · · , ∂jk+1) are
d̃ω(∂1, · · · , ∂k, ∂r) = (−1)k∂r(f).
It follows that
d̃ω =
∑
r>k
(−1)k∂r(f)dx1 ∧ · · · ∧ dxk ∧ dxr =
∑
∂r(f)dx
r ∧ dx1 ∧ · · · ∧ dxk,
which is exactly the local expression in Definition 1.1. �
4 LECTURE 22: THE EXTERIOR DERIVATIVE
¶ Properties of the exterior derivative.
The following properties will be used frequently later in this course:
Proposition 1.3. Suppose ω ∈ Ωk(U), η ∈ Ωl(U), X ∈ Γ∞(TU) and ϕ ∈ C∞(U ′, U).
Then
(1) d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη.
(2) d ◦ d = 0.
(3) ϕ∗ ◦ d = d ◦ ϕ∗.
Proof. (1): Since d is linear, it is enough to assume ω = fdxi1 ∧ · · · ∧ dxik and η =
gdxj1 ∧ · · · ∧ dxjl , with indices set I ∩ J = ∅. Then the formula follows from a direct
computation:
d(ω ∧ η) =d(fgdxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxjl)
=
∑
i
∂i(fg)dx
i ∧ dxi1 ∧ · · · ∧ dxik ∧ dxj1 ∧ · · · ∧ dxjl
=
∑
i
(∂if)dx
i ∧ dxi1 ∧ · · · ∧ dxik ∧ η + (−1)kω ∧
∑
i
(∂ig)dx
i ∧ dxj1 ∧ · · · ∧ dxjl
−dω ∧ η + (−1)kω ∧ dη.
(2): We first check this for k = 0:
d(df)(X, Y ) = X(df(Y ))−Y (df(X))−df([X, Y ]) = X(Y (f))−Y (X(f))−[X, Y ]f = 0.
For k > 0, by linearity we may assume ω = fdx1 ∧ · · · ∧ dxk. Since ddf = 0 and
ddxi = 0, we get
d(dω) = d(df ∧ dx1 · · · ∧ dxk)
= d(df) ∧ dx1 ∧ · · · ∧ dxk +
∑
i
(−1)idf ∧ dx1 ∧ · · · ∧ d(dxi) ∧ · · · ∧ dxk
= 0.
(3): Again we first check this for k = 0:
(ϕ∗df)p(Xp) = dfϕ(p)(dϕp(Xp)) = d(ϕ
∗f)p(Xp).
In general, assume ω = fdx1 ∧ · · · ∧ dxk. Then by (1), (2) and Proposition 3.5(2) in
Lecture 21,
ϕ∗dω = ϕ∗(df ∧ dx1 ∧ · · · ∧ dxk)
= ϕ∗(df) ∧ ϕ∗(dx1) ∧ · · · ∧ ϕ∗(dxk)
= d(ϕ∗f) ∧ d(ϕ∗x1) ∧ · · · ∧ d(ϕ∗xk)
= d(ϕ∗fd(ϕ∗x1) ∧ · · · ∧ d(ϕ∗xk))
= d(ϕ∗ω). �
LECTURE 22: THE EXTERIOR DERIVATIVE 5
2. Reading Materials: The Lie Derivatives (continued)
¶ The Lie derivative of differential forms along a vector field.
Recall that in Lecture 15, we defined the Lie derivative of functions:
The Lie derivative of a f ∈ C∞(M) with respect to X ∈ Γ∞(TM) is
LX(f) :=
d
dt
∣∣∣∣
t=0
φ∗tf
(
= lim
t→0
φ∗tf − f
t
)
,
where φt is the (local) flow generated by the vector field X.
Since we have defined pull-back φ∗ on differential forms, one can easily define the Lie
derivative of differential forms:
Definition 2.1. The Lie derivative of a ω ∈ Ωk(M) with respect to X ∈ Γ∞(TM) is
LX(ω) :=
d
dt
∣∣∣∣
t=0
φ∗tω
(
= lim
t→0
φ∗tω − ω
t
)
,
where φt is the (local) flow generated by the vector field X.
Note that LXf that we defined in Lecture 5 is just aspecial case of this definition,
since C∞(M) = Ω0(M).
We have seen in PSet 5-1-4 that
LXf = Xf and L[X,Y ]f = LxLY (f)− LYLX(f).
For the Lie derivative defined on differential forms, we have
Proposition 2.2. Suppose ω ∈ Ωk(M), η ∈ Ωl(M) and X,X1, X2 ∈ Γ∞(TM). Then
(1) dLXω = LXdω.
(2) LX(ω ∧ η) = LXω ∧ η + ω ∧ LXη.
(3) [Cartan’s Magic Formula] LXω = dιXω + ιXdω.
(4) L[X1,X2]ω = LX1LX2ω − LX2LX1ω.
(5) (LXω)(X1, · · · , Xk) = LX(ω(X1, · · · , Xk))−
∑
i ω(X1, · · · ,LXXi, · · · , Xk).
Sketch of proof (Details left as an exercise.)
(1) follows from ϕ∗d = dϕ∗, and (2) follows from ϕ∗(ω ∧ η) = ϕ∗ω ∧ ϕ∗η.
To prove (3), one proceed by induction on k. First it is not hard to prove Cartan’s
magic formula for k = 0 (where we set ιXf = 0). For a general k-form ω, by linearity,
we may assume that locally ω = fdx1∧· · ·∧dxk = dx1∧ω1, where ω1 = fdx2∧· · ·∧dxk.
Then it is not hard to check Cartan’s magic formula for ω by induction. (Do we need
to prove Cartan’s magic formula for k = 1 for the induction process to work?)
(4) also follows from a similar induction argument.
To prove (5), one can first check it for simple 1-forms like ω = dx1. Then use (2)
and Proposition 2.6 in Lecture 21. �
6 LECTURE 22: THE EXTERIOR DERIVATIVE
¶ The Lie derivative of differential forms along a vector field.
In Lecture 15 we also defined the Lie derivatives of vector fields:
The Lie derivative of Y ∈ Γ∞(TM) with respect to X ∈ Γ∞(TM) is
LX(Y ) :=
d
dt
∣∣∣∣
t=0
(φ−t)∗Y
(
= lim
t→0
(φ−t)∗Yφt(p) − Yp
t
)
.
where the push-forward ϕ∗X on N is defined by
(ϕ∗X)ϕ(p) = dϕp(Xp), ∀p ∈M.
and we have seen in PSet 5-1-4 that
LXY = [X, Y ] and L[X1,X2]Y = LX1LX2(Y )− LX2LX1(Y ).
It is then straightforward to extend the Lie derivatives to tensor fields by using
LX(T ⊗ S) := (LXT )⊗ S + T ⊗ (LXS).
One can show that for a general (l, k)-tensor T , the Lie derivative LXT is the (l, k)-
tensor defined via
(LXT )(ω1, · · · , ωl, X1, · · · , Xk) = X(T (ω1, · · · , ωl, X1, · · · , Xk))
−
∑
i
T (ω1, · · · ,LXωi, · · · , ωl, X1, · · · , Xk)
−
∑
j
T (ω1, · · · , ωl, X1, · · · ,LXXj, · · · , Xk)
LECTURE 23: INTEGRATION ON MANIFOLDS
1. Top forms and orientability
¶ Top forms on manifolds.
Let M be a smooth manifold of dimension m. As we have known, Ωk(M) = 0 for
k > m. Thus we will call any smooth m-form a top form on M . Now let p ∈ M and
(ϕ,U, V ) a local chart near p. Then dx1 ∧ · · · ∧ dxm is a top form on U . Note that for
any q ∈ U , (dx1 ∧ · · · ∧ dxm)q 6= 0 since at any q ∈ U ,
(dx1 ∧ · · · ∧ dxm)q(∂1, · · · , ∂m) = det(dxi(∂j))1≤i,j≤m = 1.
Moreover, since dim ΛmTpM = 1, we see that for any top form ω on U and any q ∈ U ,
there is a real number λq such that
ωq = λq(dx
1 ∧ · · · ∧ dxm)q.
Also by smoothness of ω, the coefficient λ, as a function on U , is smooth. So up to
multiplication by functions, the “canonical top form” is the “only essential” top form
in the chart. (However, this conclusion is not true globally on M : For two top forms
ω, η ∈ Ωm(M), it may happen that ωp = 0, ωq 6= 0 while ηp 6= 0, ηq = 0, and thus there
exists no function f ∈ C∞(M) with ω = fη or η = fω.)
In particular, if we change coordinates from (x1α, · · · , xmα ) to (x1β, · · · , xmβ ) on U ,
then we will get two top forms, dx1α ∧ · · · ∧ dxmα and dx1β ∧ · · · ∧ dxmβ . They should be
related by a smooth function on U . It is not hard to find out this coordinate-change
factor. We first prove
Lemma 1.1. If ϕ : Rm → Rm is a diffeomorphism and y = ϕ(x), then
ϕ∗(dy1 ∧ · · · ∧ dym) = det(dϕx)dx1 ∧ · · · ∧ dxm
Proof. If we denote ϕ = (ϕ1, · · · , ϕm), then ϕ∗yi = yi ◦ ϕ = ϕi. So
ϕ∗(dy1 ∧ · · · ∧ dym) = dϕ1 ∧ · · · ∧ dϕm.
But since
dϕ1 ∧ · · · ∧ dϕm(∂x1 , · · · , ∂xm) = det(dϕx),
we conclude
dϕ1 ∧ · · · ∧ dϕm = det(dϕx)dx1 ∧ · · · ∧ dxm. �
Now let (ϕα, U, Vα, ) and (ϕβ, U, Vβ) be two coordinate systems on U . Then the
coordinate change map is ϕαβ = ϕβ ◦ ϕ−1α , which maps ϕα(x) to y = ϕβ(x). So we get
(ϕαβ)
∗(ϕ−1β )
∗dx1β ∧ · · · ∧ dxmβ = det(dϕαβ)(ϕ−1α )∗dx1α ∧ · · · ∧ dxmα .
Since (ϕαβ)
∗(ϕ−1β )
∗ = (ϕ−1β ◦ ϕαβ)∗ = ϕ−1α , we arrive at
dx1β ∧ · · · dxmβ = det(dϕαβ)dx1α ∧ · · · ∧ dxmα .
1
2 LECTURE 23: INTEGRATION ON MANIFOLDS
¶ The need of orientability.
Let M be a smooth manifold of dimension n, and let ω ∈ Ωm(M) be a smooth n-
form. We want to define the integral
∫
M
ω. For simplicity let’s suppose ω is supported
on a chart (ϕ,U, V ) with coordinates {x1, · · · , xm}. Then we can write
ω = f(ϕ(x))dx1 ∧ · · · ∧ dxm,
where f is a smooth function on V . With the help of the Euclidean differential form
f(x)dx1 ∧ · · · ∧ dxm on V , it is natural to define
(1)
∫
U
ω :=
∫
V
f(x)dx1 · · · dxm.
Then as usual one need to check that the integral in the right hand side is independent
of the choice of coordinate charts.
So we let (ϕα, U, Vα, ) and (ϕβ, U, Vβ) be two coordinate systems on U , with tran-
sition map ϕαβ = ϕβ ◦ ϕ−1α : Vα → Vβ which maps xα = ϕα(x) to xβ = ϕβ(x). Then
ω = fβ(xβ)dx
1
β ∧ · · · ∧ dxmβ = fβ(ϕαβ(xα)) det(dϕαβ)dx1α ∧ · · · ∧ dxmα
So for the definition to be well-defined, we need∫
Vβ
fβ(xβ)dx
1
β · · · dxmβ =
∫
Vα
fβ(ϕαβ(xα)) det(dϕαβ(xα))dx
1
α · · · dxmα .
Unfortunately this is not always true: In calculus we learned that for the integrals of
multi-variable functions ∫
f(x)dx1 · · · dxm,
if ϕ : V1 → V2 is a diffeomorphism, then we have the change of variable formula:
(y = ϕ(x))
(2)
∫
V2
f(x) dy1 · · · dym =
∫
V1
f(ϕ(x))| det(dϕ)(x)| dx1 · · · dxm.
So we only have∫
Vβ
fβ(xβ)dx
1
β · · · dxmβ =
∫
Vα
fβ(ϕαβ(xα)) |det(dϕαβ(xα))| dx1α · · · dxmα .
In other words, for the definition to be independent of the choice of charts, we need
to assume
det(dϕαβ) > 0
for all charts. In fact, we have seen this condition in PSet 2-1-3:
Definition 1.2. Let M be a smooth manifold of dimension n.
(1) Two charts (ϕα, Uα, Vα) and (ϕβ, Uβ, Vβ) are orientation compatible if the tran-
sition map ϕαβ = ϕβ ◦ ϕ−1α satisfies
det(dϕαβ)p > 0
for all p ∈ ϕα(Uα ∩ Uβ).
LECTURE 23: INTEGRATION ON MANIFOLDS 3
(2) An orientation of M is an atlas A = {(ϕα, Uα, Vα) | α ∈ Λ} whose charts are
pairwise orientation compatible.
(3) We say M is orientable if it has an orientation.
Remark. Let U be a chart with coordinates {x1, · · · , xm}. We use the notation −U to
represent the same coordinate chart U but with “twisted” coordinates {−x1, x2, · · · , xm}.
Then −U and U are not orientation compatible. Let Ũ be any other coordinate chart
such that Ũ ∩ U 6= ∅ is connected. Then either
• Ũ and U are orientation compatible,
or
• Ũ and −U are orientation compatible.
As a consequence, we immediately see
Corollary 1.3. If M is connected and orientable, then M admits exactly two different
orientations.
Example. For the real projective space RPn, we have constructed an atlas consisting
of n+ 1 charts. We have seen that RPn is orientable for odd n. It turns out that RPn
is not orientable for even n.
2. Integrations on smooth manifolds
¶ Integrations of top forms on smooth manifolds.
Now assume M is a smooth orientable m-manifold and fix an orientation A on
M . Let ω be any smooth m-form on M . To define
∫
M
ω, we first assume that ω is
supported in a coordinate chart (ϕ,U, V ) which is orientation compatible with A. In
this case there is a function f supported in U such that
ω = f(ϕ(x))dx1 ∧ · · · ∧ dxm.
In this case we simply define
(3)
∫
U
ω :=
∫
V
f(x)dx1 · · · dxm,
where the right hand side is the Lebesgue integral on V ⊂ Rm. We will assume f is
integrable. In fact, in what follows the function f involved are compactly supported.
To integrate a general m-form ω ∈ Ωm(M), we take a locally finite cover {Uα}
of M that are compatible with the orientation A. Let {ρα} be a partition of unity
subordinate to {Uα}. Now since each ρα is supported in Uα, each ραω is supported Uα
also. We define
(4)
∫
M
ω :=
∑
α
∫
Uα
ραω.
We say that ω is integrable if the right hand side converges absolutely for any such
cover and any such P.O.U. This is true, for example, if ω is compactly supported.
4 LECTURE 23: INTEGRATIONON MANIFOLDS
One need to check that the definition (4) above is independent of the choices of
orientation-compatible coordinate charts, and is independent of the choices of partition
of unity.
Theorem 2.1. Suppose ω is compactly supported, or more generally, ω is integrable.
The expression (4) is independent of the choices of {Uα} and the choices of {ρα}.
Proof. We first show that (3) is well-defined. The argument is essentially the same as
in the Euclidian case: if ω is supported in U , and if {xiα} and {xiβ} are two orientation-
compatible coordinate systems on U , so that
ω = fαdx
1
α ∧ · · · ∧ dxmα = fβdx1β ∧ · · · ∧ dxmβ ,
then we want to prove ∫
Vα
fαdx
1
α · · · dxmα =
∫
Vβ
fβdx
1
β · · · dxmβ .
This is true, because
dx1β ∧ · · · ∧ dxmβ = det(dϕαβ)dx1α ∧ · · · ∧ dxmα
implies fα = det(dϕαβ)fβ. Since det(dϕαβ) > 0, the conclusion follows from the change
of variable formula in Rn.
To prove (4) is well-defined, we suppose {Uα} and {Uβ} are two locally finite cover
of M consisting of orientation-compatible charts, and {ρα} and {ρβ} are partitions of
unity subordinate to {Uα} and {Uβ} respectively. Then {Uα ∩ Uβ} is a new locally
finite cover of M , and {ραρβ} is a partition of unity subordinate to this new cover. It
is enough to prove ∑
α
∫
Uα
ραω =
∑
α,β
∫
Uα∩Uβ
ραρβω.
This is true because for each fixed α,∫
Uα
ραω =
∫
Uα
(
∑
β
ρβ)ραω =
∑
β
∫
Uα∩Uβ
ρβραω.
�
¶ Change of variable formula.
Finally we extend the change of variable formula form Rm to manifolds.
Definition 2.2. Let M,N be orientable smooth n-manifolds, with orientations A and
B respectively. A diffeomorphism ϕ : M → N is said to be orientation-preserving if for
each (ψβ, Xβ, Yβ) ∈ B, the chart (ψβ ◦ ϕ, ϕ−1(Xβ), Yβ) on M is orientation compatible
with A.
Suppose M,N are connected. It is easy to see that a diffeomorphism ϕ : M → N
is orientation-preserving if and only if there exists one chart (ψβ, Xβ, Yβ) ∈ B, such
that the chart (ψβ ◦ ϕ, ϕ−1(Xβ), Yβ) on M is orientation compatible with A. Similarly
if there exists one chart (ψβ, Xβ, Yβ) ∈ B, such that the chart (ψβ ◦ ϕ, ϕ−1(Xβ), Yβ)
on M is incompatible with A, then for every chart (ψβ, Xβ, Yβ) ∈ B, the chart (ψβ ◦
LECTURE 23: INTEGRATION ON MANIFOLDS 5
ϕ, ϕ−1(Xβ), Yβ) on M is incompatible with A. In this case we say ϕ is orientation-
reverting.
Now we state:
Theorem 2.3 (The change of variable formula.). Suppose M,N are n-dimensional
orientable smooth manifolds, and ϕ : M → N is a diffeomorphism.
(1) If ϕ is an orientation-preserving, then∫
M
f ∗ω =
∫
N
ω.
(2) If ϕ is an orientation-reverting, then∫
M
f ∗ω = −
∫
N
ω.
Proof. It is enough to prove this in local charts, in which case this is merely the change
of variable formula in Rm. �
Remark. If ω is a compactly supported k-form on M , where k < m = dimM , then one
cannot integrate ω over M . However, for any k-dimensional orientable submanifold
X ⊂ M , one can define
∫
X
ω by setting it to be
∫
X
ι∗ω, where ι : X ↪→ M is the
inclusion map. By this way we get a “pairing” between k-forms onM and k-dimensional
orientable submanifolds in M .
Remark. If M is not orientable, one cannot define integrals of differential forms as
above. However, we can still integrate via densities. (c.f. J. Lee, Introduction to
smooth manifolds, page 427-432.)
¶ Volume form and volume measure.
Next we show that orientability can be characterized via the existence of specific
top forms:
Theorem 2.4. An m-dimensional smooth manifold M is orientable if and only if M
admits a nowhere vanishing smooth m-form µ.
Proof. First let µ be a nowhere vanishing smooth m-form on M . Then on each local
chart (U, x1, · · · , xm) (where U is always chosen to be connected), there is a smooth
function f 6= 0 so that µ = fdx1 ∧ · · · ∧ dxm. It follows that
µ(∂1, · · · , ∂m) = f 6= 0.
We can always take such a chart near each point so that f > 0, otherwise we can
replace x1 by −x1. Now suppose (Uα, x1α, · · · , xmα ) and (Uβ, x1β, · · · , xmβ ) be two such
charts, so that on the intersection Uα ∩ Uβ one has
fdx1α ∧ · · · ∧ dxmα = µ = gdx1β ∧ · · · ∧ dxmβ = g det(dϕαβ)dx1α ∧ · · · ∧ dxmα .
where f, g > 0. It follows that det (dϕαβ) > 0. So the atlas constructed by this way is
an orientation.
6 LECTURE 23: INTEGRATION ON MANIFOLDS
Conversely, suppose A is an orientation. For each local chart Uα in A, we let
µα = dx
1
α ∧ · · · ∧ dxmα .
Pick a partition of unity {ρα} subordinate to the open cover {Uα}. We claim that
µ :=
∑
α
ραµα
is a nowhere vanishing smooth m-form on M . In fact, for each p ∈ M , there is a
neighborhood U of p so that the sum
∑
α ραµα is a finite sum
∑k
i=1 ρiµi. It follows
that near p,
µ(∂11 , · · · , ∂1m) =
k∑
i=1
(det dϕ1i)ρi > 0.
So µ 6= 0 near p. �
Definition 2.5. A nowhere vanishing smooth m-form µ on an m-dimensional smooth
manifold M is called a volume form.
Remark. If M is orientable, and µ is a volume form, then the two orientations of M
are represented by µ and −µ respectively. We denote the two orientations by [µ] and
[−µ].
Remark. Let µ be a volume form on M , and the orientation on M is chosen to be [µ].
Then we can define a linear functional
I : Cc(M)→ R, f 7→
∫
M
fµ.
(Here, Cc(M) represents the space of continuous functions with compact supports on
M . Obviously the integrals above still make sense even if f is not smooth.) Since the
orientation on M is chosen to be [µ], we see the functional I is positive, i.e. I(f) ≥ 0
for f ≥ 0. Since any manifold is both locally compact and σ-compact, the Riesz
representation theorem implies that there exists a unique Radon measure (=a locally
finite, regular measure defined on all Borel sets) mµ such that
I(f) =
∫
M
fdmµ.
Using the measure dµ, one can define function spaces like Lp(M,µ).
Remark. In particular, on any Lie group, one can define conceptions like left-invariant d-
ifferential forms. Since any Lie group is orientable (Exercise), there exists left-invariant
volume form on any Lie group G. The measures associated to left-invariant volume
form on Lie groups are called Haar measures.
LECTURE 24: THE STOKES FORMULA
1. Smooth manifolds with boundary
¶ Manifolds with boundary.
The Stokes formula (Green formula, Gauss formula) is one of the most important
formulae in calculus. We now extend it to smooth manifolds. We need some knowledge
of manifolds with boundary. Denote
Rm+ = {(x1, · · · , xn) | xm ≥ 0}.
The following definition is natural (See PSet 1-1-3):
Definition 1.1. A topological space M is called an m-dimensional topological manifold
with boundary if it is Hausdorff, second-countable, and for any p ∈ M , there is a
neighborhood U of p which is homeomorphic to either Rm or Rm+ .
Let M be a topological manifold with boundary, then we can define the boundary
of M to be
∂M = {p ∈M | p has no neighborhood in M that is homeomorphic to Rm}.
(Check: p ∈ ∂M if and only if p has a neighborhood that is homeomorphic to Rm+ such
that p gets mapped onto ∂Rm+ = {(x1, · · · , xn) | xm = 0}.) We will call
int(M) = M \ ∂M
the interior of M . As in Lecture 2, a smooth structure on a topological manifold
with boundary can be defined to be an atlas so that the transition maps between each
pairs of charts are smooth. (Note: As we defined in Lecture 11, a map defined on a
subset in Rm+ ⊂ Rm is smooth if it has a smooth extension to an open neighborhood of
the subset in Rm. By this way, we can talk about the smoothness of transition maps
between charts of either type.)
Example. The closed ball
Bn(1) = {x ∈ Rm | |x| ≤ 1}
is a smooth manifold with boundary, and its boundary is ∂Bm(1) = Sm−1.
Example. More generally, let M be any smooth manifold, and f ∈ C∞(M). If a is a
regular value of f , then the sub-level set
Ma = f
−1((−∞, a))
is a smooth manifold with boundary, and ∂Ma = f
−1(a).
1
2 LECTURE 24: THE STOKES FORMULA
Remark. In PSet 1-1-3 we showed that if both M and N are topological manifolds with
boundary, so is M ×N . This is not true for smooth manifolds with boundary: if both
M and N are smooth manifolds with(nonempty) boundary, their product M×N (with
the product charts as given at the end of Lecture 1) will no longer be a smooth manifold
with boundary (think about the example [0, 1]× [0, 1]). For an obvious reason, they are
known as smooth manifolds with corners. However, if M is a smooth manifolds with
boundary and N is a smooth manifolds without boundary, then M × N is a smooth
manifold with boundary.
¶ The boundary of a smooth manifold with boundary.
In the above examples, the boundaries are themselves smooth manifolds. In general
Lemma 1.2. Suppose M is an m-dimensional smooth manifold with boundary, then
∂M is a smooth manifold of dimension m−1 (unless ∂M = ∅) that is properly embedded
into M .
Sketch of proof. Let (U, x1, · · · , xm) be a chart near p ∈ ∂M that is homeomorphic to
Rm+ . Then
U ∩ ∂M = {(x1, · · · , xm) | xm = 0}.
Then (U ∩ ∂M, x1, · · · , xm−1) is a chart on ∂M . �
As in the case of smooth manifolds (without boundary), one can define an orien-
tation on a smooth manifold with boundary to be an atlas A so that det(dϕαβ) > 0
for any two charts Uα, Uβ ∈ A. It is also true that a smooth manifold with boundary
is orientable if and only if it admits a nowhere vanishing top form. We now prove
Theorem 1.3. If M is an orientable smooth manifold with boundary of dimension m,
then the boundary ∂M is an orientable m− 1 dimensional submanifold of M .
Proof. Let (Uα, x
1
α, · · · , xmα ) and (Uβ, x1β, · · · , xmβ ) be two orientation-compatible charts
of M near p ∈ ∂M so that M ∩ Uα is characterized by xmα ≥ 0, and M ∩ Uβ is
characterized by xmβ ≥ 0. We would like to show that the coordinate charts (Uα ∩
∂M, x1α, · · · , xm−1α ) and (Uβ ∩ ∂M, x1β, · · · , xm−1β ) of ∂M are orientation-compatible. In
fact, if we denote the transition map ϕαβ between Uα and Uβ by (ϕ
1, · · · , ϕm), then on
∂M ∩ Uα ∩ Uβ, we have xmα = xmβ = 0. In other words,
ϕm(x1, · · · , xm−1, 0) = 0
on Uα ∩ Uβ ∩ ∂M , and
ϕm(x1, · · · , xm−1, xm) > 0
on Uα ∩ Uβ ∩ int(M), i.e. for xm > 0. It follows
∂ϕm
∂xi
(x1, · · · , xm−1, 0) = 0, i = 1, · · · ,m− 1
and
∂ϕm
∂xm
(x1, · · · , xm−1, 0) ≥ 0.
LECTURE 24: THE STOKES FORMULA 3
Since (Uα, x
1
α, · · ·, xmα ) and (Uβ, x1β, · · · , xmβ ) are orientation-compatible, we have
det(
∂ϕi
∂xj
) > 0
everywhere in Uα ∩ Uβ. In particular,
det
(
∂ϕi
∂xj
(x1,· · ·,xm−1, 0)
)
=det
((
∂ϕi
∂xj
(x1,· · ·,xm−1, 0)
)
1≤i,j≤m−1
∗
0 ∂ϕ
m
∂xm
(x1,· · ·,xm−1, 0)
)
> 0.
It follows that
det
(
∂ϕi
∂xj
(x1, · · · , xm−1, 0)
)
1≤i,j≤m−1
> 0.
So the charts (Uα∩∂M, x1α, · · · , xm−1α ) and (Uβ∩∂M, x1β, · · · , xm−1β ) of ∂M are orientation-
compatible. �
Remark. The boundary of a non-orientable manifold could be either orientable (e.g.
the Mobiüs band) or non-orientable (e.g. [0, 1]×M , where M is non-orientable).
¶ The induced orientation on the boundary.
Now let M be an orientable smooth manifold with boundary, and let [µ] be an
orientation on M . The induced orientation on ∂M is defined as follows: Suppose
locally near the boundary, M is given by xm ≥ 0, and such that
µ = fdx1 ∧ · · · ∧ dxm
with f > 0. Then locally we define the induced orientation on ∂M to be the one that
is represented by the differential form
η = (−1)mdx1 ∧ · · · ∧ dxm−1.
By choosing a P.O.U., one can glue these local (m − 1)-forms on ∂M into a global
(m − 1)-form on ∂M which is nowhere vanishing (Please check!!) and thus give us a
volume form that defines an orientation on ∂M .
Note that in coordinate charts near boundary, M is defined by xm ≥ 0. So −xm
is the “outward pointing direction” of M , and this boundary orientation is chosen so
that
d(−xn) ∧ η = d(−xm) ∧ (−1)ndx1 ∧ · · · ∧ dxm−1 = dx1 ∧ · · · ∧ dxm.
As a result, to integrate ω ∈ Ωm−1(∂M) on ∂M , then locally we need to write
ω = f(x1, · · · , xm−1) · (−1)mdx1 ∧ · · · ∧ dxm−1
in a chart (ϕ,U, V ) of ∂M , and then calculate∫
U
ω :=
∫
V
f(x1, · · · , xm−1)dx1 · · · dxm−1.
4 LECTURE 24: THE STOKES FORMULA
2. Stokes formula and its applications
¶ Stokes formula.
Now we can state and prove our main theorem:
Theorem 2.1 (Stokes’ theorem). Let M be a smooth oriented m-dimensional manifold
with boundary ∂M (with the induced orientation above). For any ω ∈ Ωm−1(M) with
compact support, we have ∫
∂M
ι∗∂Mω =
∫
M
dω,
where ι∂M : ∂M ↪→M is the inclusion map.
Remark. The Stokes formula
(1) holds for manifold without boundary, in which case ∂M = ∅ and thus the left
hand side is zero.
(2) holds for manifolds with corners. For details, c.f. J. Lee, page 415-421.
Proof. We will discuss the following three cases:
(1) ω is supported in a chart U that is diffeomorphic to Rm.
(2) ω is supported in a chart U that is diffeomorphic to Rm+ .
(3) The general case.
Case (1): Since ω = 0 on ∂M , we have
∫
∂M
ι∗∂Mω = 0.
To calculate
∫
M
dω, we denote
ω =
∑
i
(−1)i−1fidx1 ∧ · · · ∧ d̂xi ∧ · · · ∧ dxm,
where fi’s are compactly supported smooth functions. Then
dω =
∑
i
∂fi
∂xi
dx1 ∧ · · · ∧ dxm.
So by definition,∫
M
dω =
∫
Rm
∑
i
∂fi
∂xi
dx1 · · · dxm =
∑
i
∫
Rm−1
(∫ ∞
−∞
∂fi
∂xi
dxi
)
dx1 · · · d̂xi · · · dxm = 0.
Case (2): We have the same formula to calculate
∫
M
dω, except for the last term (i.e.
i = m term), which, instead of giving us 0, will give us∫
Rm−1
(∫ ∞
0
∂fm
∂xm
dxm
)
dx1 · · · dxm−1 = −
∫
Rm−1
fn(x
1, · · · , xm−1, 0)dx1 · · · dxn−1.
On the other hand, since xm = 0 on ∂M , we see
ι∗∂Mω = (−1)m−1fm(x1, · · · , xm−1, 0) dx1 ∧ · · · ∧ dxm−1
= −fm(x1, · · · , xm−1, 0) · (−1)m dx1 ∧ · · · ∧ dxm−1.
LECTURE 24: THE STOKES FORMULA 5
So ∫
∂M
ι∗∂Mω =
∫
Rm−1
(−fm(x1, · · ·, xm−1, 0))dx1· · · dxm−1
=−
∫
Rm−1
fm(x
1, · · ·, xm−1, 0)dx1· · · dxm−1
and the desired conclusion follows.
Case (3): In general we cover the set supp(ω) (which, by our assumption, is compact)
by finitely many coordinate charts, and take a partition of unity as usual. Then∫
∂M
ι∗∂Mω =
∑
i
∫
∂M
ι∗∂M(ρiω) =
∑
i
∫
M
d(ρiω) =
∑
i
∫
M
dρi ∧ ω +
∫
M
dω.
Now the conclusion follows from the fact∑
i
∫
M
dρi ∧ ω =
∫
M
d(
∑
ρi) ∧ ω = 0.
�
¶ Application: Variation of volume.
Let M be a smooth manifold (without boundary), Ω ⊂ M be a domain so that Ω
is compact. Let µ be a volume form on M . Then we can define the volume of Ω with
respect to µ to be
Vol(Ω) =
∫
Ω
µ.
Now assume that the boundary ∂Ω is a smooth submanifold of M , and let X be any
complete vector field on M . Then X generates a flow, i.e. a family of diffeomorphisms
φt : M → M . Note that these φt’s are orientation-preserving. (Reason: det(dφt) 6= 0
for all t, and det(dφt) is continuous with respect to t, and det(dφ0) ≡ 1.)
Let Ωt = φt(Ω) be the “flow-out” of Ω under this flow. We have
Proposition 2.2. We have
d
dt
∣∣∣∣
t=0
Vol(Ωt) =
∫
∂Ω
ιXµ.
Proof. By definitions and the change of variable formula (note: φt’s preserve the ori-
entation!)
d
dt
∣∣∣∣
t=0
Vol(Ωt) =
d
dt
∣∣∣∣
t=0
∫
Ωt
µ =
d
dt
∣∣∣∣
t=0
∫
Ω
φ∗tµ =
∫
Ω
d
dt
∣∣∣∣
t=0
φ∗tµ =
∫
Ω
LXµ.
Now apply Cartan’s magic formula to get (note: dµ = 0 since it is an (n+ 1)-form)
LXµ = dιXµ+ ιXdµ = dιXµ.
So by Stokes’ formula,
d
dt
∣∣∣∣
t=0
Vol(Ωt) =
∫
Ω
dιXµ =
∫
∂Ω
ιXµ.
�
6 LECTURE 24: THE STOKES FORMULA
A special case: Let M = Rm (or more generally a Riemannian manifold). Let f be
a smooth function, and 0 is a regular value of f . Let Ω be the sub-level set
Ω = {p : f(p) ≤ 0}.
Take the vector field X to be the gradient vector field ∇f of f . Then ιXµ is a volume
form on ∂Ω and is known as the induced volume form on the boundary ∂Ω. It can be
characterized via the formula (check!)
df ∧ ιXµ = |∇f |2µ.
In this setting, the above formula can be interpreted as
Fact: The rate of change of the sub-level sets of a smooth function along
the gradient flow equals the surface area of the level set.
LECTURE 25: THE DE RHAM COHOMOLOGY
1. The De Rham cohomology
¶ Closed and exact forms.
We start with the following definition:
Definition 1.1. Let M be a smooth manifold, and ω ∈ Ωk(M) is a k-form.
(1) We say ω is closed if dω = 0.
(2) We say ω is exact if there exists a (k−1)-form η ∈ Ωk−1(M) such that ω = dη.
Denote the set of closed k-forms by Zk(M), and the set of exact k-forms by Bk(M):
Zk(M) = the set of closed k-forms = ker(d : Ωk(M)→ Ωk+1(M)),
Bk(M) = the set of exact k-forms = Im(d : Ωk−1(M)→ Ωk(M)).
As we have seen, the exterior derivative d : Ωk(M)→ Ωk+1(M) is a linear map so that
for any k and any ω ∈ Ωk(M),
d2ω = d(dω) = 0.
So we have the following inclusion relation (as vector spaces and as additive groups)
Bk(M) ⊂ Zk(M) ⊂ Ωk(M).
Remark. Suppose dimM = m. Then by definition we have
• For k > m: Bk(M) = Zk(M) = {0}.
• For k = 0: B0(M) = {0}, and
Z0(M) = {f ∈ C∞(M) | df = 0} ' RK .
where K is the number of connected components of M .
• For k = m: Zm(M) = Ωm(M).
Example. Consider M = R. We have
B0(R) = {0}, Z0(R) ' R and Ω0(R) = C∞(R).
For any 1-form g(t)dt on R, we have
ω = g(t)dt⇐⇒ ω = dG,where G(t) =
∫ t
0
g(τ)dτ.
It follows that Ω1(R) = B1(R) = Z1(R).
1
2 LECTURE 25: THE DE RHAM COHOMOLOGY
¶ The De Rham cohomology groups.
Since d2 = 0, we get the following de Rham cochain complex
0
d−→ Ω0(M) d−→ Ω1(M) d−→ · · · d−→ Ωm(M) d−→ 0.
(Such a sequence of vector spaces [could be an infinite sequence] connected by a se-
quence of linear maps, such that the composition of any two consecutive maps is the
zero map, is called a chain complex or a cochain complex, depending on the “direction”
of the maps.)
Definition 1.2. The quotient group (vector space)
HkdR(M) := Z
k(M)/Bk(M)
is called the kth de Rham cohomology group of M .
Example. For M = R, we easily see H0dR(R) ' R and HkdR(R) = {0} for k ≥ 1.
Given any ω ∈ Zk(M), we will denote by [ω] the corresponding cohomology class.
Remark. Suppose dimM = m. According to the above remark, we get
HkdR(M) = {0}, ∀k > m
and (we still denote by K the number of connected components of M)
H0dR(M) ' RK .
By definition, HkdR(M) is a vector space. We will see that for many smooth mani-
folds (including all compact manifolds),
dimHkdR(M) <∞
for all k. On the other hand, we have dimH0dR(Z) = ∞. (As another example: it is
not hard to prove dimH1dR(R2 \ Z2) = +∞.)
Definition 1.3. In the case dimHkdR(M) <∞ for each k, we will call the number
bk(M) = dimH
k
dR(M)
the kth Betti number of M , and the number
χ(M) =
m∑
k=0
(−1)kbk(M)
the Euler characteristic of M .
¶ The De Rham cohomology groups of S1.
Consider M = S1. As we have seen,
H0dR(S
1) ' R and HkdR(S1) = 0 for k ≥ 2.
To calculate H1dR(S
1), we argue as in the previous example. Note that on S1 =
R/2πZ, the “angle” variable θ is not a globally defined smooth function on S1, but
the translation invariance of d on R implies that the differential form dθ is a globally
defined 1-form on S1. (As a consequence, the 1-form dθ is a closed 1-form on S1, but
LECTURE 25: THE DE RHAM COHOMOLOGY 3
is NOT an exact 1-form on S1. However, it is an exact 1-form on any proper subset of
S1.) So we can write
Z1(S1) = Ω1(S1) = {fdθ | f ∈ C∞(S1)}
' {f ∈ C∞(R) | f is periodic with period 2π}.
On the other hand, by the fundamental theorem of calculus,
ω is an exact 1-form
⇐⇒ω = df, where f is periodic with period 2π
⇐⇒ω = g(θ)dθ, where g is periodic with period 2π and
∫ 2π
0
g(θ)dθ = 0.
So we conclude
H1dR(S
1) ' {f ∈ C
∞(R) | f is periodic with period 2π}
{g ∈ C∞(R) |g is periodic with period 2π, and
∫ 2π
0
g(θ)dθ = 0}
.
This implies that
H1dR(S
1) ' R,
since the linear map
ϕ : H1dR(S
1)→ R, [f ] 7→
∫ 2π
0
f(θ)dθ.
is an linear isomorphism:
• ϕ is well-defined:
[f1] = [f ] =⇒ f1 − f ∈ B1(S1) =⇒
∫ 2π
0
f1(θ)dθ =
∫ 2π
0
f(θ)dθ.
• ϕ is injective:
[f1] 6= [f ] =⇒ f1 − f 6∈ B1(S1) =⇒
∫ 2π
0
f1(θ)dθ 6=
∫ 2π
0
f(θ)dθ.
• ϕ is surjective: for any c ∈ R,
f(θ) := c ∈ Z1(S1) =⇒ ϕ([f ]) =
∫ 2π
0
f(θ)dθ = 2πc.
¶ Operations on cohomology classes.
One can extend the wedge product and pull-back operations on differential forms
to operations on cohomology classes.
First let ω ∈ Zk(M) and η ∈ Z l(M), then
d(ω ∧ η) = dω ∧ η + (−1)kω ∧ dη = 0,
i.e. ω ∧ η ∈ Zk+l(M). Moreover, for any ξ1 ∈ Ωk−1(M) and ξ2 ∈ Ωl−1(M),
(ω + dξ1) ∧ (η + dξ2) = ω ∧ η + d
[
(−1)kω ∧ ξ2 + (−1)k−1ξ1 ∧ η + (−1)k−1ξ1 ∧ dξ2
]
.
In other words, [ω ∧ η] is independent of the choice of ω and η in [ω] and [η]. So we
can define
4 LECTURE 25: THE DE RHAM COHOMOLOGY
Definition 1.4. The cup product between [ω] ∈ HkdR(M) and [η] ∈ H ldR(M) is
[ω] ∪ [η] := [ω ∧ η] ∈ Hk+ldR (M).
Similarly suppose ϕ : M → N is smooth. Then the fact dϕ∗ = ϕ∗d implies
ϕ∗(Zk(N)) ⊂ Zk(M) and ϕ∗(Bk(N)) ⊂ Bk(M).
It follows that ϕ∗ : Ωk(N)→ Ωk(M) descends to a pull-back ϕ∗ : HkdR(N)→ HkdR(M):
ϕ∗([ω]) := [ϕ∗ω].
Obviously ϕ∗ is a group homomorphism. It is easy to check
• (ψ ◦ ϕ)∗ = ϕ∗ ◦ ψ∗.
• Id∗ = Id.
As an immediate consequence, we see that the de Rham cohomology groups are invari-
ant under diffeomorphisms:
Corollary 1.5 (Diffeomorphism Invariance). If ϕ : M → N is a diffeomorphism, then
ϕ∗ : HkdR(N)→ HkdR(M)
is a linear isomorphism for all k. In particular,
bk(N) = bk(M)
for all k, and
χ(N) = χ(M).
Remark. For any smooth map ϕ : M → N , The cup product makes
H∗dR(M) =
m⊕
k=0
HkdR(M)
a graded ring, and the induced map ϕ∗ is in fact a ring homomorphism
ϕ∗ : H∗dR(N)→ H∗dR(M),
since
ϕ∗(α ∧ β) = ϕ∗α ∧ ϕ∗β.
Moreover, if ϕ is a diffeomorphism, then ϕ∗ : H∗dR(N)→ H∗dR(M) is a ring isomorphism.
2. Homotopic Invariance
¶ Homotopic Invariance of de Rham cohomology.
In this section we shall prove a much stronger result: if two manifolds are homotopy
equivalent, then they have the same de Rham cohomology groups. We first recall
Definition 2.1. Two topological spaces M and N are said to be homotopy equivalent
if there exist continuous maps ϕ : M → N and ψ : N →M so that ϕ ◦ψ is homotopic
to IdN and ψ ◦ ϕ is homotopic to IdM .
In Lecture 11 we have shown
LECTURE 25: THE DE RHAM COHOMOLOGY 5
• any continuous map between smooth manifolds is homotopic to some smooth
map.
• any two homotopic smooth maps are smoothly homotopic (i.e., the homotopy
F can be chosen to be smooth if both f0 and f1 are smooth).
Homotopy equivalence is a much weaker equivalence relation than homeomorphism
or diffeomorphism. For example (check details if you are not familiar with this)
• Sn−1 is homotopy equivalent to Rn \ {0}.
• any star-shaped region is homotopic equivalent to a single point set {x0}.
We would like to prove
Theorem 2.2 (Homotopy Invariance). Let M,N be smooth manifolds. If M and N
are homotopy equivalent, then
HkdR(M) ' HkdR(N), ∀k.
Obviously the homotopy invariance implies the topological Invariance of de Rham
cohomology groups:
If M is homeomorphic to N , then HkdR(M) ' HkdR(N) for all k.
Remark. Although in defining HkdR(M), we need to use the smooth structure on M
(to define the operator d and the space Ωk(M) etc), the topological invariance tells us
that HkdR(M) only depends on the topology of M , and is independent of the smooth
structure! In fact, for any topological space X, one can define its singular cohomology
groups Hksing(X,R) which depends only on the topology (and in fact depends only on
the homotopy class) of X. The famous theorem of de Rham claims
Theorem 2.3 (The de Rham theorem). HkdR(M) = H
k
sing(M,R) for all k.
We will not prove the theorem in this course.
Another immediate consequence of the homotopy invariance is
Corollary 2.4 (Poincare’s lemma). If U is a star-shaped region in Rm, then for any
k ≥ 1, HkdR(U) = 0. In particular,
HkdR(Rm) = 0, ∀k ≥ 1.
Since any point in a manifold has a neighborhood that is homeomorphic to a star-
like region in Rn, we conclude that any closed form is locally exact:
Corollary 2.5. Suppose k ≥ 1. Then for any closed k-form ω ∈ Zk(M) and any
p ∈M , there is a neighborhood U of p and an (k − 1)-form η ∈ Ωk−1(U) so that
ω = dη
on U .
6 LECTURE 25: THE DE RHAM COHOMOLOGY
¶ Homotopic invariance of de Rham cohomology: The proof.
The homotopy invariance is a consequence of
Theorem 2.6. Let f, g ∈ C∞(M,N) be homotopic, then
f ∗ = g∗ : HkdR(N)→ HkdR(M).
Proof of Theorem 2.2 assuming Theorem 2.6. Let ϕ : M → N and ψ : N → M be
continuousmaps so that ϕ ◦ ψ ∼ IdN and ψ ◦ ϕ ∼ IdM . Then one can find ϕ1 ∈
C∞(M,N) and ψ1 ∈ C∞(N,M) so that ϕ1 ∼ ϕ and ψ1 ∼ ψ. It follows that both
ϕ1 ◦ ψ1 and ψ1 ◦ ϕ1 are smooth, and ϕ1 ◦ ψ1 ∼ IdN , ψ1 ◦ ϕ1 ∼ IdM .
Applying Theorem 2.6, we get
ϕ∗1 ◦ ψ∗1 = Id : HkdR(M)→ HkdR(M)
ψ∗1 ◦ ϕ∗1 = Id : HkdR(N)→ HkdR(N).
So ϕ∗ and ψ∗ are linear isomorphisms. �
Theorem 2.6 can be proved by constructing a cochain homotopy :
· · · Ωk−1(N) Ωk(N) Ωk+1(N) · · ·
· · · Ωk−1(M) Ωk(M) Ωk+1(M) · · ·
................................................................................................................. .......
.....d .......................................................................................... ............d .......................................................................................... ............d ................................................................................................................. ............d
.............................................................................................................. .......
.....d ..................................................................................... ............d ..................................................................................... ............d .............................................................................................................. ............d
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
f ∗
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
g∗
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
f ∗
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
g∗
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
f ∗
...................................................................................
.....
.......
.....
...................................................................................
.....
.......
.....
g∗
.............
.............
.............
.............
.............
................
...
............
h
.............
.............
.............
.............
......................
...
............
h
.............
.............
.............
.............
......................
...
............
h
.............
.............
.............
.............
.............
.............
...
............
h
Definition 2.7. Let f, g ∈ C∞(M,N) be homotopic. A cochain homotopy between f ∗
and g∗ is a sequence of maps hk : Ω
k(N)→ Ωk−1(M) such that on Ωk(N),
g∗ − f ∗ = dMhk + hk+1dN .
Proof of Theorem 2.6 assuming the existence of the cochain homotopy .
Suppose [ω] ∈ HkdR(N). Then dω = 0 since ω is closed. It follows
g∗ω − f ∗ω = (dh+ hd)ω = dhω ∈ Bk(M)
Thus f ∗([ω]) = [f ∗ω] = [g∗ω] = g∗([ω]). �
¶ The existence of cochain homotopy.
It remains to construct the cochain homotopy. We will use the flow generated by
a vector field to complete the construction. Recall that if X is a complete vector field
on M , then X generates a flow φt : M →M . We need
Lemma 2.8. Let X be a complete vector field on M , and φt the flow generated by X.
Then there is a linear operator Q : Ωk(M)→ Ωk−1(M) so that for any ω ∈ Ωk(M),
φ∗1ω − ω = dQ(ω) +Q(dω).
LECTURE 25: THE DE RHAM COHOMOLOGY 7
Proof. If we set Qt(ω) = ιX(φ
∗
tω), then Qt : Ω
k(M)→ Ωk−1(M) and
d
dt
φ∗tω =
d
ds
∣∣∣∣
s=0
φ∗t+sω =
d
ds
∣∣∣∣
s=0
φ∗sφ
∗
tω
= LX(φ∗tω) = dιX(φ∗tω) + ιXd(φ∗tω)
= d(Qtω) + ιXφ
∗
t (dω) = d(Qtω) +Qt(dω).
So if we denote Q(ω) =
∫ 1
0
Qt(ω)dt, then Q : Ω
k(M)→ Ωk−1(M) and
φ∗1ω − ω =
∫ 1
0
(
d
dt
φ∗tω
)
dt = dQ(ω) +Q(dω). �
Construction of the cochain homotopy hk : Ω
k(N)→ Ωk−1(M) .
Let W = M × R, then X = ∂
∂t
is a complete vector field on W whose flow is
φt(p, a) = (p, a+ t).
By Lemma 2.8, there is a linear operator Q : Ωk(W )→ Ωk−1(W ) so that
φ∗1ω − ω = dQ(ω) +Q(dω).
Let F : W → N is a smooth homotopy between f and g, and let ι : M ↪→ W be the
inclusion map ι(p) = (p, 0), then
f = F ◦ ι and g = F ◦ φ1 ◦ ι,
It follows that for any ω ∈ Ωk(N),
g∗ω − f ∗ω = ι∗φ∗1F ∗ω − ι∗F ∗ω = ι∗(dQ+Qd)F ∗ω = (dι∗QF ∗ + ι∗QF ∗d)ω.
So if we denote h = ι∗QF ∗, then h : Ωk(N)→ Ωk−1(M) satisfies
g∗ω − f ∗ω = (dh+ hd)ω,
i.e. h is the cochain homotopy we are looking for. �
LECTURE 26: THE MAYER-VIETORIS SEQUENCE
The Mayer-Vietoris sequence is an algebraic tool using which one can compute
homology/cohomology groups of topological spaces from the knowledge of the homolo-
gy/cohomology groups of suitable subspaces. It is a sort of analogue of the (Seifert-)van
Kampen theorem for the fundamental group. Moreover, partly because of the coho-
mology groups are simpler (abelian) then the fundamental groups, the assumption for
the Mayer-Vietoris sequence is weaker, for example, no path-connectedness is needed.
1. The Mayer-Vietoris sequence
¶ Exact sequences.
Suppose we have a cochain complex (A, d), i.e. a sequence
· · · dk−2−→ Ak−1 dk−1−→ Ak dk−→ Ak+1 dk+1−→ Ak+2 dk+2−→ · · ·
where Ak’s are vector spaces, and dk’s are linear maps such that
dk ◦ dk−1 = 0, ∀k.
Obviously we have Im(dk) ⊂ Ker(dk+1). One can define cohomology groups for any
cochain complex as we did for the de Rham cochain complex:
Hk(A) := ker(dk)/Im(dk−1)
Such a complex (A, d) is called an exact sequence if Hk(A, d) = 0 for all k, i.e.
Im(dk−1) = ker(dk), ∀k.
Note that if an exact sequence starts with 0,
(· · · →) 0 d0−→ V 1 d1−→ V 2 d2−→ V 3 d3−→ · · · ,
then d1 : V
1 → V 2 is injective, while if an exact sequence ends with 0,
· · · dk−2−→ V k−1 dk−1−→ V k dk−→ V k+1 dk+1−→ 0 (→ · · · ),
then dk : V
k → V k+1 is surjective. In particular, if we have a short exact sequence
0−→ V 1 d1−→ V 2 d2−→ V 3 −→ 0,
then d1 is injective, d2 is surjective, and
V 2 ' ker(d2)⊕ Im(d2) ' Im(d1)⊕ Im(d2) ' V 1 ⊕ V 3.
Another useful fact about exact sequences is: if a finite sequence
0→ A1 → A2 → · · · → Ak → 0
is exact, then (The proof is left as an exercise.)∑
i
(−1)i dimAi = 0.
1
2 LECTURE 26: THE MAYER-VIETORIS SEQUENCE
¶ Short exact sequence of complexes.
A general principle in homological algebra: Given three cochain complexes A,B, C,
which form a short exact sequence 0→ A→ B → C → 0 in the sense that for any k,
0→ Ak → Bk → Ck → 0
is a short exact sequence, then one can construct a long exact sequence of the coho-
mology groups
· · · → Hk−1(C)→ Hk(A)→ Hk(B)→ Hk(C)→ Hk+1(A)→ · · · .
Now suppose M is a smooth manifold, and U, V are open sets in M so that M =
U ∪ V . Since M,U, V and U ∩ V are all smooth manifolds, we have four de Rham
complexes,
Ω∗(M) : 0→ Ω0(M)→ Ω1(M)→ Ω2(M)→ Ω3(M)→ · · ·morespacesandspaces
Ω∗(U) : 0→ Ω0(U)→ Ω1(U)→ Ω2(U)→ Ω3(U)→ · · ·morespacesandspaces
Ω∗(V ) : 0→ Ω0(V )→ Ω1(V )→ Ω2(V )→ Ω3(V )→ · · ·morespacesandspaces
and
Ω∗(U∩V ) : 0→ Ω0(U∩V )→ Ω1(U∩V )→ Ω2(U∩V )→ Ω3(U∩V )→ · · ·morespaces
It turns out that these complexes form a short exact sequence
0→ Ω∗(M)→ Ω∗(U)⊕ Ω∗(V )→ Ω∗(U ∩ V )→ 0.
To see this, we consider the inclusion maps
ι1 : U ↪→M, ι2 : V ↪→M
and the inclusion maps
1 : U ∩ V ↪→ U, 2 : U ∩ V ↪→ V.
These inclusion maps induce linear maps between the spaces of k-forms (and also induce
linear maps between the corresponding de Rham cohomology groups, which we use the
same notation)
αk : Ω
k(M)→ Ωk(U)⊕ Ωk(V ), ω 7→ (ι∗1ω, ι∗2ω)
and
βk : Ω
k(U)⊕ Ωk(V )→ Ωk(U ∩ V ),(ω1, ω2) 7→ ∗1ω1 − ∗2ω2.
The short exact sequence we alluded above is stated in the following proposition. The
proof is simple and is left as an exercise.
Proposition 1.1. For any k, the sequence
0 −→ Ωk(M) αk−→ Ωk(U)⊕ Ωk(V ) βk−→ Ωk(U ∩ V ) −→ 0
is a short exact sequence.
LECTURE 26: THE MAYER-VIETORIS SEQUENCE 3
¶ From short exact sequence of complexes to long exact sequence.
According to the general principle in homological algebra that we mentioned above,
we should be able to construct a long exact sequence consisting of corresponding de
Rham cohomology groups. Let’s explain the construction in our setting in detail.
The maps αk and βk defined above induces maps
αk : H
k
dR(M)→ HkdR(U)⊕HkdR(V )
and
βk : H
k
dR(U)⊕HkdR(V )→ HkdR(U ∩ V ).
To get a long exact sequence, we need to define a linear map (called the connecting
homomorphism)
δk : H
k
dR(U ∩ V )→ Hk+1dR (M).
The map δk can be defined by the standard “diagram chasing” method. In what follows
we will give an explicit construction: We fix a partition of unity {ρU , ρV } subordinate
to the cover {U, V } of M . For any ω ∈ Zk(U ∩ V ), we define
δk([ω]) := [η],
where η is the (k + 1)-form
η =
{
d(ρV ω) on U,
−d(ρUω) on V.
(Why should we define η like this? Well, we need a (k + 1)-form on M . The only way
to get such a smooth (k + 1)-form on M is to take the exterior derivative of a smooth
k-form on M . Unfortunately the zero extension of ω to M is not a smooth k-form on
M since ω could be “nonzero near the boundary of U ∩ V . P.O.U is the correct way
to “force ω to be zero near the boundaries”!)
Lemma 1.2. The map δk is well-defined.
Proof. There are many issues to be checked:
• ρV ω ∈ Ωk(U): since ρV ∈ C∞(M) and supp(ρV ) ∩ U ⊂ U ∩ V . (Note: (1) ω is
NOT defined on U \ U ∩ V . (2) ρV ω 6∈ Ωk(M).) Similarly ρUω ∈ Ωk(V ).
• η ∈ Ωk+1(M): Since ρU + ρV = 1 and since dω = 0 on U ∩ V , we see
d(ρV ω) = −d(ρUω)
on U ∩ V , so η is a well-defined smooth (k + 1)-form on M .
• η ∈ Zk+1(M): By definition we have dη = 0 on both U and V . So η ∈ Zk+1(M).
• [η] is independent of the choices of ρU and ρV : Let ρ̃U and ρ̃V be another
partition of unity subordinate to the cover {U, V }, and let η̃ be the resulting
(k + 1)-form. Then ρ̃V − ρV = ρU − ρ̃U is supported in U ∩ V . So if we let
ξ = (ρ̃V − ρV )ω,
then it is a smooth k-form on M , and by the construction, we have η̃ − η = dξ
on both U and V , and thus on M .
4 LECTURE 26: THE MAYER-VIETORIS SEQUENCE
• [η] is independent of the choices of ω: Suppose ω̃ = ω + dζ and denote the
resulting (k + 1)-from by η̃. Then
η̃ − η =
{
d(ρV dζ) on U,
−d(ρUdζ) on V.
We define
ξ =
{
−dρV ∧ ζ on U,
dρU ∧ ζ on V.
Then by the same argument above, ξ is a smooth k-form on M . Moreover,
dξ =
{
−dρV ∧ dζ = −d(ρV dζ) on U
−dρU ∧ dζ = −d(ρUdζ) on V
}
= η̃ − η.
So [η̃] = [η] and the conclusion follows. �
¶ The Mayer-Vietories Theorem.
Now we can state the main theorem:
Theorem 1.3 (Mayer-Vietories). Let U, V be open sets in M so that M = U ∪ V .
Then we have a long exact sequence
· · · δk−1−→ HkdR(M)
αk−→ HkdR(U)⊕HkdR(V )
βk−→ HkdR(U ∩ V )
δk−→ Hk+1dR (M)
αk+1−→ · · ·
Proof. One has to show
• Im(αk) = ker(βk),
• Im(βk) = ker(δk),
• Im(δk) = ker(αk+1).
This amounts to prove 6 inclusion relations. We will prove one of them and leave the
rest as exercises.
Proof of Im(βk) ⊂ ker(δk) : Suppose ω1 ∈ Zk(U), ω2 ∈ Zk(V ). Let
ω := βk(ω1, ω2) = 
∗
1ω1 − ∗2ω2 ∈ Ωk(U ∩ V ).
Then δk([ω]) = [η], where
η =
{
−d(ρV ω) = −d(ρV ω − ω1) on U,
−d(ρUω) = −d(ρUω + ω2) on V.
Note that on U ∩ V ,
ρV ω − ∗1ω1 = −ρUω − ∗2ω2.
So there is a smooth k-form ξ on M so that
ξ =
{
ρV ω − ω1 on U,
−ρUω − ω2 on V.
As a consequence, η = dξ and thus [η] = 0. �
Remark. As in the case of van Kampen’s theorem, in general knowing all HkdR(U),
HkdR(V ) and H
k
dR(U ∩ V )’s are not enough to determine the HkdR(M)’s. To determine
the HkdR(M)’s we also need the homomorphisms connecting them.
LECTURE 26: THE MAYER-VIETORIS SEQUENCE 5
2. Applications of Mayer-Vietoris sequence
¶ Application 1: The de Rham cohomology groups of the spheres.
Theorem 2.1. For n ≥ 1, HkdR(Sn) '
{
R, k = 0, n,
0, 1 ≤ k ≤ n− 1. .
Proof. Last time we have proven
H0dR(S
n) ' R and H1dR(S1) ' R.
In what follows we will prove
(1) For n ≥ 2, H1dR(Sn) = 0.
(2) For n ≥ 2, k ≥ 2, HkdR(Sn) ' Hk−1dR (Sn−1).
Obvious these results together imply the theorem.
For n ≥ 2, we let
U = Sn − {(0, · · · , 0,−1)} and V = Sn − {(0, · · · , 0, 1)}.
Then
• M = U ∪ V ,
• U and V are diffeomorphic to Rn,
• U ∩ V is homotopy equivalent to Sn−1.
To prove (1), we look at the beginning of the Mayer-Vietories sequence
0 −→ H0dR(Sn) −→ H0dR(U)⊕H0dR(V ) −→ H0dR(U∩V ) −→ H1(Sn)→ H1dR(U)⊕H1dR(V ),
which now becomes
0 −→ R α0−→ R2 β0−→ R δ0−→ H1dR(Sn)→ 0.
Since α0 is injective,
dim ker(β0) = dim Im(α0) = 1.
It follows that
dim Im(β0) = dimR2 − dim ker(β0) = 1,
i.e. β0 is surjective. So ker(δ0) = R, i.e. δ0 ≡ 0. But by exactness, δ0 is surjective.
This implies H1dR(S
n) = 0.
To prove (2), we look at the following part of the Mayer-Vietories sequence
Hk−1dR (U)⊕H
k−1
dR (V )
βk−1−→ Hk−1dR (U ∩ V )
δk−1−→ HkdR(Sn)
αk−→ HkdR(U)⊕HkdR(V ),
which becomes
0
βk−1−→ Hk−1dR (S
n−1)
δk−1−→ HkdR(Sn)
αk−→ 0.
By exactness, the map δk−1 has to be both injective and surjective, and thus must be
a linear isomorphism. This proves (2). �
As a consequence, we get a simple proof of the topological invariance of the dimen-
sion:
6 LECTURE 26: THE MAYER-VIETORIS SEQUENCE
Corollary 2.2 (Topological Invariance of Dimension). If m 6= n, then Rn is not home-
omorphic to Rm.
Proof. If f : Rn → Rm is a homeomorphism, then f : Rn \ {0} → Rm \ {f(0)} is a
homeomorphism. So
HkdR(Rn \ {0}) = HkdR(Rm \ {f(0)}, ∀k.
But Rn\{0} is homotopy equivalent to Sn−1, while Rm\{f(0)} is homotopic equivalent
to Sm−1. So
HkdR(S
m−1) = HkdR(S
n−1), ∀k.
This contradicts with the fact m 6= n. �
¶ Application 2: dimHkdR(M) <∞ for many smooth manifolds.
Definition 2.3. Let M be a smooth manifold and {Uα}α∈Λ an open cover of M . We
say {Uα}α∈Λ is a good cover if for any finite subset I = {α1, · · · , αk} ⊂ Λ of indices,
the intersection
UI := Uα1 ∩ Uα2 ∩ · · · ∩ Uαk
is either empty or diffeomorphic to Rn.
Example. The cover of Sn by Sn − {N} and Sn − {S} is not a good cover. However,
the cover of Sn by the 2n+ 2 hemispheres
Sni,+(−) = {(x1, · · · , xn+1) | xi > (<)0}
is a good cover (why?).
By using Riemannian geometry (more precisely, use the so-called geodesically con-
vex neighborhoods.), one can show that any open cover of any smooth manifold M
admits a refinement which is a good cover. In particular, if M is compact, then M
admits a good cover which contains only finitely many sets. Such a cover is called a
finite good cover.
Obviously any sub-cover of a good cover is still a good cover.
Theorem 2.4. If M admits a finite good cover, then HkdR(M) is finite dimensional.
Proof. We proceed by induction on the number of sets in a finite good cover of M . If
M admits a good cover that contains only one open set, then that open set has to be
M itself. In this case M is diffeomorphic to Rn and the conclusion follows.
Now suppose the theorem holds for any manifold that admits a good cover con-
taining l − 1 open sets. Let M be a manifold with an good cover {U1, · · · , Ul}. We
denote
U = U1 ∪ · · · ∪ Ul−1 and V = Ul.
Then U ∩ V admits a finite good cover {U1 ∩ Ul, · · · , Ul−1 ∩ Ul}. By the induction hy-
pothesis, all the de Rham cohomology groups of U , V and U ∩V are finite dimensional.
Now consider the Mayer-Vertoris sequence
· · · −→ Hk−1dR (U ∩ V )
δk−1−→ HkdR(M)
αk−→ HkdR(U)⊕HkdR(V ) −→ · · · .
LECTURE 26: THE MAYER-VIETORIS SEQUENCE 7
The conclusion follows since
dim Im(αk) ≤ dimHkdR(U)⊕HkdR(V ) <∞
and
dim ker(αk) = dim Im(δk−1) ≤ dimHk−1dR (U ∩ V ) <∞. �
As a consequence, we immediately get
Corollary 2.5. If M is compact or M is homotopy equivalent to a compact manifold,
then dimHkdR(M) <∞ for all k.
¶ Application 3: The Kunneth formula.
Theorem 2.6. LetM and N be manifolds with finite good covers. Then for any
0 ≤ k ≤ dimM + dimN , one has
HkdR(M ×N) '
k⊕
i=0
H idR(M)⊗Hk−idR (N).
Sketch of proof. Let πM : M ×N → M and πN : M ×N be the standard projections.
Then we get a map
Ψ : Ω∗(M)⊗ Ω∗(N)→ Ω∗(M ×N), ω1 ⊗ ω2 7→ π∗Mω1 ∧ π∗Nω2.
One can check that this map induces a map on cohomologies
Ψ : H∗dR(M)⊗H∗dR(N)→ H∗dR(M ×N), [ω1]⊗ [ω2] 7→ [π∗Mω1 ∧ π∗Nω2].
To prove that this Ψ is in fact an linear isomorphism, we do induction on the number
l of elements in a good cover of M .
If l = 1, i.e. M is diffeomorphic to Rn, then the Kunneth formula follows from
the fact that Rn × N is homotopy equivalent to N , and HkdR(Rn) = R for k = 0 and
HkdR(Rn) = 0 for other k’s.
Now suppose the Kunneth formula is proved for manifolds admitting a good cover
with no more than l− 1 open sets, and supppose now that M = U1 ∪ · · · ∪Ul is a good
cover. Again we let U = U1 ∪ · · ·Un−1 and V = Un. For simplicity we will denote
H̃k(M,N) :=
k⊕
i=0
H idR(M)⊗Hk−idR (N).
Consider the following diagram
H̃k(M,N) H̃k(U,N)⊕H̃k(V,N) H̃k(U∩V,N) H̃k+1(M,N)
HkdR(M×N) HkdR(U×N)⊕HkdR(V×N) HkdR((U∩V )×N) Hk+1dR (M×N)
............................................................................................................ .......
.....
α
.......................................................................................... .......
.....
β
................................................................................................................................................. .......
.....
δ
...................................................................... .......
.....
α
................................................ .......
.....
β
...................................................................................................................... .......
.....
δ
............................................................
.....
.......
.....
Ψ
............................................................
.....
.......
.....
Ψ
............................................................
.....
.......
.....
Ψ
............................................................
.....
.......
.....
Ψ
where the horizontal maps α, β, δ’s are the ones that is induced in the obvious way
from the αk, βk, δk’s that we defined above. For this diagram we have
• By using the Mayer-Vietoris sequence one can prove the two rows are exact.
8 LECTURE 26: THE MAYER-VIETORIS SEQUENCE
• Moreover, one can prove that the diagram is commutative, i.e. we have Ψ◦α =
α ◦ Ψ, Ψ ◦ β = β ◦ Ψ and Ψ ◦ δ = δ ◦ Ψ. [The first two equalities are easy to
prove, while the last one is much more complicated.]
• By using the induction hypothesis, the second and the third Ψ are linear iso-
morphisms.
Now the result follows from the well-known Five lemma in homological algebra. �
Lemma 2.7 (Five lemma). Suppose we have the following commutative diagram
V1 V2 V3 V4 V5
W1 W2 W3 W4 W5
................................................................................................................. .......
.....
α1
................................................................................................................. .......
.....
α2
................................................................................................................. .......
.....
α3
................................................................................................................. .......
.....
α4
................................................................................................................. .......
.....
β1
................................................................................................................. .......
.....
β2
................................................................................................................. .......
.....
β3
................................................................................................................. .......
.....
β4
..........................................................
.....
.......
.....
γ1
..........................................................
.....
.......
.....
γ2
..........................................................
.....
.......
.....
γ3
..........................................................
.....
.......
.....
γ4
..........................................................
.....
.......
.....
γ5
where each Vi and each Wi is a linear space, and each map is a linear map. Suppose
the two rows are exact, and the vertical maps γ1, γ2, γ4, γ5 are isomorphisms, then the
map γ3 is also an isomorphism.
Proof. This is a standard diagram-chasing exercise. �
As a simple consequence of the Kunneth formula, we can compute
HkdR(S
n × Sn) '
 R, k = 0 or 2n,R2, k = n,0, other k.
Similarly one can calculate HkdR(S
m × Sn) for m 6= n.
We can also compute the de Rham cohomology groups of the n-torus Tn.
Proposition 2.8. For the n-dimensional torus Tn = S1 × · · · × S1,
HkdR(Tn) ' R(
n
k).
The proof is left as an exercise.
LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY
1. Compactly Supported De Rham cohomology
¶ Compactly supported De Rham cohomology.
We have seen that if a smooth manifold is compact (or homotopic equivalent to a compact
manifold), then the de Rham cohomology groups are “simpler”: They are finite dimensional,
and nice formulae like Kunneth formula holds. As we will see today, when M is orientable,
a very useful tool to study cohomology classes, especially the top classes, is “integration
on manifolds”. Unfortunately, if M is non-compact, the integration of a top form is not a
nicely defined unless the differential form is compactly supported. Recall that for ω ∈ M ,
the support of ω is defined to be
supp(ω) = {p ∈M | ωp 6= 0}.
As usual, we say ω is compactly supported if supp(ω) is compact in M . It’s natural to let
Ωkc (M) = {ω ∈ Ωk(M) | ω is compactly supported}
be the set of all compactly supported smooth k-forms. Obviously
(1) if ω1, ω2 are compactly supported k-forms, so is c1ω1 + c2ω2;
(2) if ω is compactly supported, so is dω.
So Ωkc (M)’s are vector spaces, and the exterior derivative makes these vector spaces a cochain
complex
0→ Ω0c(M)
d−→ Ω1c(M)
d−→ Ω2c(M)
d−→ Ω3c(M)
d−→ · · · .
As in the ordinary de Rham theory, we denote
Zkc (M) = {ω ∈ Ωkc (M) | dω = 0},
Bkc (M) = {ω ∈ Ωkc (M) | ω = dη for some η ∈ Ωk−1c (M)}.
Definition 1.1. The k-th de Rham cohomology group with compact supports of M is
Hkc (M) =
Zkc (M)
Bkc (M)
.
Obviously if M is compact, then Ωkc (M) = Ω
k(M) for all k, and thus
Hkc (M) = H
k
dR(M), ∀k.
Remark. As in the ordinary case, we can define a cup product
∪ : Hkc (M)×H lc(M)→ Hk+lc (M), (ω, η) 7→ [ω ∧ η]
which makes H∗c (M) = ⊕nk=1Hkc (M) a graded ring. (In fact, the cup product is also well-
defined as ∪ : Hkc (M)×H ldR(M)→ Hk+lc (M) and as ∪ : HkdR(M)×H lc(M)→ Hk+lc (M).)
1
2 LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY
¶ Main difference between H∗c (M) and H∗dR(M).
In what follows we indicate the main differences between the compact supported de
Rham cohomology groups and the ordinary de Rham cohomology groups.
(1) By definition we have Zkc (M) = Z
k(M) ∩ Ωkc (M). However, in general,
Bkc (M) 6= Bk(M) ∩ Ωc(M). (Can you find an example?)
(2) For k = 0, by definition
H0c (M) = Z
0
c (M) = {f ∈ C∞(M) | df = 0 and supp(f) is compact}.
But df = 0 if and only if f is locally constant, i.e. f is constant on each connected
component. On the other hand, a locally constant compactly supported function has
to be zero on any non-compact connected component. So we conclude
H0c (M) ' Rmc ,
where mc is the number of compact connected components of M . In particular,
H0c (pt) = R and H0c (Rn) = 0, ∀n ≥ 1,
where pt is a one point set. Since Rn is homotopy equivalent to {pt}, we conclude
Hkc (M)’s are no longer homotopy invariants.
(3) Now let ϕ : M → N be a smooth map. Then by definition,
supp(ϕ∗ω) ⊂ ϕ−1(supp(ω)).
So if ω ∈ Ωkc (N), in general we may have ϕ∗ω 6∈ Ωkc (M). In particular,In general we can not pull back compactly-supported cohomology classes
on N to compactly-supported cohomology classes on M !
In the ordinary theory, we used the pull-back to prove the homotopy invariance and to
construct the M-V sequence. It turns out that in the “compactly-supported theory”,
we can still do two rescues, one can be used to get a partial rescue on homotopy
invariance, while the other can be used to construct a new M-V sequence.
Rescue A. If ϕ : M → N is proper, then the pull-back ϕ∗ω of a compactly supported
differential form ω ∈ Ωkc (N) is still compactly supported. So the map
ϕ∗ : Hkc (N)→ Hkc (M)
is still well-defined. In this case one can prove (exercise)
Theorem 1.2. If ϕ0, ϕ1 : M → N are proper smooth maps that are properly
homotopic1, then the induced maps
ϕ∗1 = ϕ
∗
2 : H
k
c (N)→ Hkc (M).
Note that any homeomorphism is proper. So in particular the compactly sup-
ported de Rham cohomology groups are still topological invariants:
Corollary 1.3. If M is homeomorphic to N , then Hkc (M) = H
k
c (N).
1i.e. there exists a homotopy Φ : M × [0, 1]→ N connecting ϕ0 and ϕ1 that is a proper map.
LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 3
Rescue B. If U is an open subset of M and ι : U ↪→ M is the inclusion map, then we can
define a “push-forward” map
ι∗ : Ω
k
c (U)→ Ωkc (M)
which extend a compactly supported k-form on U to a compactly supported
k-form on M by performing the “zero-extension”. It is easy to see dι∗ = ι∗d, so
ι∗ descends to a linear map
ι∗ : H
k
c (U)→ Hkc (M).
Using the push-forward, one can construct:
Theorem 1.4 (The Mayer-Vietories sequence for H∗c ). Suppose U , V are open
subsets in M so that M = U ∪ V . Then there exists linear maps δck : Hkc (M)→
Hk+1c (U ∩ V )2 so that the following sequence is exact
spacespacess · · · δk−1−→ Hkc (U∩V )
αck−→ Hkc (U)⊕Hkc (V )
βck−→ Hkc (M)
δck−→ Hk+1c (U∩V )
αck+1−→ · · · ,
where αck and β
c
k are given by
spacespaceαck([ω]) = ((1)∗[ω],−(2)∗[ω]), βck([ω1], [ω2]) = (ι1)∗[ω1] + (ι2)∗[ω2].
[Note that on each level k, the “direction” of this new M-V sequence is opposite
to that of the M-V sequences for the usual de Rham cohomology groups.]
¶ Example: Hkc (Rm) for k < m.
Suppose m ≥ 1. We have seen H0c (Rm) = 0. Now we prove Hkc (Rm) = 0 for 1 ≤ k < m.
We identify Rn with Sm−{N}, where N is the north pole. Then we get an “inclusion” map
ι : Rm → Sm, and the words “compactly supported in Rm” is equivalent to “supported in a
subset of Sm that is away from N”.
• Case 1: 1 = k < m. Take any ω ∈ Z1c (Rm). Then ι∗ω ∈ Z1(Sm) which is supported
in Sm − U for some neighborhood U of p. Since H1(Sm) = 0, the closed 1-form ι∗ω
is exact, i.e. there exists η ∈ Ω0(Sm) = C∞(Sm) so that ι∗ω = dη. Moreover, the
fact dη = ι∗ω = 0 on U implies that η equals some constant c on U . It follows that
if we take η̃ = η − c, then η̃ ∈ Ω0c(Sm − {N}) = Ω0c(Rm) and dη̃ = ω.
• Case 2: 1 < k < m Again we take ω ∈ Zkc (Rm) and consider ι∗ω ∈ Zk(Sm), which
is supported in some Sm − U . Since HkdR(Sm) = 0, one can find η ∈ Ωk−1(Sm) such
that ι∗ω = dη. By shrinking the neighborhood U of p, we can assume that U is
contractible. Then the fact dη = ι∗ω = 0 in U implies that η is exact in U , i.e. one
can find a µ ∈ Ωk−2(U) such that η = dµ. Now one pick a bump function ρ on Sm
which vanishes on Sm − U and equals 1 near p. Then η̃ = η − d(ρµ) ∈ Ωk−1(Sm)
and η̃ = 0 near p, i.e. it defines a compactly supported (k − 1)-form on Rm. By
construction, dη̃ = dη = ω.
2The coboundary morphism δck is defined as follows: For any ω ∈ Zkc (M), we have ρUω ∈ Ωkc (U), ρV ω ∈
Zkc (V ). Since d(ρUω) = −d(ρV ω), we have d(ρUω) ∈ Zk+1c (U ∩ V ) and one just defines δck([ω]) := [d(ρUω)].
4 LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY
2. The de Rham cohomology groups of top degree
In this section we would like to calculate Hmc (M) and H
m
dR(M) for m-manifold M .
¶ Example: H1c (R).
We start with an example.
Example. Let’s try to compute H1c (R). To do so we consider the integration map∫
R
: Z1c (R) = Ω1c(R)→ R, ω 7→
∫
R
ω.
This map is clearly linear and surjective. Moreover, it vanishes on B1c (R) (by the fundamental
theorem of calculus), so it induces a surjective linear map∫
R
: H1c (R)→ R.
Moreover, if
∫
R f(t)dt = 0, where f ∈ C
∞
c (R), then the function g(t) =
∫ t
−∞ f(τ)dτ is smooth
and compactly supported and dg = f(t)dt. In other words, f(t)dt ∈ B1c (R), i.e. [f(t)dt] = 0
in H1c (R). So
∫
R is an isomorphism between H
1
c (R) and R, i.e.
H1c (R) ' R.
[Compare with the example at the bottom of page 1 in Lecture 25.]
¶ The integration map.
Essentially the same method works in higher dimension. Let M be any m-dimensional
connected oriented (= orientable with an orientation chosen) manifold, and ω ∈ Ωmc (M) a
compactly supported top form. Then ω is closed, and we have defined the integral
∫
M
ω. So
we get a map ∫
M
: Ωmc (M)→ R, ω 7→
∫
M
ω.
Now suppose ω ∈ Bmc (M), i.e. ω = dη for some η ∈ Ωm−1c (M). We can take a compact
set K in M with smooth boundary, so that K ⊃ supp(η). Then by the Stokes’ formula,∫
M
ω =
∫
M
dη =
∫
K
dη =
∫
∂K
η = 0.
So
∫
M
induces a linear map ∫
M
: Hnc (M)→ R, [ω] 7→
∫
M
ω.
Proposition 2.1. Suppose M is oriented. Then the map
∫
M
: Hmc (M)→ R described above
is surjective.
Proof. Fix a volume form ω on M . For any c, one can find a smooth function f that is
compactly supported in a coordinate chart U , such that
∫
fω = c. �
LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 5
¶ Poincaré lemma for compactly supported de Rham cohomology.
The surjectivity of the integration map has an important consequence:
Corollary 2.2. If ω ∈ Ωm(Sm) and
∫
Sm
ω = 0, then ω is exact.
Proof. We have seen that the map
∫
M
is linear and surjective. But
Hmc (S
m) = Hm(Sm) ' R.
So
∫
M
must be a linear isomorphism. In other words, if
∫
Sm
ω = 0, then [ω] = 0, i.e. ω is
exact. �
Using this, we can prove:
Theorem 2.3 (Poincaré lemma for Hkc ). H
k
c (Rm) =
{
R, k = m,
0, k 6= m.
Proof. It remains to prove: for m = k ≥ 2, the surjective linear map∫
Rm
: Hmc (Rm)→ R.
is an isomorphism. More precisely, we need to prove that if
∫
Rm ω = 0 for some ω ∈
Ωmc (Rm)= Zmc (Rm), then ω ∈ Bmc (Rn). As before we consider the “inclusion” map ι : Rm →
Sm. Then for any ω ∈ Ωmc (Rm), we have ι∗ω ∈ Ωm(Sm). Since∫
Sm
ι∗ω =
∫
Rm
ω = 0,
By Corollary 2.2 we see ι∗ω = dη for some η ∈ Ωm−1(Sm).
The rest of proof is identically the same as before: We take an open contractible neigh-
borhood U of p in Sm on which ω vanishes. Then we adjust η to η̃ = η − d(ρµ), where ρ
is a bump function and µ ∈ Ωm−2(U) is such that dµ = η on U . Then η̃ is a compactly
supported (m− 1)-form on Rm and dη̃ = ω. �
As an immediate corollary, we get
Corollary 2.4. If M admits a finite good cover, then dimHkc (M) <∞ for all k.
Proof. Use Mayer-Vertoris sequence and induction and the number of open sets in a good
cover. The same as the proof for the ordinary de Rham cohomology that we did last time. �
¶ Hmc (M) for orientable M .
Now we prove
Theorem 2.5. For any m-dimensional connected orientable manifold M , the map
∫
M
:
Hmc (M)→ R is an isomorphism. In particular, Hmc (M) ' R.
6 LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY
Proof. Since
∫
M
is linear and surjective, it remains to prove
∫
M
is injective, i.e.
(1) ω ∈ Ωmc (M) such that
∫
M
ω = 0 =⇒ ω = dµ for some µ ∈ Ωm−1c (M).
To prove this, we can use induction on the number of open sets that is needed to cover the
support of ω by a good cover. More precisely, since M is connected and supp(ω) is compact,
we can take a connected compact set Kω ⊃ supp(ω). If we can cover Kω by a good cover
which contains only one chart, then the Poincaré lemma implies (1). Now suppose (1) holds
for any ω ∈ Ωmc (M) such that Kω can be covered by k − 1 “good charts”, and suppose
ω ∈ Ωmc (M) satisfies the property that Kω admits a good cover {U1, · · · , Uk}. There exists
one Ui, say Uk for simplicity, such that both U = U1 ∪ · · · ∪Uk−1 and V = Uk are connected.
(Reason: Onecan form a graph G of k vertices v1, · · · , vk, corresponding to U1, · · · , Uk
respectively, so that vi is connected to vj if and only if Ui ∩ Uj 6= ∅. Then this graph G
is a connected graph. By graph theory, one can always delete one vertex of a connected
graph so that the remaining graph is still connected: one only need choose a spanning tree
of the graph, and delete a “leaf” of the tree.) We pick a partition of unity {ρU , ρV } of U ∪V
subordinate to the cover {U, V }, and let
ωU = ρUω, ωV = ρV ω.
Since Kω is connected, U ∩ V 6= ∅. We pick an m-form ω0 compactly supported in U ∩ V so
that ∫
M
ω0 =
∫
M
ωU .
Then ωU − ω0 is compactly supported in U which is connected and admits a good cover of
k − 1 good charts, and
∫
M
(ωU − ω0) = 0. So by the induction hypothesis,
ωU − ω0 = dηU
for some ηU ∈ Ωm−1c (M). Similarly
∫
M
(ωV + ω0) = −
∫
M
ωU +
∫
M
ω0 = 0 implies
ωV + ω0 = dηV
for some ηV ∈ Ωm−1c (M). It follows that
ω = ωU + ωV = d(ηU + ηV ),
where ηU + ηV ∈ Ωm−1c (M). �
As a corollary, we get
Corollary 2.6. Let M be compact, connected, orientable and dimM = m. Then
HmdR(M) ' R.
LECTURE 27: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 7
¶ HmdR(M) for orientable non-compact M .
For orientable non-compact manifolds, one has
Theorem 2.7. For any non-compact connected orientable manifold M of dimension m,
HndR(M) = 0
Proof. In PSet 1-2-4 we have seen that for any smooth manifold M , there exists a positive
smooth exhaustion function f . In other words, f is a smooth function on M so that for any
c, the sub-level set f−1((−∞, c]) is compact in M . By adding a constant, we may assume
inf f = 0. Since M is connected and non-compact, we must have f(M) = [0,∞). Now for
each positive integer k we let Vk = f
−1((k−2, k)). Then {Vk} is a locally finite open covering
of M . Let {ρk} be a partition of unity subordinate to this covering. For each k, we choose
some ηk ∈ Ωmc (Vk ∩ Vk+1) ⊂ Ωmc (M) so that
∫
M
ηk = 1.
For any ω ∈ Ωm(M), we let ωk = ρkω ∈ Ωmc (Vk) and c1 =
∫
V1
ω1. Then ω1−c1η1 ∈ Ωmc (V1)
and ∫
V1
(ω1 − c1η1) = 0.
So by Theorem 2.5, one can find µ1 ∈ Ωm−1c (V1) so that
dµ1 = ω1 − c1η1.
Next we choose c2 ∈ R so that
∫
V2
(ω2 + c1η1 − c2η2) = 0, and conclude that there exists
µ2 ∈ Ωm−1c (V2) such that
dµ2 = ω2 + c1η1 − c2η2.
Continuing this process, we can find a sequence ck ∈ R and µk ∈ ωm−1c (Vk) so that
dµk = ωk + ck−1ηk−1 − ckηk.
Let µ =
∑
µk. Note that this is a locally finite sum, and thus defines an element in Ω
m−1(M).
Moreover, by construction we have
dµ = d
∑
µk =
∑
ωk = ω. �
Remark. If M is a non-orientable connected smooth manifold of dimension m, then
HmdR(M) = 0 and H
m
c (M) = 0.
For a proof, c.f. Lee, page 456-457.
In summary: suppose M is a connected smooth manifold of dimension m. Then
HmdR(M) '
{
R, M is compact and orientable,
0, M is non-compact or non-orientable
while
Hmc (M) '
{
R, M is orientable,
0, M is non-orientable.
LECTURE 28: APPLICATIONS OF DE RHAM THEORY
1. The mapping degree and its applications
¶ The mapping degree.
Let M,N be m-dimensional connected oriented manifolds, and f : M → N a proper
smooth map. Then the pull-back map
f ∗ : R ' Hmc (N) −→ Hmc (M) ' R
is linear, and thus is a map of the form λ 7→ cλ. The constant c is called the degree of f :
Definition 1.1. The degree of a proper smooth map f : M → N is the number deg(f) s.t.∫
M
f ∗ω = deg(f)
∫
N
ω, ∀ω ∈ Ωmc (N).
Remark. The isomorphism Hmc (N) ' R is induced by the integral
∫
N
and thus does depend
on the choices of orientations: with respect to opposite orientations, the same ω will give us
opposite real numbers. Thus to define deg(f) we need to fix orientations on M and N first.
(However, if M is orientable, and f : M → M is a map from M to itself, then deg(f) is
independent of the choices of the orientations on M . )
We have
Proposition 1.2. Let M,N,P be connected oriented manifolds of the same dimension.
(1) If f : M → N and g : N → P are both proper smooth maps, then g ◦ f is proper, and
deg(g ◦ f) = deg(f)deg(g).
(2) If f : M → N and g : M → N are properly homotopic maps , then deg(f) = deg(g).
(3) If f : M → N is a diffeomorphism, then
deg(f) =
{
1, f is orientation preserving,
−1, f is orientation reversing.
Proof. (1) follows from definition, (2) follows from the fact f ∗ = g∗ on compactly supported
cohomology groups, and (3) follows from the change of variable formula. �
Remark. We have seen that any continuous map is homotopic to a smooth map, and homo-
topic smooth maps are smoothly homotopic. Moreover, under the compactness assumption,
any homotopy is a proper homotopy. So if the manifolds under consideration are compact,
we can define the degree of a continuous map to be the degree of the corresponding smooth
maps, which is independent of the choices of the smooth approximation. In fact, in algebraic
topology, the mapping degree theory is developed for continuous maps: in that case there is
no smoothness at all.
1
2 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
¶ Computing the mapping degree.
We give a couple examples to compute the mapping degree.
Example. Consider the antipodal map
f : Sn → Sn, p 7→ f(p) = −p.
Then according to PSet 7-1-5(e),
deg(f) =
{
1, n is odd,
−1, n is even.
It turns out that deg(f) = 0 as long as f is not surjective:
Proposition 1.3. If a proper smooth map f : M → N is not surjective, then deg(f) = 0.
Proof. According to PSet 3-1-5(a), f is a closed map. So if q 6∈ f(M), then one can find an
open neighborhood Ũ of q so that Ũ ∩ f(M) = ∅. We pick an m-form ω supported in Ũ so
that
∫
N
ω = 1. Since by our construction, f ∗ω = 0, we conclude deg(f) = 0. �
In fact, we have the following remarkable property of the degree:
deg(f) is always an integer!
To see this, we may assume that f is surjective. Let q ∈ N be a regular value of f . According
to PSet 3-1-5(c), f−1(q) = {p1, · · · , pk} is a finite set, and there exists a neighborhood Ũ of
q and neighborhoods Ui of each pi, so that
• For i 6= j, Ui ∩ Uj = ∅, and f−1(Ũ) = ∪ki=1Ui.
• f maps Ui diffeomorphically to Ũ .
Obviously we can choose U and Ui’s to be small connected oriented charts. We let
σi =
{
1, if f : Ui → Ũ is orientation preserving,
−1, if f : Ui → Ũ is orientation reversing.
The fact deg(f) is an integer follows from
Theorem 1.4. The degree of f is the integer
deg(f) =
k∑
i=1
σi.
Proof. Take ω ∈ Ωnc (Ũ) so that
∫
N
ω = 1. Then f ∗ω is supported in f−1(Ũ) = ∪ki=1Ui, and∫
M
f ∗ω =
k∑
i=1
∫
Ui
f ∗ω =
k∑
i=1
σi
∫
Ũ
ω =
k∑
i=1
σi.
�
Example. The degree of the map
f : C→ C, z 7→ f(z) = zk
is deg(f) = k.
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 3
¶ Application 1: The Hairy Ball Theorem.
Theorem 1.5. Even dimensional spheres do not admit non-vanishing smooth vector fields.
Proof. Suppose X is a non-vanishing smooth vector field on S2n ⊂ R2n+1. By normalizing
the vectors, we may assume |Xp| = 1 for all p ∈ S2n. We will think of p and Xp as vectors
in R2n+1, and consider the map F : S2n × [0, 1]→ S2n defined by
F (p, t) = p cos(tπ) +Xp sin(tπ).
Then for each t ∈ [0, 1], F (·, t) is a map from S2n to S2n. (Note: |p| = |Xp| = 1 and
p ⊥ Xp.) So F is a homotopy between the identity map F (·, 0) = IdS2n and the antipodal map
F (·, 1) = f : S2n → S2n, f(x) = −x. So −1 = deg(f) = deg(IdS2n) = 1, a contradiction. �
As a consequence, we get
Corollary 1.6. For any n ≥ 1, S2n admits no Lie group structure.
¶ Application 2: The Brouwer Fixed Point Theorem.
The degree is a topological obstruction to extend a map defined on the boundary to a
map defined in the interior:
Proposition 1.7. Suppose M is an m-dimensional oriented compact manifold with smooth
and connected boundary ∂M , X a connected oriented (m− 1)-manifold, and f : ∂M → X a
smooth map that extends to a smooth map g : M → X. Then deg(f) = 0.
Proof. Pick ω ∈ Ωm−1(X) so that
∫
X
ω = 1. Then
deg(f) = deg(f)
∫
X
ω =
∫
∂M
f ∗ω =
∫
∂M
ι∗g∗ω =
∫
M
d(g∗ω) =
∫
M
g∗dω = 0,
where ι : ∂M →M is the inclusion map, and in the last step we used thefact dω = 0. �
Corollary 1.8 (Brouwer Fixed Point Theorem). Every continuous map from Bm (=the unit
ball in Rm) to itself has a fixed point.
Proof. Let F0 : B
m → Bm be a continuous map without fixed point. Take a positive number
0 < r < inf
p∈Bn
|p− F0(p)|/3.
Then by Whitney approximation theorem, there exists a smooth map F1 : B
m → Rm s.t.
|F0(p)− F1(p)| < r, ∀p ∈ Bm.
Define F : Bm → Bm by F (p) = F1(p)/(1 + r). Then F : Bm → Bm is a smooth map. F
has no fixed point since
|F (p)− F0(p)| ≤ |F (p)− F1(p)|+ |F1(p)− F0(p)| < |F (p)|r + r ≤ 2r < |F0(p)− p|.
It follows that the smooth map
G : Bn → Sn−1, p 7→ p− F (p)
|p− F (p)|
4 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
is the extension of the smooth map
g = G|Sn−1 : Sn−1 → Sn−1.
By Proposition 1.7, deg(g) = 0.
On the other hand, the map
H : Sn−1 × [0, 1]→ Sn−1, (p, t) 7→ p− tF (p)
|p− tF (p)|
is a homotopy between the identity map and g. So deg(g) = deg(Id) = 1. Contradiction. �
¶ More Applications.
Mapping degree has many other applications. For example,
• Given any two non-intersecting smooth curve γi : S1 → R3(i = 1, 2), we can define
the linking number Link(γ1, γ2) to be
Link(γ1, γ2) := deg(Γγ1,γ2),
where Γγ1,γ2 is the Gauss map
Γγ1,γ2 : T2 → S2, (eis, eit) 7→
γ1(e
is)− γ2(eit)
|γ1(eis)− γ2(eit)|
.
Geometrically, the linking number represents the number of times that each curve
winds around the other, which may be positive or negative since we count the orien-
tation of the two curves.
• One can also use the mapping degree to prove the fundamental theorem of algebra:
f(z) = zn + a1z
n−1 + · · ·+ an−1z + an
be any polynomial (with complex coefficients). We can regard f as a map S2 from
S2. (how?) Then one can prove that the mapping degree of f is n (by showing that
f ∼ f0 and computing deg(f0), where f0(z) = zn. ) So f admits at least one complex
root. (Why?)
• We know that two properly homotopic maps have the same degree. Conversely degree
can be used to characterize whether two maps are homotopic:
Theorem 1.9 (Hopf degree theorem). Let M be a compact connected oriented man-
ifold of dimension m. Then two maps f, g : M → Sm are homotopic if and only if
deg(f) = deg(g).
• One can prove that if f : Sm → Sm is an odd map, namely f(−p) = −f(p) for all
p ∈ Sm, then the degree of f is odd. This implies
Theorem 1.10 (Borsuk-Ulam theorem). For any continuous map f : Sm → Rm,
there exists x0 such that f(−x0) = f(x0).
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 5
2. The Poincaré duality and its applications
¶ The Poincaré duality.
Let M be an oriented manifold of dimension n. We have the following maps
• ∪ : HkdR(M)×H lc(M)→ Hk+lc (M), ([ω], [η]) 7→ [ω ∧ η].
•
∫
M
: Hnc (M)→ R, [ω] 7→
∫
M
ω.
For any 0 ≤ k ≤ n, consider the bilinear pairing map
P kM : H
k
dR(M)×Hn−kc (M)→ R, P kM([ω], [η]) =
∫
M
ω ∧ η.
The map P kM induces the following Poincaré duality operator
PkM : HkdR(M)→
(
Hn−kc (M)
)∗
, Pk([ω]) =
{
η 7→
∫
M
ω ∧ η
}
.
For example, if M is connected, P0M maps the element 1 ∈ R ' H0dR(M) to the linear map∫
M
: Hnc (M)→ R, η 7→
∫
M
η
on Hnc (M), so that one can think of
∫
M
as an element in (Hnc (M))
∗.
The major theorem we would like to discuss in this section is
Theorem 2.1 (Poincaré duality). For any oriented manifold M and any k, the Poincaré
duality map PkM is a linear isomorphism from HkdR(M) to
(
Hn−kc (M)
)∗
.
Remark. If dimHm−kc (M) <∞, then
(
Hn−kc (M)
)∗
is isomorphic to Hn−kc (M). So we get
HkdR(M) ' Hn−kc (M).
Example. For M = Rn, we have
HkdR(Rn) '
{
R, k = n,
0, k 6= n, and H
k
c (Rn) '
{
R, k = 0,
0, k 6= 0.
Example. For M = Sn, we have Hkc (S
n) = HkdR(S
n) '
{
R, k = 0, n,
0, k 6= 0, n.
Example. For any connected oriented manifold of dimension n, we have already seen
H0dR(M) ' R ' Hnc (M)
and
HndR(M) '
{
R, M is compact
0, M is non-compact
}
' H0c (M).
Example. Let M = ∪i∈N(i, i+ 1) be a countable union of disjoint open intervals. Then
H0dR(M) '
∏
i∈N
R = {(a1, a2, · · · ) | ai ∈ R}
6 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
and (Note: H1c ((i, i+ 1)) ' R.)
H1c (M) '
⊕
i∈N
R = {(a1, a2, · · · ) | ai ∈ R, all but finitely many a′is are zero }.
A well-known fact (but non-trivial) in algebra:(⊕
i∈N
R
)∗
=
∏
i∈N
R while
(∏
i∈N
R
)∗
6=
⊕
i∈N
R.
So we have Poincaré duality
HkdR(M) '
(
Hn−kc (M)
)∗
.
But in general (
HkdR(M)
)∗ 6' Hn−kc (M).
¶ The Poincaré duality: Sketch of a proof.
In what follows we will sketch a proof of Poincaré duality for oriented manifolds admitting
a finite good cover, although the theorem holds for any oriented manifold. We need
Lemma 2.2. The following diagram commutes:
HkdR(M) H
k
dR(U)⊕HkdR(V ) HkdR(U ∩ V ) Hk+1dR (M)· · · · · ·
Hn−kc (M)
∗ Hn−kc (U)
∗ ⊕Hn−kc (V )∗ Hn−kc (U ∩ V )∗ Hn−k−1c (M)∗· · · · · ·
.................................................................................................... .......
.....α ........................................................................... ............
β
.................................................................................................................................. .......
.....
(−1)k+1δ
............................................................... .......
..... ....................................................... .......
.....
.............................................................................................
.....
.......
.....
PkM
.............................................................................................
.....
.......
.....
PkU ⊕ PkV
.............................................................................................
.....
.......
.....
PkU∩V
.............................................................................................
.....
.......
.....
Pk+1M
....................................................... .......
.....
α∗
................................. .......
.....
β∗
.................................................................................................. .......
.....
δ∗
................................................ .......
..... ................................... .......
.....
where the bottom row is the dual of the Mayer-Vietoris sequence for compactly supported de
Rham cohomology groups.
Sketch of proof of Poincaré duality for oriented manifolds M admitting a finite good cover.
We proceed by induction. The theorem holds for M admitting one good chart, in which case
M ' Rn, and the isomorphism follows from the two versions of Poincaré lemma that we
proved.
Now suppose the theorem holds for manifolds admitting a good cover of no more than
k − 1 open sets, and suppose M admits a good cover {U1, · · · , Uk}. We let
U = U1 ∪ · · · ∪ Uk−1 and V = Uk.
Then U, V and U ∩ V all admit a good cover of no more than k− 1 open sets. By induction
hypothesis, PkU , PkV and PkU∩V are all isomorphisms. By the above lemma and the Five
Lemma (see Lecture 26), PkM is an isomorphism. �
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 7
¶ Application 1: The Künneth formula for Hkc (M ×N).
As a corollary, we get (the result actually holds for any manifold M)
Corollary 2.3. For any oriented manifold M whose compact supported cohomology groups
are finite dimensional,
Hk+lc (M × Rl) ' Hkc (M).
Proof. By Poincaré duality and homotopy invariance of the de Rham cohomology groups,
Hk+lc (M × Rl) ' Hn+l−k−ldR (M × R
l) ' Hn−kdR (M) ' H
k
c (M). �
More generally, one can prove (the result actually holds for any manifold M)
Corollary 2.4 (Künneth formula). If M,N are orientable with finite good cover, then
Hkc (M ×N) '
k⊕
i=0
H ic(M)⊗Hk−ic (N).
Proof. By Poincaré Duality and the Künneth formula for the de Rham cohomology groups,
Hkc (M ×N) ' Hm+n−kdR (M ×N) '
m+n−k⊕
i=0
H idR(M)⊗Hm+n−k−idR (N).
The indices i should satisfy i ≤ m and m+ n− k − i ≤ n, i.e. m− k ≤ i ≤ m. Thus
Hkc (M ×N) '
m⊕
i=m−k
H idR(M)⊗Hm+n−k−idR (N) '
k⊕
i=0
Hm−idR (M)⊗H
n−k+i
dR (N).
Now the result follows from the Poincaré duality. �
¶ Application 2: The Betti numbers andthe Euler characteristics.
We also give a couple applications to the Betti numbers and the Euler characteristics.
Recall that the Betti number
bk = dimH
k
dR(M),
and the Euler characteristic
χ(M) =
∑
k
(−1)kbk.
We have
Proposition 2.5. Let M be a compact orientable manifold of dimension m, then
(1) For any k, bk = bm−k.
(2) If m = 4n+ 2, then b2n+1 is even.
8 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
Proof. (1) This follows from the fact
HkdR(M) ' Hn−kc (M) = Hn−kdR (M).
(2) follows from the non-degeneracy of the pairing
P 2n+1M : H
2n+1
dR (M)×H
2n+1
dR (M)→ R.
If fact, for any [ω], [η] ∈ H2n+1dR (M), we have
P 2n+1M ([ω], [η]) =
∫
M
ω ∧ η =
∫
M
(−1)(2n+1)(2n+1)η ∧ ω = −P 2n+1M ([η], [ω]).
It follows that the matrix for the pairing P 2n+1M : H
2n+1
dR (M) × H
2n+1
dR (M) → R is an anti-
symmetric b2n+1 × b2n+1 matrix. So
det(P 2n+1M ) = det((P
2n+1
M )
T ) = (−1)b2n+1 det(P 2n+1M ).
So b2n+1 must be an even number, otherwise det(P
2n+1
M ) = 0 and thus P
2n+1
M is not a non-
degenerate pairing. �
As consequences, we get Finally we prove
Theorem 2.6. Let M be a compact oriented manifold.
(1) If dimM = 2n+ 1, then χ(M) = 0.
(2) If dimM = 4n+ 2, then χ(M) is even.
Proof. (1) If dimM = 2n+ 1, we have
χ(M) =
2n+1∑
k=0
(−1)kbk =
n∑
k=0
(−1)kbk +
2n+1∑
k=n+1
(−1)kb2n+1−k
=
n∑
k=0
((−1)k + (−1)2n+1−k)bk
= 0.
(2) Suppose dimM = 4n+ 2, then the same argument yields
χ(M) =
4n+2∑
k=0
(−1)kbk =
2n∑
k=0
((−1)k + (−1)4n+2−k)bk + b2n+1.
Since (−1)k + (−1)4n+2−k = ±2 and since b2n+1 is even, the result follows. �
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 9
¶ The Poincaré dual of a submanifold.
Now let M be an oriented manifold of dimension m and let ι : S ↪→ M be a closed
oriented submanifold of dimension k ≤ m. Then the map (why it is well-defined?)∫
S
: Hkc (M)→ R, [η] 7→
∫
S
ι∗η
defines an element
∫
S
in
(
Hkc (M)
)∗
, and thus by Poincaré duality, it defines an element
PdM(S) in H
m−k
dR (M), characterized by the following property: if ω ∈ Zm−k(M) is a repre-
sentative of PdM(S) ∈ Hm−kdR (M), then for any [η] ∈ Hkc (M), we have∫
S
ι∗η =
∫
M
η ∧ ω.
Definition 2.7. We call PdM(S) the Poincaré dual of S in M .
Example. Let M be closed and oriented. Then M itself is a closed oriented submanifold of
M . By definition, the element in H0dR(M) corresponding to the (sub-)manifold M is [1], i.e.
PdM(M) = [1].
Example. Let M be a closed oriented manifold of dimension m, and let
∆ = {(x, x) | x ∈M} ⊂M ×M
be the diagonal submanifold. We would like to calculate PdM×M(∆) ∈ HmdR(M ×M). We
denote by pr1 : M ×M → M and pr2 : M ×M → M the two canonical projections of
M ×M to the two components. Let
{[ωji ] | 1 ≤ i ≤ bj}
be a basis of HjdR(M), where bj = dimH
j
dR(M) is the j, and let
{[νm−ji ] | 1 ≤ i ≤ bm−j = bj}
be the dual basis of Hm−jdR (M) under Poincaré duality, namely,∫
M
ωji ∧ ν
m−j
k = δik, ∀1 ≤ i, k ≤ bj.
Then we have∫
M×M
(pr∗1ω
j
i ∧pr∗2ν
m−j
k )∧(pr
∗
1ν
m−r
s ∧pr∗2ωrt ) = (−1)m(m−j)
∫
M×M
pr∗1(ω
j
i ∧νm−rs )∧pr∗2(ωrt ∧ν
m−j
k )
= (−1)m(m−j)δjr
∫
M
(ωji ∧ νm−js )
∫
M
(ωjt ∧ ν
m−j
k )
= (−1)m(m−j)δrjδisδtk,
which implies that
{[pr∗1ω
j
i ∧ pr∗2ν
m−j
k ] | 0 ≤ j ≤ m, 1 ≤ i, k ≤ bj}
10 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
are linearly independent. By Künneth formula and dimension counting, they form a basis
of HmdR(M ×M). Thus we can write
PdM×M(∆) =
∑
i,j,k
ci,kj [pr
∗
1ω
j
i ∧ pr∗2ν
m−j
k ].
To calculate the coefficients, we let ω ∈ Zm(M ×M) be a representative of PdM×M(∆) and
for each r, s, t, calculate both sides of∫
M×M
(pr∗1ν
m−r
s ∧ pr∗2ωrt ) ∧ ω =
∫
∆
ι∗(pr∗1ν
m−r
s ∧ pr∗2ωrt ).
Since [ω] = PdM×M(∆), the left hand side gives
LHS = (−1)m
∑
i,j,k
ci,kj
∫
M×M
(pr∗1ω
j
i ∧ pr∗2ν
m−j
k ) ∧ (pr
∗
1ν
m−r
s ∧ pr∗2ωrt )
= (−1)m
∑
i,j,k
ci,kj (−1)m(m−j)δrjδisδtk
= (−1)mrcs,tr .
For the right hand side, if we denote ϕ : M → ∆ be the orientation-preserving (why)
diffeomorphism given by ϕ(x) = (x, x), then pr1 ◦ ι ◦ ϕ = Id = pr1 ◦ ι ◦ ϕ and thus
RHS =
∫
M
ϕ∗ι∗(pr∗1ν
m−r
s ∧ pr∗2ωrt ) =
∫
M
νm−rs ∧ ωrt = (−1)r(m−r)δst.
So we conclude cs,tr = (−1)r
2
δst = (−1)rδst, and thus
PdM×M(∆) =
∑
i,j
(−1)j[pr∗1ω
j
i ∧ pr∗2ν
m−j
i ].
¶ The Lefschetz fixed point theorem.
More generally, let f : M →M be a smooth map, and
Γf = {(x, f(x)) | x ∈M} ⊂M ×M
be its graph. Then the pull-back f ∗ : H∗dR(M)→ H∗dR(M) is linear and thus has the form
f ∗(ωji ) =
∑
k
Aikj ω
j
k.
By the same computation as above, one can show (exercise)
PdM×M(Γf ) =
∑
i,j,k
(−1)jAjik[pr
∗
1ω
j
i ∧ pr∗2ν
m−j
k ].
As a consequence, we have
Proposition 2.8. Let ι : ∆→M ×M be the inclusion map. Then∫
∆
ι∗PdM×M(Γf ) =
∑
j
(−1)jtr(f ∗|HjdR(M)).
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 11
Proof. Let ϕ : M → ∆ be given by ϕ(x) = (x, x) as above. Then∫
∆
ι∗PdM×M(∆) =
∫
M
ϕ∗ι∗PdM×M(∆) =
∑
j
(−1)j
∑
i,k
Ajik
∫
M
ωji ∧ ν
m−j
k
=
∑
j
(−1)j
∑
i,k
Ajii
=
∑
j
(−1)jtr(f ∗|HjdR(M)). �
This result has a very important application. Recall from PSet 4-1-4:
Let f : M →M be a smooth map. A point p ∈M is a called a fixed point of
f if f(p) = p. We say f is a Lefschetz map if for each fixed point p of f , 1 is
not an eigenvalue of dfp : TpM → TpM . The local Lefschetz number Lp(f) of
a Lefschetz map at a fixed point p is the sign of the determinant det(dfp− Id),
i.e. Lp(f) := 1 if det(dfp − Id) > 0, and Lp(f) := −1 if det(dfp − Id) < 0.
Theorem 2.9 (Lefschetz fixed point theorem). Let f : M →M be a Lefschetz map. Then∑
p∈Fix(f)
Lp(f) =
∑
j
(−1)jtr(f ∗|HjdR(M)).
I will not give a detailed proof since it was assigned as an “A+-project”.
¶ The Poincaré-Hopf theorem.
Finally we let M be a compact oriented smooth manifold, and X a smooth vector field
on M . Then for t small enough, the fixed points of the flow ϕt : M → M generated by
X are exactly the zeroes of X. Let x be a zero of X, then we get a family of linear maps
dϕtx : TxM → TxM , and thus a linear map
Ax :=
d
dt
∣∣∣∣
t=0
(dϕtx).
Definition 2.10. We say x is a nondegenerate zero of X if detAx 6= 0. In this case we call
the number
Ind(X, x) = sgn(detAx)
the index of X at x.
Remark. Let x0 be a non-degenerate zero of a smooth vector field X on M . Choose a local
coordinate system near x0 so that x0 is the only zero of X. Then X defines a vector field
X̃ on V ⊂ Rn with a non-degenerate zero at 0. In particular, for r small enough, X̃ has no
zero on the sphere S(r) = {x ∈ V | |x| = r}. Consider the map Fr : S(r)→ Sn−1 by
Fr(x) = X̃(x)/|X̃(x)|.
Then one can prove: degFr = Ind(X, x0). In other words, the index of a vector field at each
non-degenerate critical point is a mapping degree.
12 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
Note that by definition
det(dϕtx − I) = det(tAx + h.o.t.) = tm detAx + h.o.t.
and thus Lx(ϕ
t) = Ind(X, x). So we get∑
j
Ind(X, xj) =
∑
x∈Fix(ϕt)
Lx(ϕ
t) =
∑
j
(−1)jtr((ϕt)∗|HjdR(M)).
But ϕt is homotopic to the identity map, and thus (ϕt)∗ = Id, and thus we proved the
following famous theorem that I promised to some of you last semester:
Theorem 2.11 (Poincaré-Hopf theorem). Let M be a compact oriented smooth manifold,
and X a smooth vector field on M , with only non-degenerate zeroes x1, · · · , xk. Then∑
j
Ind(X, xj) = χ(M).
3. Reading: Gauss-Bonnet-Chern Theorem
¶ Vector bundles.
We need some definitions:
Definition 3.1. Let E,M be smooth manifolds, and π : E →M a surjective smooth map.
We say (π,E,M) is a vector bundle of rank r if for every p ∈ M , there exits an open
neighborhood Uα of p and a diffeomorphism (called the local trivialization)
Φα : π
−1(Uα)→ Uα × Rr
so that
(1) Ep = π
−1(p) is a r dimensional vector space, and Φα|Ep : Ep → {p} × Rr is a linear
map.
(2) For Uα ∩ Uβ 6= ∅, there is a smooth map gβα : Uα ∩ Uβ → GL(r,R) so that
Φβ ◦ Φ−1α (m, v) = (m, gβα(m)(v)), ∀m ∈ Uα ∩ Uβ, v ∈ Rr.
We will call E the total space, M the base and π−1(p) the fiber over p. (In the case there is
no ambiguity about the base, we will denote a vector bundle by E for short.)
For example, we have seen the following vectorbundles:
• the tangent bundle TM of a smooth manifold M .
• the cotangent bundle T ∗M of a smooth manifold M .
• the normal bundle NSM of a submanifold S in M
One may define the conception of vector bundle isomorphism (try to define this by yourself),
and prove (exercise)
Lemma 3.2. Let ∆ ⊂M×M be the diagonal submanifold. Then N∆ is isomorphic to T∆.
Roughly speaking, a vector bundle E over M is “a smooth varying family of vector
spaces parameterized by a base manifold M”. It is not hard to get
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 13
Proposition 3.3. For any vector bundle E over M , one has
HkdR(E) = H
k
dR(M), ∀k.
Proof. This is a consequence of the homotopy invariance: E is homotopy equivalent to M ,
since if we let s0 : M → E be the zero section, then π ◦ s0 = IdM , and s0 ◦ π ∼ IdE via the
homotopy
F : E × R→ E, (x, v, t) 7→ (x, tv). �
Again by Poincaré duality, we get the following generalization of Corollary 2.3:
Theorem 3.4 (Thom isomorphism). Let E be a rank r vector bundle over M . Assume
both E and M are oriented with finite dimensional compact supported de Rham cohomology
groups, then
Hkc (E) ' Hk−rc (M), ∀k.
Proof. We apply Poincaré duality twice:
Hkc (E) ' Hn+r−kdR (E) ' H
n+r−k
dR (M) ' H
k−r
c (M). �
¶ The Thom class and the Euler class.
Now let M be an oriented, compact and connected smooth n-manifold, and E an oriented
vector bundle of rank r over M . Then the zero section
s0 : M → E, x 7→ (x, 0)
defines an embedding of M into E.
Definition 3.5. We call τ(E) := PdE(M) ∈ Hrc (E) the Todd class of (π,E,M).
Remark. The constant function 1 on M gives us a degree-0 cohomology class [1] ∈ H0dR(M).
So the Thom isomorphism above produces an element
τ(E) ∈ Hrc (E).
One can show that this is the same as the Poincaré dual of M in E.
Let E be an oriented vector bundle of rank r over an oriented connected compact smooth
manifold M . Let s : M → E be any global section of E, that is, a smooth map such that
π ◦ s = IdM .
Then by using the pull-back, we get an element
s∗(τ(E)) ∈ HrdR(M).
Proposition 3.6. The de Rham cohomology class s∗(τ(E)) is independent of the choices of s.
Proof. Let s0 : M → E be the zero section. Then s0 ∼ s via the homotopy
F : M × R, (m, t) 7→ (m, ts(m)).
So s∗(τ(E)) = s∗0(τ(E)). �
14 LECTURE 28: APPLICATIONS OF DE RHAM THEORY
Definition 3.7. The de Rham cohomology class
e(E) := s∗(τ(E)) ∈ HrdR(M)
is called the Euler class of E.
¶ Gauss-Bonnet-Chern theorem.
The following theorem is an important part of Gauss-Bonnet-Chern theorem:
Theorem 3.8. Let M be a compact oriented manifold Then∫
M
e(TM) = χ(M).
Proof. Since ∆ is diffeomorphic to M , and since T∆ ' N∆ as vector bundles over ∆,∫
M
e(TM) =
∫
∆
e(T∆) =
∫
∆
e(N∆).
But by definition and the characterization formula of Poincaré dual,∫
∆
e(N∆) =
∫
∆
s∗0τ(N∆) =
∫
∆
s∗0PDN∆(∆) =
∫
N∆
PDN∆(∆) ∧ PDN∆(∆).
By tubular neighborhood theorem, we get∫
N∆
PDN∆(∆) ∧ PDN∆(∆) =
∫
M×M
PDM×M(∆) ∧ PDM×M(∆) =
∫
∆
ι∗PDM×M(∆),
where ι : ∆ → M ×M is the canonical embedding. But if we apply Proposition 2.8 to the
identity map, we have ∫
∆
ι∗PdM×M(∆) = χ(M).
So we get the desired formula. �
There is a geometric way to define the Euler class: One first fix a connection ∇ on E,
and let R∇ be the curvature 2-form of ∇. Then according to the famous Chern-Weil theory,
if the rank of E is even, then χ(E) can be calculated via R∇,
χ(E) = [Pf(R∇)]
where Pf is defined as follows: For any A = (aij) ∈ so(r),
Pf(A) =
1
(4π)r/2(r/2)!
∑
σ∈Sr
(−1)σaσ(1)σ(2)a
σ(3)
σ(4) · · · a
σ(r−1)
σ(r) .
It is homogeneous polynomial of degree r/2.
LECTURE 28: APPLICATIONS OF DE RHAM THEORY 15
Finally we assume M is a closed oriented smooth manifold of dimension m = 2n. Endow
with M a Riemannian metric, and consider the tangent bundle E = TM . Then we have a
Levi-Civita connection ∇ on TM , and we arrive at the famous
Theorem 3.9 (Gauss-Bonnet-Chern). We have∫
M
Pf(R∇) = χ(M).
Dedicated to
Shiing-Shen Chern (1911-2004)
for his 110th Birth Anniversary