Solução_Topologia_Diferencial _de_Guillemin_Pollack

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Differential Topology by Guillemin & Pollack Solutions
Christopher Eur
May 15, 2014
In the winter of 2013-2014, I decided to write up complete solutions to the starred exercises in
Differential Topology by Guillemin and Pollack. There are also solutions or brief notes on non-
starred ones. Please email errata to ceur@college.harvard.edu.
Notation: A neighborhood is always assumed to be an open neighborhood. A graph of a function
f is denoted Γ(f).
Contents
1 Chapter 1: Manifolds and Smooth Maps 2
1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Derivatives and Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3 The Inverse Function Theorem and Immersions . . . . . . . . . . . . . . . . . . . . . 6
1.4 Submersions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.6 Homotopy and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.7 Sard’s Theorem and Morse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.8 Embedding Manifolds in Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Chapter 2. Transversality and Intersection 16
2.1 Manifolds with Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2 Transversality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1
1 Chapter 1: Manifolds and Smooth Maps
1.1 Definitions
Exercise 1 (1.1.2). If X ⊂ RN , Z ⊂ X, and f : X → Rm is smooth / diffeomorphic, then f |Z is
also smooth / diffeomorphic.
Solution) It suffices to show the smooth part (then apply it to the inverse map to get diffeomor-
phism). Fix any z ∈ Z ⊂ X. Since f : X → Rm is smooth, there exists z ∈ U ⊂ RN open (i.e.
X ∩U a neighborhood of z ∈ X) and F : U → Rm smooth such that F = f on X ∩U . Now, U ∩Z
is a neighborhood of z ∈ Z such that F = f |Z on Z ∩ U . �
Exercise 2 (1.1.3). Let X ⊂ RN , Y ⊂ RM , Z ⊂ RL, and let f : X → Y, g : Y → Z. Then:
f and g are smooth / diffeomorphism ⇒ g ◦ f : X → Z is smooth / diffeomorphism.
Solution) It suffices to show the smooth part (diffeomorphism part follows easily). Note that
this is true when X, Y , Z are open subsets of RN , RM , RL (Chain Rule from calculus). Now fix
any x ∈ X, y := f(x). We have y ∈ V ⊂ RM open with G : V → RL smooth, and x ∈ U ⊂ RN open
with F : U → V smooth (if necessary by replacing U with U ∩f−1(V )), and F = f on X ∩U , G = g
on Y ∩ V . Then G ◦ F : U → RL is smooth, and G ◦ F = g ◦ f on X ∩U since f(X ∩U) ⊂ (Y ∩ V )
by construction. �
Exercise 3 (1.1.4). Show that any open ball Br(0) in Rk is diffeomorphic to Rk, and hence, if X
is a k-dimensional manifold then every point in X has a neighborhood diffeomorphic to Rk.
Solution) Consider the maps
Br(0)→ Rk, x 7→
rx√
r2 − ‖x‖2
and Rk → Br(0), y 7→
ry√
r2 + ‖y‖2
They are mutual inverses, and by the previous exercise both are smooth (lots of composition of
smooth maps). Lastly, if x ∈ X and φ : U → X is local parametrization at x, then take (for
small enough r > 0) Br(φ
−1(x)) ⊂ U so that φ|Br is also a local parametrization at x. With
diffeomorphism ψ : Rk → Br, we have a local parametrization φ|Br ◦ ψ : Rk → X at x. �
Exercise 4 (1.1.5). Every k-dimensional vector subspace V of RN is a manifold diffeomorphic to Rk,
all linear maps on V are smooth, and if φ : Rk → V is a linear isomorphism, then the corresponding
coordinate functions are linear functionals on V (called linear coordinates).
Solution) Lemma: every linear transformation L : Rn → Rm is smooth (∵ dLx = L ∀x ∈ Rn),
and note that any linear map on V is smooth since it extends to a linear map on RN .
Now, by choosing a basis of V we have an isomorphism φ : V → Rk, which we can extend to a
linear map φ̃ : RN → Rk. So, φ is smooth, φ−1 : Rk → V ⊂ RN (linear) is also smooth, and thus V
is a manifold diffeomorphic to Rk.
Let φ := (x1, . . . , xk). Since xi = πi ◦ φ, each xi is a linear functional on V (projections
πi : RN → R is linear). �
2
Exercise 5 (1.1.6, 7, 8). Smooth bijection need not be diffeomorphism. Union of two coordinate
axes in R2 is not a manifold; hyperboloid in R3 defined by x2 + y2 − z2 = a, a > 0 is a manifold
(not when a = 0).
Solution) f : R→ R via f(x) = x3 is smooth and bijective but f−1(x) = 3
√
x is not smooth at x =
0. When (0, 0) is removed from {x = 0}∪{y = 0} we get four disconnected components,but R−{∗}
has two components, and Rn − {∗} is connected for n ≥ 2. Lastly, if H is the hyperboloid, H+ and
H− can be parametrized by R2 −Ba(0)→ R3 via(u, v) 7→ (u, v,±
√
u2 + v2 − a), and for (x, y, 0) ∈
H ∩ {z = 0}, it also has a local parametrization Ba(0) → R3 via (u, v) 7→ (±
√
a+ u2 − v2, u, v)
(switch x, y if necessary). When a = 0, (0, 0, 0) has no neighborhood diffeomorphic to R2 (again use
connectivity argument).
Exercise 6 (1.1.12,13). Let Sk ⊂ Rk+1 sphere, and N = (0, . . . , 0, 1) the north pole. Then the
stereographic projection π : Sk − {N} → Rk is a diffeomorphism.
Solutions) Direct computation yields the maps explicitly:
π : (u1, . . . , uk+1) 7→
(
u1
1− uk+1
, . . . ,
uk
1− uk+1
)
π−1 : (x1, . . . , xk) 7→
(
2x1
x21 + · · ·x2k + 1
, . . . ,
2xk
x21 + · · ·x2k + 1
,
x21 + · · ·x2k − 1
x21 + · · ·x2k + 1
)
and that π, π−1 are smooth follows from [1.1.3] and that they are mutual inverses is checked by
direct computation. (This also shows that Sk is a k-dimensional manifold). �
Exercise 7 (1.1.14,15). If f : X → X ′, g : Y → Y ′ smooth, then the product map f×g : X×Y →
X ′ × Y ′ defined by (x, y) 7→ (f(x), g(y)) is smooth. Hence, if f, g are diffeomorphisms, f × g is also
a diffeomorphism. Lastly, projection map X × Y → X is smooth.
Solution) Note that for A,A′ ⊂ X and B,B′ ⊂ Y , (A×B)∩(A′×B′) = (A∩A′)×(B∩B′) ⊂ X×Y
(NOT true for unions). Also, it is easy to check that f × g is smooth when X,Y are open subsets
(∵ f × g = (f1, . . . , fN ′ , g1, . . . , gM ′)), so taking neighborhoods U , V of X,Y with F,G smooth,
F ×G = f × g on (X ∩ U)× (Y ∩ V ).
If f, g are diffeomorphisms, f−1× g−1 is smooth inverse of f × g. Finally, X×Y → X is smooth
since RN × RM → RN is smooth (in fact the Jacobian looks like [IN | 0]). �
Exercise 8 (1.1.16,17). Let f : X → Y be smooth, and define f̃ : X → Γ(f) by x 7→ (x, f(x)).
Then f̃ is a diffeomorphism (hence if X a manifold so is Γ(f)).
Solution) Lemma: the diagonal map ∆ : X → X × X, x 7→ (x, x) is a diffeomorphism (∵
∆ : RN → RN × RN has Jacobian [IN | IN ]t so it is smooth; the inverse map is same as projection
so it is smooth). Now, using the lemma and the previous exercise, f̃ = (Id×f) ◦∆ is smooth. The
inverse map is same as the projection map, so it is smooth, so f̃ is a diffeomorphism. �
3
Exercise 9 (1.1.18). Let 0 < a < b. The function f : R→ R defined by
f(x) =
{
e−1/x
2
x > 0
0 x ≤ 0
is smooth. g(x) := f(x − a)f(b − x) is a smooth map, and is positive on (a, b) and zero elsewhere.
Moreover, h(x) :=
∫ x
−∞ gdx∫∞
−∞ gdx
is also smooth with the property that h(x) = 0 for x ≤ a, h(x) = 1
for x ≥ b, and 0 < h(x) < 1 for x ∈ (a, b) (h is also non-decreasing). Lastly, construct a smooth
function H : Rk → R that is 1 on Ba, 0 on Rk −Bb, and (0, 1) otherwise.
Solution) f is smooth: for x > 0 the ith derivative of f is of the form (rational polynomial)e−1/x
2
hence limx→0 f
(i)(x) = 0, ∀i = 0, 1, . . .. Addition/multiplication by a constant and (x, y) 7→ xy are
smooth maps, so g is smooth with desired property. Lastly, h is smooth with desired property by
Fundamental Theorem of Calculus. Now note that Rk → R, x 7→ ‖x‖ is smooth, so x 7→ 1− h(‖x‖)
is the desired function H. �
1.2 Derivatives and Tangents
Note: When the context is clear, X,Y are assumed to be manifolds (of dimension Rk,Rl residing
in RN ,RM ).
Exercise 10 (1.2.1,2). Let X ⊂ Y be a submanifold, and j : X ↪→ Y be the inclusion map. Then
∀x ∈ X, djx : Tx(X) ↪→ Tx(Y ) is injective—in factit is an inclusion. If Ũ is a open subset of a
manifold X, Tx(Ũ) = Tx(X) for x ∈ Ũ .
Solution) Lemma: if X is a manifold, x ∈ X, and φ : U → X is a local parametrization with
φ(0) = x, then (dφ0)
−1 = d(φ−1)x (∵ Exercise [1.2.4]). Now, let x ∈ X, φ : U → X, ψ : V → Y be
local parametrizations (U ⊂ Rk, V ⊂ Rl open, φ(0) = x = ψ(0)). Note that ψ−1 ◦ j ◦ φ = ψ−1 ◦ φ,
and so we have djx = dψ0 ◦ d(ψ−1 ◦ φ)0 ◦ (dφ0)−1 = dψ0 ◦ d(ψ−1)y ◦ dφ0 ◦ (dφ0)−1 = Id (∵ first
equality by definition, second by chain rule, third by lemma). �
Exercise 11 (1.2.3+α). Let V ⊂ RN be a vector subspace. Then ∀x ∈ V , Tx(V ) = V . Moreover,
if L : V → RM is a linear map, then for x ∈ V , dLx = L.
Solution) By choosing basis we have linear isomorphism φ : Rk → V , and φ = dφu for φ(u) ∈ V ,
hence Tx(V ) = V . The remaining statement follows from [Exercise 17]; extend L to a linear map
on the whole RN , then apply [Exercise 17] and Tx(V ) = V . �
Exercise 12 (1.2.4). Let f : X → Y is a (local) diffeomorphism and x ∈ X, y := f(x), then the
linear map dfx : Tx(X)
∼→ Ty(Y ) is an isomorphism. In fact, (dfx)−1 = d(f−1)x.
Solution) Let f−1 be the smooth (local) inverse map. Obviously, d Id = Id, so from the chain
rule we have Id = d(f ◦ f−1)y = dfx ◦ d(f−1)y, and likewise, d(f−1)y ◦ dfx = Id. �
4
Exercise 13 (1.2.8). What is the tangent space to the hyperboloid defined by x2 + y2 − z2 = a at
(
√
a, 0, 0) (a > 0)?
Solution) As in [1.1.8], we have a local parametrization φ : Ba(0) → R3 given by (u, v) 7→
(
√
a+ u2 − v2, u, v), and [dφ] =
u(a+ u2 − v2)−1/2 −v(a+ u2 − v2)−1/21 0
0 1
. And so [dφ(√a,0,0)] =0 01 0
0 1
, and thus the tangent space is the y, x-plane, as expected. �
Exercise 14 (1.2.9). Let X,Y manifolds, π : X × Y → X be the projection map, and for y ∈ Y let
fy : X → X × Y be a smooth injection defined by x→ (x, y). Then
1. T(x,y)(X × Y ) = Tx(X)× Ty(Y )
2. dπ(x,y) : Tx(X)× Ty(Y )→ Tx(X) is also a projection (v, w) 7→ v.
3. fy is diffeomorphism onto its image and d(fy)x : v 7→ (v, 0).
Lastly, if f : X → X ′, g : Y → Y ′ are smooth maps, then d(f × g)(x,y) = dfx × dgy.
Solution) Let (x, y) ∈ X × Y and φ × ψ : U × V → X × Y be local parametrization. Now (1.)
follows from the following lemma:
Lemma: Let U1, U2 be open subset of Rn1 ,Rn2, and g1 × g2 : U1 ×U2 → Rm1 ×Rm2 is smooth, then
dg(u1,u2) = d(g1)u1 × d(g2)u2 (∵ [dg(u1,u2)] =
[
[d(g1)u1 ] 0
0 [d(g2)u2 ]
]
). For (2.),
X × Y π−−−−→ X
φ×ψ
x xφ
U × V h−−−−→ U
we note that the map h := φ−1 ◦ π ◦ (φ × ψ) : U × V → U is also the projection map (u, v) 7→ u.
So, dπ(x,y) = dφ0 ◦ dh(0,0) ◦ d(φ× ψ)−10 is easily computed to be the projection, as desired. Now for
(3.), fy is diffeomorphism since (fy)−1 is just projection, and we have the diagram
X
fy−−−−→ X × y ⊂ X × Y
φ
x xφ×ψ
U
h−−−−→ U × 0 ⊂ U × V
It’s easy to see that dh0 = Idk×0. So, d(fy)x(v) = ((dφ0×dψ0)◦dh0 ◦dφ−10 )(v) = (v, 0), as desired.
Finally, we show d(f × g)(x,y) = dfx × dgy. Let π1 : X × Y → X, π2 : X × Y → Y be
projections, and j1 : X
′ → RN ′+M ′ , j2 : Y ′ → RN
′+M ′ be smooth maps x′ 7→ (x′, 0), y′ 7→ (0, y′).
Now, note that f × g = j1 ◦ f ◦ π1 + j2 ◦ g ◦ π2. So, using (1., 2., 3.) and chain rule, we have
d(f × g)(x,y)(u, v) = (dfx(u), dgy(v)), as desired. �
5
Exercise 15 (1.2.10,11). Let f : X → Y smooth map of manifolds and define f̃ : X → X × Y by
x → (x, f(x)), then df̃x(v) = (v, dfx(v)), and hence T(x,f(x))(Γ(f)) is the graph of dfx : Tx(X) →
Tf(x)(Y ).
Solution) Lemma: if ∆ : X → X × X is the diagonal map, then d∆x(v) = (v, v) (∵ h : U →
U × U has dh0 = [Ik|Ik]). Now, f̃ = (Id×f) ◦ ∆, so the lemma and [Exercise 1.2.9] implies that
f̃x(v) = (v, dfx(v)), as desired. �
Exercise 16 (1.2.12). A curve on a manifold X is a smooth map c : I → X, t 7→ c(t), where
I ⊂ R interval. The velocity vector of the curve c at time t0, denoted dcdt (t0) or ċ(t0), is defined
as dct0(1) ∈ Tx0(X) where x0 = c(t0). Note that when X = Rk and c(t) = (c1(t), . . . , ck(t)), then
dc
dt (t0) = (c
′
1(t0), . . . , c
′
k(t0)). Every vector in Tx(X) is the velocity vector of some curve in X, and
conversely.
Solution) First, it is obvious that tangent space at a point on an interval is R, so dct0 : R →
Tx0(X), and we thus we have the converse.
Now fix any x ∈ X and v ∈ Tx(X). Let φ : U → X be local parametrization around x, φ(0) = x,
and WLOG U = R (∵ [Exercise 1.1.4]). By definition v = dφ0(w) for some w = (w1, . . . , wk) ∈ Rk.
Now, define c̃ : R → Rk by t 7→ (w1t, . . . , wkt), and c := φ ◦ c̃. Now, dc0 = dφ0 ◦ c̃0 and so
dc0(1) = dφ0(w), as desired. �
Exercise 17 (made-up). Let f : X → Y be smooth map of manifolds, x ∈ X, and let x ∈ U ⊂ RN
open and F : U → RM smooth such that F = f on X ∩U . Then dfx = dFx|Tx(X) : Tx(X)→ Ty(Y ).
Solution) WLOG, we have φ : Ũ → X ∩ U and ψ : Ṽ → Y ∩ V local parametrizations (diffeo-
morphisms) such that the following commutes (with h := ψ−1 ◦ f ◦ φ, φ(u) = x, ψ(v) = f(x))
X ∩ U f−−−−→ Y ∩ V
φ
x xψ
Ũ
h−−−−→ Ṽ
Now since f = F on X ∩ U , h = ψ−1 ◦ F ◦ φ, and hence dhu = dψ−1v ◦ dFx ◦ dφu, and plugging
this into dfx = dψu ◦ dhu ◦ dφ−1u , we have dfx = IdRM ◦dFx ◦ IdTx(X), and hence dfx = dFx|Tx(X) as
desired. In fact, dFx is a linear extension of the linear map dfx. �
1.3 The Inverse Function Theorem and Immersions
Exercise 18 (1.3.2). Suppose Z ⊂ X l-dimensional submanifold, z ∈ Z. Then there exists a local
coordinate system {x1, . . . , xk} on a neighborhood U of z in X such that Z ∩ U is defined by the
equations xl+1, . . . , xk = 0.
6
Solution) Since the inclusion map j : Z ↪→ X is an immersion, by Local Immersion Theorem,
we have parametrizations around z such that:
Z
j−−−−→ X
φ
x xψ
V
can. imm.−−−−−−→ Ũ
commutes, with φ(0) = z = ψ(0) and Ũ = V ×V ′. Now, let ψ(Ũ) := U ⊂ X and ψ−1 = (x1, . . . , xk),
then Z ∩ U = {v ∈ U | xl+1(v) = 0, . . . , xk(v) = 0}, as desired. �
Exercise 19 (1.3.3,4,5). If f : R→ R is a local diffeomorphism, then Im(f) is an open interval and
f : R→ Im(f) is a diffeomorphism. Such is not true for h : R2 → R2 local diffeomorphism.
Solution) Lemma[1.3.5]: if f is injective local diffeomorphism, its image is open in Y , and it
is diffeomorphism onto its image (∵ open: local homeomorphisms are open maps, and bijective
local diffeomorphism admits smooth inverse) . R is connected thus so is Im(f), hence Im(f) is an
interval, and f local diffeomorphism implies that f is an open map and that f ′(x) 6= 0 (and hence
f is injective).
This is not true for R2; consider h = g × arctan where g : R → S1, t 7→ (cos 2πt, sin 2πt) (h is
not injective). �
Exercise 20 (1.3.6). Let f : X → Y, g : Y → Y ′ be immersions, Z ⊂ X submanifold. Then f × g,
g ◦ f , and f |Z are immersions, and if dimX = dimY , then f is in fact a local diffeomorphism.
Solution) f × g is immersion by [Exercise 1.2.9] and g ◦ f immersion by chain rule. If j : Z ↪→ X
is inclusion map, f |Z = f ◦ j, so f |Z is immersion. If dimX = dimY , f is local diffeomorphism by
Local Immersion Theorem. �
Exercise 21 (1.3.7). Define g : R → S1, t 7→ (cos 2πt, sin 2πt) (local diffeomorphism) and G :=
g × g : R2 → S1 × S1, and let L ⊂ R2 a line with irrational slope. Then G|L is an injective local
diffeomorphism, but its image is not a submanifold of S1 × S1.
Solution) G : (s, t) 7→ (cos 2πs, sin 2πs, cos 2πt, sin 2πt), and WLOG let L be defined by t =
αs, α ∈ R − Q. If G(s1, αs1) = G(s2, αs2), then s1 − s2 ∈ Z and α(s1 − s2) ∈ ZZ, which implies
that s1 = s2 since α is irrational. Moreover, since {nα}n∈N dense in R/Z, we have that Im(G|L) is
dense in S1 × S1, so Im(G|L) cannot be a submanifold. �
Exercise 22 (1.3.9). Let (x1, . . . , xN ) be standard coordinate functions on RN , and X ⊂ RN be a
k-dimensional manifold. Then any x ∈ X has a neighborhood on which the restrictions of some k
coordinate functions xi1 , . . . , xik form a local coordinate system.
Now for simplicity assume that x1, . . . , xk form a local coordinate on a neighborhood V of x ∈ X.
Then ∃gk+1, . . . , gN : U ⊂ Rk → R smooth such that V = Γ(g) where g = (gl+1, . . ., gN ) : U →
RN−k, and thus every manifold is locally a graph of a smooth function.
7
Solution) Choose a basis v1, . . . , vk ∈ RN of Tx(X), then the matrix [v1 · · · vk] has rank k, so
it has k linearly independent rows, say i1, . . . , ik. Now, let π : RN → Rk be projection defined
by (x1, . . . , xN ) 7→ (xi1 , . . . , xik), then dπx|Tx(X) : Tx(X) → Rk is an isomorphism by construction,
hence by IVT π|X : X → Rk is a local diffeomorphism. Now, assume i1, . . . , ik = 1, . . . , k and let
π̃ := π|X . π̃ : X → Rk is a diffeomorphism on x ∈ V ⊂ X, and the smooth inverse π̃−1 : U → V is
of form Id×g, and hence the result as desired. �
Exercise 23 (1.3.10). Generalized IVT: Let f : X → Y be smooth map that is injective on
a compact submanifold Z ⊂ X, and suppose ∀x ∈ Z, dfx : Tx(X) → Tf(x)(Y ) is isomorphism.
Then f maps Z diffeomorphically on f(Z), and in fact, maps an open neighborhood of Z in X
diffeomorphically onto an open neighborhood of f(Z) in Y . IFT is when Z is a single point.
Solution) Since dfx is an isomorphism for all x ∈ Z, for each x ∈ Z there exists Ux on which
f |Ux is a diffeomorphism. {Ux} is a cover of X, hence we choose a finite subcover U =
⋂
Ui ⊃ Z (X
is compact). On U , f is a local diffeomorphism, so only need show that f is injective on some open
set V containing Z (then f is injective local diffeomorphism on V ∩ U , hence a diffeomorphism).
Suppose V does not exist; then taking consecutively smaller �-neighborhoods Z� of Z, we obtain
a sequence {ai}, {bi} such that ai 6= bi but f(ai) = f(bi). Passing through subsequences, ai → a and
bi → b (both converge) since WLOG they all belong to some Z� for which Z� is compact. Moreover,
by construction the limit point is on Z and hence a = b = z̃ since f is injective on Z. However, this
implies that f cannot be a local diffeomorphism at z̃. Contradiction. �
1.4 Submersions
Exercise 24 (1.4.1). If f : X → Y is a submersion and U is open in X, then f(U) is open in Y .
Solution) Fix any y ∈ f(U); we need show ∃V ∈ y open in Y such that V ⊂ f(U). Pick
x ∈ f−1(y), then by local submersion theorem we have the following commutative diagram
X
f−−−−→ Y
φ
x xψ
Ṽ × Ũ π−−−−→ Ṽ
with φ(Ṽ × Ũ) ⊂ U and ψ(Ũ) := V , and V is as desired. �
Exercise 25 (1.4.2). Let X compact and Y connected, then every submersion f : X → Y is
surjective; hence, there exist no submersions of compact manifolds into Euclidean spaces.
Solution) Clearly, Y = f(X) t (Y − f(X)), so it suffices to show that Y − f(X) is both open
and closed in Y . f(X) is compact in Y since X is compact, hence f(X) is closed in Y ; X is open
in X, of by previous problem f(X) is open in Y . Now, if f : X → Rm is a submersion then f is
surjective, which is contradiction to Rm not being compact. �
8
Exercise 26 (1.4.5,6). Example: Let f : R3 → R, (x, y, z) 7→ x2 + y2 − z2, a, b both positive or
negative. Then 0 is the only critical value of f , and f−1(a), f−1(b) are diffeomorphic.
More generally, let p be any homogeneous polynomial in k − variables, then p−1(a) (a 6= 0) is a
k − 1-dimensional submanifold of Rk, and a > 0 ones are all diffeomorphic, as are a < 0 ones.
Solution) Lemma: p(x1, . . . , xk) homogeneous of order m, then
k∑
i=1
xi
∂p
∂xi
= m · p. The lemma
implies that 0 is the only critical value of p. Diffeomorphisms are made by scaling (note how m
being odd or even makes a small difference). �
Exercise 27 (1.4.7). Suppose y is a regular value of f : X → Y , X compact, and dimX = dimY .
Then f−1(y) = {x1, . . . xn} (finite), and there exists U ∈ y open in Y such that f−1(U) = V1t· · ·tVn
where Vi is open neighborhood of xi and f |Vi : Vi → U is a diffeomorphism (∀i = 1, . . . , n).
Solution) Note that y regular and dimX = dimY implies that f is local diffeomorphism at
any x ∈ f−1(y). If f−1(y) is not finite, it contains a limit point (∵ X compact), say x′ ∈ f−1(y),
but f cannot be local diffeomorphism at x′ (not injective). Now since f is local diffeomorphism at
x1, . . . , xn, we can construct desired U and Vi’s by taking finite intersections. �
Exercise 28 (1.4.10,11). Tangent space to O(n) at identity I is the space of skew symmetric ma-
trices. The group SL(n) is a (sub)manifold of M(n) and is moreover a Lie group, and the tangent
space to SL(n) at at the identity I is {H ∈M(n) : Tr(H) = 0}
Solution) Recall O(n) = f−1(I) where f : M(n) → S(n), A 7→ AAt. So, TI(O(n)) = ker dfI ,
and since dfI : M(n)→ S(n), H 7→ IHt +HIt = Ht +H, TI(O(n)) = {H ∈M(n) : H = −Ht}, as
desired. SL(n) = det−1(1) where det : M(n) → R, H 7→ detH, and 1 is a regular value of det by
[Exercise 1.4.6]. Moreover, matrix multiplication and inversion is a smooth map (Cramer’s rule), so
SL(n) is a Lie group. Finally, we compute ddetI : M(n)→ R:
lim
t→0
det(I + tH)− det(I)
t
= lim
t→0
O(t2) +
∑n
i=0(tH)ii + 1− 1
t
= Tr(H)
and hence the desired result about tangent space to SL(n) at I. �
Exercise 29 (1.4.13). (skip: this is linear algebra)
1.5 Transversality
Note: When the context is clear given x on X a manifold, we will not distinguish X as a whole
manifold and the open neighborhood of x in X since the tangent space at x turns out same (will
write X for both).
Exercise 30 (1.5.1,2). Examples of transversal & non-transversal intersection of linear spaces:
a. the xy plane and the z-axis in R3: transversal
b. the xy plane and the plane spanned by {(3, 2, 0), (0, 4,−1)} in R3: transversal
9
c. the plane spanned by {(1, 0, 0), (2, 1, 0)} and the y axis in R3: not transversal
d. Rk × {0} and {0} × Rl in Rn: transversal if n ≤ k + l.
e. V × {0} and the diagonal in V × V : transversal
f. symmetric and skew symmetric matrices in M(n): transveral
Solution) All of the above are easy to check with the following lemma:
Lemma: if V and W are linear subspace of Rn, then V t̄W means V +W = Rn (∵ [Exercise 1.2.3]).
Exercise 31 (1.5.4). Let X and Z be transversal submanifolds of Y , then for y ∈ X ∩ Z,
Ty(X ∩ Z) = Ty(X) ∩ Ty(Z)
Solution) We have V ⊂ Y open such that X ∩V = g−1(0), Z ∩V = h−1(0), for g = (g1, . . . , gk) :
V → Rk, h = (h1, . . . , hl) : V → Rl. Then (X ∩Z) ∩ V = f−1(0) where f = (g × h) ◦∆ : V → Rk+l
(N.B. 0 is regular value for f by transversality). Moreover, dfy : Ty(Y )→ Rk+l, v 7→ (dgy(v), dhy(v))
(∵ [Exercise 1.2.9,10]). And since ker dfy = ker dgy ∩ ker dhy, we have Ty(X ∩ Z) = ker dfy =
ker dgy ∩ ker dhy = Ty(X) ∩ Ty(Z), as desired. �
Exercise 32 (1.5.5). Let f : X → Y , Z ⊂ Y submanifold, f t̄Z, and W := f−1(Z). Then Tx(W )
is the preimage of Tf(x)(Z) under dfx : Tx(X)→ Tf(x)(Y ), i.e. Tx(f−1(Z)) = df−1x (Tf(x)(Z))
Solution) As in the proof, we have open neighborhoods U, V in X,Y around x, f(x) such that
U
f→ V g→ Rl and Z ∩V = g−1(0), f−1(Z)∩U = (g ◦f)−1(0). Now, noting Tf(x)(Z) = ker dgf(x), we
have Tx(Z) = ker d(g ◦ f)x = ker(dgf(x) ◦ dfx) = {v ∈ Tx(X) : dfx(v) ∈ ker dgf(x)} = df−1x (Tf(x)(Z)),
as desired. (This implies [Exercise 1.5.4] f = i : X ↪→ Y and dix is just inclusion). �
Exercise 33 (1.5.7). X
f→ Y g→ Z smooth maps of manifolds, W ⊂ Z submanifold such that gt̄W .
Then f t̄g−1(W ) if and only if (g ◦ f)t̄W .
Solution) Fix any x ∈ (g◦f)−1(W ), and y := f(x), z := g(y). Note that since gt̄W , dgy(Ty(Y ))+
Tz(W ) = Ty(Z).
f t̄g−1(W ) ⇒ (g ◦ f)t̄W : we need show Im(dgy ◦ dfx) + Tz(W ) = Tz(Z). Let w̃ ∈ Tz(Z) be
given. gt̄W implies ∃u ∈ Ty(Y ), v ∈ Tz(W ) such that dgy(u) + v = w̃. Moreover, f t̄g−1(W ),
so Ty(Y ) = Im(dfx) + Ty(g
−1(W )) = Im(dfx) + dg
−1
y (Tz(W )) (∵ [Exercise 1.5.5]), hence there
exists ũ ∈ Tx(X) and v′ ∈ dg−1y (Tz(W )) such that dfx(ũ) + v′ = u. Finally, we see that then
(dgy ◦ dfx)(ũ) + dgy(v′) + v = w̃, as desired.
(g ◦ f)t̄W ⇒ f t̄g−1(W ): we need show Im(dfx) + dg−1y (Tz(W )) = Ty(Y ). Again, fix a w̃ ∈
Ty(Y ). Note the existence of u ∈ Tx(X) such that (dgy ◦ dfx)(u) + w = dgy(w̃), and moreover,
dgy(dfx(u)− w̃) ∈ Tz(W ) is guaranteed by (g ◦ f)t̄W . �
Exercise 34 (1.5.9). Let V be a vector space, ∆ the diagonal of V × V , A : V → V linear map,
and W = Γ(A). Then W t̄∆ if and only if 1 is not an eigenvalueof A.
Solution) Since ∆ and W are both vector subspaces of V × V , W t̄∆ is equivalent to W + ∆ =
V ×V . Fix an arbitrary (u, v) ∈ V ×V , then if we can always find (v1, v1) ∈ ∆, (v2, Av2) ∈W such
that v1 + v2 = u, v2 +Av2 = v if and only if (A− I) is invertible. �
10
Exercise 35 (1.5.10). Let f : X → X be a smooth map with fixed point x (i.e. f(x) = x). If 1 is
not an eigenvalue of dfx : Tx(X)→ Tx(X), then x is called a Lefschetz fixed point of f, and f is
Lefschetz map if all its fixed points are Lefschetz. If X is compact and f is Lefschetz, then f has
only finitely many fixed points.
Solution) Let ∆X be the diagonal of X ×X, respectively. We wish to show that ∆X ∩ Γ(f) is
finite. Claim: it suffices to show ∆X t̄Γ(f). If it is so, then ∆X ∩ Γ(f) ⊂ X ×X is a submanifold
of dimension 0, since dim ∆X = dim Γ(f) = dimX (∵[Exercise 1.1.16,17]). Now, X ×X is compact
so its 0-dimensional submanifold is finite (if not then it has a limit point, which does not admit a
neighborhood diffeomorphic to a point).
Now we show ∆X t̄Γ(f). Fix any x ∈ ∆X ∩Γ(f), and let ∆V be the diagonal of Tx(X)×Tx(X).
Then by [Exercise 1.5.9] we have T(x,x)(∆X) + T(x,x)(Γ(f)) = ∆V + Γ(dfx) = Tx(X) × Tx(X) =
Tx(X ×X) (first equality by [Exercise 1.2.10,10]). �
1.6 Homotopy and Stability
Exercise 36 (1.6.1,2). Suppose f0 ∼ f1 (homotopic), then ∃F̃ : X × I → Y homotopy such that
F̃ (x, t) = f0(x) ∀t ∈ [0, 1/4] and F̃ (x, t) = f1(x) ∀t ∈ [3/4, 1]. Hence, homotopy is an equivalence
relation.
Solution) By [Exercise 1.1.18], there exists a function h : R→ R such that h(x) = 0 on x ≤ 1/4
and h(x) = 1 on x ≥ 3/4. Now, if F is the homotopy between f0, f1, set F̃ = F ◦ (Id×h).
Now we show equivalence relation. Let F,G be homotopies for f ∼ g, g ∼ h. Then define
H : X × [0, 7/4]→ Y by
H(x, t) =
{
F̃ (x, t) t ∈ [0, 1]
G̃(x, t− 3/4) t ∈ [3/4, 7/4]
Then H is smooth, and H̃ := H(x, 47 t) is the desired homotopy for f ∼ h. �
Exercise 37 (1.6.3). Every connected manifold X is arcwise connected, i.e. ∀x0, x1 ∈ X, ∃f :
I → X smooth with f(0) = x0, f(1) = x1.
Solution) We first note that arcwise connectedness (x0 ∼ x1) is a equivalence condition since
x0 ∼ x1 ⇔ f0 ∼ f1 where f0 : {∗} → X, ∗ 7→ x0, f1 : {∗} → X, ∗ 7→ x1, and homotopy is a
equivalence relation. By going through the local parametrizations, it is easy to show that arcwise
connected components are clopen in X, and hence if X is connected it is arcwise connected. �
Exercise 38 (1.6.7). The antipodal map α : Sk → Sk, x→ −x is homotopic to the identity if k is
odd.
Solution) Note that for any fixed θ ∈ R, the map Lθ : R2 → R2 given by multiplying the matrix[
cos θ − sin θ
sin θ cos θ
]
preserves the norm: i.e. ‖(x, y)‖2 = ‖L(x, y)‖2. So, for S2k−1 ⊂ R2k, the map
L := Lπt × · · ·
k times
× Lπt : R2k × I → R2k is the smooth homotopy. �
11
Exercise 39 (1.6.8). Diffeomorphisms of compact manifolds constitute a stable class.
Solution) Let f0 : X → Y be diffeomorphism, X,Y compact, ft homotopy of f0. We need
produce � > 0 such that ft is diffeomorphism for all t < �. WLOG, we assume that X is connected,
and also Y connected. (X is compact, so it has finitely many connected components, so we can find
� for each and take the minimum).
Local diffeomorphism and embedding is stable, so ∃� > 0 such that∀t < �, ft is a local diffeo-
morphism and an embedding. Fix any t < �. We are done if ft(X) = Y , but ft(X) is open in Y
since ft is local diffeomorphism, and it is closed since it is image of compact set in Y also compact.
Hence, Y connected implies ft(X) = Y . �
Exercise 40 (1.6.9). Let ρ : R → R be a function with ρ(s) = 1 if |s| < 1, ρ(s) = 0 if |s| > 2,
and define ft : R → R by ft(x) = xρ(tx). This is a counterexample to all the parts of the stability
theorem when X is not compact.
Solution) We simply need verify. f0(x) = xρ(0) = x, hence f0 is a diffeomorphism and any
submanifold of R is transversal to identity. Now, for any t > 0, note that |tx| > 2 for all |x| > 2/|t|,
so ft cannot be local diffeomorphism/immersion/submersion/embedding/diffeomorphism. And for
|x| > 2/|t|, ft(x) = 0, so clearly, {0} is not transversal to ft. �
Exercise 41 (1.6.10). A deformation of a submanifold Z ⊂ Y is a smooth homotopy it : Z → Y
where i0 is the inclusion map and each it is an embedding (and thus, Zt = it(Z) is a smoothly
varying submanifold of Y with Z0 = Z). If Z is compact, then any homotopy it of its inclusion map
is a deformation for small t.
Solution) Since Z is compact, embedding is stable class. �
1.7 Sard’s Theorem and Morse Functions
Exercise 42 (1.7.5). Exhibit a smooth map f : R→ R whose set of critical values is dense.
Solution) From [Exercise 1.1.18], there is a function g : R → R such that g(x) = 1 if |x| ≤ 1/4
and g(x) = 0 if |x| ≥ 1/2. Now, write Q = {q1, q2, . . .}, and then for i ∈ N, define gi : R → R by
gi(x) = qig(x− i). Now define f :=
∑
i gi, then all the rationals are critical values for f and dense
in R.
Remark: Measure zero implies empty interior, but the converse is false. �
Exercise 43 (1.7.6). The sphere Sk is simply connected if k > 1.
Solution) Since dimS1 < dimSk (k > 1), p ∈ Sk is a regular value iff p /∈ f(S1). That is, Sard’s
Theorem implies that f(S1) is measure zero in Sk, hence ∃p /∈ f(S1). Now under Sk − {p} ' Rk,
and with Rk being contractible, f(S1) is homotopic to a constant map. �
Exercise 44 (1.7.8). Analyze the critical behavior at the origin in the following functions:
a. f(x, y) = x2 + 4y3
b. f(x, y) = x2 − 2xy + y2
12
c. f(x, y) = x2 + y4
d. f(x, y) = x2 + 11xy + y2/2 + x6
e. f(x, y) = 10xy + y2 + 75y3
Solution) In the order of nondegenerate?/isolated?/local min or max?(non strict)?, we have:
a. N/N/N, b. N/N/Y(min), c. N/Y/Y(min), d. Y/Y/N, e. Y/Y/N �
Exercise 45 (1.7.11,12). If a ∈ Rn is a non degenerate critical point of f : Rn → R, there exists a
local coordinate system (x1, . . . , xn) around a such that
f = f(a) +
n∑
i=1
�ix
2
i , �i = ±1
And hence derive the usual second derivative test.
Solution) Note that Hessian matrix is always real symmetric and hence diagonalizable by or-
thogonal matrix P . Let H be Hessian matrix of f at a, and P be the orthogonal matrix that
diagonalizes H. Morse Lemma gives us local coordinates ~x′′ := (x′′1, . . . , x
′′
n) : Rn → Rn around a
such that f = f(a) + ~x′′
t
H ~x′′. Then ~x′ := (x′1, . . . , x
′
n) := P
t ~x′′ is a new local coordinates around a
such that f = f(a) + ~x′′
t
PP tHPP t ~x′′ = f(a) + ~x′
t
(P tHP )~x′. So,
f = f(a) +
n∑
i=1
λix
′2
i , λi’s are eigenvalues of H
Finally, let φ : Rn → Rn be defined by (y1, . . . , yn) 7→ (
√
|λ1|y1, . . . ,
√
|λn|yn), and ~x := (x1, . . . , xn) :=
g ◦ ~x′ be the new local coordinates around a. Then
f = f(a) +
n∑
i=1
�ix
2
i , �i = ±1
as desired, and since (x1, . . . , xn)(a) = 0 the second derivative test immediately follows. �
Exercise 46 (1.7.14). Check that the ”height function” (x1, . . . , xk) 7→ xk on the sphere Sk−1 is a
Morse function with two critical points, the poles (one max, one min).
Solution) Let f : Sk−1 → R be the restriction of the projection π : Rk → R, (x1, . . . , xk) 7→ xk.
Then for x ∈ Sk−1, dfx : Tx(Sk−1) → R is the restriction of dπx = π. Thus, x ∈ Sk−1 is a critical
value iff Tx(S
k−1) = Rk−1 × {0}. Since, Sk−1 = g−1(1) where g : x → ‖x‖2, Ta(Sk−1) = ker(x 7→
2atx), Ta(S
k−1) = Rk−1 × {0} exactly when a = (0, . . . , 0, 1) := N or (0, . . . , 0,−1) := S. Now,
φ± : Rk−1 ⊃ B1(0)→ R, x 7→ x×±
√
1− ‖x‖2 are local parametrizations of N,S. And calculating
H(f ◦ φ+)0, we have −I (hence max), and for the − case we have I.
Exercise 47 (1.7.16). Let U ⊂ Rk open, f : U → R smooth, and H(x) be the Hessian of f for
x ∈ U . Then f is Morse if and only if
det(H)2 +
k∑
i=1
(
∂f
∂xi
)2
> 0 on U
13
Solution) x ∈ U is a critical point of f iff ∇f = 0, and hence iff
∑k
i=1
(
∂f
∂xi
)2
= 0. The rest is
follows immediately from definition of Morse. �
Exercise 48 (1.7.17,18). (Stability of Morse Functions) Suppose ft is a homotopic family of func-
tions on Rk. Show that if f0 isMorse in some neighborhood f a compact set K, then so is every ft
for t sufficiently small. And thus, Morse function is stable class.
Solution) Let U ⊂ Rk be open containing K such that f0 : U → R is Morse. Now, denote the
Hessian of ft at x by Hft|x, and define h : Rk × I → R by (x, t) 7→ det(Hft|x)2 +
∑k
i=1
(
∂ft
∂xi
)2
.
Clearly, h is smooth, and by [Exercise 1.7.16], we know that h > 0 on U × {0}, hence on K × {0}.
Since K × {0} is compact, h ≥ 2δ for some δ > 0. By continuity of h, there exists an open set
U ′ ⊃ K × {0} such that h > δ on U ′. By Tube Lemma, ∃� > 0 such that h > δ on K × [0, �]. Using
continuity of h again, for any fixed t ∈ [0, �] there exists open V ⊃ K such that h > 0 on V × {t},
as desired.
Now let X be a compact manifold, f0 : X → R Morse, and ft homotopic family of functions.
Suppose for any x ∈ X, there exists Ux, neighborhood of x, and �x > 0 such that ft is Morse on Ux
for t ∈ [0, �x]; then
⋃
x Ux form an open over of X, so choosing a finite sub cover Ux1 ∪ · · · ∪ Uxn ,
we have that ft is Morse on X for any t ∈ [0,min(�x1 , . . . , �xn)]. Existence of Ux and �x is given by
[Exercise 1.7.17]: for φ : Vx → X local parametrization around x with φ(0) = x, set Ux = φ(Br(0))
where Br(0) ⊂ Vx. �
1.8 Embedding Manifolds in Euclidean Space
A few practice with tangent bundles:
Exercise 49 (1.5.2). Let g : X → R be smooth, everywhere-positive. Then the map f : T (X) →
T (X) given by (x, v) 7→ (x, g(x)v) is smooth.
Solution) If m : R×RN → RN is scalar multiplication, ∆ : RN → RN ×RN diagonal map, then
f is just restriction of (Id×m) ◦ (Id×g × Id) ◦ (∆× Id). �
Exercise 50 (1.5.3,4). T (X × Y ) is diffeomorphic to T (X) × T (Y ). T (S1) is diffeomorphic to
S1 × R.
Solution) First statement is immediate from T(x,y)(X ×Y ) = Tx(X)×Ty(Y ). The map (x, v) 7→
(x, ‖v‖) is a diffeomorphism of T (Sk) and S1 × R. �
Exercise 51 (1.5.6). A vector field ~v on a manifold X in RN is a smooth map ~v : X → RN such
that ~v(x) ∈ Tx(X). Equivalently, a vector field ~v on X is a cross section of T (X)—i.e. a smooth
map ~v : X → T (X) such that p ◦ ~v is identity on X. Lastly, x ∈ X is zero of a vector field ~v if
~v(x) = 0.
Solution) (Just definitions)
14
Exercise 52 (1.5.7,8). If k is odd, there exists a nonvanishing vector field ~v on Sk. If Sk has a
nonvanishing vector field ~v, then its antipodal map is homotopic to the identity.
Solution) If k is odd, then the map (x1, . . . , xk+1) 7→ (−x2, x1,−x4, x3, . . . ,−xk+1, xk) is a (linear)
smooth map, i.e. a nonvanishing vector field on Sk. Now, let ~v is a nonvanishing vector field on
Sk, and WLOG ‖~v(x)‖ = 1, ∀x ∈ Sk (define new vector field by x 7→ ~v(x)/‖~v(x)‖), and thus
~v : Sk → Sk smooth and x⊥~v(x). Now, H : Sk × I → Sk defined by (x, t) 7→ x cosπt+ ~v(x) sinπt is
a homotopy between the identity and antipodal map. �
15
2 Chapter 2. Transversality and Intersection
2.1 Manifolds with Boundary
Exercise 53 (Made-up). Let f : X → Y be a smooth map of manifolds with boundary, Z ⊂ X a
submanifold with boundary, and define g := f |Z . Then for x ∈ Z, dgx : Tx(Z) → Tf(x)(Y ) is equal
to dfx|Tx(Z).
Solution) The proof reads exactly like [Exercise 17]. �
Exercise 54 (2.1.2). Let f : X → Y be a diffeomorphism of manifolds with boundary. Then ∂f
maps ∂X diffeomorphically onto ∂Y .
Solution) A diffeomorphism X
f→ Y is locally equivalent to U Id→ U where U ⊂ Hk. So, it follows
that ∂f(x) ∈ ∂Y if and only if x ∈ ∂X. �
Exercise 55 (2.1.3). S := [0, 1]× [0, 1] is a not a manifold with boundary.
Solution) Suppose S is a manifold with boundary, then s = (0, 0) ∈ S has a neighborhood
U ⊂
open
I2 diffeomorphic to V ⊂
open
Hk. Let f : U → V be the diffeomorphism, and shrinking if U
if necessary, let F : Ũ → R2 be the smooth extension of f where U ⊂ Ũ ⊂
open
R2. Then dFs is an
isomorphism.
Now, note that ∂U maps to ∂Hk, so if F = (F1, F2) then F2(x, 0) = 0 = F2(0, y) for any
(x, 0), (0, y) ∈ U . Thus, ∂F2∂x (s) = 0 an
∂F2
∂y (s) = 0. But this implies that dFs(e1), dFs(e2) ∈ R× {0}
and thus not linearly dependent. �
Exercise 56 (2.1.4). The solid hyperboloid defined by x2 +y2− z2 ≤ a is a manifold with boundary.
Solution) Define π : R3 → R by (x, y, z) 7→ a − (x2 + y2 − z2). Since a > 0, it is easily checked
that 0 is regular value of π. Hence, π−1([0,∞)) is a manifold with boundary. �
Exercise 57 (2.1.7). Let X be a manifold with boundary, x ∈ ∂X, φ : U → X local parametrization
with φ(0) = x (so dφ0 : Rk → Tx(X) is isomorphism). We define the upper half space Hx(X) in
Tx(X) by Hx(X) := dφ0(H
k). Hx(X) is independent of the choice of parametrization.
Solution)
Exercise 58 (2.1.8).
Solution)
2.2 Transversality
16