sol_HW5

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MATH 4030 Differential Geometry
Homework 5
Suggested solutions
1. (1 point)
d
dt
〈X(t), Y (t)〉 = 〈Dα′(t)X(t), Y (t)〉+ 〈X(t), Dα′(t)Y (t)〉
= 〈∇α′(t)X(t) + (Dα′(t)X(t))⊥, Y (t)〉+ 〈X(t),∇α′(t)Y (t) + (Dα′(t)Y (t))⊥〉
= 〈∇α′(t)X(t), Y (t)〉+ 〈X(t),∇α′(t)Y (t)〉
If X and Y are parallel, then ∇α′(t)X(t) = ∇α′(t)Y (t) = 0, and hence
d
dt
〈X(t), X(t)〉 = d
dt
〈X(t), Y (t)〉 = d
dt
〈Y (t), Y (t)〉 = 0.
It follows that the angle between X and Y is cos−1
[
〈X(t), Y (t)〉
|X(t)||Y (t)|
]
which is constant.
2. (2 points) By putting i = j = 1, k = ` = 2 into the Gauss equation ∂kΓ
`
ij−∂jΓ`ik + Γ
p
ijΓ
`
pk−
ΓpikΓ
`
pj = g
`p(AijAkp − AikAjp), we have
det(Aij) =
1
g22
(∂2Γ
2
11 − ∂1Γ212 + Γ
p
11Γ
2
p2 − Γ
p
12Γ
2
p1).
We compute 
Γ111 =
Eu
2E
Γ112 =
Ev
2E
Γ122 = −
Gu
2E
and

Γ211 = −
Ev
2G
Γ212 =
Gu
2G
Γ222 =
Gv
2G
,
and then
det(Aij) = G
[(
− Ev
2G
)
v
−
(
Gu
2G
)
u
+
(
Eu
2E
)(
Gu
2G
)
+
(
− Ev
2G
)(
Gv
2G
)
−
(
Ev
2E
)(
− Ev
2G
)
−
(
Gu
2G
)(
Gu
2G
)]
=
G
2
{
−
(
Ev
G
)
v
−
(
Gu
G
)
u
− 1
2
[
Gv
G
− Ev
E
](
Ev
G
)
− 1
2
[
Gu
G
− Eu
E
](
Gu
G
)}
= −
√
EG
2
[(√
G
E
)(
Ev
G
)
v
+
(√
G
E
)(
Gu
G
)
u
+
(√
G
E
)
v
(
Ev
G
)
+
(√
G
E
)
u
(
Gu
G
)]
= −
√
EG
2
{[(√
G
E
)(
Ev
G
)
v
+
(√
G
E
)
v
(
Ev
G
)]
+
[(√
G
E
)(
Gu
G
)
u
+
(√
G
E
)
u
(
Gu
G
)]}
= −
√
EG
2
[(
Ev√
EG
)
v
+
(
Gu√
EG
)
u
]
.
1
It follows that
K =
det(Aij)
det(gij)
= − 1
2
√
EG
[(
Ev√
EG
)
v
+
(
Gu√
EG
)
u
]
.
Now suppose E = G = λ(u, v), then
K = − 1
2λ
[(
λv
λ
)
v
+
(
λu
λ
)
u
]
= − 1
2λ
[
(log λ)vv + (log λ)uu
]
= − 1
2λ
∆(log λ).
3. (2 points) X(u, v) = (u cos v, u sin v, log u)
Xu =
(
cos v, sin v,
1
u
)
Xv = (−u sin v, u cos v, 0)
g =
(
〈Xu, Xu〉 〈Xu, Xv〉
〈Xu, Xv〉 〈Xv, Xv〉
)
=
(
1 + 1
u2
0
0 u2
)
By Q2 above,
K = − 1
2
√
1 + u2
[(
0√
1 + u2
)
v
+
(
2u√
1 + u2
)
u
]
= − 1
(1 + u2)2
.
X̃(u, v) = (u cos v, u sin v, v)
X̃u = (cos v, sin v, 0)
X̃v = (−u sin v, u cos v, 1)
g =
(
〈X̃u, X̃u〉 〈X̃u, X̃v〉
〈X̃u, X̃v〉 〈X̃v, X̃v〉
)
=
(
1 0
0 1 + u2
)
K̃ = − 1
2
√
1 + u2
[(
0√
1 + u2
)
v
+
(
2u√
1 + u2
)
u
]
= − 1
(1 + u2)2
= K.
Since the first fundamental forms of X and X̃ are different, it follows that X̃ ◦X−1 is not
an isometry.
2
4. (2 points) X(u, v) = (f(v) cosu, f(v) sinu, g(v))
Xu = (−f(v) sinu, f(v) cosu, 0)
Xv = (f
′(v) cosu, f ′(v) sinu, g′(v))
(gij) =
(
f 2(v) 0
0 (f ′(v))2 + (g′(v))2
)
and (gij) =
(
1
f2(v)
0
0 1
(f ′(v))2+(g′(v))2
)
Now recall
Γkij =
1
2
gk`(∂ig`j + ∂jgi` − ∂`gij).
We have
Γ111 =
1
2f 2(v)
(∂1g11 + ∂1g11 − ∂1g11) = 0
Γ112 =
1
2f 2(v)
(∂1g12 + ∂2g11 − ∂1g12) =
f ′(v)
f(v)
Γ122 =
1
2f 2(v)
(∂2g12 + ∂2g21 − ∂1g22) = 0
Γ211 =
1
2[(f ′(v))2 + (g′(v))2]
(∂1g21 + ∂1g12 − ∂2g11) =
−f(v)f ′(v)
(f ′(v))2 + (g′(v))2
Γ212 =
1
2[(f ′(v))2 + (g′(v))2]
(∂1g22 + ∂2g12 − ∂2g12) = 0
Γ222 =
1
2[(f ′(v))2 + (g′(v))2]
(∂2g22 + ∂2g22 − ∂2g22) =
f ′(v)f ′′(v) + g′(v)g′′(v)
(f ′(v))2 + (g′(v))2
.
Alternatively, observe that X is an orthogonal chart, i.e. 〈Xu, Xv〉 = 0 which means that
finding Γkij directly from its definition will not be too complicated, since in this case we
have
Γkij =
〈Xij, Xk〉
〈Xk, Xk〉
.
Try it yourself!
5. (1 point) By HW3 Q7, the Gauss curvature of the saddle surface {z = x2 − y2} at (0, 0, 0)
is −4 < 0. On the other hand, we know that any round sphere of radius R > 0 has Gauss
curvature 1
R
> 0 everywhere, and any cylinder has zero Gauss curvature everywhere (see
for example Tut6 Q1a). Since local isometry between two surfaces preserves their Gauss
curvatures, it follows that the saddle surface cannot be locally isometric to any round sphere
or cylinder.
6. (2 points)
(a) Since all gij’s are constant, we have Γ
k
ij ≡ 0 for all i, j, k = 1, 2. It follows that
∂2Γ
2
11 − ∂1Γ212 + Γ
p
11Γ
2
p2 − Γ
p
12Γ
2
p1 = 0 (i = j = 1, k = ` = 2)
g2p(A11A2p − A12A1p) = det(Aij) = −1.
Since the pair (gij), (Aij) does not satisfy the Gauss equation, we conclude that there
does not exist any parametrization such that the first and second fundamental forms
are (gij) and (Aij) respectively.
3
(b) Assume cosu 6= 0 on U . Observe that the only term ∂kgij which may be non-zero is
∂1g22 = −2 sinu cosu. It follows that
Γ212 =
1
2 cos2 u
(−2 sinu cosu) = − tanu
Γ122 =
1
2
(2 sinu cosu) = sinu cosu
,
and hence
∂1A22 − ∂2A21 + Γp22Ap1 − Γ
p
21Ap2 = cos
3 u sinu+ tanu (i = j = 2, k = 1).
Since the pair (gij), (Aij) does not satisfy the Codazzi equation, we conclude that there
does not exist any parametrization such that the first and second fundamental forms
are (gij) and (Aij) respectively.
4