HOMEWORK 1 - RIEMANNIAN GEOMETRY
ANDRÉ NEVES
1. Exercises
1. Let M , N be two smooth manifolds. Show that the set
M × N = {(x, y) | x ∈ M, y ∈ N }
is also a smooth manifold.
This one is simple. The thing to check is that if (φα , Uα )α∈Λ and (ψβ , Vβ )β∈Λ̄
are the charts for M and N respectively, then (fα,β , Uα,β ×Vβ )(α,β)∈Λ×Λ̄ with
fα,β : Uα,β × Vβ :−→ M × N
fα,β (x, y) = (φα (x), ψβ (y))
are admissible charts for M × N .
2. Consider X : Cp −→ R. Show that X ∈ Tp M if and only if
i) X(f + cg) = X(f ) + cX(g) for every f, g ∈ Cp , c ∈ R;
ii) X(f g) = f (p)X(g) + g(p)X(f ) for every f, g ∈ Cp ;
Let’s show the easy direction first. If X ∈ TP M then, by definition, there is
d
(f ◦ γ)(0). Thus
a curve γ in M with γ(0) = p such that X(f ) = dt
X(f +cg) =
d
d
d
((f +cg)◦γ)(0) = (f ◦γ)(0)+c (g ◦γ)(0) = X(f )+cX(g)
dt
dt
dt
and
d
((f g) ◦ γ)(0)
dt
d
d
= g(γ(0)) (f ◦ γ)(0) + f (γ(0)) (g ◦ γ)(0) = f (p)X(g) + g(p)X(f ).
dt
dt
Now the other direction. Pick a chart (φα , Uα ) with φα (0) = p and consider
the coordinate functions x̃i = xi ◦ φ−1
α defined on φα (Uα ), where xi are the
coordinates of Uα ⊂ Rn , i = 1, ..., n. Now set ai = X(x̃i ) and we are going
to show that
n
X
∂
X=
ai
∂xi
X(f g) =
i=1
1
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ANDRÉ NEVES
meaning that for every function f on M we have
n
X
∂
X(f ) =
(f ◦ φα )(0)
ai
∂xi
i=1
or, equivalently, that for every function f on Uα ⊂ Rn we have
n
X
∂f
X(f ◦ φ−1
)
=
(0).
ai
α
∂xi
i=1
~ (0) = 0. Then by Taylor’s formula
Say that f ∈ C ∞ (Uα ) is such that ∇f
∞
we can find functions bij ∈ C (Uα ) such that
f (x) = f (0) +
n
X
bij xi xj .
i≤j
Hence, from property ii) we get that by setting b̃ij = bij ◦ φ−1
α
X(f ◦
φ−1
α )
= X(f (0)) +
n
X
X(b̃ij x̃i x̃j )
i≤j
= X(f (0)) +
n X
x̃j (p)X(b̃ij x̃i ) + b̃ij (p)x̃i (p)X(x̃j ) = X(f (0))
i≤j
because x̃i (p) = 0 for all i = 1, ..., n. Furthermore we have form the linearity
of property i) and property ii) that
X(f (0)) = f (0)X(1) = f (0)X(1∗1) = f (0)(X(1)+X(1)) =⇒ X(f (0)) = 0.
~ (0) = 0 we have X(f ◦ φ−1 ) = 0.
Thus for every f with ∇f
Pα ∂f
(0)xi is such that
Pick any f ∈ C ∞ (Uα ). Then g(x) = f (x) − ni=1 ∂x
i
~
∇g(0) = 0 and hence
X(f ◦
φ−1
α )
= X(g ◦
φ−1
α
n
n
X
X
∂f
∂f
+
(0)x̃i ) = X(
(0)x̃i )
∂xi
∂xi
i=1
i=1
n
n
X
X
∂f
∂f
=
(0)X(x̃i ) =
ai
(0).
∂xi
∂xi
i=1
i=1
3. On S 3 = {x ∈ R4 | |x|2 = 1} consider the vector fields
E1 (x) = (−x2 , x1 , x4 , −x3 ),
E2 (x) = (x3 , x4 , −x1 , −x2 )
and
E3 (x) = (−x4 , x3 , −x2 , x1 ).
i) Show that each these vectors belong to the tangent space of S 3 , i.e.,
for each x ∈ S 3 , Ei (x) ∈ Tx S 3 .
ii) Show that {E1 (x), E2 (x), E3 (x)} forms a basis for Tx S 3 .
HOMEWORK 1 - RIEMANNIAN GEOMETRY
3
iii) Conclude that T S 3 is diffeomorphic to S 3 × R3 .
Before we do this exercise that is some facts we need to establish. Say that
zero is a regular value of a smooth map F : Rn+k −→ Rk which implies that
M = F −1 (0) is a smooth n-dimensional manifold as described in Example
4.3. page 18 of Do Carmo’s book. We can consider the inclusion map
i : M −→ Rn+k which, using the charts described by Do Carmo in Example
4.3. one can see that it is a smooth map with Dip : Tp M −→ Ti(p) Rn+k
injective for all p in M . For this reason we set Lp = Dip (Tp M ) which is a n
dimensional linear subspace of Ti(p) . For this reason, we identify vectors in
Tp M with vectors in Lp . We note that these vectors are based at i(p) and so
we translate them to the origin so that Lp is nothing but a linear subspace
of Rn+k .
Now the question becomes which subspace is this? We argue that
(1)
Lp = {~v ∈ Rn+k | DFi(p) (~v ) = 0}
or, more correctly but more cumbersome as well,
Lp = {i(p) + ~v ∈ Rn+k | DFi(p) (~v ) = 0}.
For simplicity, we will always use the first description instead of the latter,
meaning I write the first but I know what I really mean is the second. More
often that not I will just refer to a vector ~v in Lp as being a vector in Tp M
where, technically speaking, I should refer to (Dip )−1 (~v ) instead.
Set
Qp = {~v ∈ Rn+k | DFi(p) (~v ) = 0}.
It suffices to show that LP ⊂ Qp because Qp is n-dimensional and so if the
inclusion holds then equality must hold as well.
Denote F = (F1 , ..., Fk ) and consider ~v = Dip (X) where, by definition,
X = dγ
dt (0) ∈ Tp M for some curve γ in M with γ(0) = p. We have for every
j = 1, ..., k
~ j (i(p))i = ~v (Fj )(i(p)) = X(Fj ◦ i)(p) = d (Fj ◦ i ◦ γ)(0) = 0
h~v , ∇F
dt
because Fj ◦ i ◦ γ(t) is zero for all t (recall that M = F −1 (0)). Thus
~ j (i(p))i = 0 for all j = 1, ..., k and so DFi(p) (~v ) = 0 as we wanted
h~v , ∇F
to prove.
We note that using this description we see that M has a naturally induced
metric from Rn+k that we define as
gp (X, W ) = hDip (X), Dip (Y )i.
We can now solve the first question but first a disclaimer. If i denotes
the inclusion map of S 3 into R4 I will just identify x in S 3 with i(x) in R4
P
without further care. In this case n = 3, k = 1, F (x) = 4i=1 x2i − 1 for
every x ∈ R4 , and the linear map DFx is simply
~ (x), ~v i = 2hx, ~v i.
DFx (~v ) = h∇F
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ANDRÉ NEVES
Thus to make sure that Ei (x) are tangent vectors wioth i = 1, 2, 3 we need
to show that Ei (x) ∈ Lx using the notation of (1). More precisely we need
to see that
DFx (Ei (x)) = 2hx, Ei (x)i = 0.
I only do this when i = 1 and the others is just more of the same.
hx, E1 (x)i = h(x1 , x2 , x3 , x4 ), (−x2 , x1 , x4 , −x3 )i
= −x1 x2 + x1 x2 + x3 x4 − x3 x4 = 0.
Now we do the second question. Let g denote the induced metric on S 3
and set gi,j (x) = hEi (x), Ej (x)i for i, j = 1, 2, 3. An exhaustive computation
shows that for every x ∈ S 3 , gi,j (x) is 1 if i = j and zero otherwise. Suppose
now that the vectors E1 , E2 , E3 are not linearly independent for some x0 ∈
S 3 . Then there are a1 , a2 , a3 so that the vector V = a1 E1 (x0 ) + a2 E2 (x0 ) +
a3 E3 (x0 ) is zero. By what I have just said in the beginning of this paragraph
we have that
0 = hV, Ei (x0 )i = ai
and so all the ai have to be zero.
For the third question consider the map Φ : S 3 × R3 −→ T S 3 so that
Φ(x, (a1 , a2 , a3 )) = (x, a1 E1 (x) + a2 E2 (x) + a3 E3 (x)).
I will just check that the map is one to one but using the charts for S 3 you
should be able to see that the maps is actually smooth and its differential
is injective. If (Φ(x, (a1 , a2 , a3 )) = Φ(y, (b1 , b2 , b3 )), then y = x and so
a1 E1 (x) + a2 E2 (x) + a3 E3 (x) = b1 E1 (x) + b2 E2 (x) + b3 E3 (x).
By question ii), we have that the vectors (Ei )3i=1 form a basis of Tx S 3
and so indeed ai = bi . By the same token if X in Tx S 3 , then X is a
linear combination of E1 , E2 , E3 and thus we can find a1 , a2 , a3 so that
Φ(x, (a1 , a2 , a3 )) = (x, X).
4. Show that the tangent manifold of S 1 = {x ∈ R2 | |x|2 = 1}, T S 1 , is
diffeomorphic to S 1 × R. Show that the tangent manifold of T 2 = S 1 × S 1
is diffeomorphic to T 2 × R2 .
To do the first part we consider the vector field E1 on S 1 so that E1 (x, y) =
(y, −x) for every (x, y) ∈ S 1 . The first thing to check is that E1 is indeed a
vector field on S 1 . For this we just need to see that E1 (x, y) is perpendicular
to the gradient of F (x, y) = x2 + y 2 − 1 which is trivially true. Having that
we can now define
Φ : S 1 × R −→ T S 1
Φ((x, y), a) = ((x, y), aE1 (x, y)).
We need to check that Φ is a diffeomorphism. This one is pretty easy and
so I will do it. To check that Φ is one to one is easy. Let’s check that
HOMEWORK 1 - RIEMANNIAN GEOMETRY
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it is smooth and that its differential is always injective. Pick a point p in
S 1 which, without loss of generality, we assume to be p = (1, 0). Then
we have the chart φ : (−π/2, π/2) −→ S 1 where φ(θ) = (cos θ, sin θ). For
∂
these coordinates we have that ∂θ
= (− sin θ, cos θ). Hence, we obtain the
local diffeomorphism φ̃ : (−π/2, π/2) × R −→ T S 1 such that φ̃(θ, t) =
(φ(θ), (−t sin θ, t cos θ)). We can check that
G(θ, a) = φ̃ ◦ Φ(φ(θ), a) = (θ, −a)
which is clearly smooth with injective differential.
For the second question the shortest way is probably to note that T (M ×
N ) is diffeomorphic to T M × T N for every M, N smooth manifolds and
thus using what we have just done we obtain that, up to diffeomorphisms,
T (S 1 × S 1 ) = T S 1 × T S 1 = S 1 × R × S 1 × R and this is clearly diffeomorphic
to S 1 × S 1 × R2 . The long way would be to note that S 1 × S 1 = F −1 (0)
where F : R4 −→ R2 is given by F (x, y, z, w) = (x2 + y 2 − 1, z 2 + w2 − 1)
and to consider the vector fields
E1 (x, y, z, w) = (−y, x, 0, 0)
and E2 (x, y, z, w) = (0, 0, −w, z).
Then, one would use the machinery of exercise 3 to see that E1 and E2
belong to T (S 1 × S 1 ), i.e., they lie on the kernel of the linear map
DF(x,y,z,y) (v1 , v2 , v3 , v4 ) = 2(xv1 + yv2 , zv3 + wv4 ).
After that, we would notice that E1 , E2 for a basis for T (S 1 × S 1 ) and then
define the map
Φ : S 1 × S 1 × R2 −→ T (S 1 × S 1 )
by setting Φ(p, q, (u, v)) = (p, q, uE1 (p, q)+vE2 (p, q)). If we aren’t exhausted
already we would still see that Φ is indeed a diffeomorphism.
5. Given a properly discontinuous action F : G × M −→ M on a smooth
manifold M , show that M/G is orientable if and only if M is orientable and
F (g, ·) preserves the orientation.
Let’s assume first that M/G is orientable. M being orientable is equivalent
to have a well defined way of saying whether any given basis {ei }ni=1 of Tp M
is positively oriented or not. Consider the projection map π : M −→ M/G
which we know it is a local diffeomorphism. We define an orientation on
M by saying that a basis {ei }ni=1 of Tp M is positively oriented if and only
if {Dπp (ei )}ni=1 forms a positive basis of T[p] M/G. We now need to check
that Fg = F (g, ·) preserve the orientation, meaning that if {ei }ni=1 is a
positive basis of Tp M then {D(Fg )p (ei )}ni=1 is a positive basis of TFg (p) M .
This follows at once because π ◦ Fg = π and thus DπFg (p) (D(Fg )p (ei )) =
D(π ◦ Fg )p (ei ) = Dπp ei .
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ANDRÉ NEVES
Now for the other direction. Pick {ei }ni=1 a basis of T[p] M/G. Then, π
is a diffeomorphism in a neighborhood U of p and so we can consider the
basis {Dπp−1 (ei )}ni=1 of Tp M . We say that {ei }ni=1 is positively oriented if
and only if {Dπp−1 (ei )}ni=1 is positively oriented. To see that this is well
defined we need to make sure that if [p] = [q] on M/G then {Dπp−1 (ei )}ni=1
is positive if and only if {Dπq−1 (ei )}ni=1 is positive. This follows at once
from the fact that if Fg (p) = q then this map preserved the orientation and
D(Fg )p (Dπp−1 (ei )) = Dπq−1 (ei ).
i) Show that the Mobius band is not orientable and that RPn = S n /Z2
is orientable if and only if n is odd.
The Mobius band was defined as R2 /Z where for every n ∈ Z, Fn (x, y) = (x+
n, (−1)n y) and we identify (x, y) with Fn (x, y). As we saw in the class, when
n = 1 the map F1 does not preserve orientation and so the Mobius band it
is not orientable. Likewise, the antipodal map of S n preserves orientation if
and only if n is odd as we saw in class and so the result follows.
6. Given any nonorientable manifold M show the existence of a smooth
orientable manifold M̄ which is a double covering of M , i..e, there is a local
diffeomorphism π : M̄ −→ M such that π −1 (p) is a set with two elements
for every p in M . (HINT: Follow the leads of exercise 12, page 36, of Do
Carmo’s book.)
i) Find M̄ when M is either RP2 or the Mobius band.
The leads of Do Carmo’s book, page 36 are pretty much self explanatory. If
you are confused with something just ask me. For the second part, well, S 2
is a double cover of RP2 and is orientable, hence it is the orientable double
covering. For the Mobius band we consider the cylinder M1 = R2 /Z where
the Z action is given by Gn (x, y) = (x + 2n, y), i..e, on M1 we identify
(x, y) with Gn (x, y) for all n ∈ Z. Denoting the Mobius band by M2 (see
construction in previous exercise) we can consider the map π : M1 −→ M2
given by π[(x, y)] = [(x, y)]. We need to check that this is well defined and
that it is a double covering. It is well defined because if Gn (x, y) = F2n (x, y)
is equivalent to (x, y) on M2 . It is a double covering because if π[(x, y)] =
π[(u, v)] then this means Fn (x, y) = (u, v) for some n. If n is even than
they are also the same point on M1 and if n is odd we have that (u, v) is
equivalent to (x + 1, −y).
HOMEWORK 1 - RIEMANNIAN GEOMETRY
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7. Set B1 to be the open ball in R3 centred at the origin. Let (M g) be a
Riemannian three manifold with two charts φ1 : B1 −→ M , φ2 : B1 −→ M ,
both with φ1 (0) = p, φ2 (0) = p. Furthermore, near the origin, we know that
2
φ−1
2 ◦ φ1 (x, y, z) = (2x + y , exp(y) − 1 − z, y(x + 1) + z)
and that, with respect to the chart φ1 , the metric at the origin can be written
as (gij ) where
g11 = g22 = 4, g12 = g21 = 1, g13 = g31 = g23 = g32 = 0, g33 = 1
i) If X is a vector in Tp M such that its coordinate representation with
respect to φ2 is (2, 0, 4), find its coordinate representation with respect to φ1 .
v = (2, 0, 4)
Let h = φ−1
2 ◦ φ1 . The formula we saw in class is that if ~
−1
then the vector we are looking for is Dh0 (~v ). The matrix Dh−1
0 is
the inverse of the matrix Dh0 where the j-column and i-row is given
∂hi
by ∂x
. Hence
j


2 0 0
Dh0 =  0 1 −1 
0 1 1
and so

1
2
0
(Dh0 )−1 =  0 12
0 − 12
0

1
2
1
2
.
Therefore, the coordinate representations are (1, 2, 2).
ii) Show that the metric described is indeed a metric, i..e, a symmetric
positive definite matrix.
The metric is represented by the

4 1
g= 1 4
0 0
matrix

0
0 .
1
It is trivially symmetric and we need to check that it has positive
eigenvalues to make sure it is positive definite. An easy computation
shows that the eigenvalues are 1, 3 and 5.
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ANDRÉ NEVES
iii) Find the metric with respect to the coordinates induced by φ2 and
compute g(X, X) where X is given as above.
The formula derived in the class was that the matrix we are looking
for is ḡ = AT gA, where A = (Dh0 )−1 . A simple computation shows
that


4 1 1
1
1 5 3 .
ḡ =
4
1 3 5
The norm g(X, X) is given by h(2, 0, 4), ḡ(2, 0, 4)T i = 28.