HOMEWORK 2 - RIEMANNIAN GEOMETRY
ANDRÉ NEVES
1. Problems
In what follows (M, g) will always denote a Riemannian manifold with a
Levi-Civita connection ∇.
1 Let X, Y, Z be vector fields on M so that X(p) = Z(p) for some p in
M . Show that (∇X Y ) (p) = (∇Z Y ) (p). Find an example that shows that
(∇Y X) (p) might be different from (∇Y Z) (p).
By linearity it suffices to show that if X is a vector field with X(p) = 0 then
(∇X Y ) (p) = 0.
Choose coordinates centered at p, in which case we have, with respect to
those coordinates,
n
X
∂
X(x) =
ui (x)
.
∂xi
i=1
The condition that X(p) = 0 becomes ui (0) = 0 for all i = 1, ..., n because
we can assume without loss of generality that p becomes the origin when
using these coordinates. From linearity we have that
∇X Y (x) =
n
X
i=1
ui (x)∇
∂
∂xi
Y.
Hence, direct substitution shows ∇X Y (0) = 0.
∂
Consider the connection D on R2 and set Y = ∂x
= (1, 0), Z = (0, 0),
∂
and X = x ∂x = (x, 0). At the origin we have X(0) = Z(0) = (0, 0). Now
DY X(0, 0) = (1, 0)
and DY Z(0, 0) = (0, 0)
which are different.
2 Suppose that M = F −1 (0), where F : Rn+k −→ Rk and 0 is a regular
value of F . Following the discussion I wrote on the solutions to HW 1,
1
2
ANDRÉ NEVES
problem 3, we see that we can identify Tp M with a subspace of Rn+k . More
precisely
Tp M = {~v ∈ Rn+k | DFp (~v ) = 0.}
Thus for every point x in M we can consider the map Px : Rn+k −→ Tx M
where Px (~v ) is the projection of the vector ~v on Tp M . Define a connection
on M to be
(∇X Y )(x) = Px (DX Y ),
where X and Y are vector fields on M and D denotes the Euclidean connection (i.e., standard differentiation). Show that this is the Levi-Civita
connection for the induced metric on M .
We need to check that ∇ satisfies all the axioms of being a Levi-Civita
connection. Uniqueness implies that ∇ is indeed the Levi-Civita connection.
For the sake of simplicity we denote Px (X) simply by X T .
The axioms regarding linearity and Leibniz rule are easy. I will just check
the symmetry of the connection and the compatibility with the induced
metric. Let φ : U −→ M be a coordinate chart where U is a subset of Rn .
Denote by ∂y∂ i , i = 1, ..., n the coordinate vectors. Need to see that
∇
∂
∂yj
∂
∂
=∇ ∂
.
∂yi ∂yj
∂yi
Denote by i : M −→ Rn+k the inclusion map. Using this we can regard
the coordinate vectors ∂y∂ i (y) as being a vector in Rn+k for every y ∈ U by
identifying it with
∂(i ◦ φ)
∂(i ◦ φ)1 (y) ∂(i ◦ φ)2 (y)
∂(i ◦ φ)n+k (y)
(y) =
,
, ...,
∂yi
∂yi
∂yi
∂yi
=
n+k
X
j=1
∂(i ◦ φ)j (y) ∂
.
∂yi
∂xj
What this mean is that if f is function defined in a neighborhood of p =
φ(y) on M and if we consider another function f¯ which is defined in a
neighborhood of p on Rn+k such that f = f¯ on M then
∂(f ◦ φ)
∂(i ◦ φ)
~ f¯(p) .
(y) =
(y), ∇
∂yi
∂yi
We can now finish what we are trying to do, i.e., check that ∇
∇
∂
∂yi
∂
∂yj .
∂
∂yj
∂
∂yi
=
Set fl to be the function defined in a neighborhood of p = φ(y)
on M such that fl ◦ φ = (i ◦ φ)l for l = 1, ..., n + k. Likewise, define ∂j fl
on a neighborhood of p = φ(y) on M such that (∂j fl ) ◦ φ =
j = 1, ..., n and l = 1, ..., n + k.
∂(i◦φ)l
∂yj ,
where
HOMEWORK 2 - RIEMANNIAN GEOMETRY
3
Lemma 1.1.
∂
∂
(∂j fl ) =
(∂i fl )
∂yi
∂yl
Proof. By definition, we have that
∂(∂j fl ◦ φ)
∂((∂j fl ) ◦ φ)
∂ 2 (i ◦ φ)l
∂
(∂j fl ) =
=
=
∂yi
∂yi
∂yi
∂yi ∂yj
=
∂ 2 (i ◦ φ)l
∂
=
(∂i fl )
∂yj ∂yi
∂yj
Denote by ∂j fl a function defined on a neighborhood of p in Rn+k which
coincides with ∂j fl when restricted to M . In view of the lemma above we
have
!T
n+k
X
∂
∂(i ◦ φ) T
∂(i ◦ φ)r ∂
= D ∂(i◦φ)
D ∂(i◦φ)
∇ ∂
=
∂yi ∂yj
∂yj
∂yj ∂xr
∂yi
∂yi
r=1
!T
!T
n+k
n+k
X
X ∂(i ◦ φ)
∂
∂
~ j fr
D ∂(i◦φ) ∂j fr
=
, ∇∂
=
∂xr
∂yi
∂xr
∂yi
r=1
r=1
!
!
T
T
n+k
n+k
X ∂((∂j fr ) ◦ φ) ∂
X ∂
∂
(∂j fr )
=
=
∂yi
∂xr
∂yi
∂xr
r=1
r=1
!
T
n+k
X ∂
∂
∂
(∂j fr )
.
=
=∇ ∂
∂yj ∂yi
∂yj
∂xr
r=1
We need to check compatibility of the connection with the Riemannian
metric. Given X, Y, Z vector fields on M we can see them as vector fields
in Rn+k defined on M . For the sake of rigour denote them by X̄, Ȳ , and Z̄.
X(g(Y, Z)) = X(hȲ , Z̄i) = X̄(hȲ , Z̄i) = hDX̄ Ȳ , Z̄i + hȲ , DX̄ Z̄i
= h(DX̄ Ȳ )T , Z̄i + hȲ , (DX̄ Z̄)T i = g(∇X Y, Z) + g(Y, ∇X Z)
3 Let hX, Y i denote the standard Euclidean inner product. On a unit ball
B n = {|x| < 1} consider the metric given by
g(X, Y ) =
4
hX, Y i,
(1 − |x|2 )2
where X, Y are vectors at Tx B n . This metric is called the hyperbolic metric
and it will show up countless many times in the future.
4
ANDRÉ NEVES
i) Compute the Christoffel symbols for this metric with respect to the
Euclidean coordinates.
With respect to the coordinates (x1 , ..., xn ), the matrix that stands
for the metric becomes gij (x) = f (x)δij , where f (x) = 4(1 − |x|2 )−2 .
In this case g ij = f1 δij and so the Christoffel symbols are
∂gij
∂gik
1
∂f
∂f
∂f
1 ∂gjk
k
+
−
=
δjk +
δik −
Γij =
δij
2f ∂xi
∂xj
∂xk
2f ∂xi
∂xj
∂xk
= 2(1 − |x|2 )−1 (δjk xi + xj δik − xk δij )
ii) Show that the line γ(t) = t(1, 0, ..., 0) with 0 ≤ t < 1 can be
parametrized so that it becomes a geodesic for the hyperbolic metric.
What is the length of this geodesic?
First we parametrize the geodesic by unit length. We find t = f (s)
so that α(s) = γ(f (s)) = (f (s), 0, ..., 0) has |α0 (s)| = 1. Because
|α0 (s)| = 2|f 0 |(1 − f 2 )−1 we find f so that
f 0 (s) =
1 − (f (s))2
2
and f (0) = 0.
s
Integration shows that the solution is f (s) = ees −1
+1 . The equation
above also implies that f 00 = −f f 0 by differentiating both sides.
Therefore
0
0
0
0
0 2
∇ α = ∇ (f ∂x1 ) = α (f )∂x1 + (f )
α0
α0
n
X
Γk11 ∂xk
k=1
00
0 2
2 −1
= f ∂x1 + 2(f ) f (1 − f )
∂x1 = (−f f 0 + 2(f 0 )2 f (1 − f 2 )−1 )∂x1
= f 0 f (−1 + 2f 0 (1 − f 2 )−1 )∂x1 = 0.
The parameter s ranges from 0 to infinity and so the length γ with
respect to the hyperbolic metric is infinite.
4* Consider a curve α(s) = (r(s), z(s)) on the plane R2 where 0 < s < b and
α is parametrized by arc length, i.e., |α0 (s)| = 1 for all 0 < s < b. Moreover
r(s) is always positive and the curve α has no self-intersections. With this
curve we can consider a surface in R3 obtained by rotating this curve around
the z-axis which will be parametrized by
S : (0, b) × [0, 2π] −→ R3
where S(s, θ) = (r(s) cos θ, r(s) sin θ, z(s)).
Make yourself a drawing to make sure you understand what this is.
HOMEWORK 2 - RIEMANNIAN GEOMETRY
5
i) Show that, with respect to the coordinates (s, θ), the induced metric
becomes gss = 1, gsθ = 0, gθθ = r2 (s).
Consider the coordinates (s, θ). The coordinate vectors are
∂s = (r0 cos θ, r0 sin θ, z 0 )
and ∂θ = (−r sin θ, r cos θ, 0).
Hence we have that gsθ = h∂s , ∂θ i = r0 r cos θ sin θ − r0 r cos θ sin θ = 0
and gθθ = h∂θ , ∂θ i = r2 . Likewise gss = h∂s , ∂s i = (r0 )2 + (z 0 )2 =
|α0 |2 = 1.
ii) If γ is a geodesic on S which, using the coordinates (s, θ), becomes
γ(t) = (s(t), θ(t)), then the following equations hold:
d2 θ
r0 dθ ds
+
2
=0
dt2
r dt dt
2
d2 s
0 dθ
− rr
= 0,
dt2
dt
where r0 denotes differentiation with respect to the s-variable. Moreover, when you see something like rr0 what this really means is
r(s(t))r0 (s(t)). I did not include this because then the formulas
would become a mess.
Let’s compute the Christoffel symbols first. The formula to use is
Γkij =
n
X
1
l=1
2
g kl (∂i gjl + ∂j gil − ∂l gij ) .
Using the fact that g ss = 1, g sθ = 0, and g θθ =
Γsss = 0
Γθss = 0
Γssθ = 0
Γθsθ =
r0
r
1
,
r2
Γsθθ = −rr0
we have
Γθθθ = 0
which means that
∇∂s ∂s = 0,
∇∂s ∂θ = ∇∂θ ∂s =
r0
∂θ
r
∇∂θ ∂θ = −r0 r∂s .
Hence
ds
dθ
ds
dθ
0 ds
0 dθ
∇γ 0 γ = ∇γ 0
∂s + ∂θ = γ
∂s +γ
∂θ + ∇γ 0 ∂s + ∇γ 0 ∂θ
dt
dt
dt
dt
dt
dt
2
2
ds
dθ
ds
dθ ds
dθ
= γ0
∂s + γ 0
∂θ +
∇ ∂s ∂ s + 2
∇∂s ∂θ +
∇∂θ ∂θ
dt
dt
dt
dt dt
dt
2
ds
dθ
dθ ds r0
dθ
= γ0
∂s + γ 0
∂θ + 2
∂θ −
r0 r∂s
dt
dt
dt dt r
dt
0
6
ANDRÉ NEVES
d2 s
0 dθ = d2 θ . This has been some
We note that γ 0 ds
dt = dt2 and γ
dt
dt2
source of confusion so I will explain why. When we write γ 0 ds
dt ,
what we mean is that we consider a function f defined on a neighborhood of γ(t0 ) in S so that f (γ(t)) = ds
dt for all t close to t0 . This
0
is because, technically speaking, γ is a vector and so it differentiates
functions on S. Thus by definition of vector we obtain that
ds
d
d(f
◦
γ)
d2 s
dt
γ 0 (f ) =
=
= 2.
dt
dt
dt
Having said that we finally have
2 !
2
2s
dθ
d θ
r0 dθ ds
d
0
0
−rr
∂s +
+2
∂θ .
∇γ 0 γ =
dt2
dt
dt2
r dt dt
Thus ∇γ 0 γ 0 being zero implies the desired result.
iii) Conclude that the meridian lines (i.e. curves with θ constant) are
geodesics and that the parallels (curves with s constant) are geodesics
when r0 (s) = 0. Draw a picture to convince yourself that this makes
sense.
(NOTE: I am freely identifying a geodesic with its image. More
precisely, when I ask you to show that meridians are geodesics what I
mean is that you need to find a parametrization so that they become
geodesics (same thing for parallels)).
Parametrize a meridian by γ(t) = (t, θ0 ). Then it is straightforward
to see that the equation is satisfied. Regarding the parallels, parametrize them by γ(t) = (s0 , t). Because r0 (s0 )r(s0 ) = 0, it is simple to
see that γ is a geodesic as well.
iv) Conclude from ii) that
r2
dθ
is constant as a function of t
dt
and that
2
2
ds
2 dθ
+r
is constant as a function of t.
dt
dt
For the first identity we use
2
2
d
dθ
dr dθ 2 d2 θ
d θ
r0 dθ ds
0 ds dθ
2d θ
r2
= 2r
+r
=
2rr
+r
=
r
+
2
= 0.
dt
dt
dt dt
dt2
dt dt
dt2
dt2
r dt dt
HOMEWORK 2 - RIEMANNIAN GEOMETRY
7
Thus f (t) = r2 (s(t)) dθ
dt is a constant function and the result follows.
For the second one we argue in the same way.
2
2 !
ds
d
d2 s ds
dr dθ 2
d2 θ dθ
2 dθ
+r
=2 2
+ 2r
+ 2r2 2
dt
dt
dt
dt dt
dt dt
dt dt
2 !
2
2
2s
0
dθ
dθ
d
dθ
d2 s ds
ds
r
ds
ds
=2 2
+ 2rr0
− 4r2
=2
− rr0
2
dt dt
dt dt
r dt
dt
dt dt
dt
=0
and thus f (t) =
ds 2
dt
+ r2 (s(t))
dθ 2
dt
is constant as a function of t.
What this shows is that the geodesics oscillate between two parallels (case A) except in the case that one of these parallels has
r0 (s) = 0, in which case the geodesic approximates this parallel as
time t goes to infinity (case B), which itself is a geodesic. Make two
nice little pictures where you show case A and case B happening.
Let α(s) = (r(s), z(s)) be a curve which is defined for all s, the
function r has exactly three critical points at s = −1, s = 0, and
s = 1 with r(0) = 1, and r(1) = r(−1) = 1/10. Moreover
lim r(s) = +∞,
s→±∞
lim z(s) = ±∞.
s→±∞
Note that s = 0 is necessarily a local maximum of r, while s = ±1
are absolute minimums of r.
Pick a geodesic
γ with initial conditions γ(0) = (0, 0) and γ 0 (0) =
√
(∂s + ∂θ )/ 2. Note that γ 0 (0) is a unit vector. Hence, we must have
for every t
2
2
2
√
ds
ds
1
2 dθ
2 dθ
(1) r
= 1/ 2 and
+r
= 1 =⇒
+ 2 =1
dt
dt
dt
dt
2r
Note that (1) implies that 2r12 ≤ 1, which means that r2 (s(t)) ≥ 1/2
for all t. Furthermore
there is exactly
one 0 < a < 1 and −1 < b < 0
√
√
so that r(a) = 2/2 and r(b) = 2/2. It follows from what we have
just said that the s component of γ must belong to [a, b] and we
argue that indeed γ oscillates between the parallels s = a and s = b.
We start by arguing that there is a first time t0 for which s(t0 ) = b
(which implies from (1) that ds
dt (t0 ) = 0). If not, then this would
mean that ds
is
never
zero
(because
of (1)) and thus s(t) would be
dt
an increasing function
√ of t (it has to be increasing and not decreasing
because s0 (0) = 1/ 2). Because s(t) is bounded we must have that
d2 s
limt→∞ ds
dt = 0 and limt→∞ dt2 = 0 (this is an easy fact of one
variable calculus). Hence, we can make t tend to infinity in the
8
ANDRÉ NEVES
second equation of ii) and conclude that limt→∞ r0 (s(t)) = 0. This
is impossible because then s(t) should converge to 1 (we have argue
previously than s(t) ≤ b < 1 for all t.) The conclusion is then that
there is a first time t0 for which s(t0 ) = b.
It is not hard to recognize that for times t immediately after t0 the
function s(t) must be decreasing and so we could repeat the above
arguments with some obvious modifications to conclude the existence
of some other time t1 for which s(t1 ) = a. Arguing inductively one
can then see an infinite sequence (tn )n∈N where s(tn ) = b if n is even
and s(tn ) = a if n is odd. In other words, γ oscillates between s = a
and s = b.
To find a geodesic which satisfies case B) we consider γ with initial
condition γ(0) = (0, 0) and γ 0 (0) = cos α∂s + sin α∂θ (previously
α = π/4) where α is chosen so that sin α = 1/10. In this case, (1)
becomes
(2)
r
2 dθ
dt
= sin α
and
ds
dt
2
+r
2
dθ
dt
2
= 1 =⇒
ds
dt
2
+
sin2 α
= 1.
r2
2
This equations imply that sinr2 α ≤ 1, which means that r2 (s(t)) ≥
sin2 (α) = 1/100 for all t. Hence the s component of γ must belong
to [−1, 1] because r(±1) = 1/10. Note that is this case ds
dt can
never be zero in finite time because that would imply that r would
be 1 or −1 which would mean that γ would intersect tangentially
the geodesic given by the parallel r = ±1 and thus contradict the
uniqueness of geodesics for the same initial conditions. If ds
dt can
never be zero in finite time, this means that (like we argued before)
γ should approach the parallel s = 1 at infinity.
5 Show that all geodesics on a sphere S n are just great circles, i.e., intersection of S n with a plane passing through the origin.
Pick a point p in S n and a unit vector ~v ∈ Tp M ⊂ Rn+1 . We are going to
show that the geodesic with initial condition p and ~v is given by
γ(t) = cos tp + sin t~v .
First, ~v ∈ Tp M ⊂
S n because
Rn+1
means that hv, pi = 0. Thus γ is indeed a curve in
hγ(t), γ(t)i = cos2 t|p|2 + sin2 t|~v |2 = 1.
HOMEWORK 2 - RIEMANNIAN GEOMETRY
9
We need to check that ∇γ 0 γ 0 = 0. Denoting by D the Euclidean connection
we have that
∇γ 0 γ 0 = (Dγ 0 γ 0 )T = (γ 00 )T = (− cos tp − sin t~v )T = −(γ)T = 0.
We now justify somewhat carefully the second and fourth equality above.
The second inequality comes from the fact that if X = γ 0 (t0 ) is a vector in
Tγ(t0 ) Rn and f a function defined on a neighborhood of γ(t0 ) in Rn so that
f (γ(t)) = a(t) for all t close to t0 , then
d(f ◦ γ)
da
(t0 ) =
(t0 ) = a0 (t0 ).
dt
dt
If we apply this with a(t) = x0i (t), where γ(t) = (x1 (t), ..., xn (t)), we see that
γ 0 (x0i ) = x00i and so Dγ 0 γ 0 = (γ 0 (x01 ), ..., γ 0 (x0n )) = γ 00 .
For the fourth inequality we recall that the tangent space Tq S n is nothing
but all the vectors which are orthogonal to the vector q. Thus the tangential
projection of q on Tq S n is zero.
A moment of thought shows γ(t) as above is just the intersection of S n
with the plane spanned by p and ~v .
X(f )(γ(t0 )) =
6* We can consider S 3 as the following subset of C2 (which as a real vector
space is the same thing as R4 ).
S 3 = {(z, w) ∈ C2 | z z̄ + ww̄ = 1}.
The advantage of using complex notation is that for every prime p we can
consider the diffeomorphisms
F : S 3 −→ S 3
where
F (z, w) = (e2iπ/p z, e2iπ/p w).
i) Show that G = {I, F, F 2 , ..., F p−1 } is a group isomorphic to Zp ,
where F j denotes composition with itself j-times. Hence, we can
consider the manifold Lp = S 3 /Zp , where x is identified with y if
F k (x) = y for some k. When p = 2, L2 is just the projective plane
RP3 .
The map F is a linear map of C2 with determinant non-zero and
hence it is an isomorphism when restricted to S 3 . In order to conclude the desired result we have to see that F p is the identity map
and that F j being the identity map implies that j is a multiple of p.
It is simple to see that
F j (z, w) = (e2iπj/p z, e2iπj/p w)
and thus F j is the identity map if and only if j is indeed a multiple
of j.
10
ANDRÉ NEVES
ii) Show that F is an isometry for the induced metric on S 3 . This
implies that we can induce this metric on Lp as well
It is enough to see that F is an orientation preserving isometry of
C2 = R4 because if F preserves the Euclidean metric then it will
preserve the induced metric on S 3 as well. Because F is a linear map,
this corresponds to see that if A denotes its matrix representation
then AT = A−1 and det A is positive (in which case it has to be one.)
The matrix A is given by


cos(2π/p) − sin(2π/p)
0
0
 sin(2π/p) cos(2π/p)

0
0

A=

0
0
cos(2π/p) − sin(2π/p) 
0
0
sin(2π/p) cos(2π/p)
Direct computation shows that AT = A−1 and that det A=1.
iii) Show that all the geodesics of Lp are closed curves. On L3 find two
closed geodesics with different lengths.
Pick a point [x] in Lp and a unit vector ~v ∈ T[x] Lp . Denote by α
the geodesic which has α(0) = [x] and α0 (0) = ~v . We will show that
α(0) = α(2π) and so α is closed. The projection map π : S 3 −→ Lp
is a local diffeomorphism and so we can identify ~v ∈ T[x] Lp with a
unit vector ~v ∈ Tx S 3 ⊆ R4 . Consider the curve given by
γ(t) = x cos t + sin t~v
which, as we saw in exercise 6, is a geodesic in S 3 . Because the
map π is a local isometry then [γ(t)] = π ◦ γ(t) is a geodesic in Lp
as well. Uniqueness of geodesics for a given initial condition and
velocity shows that α(t) = [γ(t)]. Because γ(2π) = γ(0) we have
α(0) = α(2π).
We now do the second part of the exercise. We first remark that if
the geodesic is parametrized by arclength and t0 is the first time for
which α(t0 ) = α(0), then the length of α is t0 . Set [x] = [(1, 0, 0, 0)]
and ~v = (0, 0, 1, 0). Consider the geodesic
α(t) = [cos tx + sin t~v ] = [(cos t, 0, sin t, 0)].
We claim that if α(t0 ) = α(0) then t0 is a multiple of 2π. Indeed
we must have (1, 0, 0, 0) = F j (α(t0 )) = (cos t0 e2iπj/3 , sin t0 e2iπj/3 ),
in which case sin t0 e2iπj/3 = (0, 0). Thus t0 is a multiple of π. If t0 is
not a multiple of 2π then cos t0 e2iπj/3 = −e2iπj/3 and this can never
be (1, 0). Therefore the length of the closed geodesic is 2π.
HOMEWORK 2 - RIEMANNIAN GEOMETRY
11
Let us find now a closed geodesic with length less than 2π. Set
[x] = [(2−1/2 , 0, 2−1/2 , 0)] and ~v = (0, 2−1/2 , 0, 2−1/2 ). Consider the
geodesic
α(t) = [cos tx + sin t~v ] = [2−1/2 (cos t, sin t, cos t, sin t)].
We will show that α(4π/3) = α(0) and so the length of α is less than
4π/3. This is because
F (α(4π/3)) = 2−1/2 F (e4πi/3 , e4πi/3 ) = 2−1/2 (e4πi/3 e2πi/3 , e4πi/3 e2πi/3 )
= 2−1/2 (1, 0, 1, 0) = α(0),
where we are freely identifying α(t) with cos tx + sin t~v (technically
speaking they are different things because one is a curve in L3 and
the other a curve in S 3 ).
7 Given a smooth function f on a compact manifold with consider its gradient defined in the following way. Given p in M we define ∇f (p) to be the
only vector for which
d(f ◦ γ)
g ∇f (p), γ 0 (0) =
(0)
dt
for every curve γ on M with γ(0) = p.
i) Show that ∇f is well defined and compute its expression on charts
(φα , Uα ). Show that there is some point q such that ∇f (q) = 0.
Given a chart (φα , Uα ) we consider the function fα = f ◦ φα . Set
ai (x) =
n
X
g ij (x)
j=1
∂fα
(x)
∂xj
We are going to check that if
X=
n
X
ai
i=1
∂
∂xi
then for every curve γ with γ(0) = φα (x) we have that
(3)
d(f ◦ γ)
(0).
dt
P
∂
In coordinates, we have γ 0 (0) = nj=1 bi ∂x
which means that for
i
evert function h on M we have
n
X
∂(h ◦ φα )
γ 0 (0)(h) =
(x).
bi
∂xi
gφα (x) (X, γ 0 (0)) =
i=1
12
ANDRÉ NEVES
Then
gφα (x) (X, γ 0 (0)) =
n
X
ai (x)bj gij (x) =
i,j=1
n
X
∂fα
(x)bj g ki gij (x)
∂xk
i,j,k=1
=
n
X
∂fα
k=1
∂xk
(x)bk = γ 0 (0)(f ) =
The expression for X is unique because if Y =
other vector satisfying property (3) we have that
n
X
i=1
d(f ◦ γ)
(0).
dt
Pn
∂
i=1 ci ∂xi
is any
∂
∂fα
ci gij = gφα (x) Y,
(x)
=
∂xj
∂xj
and so
ci (x) =
n
X
j=1
g ij (x)
∂fα
(x).
∂xj
Let q be a point where the absolute maximum of f is achieved.
Choosing an appropriate chart, we see that fα defined as above will
have an absolute maximum at the origin, in which case all of its partial derivatives vanish. From the expression for ∇f (q) given above
we see that this vector must be zero.
ii) For any p ∈ M , define the Hessian ∇2 f to be
∇2 f : Tp M × Tp M −→ R
where
∇2 f (X, Y ) = g ((∇X ∇f ) (p), Y ) .
Show that ∇2 f is symmetric (∇2 f (X, Y ) = ∇2 f (Y, X)) and bilinear.
The bilinearity is easy. We show that is symmetric. By bilinearity
it suffices to see that for any coordinate vectors of a given chart
(U, φ). For short, we denote them by ∂xi , i = 1, ..., n. We also use
the notation f¯ = f ◦ φ. Note that by the definition of ∇f we have
that
g(∇f, ∂xi )(φ(x)) =
∂ f¯
d ¯
(f (x + t(0, ..., 0, 1, 0, ..., 0))) =
(x)
dt
∂xi
HOMEWORK 2 - RIEMANNIAN GEOMETRY
13
Thus
∇2 f (∂xi , ∂xj ) = g ∇∂xi ∇f, ∂xj = ∂xi (g(∇f, ∂xj )) − g(∇f, ∇∂xi ∂xj )
¯
∂f
− g(∇f, ∇∂xj ∂xi )
= ∂xi (g(∇f, ∂xj )) − g(∇f, ∇∂xj ∂xi ) = ∂xi
∂xj
∂ 2 f¯
∂ 2 f¯
− g(∇f, ∇∂xj ∂xi ) =
− g(∇f, ∇∂xj ∂xi )
=
∂xi ∂xj
∂xj ∂xi
= ∇2 f (∂xj , ∂xi ).
iii) Suppose that p is an absolute minimum of f . Show that ∇2 f is
nonnegative definite at p
Pick a vector V = γ 0 (0) ∈ Tp M . We need to see that ∇2 f (V, V ) ≥ 0.
The function a(t) = f ◦ γ(t) has an absolute minimum at the origin
and so its second derivative a00 (0) ≥ 0. From the first question we
have that ∇f (p) = 0. Using this we have
∇2 f (V, V ) = g(∇V ∇f, V ) = V (g(∇f, V ))−g(∇f (p).∇V V ) = V (g(∇f, V ))
=
d2 a(t)
d
(g(∇f, V ) ◦ γ)(0) =
(0) = a00 (0) ≥ 0
dt
dt2
iv) Define the Laplacian of f to be
∆f (p) =
n
X
∇2 f (ei , ei )
i=1
where e1 , ...en is an orthonormal basis for Tp M . Show that this is
well defined, i.e., does not depend on the basis chosen and compute
this for the Euclidean metric on Rn .
Say that (uj )nj=1 is another orthonormal basis for Tp M . Then ui =
Pn
j=1 aij ej , where the fact that both bases are orthonormal implies
that the matrix A = (aij )i,j=1,...,n has A−1 = AT . Hence, from
bilinearity we have
!
n
n
n
n
X
X
X
X
2
2
∇ f (ui , ui ) =
aik aij ∇ f (ej , ek ) =
aik aij ∇2 f (ej , ek )
i=1
i,j,k=1
j,k=1
=
n
X
j,k=1
i=1
δjk ∇2 f (ej , ek ) =
n
X
j=1
∇2 f (ej , ej ).
14
ANDRÉ NEVES
In Rn , we have that ∇f =
2
∇ f
∂
∂
,
∂xi ∂xi
=
Pn
∂f ∂
i=1 ∂xi ∂xi
∂
D ∂ ∇f,
∂xi
∂xi
and so
n X
∂2f
∂
∂
∂2f
=
,
=
.
∂xj ∂xi ∂xj ∂xi
∂xi ∂xi
j=1
Thus
n
X
∂2f
.
∆f =
∂xi ∂xi
i=1
8* Consider the exponential map expp : Bε (0) −→ M which we have seen
in class that it is a diffeomorphism onto its image and hence it is a chart.
∂
Denote by ∂x
, i = 1, ..., n the coordinate vectors. (See at the end to see
i
what I mean with this.)
i) Show that the matrix that represents
the
metric at origin is the
∂
∂
identity matrix, i..e, if gij = g ∂xi , ∂xi , then gij (0) is 1 if i = j
and 0 otherwise.
∂
For simplicity we denote ∂x
by ∂xi . We proved in class that D(expp )|0 :
i
TP M −→ Tp M is the identity map. The argument is super simple and so I will do it again. Recall that if γv (t) = γ(v, t) denotes the geodesic with γv (0) = p and γv0 (0) = v ∈ Tp M , then
expp (tv) = γ(tv, 1) = γ(v, t). Hence
D(expp )|0 (v) =
d
d
(expp (tv)) =
γv (t) = γv0 (0) = v.
dt |t=0
dt |t=0
If ēi denotes the vector in Rn with zeros everywhere expect ith position, we have by definition that ∂xi = Dφ(ēi ), where φ is
defined at the end of this exercise. Keeping with the notation at
the end of this exercise we have DF0 (ēi ) = F (ēi ) = ei (F is a linear
map!). Thus ∂xi = Dφ(ēi ) = D(expp )0 (DF0 (ēi )) = D(expp )0 (ei ) =
ei . Because (ei )ni=1 was chosen to be an orthonormal basis we have
that gij (0) = g(ei , ej ) = δij .
∂g
ii) Show that ∂xijk (0) = 0 for all i, j, k = 1, ..., n. Conclude that the
Christoffel symbols Γkij vanish at the origin for every i, j, k = 1, ..., n.
∂
HINT: First show that ∇ ∂ ∂x
(0) for every i. Then understand
i
∂xi
why does this imply that
follow likewise...
∂gii
∂xi (0)
= 0 for every i. The others should
HOMEWORK 2 - RIEMANNIAN GEOMETRY
15
P
Pick any vector V = i=1 ai ∂xi , where ai are constants (we will call this
a“constant” vector). We are going to show first that ∇V V (0) = 0. Indeed
γ(t) = φ(tV ) = expp (tV ) is a geodesic with γ 0 (0) = V . Thus, by definition,
∇V V (0) = 0. This implies that for any “constant” vector V and W we have
∇V W (0) = 0 because ∇V W = ∇W V (easy to see because it is a “constant”
vector) and 0 = ∇V +W (V + W )(0) = ∇V V (0) + ∇W W (0) + 2∇V W (0) =
2∇V W (0). Therefore we have that V (g(X, Y ))(0) = g(∇V X, Y )+g(X, ∇V Y ) =
0 if X, Y, Z are “constant” vectors. Applying this with V = ∂xk , X = ∂xi ,
∂g
and Y = ∂xj we obtain ∂xijk (0) = 0. To see that the Christoffel symbols are
P
zero use V = ∂xi and W = ∂xi and 0 = ∇V W (0) = nk=1 Γkij (0)∂xk .
What the first and second question showed is that the metric in the coordinates induced by the exponential map coincides with the Euclidean metric
up to order two, i.e., not only the matrix (gij )i,j=1,...,n is the identity at the
origin as all its first derivatives are zero at the origin as well.
iii) Show that if φα is a coordinate chart of M with Uα = {x ∈ Rn | |x| <
1} such that the metric (gij )i,j=1,...,n is the identity at the origin and
the Christoffel symbols Γkij (x) are all zero for every x ∈ Uα , then
the metric (gij )i,j=1,...,n (x) is the identity for every x ∈ Uα , i.e., the
manifold φα (Uα ) is isometric to the unit ball with the Euclidean
metric.
Because the Christoffel symbols are all zero we have ∇∂xi ∂xj = 0 for all
x ∈ U . Hence
∂gij
= g(∇∂xk ∂xi , ∂xj ) + g(∂xi , ∇∂xk ∂xj ) = 0.
∂xk
Therefore, the function gij is constant and so gij (x) = gij (0) = δP
ij . Theren
∗ g is the Euclidean metric because if U =
fore, we have
that
φ
α
i=1 ui ∂xi
Pn
and V = i=1 vi ∂xi then
φ∗α g(U, V
)=
n
X
i,j=1
ui vj gij =
n
X
ui vj .
i,j=1
Let’s be more precise on the definition of coordinates on the exponential
map. A priori what we have is that expp is defined on a subset
Bε (0) = {V ∈ Tp M | gp (V, V ) < ε2 }
of the tangent plane Tp M . In order to see expp as a chart I need it to be
define on an open subset of Rn . We do this as following. Pick an orthonormal
16
ANDRÉ NEVES
basis e1 , ..., en for Tp M , where being orthonormal is defined with respect to
the dot product gp . Consider the map
F : Rn −→ Tp M,
F (x) =
n
X
xi ei
i=1
and set U to be F −1 (Bε (0)). Now we consider the chart φ : U −→ M where
φ(x) = expp (F (x)).
∂
As soon we have a chart we can talk about the coordinate vectors ∂x
and
i
the metric gij as we did in the beginning of the course. What I want you to
see is that at the origin (which is contained in U ), this matrix is the identity
matrix. There is no reason to expect the matrix gij to be the identity at
any other point.