MATH 782 Differential Geometry : solutions to homework assignment five
1. A ray is a geodesic γ : [0, ∞) → M which minimizes the distance from γ(0) to γ(t) for
all t ∈ [0, ∞). If M is complete and non-compact, prove that there is a ray starting at
γ(0) = p for all p ∈ M .
Choose a point p ∈ M . If expp : Tp M → M maps a ball B0 (n) onto M for some n ∈ R,
then M would be compact (the continuous image of a compact set B0 (n)). Since this is not
the case, for every n there exists a point pn ∈ M with d(p, pn ) = n. Moreover, for every n
there is a length-minimizing geodesic segment γn from p to pn , because M is complete. We
can take γn (t) = exp(tXn ) for some unit vector Xn ∈ Tp M ; then pn = γn (n) = exp(nXn ).
Since the unit sphere in Tp M is compact, the sequence {Xn } has a convergent subsequence
with limit X ∈ Tp M . We claim that γ(t) = exp(tX) is a ray.
If γ were not a ray, then there would exist some m such that the distance from p to
exp(mX) is strictly smaller than m, say d(p, exp(mX)) = m − . Fixing m, by continuity
of the exponential map there exists δ > 0 such that
d(exp(mX), exp(mY )) < for Y ∈ Tp M with kX − Y k < δ. In particular, since {Xn } has a subsequence converging
to X, there exists n ≥ m such that kX − Xn k < δ and
d(exp(mX), exp(mXn )) < .
Therefore
d(p, exp(nXn )) ≤ d(p, exp(mX)) + d(exp(mX), exp(mXn )) + d(exp(mXn ), exp(nXn ))
< (m − ) + + (n − m)
= n.
This is a contradiction, since d(p, exp(nXn )) = d(p, pn ) = n. Therefore γ is a ray.
2. Let M and M be Riemannian manifolds and let f : M → M be a diffeomorphism.
Assume that M is complete and that there exists a constant c > 0 such that
|v| ≥ c|dfp (v)|
for all p ∈ M and all v ∈ Tp M . Prove that M is complete.
We use the Hopf-Rinow Theorem, which states that M is geodesically complete if and
only if (M, dM ) is complete as a metric space, where
Z 1
dM (p, q) := inf γ;γ(0)=p,γ(1)=q
|γ 0 (t)|dt
t=0
is the infinum of the lengths of all paths joining p to q (and similarly for M ). Now if γ is
a path in M joining p to q then f (γ) is a path in M joining f (p) to f (q), and
Z 1
`(f (γ)) =
|f (γ)0 (t)|dt
t=0
1
Z
1
=
Zt=0
1
≤
t=0
|df (γ 0 (t))|dt
1 0
|γ (t)|dt
c
1
=
`(γ).
c
Taking the infinum over all paths (and noting that f is a diffeomorphism, so every path
in M between f (p) and f (q) will look like f (γ) for some path γ in M between p and q),
we find
1
dM (f (p), f (q)) ≤ dM (p, q).
c
Suppose that {pn } is a Cauchy sequence in M . Thus for all > 0 there exists N ∈ Z such
that dM (pn , pm ) < for n, m > N . The inequality above implies that dM (f (p), f (q)) < 1c for n, m > N , and therefore {f (pn )} is a Cauchy sequence in M (c is a constant). Because
M is complete, {f (pn )} will converge to some point q ∈ M . Let p = f −1 (q). It remains
to prove that {pn } converges to p. Given > 0, let
B (p) := {r | dM (p, r) < }
be a metric ball of radius around p. Because f is a diffeomorphism, the image f (B (p))
is an open set in M , and it obviously contains q = f (p). Therefore there exists δ > 0
such that the metric ball Bδ (q) is contained in f (B (p)). Because {f (pn )} converges to
q, there exists N ∈ Z such that f (pn ) ∈ Bδ (q) for n > N . Therefore
pn ∈ f −1 (Bδ (q)) ⊂ B (p)
for n > N , proving that {pn } converges to p. We have shown that the Cauchy sequence
{pn } must converge in M , and thus M is complete, as required.
3. Let M be the upper half-plane R2+ with the metric
ds2 =
dx2 + dy 2
.
yk
For which values of k is M complete?
Case 1 : k > 2. Write k = 2m + 2 for some positive real number m. Let c : [0, b) → M
be the vertical segment c(t) = (0, y(t)), starting at (0, 1) and heading upward. It is easy
to check that c will be a geodesic up to parameterization. We can choose y(t) in such a
way that c is an actual geodesic; we just need to make sure that the velocity vector has
constant length, i.e.,
y 0 (t)
|c0 (t)| = |(0, y 0 (t))| =
=1
y(t)m+1
say. Solving for y(t) gives
y(t) =
2
1
1
(1 − mt) m
which will be defined for t ∈ [0, 1/m). So we see that the geodesic
!
1
c(t) = 0,
1
(1 − mt) m
cannot be extended beyond t = 1/m. Thus M is not geodesically complete.
Case 2 : k < 2. Write k = −2m + 2 for m > 0. This time we take a vertical segment
starting at (0, 1) and heading downward. We obtain a geodesic c : [0, 1/m) → M given
by
1
c(t) = 0, (1 − mt) m
which cannot be extended beyond t = 1/m. Again M is not geodesically complete.
Case 3 : k = 2. In this case both vertical geodesics starting at (0, 1) are defined for all
t ∈ [0, ∞). Specifically, we have
c1 (t) = (0, et )
heading upward and
c2 (t) = (0, e−t )
heading downward. Of course M is the hyperbolic space, and as we have seen all geodesics
are obtained from the vertical geodesics by isometries. This means that all geodesics are
defined for all t ∈ [0, ∞).
4. a) Let
A=
a b
c −a
∈ sl(2, R)
be an arbitrary matrix in the Lie algebra of SL(2, R), and let d = −detA = a2 + bc. Prove
that
√

√
ω
2
A
if
d
>
0,
where
ω
=
d
=
 cosh ωI + sinh
ω
√ a + bc,
√
A
sin
ω
e =
if d < 0, where ω = −d = −a2 − bc,
cos ωI + ω A

I +A
if d = 0.
First observe that A2 = dI. Therefore
1
1
1
1
eA = I + A + A2 + A3 + A4 + A5 + . . .
2!
3!
4! 5!
1 2 1 4
1 3 1 5
=
I + A + A + ... + A + A + A + ...
2!
4!
3!
5!
1
1 2
1
1 2
=
I + dI + d I + . . . + A + dA + d A + . . .
2!
4!
3!
5!
1
1 2
1
1 2
=
1 + d + d + . . . I + 1 + d + d + . . . A.
2!
4!
3!
5!
3
If d > 0 and ω =
√
d then we get
1 2 1 4
1 2 1 4
=
1 + ω + ω + ... I + 1 + ω + ω + ... A
2!
4!
3!
5!
1 3
ω + 3! ω + 5!1 ω 5 + . . .
1 2 1 4
=
1 + ω + ω + ... I +
A
2!
4!
ω
sinh ω
A.
= cosh ωI +
ω
√
Similarly, when d < 0 and ω = −d then we get
1 2 1 4
1 2 1 4
=
1 − ω + ω − ... I + 1 − ω + ω − ... A
2!
4!
3!
5!
1 3
ω − 3! ω + 5!1 ω 5 − . . .
1 2 1 4
A
=
1 − ω + ω − ... I +
2!
4!
ω
sin ω
= cos ωI +
A.
ω
Finally, when d = 0 we get
= I + A.
b) Using part a), prove that the image of the exponential map applied to sl(2, R) is
precisely the subset
{B ∈ SL(2, R) | traceB > −2 or B = −I}
of SL(2, R).
Suppose that B = eA for some A ∈ sl(2, R). Then by part a), traceB is either 2 cosh ω ≥ 2,
2 cos ω ≥ −2, or 2. Moreover, traceB = −2 only in the second case and when ω = (2k+1)π
for some integer k; but then sin ω = 0 and
B = eA = cos ωI +
sin ω
A = −I.
ω
This proves that the image of the exponential map is contained in
{B ∈ SL(2, R) | traceB > −2 or B = −I}.
Conversely, suppose B ∈ SL(2, R) satisfies traceB > 2. Let ω = cosh−1
let
ω
traceB
A=
B−
I .
sinh ω
2
Then A is trace-free so it lies in sl(2, R). Moreover
d = −detA
−ω 2
traceB
=
det B −
I
2
sinh2 ω
4
traceB
2
> 0 and
−ω 2
=
sinh2 ω
detB −
traceB
2
2 !
−ω 2
2
=
2 (1 − cosh ω)
sinh ω
= ω 2 > 0.
Thus we are in the first case and
sinh ω
eA = cosh ωI +
A
ω traceB
traceB
=
I+ B−
I
2
2
= B,
as required.
Next, suppose B ∈ SL(2, R) satisfies −2 < traceB < 2. Let ω = cos−1
traceB
ω
B−
I .
A=
sin ω
2
traceB
2
and let
Note that sin ω 6= 0 as cos ω 6= ±1. A similar argument as above shows that d = −detA =
−ω 2 < 0, and thus we are in the second case, and eA = B as required.
Next, suppose B ∈ SL(2, R) satisfies traceB = 2. Then we are in the third case and can
let A = B − I.
0 π
A
Finally, we have already seen that B = −I = e when A =
, for instance.
−π 0
Thus we have shown that every matrix in
{B ∈ SL(2, R) | traceB > −2 or B = −I}
is in the image of the exponential map, completing the proof.
5