[Update: A counter example when is dropped and another version of this theorem by Pong Ting Kei (15-10-2011)]

In this post I will give a characterization when a complete connected Riemannian manifolds is isometric to the hyperbolic space. Of course this is the analogue of Obata’s theorem for the necessary and sufficient condition for a manifold to be a sphere.

Recall that is the Minkowski spacetime with metric and the hyperbolic space is

with induced metric. In geodesic polar coordinates, the metric can be expressed as

where is the distance function from the origin and is the metric on the standard unit ball. It can be checked that has constant sectional curvature , i.e. a space form.

We can now state the Obata-type theorem for hyperbolic space:

Theorem 1Let be a complete connected Riemannian manifold. Then it is isometric to the hyperbolic space if and only if there is a non-constant smooth function on such that

The proof is quite similar to Obata’s proof, with some modification.

*Proof:* The necessary condition is clear. Indeed, for the position function of , note that can be regarded as the normal of the hypersurface , using the fact that for any tangent vector of , together with , it is easy to see that the second fundamental form and so

Thus if we take , then and .

For the converse, suppose there is such . Take any such that . Then for any arclength-parametrized geodesic starting from this point, the equation is reduced to

so the solution is

where and . Using the compound angle formula, it can also be written as

where . [This is where we have to use the condition . ] From this form we see that when we choose and when , then is actually the unique minimum of over . (The uniqueness of the minimum can also be seen from the fact that is strictly convex in view of (1). )

Now let , we can further rescale such that for any arclength-parametrized geodesic from this point

It is not hard to see from (2) that the level set is exactly the geodesic sphere centered at of radius . Furthermore, we claim that there is only one geodesic from each point to . For, if there are two distinct geodesics joining to , then they must be of the same length (why?), i.e. both have starting point and endpoint . As is complete, we can extend on . (2) then shows that . But then since is a broken geodesic joining and , the distance must be strictly smaller than , but then (2) shows that , a contradiction.

The above argument shows that is a bijection, and clearly and its inverse (if exists) are smooth, thus is actually a diffeomorphism.

We will show that then is pulled back to under this map and using the usual spherical coordinates (i.e. normal coordinates around ), the pulled back metric is , and thus is isometric to the hyperbolic space. For this purpose we make use of the Jacobi field.

First of all, fixed any with length . Let . Fix and let such that is orthonormal. Parallel transport along , which we denote by . We then claim that

As

By the Jacobi field equation

To see this, note that (1) implies

Also (2) implies , so

(6)–(7)–(8) then gives the RHS of (5):

This proves (4). To finish the proof, for any vector , , where is tangential to the sphere and is a multiple of . By Gauss lemma, and so . If we denote the Euclidean metric on by and push forward it to by , which we still denote by , then by Gauss lemma, . Let be the normal coordinates around and so in this coordinates, . Take a Jacobi field on such that and (it exists because there is no pair of conjugate points on (why?)). Then on one hand, it is well known that

On the other hand, we know from (3) that where is parallel along , this implies , thus

Using (9) and the fact that , we know that . So the above becomes

since agrees with at . We conclude that

for .

Finally, we know that the Euclidean metric is given by

where is the standard metric on the unit sphere. So we can transform the above to

which is the hyperbolic metric.

Remark 1We will show by an example that the condition cannot be dropped.Consider the Riemannian manifold where is a complete Riemannian manifold. Let . We first show that this function satisfies

Choose local coordinates on . Denote by . Then

and

We now show that is complete. We will denote by and the distance (resp. diameter) function corresponding to by (resp. ), etc.

Let be a closed and bounded subset in , it suffices to show that it is compact. Let be a sequence in . It is easy to see that is bounded and thus by passing into subsequence, we can w.l.o.g assume that . Here is the projection. Also this shows that . Obviously we have for . We now claim that is bounded in . Indeed, for , . So . As is complete, we can pick a subsequence as . It remains to show that in and . But the second assertion follows from the first because is closed in . To show the first claim, note that

as .

Remark 2As remarked by Pong Ting Kei, the condition can also be replaced by the conditionfor some constant . Indeed, this condition together with implies that , i.e. is strongly convex. By some convex analysis, this will show that must have a (necessarily unique) minimum, as in the proof of Theorem 1, and the proof can proceed as before.

We now give some details of the convex analysis. First we can w.l.o.g. assume that . Thus we have . Fix any , let be the distance between them and be a unit speed geodesic joining to . We let . From , we immediately have on . By fundamental theorem of calculus, for , . So

Similarly

Subtract the second inequality from the first one, we have

We have showed that for any minimal geodesic joining ,

We now prove the each sublevel set is bounded, with the diameter depending on . Indeed, if , join them by a geodesic of length , then

Thus . Therefore . The sublevel sets are all closed and bounded, and so is compact, therefore the minimum of must be the minimum of in .

Therefore we have

Theorem 2 (Pong Ting Kei’s theorem, 2011)Let be a complete connected Riemannian manifold. Then it is isometric to the hyperbolic space if and only if there is a smooth function on such thatand

Hi, right below equation 2, should that read is the sphere ? [Oh, yes. Corrected, thanks! -KKK]

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