MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it true that any totally geodesic hypersurface in a nonpositively curved manifold has a tubular neighborhood such that the metric on the neighborhood is a warped product? At least, if the manifold is simply connected?

share|cite|improve this question
I don't think so -- you need some additional hypothesis on the manifold, for instance constant curvature? Or low dimension? – Jean-Marc Schlenker Aug 17 '11 at 14:40
If you google "warped product", you can find a nice set of notes by Peter Petersen of UCLA. I agree with Jean-Marc that it seems unlikely to be true. Why do you think it might be? – Deane Yang Aug 17 '11 at 15:59
Assume "yes" for a hypesurface $\Sigma$. Then you have to have warped product of $\Sigma$ with an interval. The horizontal lines have to be geodesics perpendicular to $\Sigma$. Then easy calculations show that all principle curvatures of $\Sigma$ have to be the same. So $\Sigma$ is (a domain in) a hyperplane or round sphere. – Anton Petrunin Aug 19 '11 at 10:30

The paper "Lower bounds on Ricci curvature and almost rigidity of warped products" by Cheeger and Colding might be useful to you. In short, they proved that a locally warped product piece of a manifold must satisfy $Hess(F)=kg$ for some functions $F$ and $k$, where $k=F''$. (See page 192. We can also see that $F'$ is the warping function.) And this condition is actually sufficient.

It seems to me that your totally geodesic hypersurface, say $S$, must be a level set of $F$. In this case, we have $Hess(F)=0$ because the second fundamental form of $S$ is $Hess(F)/|\nabla F|$. This implies that $k=0$ (otherwise we have $g=0$, it's impossible.) In this case, you can find that $F$ is a linear function in the radial direction and in fact you get a (non-warped) product manifold. So your hope could be true only if your manifold could be of product type at least. (Then by Anton's comment, it is indeed rigid.)

On the other hand, I don't think the condition "simply-connected" is crucial because you can always assume it by passing to the universal cover.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.