# Warped product neighborhoods of geodesic hypersurfaces

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?

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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.)