## Is any Morse trajectory contained in a contractible open set?

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Suppose $f$ is a Morse function on a Riemannian Hilbert manifold $M$. Let $p_{\pm}\in \text{Crit}(f)$ be given and fix some $u:R\rightarrow M$ which is an integral curve of $-\nabla f$ connecting $p_-$ and $p_+.$ Is it true that there is an open contractible set $U\subseteq M$ containing the image of $u$?

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you may want to email Lizhen Qin about this question ( qinl@math.purdue.edu ).  He has though a lot about Morse theory on Hilbert manifolds.  – John Klein May 3 2012 at 1:43

## 1 Answer

Yes, this follows from what's usually called the "$\epsilon$-neighbourhood theorem" in textbooks like Guillemin and Pollack's Differential Topology.

Specifically, given a submanifold $N$ of a manifold $M$ there is an open neighbourhood $V$ of $N$ in $M$ which is diffeomorphic to a vector bundle over $N$. So $V$ is open in $M$.

I believe versions of this theorem appears in Milnor's Topology from a differentiable viewpoint and Hirsch's Differential Topology.

The only restriction on this theorem is that $N$ can not be a manifold with boundary. It can be a non-compact manifold with empty boundary. But it's perfectly fine for $N$ not to be closed in $M$. The proof of this theorem is basically the same as the tubular neighbourhood theorem, except you give up on the idea of having a uniform injectivity radius for the normal bundle's exponential map, and you let the injectivity radius vary smoothly along $N$.

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@Ryan: The word Hilbert in question suggests that Orbicular is asking about infinite-dimensional manifolds modelled on Hilbert spaces. I think the idea of tubular neighborhood still works, but one may need a better control on the metric tensor so that geodesics are locally unique. – Misha Apr 18 2012 at 14:44
Ah, I didn't notice it was meant to be infinite-dimensional. – Ryan Budney Apr 18 2012 at 21:11
Lang's book Differential and Riemannian Manifolds discusses in great detail tubular neighborhoods of submanifolds in a Hilbert manifold (Sec. IV.5, VII.4) and I think that Ryan's argument extends to Hilbert manifolds as well. – Liviu Nicolaescu May 17 2012 at 15:55
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