Is any Morse trajectory contained in a contractible open set? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T17:36:20Z http://mathoverflow.net/feeds/question/94397 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94397/is-any-morse-trajectory-contained-in-a-contractible-open-set Is any Morse trajectory contained in a contractible open set? Orbicular 2012-04-18T12:55:16Z 2012-05-16T17:22:00Z <p>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 $u$?</p> http://mathoverflow.net/questions/94397/is-any-morse-trajectory-contained-in-a-contractible-open-set/94408#94408 Answer by Ryan Budney for Is any Morse trajectory contained in a contractible open set? Ryan Budney 2012-04-18T14:33:45Z 2012-04-18T14:33:45Z <p>Yes, this follows from what's usually called the "$\epsilon$-neighbourhood theorem" in textbooks like Guillemin and Pollack's <em>Differential Topology</em>. </p> <p>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 $M$. </p> <p>I believe versions of this theorem appears in Milnor's <em>Topology from a differentiable viewpoint</em> and Hirsch's <em>Differential Topology</em>. </p> <p>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 $N$. </p>