Contractible manifold with boundary - is it a disc? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:22:02Z http://mathoverflow.net/feeds/question/18797 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18797/contractible-manifold-with-boundary-is-it-a-disc Contractible manifold with boundary - is it a disc? Sergei Ivanov 2010-03-19T22:03:09Z 2010-03-27T19:12:39Z <p>I'm sure this is standard but I don't know where to look. Let $M$ be a contractible compact smooth $n$-manifold with boundary. Does it have to be homeomorphic to $D^n$? What about diffeomorphic?</p> <p>[UPDATE: the answer is well-known to be negative as many people kindly pointed out. But actually I assume more about the manifold, namely the following:]</p> <p>There is a Riemannian metric on $M$ such that every two points are connected by a unique shortest path. So $M$ can be contracted to a point $p\in M$ by sending every point along a shortest path to $p$. These paths can bend along the boundary and can merge because of this. But they are relatively nice (namely $C^{1,1}$) curves and their first derivatives depend continuously on their endpoints. Given all this, can one conclude that $M$ is a disc?</p> <p>ADDED: These curves are of course gradient curves of a function (the distance to $p$) which is $C^1$ and has no critical points in the interior of $M$, except at $p$.</p> http://mathoverflow.net/questions/18797/contractible-manifold-with-boundary-is-it-a-disc/18802#18802 Answer by Petya for Contractible manifold with boundary - is it a disc? Petya 2010-03-19T22:32:57Z 2010-03-19T22:32:57Z <p>If $M$ is contractible and the boundary of $M$ is simply-connected and $n\ge 6$ then $M$ is diffeomorphic to $D^n$. See Milnor's "Lectures on the h-cobordism theorem".</p> http://mathoverflow.net/questions/18797/contractible-manifold-with-boundary-is-it-a-disc/18803#18803 Answer by Igor Belegradek for Contractible manifold with boundary - is it a disc? Igor Belegradek 2010-03-19T22:43:35Z 2010-03-19T23:45:12Z <p>Sergei, there are lots of compact contractible smooth manifolds; see e.g. my answer <a href="http://mathoverflow.net/questions/18569/circle-action-on-sphere/18577#18577" rel="nofollow">here.</a> </p> <p>I am a bit confused about what you say next. Are you claiming that any compact contractible manifold admits the metric as you describe? </p> <p>You might be interested in a paper of Ancel-Guilbaut who put a negatively curved (in the comparison sense) metric on the interior of any compact contractible manifold; see also discussion of this paper on the bottom of page 4 of the paper by Alexander-Bishop <a href="http://www.math.uiuc.edu/~sba/wp.pdf" rel="nofollow"> here.</a></p> http://mathoverflow.net/questions/18797/contractible-manifold-with-boundary-is-it-a-disc/19522#19522 Answer by Anton Petrunin for Contractible manifold with boundary - is it a disc? Anton Petrunin 2010-03-27T16:54:10Z 2010-03-27T19:12:39Z <p>Given a function $\psi:\mathbb R\to \mathbb R$, set $$\Psi=\psi\circ\mathrm{dist}_ {\partial M},\ \ \ \ \ f=\Psi\cdot(R-\mathrm{dist}_ p)$$ for some fixed $R>\mathrm{diam}\, M$.</p> <p>Further, $$d\,f= (R-\mathrm{dist}_ p)\cdot d\,\Psi-\Psi\cdot d\,\mathrm{dist}_ p$$ Thus, we may choose smooth increasing $\psi$, such that $\psi(0)=0$ and it is constant outside of little nbhd of $0$ so that $\Psi$ is smooth. (It is possible since the function $\mathrm{dist}_ {\partial M}$ is smooth and has no critical points in a small neighborhood of $\partial M$.) Note that $d\,\Psi$ is positive muliple of $d\,\mathrm{dist}_ {\partial M}$. Thus $d_x\,f=0$ means that geodesic from $x$ to $p$ goes directly in the direction of minimizing geodesic from $x$ to $\partial M$, which can not happen.</p> <p>Now we can apply Morse theory for $f$...</p>