# Contractible manifold with boundary - is it a disc?

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?

[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:]

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?

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 $p$.

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No. See the Wikipedia page on Mazur manifolds for examples and relevant history of your question. – Ryan Budney Mar 19 '10 at 22:42
I see. By the way, are there examples bounded by the standard sphere? – Sergei Ivanov Mar 19 '10 at 23:21
If the boundary is a standard sphere then the contractible manifold has to be the standard disc. This is an h-cobordism theorem type argument -- puncture your contractible manifold, this gives an h-cobordism, etc. – Ryan Budney Mar 19 '10 at 23:25

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 $R>\mathrm{diam}\\, M$.

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.

Now we can apply Morse theory for $f$...

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Is $\Phi$ needed here? I don't see why. – Sergei Ivanov Mar 27 '10 at 18:37
ОЙ, right, I'll remove it :) – Anton Petrunin Mar 27 '10 at 19:08

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".

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Sergei, there are lots of compact contractible smooth manifolds; see e.g. my answer here.

I am a bit confused about what you say next. Are you claiming that any compact contractible manifold admits the metric as you describe?

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 here.

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No the metric is assumed to exist and this implies that the manifold is contractible. – Sergei Ivanov Mar 19 '10 at 22:47
Since you have a $C^1$ function with no critical points outside $p$, should not the standard arguments (like in the Soul Theorem) imply that the manifold is smoothly a disk. – Igor Belegradek Mar 19 '10 at 23:06
Formally no, because you can remove any subregion and the function is still there. The fact that the gradient never points outwards could help but I don't know how. – Sergei Ivanov Mar 19 '10 at 23:19
Am I correct that when you start with a small metric sphere around $p$ and flow it by the gradient flow, it stays a topological sphere for a while, and then it hits the boundary and possibly gets messed up, and stops beings a topological sphere, and this is what makes you unsure the manifold is a disk? – Igor Belegradek Mar 19 '10 at 23:43
Yes. And actually there is no gradient flow away from p, only towards p. – Sergei Ivanov Mar 20 '10 at 8:41