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noticed that the first question is answered
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Sergei Ivanov
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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?

If not, there[UPDATE: the answer is morewell-known to saybe negative as many people kindly pointed out. But actually I assume more about $M$the manifold, namely the following: there]

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 $p$.

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?

If not, there is more to say about $M$: 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 $p$.

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 $p$.

added a remark about distance function
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Sergei Ivanov
  • 32.4k
  • 2
  • 99
  • 154

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?

If not, there is more to say about $M$: 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 $p$.

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?

If not, there is more to say about $M$: 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?

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?

If not, there is more to say about $M$: 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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Sergei Ivanov
  • 32.4k
  • 2
  • 99
  • 154

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?

If not, there is more to say about $M$: 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?