Is a manifold-with-boundary with given interior and non-empty boundary essentially unique? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Contractible manifold with boundary - is it a disc?Manifolds with two coordinate chartsDoes a *topological* manifold have an exhaustion by compact submanifolds with boundary?If 2-manifolds are homeomorphic and smooth, are they diffeomorphic?Exotic line arrangementsVolume form on a hyperbolic manifold with geodesic boundaryOn compact, orientable 3-manifolds with non-empty boundaryExtension of a group action beyond the boundaryFinding a specific Global Smooth FunctionRemove a disc from a manifold. When is the resulting sphere nullhomotopic?
Is a manifold-with-boundary with given interior and non-empty boundary essentially unique?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Contractible manifold with boundary - is it a disc?Manifolds with two coordinate chartsDoes a *topological* manifold have an exhaustion by compact submanifolds with boundary?If 2-manifolds are homeomorphic and smooth, are they diffeomorphic?Exotic line arrangementsVolume form on a hyperbolic manifold with geodesic boundaryOn compact, orientable 3-manifolds with non-empty boundaryExtension of a group action beyond the boundaryFinding a specific Global Smooth FunctionRemove a disc from a manifold. When is the resulting sphere nullhomotopic?
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Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?
(I have asked this question before here, but there were no replies.)
differential-topology manifolds
edited 2 hours ago
YCor
29.1k486140
asked 4 hours ago
kabakaba
1111
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Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?
(I have asked this question before here, but there were no replies.)
differential-topology manifolds
edited 2 hours ago
YCor
29.1k486140
asked 4 hours ago
kabakaba
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?
(I have asked this question before here, but there were no replies.)
differential-topology manifolds
edited 2 hours ago
YCor
29.1k486140
asked 4 hours ago
kabakaba
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $M$ be a compact connected manifold-with-boundary such that $circ M neq emptyset$, where $circ M$ is the boundary of $M$. Let $N$ be a compact connected manifold-with-boundary such that $circ N neq emptyset$ and $bullet M approx bullet N$, where $bullet M$ denotes the interior of $M$ and $approx$ denotes homeomorphic. Does it necessarily hold that $N approx M$?
(I have asked this question before here, but there were no replies.)
differential-topology manifolds
differential-topology manifolds
edited 2 hours ago
YCor
29.1k486140
asked 4 hours ago
kabakaba
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
YCor
29.1k486140
asked 4 hours ago
kabakaba
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
YCor
29.1k486140
edited 2 hours ago
YCor
29.1k486140
edited 2 hours ago
YCor
29.1k486140
29.1k486140
asked 4 hours ago
kabakaba
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
kabakaba
1111
asked 4 hours ago
kabakaba
1111
1111
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
kaba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
1 Answer
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No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
$endgroup$
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
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– kaba
53 mins ago
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Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
$endgroup$
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
add a comment |
$begingroup$
No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
$endgroup$
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
add a comment |
$begingroup$
No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
$endgroup$
No, there are examples detected by Whitehead torsion. If $P$ is a compact connected $(n-1)$-manifold with empty boundary, then (assuming $nge 6$) for every element $tau$ of the Whitehead group of $pi_1(P)$ there is an $h$-cobordism $M$ on $P$ such that $tau$ is the Whitehead torsion of the pair $(M,P)$. The interior of $M$ will be isomorphic to $Ptimesmathbb R$, but if $tau$ is nontrivial then $M$ will not be isomorphic to $Ptimes I$.
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
edited 2 hours ago
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
answered 3 hours ago
Tom GoodwillieTom Goodwillie
40.4k3110200
40.4k3110200
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
add a comment |
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Thank you. I do not yet understand the answer; I will have to read about Whitehead torsion tomorrow. Would the same answer apply if $M$ and $N$ were orientable?
$endgroup$
– kaba
53 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
$begingroup$
Yes, it has nothing to do with orientability.
$endgroup$
– Tom Goodwillie
8 mins ago
add a comment |
kaba is a new contributor. Be nice, and check out our Code of Conduct.
kaba is a new contributor. Be nice, and check out our Code of Conduct.
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