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When is the topological closure of a submanifold a submanifold with boundary?
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Let $M$ be a smooth manifold and $Nsubset M$ a connected embedded submanifold which is not topologically closed. Assume that $barNsetminus N$ is a smooth embedded submanifold (where $barN$ is the closure of $N$).
Question: Then is $barN$ a smooth embedded submanifold with boundary? If not, what assumptions can be added to ensure this?
The answer is yes in the special case that $N$ is an open subset of $M$, although this question contains an example showing that if $N$ is not connected, then the manifold boundary of $N$ need not coincide with $barNsetminus N$, which is equal to $barNsetminus textint(N)$ in this case.
Edit: as pointed out by Anubhav with the cone counterexample, the answer is negative for the question as stated. But the cone counterexample seems to produce a “closure boundary†which is, in some sense, inconsistent with what the manifold boundary “should beâ€Â.
So, what if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
Edit 2: I thought of an example. Consider a trajectory with nonzero initial condition of the dynamical system $dotz = (a+ib)z$ on $mathbbC$ with $a < 0$ and $b neq 0$. This trajectory spirals towards $0$, and I believe it is an embedded submanifold. Taking the closure of this trajectory adds $0$ (which consistent with what the manifold boundary "should be"), and I believe the closure is locally connected and homeomorphic to $[0,1]$. However the closure cannot be diffeomorphic to $[0,1]$ because any such homeomorphism cannot be differentiable at $1$ (the difference quotient vectors spiral infinitely fast near $1$ and hence their angles cannot converge).
This is consistent with a result John mentioned in the comments holding only in the topological category.
differential-topology
asked Jul 25 at 15:18
Matthew Kvalheim
666416
 |Â
show 8 more comments
up vote
2
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Let $M$ be a smooth manifold and $Nsubset M$ a connected embedded submanifold which is not topologically closed. Assume that $barNsetminus N$ is a smooth embedded submanifold (where $barN$ is the closure of $N$).
Question: Then is $barN$ a smooth embedded submanifold with boundary? If not, what assumptions can be added to ensure this?
The answer is yes in the special case that $N$ is an open subset of $M$, although this question contains an example showing that if $N$ is not connected, then the manifold boundary of $N$ need not coincide with $barNsetminus N$, which is equal to $barNsetminus textint(N)$ in this case.
Edit: as pointed out by Anubhav with the cone counterexample, the answer is negative for the question as stated. But the cone counterexample seems to produce a “closure boundary†which is, in some sense, inconsistent with what the manifold boundary “should beâ€Â.
So, what if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
Edit 2: I thought of an example. Consider a trajectory with nonzero initial condition of the dynamical system $dotz = (a+ib)z$ on $mathbbC$ with $a < 0$ and $b neq 0$. This trajectory spirals towards $0$, and I believe it is an embedded submanifold. Taking the closure of this trajectory adds $0$ (which consistent with what the manifold boundary "should be"), and I believe the closure is locally connected and homeomorphic to $[0,1]$. However the closure cannot be diffeomorphic to $[0,1]$ because any such homeomorphism cannot be differentiable at $1$ (the difference quotient vectors spiral infinitely fast near $1$ and hence their angles cannot converge).
This is consistent with a result John mentioned in the comments holding only in the topological category.
differential-topology
asked Jul 25 at 15:18
Matthew Kvalheim
666416
1
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
2
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
1
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
1
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07
 |Â
show 8 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $M$ be a smooth manifold and $Nsubset M$ a connected embedded submanifold which is not topologically closed. Assume that $barNsetminus N$ is a smooth embedded submanifold (where $barN$ is the closure of $N$).
Question: Then is $barN$ a smooth embedded submanifold with boundary? If not, what assumptions can be added to ensure this?
The answer is yes in the special case that $N$ is an open subset of $M$, although this question contains an example showing that if $N$ is not connected, then the manifold boundary of $N$ need not coincide with $barNsetminus N$, which is equal to $barNsetminus textint(N)$ in this case.
Edit: as pointed out by Anubhav with the cone counterexample, the answer is negative for the question as stated. But the cone counterexample seems to produce a “closure boundary†which is, in some sense, inconsistent with what the manifold boundary “should beâ€Â.
So, what if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
Edit 2: I thought of an example. Consider a trajectory with nonzero initial condition of the dynamical system $dotz = (a+ib)z$ on $mathbbC$ with $a < 0$ and $b neq 0$. This trajectory spirals towards $0$, and I believe it is an embedded submanifold. Taking the closure of this trajectory adds $0$ (which consistent with what the manifold boundary "should be"), and I believe the closure is locally connected and homeomorphic to $[0,1]$. However the closure cannot be diffeomorphic to $[0,1]$ because any such homeomorphism cannot be differentiable at $1$ (the difference quotient vectors spiral infinitely fast near $1$ and hence their angles cannot converge).
This is consistent with a result John mentioned in the comments holding only in the topological category.
differential-topology
asked Jul 25 at 15:18
Matthew Kvalheim
666416
Let $M$ be a smooth manifold and $Nsubset M$ a connected embedded submanifold which is not topologically closed. Assume that $barNsetminus N$ is a smooth embedded submanifold (where $barN$ is the closure of $N$).
Question: Then is $barN$ a smooth embedded submanifold with boundary? If not, what assumptions can be added to ensure this?
The answer is yes in the special case that $N$ is an open subset of $M$, although this question contains an example showing that if $N$ is not connected, then the manifold boundary of $N$ need not coincide with $barNsetminus N$, which is equal to $barNsetminus textint(N)$ in this case.
Edit: as pointed out by Anubhav with the cone counterexample, the answer is negative for the question as stated. But the cone counterexample seems to produce a “closure boundary†which is, in some sense, inconsistent with what the manifold boundary “should beâ€Â.
So, what if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
Edit 2: I thought of an example. Consider a trajectory with nonzero initial condition of the dynamical system $dotz = (a+ib)z$ on $mathbbC$ with $a < 0$ and $b neq 0$. This trajectory spirals towards $0$, and I believe it is an embedded submanifold. Taking the closure of this trajectory adds $0$ (which consistent with what the manifold boundary "should be"), and I believe the closure is locally connected and homeomorphic to $[0,1]$. However the closure cannot be diffeomorphic to $[0,1]$ because any such homeomorphism cannot be differentiable at $1$ (the difference quotient vectors spiral infinitely fast near $1$ and hence their angles cannot converge).
This is consistent with a result John mentioned in the comments holding only in the topological category.
differential-topology
asked Jul 25 at 15:18
Matthew Kvalheim
666416
edited Jul 28 at 23:55
asked Jul 25 at 15:18
Matthew Kvalheim
666416
asked Jul 25 at 15:18
Matthew Kvalheim
666416
asked Jul 25 at 15:18
Matthew Kvalheim
666416
666416
1
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
2
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
1
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
1
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07
 |Â
show 8 more comments
1
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
2
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
1
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
1
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07
1
1
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
2
2
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
1
1
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
1
1
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07
 |Â
show 8 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $mathbb R^n$. And now consider $N= C(X)- cone point$. Then $barN= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $partial D^2 to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $mathbb R^n$. And now consider $N= C(X)- cone point$. Then $barN= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $partial D^2 to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
 |Â
show 3 more comments
up vote
3
down vote
accepted
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $mathbb R^n$. And now consider $N= C(X)- cone point$. Then $barN= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $partial D^2 to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
 |Â
show 3 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $mathbb R^n$. And now consider $N= C(X)- cone point$. Then $barN= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $partial D^2 to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
It is not true. Take any smooth manifold (which is not a sphere) $X$. Consider the cone C(X). Embed it into some higher $mathbb R^n$. And now consider $N= C(X)- cone point$. Then $barN= C(X)$ which is not a manifold (WHY?).
Using this similar idea, you can construct many more examples whose closure's boundary is a smooth submanifold but not the closure one.
I don't think it is even true in low dimension i.e 3. For example consider an open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by annulus, and their closure is no longer a surface.
EDIT Another explicit example: Consider the lens space $L(3,1)$. It can be build by gluing a $D^2$ to $S^1$ by a degree three map form $partial D^2 to S^1$. Then glue on an B^3. It is not hard to see that the interior of D^2 is a submanifold of $L(3,1)$ with one end. It’s closer is not a submanifold.
Some geometric curvature bound can ensure it to be an embedded manifold.
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
edited Jul 26 at 13:15
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
answered Jul 25 at 21:56
Anubhav Mukherjee
4,6311826
4,6311826
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
 |Â
show 3 more comments
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
Nice example. I didn’t think of something like the cone construction. That means I need another assumption, for there to be any hope. What if the “closure boundary†is required to be consistent with what the manifold boundary should be? E.g., if I have an embedded open 2-disk, what if I require its closure boundary to be a circle? Would that guarantee that the closure is a smooth embedded closed 2-disk?
– Matthew Kvalheim
Jul 25 at 22:55
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example...
– Anubhav Mukherjee
Jul 25 at 22:57
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
There is a small hope for a very particular case which I'm thinking currently...if I get any better positive result from there, I'll update
– Anubhav Mukherjee
Jul 25 at 22:58
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
Ok, thanks a lot. I’ll have to look up the book decomposition. I also edited my question (incidentally, the case of an open 2-disk immersed in $mathbbR^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question).
– Matthew Kvalheim
Jul 25 at 23:02
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
My second counter example is giving that answer as well
– Anubhav Mukherjee
Jul 25 at 23:29
 |Â
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1
In dimension $3$ or less, the closure being locally connected will be sufficient by some results of Bing, and earlier Schoenflies et al., based on the concept of two-sided accessibility. But I am not sure for higher dimensions. I am referring to the topological category, I should say.
– John Samples
Jul 25 at 21:05
@JohnSamples: interesting... I’m having a hard time imagining a case that wouldn’t be locally connected. I’m also more interested in the smooth case and will make a small edit to clarify.
– Matthew Kvalheim
Jul 25 at 21:31
2
@JohnSamples I don't think it is true, for example open book decomposion of a 3 manifold. We could take three pages of the open book, connect them by an annulus, and their closure is no longer a surface.
– Anubhav Mukherjee
Jul 25 at 22:07
1
The problem is that compactifications of a manifold are very far from unique. Take $S^1 times Bbb R^2$. You could compactify that to $S^1 times D^2$ (by adding a 2-torus boundary), or you could compactify to $S^1 times S^2$ (by adding a 'circle at infinity'. You could also take a one-point compactification, which is not a manifold, and is homemorphic to $S^1 times S^2/(S^1 times *)$.
– Mike Miller
Jul 26 at 15:23
1
No no, I am not referring to 3-manifolds, I am referring to (tamely) embedded manifolds in threespace. Regarding an example where local connectedness fails, how about an arc spiraling toward a circle.
– John Samples
Jul 28 at 22:07