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Let $M$ be a smooth manifold and $N\subset M$ a connected embedded submanifold which is not topologically closed. Assume that $\bar{N}\setminus N$ is a smooth embedded submanifold (where $\bar{N}$ is the closure of $N$).

Question: Then is $\bar{N}$ 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 $\bar{N}\setminus N$, which is equal to $\bar{N}\setminus \text{int}(N)$ 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 be”.

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 $\dot{z} = (a+ib)z$ on $\mathbb{C}$ 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.

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    $\begingroup$ 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. $\endgroup$ – John Samples Jul 25 '18 at 21:05
  • $\begingroup$ @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. $\endgroup$ – Matthew Kvalheim Jul 25 '18 at 21:31
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    $\begingroup$ @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. $\endgroup$ – Anubhav Mukherjee Jul 25 '18 at 22:07
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    $\begingroup$ 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 $S^1 \times S^2/(S^1 \times *)$. $\endgroup$ – user98602 Jul 26 '18 at 15:23
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    $\begingroup$ 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. $\endgroup$ – John Samples Jul 28 '18 at 22:07
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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 $\bar{N}= C(X)$ 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.

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  • $\begingroup$ 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? $\endgroup$ – Matthew Kvalheim Jul 25 '18 at 22:55
  • $\begingroup$ See my second counter example for 3 manifold...there is something called open book decomposition...with using it, I constructed a better example... $\endgroup$ – Anubhav Mukherjee Jul 25 '18 at 22:57
  • $\begingroup$ 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 $\endgroup$ – Anubhav Mukherjee Jul 25 '18 at 22:58
  • $\begingroup$ 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 $\mathbb{R}^3$ with closure boundary known to be a smooth $S^1$ is the situation that inspired my question). $\endgroup$ – Matthew Kvalheim Jul 25 '18 at 23:02
  • $\begingroup$ My second counter example is giving that answer as well $\endgroup$ – Anubhav Mukherjee Jul 25 '18 at 23:29

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