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I am reading some basic context books about topology (i.e. The Poincaré Conjecture, by Donal O'Shea between others) and following this open Topology and Geometry video lectures of the brilliant professor Tadashi Tokieda in the African Institute for Mathematical Sciences (for beginners in the matter, if you have time I would suggest you to have a look to them!).

I would like to ask the following question:

Is every $n$-manifold the boundary of an $(n+1)$-manifold? Is every compact $n$-manifold the boundary of a compact $(n+1)$-manifold?

Thank you!

p.s. This question was rewritten according to the suggestions in the comments and Meta (here), I hope now will be more accurate. Thanks to everybody for the suggestions!

anon
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iadvd
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    I didn't read the physics content of your question. The boundary of an $n$-manifold is an $(n-1)$-manifold. Not every $(n-1)$-manifold is the boundary of an $n$-manifold. In the simplest case (not asking about orientations, structures on the manifold, etc) the question of what $(n-1)$-manifolds bound an $n$-manifold is completely answered by the Stiefel-Whitney classes. –  Aug 03 '15 at 01:42
  • @MikeMiller thanks for the information and the confirmation about the Topological side of the question! that is exactly the reason why I do not understand why it is possible to state that there are only $11$ dimensions in the Superstring theory... because a $n$-manifold understood as a boundary, should imply the possibility of a $(n+1)$-manifold... so I wonder which restrictions are applied to obtain only 11 dimensions and no more when theoretically if a $11$-manifold is possible, then a $12$-manifold is possible too. – iadvd Aug 03 '15 at 02:00
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    The choice of 11 dimensions is so that the principles don't violate any laws of physics. Any more or any less would result in string theory, as a theory of physics, being completely useless. –  Aug 03 '15 at 02:23
  • @JakeLebovic thank you for the comment! I understand that the restrictions are related with the laws of physics, but, how is it possible to make compatible those restrictions with the theoretical possibility of the existence of a 11-dimensional manifold that implies a boundary of a hypothetical 12-dimensional manifold? Please if you had time, and could kindly elaborate a little bit more that side of the question with a little answer I would accept it as an answer (some external reference to papers or docs would be very welcomed). – iadvd Aug 03 '15 at 02:31
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    "How is it possible to make compatible those restrictions with [existence of 12 dimensional object]" is similar to the question "how is it possible to reconcile the complex numbers being 2 dimensional with the existence of 3d space?" The existence of some 12d spaces has no bearing on however the string theory works out. – Mark S. Aug 04 '15 at 01:57
  • @MarkS. I think it would be slightly different.The existence of the 2D complex n. does not imply any kind of relationship with any type of 3D object by definition,there is no cause-effect between the 2D existing complex n. and an object in the 3D space.But the existence of a 11d manifold implies by definition the existence of a 12d boundary and implies (if a boundary exist) the existence of a 12d manifold, so a 12d space must exist (cause-effect).But the Superstring theory restricts the dimensions up to 11,and that restriction does not exist in the theory of Topology.So how is it possible? – iadvd Aug 04 '15 at 02:19
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    Mathematical "existence" is very different from both physical existence and the relevant physical theoretical possibility. Without knowing the relevant physical details myself, imagine if one restriction for the string theory were something like "the number of dimensions must be squarefree" then even if the 11d candidate were the boundary of a 12d manifold, that 12d manifold couldn't be a candidate. – Mark S. Aug 04 '15 at 02:32
  • the question was rewritten and simplified, hope it will be valid now :) – iadvd Aug 05 '15 at 00:04
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    Your new phrasing does not mean anything, as far as I can tell. Is the question you are trying to ask «Is every $n$ manifold the boundary of an $(n+1)$-manifold?»? – Mariano Suárez-Álvarez Aug 05 '15 at 00:11
  • @MarianoSuárez-Alvarez basically this chain: if a n-manifold exists (a) then it implies that a (n+1)-boundary exists (b) (which is the n-manifold) and that implies that if a (n+1) boundary exists, then (c) a (n+1)-manifold exists (which is the manifold associated to the boundary). Is that chain (a)->(b)->(c)correct? – iadvd Aug 05 '15 at 00:41
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    You have repeatedly said the boundary of an n-manifold is an (n+1)-manifold. This is wrong - it is the other way around. For instance, the closed disk is 2-dimensional, and its boundary circle is 1-dimensional. We've all been trying very hard here to get you to pin down what your real question is - you need to work on your end too. I am also voting to close - if you ever figure out what your real question is I will be inclined to reopen. – anon Aug 05 '15 at 01:35
  • @anon then I explained myself wrongly, I have simplified the question (please be aware that I am non English native speaker)... what I am trying to say is: if a n-manifold exists (named "M"), then it can be seen as a boundary (named "B") of another (n+1) manifold, so if that (n+1)-boundary "B" exists, then there must exists that another (n+1) manifold (named "N" for instance) whose boundary is exactly "B" (which is indeed "M"). That is what I want to express. – iadvd Aug 05 '15 at 01:45
  • @anon sorry does it look now correctly expressed? thank you :) – iadvd Aug 05 '15 at 01:48
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    If you're saying an n-manifold M is the boundary of an (n+1)-manifold N, then do not call N an "(n+1)-boundary." M is the boundary of N, not the other way around! A circle is the boundary of a disk, but a disk is not the boundary of a circle. Do not get mixed up. – anon Aug 05 '15 at 02:12
  • @anon again thank you for taking time for this, I am giving you too much trouble today, sorry. I see, that was the failure!! I think I understand now... I have rewritten that part of the question to avoid calling N an "(n+1)-boundary", please if you find it now correct, may I propose removing the on-hold? – iadvd Aug 05 '15 at 02:25
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    Your question now reads, essentially "if an n-manifold M is the boundary of an (n+1)-manifold N, then there exists an (n+1)-manifold N that M is the boundary of." This is trivially, obviously true isn't it? If something exists, then it exists. – anon Aug 05 '15 at 03:03
  • @anon my English is horrible, I have added [always] to the question. If an n-manifold M exists, is it always the boundary of a (n+1)-manifold? According to the answer (which I did not know) it is not true in general that every compact n-manifold is the boundary of some compact (n+1)-manifold with boundary. Hope now will be fine. – iadvd Aug 05 '15 at 03:15
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    Yes it's fine now. Knowing what your two underlying questions are for sure now, I decided to write them my way with an edit and voted to reopen. Additionally upvoted to reward your patience, effort and a good attitude and to mitigate the punishment of the downvotes. – anon Aug 05 '15 at 03:35
  • @anon very kind, thanks again for your time, and I am sorry for so many changes. :) – iadvd Aug 05 '15 at 03:36
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    iadvd: You may be interested to know that MSE has a chat room where, if problems like this come up in the future, you can get a faster feedback cycle. – Eric Stucky Aug 08 '15 at 15:51
  • A similar question. – Michael Albanese Jun 01 '22 at 13:06

1 Answers1

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I am not entirely sure what your question is, but here is my interpretation of it: is every $n$-manifold $X$ (without boundary) the boundary of some $(n+1)$-manifold with boundary $Y$? The answer is yes: just take $Y=X\times [0,\infty)$, identifying $X$ with $\partial Y=X\times\{0\}$.

(Mike Miller gave an answer to the contrary in the comments; however, his answer applies only if you demand that the manifolds be compact. That is, it is not true in general that every compact $n$-manifold is the boundary of some compact $(n+1)$-manifold with boundary. In particular, my answer does not work in that case, since $X\times [0,\infty)$ is never compact if $X$ is nonempty.)

Eric Wofsey
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    An explicit example of a compact $4$-fold which does not bound a compact $5$-fold, and can be understood with only a first course in algebraic topology, is $\mathbb{CP}^2$. It has Euler characteristic $3$; the Euler characteristic of a boundary is always even. See http://math.stackexchange.com/questions/99898/euler-characteristic-of-a-boundary – David E Speyer Aug 05 '15 at 00:27
  • @EricWofsey, thank you! this time it was hard to explain it, I apologize! – iadvd Aug 05 '15 at 00:46