# If a $n$-manifold exists, then is it the boundary of an existing $(n+1)$-manifold?

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!

• 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. – user98602 Aug 3 '15 at 1:42
• 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. – user204299 Aug 3 '15 at 2:23
• "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 4 '15 at 1:57
• 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 4 '15 at 2:32
• 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 5 '15 at 3:03

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.)
• 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 math.stackexchange.com/questions/99898/… – David E Speyer Aug 5 '15 at 0:27