When can a manifold written as boundary of a higher dim manifold?

Question

From Stoke's we have, we have:

$$\int_{\partial \Omega} w = \int_{\Omega} dw$$

Does it make sense to write, $\Omega= \partial \phi$:

$$\int_{\partial \phi^2} w = \int_{ \partial \phi} dw =\int_{\phi} d^2 w $$

We know that $d^2$ is zero, so the right most would be integrating zero over a set. If the last equality was true then it would be equal to saying that any possible integral is zero... so that's clearly wrong. So I guess the non sensical part was thinking that my manifold could be written as a boundary of a higher dimensional manifold. Could it be explained what exactly issue with this thinking is? Why can't we think of it in such a way?

$d^2=0$ is true. It is also true that $\partial^2=0$ (boundary of a boundary of manifold is the empty set), and using Stokes theorem as you did is one way of proving this (or it's also immediate from the definitions). However what is not true is that every manifold is the boundary of some other manifold; regarding this you want to look up 'cobordism' — peek-a-boo, May 02 '22 at 11:18
How could we know for sure that a manifold can't be written as another? Are there any characterizing property for the case when a manifold can be written as boundary of another? @peek-a-boo — tryst with freedom, May 02 '22 at 11:20

Overflowian · Answer 1 · 2022-05-02T12:51:27.020

If a closed, oriented manifold $M^n$ is oriented boundary of higher dimensional manifold $X^{n+1}$ then it is said to be null-cobordant. Not all manifolds have this property. Clearly, all oriented 2-manifolds are null-cobordant and it can be shown that also oriented 3-manifolds are. However, in dimension 4, it can be shown that $M^4$ is null-cobordant if and only if its signature vanishes $\sigma(M) = 0$ this is due to Rochlin (as a reference see e.g. Scorpan's book the wild world of 4-manifolds). It is a result of Thom, Pontrjagin and Wall that a manifold is null-cobordant if and only if all its characteristic numbers (Pontrjagin and Steifel-Whitney) vanish. This is explained also in Milnor and Stasheef's book on Characteristic classes (see pg 217).

EDIT to answer the comment. No, you can always find an $n$-form with non-zero integral on any orientable $n$-manifold, $n>0$ just take its volume form.

What is true is that if $M^n= \partial X^{n+1}$ and $\omega^n\in \Omega^n(M)$ is restriction of a form $\hat \omega\in \Omega^n(X)$ which is closed over $X$ i.e. $d\hat\omega = 0$, then you can apply Stokes to conclude that $\int_M \omega = \int_X d\hat \omega = 0$. More abstractly, there are relative de Rham cohomology groups (see Bott-Tu's book pg. 78) $H^\bullet(X,M)$ and a map $\alpha: H^{n}(M)\to H^{n+1}(X,M)$ (giving the exact sequence of the pair $(X,M)$). Then Stokes tells you, that if $\omega^n\in \Omega^n(M)$, is closed in $M$, i.e. $d\omega = 0$, and $\alpha([\omega]) = 0 \in H^{n+1}(X,M)$, then $\int_M \omega = 0$ as well. In other words, the forms to which you can apply the Stokes argument above, are exactly those in the kernel of $\alpha$.

It seems that every natural question one could ask about these things turns out to be books worth of knowledge. I very much appreciate your answer, could you perhaps give a simpler explanation on how one could possibly show that 2 manifold can't be written as boundary of higher manifold? — tryst with freedom, May 02 '22 at 11:25
Does this mean that there are some 4 dimensional manifold for which any form integrated over them is zero? O_O — tryst with freedom, May 02 '22 at 11:38
@Buraian oriented 2-manifolds are boundary of 3-manifolds. Infact the orientable surface with $g$ handles is boundary of the solid handlebody with $g$-handles. — Overflowian, May 02 '22 at 11:40
I did upvote due to the effort but I do not understand most of the terminology to accept it — tryst with freedom, May 02 '22 at 16:09

When can a manifold written as boundary of a higher dim manifold?

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