# What are the possible boundaries of connected compact manifolds?

I'm trying to understand what types of manifolds can occur as the boundary of a compact manifold $M$. When $M$'s dimension $d$ is 1, the boundary is always either empty or two points (corresponding to the two options of $M \cong S^1$ and $M \cong [0,1]$). When $d\ge 2$, the question seems .. hard? All I've really gotten is that the question of "what manifolds are boundaries of other manifolds" is the subject of cobordism. I'm having trouble getting too far into learning about that, though -- I've just seen someone mention it was an equivalence relation which I was able to prove to myself, and I'm trying to go off of that. My thoughts so far, which I would like confirmed:

• Under the addition operation of disjoint union, I think cobordism forms an Abelian group. I think that all elements in this group are order 2, since any manifold $M$ is the boundary of $M\times S^1$.
• The boundaries of a compact manifold are, in general, the elements of the "0" cobordism class -- the equivalence class containing the empty set.
• The boundaries of a compact connected manifold are, in all dimensions higher than 1, exactly the same boundaries of a compact arbitrary manifold. Because given a manifold of dimension $n$ with a given boundary and two distinct components, I can cut out an $n$-disk from each, and attach a $S^{n-1}\times [0,1]$ at the boundary to get a new manifold with the same boundary and one less component. (And repeat.) This only fails in $n=1$ because then $S^{n-1}$ is then no longer connected.

Ideally I could find some sort of classification of what the 0 cobordism class looks like in different dimensions; several sources say that cobordisms are "comparatively easy to classify", but I can't find anything.

Other than that, is it correct to say that the boundaries of connected compact manifolds are:

• Either $\{\}$ or $S^0$ when $d=1$
• The 0 cobordism class when $d>1$

?

Well...its not clear exactly what constitutes an answer for your question, but I can say one or two things.

1. "$M$ is null-cobordant" is the same as "$M$ is the boundary of some manifold $N$" (straight from the definition).

2. The boundaries of all null-cobordant manifolds are empty. For instance, the unit disk is not the boundary of anything.

3. Cobordism classes form a group under disjoint union. Every element has order 2, because $M$ is the boundary (at both ends) of $M \times I$. But if we discuss oriented cobordism groups, then we have to say that that the inverse of an oriented manifold $M$ is $-M$, the same manifold with the opposite orientation.

After that, it gets tougher. In fact, the main things that are known are known through the theory of "characteristic cohomology classes". All this is beautifully developed in Milnor and Stasheff, Characteristic Classes, but that's a demanding reading assignment.

If we let $\Omega_i$ denote the oriented bordism group, then a few basic facts are these:

1. $\Omega_0 = \Bbb Z$, with the generator being a single point.

2. $\Omega_1 = 0$, because all closed compact 1-manifolds are unions of circles, and each circle can be made the boundary of a disk/

3. $\Omega_2 = 0$, because the classification of compact surfaces says that they all look like $n$-holed tori, which bound $n$-holed solid tori.

4. $\Omega_3 = 0$. I seem to recall that this is a nontrivial result, but I may just be forgetting something obvious.

5. $\Omega_4 = \Bbb Z$, generated by $\Bbb CP^2$.

There's something called the "signature" of a $4k$-manifold, and it turns out that $M^4$ is null-bordant exactly when its signature is zero.

And after that, things get more complicated, with various Stieffel-Whitney numbers and Pontriagin numbers being the things that detect null-bordism.

Sorry not to be more informative, but it did take Milnor and Stasheff an entire book to describe and explain the answers to the general question of cobordism groups. :)

• This is great, thank you! Can you just confirm my understanding -- null-cobordant manifolds of dimension d are also necessarily the boundaries of connected compact manifolds in all dimensions d>0? – Alex Meiburg Dec 21 '17 at 19:33
• If a compact manifold $M$ of dimension $d>0$ is the boundary of a compact manifold-with-boundary $N^{d+1}$ of finitely many components (and infinitely many seems like it'd violate compactness, right?), then it's also the boundary of a connected one, for one can perform surgery (as described in the third of your triple of bullets) on $N$ to join any two components, and then proceed by induction. – John Hughes Dec 21 '17 at 20:16

A manifold is a boundary if and only if its Stiefel-Whitney numbers vanish. An oriented manifold is a boundary if and only if its Stiefel-Whitney numbers and its Pontryagin numbers vanish. This is a deep theorem due to Thom.