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I'm just disovering cobordism theory and piecing together the subject from various resources, and the concept of explicitly constructing cobordisms between 3-manifolds is confusing me. Here's my situation:

Background: Define a cobordism between two compact 3-manifolds $M_0, M_1$ as a 4-manifold $W$ whose boundary consists of the disjoint union of the $M_i$, i.e. $\partial W = M_0 \sqcup M_1$. A standard way to construct cobordisms is as the "trace" of a surgery:

Theorem 1: If $M'$ is obtainable from $M$ through surgery, then $M$ and $M'$ are cobordant

I've seen the proof of this as Theorem 3.12 in Milnor's "Lectures on the h-cobordism theorem". It seems like more recent literature (I can't remember where) simplifies Milnor's proof by considering the manifold $W = M \times [0,1]$ and performing surgery on the boundary $M \times \{1\}$. Then $W$ will be the desired cobordism, after some smoothing process is undertaken.

As well as this, I am aware that cobordism theory was initally developed as an algebraic theory, and I've seen the following result (It should be noted here that I am merely parrotting a result that I don't understand):

Theorem 2: For manifolds $M, M'$ of the same dimension, they are cobordant iff their Steifel-Whitney numbers agree

A corollary of this is that since the S-W numbers of any closed 3-manifolds are all zero, any pair of closed 3-manifolds are cobordant.

My Confusion: Take a 3-manifold $M$ and it's disjoint union $M \sqcup M$. We know from Thm 2 above that $M$ and $M \sqcup M$ must be cobordant. If I try the approach outlined in Thm 1, I can only do this by constructing the cobordism between $M \sqcup M$ and the connected sum $M \# M$ (details found here), and then requiring that $M \cong M \# M$. The problem is that this requirement is only true in the case when $M$ is the sphere $S^3$, as can be found here.

My Question: What is going on here? Is there another way to construct cobordisms between 3-manifolds, or can we only explicitly construct manifolds between copies of 3-spheres?

Doc
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  • What are you trying to do? Not at all clear to me. – Anubhav Mukherjee Apr 07 '18 at 15:25
  • Explicitly construct cobordisms from manifolds that aren't $S^3$. It seems that you can only use Milnor's method on 3-manifolds that are homeomorphic to $S^3$, so how are other ones constructed? – Doc Apr 07 '18 at 16:42

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As you say $M$ is nullbordant, because it is a threefold (you can also obtain $M$ from surgery along links in $S^3$ and $S^3$ is obviously nullbordant). Thus there exist a four fold $W$ with $\partial W=M$. Then $M\times[0,1]+ W$ is a cobordism between $M\coprod M$ to $M$.

To see non-trivial examples of cobordisms of threefolds you can take your favourite four fold and remove neighborhoods of embedded three dimensional manifolds.

Thomas Rot
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