Are there Kirby diagrams with 3-handles?

Question

Let $M\colon \partial_- M \to \partial_+ M$ be an oriented, compact cobordism. Assume that there is a handle decomposition with at most one 0-handle, and denote the handle bodies by $M_i, i \in \{0,\dots 4\}$.

It is known ("On attaching 3-handles to a 1-connected manifold" by Bruce Trace) that if $M$ is simply connected and $\partial_+ M$ is empty or connected, $M_4$ is essentially determined by $M_2$. Or put differently, $M$ is completely determined by $\partial_- M$, $\partial_+ M$ and $M_2$.

What can we say for general $M$? Does the attaching map of a 3-handle contain nontrivial data? Are there Kirby diagrams with 3-handle attachments? (I've never seen 3-handles represented in Kirby diagrams, probably because in many cases, only their number matters.)

Answer 1

3-handles do matter. The reason why they're not usually there is because people often care about closed 4-manifolds; in this case, there is an essentially unique way of attaching all 3-handles to the boundary of (in your notation) $M_2$, which is just $\#^n(S^1\times S^2)$ for some $n$. The result is due to Laudenbach and Poenaru, based on Cerf theory, if I recall correctly. You can find more precise references and more details in Gompf and Stipsicz's 4-manifolds and Kirby calculus.

For open 4-manifolds, attaching 3-handles does create differences. A 3-handle is attached along a (0-framed) 2-sphere in a 3-manifold; such a sphere can be either separating or non-separating (in the boundary). Attaching a 3-handle along a separating one creates an extra boundary component (and creates $H_3$); attaching along a non-separating one kills an $S^1\times S^2$ summand in the boundary.

Answer 2

If you want a specific example where $M$ is not determined by $\partial_- M$, $\partial_+ M$, and $M_2$, here is a simple one.

Write the 4-disc $D^4$ as a $0$-handle, a $2$-handle, and a $3$-handle that cancels the 2-handle. And write $(S^1 \times S^3) \setminus D^4$ as a $0$-handle, a $1$-handle, and a $3$-handle (the missing $D^4$ is the $4$-handle removed from the usual presentation of $S^1 \times S^3$).

Then you get the boundary-connect-sum of these two 4-manifolds

$$D^4 \natural ((S^1 \times S^3) \setminus D^4) = (S^1 \times S^3) \setminus D^4$$

by taking the boundary-connect-sum of the two $0$-handles away from where the handle attachments take place.

So we've written $(S^1 \times S^3) \setminus D^4$ as a $0$-handle, a $1$-handle, a $2$-handle, and 2 $3$-handles.

Removing the first of these 2 $3$-handles gives a presentation of $(S^2 \times D^2) \natural ((S^1 \times S^3) \setminus D^4)$, while removing the second gives a presentation of $D^4 \natural (S^1 \times D^3) = S^1 \times D^3$. In both cases the boundary is $S^1 \times S^2$.

The betti numbers of the first manifold are $b_0 = b_1 = b_2 = b_3 = 1$, while for the second $b_0=b_1=1$ and $b_2 = b_3 = 0$.