4
$\begingroup$

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.)

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ By open you mean "with nonempty, compact boundary" or even noncompact? Let's assume that $M$ is compact. $\endgroup$ – Manuel Bärenz Dec 11 '17 at 15:47
  • 1
    $\begingroup$ I had "compact with non-empty boundary" in my head. But the discussion about attaching a single 3-handle has nothing to do with compactness. $\endgroup$ – Marco Golla Dec 11 '17 at 20:13
  • 1
    $\begingroup$ Right, I'm just wondering for another reason, since allowing for noncompact manifolds opens a can of worms with infinitely many handles and that kind of things. $\endgroup$ – Manuel Bärenz Dec 11 '17 at 22:06
4
$\begingroup$

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 $b_2 = b_3 = 0$.

$\endgroup$
  • $\begingroup$ At first I was confused, but then I realised that $D^4 \natural -$ can be thought of attaching a 0-handle and a 1-handle. Great example, thanks! $\endgroup$ – Manuel Bärenz Dec 13 '17 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.