# Cancelling 3-handles in Kirby diagrams

Recently been trying to understand the proofs of Gompf and Akbulut that certain 4-manifolds are $S^4$ (these 2 papers: Gompfs paper in Topology Vol. 30 Issue. 1, Akbulut). In which they use a clever 2 and 3 handle cancelling pair to reduce the Kirby diagrams associated to these manifolds.

The part of the proof that I don't quite understand is the way the 2 handle is added.

In an unpublished book on Akbulut's web page Book.pdf he states (on p.14/15),

"Notice that since 3-handles are attached uniquely, introducing a canceling 2- and 3- handle pair is much simpler operation. We just draw the 2-handle as 0-framed unknot, which is $S^2 × B^2$, and then declare that there is a canceling 3-handle on top of it. In a handle picture of 4-manifold, no other handles should go through this 0-framed unknot to be able to cancel it with a 3-handle."

A similar definition for cancelling a 2-3 handle pair is given on p.146 of 4 Manifolds and Kirby Calculus by Gompf and Stipsicz.

My problem is that in both the papers above they add in a 2-3 handle pair but the 2 handle is drawn in such a way that is goes through a 1 handle, which seems to contradict the above cancelling criteria. If anyone could explain this to me it would be greatly appreciated.

• Could you please specify which figure of Akbulut's Book.pdf, and in the paper, seems wrong? Book Fig 1.12 is a 1-2 cancellation, and I don't see a figure of 2-3 cancellation, only the statement that nothing should be linked with the 2-handle so that a 3-handle actually can be put on top of it. – gmoss Feb 10 '13 at 4:58
• Unfortunately there isn't a diagram in his book but in the paper it is attached in figure. 4 along the curve $\alpha$ specified in figure. 3. In the book by Gompf and Stipsicz the diagram is as simple as a circle with framing 0. – Rob Smith Feb 10 '13 at 9:34
• In Figs 3-4, it looks like he's claiming that the black knot is the unknot, so the resulting manifold of a 1-framing on it is $S^3$, and therefore $\alpha$ and $\beta$ are isotopic, and also unlinked from the knot, so there's nothing passing through them; then apparently there's also an isotopy between them which increases $m$? I don't see either of these things easily, you'll probably have to play around with it, but it looks like that's what's going on. – gmoss Feb 10 '13 at 10:14
• So I'm thinking that I can look at it as follows, since the knot is the unknot we can add in out 0 framed unknot somewhere in the diagram and can then move it until it coincides with the curve $\alpha$ after doing this we reverse all the surgery do get back the original manifold with an extra 2 handle where the curve $\alpha$ was. Is this the right way to think about it? – Rob Smith Feb 10 '13 at 13:15

Something like that. If you believe that the knot is the unknot, then you may attach a 2-handle and immediately cancel it with a 3-handle to either $\alpha$ or to $\beta$, and get the diffeomorphism indicated on the 6th-to-last line of page 2. Then the claim is that if you attach the 2-handle to $\beta$, and then do a handle slide, you untwist one of the $m$-twists and wind up with $\alpha$ with a 2-handle, which can then be cancelled by the previous statement, and therefore $\Sigma_m\simeq\Sigma_{m-1}$. You might need to go digging through the references to see why some of those claims are true (the ones I'd want to verify are that that thing is the unknot, and the handleslide untwisting $m$).