It is always referred as the "classical argument" that one only needs to specify the 1 and 2-handles in the Kirby diagram of a 4-manifold $X$ since by Laudenbach-Poénaru there is a unique way to attach 3 and 4-handles to get a closed 4-manifold, as mentioned here as well. The theorem by Laudenbach-Poénaru says
Every diffeomorphism of $\sharp^k (S^1\times S^2)$ extends to $\natural^k (S^1\times D^3)$.
I have two questions about this:
- Fix a closed 4-manifold $X$ and let $X_2$ denote the union of 1 and 2-handles in its handle decomposition. By considering a lower dimensional analogue of it, I think I was able to understand how attaching a 3-handle "cancels" an $S^1\times S^2$. But conversely, why is it the case that if $X$ is closed then $\partial X_2$ must be $\sharp^k (S^1\times S^2)$?
- How exactly are we using Laudenbach-Poénaru to justify that specifying 1 and 2-handles in the Kirby diagram is enough?
I think I see why (1) is true. Since $-f$ is a Morse function such that a $k$-handle of $f$ corresponds to a $(4-k)$-handle of $-f$, the union of 3 and 4-handles should be $\natural^k(S^1\times D^3)$. Now, assuming that $X$ is closed, $$\partial X_2 = \partial\big(\natural^k(S^1\times D^3)\big) = \sharp^k(S^1\times S^2)$$ Armed with this fact, we can now answer (2). Once we specify the 1 and 2-handles in the Kirby diagram, we actually also specify a smooth structure on $X_2$ because there is a predetermined way to smooth out the regions where handles are attached. Hence, restricting this smooth structure to the boundary, we have a canonical smooth structure on $\partial X_2$. Now, no matter how we attach the 3 and 4-handles, the corresponding diffeomorphism on $\partial X_2$ will extend to $X$ by Laudenbach-Poénaru.
Does this seem correct? I'd appreciate if someone can check my reasoning!