# Does there always exist a sequence of handle moves between handle decompositions that does not increase index? (+ ref. request)

Reference request: Firstly, I'm looking for a proof of the following well-known result about handle decompositions:

($\ast$) Given two handle decompositions of a smooth $n$-manifold $M$, there exists a sequence of handle pair creations, cancellations, and handle slides (as well as isotopies) that takes one decomposition to the other.

I think I broadly understand the conceptual idea behind the proof, namely that handle decompositions correspond to Morse functions on $M$, and homotopies of the Morse functions give such sequences of handle moves, but I'd like to be able to work through this more rigorously. So far the only lead I've been able to find is Gompf and Stipsicz's 4-Manifolds and Kirby Calculus (Theorem 4.2.12), where they attribute the result to Cerf and simply reference his paper La stratification naturelle des espaces de fonctions différentiables réelles et le théorème de la pseudo-isotopie. Unfortunately my French is extremely rudimentary so I haven't been able to locate the relevant part(s) of the paper. Would anyone be able to point me in the right direction, or know where I can find a proof of this theorem?

Main question:

If an $n$-dimensional smooth manifold $M$ has handle decompositions $\mathscr{H}, \mathscr{H}^\prime$, both in terms of $0,1,\dots, k$-handles where $k <n$, does there exist a sequence of handle moves taking $\mathscr{H}$ to $\mathscr{H}^\prime$ which only involves handles of index at most $k$?

I expect the answer is "yes", and that it follows immediately from the proof of ($\ast$), but as I have not yet seen such a proof written down, I am not sure.

• For the first question, I would take a look at Milnor's books: "Lectures on the h-cobordism theorem" and "Morse theory". (And even if the answer doesn't lie there, it's still worth taking a look.) – Marco Golla Dec 9 '14 at 14:29
• I don't think that (*) immediately implies an affirmative answer to your Main question. Cerf's theorem would allow for births/deaths of handles of arbitrarily high index. A good example to study would be Gompf's proof that the Cappell-Shaneson $S^4$ is $S^4$ by sliding handles. At some point, he has to introduce a canceling pair of index 2 and 3 handles. There's no proof that this is necessary, but no expert would guess that you could avoid it. I'd bet there's a counterexample in the non-simply-connected world. – Danny Ruberman Dec 10 '14 at 22:38
• A comment on my comment; it should really be about proving that the homotopy $B^4$ you get from the CS $S^4$ is $B^4$ to match your hypothesis; the problems are equivalent (by another theorem of Cerf). – Danny Ruberman Dec 10 '14 at 22:40

Your main question has an obvious positive answer when $k=1$, but when $k=2$ in dimension $n=4$ it is either open or false (as far as I know). There are handle-decompositions of the smooth four-ball made of 0-, 1-, and 2-handles that no-one knows how to simplify without using creations and cancellations of 2-/3-handles pairs.
In dimension $n\geqslant 5$ the question with $k=2$ for contractible manifolds should be more or less equivalent to the Andrew-Curtis conjecture, which is usually believed to be false by experts (there are explicit potential counterexamples). Every finite two-complex embeds in $\mathbb R^n$ when $n\geqslant 5$ and thickens to a handle-decomposition with 0-, 1-, 2-handles, and the moves that involve only 0-, 1-, 2-handles should be equivalent to the Andrew-Curtis moves (as far as I remember...)