Manifold with 3 nondegenerate critical points

Question

Suppose $M$ is a n-dimensional (compact) manifold and $f$ is a differentiable function with exactly three (non-degenerate) critical points. Then one can show, using Morse theory, that $M$ is homeomorphic to a $\frac{n}{2}$ sphere with an $n$ cell attached.

I understand why we will have critical points with index $0$ and $n$ (since $M$ is compact and it achieves its max and min). My question is exactly why there one of index $\frac{n}{2}$?

By Poincare duality, we have that $H^{k}(M) \cong H_{n-k}(M)$. So if $k$ is not $\frac{n}{2}$, then $n-k \neq k$ so we will have two additional nonzero (co)homology classes. Does this somehow correspond to having two additional critical points, contradicting that there are only 3?

Answer 1

You can always do $\mathbb{Z}/2$ Morse homology, because then your manifold is guaranteed to be orientable with respect to your coefficient system, and so it's got to be that $H_0(M;\mathbb{Z}/2)=\mathbb{Z}/2$ and $H_n(M;\mathbb{Z}/2)=\mathbb{Z}/2$. So not only do we know we have critical points of index $0$ and $n$, but we know that they actually have to descend to generators in homology. This means that all differentials in our Morse complex will be zero, so generators in the complex are the same as generators for homology. This proves that the last guy has to be in dimension $n/2$.* As you point out, this means that $M$ is homeomorphic to $S^{n/2}$ with an $n$-cell attached.

*** Just to put what I said in the comments right here in the answer: This last critical point has some index $k\in [0,n]$, and it becomes a generator of $H_k(X)$ (with coefficients, if you'd like). But it's trivial to see that when $f$ is a Morse function then $-f$ is a Morse function too, and that this critical point of index $k$ will become a critical point of index $n-k$ for $-f$. And $-f$ equally well satisfies what I just said, i.e. all its critical points become homology generators too. So if you believe that (Morse) homology is a topological invariant, then it must be true that $n-k=k$.

Edit: As Jason explains in answer to my related question, the only manifolds this question can even apply to are homotopy equivalent to $\mathbb{R}P^2$, $\mathbb{C}P^2$, $\mathbb{H}P^2$, and $\mathbb{O}P^2$! Crazy.

Answer 2

Warning: I don't really know what I'm talking about.

That said: It seems like $-f$ will also be a morse function with 3 critical points. Clearly the index 0 critical point from $f$ will correspond to an index $n$ one for $-f$ and similarly for the index $n$ critical point from $f$. We are left with our critical point of unknown index... But I think it would make sense that if we put two morse functions on a manifold, $f$ and $g$, and they have the same number of critical points and the indices all match except for one unknown index for each function, then these remaining two must match. So it seems like the unknown critical point must have the same index in $f$ and $-f$, which means it should have index $n/2$.

In fact... the statement that if $f$ is a Morse function so is $-f$ is kind of a baby version of Poincare duality (it gives the correspondence between cells that you want, but it doesn't immediately give the map with the cap product). So maybe this argument is pretty much what you were saying...