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?

  • $\begingroup$ How are you using Poincare duality if M is not orientable (RP^2) ? $\endgroup$ – user641 Feb 10 '11 at 5:07
  • $\begingroup$ M is $\mathbb{Z}/2$ orientable in that case $\endgroup$ – confusedmath Feb 10 '11 at 17:51

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.

  • $\begingroup$ I don't know much about Morse homology, but this same sort of argument should work with Cellular homology, correct? $\endgroup$ – confusedmath Feb 10 '11 at 17:53
  • $\begingroup$ that is actually the end step (I think). The 3 critical points tell you something about how many (co)homology groups can be non trivial. Then Poincare duality and the UCT helps eliminate the cases where that non-zero group is not in the middle dimension. Now we have that the Morse homology is cellular homology and that tells you about how to build the manifold. Does that seem right Aaron? $\endgroup$ – Sean Tilson Feb 10 '11 at 18:24
  • $\begingroup$ @Patrick: You should know a little Morse homology, it's nice! In a tiny nutshell, you take a Morse function (wiki it) on $M$, and let the critical points (w.i.) generate a chain complex, where index (w.i.) determines grading. The boundary map takes a critical point of index $k$ and spits out various critical points of index $k-1$, each with coefficient the number of unparametrized flow lines of $-\nabla f$ to that critical point. The theorem is that $H_*^{Morse}(M,f)\cong H_*^{sing}(X)$ (and this still holds if you use other coefficients. (cont.) $\endgroup$ – Aaron Mazel-Gee Feb 10 '11 at 19:46
  • $\begingroup$ Another direction to take this is that you can reconstruct a CW complex with the same homotopy type as $M$ by moving up along the level sets of $f$ and attaching a $k$-cell every time you hit a critical point of index $k$. So in our example, we have a 0-cell, and then an $n/2$-cell, and then an $n$-cell. So actually, applying cellular homology here is really just another name for applying Morse homology! This should roughly explain everything I've said -- let me know if you still have questions. $\endgroup$ – Aaron Mazel-Gee Feb 10 '11 at 19:57
  • 1
    $\begingroup$ @Aaron: the classic examples are the lens spaces: you will find the classification up to homeo and up to equivalence in the corresponding Wikiepedia page, and using it you can find pairs of non homeomorphic but equivalent closed manifolds. $\endgroup$ – Mariano Suárez-Álvarez Feb 20 '11 at 6:08

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...


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.