MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index <= k? I believe this is not true, so I would like to see a counterexample.

share|cite|improve this question
up vote 10 down vote accepted

Take a contractible $3$-manifold which is not homeomorphic to $\mathbb R^3$ -- like the Whitehead manifold.

If such a Morse function existed on the Whitehead manifold, it would be a Morse function with only one critical point, the minimum, and therefore the Whitehead manifold would be an open 3-ball. The proof of this has two steps: (1) $f$ can have at most one critical point, WLOG a minimum by homotopy-type considerations, and it can't have less than one critical point by the "edit" below. Step (2) if it has one critical point let it be $f(p)=0$, the minimum. By the Morse Lemma, $f^{-1}[0,\epsilon]$ is diffeomorphic to a closed 3-ball. The flow the the gradient of $f$ gives a diffeomorphism between $f^{-1}[\epsilon,\infty)$ and $S^2 \times [\epsilon, \infty)$. Pasting these two diffeomorphisms together gives you a diffeomorphism between the manifold and $\mathbb R^3$.

edit: Well, I guess there's the special case to consider that the Morse function could have no critical points but then you could deal with this by the argument that the Whitehead manifold isn't a product of a surface and $\mathbb R$, which by the classification of surfaces amounts to saying that the Whitehead manifold isn't homeo/diffeomorphic to $\mathbb R^3$.

Welcome to MO, Victor!

share|cite|improve this answer
Thanks, Ryan! That's what I needed :) – Victor Oct 5 '10 at 18:57
Can one find a counterexample among interiors of compact manifolds with boundary? – Victor Oct 7 '10 at 14:40
I believe it's unknown whether or not there are exotic smooth structures on open 4-balls that extend to smooth structures on a compact 4-ball so that would be a potential counter-example for $k=0$. Less hypothetically, Mazur manifolds are counter-examples. In high dimensions if your manifold and its boundary are simply-connected the minimal handle presentation theorem would kick in and say there are no counter-examples. But once you have $\pi_1$ in either the manifold or the boundary there are potential obstructions. I wrote up a little thing on Mazur manifolds on Wikipedia. – Ryan Budney Oct 7 '10 at 16:12
In full generality there's likely a natural Whitehead-torsion obstruction to doing what you want to do, given by the s-cobordism theorem. That would require more thought than I'm prepared for this morning. – Ryan Budney Oct 7 '10 at 16:28
Thank you Ryan. As usual you give a very interesting answer. In relation with that, don't you know what is known about the configuration spaces of these manifolds? And what about their embeddings into $R^3$, $R^4$, $R^5$...? – Victor Oct 8 '10 at 16:44

Suppose that $M$ has a proper Morse function $f\ge 0$ with all critical points of index at most $k$. Then homotopically $M$ can be made of low-dimensional cells: it has homotopical dimension at most $k$. But also $M$, relative to its boundary $M^{\ge c}$, can be made by attaching high-dimensional cells. Thus the pair $(M,M^{\ge c})$ is $(dim(M)-k-1)$-connected. So for example if $k\le dim(M)-3$ and $M$ is simply connected then $M$ must also be "simply connected at infinity".

share|cite|improve this answer
Which is consistent with Ryan's answer. – Jim Conant Oct 6 '10 at 9:57
Does it mean that the one-point compactification of any manifold is always (dim(M)-k-1)-connected? By "any" I mean not only those which are interiors of compact manifolds with boundary. Perhaps my question should be formulated as "is the homotopy inverse limit of m-connected spaces m-connected provided all maps are m-connected"? – Victor Oct 7 '10 at 14:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.