Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Q1: If a Morse function on a smooth closed $n$-manifold $X$ has critical points of only index $0$ and $n$, does it follow that $X\approx \mathbb{S}^n\coprod\ldots\coprod\mathbb{S}^n$?

I think the following question is essential in regard to the one above:

Q2: If $f$ is a Morse function on a closed connected smooth $n$-manifold $X$ that has critical points of only index $0$ and $n$ and $f(X)\!=\![a,b]$, can a critical point of index $0$ or $n$ be mapped into $(a,b)$?

share|improve this question
    
Yes and no. Think of two disjoint spheres one of which has its "bottom" point at the halfway point of the other. If you pick a Morse function that is just the "height", then you will get critical values in the interior of the interval. On the other hand, you can alter critical values and get an equivalent Morse function such that $f(X)=[a_1, b_1]\cup \cdots \cup [a_m, b_m]$ so that the index $0$ happen at the $a_i$ and index $n$ happen at the $b_i$. –  Matt Aug 31 '12 at 15:50
    
Are you saying the answer to the first question is in the affirmative? Please prove it. And as far as the second question goes, I meant connected manifold. –  Leon Lampret Aug 31 '12 at 20:06
    
Woops. I meant "yes and no" to the second question. The point is that for the purposes of Morse theory you can alter the Morse function so that all critical points have distinct critical values. Then the previous argument shows that if they have index $0$ and $n$ and $X$ is connected, then there are only 2 critical points and the correspond to the max and min, so they must happen at the endpoints. The answer to the first question is yes, but I only know a complicated proof by attaching handlebodies. –  Matt Aug 31 '12 at 20:49
    
Actually. I'm being somewhat circular. Let me think about this more. I guess I'm using that if you only have index $n$ and $0$, then you keep attaching $0$ and $n$ handlebodies and hence get a bunch of spheres. Once you know they are spheres you know you can spread out the critical values enough to get disjoint intervals. A priori you can only spread out the critical values a small amount. –  Matt Aug 31 '12 at 20:59

2 Answers 2

As Matt mentioned, a Morse function gives you a handle decomposition.

In the connected case when $n>1$, Poincare duality forces you to have exactly two critical points. The handle decomposition then means that you can form your smooth manifold by taking two $n$-discs and gluing them together along their boundary.

As a result, the actual answer to your question depends on what you mean by $\approx$.

If you mean homeomorphism, the answer is yes. To pull out a heavy hammer, the Poincare conjecture tells you that the resulting topological manifold has to be the $n$-sphere $S^n$, because it's $(n-1)$-connected.

If you mean diffeomorphism, then I suspect that the answer is no. After all, this procedure of taking two discs and gluing them together along a diffeomorphism of their boundaries, which is what the handle decomposition gives you, is also how Milnor originally produced exotic spheres.

share|improve this answer

Say $M$ has $j$ connected components and you only have critical points of index 0 and $n$, then you must have $j$ of each index. You can probably also get this from the Morse inequalities but a very easy way to see this is because the homology of $M$ can be computed from the complex whose $k$th degree group is the free $\mathbb Z$ module generated by the critical points of index $k$. Then since the chain complex is zero in degrees $n-1$ or $n+1$, the $n$th homology must be $\mathbb Z^m$ where $m$ is the number of critical points of index $n$. But an $n$-manifold with $j$ components must have $\mathbb Z^j$ for $n$th homology. Similarly using Poincare duality (i.e. replacing $f$ by $-f$), shows that there must be $j$ critical points of index 0 as well.

Now $f$ restricted to each connected component has exactly two critical points and it is a theorem of Reeb that any such manifold must be homeomorphic to $S^n$. An easy proof of this is in Milnor's Morse Theory.

share|improve this answer

Your Answer

 
discard

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.