We now come to the main point of all these Morse theory posts. We want to somehow figure out what a closed manifold looks like based a Morse function that it admits (who knows how long I’ll develop this theory, maybe we’ll even get to how Smale proved the Poincare Conjecture in dimensions greater than or equal to 5).

Suppose is closed and a Morse function. We’ll use the convenient notation . So again, with the height analogy, as t increases, we will be looking at the entire manifold up to that height. Since M is compact, there is some finite interval such that and .

Note that with essentially no modification, we have already proved the Theorem that if contains no critical values, then . So really, the point is to now figure out what happens as we pass through the critical values.

First off, there are only finitely many critical points, and we can assume that each of these has distinct critical values by raising and lowering critical values. So if are the critical points and , we can order the indices so that .

To be explicit, is the min, so for and for t greater than , since is the max (also, wordpress hates inequalities, or me, I haven't decided yet, but it always cuts out lots of stuff and I just have to write the inequality in words).

These two critical points would be a nice place to start our examination. By the Morse lemma and the fact that a min has index 0, we know that there exists a neighborhood of on which (Alright, I’m sorry about that, but I just realized I have n critical points, so the dimension of my manifold is now m).

More explicitly there is some such that . So if we are thinking of height (of a 2-dim manifold), we’ll want to visualize this as a “bowl” where you have the bottom of the bowl the min and then it slopes upward along a sphere, and then you have the boundary circle at height .

So note that the only thing we used about this critical point is that it had index 0. This shape is called a (m-dimensional) 0-handle.

The reverse happens at our max. We have , since the critical point has index m. This is an -handle and thinking in 2-d height, it is a downward facing bowl.

Again, there is nothing special about being the absolute max, any index m critical point will locally be an -handle.

Index k critical points where are more complicated so I’ll leave those for next time.

Now we have a nice overview of how this will work. We just need to figure out what a -handle looks like, then as t increases through a critical value with index k, will “attach a k-handle”. When we are not near a critical value, the will not change diffeomorphism-type. We just need to make this a little more precise next time (or maybe even the time after).

September 29, 2009 at 2:08 pm

“also, wordpress hates inequalities, or me, I haven’t decided yet, but it always cuts out lots of stuff and I just have to write the inequality in words).”

Do you use Luca Trevisan’s LaTeX to WordPress software? I don’t believe I’ve ever had this problem with that.

September 30, 2009 at 9:08 pm

No. I just do it all by hand. Maybe I’ll try that next time.

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