Now we want to start building some technique that will allow us to figure out what our closed manifold looks like based on the Morse functions it admits.

We’ll call a vector field , a gradient-like vector field for f, if away from critical points, and if is a critical point of index , then there is a coordinate neighborhood about such that f has the standard form as in the Morse lemma, and (i.e. it is the gradient in this neighborhood).

Intuitively, if we think back to our example, we visualize Morse functions as “height functions”. So we are attempting to construct in some sense an everywhere “upward” pointing vector field. If we’re thinking of the entire manifold flowing along this, then the only places where it is allowed to get “stuck” is at the critical points of .

The theorem is that there always exists a gradient-like vector field for a Morse function on a compact manifold.

Proof: As before, let be a finite subcover of coordinate charts, and be a compact refinement. Since the critical points are isolated (immediate corollary to the Morse lemma), there can only be finitely many since our manifold is compact. So we can assume that each critical point has a neighborhood small enough so that it is entirely contained in exactly one of the , and that the were chosen so that has standard form in those coordinates.

Let be a bump function for supported in . Then we get a smooth function on the entire manifold by letting outside of .

Let be the gradient of on . Let . The claim is that this is our gradient-like vector field for .

Let’s check at non-critical points. If is not a critical point, and , then since is the gradient and . All other terms of the sum are 0 since for any such that . Thus .

The other condition we have set up to work since each critical point has a neighborhood that is contained in precisely one of the , thus on that neighborhood is in standard form, and which is of the correct form. Thus is gradient-like for .

As a preview of things to come, I’ll prove our first result about what our manifold looks like using Morse functions. This is often called the Regular Interval Theorem.

Suppose that has no critical value in , then is diffeomorphic to .

Let be gradient-like for . Define which is smooth off of the critical points of , but since contains no critical points it is a smooth vector field there (in fact, on an open set containing ).

Let be an integral curve for starting at . But now

.

Thus, the integral curve continues along at constant speed 1 for the entire time it is in . But it starts at at time 0, so it reaches at time .

Thus by is a diffeomorphism. But rescaling gives the diffeo to .

This basically says that between critical points of a Morse function, we must have the manifold looking like cylinder built off of a single slice of the function (if we’re thinking in terms of height, we can pick any height, and at anywhere between the two nearest critical heights, all the level sets will look the same).

I’m curious about one thing. I’ve seen introductions to Morse theory which construct the diffeomorphism between the different level sets by first fixing a metric and then using the gradient and the metric to construct a vector field with the desired flows. As far as I can tell it seems as though you are doing the same thing here, simultaneously defining X and a metric on which X=grad f.

This is my first time studying Morse theory… is there a reason for keeping everything metric independent (at least on the surface)? Do gradient-like vector fields come in useful later, or is it all a matter of taste?

I’ll probably come back and give a better response later, but my guess would be it is a matter of taste. To me, working with a metric probably obscures the fact that we are aiming at topological results, so everything better be independent of metric choice.

The thing I want to look up and give a better reply about is that there is infinite-dimensional Morse theory (which I haven’t learned any of), and here it might be useful to have finite-dimensional analogues that can serve as intuition. I can’t say for sure, though.