A Mind for Madness

Musings on art, philosophy, mathematics, and physics

Altering the Critical Points

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I officially have a new favorite search for which someone found this blog: How to write a Japanese satire.

Let’s introduce a new term. Two Morse functions are considered equivalent if they have the same critical points and same index at each critical point.

The hope here is that two equivalent Morse functions will give the same topological data about our manifold, and so we want to develop techniques of altering our Morse function to something extremely nice to work with, but having it be equivalent to the origin one.

Our first excursion into this technique is the following: If M is a compact manifold and f is a Morse function on M, then we can find an equivalent Morse function g such that all the critical values are distinct.

If we’re going back to the height intuition, this is the technique that corresponds to “raising” or “lowering” critical points. So if you have two strange things happening at the same height (two mountain peaks that have the same height), the idea is sort of that you can slightly move the manifold around so that one is now higher than the other. Of course, we won’t actually move the manifold in any real sense, we’re going to construct the function.

This is going to be really nice, because it says that we can always get a Morse function in which only a single “change” can happen at any given height.

We’ll do this by first proving a Lemma which does all the work for us. Let f be our Morse function, and p a critical point. Then there is some \varepsilon>0 such that for all c\in (-\varepsilon, \varepsilon) there is an equivalent Morse function h that has the same critical values as f, except for h(p)=f(p)+c.

The arguments here are essentially the same as in previous posts, so I’ll be a little looser and only outline the proof.

Since the critical points are isolated we can take a small coordinate chart centered at p that contains no other critical points. Now let \psi be a bump function that is 1 on some small neighborhood of p and dies to zero before getting to the edge of the chart.

Then we define h_c=f+c\psi. We definitely have that all the critical points of f are still critical points of h_c and since on a neighborhood of any of those points the functions either agree or differ by adding a constant, they have the same index. Also, h_c(p)=f(p)+c, so we have constructed our desired function as long as we don’t have any extra critical points.

But in the same was as before, \Big|Dh_c\Big|=\Big|Df+cD\psi\Big|\geq \delta-ca>0 for all |c|<\varepsilon where \varepsilon=\delta/a, since we're only concerned with the compact set on which \psi is decaying, Df has a positive min \delta, and D\psi has a finite max a. Thus we do not gain any critical points in that set and we are done.

To get to the whole theorem all we need to do is note that there are only finitely many critical points (since compact). So if any of the values are shared, we can use the lemma to give an equivalent Morse function with shifted critical value, where we shift by a small enough value that it can't make it to any other critical value. We only have to apply this a finite number of times.

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Author: hilbertthm90

I write about math, philosophy, literature, music, science, computer science, gaming or whatever strikes my fancy that day.

One thought on “Altering the Critical Points

  1. Pingback: Handlebodies I « A Mind for Madness

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