A Mind for Madness

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The Morse Lemma

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Today we prove what is known as The Morse Lemma. It tells us exactly what our Morse function looks like near its critical points.

Let p\in M be a non-degenerate critical point of f:M\to \mathbb{R}. Then we can choose coordinates about p, (x_i), such that in these coordinates f=-x_1^2-x_2^2-\cdots -x_\lambda^2+x_{\lambda+1}^2+\cdots +x_n^2+f(p). Moreover, \lambda is the index of the critical point. (Note that 0\mapsto f(p)).

Proof: Choose local coordinates, (x_i), centered at p. Without loss of generality f(p)=0 by replacing f with f-f(p). Thus in coordinates, since p corresponds to 0, f(0)=0 (it is a little sloppy, but I’ll probably call the actual function and the function in coordinates the same thing and go back and forth).

By a general theorem of multi-variable calculus (I don’t know if it has a name, it might be Taylor’s theorem? I always get confused at how much is actually included in that), we have smooth functions g_1, \ldots, g_n such that f(x_1, \ldots, x_n)=\sum_{k=1}^n x_ig_i(x_1, \ldots, x_n) and \displaystyle \frac{\partial f}{\partial x_i}\Big|_0=g_i(0).

But 0 is a critical point of f, so g_i(0)=0 and we can apply the theorem again to each g_i. We’ll suggestively call the smooth functions g_k(x_1, \ldots, x_n)=\sum_{i=1}^n x_i h_{ki}(x_1, \ldots, x_n).

Thus, we now have \displaystyle f=\sum_{k,i}x_kx_i h_{ki}. Let \displaystyle H_{ki}=\frac{(h_{ki}+h_{ik})}{2}.

Then \displaystyle f=\sum_{k, i}x_kx_i H_{ki}, and H_{ki}=H_{ik}.

But in that form we see that the second partial derivatives are \displaystyle \frac{\partial^2 f}{\partial x_k \partial x_i}\Big|_0=2H_{ki}(0).

By assumption 0 is a non-degenerate critical point, so det(H_{ki}(0))\neq 0 and hence we can apply a linear transformation to our current coordinates and get that \frac{\partial^2 f}{\partial x_1^2}\Big|_0\neq 0. Thus H_{11}(0)\neq 0.

Now H_{11} is continuous, so that means it is non-zero in a neighborhood of 0.

Let (y_1, x_2, \ldots, x_n) be a new coordinate neighborhood where y_1=\sqrt{|H_{11}|}\left(x_1+\sum_{i=2}^n x_i\frac{H_{1i}}{H_{11}}\right). (Note this is actually a coordinate system, since the determinant of the Jacobian of the transformation from this one to the old one is non-zero).

Now \displaystyle y_1^2=|H_{11}|\left(x_1+\sum_{i=2}^nx_i \frac{H_{1i}}{H_{11}}\right)^2
= H_{11}x_1^2 + 2\sum_{i=2} x_1x_i H_{1i} +\left(\sum_{i=2} x_i H_{1i}\right)^2/H_{11} if H_{11}>0, and the same thing with minus signs everywhere if H_{11} is negative.

Thus the function is y_1^2+\sum_{i,j=2}x_ix_jH_{ij}-\left(\sum_{i=2} x_i H_{1i}\right)^2/H_{11} if H_{11}>0 or
-y_1^2 +\sum_{i,j=2} x_ix_j H_{ij} -\left(\sum_{i=2}x_i H_{1i}\right)^2/H_{11} otherwise.

(I awkwardly wrote this with words, because I couldn’t get cases to look right, and was having weird errors I couldn’t figure).

Now just isolate the stuff after the \pm y_1^2. It satisfies the same conditions as f, but has fewer variables, so we can induct on the number of variables until we have f(y_1, \ldots , y_n)=-y_1^2-\cdots - y_\lambda^2 +y_{\lambda +1}^2+\cdots +y_n^2.

And since the plus and minus signs came from changing basis to put the Hessian into diagonal form with plus and minus 1′s, the number of minus signs is indeed the index.

The proof of this tended to be sort of tedious to check everything, so don’t worry if you didn’t go through it. I don’t think there is really insight you get from going through it. This is one of those rare instances that I think the result is more important than the proof.

Now we have real good reason to believe the index will be n or 0 if we are at a local max or min. What does a max or min look like near the point? Well, it slopes all in the same direction, i.e. it will locally look like a sphere. But this is exactly what the Morse lemma tells us about index n and 0 critical points. We’ll make this more precise later.

I wasn’t sure how I was going to proceed. My two options seemed to be to build the Morse theory I need for Lefschetz, and then do Lefschetz, then come back to Morse theory. But I think I’m just going to continue as far as I want to go ignoring what is needed for the Hyperplane Theorem, then reference what I need.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

2 thoughts on “The Morse Lemma

  1. Pingback: Gradient-Like Vector Fields Exist « A Mind for Madness

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