A Morse function is a smooth function from a smooth manifold, , to that is in some sense “non-degenerate.”

Suppose , then define the Hessian of f at p, to be the bilinear form that sends .

So picking a basis, the Hessian is just the matrix of second partial derivatives.

Now we call a Morse function if for any we have is compact, and for any critical point of f (the derivative is 0), then is non-singular. So in matrix form, it would have non-zero determinant. In bilinear form terms, it is non-degenerate or zero is not an eigenvalue.

The index at p, is the index of the Hessian at p as a bilinear form. Recall that the index of a bilinear form is the maximal dimension of a linear subspace such that the form is negative definite. (This is sort of backwards of the intuition of counting how big the positive dimension can be. So note that a form is positive semidefinite iff it has index 0).

We really have to check that the property of being a “Morse function” actually is a well-defined concept for smooth manifolds. i.e. is it a diffeomorphism invariant?

We’ll work locally in coordinates. Suppose we have a diffeomorphism where . Define (i.e. the change of coordinates of our so-called Morse function). The well-defined claim is that is a critical point of and that the Hessian of g at p is non-singular.

Well, the critical point claim is just the chain rule. Now we’ll actually compute the Hessian. I propose that it is to make it easier to follow.

We’ll do the right hand side first. The j-th column .

Thus the i-j entry of the right side is multiplying on the left by the ith row of , which gives .

Now we’ll calculate the i-j entry of the left side and see if it is the same. So we’ll need the chain rule for partial derivatives.

.

But that last line is the same as the right side with that extra plus stuff. But since f is critical at p, that term is zero, and the two sides are equal.

So there is no problem calling a smooth function Morse, but I also introduced the idea of the index of f at a point. Hopefully this doesn’t change under diffeomorphism. Let’s check.

Suppose . Then since is a diffeo, is non-singular. But the index is a well-defined notion of a bilinear form at a point, so it is independent of choice of basis. Our previous calculation showed that which is just a change of basis, so as well.

I don’t want to leave you without some sort of concrete idea of what is going on. So define to be the “height function” . If I’m not at the north or south pole, then I can write this function in one of the “side” coordinate patches, i.e. . Hence the Jacobian is non-singular. So every point that is not the north or south pole is a regular value.

The north and south poles are critical values. Now write in the “north patch.” Then at the north pole we are at . Thus . Not only does this tell us that the critical point is non-degenerate, but it tells us the index is 2. In fact, the index of the south pole is 0.

Some final points that our example might have just revealed. The index seems to actually give us some information. Note that we could have done the same thing for . In this case, the two critical points are the same, but the indexes are 0 and n. Is it in fact the case that local mins of Morse functions have index 0 and local maxes on an n-manifold have index n? What does it even mean for a critical point to be something other than a local max or min (i.e. if the previous conjecture holds what is the meaning of an index strictly between 0 and n)? Non-critical points are regular values, and since f is smooth to a 1-manifold (), the level sets are codimension 1 properly embedded submanifolds (“hypersurfaces?”). What happens are the relations of these families of submanifolds as we cross critical values?

Alright. I think that is enough of a preview of what is coming up.

May 20, 2011 at 12:35 am

Nice post.

Thanks sir for writing about Morse function , I am a high school student and I enjoy reading blogs of math professionals. Thanks again for the Explanation on Morse Function.

May 21, 2011 at 4:02 pm

No problem. I’m glad you enjoyed it.