Handlebodies II

Let’s think back to our example to model our \lambda-handle (where \lambda is not a max or min). Well, it was a “saddle point”. So it consisted of a both a downward arc and upward arc. If you got close enough, it would probably look like D^1\times D^1.

Well, generally this will fit with our scheme. An n-handle looked like D^n … or better yet D^n\times 0, and a 0-handle looked like 0\times D^n, so maybe it is the case that a \lambda-handle looks like D^\lambda\times D^{n-\lambda}. Let’s call D^\lambda\times 0 the core of the handle, and D^{n-\lambda} the co-core.

By doing the same trick of writing out what our function looks like at a critical point of index \lambda in some small enough neighborhood using the Morse lemma, we could actually prove this, but we’re actually more interested now in how to figure out what happens with M_t as t crosses this point.

By that I mean, it is time to figure out what exactly it is to “attach a \lambda-handle” to the manifold.

Suppose as in the last post that c_i is a critical value of index \lambda. Then I propose that M_{c_i+\varepsilon} is diffeomorphic to M_{c_i-\varepsilon}\cup D^\lambda\times D^{m-\lambda} (sorry again, recall my manifold is actually m-dimensional with n critical values).

I wish I had a good way of making pictures to get some of the intuition behind this across. I’ll try in words. A 1-handle for a 3-manifold, will be D^1\times D^2, i.e. a solid cylinder. So we can think of this as literally a handle that we will bend the cylinder into, and attach those two ends to the existing manifold. This illustration is quite useful in bringing up a concern we should have. Attaching in this manner is going to create “corners” and we want a smooth manifold, so we need to make sure to smooth it out. But we won’t worry about that now, and we’ll just call the smoothed out M_{c_i-\varepsilon}\cup D^\lambda\times D^{m-\lambda}, say M'.

Let’s use our gradient-like vector field again. Let’s choose \varepsilon small enough so that we are in a coordinate chart centered at p_i such that f=-x_1^2-\cdots - x_\lambda^2 + x_{\lambda +1}^2+\cdots + x_m^2 is in standard Morse lemma form.

Let’s see what happens on the core D^\lambda\times 0. At the center, it takes the critical value c_i and it decreases everywhere from there (as we move from 0, only the first \lambda coordinates change). This decreasing goes all the way to the boundary where it is c_i-\varepsilon. Thus it is the upside down bowl (of dimension \lambda). Likewise, the co-core goes from the critical value and increases (as in the right side up bowl) to the boundary of a m-\lambda disk at a value c_i+\delta (where 0<\delta<\varepsilon).

Let's carefully figure out the attaching procedure now. If we think of our 3-manifold for intuition, we want to attach D^\lambda\times D^{m-\lambda} to M_{c_i-\varepsilon} by pasting \partial D^\lambda\times D^{m-\lambda} along \partial M_{c_i-\varepsilon}.

So I haven't talked about attaching procedures in this blog, but basically we want a map \phi: \partial D^\lambda\times D^{m-\lambda}\to \partial M_{c_i-\varepsilon} and then forming the quotient space of the disjoint union under the relation of identifying p\in \partial D^\lambda\times D^{m-\lambda} with \phi (p). Sometimes this is called an adjunction space.

So really \phi is a smooth embedding of a thickened sphere S^{\lambda - 1}, since \partial D^\lambda=S^{\lambda-1}. And the dimensions in which it was thickened is m-\lambda. Think about the "handle" in the 3-dimensional 1-handle case. We gave the two endpoints of line segment (two points = S^0) a 2-dimensional thickening by a disk.

Now it is the same old trick to get the diffeo. The gradient-like vector field, X, flows from \partial M' to \partial M_{c_i+\varepsilon}, so just multiply X by a smooth function that will make M' match M_{c_i+\varepsilon} after some time. This is our diffoemorphism and we are done.


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