# 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.

# The Morse Lemma

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.