# Handle Decomposition

Today I’ll just prove that a Morse function will give a handle decomposition of a closed manifold. Let’s use all the notation already set up (meaning critical points, values, attaching maps, dimension, Morse function, gradient-like vector field, etc).

We just induct on the subscripts of critical points. We’ve already done the base case (it is a min and hence a 0-handle from here). So we just need to show that if $M_{t}$ is a handlebody for $t\in (c_{i-1}, c_i)$, then $M_{c_i+\varepsilon}$ is a handlebody with the appropriate handle attached.

So we’ve assumed that we have some decomposition $M_{c_{i-1}+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$. We also know that we attach a handle of index $\lambda_i$ when crossing $c_i$, so we do have a diffeo to a manifold $M_{c_i-\varepsilon}$ with a $\lambda_i$-handle attached with attaching map $\phi: \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial M_{c_i-\varepsilon}$.

Note that $[c_{i-1}+\varepsilon, c_i-\varepsilon]$ contains no critical values, so by flowing along $X$ we get a diffeo $M_{c_{i-1}+\varepsilon}\cong M_{c_i-\varepsilon}$. Let $\psi:M_{c_{i-1}+\varepsilon}\to M_{c_i-\varepsilon}$ be this diffeo.

So by inductive hypothesis, $M_{c_i-\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1})$, so we can assume $\psi$ actually maps from the handlebody to $M_{c_i-\varepsilon}$. Now by composing we get our actual attaching map (note that before now the handle was attached to $M_{c_i-\varepsilon}$ and not the handlebody itself).

i.e. $\psi^{-1}\circ \phi : \partial D^{\lambda_i}\times D^{m-\lambda_i}\to \partial (\mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}))$. So let $\phi_i=\psi^{-1}\circ \phi$, and we get that $M_{c_i+\varepsilon}\cong \mathcal{H}(D^m;\phi_1, \ldots , \phi_{i-1}, \phi_i)$, so we are done.

So I sort of dragged on longer than probably necessary there, since there was essentially nothing new. It was just being pedantic about the diffeo of the manifold and the handlebody.

There are some subtleties that should be pointed out, though. The index of the critical point did determine the index of the handle, and we went in “ascending” order. The other much more important and also more subtle point is that the choice of gradient-like vector field was how we constructed the attaching map. So even the same Morse function with a different choice of gradient-like vector field could actually give a “different” handle decomposition when considering attaching maps as part of the data.

# Handlebodies III

I keep naming my posts “handlebodies”, so I think it is officially time to define what one is. A handlebody is a manifold obtained from $D^m$ by attaching various $\lambda$-handles successively. Thus a general handlebody will look like $D^m\cup D^{\lambda_1}\times D^{m-\lambda_1}\cup \cdots \cup D^{\lambda_n}\times D^{m-\lambda_n}$.

If you’re familiar with how to construct a CW-complex, this is pretty similar. You just inductively attach the handles using smooth maps, and then smooth out the manifold so that at each step we have a legitimate smooth manifold. It may be useful to introduce a notation for this. The first attaching $D^m\cup_{\phi_1} D^{\lambda_1}\times D^{m-\lambda_1}$ with attaching map $\phi_1: \partial D^{\lambda_1}\times D^{m-\lambda_1}\to \partial D^m$ will be denoted $\mathcal{H}(D^m; \phi_1)$. So inductively denote the i-th attaching by $\mathcal{H}(D^m ; \phi_1, \ldots , \phi_{i-1}, \phi_i)$.

After i steps, we will always have i attaching maps even if some are formally meaningless (attaching a 0-handle is a disjoint union, so there is no attaching).

If we express M as a handlebody we call that a handle decomposition of the manifold.

Next time I’ll prove the result that everyone that has been reading the posts will have already guessed. Given a Morse function $f: M\to \mathbb{R}$ on a closed manifold, $f$ determines a handle decomposition of $M$. Moreover the handles of this handlebody correspond to the critical points of $f$, and the indices of the handles coincide with the indices of the corresponding critical points.

I’m short on time today, so I’m going to put off proving it.

I’m not sure how much to say about my other news, since it is still sort of up in the air. I passed two of my qualifying exams (which was all that was necessary for now), and I may officially lock myself into the path of algebraic geometry as my field in the next couple of days. Before I say too much about this, I’ll just say that I should have more information on Monday about what I’m officially doing.

# 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.

# Handlebodies I

We now come to the main point of all these Morse theory posts. We want to somehow figure out what a closed manifold looks like based a Morse function that it admits (who knows how long I’ll develop this theory, maybe we’ll even get to how Smale proved the Poincare Conjecture in dimensions greater than or equal to 5).

Suppose $M$ is closed and $f:M\to\mathbb{R}$ a Morse function. We’ll use the convenient notation $M_t=\{p\in M : f(p)\leq t\}$. So again, with the height analogy, as t increases, we will be looking at the entire manifold up to that height. Since M is compact, there is some finite interval $[a,b]$ such that $M_a=\emptyset$ and $M_b=M$.

Note that with essentially no modification, we have already proved the Theorem that if $[c,d]$ contains no critical values, then $M_c\cong M_d$. So really, the point is to now figure out what happens as we pass through the critical values.

First off, there are only finitely many critical points, and we can assume that each of these has distinct critical values by raising and lowering critical values. So if $p_0, \ldots, p_n$ are the critical points and $c_k=f(p_k)$, we can order the indices so that $c_0 < c_1 < \cdots < c_n$.

To be explicit, $c_0$ is the min, so $M_t=\emptyset$ for $t < c_0$ and $M_t=M$ for t greater than $c_n$, since $c_n$ is the max (also, wordpress hates inequalities, or me, I haven't decided yet, but it always cuts out lots of stuff and I just have to write the inequality in words).

These two critical points would be a nice place to start our examination. By the Morse lemma and the fact that a min has index 0, we know that there exists a neighborhood of $p_0$ on which $f=x_1^2+\cdots + x_m^2+c_0$ (Alright, I’m sorry about that, but I just realized I have n critical points, so the dimension of my manifold is now m).

More explicitly there is some $\varepsilon>0$ such that $M_{c_0+\varepsilon}=\{(x_1, \ldots , x_m) : x_1^2+\cdots + x_m^2\leq \varepsilon\}\cong B^m$. So if we are thinking of height (of a 2-dim manifold), we’ll want to visualize this as a “bowl” where you have the bottom of the bowl the min and then it slopes upward along a sphere, and then you have the boundary circle at height $c_0+\varepsilon$.

So note that the only thing we used about this critical point is that it had index 0. This shape is called a (m-dimensional) 0-handle.

The reverse happens at our max. We have $M_{c_n-\varepsilon}=\{(x_1, \ldots , x_m) : x_1^2+\cdots +x_m^2\geq \varepsilon\}$, since the critical point has index m. This is an $m$-handle and thinking in 2-d height, it is a downward facing bowl.

Again, there is nothing special about being the absolute max, any index m critical point will locally be an $m$-handle.

Index k critical points where $k\neq 0,m$ are more complicated so I’ll leave those for next time.

Now we have a nice overview of how this will work. We just need to figure out what a $k$-handle looks like, then as t increases through a critical value with index k, $M_t$ will “attach a k-handle”. When we are not near a critical value, the $M_t$ will not change diffeomorphism-type. We just need to make this a little more precise next time (or maybe even the time after).