# 2012-01-13

Further addendum to homework problem: As a warmup, do the following one-lower-dimensional version: Let $\Sigma$ be the embedding of $S^1$ in $\mathbb{R}^2$ shown at right and now, for each $v \in S^1$, define the analogous $f_v : S^1 \to \mathbb{R}$. Then draw a graph in $S^1 \times \mathbb{R}$ of the critical values and their indices as a function of $v$. Now the indices will only be $0$ and $1$. The critical events to note are births and deaths of pairs of critical points and crossings of critical values (one critical value rising above or below another one).

Now continuing our proof: We need a Riemannian metric on $X$, so here is the quick proof that Riemannian metrics exist.  Cover $X$ with coordinate charts $\{U_i\}$ with a corresponding partition of unity $\{ \mu_i \}$. In each coordinate chart choose the standard Euclidean inner product $g_i$. Then let $g = \sum \mu_i g_i$. This works because convex combinations of positive definite symmetric matrices are positive definite symmetric matrices.

Now note that a metric $g$, at each point $p$, is a non-degenerate bilinear form $g_p : T_p X \times T_p X \to \mathbb{R}$ and can thus equivalently be thought of as an isomorphism $g_p : T_p X \to T_p^* X$. Then using this isomorpism, we construct a vector field $W$ by $W_p = g_p^{-1}(df_p)$. Because $f$ has no critical points, $W$ is nowhere $0$. This is the gradient vector field for $f$ with respect to the metric $g$, denoted $\nabla_g f$. As a basic exercise you should verify that, when $g$ is the standard inner product on $\mathbb{R}^n$ and $f : \mathbb{R}^n \to \mathbb{R}$, then $\nabla_g f$ is the usual gradient $\nabla f$.

Now because $W$ is never $0$, $df(W) = g(W,W)$ is never $0$, so we can let $V = (1/df(W)) W$, so that $df(V) \equiv 1$. This is the vector field we wanted. Now we construct a diffeomorphism $\phi : [0,1] \times M_0 \to X$ by making $\phi(t,p)$ equal to the point $q$ you get to by starting at $p \in M_0$ and flowing forward along $V$ for time $t$. The fact that $df(V) \equiv 1$ means that $f(q) = t$, and from this and the existence and uniqueness of solutions to ordinary differential equations shows that $\phi$ is a diffeomorphism. $\Box$

So now what if there are critical points?

Suppose that $X$ is a cobordism from $M_0$ to $M_1$ with a Morse function $f : X \to [0,1]$  with one single critical point $p \in X$ of index $k$, as in the picture at right. We will again use a gradient vector field $W = \nabla_g f$ to understand the topology of $X$ in terms of the topology of $W_0$ but now, because $df_p = 0$, we cannot rescale﻿ $W$ to get a vector field $V$ with $df(V) \equiv 1$ on all of $X$. So instead we will divide $X$ into four parts, on three of which we will rescale $W$. But before we do this we need to construct our metric $g$ a little more carefully: We want there to be a coordinate chart $U$ around $p$ with respect to which $g$ is the standard Euclidean inner product and $f$ is the standard Morse local model $\sum \pm x_i^2$, so that $W = \sum \pm 2 x_i \partial_{x_i}$. This is possible because we can start with a standard Morse chart around $p$ as one of the charts in our partition of unity construction and then arrange that, in a ball neighborhood around $p$, one of the $\mu_i$'s is identically $1$ and all the others are $0$.

So now, assuming that $g$, $f$ and $W$ are standard inside a neighborhood $U$ of $p$, we draw a picture of $U$ with the level sets of $f$ and the flow lines of $V$ to the right. We choose an $\epsilon > 0$ so that $f^{-1}(f(p) - \epsilon)$ and $f^{-1}(f(p)+\epsilon)$ intersect $U$ as shown. Then our four pieces of $X$, which we will study more carefully next time, are:

1. $f^{-1}[0,f(p)-\epsilon]$, which is diffeomorphic to $[0,f(p)-\epsilon] \times M_0$ using flow along $(1/df(W))W$.
2. $f^{-1}([f(p)+\epsilon,1]$, which is diffeomorphic to $[f(p)+\epsilon,1] \times M_1$ using backward flow along $(1/df(W))W$.
3. The intersection of $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$ with the closure of the union of all flow lines for $W$ which start in some tubular neighborhood of $x_1^2+\ldots+x_k^2$ in $f^{-1}(f(p)-\epsilon)$. This is the "mystery piece" that we will understand better soon.
4. The rest of $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$, which is a product that we will also discuss next time.