# 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:

- $f^{-1}[0,f(p)-\epsilon]$, which is diffeomorphic to $[0,f(p)-\epsilon] \times M_0$ using flow along $(1/df(W))W$.
- $f^{-1}([f(p)+\epsilon,1]$, which is diffeomorphic to $[f(p)+\epsilon,1] \times M_1$ using backward flow along $(1/df(W))W$.
- 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.
- The rest of $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$, which is a product that we will also discuss next time.