2012-01-18

Exercise: Let us say that a Riemannian metric $g$ is adapted to a Morse function $f$ if, for each critical point $p$ of $f$, there exist coordinates around $p$ with respect to which $f = \sum \pm x_i^2$ and $g$ is the standard inner product. Show that the space of metrics adapted to a fixed Morse function is connected. I.e. if $g_0$ and $g_1$ are adapted to $f$ then they are connected by a smooth family $g_t$, adapted to $f$ for each $t$. It might be helpful to show that any two coordinate charts near $p$, with the same orientation, for which $f$ is standard can be connected by a smooth path of such coordinate charts. (Thanks to Bruce Bartlett and Eric Burgess for pointing out the importance of the orientation here, since $O(n)$ is disconnected.) It is also helpful to show that the space of inner products on $\mathbb{R}^n$ is connected.

Now back to the main thread:OneCriticalPoint We are thinking about the situation where $X$ is a cobordism from $M_0$ to $M_1$ with a Morse function $f : X \to [0,1]$ with a single critical point $p$ of index $k$, and we want to understand what this says about the topology of $X$. Refer again to the figure at right. Where we are going is: we want to describe $X$ as built as a product on $M_0$ at the bottom, with some kind of "handle" attached going over the critical point $p$, followed by another product on $M_1$ at the top.

For our first approach to making this precise,HandleAsCoborism we break $X$ into four pieces: $f^{-1}[0,f(p)-\epsilon]$, $f^{-1}[f(p)+\epsilon,1]$ (both of which are products) and two pieces making up $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$. The small $\epsilon>0$ is chosen so that there is a coordinate chart $U$ around $p$ making $f$ standard, with coordinates $(x_1, \ldots, x_n)$, such that the closed ball $\sum x_i^2 \leq \epsilon$ is contained in $U$. Then, for some $\epsilon'>\epsilon$, we can take our coordinate chart $U$ to be the open ball $\sum x_i^2 < \epsilon'$, and $U$ and $f$ look like the figure at right. Note that $f^{-1}(f(p)-\epsilon) \cap \{x_{k+1} = \ldots = x_n = 0\}$ is a sphere $S^{k-1}$. Pick some small $\delta > 0$ so that the $\delta$-neighborhood $S^{k-1} \times B^{n-k}$ of this $S^{k-1}$ in $f^{-1}(f(p)-\epsilon)$ is contained in $U$, and then let $A$ be the closure of the union of the flow lines for $\nabla_g f$ which pass through this $S^{k-1} \times B^{n-k}$, intersected with $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$. This is the region shaded in blue. We want to think of $A$ as a cobordism from $S^{k-1} \times B^{n-k}$ to $B^k \times S^{n-k-1}$, where the $B^k \times S^{n-k-1}$ is $A \cap f^{-1}(f(p)+\epsilon)$, which is a $\delta$-neighborhood of $f^{-1}(f(p)+\epsilon) \cap \{x_1 = \ldots = x_k = 0\}$ in $f^{-1}(f(p)+\epsilon)$. In the preceding figure of the whole cobordism $X$, the region $A$ is also outlined in blue.

Seeing $A$ as a cobordism between manifolds with boundary means enlarging the definition of a cobordism to include manifolds with boundary and corners, with the corners separating "vertical" boundary (which is a product) and "horizontal" boundary (the top and the bottom). If we allow this, and the definitions are natural, then the complement of $A$ $f^{-1}[f(p)-\epsilon,f(p)+\epsilon]$ is also a cobordism, but this time a product. Thus we can characterize $X$ as follows: $X$ is built from $M_0$ by first constructing a product $[0,1] \times M_0$ (we replace $[0,f(p)-\epsilon]$ with $[0,1]$ for simplicity). Then we attach $A$ to $\{1\} \times M_0$ via an embedding of $S^{k-1} \times B^{n-k}$ into $\{1\} \times M_0$. At this point we do not have a smooth manifold but we make it smooth by also attaching a product cobordism to the complement of this embedding, and gluing the sides of the product cobordism to the sides of $A$. Finally we complete with another product cobordism $[0,1] \times M_1$, but since this doesn't "do anything" we can just ignore that step.

Next time I hope to say this a little more carefully, so I'll leave out the rest of my waffle from this lecture and clarify in my next post. Here's the video (thanks to Eddie Beck):