# 2012-01-25

**Exercise**: Let $\Sigma$ be a torus with two boundary components embedded in $\mathbb{R}^3$ as in the top picture at right, and let $f : \Sigma \to [0,1]$ be projection onto the vertical direction, a Morse function with two critical points of index $1$ and with $a \cup b$ being a level set. Identify $\Sigma$ with the square-with-opposite-sides glued minus 2 disks shown below the embedded picture, so that the indicated curves $a$, $b$ and $c$ match up. On the square picture, we have an automorphism $\phi : \Sigma \to \Sigma$ obtained by rotating the square $90^\circ$. Let $f_0 = f$ and let $f_1 = f \circ \phi$. Find a generic homotopy $f_t$ from $f_0$ to $f_1$, constant on $f^{-1}(0)$ and $f^{-1}(1)$. Generic means that $f_t$ is Morse for all but finitely many times $t$, and when not Morse, we have a simple birth or death of a pair of cancelling critical points of successive index. Also, if you don't consider distinct critical points with the same critical value as being Morse, then you should also allow finitely many times when two critical points cross. Draw the *Cerf graphic* for this homotopy, i.e. the graph in $[0,1] \times [0,1]$ of the critical values with their indices for all $t \in [0,1]$.

**Heegaard splittings and Heegaard diagrams**: We need three facts here that will be proved later:

- For any closed connected $n$-manifold $X$ there is a Morse function on $X$ with exactly one index $0$ critical point (minimum) and one index $n$ critical point (maximum).
- Given any Morse function $f : X \to \mathbb{R}$, there is a homotopy of maps $f_t: M \to \mathbb{R}$, with $f_0 = f$, and $f_t$ Morse for all $t$, such that the critical values of $f_1$ are ordered by index. In other words, if the indices of critical point $p$ and $q$ are $i$ and $j$, resp., and if $i < j$, then $f(p) < f(q)$. (Note that there if $f$ does not satisfy this property then there will necessarily be times $t$ where $f_t$ has two critical points with the same critical value, and these times should be isolated. We call these times "critical value crossing times".)
- Given a Morse function $f : X \to \mathbb{R}$ with critical values ordered by index, a generic choice of an adapted metric allows us to assume that all the $k$-handles are attached simultaneously to the boundary of the union of the handles of index $< k$. In other words, if $p$ and $q$ are critical points of index $k$ with $f(p) < f(q)$ and no critical values in $(f(p),f(q))$, we can use the gradient flow to compare the free region of the boundary of the handle for $p$ and the attaching region of the boundary of the handle for $q$ inside an intermediate level set $f^{-1}(y)$ for $y \in (f(p),f(q))$. When the metric is chosen generically these will be disjoint, and thus we can flow from the attaching region of the handle for $q$ down along the gradient field past the handle for $p$ and see them as both attached to a level set below $f(p)$.

*self-indexing Morse functions*, functions with the property that, for a critical point $p$, the index of $p$ equals $f(p)$. Note that these are not quite generic because distinct critical points do not have distinct critical values, so some might consider these not to be Morse functions, but that might be being too nitpicky. I will not use that terminology much, but you see it a lot in $3$-manifold topology.)

Now consider a closed connected oriented $3$-manifold $X^3$ with a Morse function $f : X \to \mathbb{R}$ and corresponding handle decomposition as in 3 above. Let $y$ be a regular value between the index $1$ and index $2$ critical values, and let $A = f^{-1}(-\infty,y]$, $B = f^{-1}[y,\infty)$ and $\Sigma = f^{-1}(y)$. Then $A$ is the result of attaching some number $g$ of $1$-handles to a ball ($0$-handle) and $\Sigma = \partial A$ is a genus $g$ surface. On $B$, consider the Morse function $-f$; the index $2$ critical points of $f$ become index $1$ criitical points for $-f$ and thus $B$ is the result of attaching $g'$ $1$-handles to a ball and $\Sigma = \partial B$ is a genus $g'$ surface. Therefore $g = g'$, so $f$ has the same number of index $1$ and $2$ critical points, and both $A$ and $B$ are diffeomorphic to the standard genus $g$ handlebody (the solid object in $\mathbb{R}^3$ bounded by the standard embedding of a genus $g$ surface in $\mathbb{R}^3$). This decomposition of $X$ into two solid handlebodies is called a *Heegaard splitting* of $X$.

Thinking now of constructions of manifolds, rather than decompositions of manifolds, we get the related notion of a *Heegaard diagram*. In the above figure, noting that $\Sigma$ is diffeomorphic to the standard genus $g$ surface $\Sigma_g$, we can now instead consider the Morse functions $-f$ on $A$ and $f$ on $B$, in which case we see both $A$ and $B$ as built by attaching $g$ $2$-handles and a $3$-handle to $\Sigma_g$. In other words, $X$ is built (or, rather, a $3$-manifold diffeomorphic to $X$ is built) by starting with $[-1,1] \times \Sigma_g$ and attaching $g$ $2$-handles and a $3$-handle to $1 \times \Sigma_g$ (producing $B$) and then turning things upside down and attaching $g$ more $2$-handles and a $3$-handle to $-1 \times \Sigma_g$ (producing $A$). This construction is completely determined by the $2g$ attaching circles (simple closed curves), often labelled $\alpha_1, \ldots, \alpha_g$ for $A$ and $\beta_1, \ldots, \beta_g$ for $B$. The $\alpha$ curves must be mutually disjoint and their complement in $\Sigma_g$ must be a $2g$-punctured sphere; ditto for the $\beta$ curves. Any such collection of simple closed curves in $\Sigma_g$ determines a closed $3$-manifold; this is a Heegaard diagram. The example below is $S^1 \times S^2$, again:

This is the data that is used to compute the famous Heegaard-Floer invariants of $3$-manifolds, but that story is beyond the scope of this course.

**Surgery:** The comment above that, in a Heegaard diagram, the complement of the $\alpha$ curves (resp. $\beta$ curves) should be a $2g$-punctured sphere merits further discussion. This is a condition that guarantees that, after attaching the $2$-handles along these curves, the new boundary is $S^2$ and so we can cap off with a $3$-handle. *Surgery* is the process by which the boundary of a manifold changes when one attaches a handle along that boundary. If $X^n$ is a cobordism from $M_0$ to $M_1$ and we attach an $n$-dimensional $k$-handle along an embedding $\phi : S^{k-1} \times B^{n-k} \hookrightarrow M_1$, we get a new cobordism from $M_0$ to $M_2$, and $M_2$ is diffeomorphic to $(M_1 \setminus \phi(S^{k-1} \times B^{n-k})) \cup_\phi B^k \times S^{n-k-1}$, because we "cover up" the image of the attaching region of the handle and "expose" the free region of the handle. The free region $B^k \times S^{n-k-1}$ is glued via $\phi : S^{k-1} \times S^{n-k-1} \hookrightarrow M_1$. This is $(n-1)$-dimensional $(k-1)$-surgery.

In the Heegaard diagram case, we have $n=3$ and $k=2$, so that we are doing $2$-dimensional $1$-surgery by removing a $S^1 \times B^1$ and replacing with $B^2 \times S^0$. In other words, we cut open $\Sigma$ along the attaching curve, leaving two new boundary components, and then we cap off each component with a disk.

We just touched on $4$-manifolds in this lecture, but in the notes I'll start that in the next post. Here's the video (thanks Eddie)