Stabilization of Heegaard splittings

May 9, 2011

In the last lecture of a course on Heegaard splittings, professor Gabai sketched an example due to Hass-Thompson-Thurston of two genus $g$ Heegaard splittings of a $3$-manifold that requires at least $g$ stabilization to make them equivalent. The argument is, in my opinion, very metric-geometric. The connection is so striking (to me) so that I feel necessary to give a brief sketch of it here.

(Side note: This has been a wonderful class! Although I constantly ask stupid questions and appear to be confused from time to time. But in fact it has been very interesting! I should have talked more about it on this blog…Oh well~)

The following note is mostly based on professor Gabai’s lecture, I looked up some details in the original paper ( Hass-Thompson-Thurston ’09 ).

Recall: (well, I understand that I have not talked about Heegaard splittings and stabilizations here before, hence I’ll *try to* give a one minute definition)

A Heegaard splitting of a 3-manifold $M$ is a decomposition of the manifold as a union of two handlebodies intersecting at the boundary surface. The genus of the Heegaard splitting is the genus of the boundary surface.

All smooth closed 3-manifolds has Heegaard splitting due the mere existence of a triangulation ( by thicken the 1-skeleton of the triangulation one gets a handlebody perhaps of huge genus, it’s easy exercise to see its complement is also a handlebody). However it is of interest to find what’s the minimal genus of a Heegaard splitting of a given manifold.

Two Heegaard splittings are said to be equivlent if there is an isotopy of the manifold sending one splitting to the other (with boundary gluing map commuting, of course).

A stabilization of a Heegaard splitting $(H_1, H_2, S)$ is a surgery on $S$ that adds genus (i.e. cut out two discs in $S$ and glue in a handle). Stabilization will increase the genus of the splitting by $1$)

Let $M$ be any closed hyperbolic $3$-manifold that fibres over the circle. (i.e. $M$ is $F_g \times [0,1]$ with the two ends identified by some diffeomorphism $f: F_g \rightarrow F_g$, $g\geq 2$):

Let $M'_k$ be the $k$ fold cover of $M$ along $S^1$ (i.e. glue together $k$ copies of $F_g \times I$ all via the map $f$:

Let $M_k$ be the manifold obtained by cut open $M'_k$ along $F_g$ and glue in two handlebodies $H_1, H_2$ at the ends:

Since $M$ is hyperbolic, $M'_k$ is hyperbolic. In fact, for any $\varepsilon > 0$ we can choose a large enough $k$ so that $M_k$ can be equipped with a metric having curvature bounded between $1-\varepsilon$ and $1+\varepsilon$ everywhere.

( I’m obviously no in this, however, intuitively it’s believable because once the hyperbolic part $M'_k$ is super large, one should be able to make the metric in $M'_k$ slightly less hyperbolic to make room for fitting in an almost hyperbolic metric at the ends $H_1, H_2$). For details on this please refer to the original paper. :-P

Now there comes our Heegaard splittings of $M_k$!

Let $k = 2n$, let $H_L$ be the union of handlebody $H_1$ together with the first $n$ copies of $M$, $H_R$ be $H_2$ with the last $n$ copies of $M$. $H_L, H_R$ are genus $g$ handlebodies shearing a common surface $S$ in the ‘middle’ of $M_k$:

Claim: The Heegaard splitting $\mathcal{H}_1 = H_L \cup H_R$ and $\mathcal{H}_2 = H_L \cup H_R$ cannot be made equivalent by less than $g$ stabilizations.

In other words, first of all one can not isotope this splitting upside down. Furthermore, adding handles make it easier to turn the new higher genus splitting upside down, but in this particular case we cannot get away with adding anything less than $g$ many handles.

Okay, not comes the punchline: How would one possible prove such thing? Well, as one might have imagined, why did we want to make this manifold close to hyperbolic? Yes, minimal surfaces!

Let’s see…Suppose we have a common stabilization of genus $2g-1$. That would mean that we can sweep through the manifold by a surface of genus (at most) $2g-1$, with $1$-skeletons at time $0, 1$.

Now comes what professor Gabai calls the ‘harmonic magic’: there is a theorem similar to that of Pitts-Rubinstein

Ingredient #1: (roughly thm 6.1 from the paper) For manifolds with curvature close to $-1$ everywhere, for any given genus $g$ Heegaard splitting $\mathcal{H}$, one can isotope the sweep-out so that each surface in the sweep-out having area $< 5 \pi (g-1)$.

I do not know exactly how is this proved. The idea is perhaps try to shrink each surface to a ‘minimal surface’, perhaps creating some singularities harmless in the process.

The ides of the whole arguement is that if we can isotope the Heegaard splittings, we can isotope the whole sweep-out while making the time-$t$ sweep-out harmonic for each $t$. In particular, at each time there is (at least) a surface in the sweep-out family that divides the volume of $M'_n$ in half. Furthermore, the time 1 half-volume-surface is roughly same as the time 0 surface with two sides switched.

We shall see that the surfaces does not have enough genus or volume to do that. (As we can see, there is a family of genus $2g$ surface, all having volume less than some constant independent of $n$ that does this; Also if we have ni restriction on area, then even a genus $g$ surface can be turned.)

Ingredient #2: For any constant $K$, there is $n$ large enough so no surfaces of genus $ and area $ inside the middle fibred manifold with boundary $M'_n$ can divide the volume of $M'_n$ in half.

The prove of this is partially based on our all-time favorite: the isoperimetric inequality:

Each Riemannian metric $\lambda$ on a closed surface has a linear isoperimetric inequality for 1-chains bounding 2-chains, i.e. any homologically trivial 1-chain $c$ bounds a $2$ chain $z$ where

$\mbox{Vol}_2(z) \leq K_\lambda \mbox{Vol}_1(c)$.

Fitting things together:

Suppose there is (as described above) a family of genus $2g-1$ surfaces, each dividing the volume of $M_{2n}$ in half and flips the two sides of the surface as time goes from $0$ to $1$.

By ingredient #1, since the family is by construction surfaces from (different) sweep-outs by ‘minimal surfaces’, we have $\mbox{Vol}_2(S_t) < 5 \pi (2g-2)$ for all $t$.

Now if we take the two component separated by $S_t$ and intersect them with the left-most $n$ copies of $M$ in $M'_{2n}$ (call it $M'_L$), at some $t$, $S_t$ must also divide the volume of $M'_L$ in half.

Since $S_t$ divides both $M'_2n$ and $M'_L$ in half, it must do so also in $M'_R$.

But $S_t$ is of genus $2g-1$! So one of $S_t \cap M'_L$ and $S_t \cap M'_R$ has genus $< g$! (say it's $M'_L$)

Apply ingredient #2, take $K = 5 \pi (2g-2)$, there is $n$ large enough so that $S_t \cap M'_L$, which has area less than $K$ and genus less than $g$, cannot possibly divide $M'_L$ in half.