## Archive for May, 2011

### A report from the Workshop in Geometric Topology @ Utah (part 1)

May 29, 2011

I went to Park City this passed week for the Workshop in Geometric Topology. It was a quite cool place filled with ski-equipment stores, Christmas souvenir shops, galleries and little wooden houses for family winter vacations. Well, as you may have guessed, the place would look very interesting in summer. :-P

As the ‘principal speaker’, Professor Gabai gave three consecutive lectures on his ending lamination space paper (this paper was also mentioned in my last post). I would like to sketch some little pieces of ideas presented in perhaps couple of posts.

Classification of simple closed curves on surfaces

Let $S_{g,p}$ denote the (hyperbolic) surface of genus $g$ and $p$ punchers. There is a unique geodesic loop in each homotopy class. However, given a geodesic loop drew on the surface, how would you describe it to a friend over telephone?

Here we wish to find a canonical way to describe homotopy classes of curves on surfaces. This classical result was originally due to Dehn (unpublished), but discovered independently by Thurston in 1976. For simplicity let’s assume for now that $S$ is a closed surface of genus $g$.

Fix pants decomposition $\mathcal{T}$ of $S$, $\mathcal{T} = \{ \tau_1, \tau_2, \cdots, \tau_{3g-3} \}$ is a disjoint union of $3g-3$ ‘cuffs’.

As we can see, any simple closed curve will have an (homology) intersection number with each of the cuffs. Those numbers are non-negative integers:

Around each cuff we may assign an integer twist number, for a cuff with intersection number $n$ and twist number $z$, we ‘twist’ the curve inside a little neighborhood of the cuff so that all transversal segments to the cuff will have $z$ intersections with the curve.

Negative twists merely corresponds to twisting in the other direction:

Theorem: Every simple closed curve is uniquely defined by its intersection number and twisting number w.r.t each of the cuffs.

Conversely, if we consider multi-curves (disjoint union of finitely many simple closed curves) then any element in $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$ describes a unique multi-curve.

To see this we first assume that the pants decomposition comes with a canonical ‘untwisted’ curve connecting each pairs of cuffs in each pants. (i.e. there is no god given ‘0’ twist curves, hence we have to fix which ones to start with.)

In the example above our curve was homotopic to the curve $((1,2), (2,1), (1,-4))$.

In other words, pants decompositions (together with the associated 0-twist arcs) give a natural coordinate chart to the set of homotopy class of (multi) curves on a surface. i.e. they are perimetrized by $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$.

For the converse, we see that any triple of integers can be realized by filling the pants with a unique set of untwisted arcs:

In fact, this kind of parametrization can be generalized from integers to real numbers, in which case we have measured laminations instead of multi-curves and maximal train trucks on each pants instead of canonical untwisted arcs. i.e.

Theorem: (Thurston) The space of measured laminations $\mathcal{ML}(S)$ on a surface $S$ of genus $g$ is parametrized by $\mathbb{R}^{3g-3} \times \mathbb{R}_{\geq 0}^{3g-3}$. Furthermore, the correspondence is a homeomorphism.

Here the intersection numbers with the cuffs are wrights of the branches of the train track, hence it can be any non-negative real number. The twisting number is now defined on a continuous family of arcs, hence can be any real number, as shown below:

As we can see, just as in the case of multi-curves, any triple of real numbers assigned to the cuffs can be realized as the weights of branches of a train track on the pants.

### Gromov boundary of hyperbolic groups

May 16, 2011

As we have seen in pervious posts, the Cayley graphs of groups equipped with the word metric is a very special class of geodesic metric space – they are graphs that have tons of symmetries. Because of that symmetry, we can’t construct groups with any kind of Gromov boundary we want. In fact, there are only few possibilities and they look funny. In this post I want to introduce a result of Misha Kapovich and Bruce Kleiner that says:

Let $G$ be a hyperbolic group that’s not a semidirect product $H \ltimes N$ where $N$ is finite or virtually cyclic. (In those cases the boundary of $G$ can be obtained from the boundary of $H$$). Theorem: When $G$ has $1$-dimensional boundary, then the boundary is homeomorphic to either a Sierpinski carpet, a Menger curve or $S^1$. OK. So what are those spaces? (don’t worry, I had no clue about what a ‘Menger curve’ is before reading this paper). The Sierpinski carpet (I believe most people have seen this one) Start with the unit square, divide it into nine equal smaller squares, delete the middle one. Repeat the process to the eight remaining squares. and repeat… Of course we then take the infinite intersection to get a space with no interior. Proposition: The Sierpinski carpet is (covering) 1-dimensional, connected, locally connected, has no local cut point (meaning we cannot make any open subset of it disconnected by removing a point). Theorem: Any compact metrizable planar space satisfying the above property is a Sierpinski carpet. The Menger curve Now we go to $\mathbb{R}^3$, the Menger curve is the intersection of the Sierpinski carpet times the unit interval, one in each of the $x, y, z$ direction. Equivalently, we may take the unit cube $[0,1]^3$, subtract the following seven smaller cubes in the middle: In the next stage, we delete the middle ‘cross’ from each of the remaining 20 cubes: Proceed, take intersection. Proposition: The Menger curve is 1-dimensional, connected, locally connected, has no local cut point. Note this is one dimensional because we can decompose the ‘curve’ to pieces of arbitrary small diameter by cutting along thin rectangular tubes, meaning if we take those pieces and slightly thicken them there is no triple intersections. Theorem: Any compact metrizable nowhere planar (meaning no open set of it can be embedded in the plane) space satisfying the above property is a Menger curve. Now we look at our theorem, infact the proof is merely a translation from the conditions on the group to topological properties of the boundary and then seeing the boundary as a topological space satisfies our universal properties. A group being Gromov hyperbolic implies the boundary is compact metrizable. No splitting over finite or virtually cyclic group implies the boundary is connected, locally connected and if it’s not $S^1$, then it has no local cut point. Now what remains is to show, for groups, if the boundary is not planar then it’s nowhere planar. This is an easy argument using the fact that the group acts minimally on the boundary. Please refer to first part of their paper for details and full proof of the theorem. Remark: When study classical Polish-school topology, I never understood how on earth would one need all those universal properties (i.e. any xxx space is a xxx, usually comes with a long condition include ten or so items >.<). Now I see in fact such thing can be powerful. i.e. sometimes this allows us to actually get a grib on what does some completely unimaginable spaces actually look like! Another wonderful example of this is the recent work of S. Hensel and P. Przytycki and the even more recent work of David Gabai which shows ending lamination spaces are Nobeling curves. ### 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.