# A train track on twice punctured torus

This is a non-technical post about how I started off trying to prove a lemma and ended up painting this:

One of my favorite books of all time is Thurston‘s ‘Geometry and Topology of 3-manifolds‘ (and I just can’t resist to add here, Thurston, who happen to be my academic grandfather, is in my taste simply the coolest mathematician on earth!) Anyways, for those of you who aren’t topologists, the book is online and I have also blogged about bits and parts of it in some old posts such as this one.

I still vividly remember the time I got my hands on that book for the first time (in fact I had the rare privilege of reading it from an original physical copy of this never-actually-published book, it was a copy on Amie‘s bookshelf, which she ‘robbed’ from Benson Farb, who got it from being a student of Thurston’s here at Princeton years ago). Anyways, the book was darn exciting and inspiring; not only in its wonderful rich mathematical content but also in its humorous, unserious attitude — the book is, in my opinion, not an general-audience expository book, but yet it reads as if one is playing around just to find out how things work, much like what kids do.

To give a taste of what I’m talking about, one of the tiny details which totally caught my heart is this page (I can’t help smiling each time when flipping through the book and seeing the page, and oh it still haunts me >.<):

This was from the chapter about Kleinian groups, when the term ‘train-track’ was first defined, he drew this image of a train(!) on moving on the train tracks, even have smoke steaming out of the engine:

To me such things are simply hilarious (in the most delightful way).

Many years passed and I actually got a bit more into this lamination and train track business. When Dave asked me to ‘draw your favorite maximal train track and test your tube lemma for non-uniquely ergodic laminations’ last week, I ended up drawing:

Here it is, a picture of my favorite maximal train track, on the twice punctured torus~! (Click for larger image)

Indeed, the train is coming with steam~

Since we are at it, let me say a few words about what train tracks are and what they are good for:

A train track (on a surface) is, just as one might expect, a bunch of branches (line segments) with ‘switches’, i.e. whenever multiple branches meet, they must all be tangent at the intersecting point, with at least one branch in each of the two directions. By slightly moving the switches along the track it’s easy to see that generic train track has only switches with one branch on one side and two branches on the other.

On a hyperbolic surface $S_{g,p}$, a train track is maximal if its completementry region is a disjoint union of triangles and once punctured monogons. i.e. if we try to add more branches to a maximal track, the new branch will be redundant in the sense that it’s merely a translate of some existing branch.

As briefly mentioned in this post, train tracks give natural coordinate system for laminations just like counting how many times a closed geodesic intersect a pair of pants decomposition. To be slightly more precise, any lamination can be pushed into some maximal train track (although not unique), once it’s in the track, any laminations that’s Hausdorff close to it can be pushed into the same track. Hence given a maximal train track, the set of all measured laminations carried by the train track form an open set in the lamination space, (with some work) we can see that as measured lamination they are uniquely determined by the transversal measure at each branch of the track. Hence giving a coordinate system on $\mathcal{ML})(S)$.

Different maximal tracks are of course them pasted together along non-maximal tracks which parametrize a subspace of $\mathcal{ML}(S)$ of lower dimension.

To know more about train tracks and laminations, I highly recommend going through the second part of Chapter 8 of Thurston’s book. I also mentioned them for giving coordinate system on the measured lamination space in the last post.

In any case I shall stop getting into the topology now, otherwise it may seem like the post is here to give exposition to the subject while it’s actually here to remind myself of never losing the Thurston type childlike wonder and imagination (which I found strikingly larking in contemporary practice of mathematics).

# Filling and unfilling measured laminations

(images are gradually being inserted ~)

I’m temporarily back into mathematics to (try) finish up some stuff about laminations. While I’m on this, I figured maybe sorting out some very basic (and cool) things in a little post here would be a good idea. Browsing through the blog I also realized that as a student of Dave’s I have been writing surprisingly few posts related to what we do. (Don’t worry, like all other posts in this blog, I’ll only put in stuff anyone can read and hopefully won’t be bored reading :-P)

Here we go. As mentioned in this previous post, my wonderful advisor has proved that the ending lamination space is connected and locally connected (see Gabai’08).

Definition: Let $S_{g,p}$ be a hyperbolic surface of genus $g$ and $p$ punctures. A (geodesic) lamination $L \subseteq S$ is a closed set that can be written as a disjoint union of geodesics. i.e. $L = \sqcup_{\alpha \in I} \gamma_\alpha$ where each $\gamma_\alpha$ is a (not necessary closed) geodesic, $\gamma$ is called a leaf of $L$.

Let’s try to think of some examples:

i) One simple closed geodesic

ii) A set of disjoint simple closed geodesics

iii) A non-closed geodesic spirals onto two closed ones

iV) Closure of a single simple geodesic where transversal cross-sections are Cantor-sets

An ending lamination is a lamination where
a) the completement $S \backslash L$ is a disjoint union of discs and once punctured discs (filling)
b) all leaves are dense in $L$. (minimal)

Exercise: example i) satisfies b) and example iv) as shown satisfies both a) and b) hence is the only ending lamination.

It’s often more natural to look at measured laminations, for example as we have seen in the older post, measured laminations are natural generalizations of multi-curves and the space $\mathcal{ML}(S)$ is homeomorphic to $\mathbb{R}^{6g-6+2p}$ (Thurston) with very natural coordinate charts (given by train-tracks).

Obviously not all measured laminations are supported on ending laminations (e.g. example i) and ii) with atomic measure on the closed curves.) It is well known that if a lamination fully supports an invariant measure, then as long as the base lamination satisfies a), it automatically satisfies b) and hence is an ending lamination. This essentially follows from the fact that having a fully supported invariant measure and being not minimal implies the lamination is not connected and hence won’t be filling.

Exercise:Example iii) does not fully support invariant measures.

Scaling of the same measure won’t effect the base lamination, hence we may eliminate a dimension by quotient that out and consider the space of projective measured laminations $\mathcal{PML}(S) \approx \mathbb{S}^{6g-7+2p}$. Hence we may decompose measured laminations into filling and unfilling ones. i.e.

$\mathcal{PML}(S) = \mathcal{FPML}(S) \sqcup \mathcal{UPML}(S)$

where $\mathcal{FPML}(S)$ projects to the ending laminations via the forgetting measure map $\pi$.

This decomposition of the standard sphere $\mathbb{S}^{6g-7+2p}$ is mysterious and very curious in my opinion. To get a sense of this, let’s take a look at the following facts:

Fact 1: $\mathcal{UPML}$ is a union of countably many disjoint hyper-discs (i.e. discs of co-dimension $1$).

Well, if a measured lamination is unfilling, it must contain some simple closed geodesic as a leaf (or miss some simple closed geodesic). For each such geodesic $C$, there are two possible cases:

Case 1: $C$ is non-separating. The set of measured laminations that missed $C$ is precisely the set of projective measured laminations supported on $S_{g-1, p+2}$, hence homeomorphic to $\mathbb{S}^{6g-13+2p+4} = \mathbb{S}^{(6g-7+2p)-2}$ we may take any such measured lamination, disjoint union with $C$, we may assign any ratio of wrights to $C$ and the lamination. This corresponds to taking the cone of $\mathbb{S}^{(6g-7+2p)-2}$ with vertex being the atomic measure on $C$. Yields a disc of dimension $(6g-7+2p)-1$.

Case 2: $C$ is separating. Similarly, the set of measured laminations missing $C$ is supported on two connected surfaces with total genus $g$ and total punctures $p+2$.

To describe the set of projective measured laminations missing $C$, we first determine the ratio of measure between two connected components and then compute the set of laminations supported in each component. i.e. it’s homeomorphic to $[0,1] \times \mathbb{S}^{d_1} \times \mathbb{S}^{d_2}/\sim$ where $d_1+d_2 = 6g-2*7+2(p+2) = 6g-10+2p$ and $(0, x_1, y) \sim (0, x_2, y)$ and $(1, x, y_1) \sim (1, x, y_2)$.

Exercise: check this is a sphere. hint: if $d_1 =d_2 = 1$, we have:

Again we cone w.r.t. the atomic measure corresponding to $C$, get a hyper disc.

At this point you may think ‘AH! $\mathcal{UPML}$ is only a countable union of hyper-discs! How complicated can it be?!’ Turns out it could be, and (unfortunately?) is, quite messy:

Fact 2: $\mathcal{UPML}$ is dense in $\mathcal{PML}$.

This is easy to see since any filling lamination is minimal, hence all leaves are dense, we may simply take a long segment of some leaf where the beginning and end point are close together on some transversal, close up the segment by adding a small arc on the transversal, we get a simple closed geodesic that’s arbitrarily close to the filling lamination in $\mathcal{PML}$. Hence the set of simple closed curves with atomic measure are dense, obviously implying $\mathcal{UPML}$ dense.

So how exactly does this decomposition look like? I found it very mysterious indeed. One way to look at this decomposition is: we know two $\mathcal{UPML}$ discs can intersect if and only if their corresponding curved are disjoint. Hence in some sense the configuration captures the structure of the curve complex. Since we know the curve complex is connected, we may start from any disc, take all discs which intersect it, then take all discs intersecting one of the discs already in the set, etc.

We shall also note that all discs intersecting a given disc must pass through the point corresponding to the curve at the center. Hence the result will be some kind of fractal-ish intersecting discs:

(image)

Yet somehow it manages to ‘fill’ the whole sphere!

Hopefully I have convinced you via the above that countably many discs in a sphere can be complicated, not only in pathological examples but they appear in ‘real’ life! Anyways, with Dave’s wonderful guidance I’ve been looking into proving some stuff about this (in particular, topology of $\mathcal{FPML}$). Hopefully the mysteries would become a little clearer over time~!

# Haken manifolds and virtual Haken conjecture

Hi people~ My weekends have been unfortunately filled up with grading undergrad assignments for the last couple of weeks >.< I'll try to catch up on blogging by finding some other time slot during the week.

As a grand-student of Thurston's I feel obligated to end my ignorance regarding Haken manifolds. I guess it's a good idea to start by writing my usual kids-friendly exposition here.

In the rest of the post, $M$ is a compact (so perhaps with boundary), orientable, irreducible (meaning each embedded 2-sphere bounds a ball) 3-manifold.

Definition: A properly embedded oriented surface $S \subseteq M$ is incompressible if $S$ is not the 2-sphere and any simple closed curve on $S$ which bounds an embedded disc in $M \backslash S$ also bounds one in $S$.

Figure 1

In other words, together with Dehn’s lemma this says the map $\varphi: \pi_1(S) \rightarrow \pi_1 (M)$ induced by the inclusion map is injective.

Note that the surface $S$ could have boundary, for example:

Figure 2

Definition: $M$ is Haken if it contains an incompressible surface.

Okay, at this point you should be asking, what’s good about Haken manifolds? The beauty about it is that, roughly speaking, once you find one incompressible surface in the manifold, you can just keep finding them until the manifold is completely chopped up into balls by incompressible surfaces.

Theorem: (Haken) Any Haken 3-manifold $M$ contains a hierarchy $S_0 \subseteq S_1 \subseteq \cdots \subseteq S_n$ where

1.$S_0$ is an incompressible surface in $M$
2.$S_i = S_{i-1} \cup S$ where $S$ is an incompressible surface for the closure of some connected component $K$ of $latex $M \backslash S_{i-1}$ 3.$M \backslash S_n$ is a union of 3-balls Sketch of proof: This is much simpler than it might appear to be. The point is (at least in my opinion), except for trivial cases as long as a manifold has boundary it must be Haken. Lemma: If $\partial M$ has a component that’s not $\mathbb{S}^2$ then $M$ is Haken. The proof of the lemma is merely that any such $M$ will have infinite $H_2$ hence by the sphere theorem it will contain an embedded surface with non-trivial homology, if such surface is compressible then we just cut along the boundary of the compressing disc and glue two copies of it. This does not change the homology. Hence at the end we will arrive at a non-trivial incompressible surface. Figure 3 Now back to proving of the theorem, so we start by setting $S_0$ to be an incompressible surface given by $M$ being Haken. Now since $M$ is irreducible, we cut along $S_0$, i.e. take the closure of each component (may have either one or two components) of $M\backslash S_0$. Those will have a non-spherical boundary component, hence by lemma containing homologically non-trivial incompressible surface. This process continuous as long as some pieces has non-spherical boundary components. But since $M$ is irreducible, any sphere bounds a 3-ball in $M$, hence all components with sphere boundary are 3-balls. (In particular, the case where a component have multiple sphere boundary components cannot occur since the first boundary component bounds a 3-ball hence it can’t have any non-trivial incompressible surfaces on both sides.) Now the only remaining piece is to show that this process terminates. We apply a standard ‘normal surface argument’ for this. Essentially if we fix a triangulation of $M$, A normal surface in $M$ is one that intersects each 3-simplex in a disjoint union of following two shapes: Figure 4 There can’t be infinitely many non-parallel disjoint normal surfaces in $M$ (in fact there can be no more than 6 times the number of 3-simplexes since each complementry component need to contain at least one non-I-bundle part from one 3-simplex). Figure 5 However, if the above process do not terminate, we would obtain a sequence of non-parallel non-spherical boundary components: Figure 6 They represent different homology classes hence can be represented by disjoint normal which results in a contradiction. In general, this gives a way to prove theorems about Haken manifolds by using inductionL i.e. one may hope to just show the property trivially holds for 3-balls and is invariant under gluing two pieced along an incompressible surface. Note that the gluing surface being incompressible is in fact quite strong hence making the induction step possible in many cases. For example, by applying an incredible amount of brilliant techniques, Thurston was able to prove his revolutionary result: Hyperbolization theorem for Haken manifolds: Any Haken manifold $M$ with tori boundary components that does not contain incompressible tori admits a complete hyperbolic structure of finite volume in its interior. In other words, this is saying that given a Haken manifold, we cut along any incompressible tori, the resulting manifold with tori boundary must have a complete hyperbolic structure with cusps near each boundary component, This is the best we could hope for since manifolds with incompressible tori would have their fundamental group split over $Z^2$ which of course imply they can’t be hyperbolic. Now the more manifolds being Haken means the better this theorem is. Many evidences show that in fact a lot of manifolds are indeed Haken, in perticular we have: Virtual Haken Conjecture: $M$ is finitely covered by a Haken manifold as long as $\pi_1(M)$ is infinite. We can see that together with Thurston’s hyperbolization theorem, this would give full solution to the geometrization conjecture for general 3-manifolds. However, although now Perelman has proved the geometrization conjecture, the virtual Haken conjecture remains open. But in light of Perelman’s result now we are able to try to ‘back-solve’ the puzzle and only prove the virtual Haken conjecture for hyperbolic manifolds. (to be continued) # A report from the Workshop in Geometric Topology @ Utah (part 1) 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 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.