## Posts Tagged ‘3-manifolds’

### A train track on twice punctured torus

April 22, 2012

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).

### Haken manifolds and virtual Haken conjecture

November 21, 2011

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) ### Graph of groups in relation to 3-manifolds October 24, 2011 (some images might appear soon) Somehow I decided to wake up at 6:30 a.m. every Thursday to attend Bruce Kleiner‘s 9:30 course in NYU this semester. So far it’s been fun~ I learned about this thing called graph of groups. If you have been reading posts on this blog regarding any geometric group theory stuff (especially those posts related to Kleiner), then warning: this ‘graph’ has nothing to do with the Cayley graph. It’s not much about geometry but a rather ‘category-theoretical’ thing. Well, at this point you may think that you hate those algebra prople and is ready to leave…just don’t do that yet, because I hated them too, and now I finally got a tiny bit of understanding and appreciation on what those abstract non-sense was all about! :-P So how do we connect cool 3-manifold stuff (incompressible surfaces, loops, embedded discs Heegaard splittings etc.) to groups? Well, one handy thing is of course the Dehn’s lemma: Theorem: For 3-manifold $M$ with boundary, if the inclusion map $i: \pi_1(\partial(M)) \rightarrow \pi_1(M)$ is not injective, then there exists a simple non-trivial loop in $\partial M$ bounding an embedded disc in $M$. Note: Dehn’s theorem was proved by Papakyriakopoulos, I talked about it in this pervious post, although not exactly stated in this form, we can see that Dehn’s lemma follows easily from the loop theorem. This means we can say things about the 3-manifold by only looking solely at maps between groups! That’s cool, but sometimes we find groups and just one map between two groups are not enough, and that’s when graph of groups comes in: Definition: A graph of groups is a graph with vertice set $V$, edge set $E$, to each vertex $v$ we associate a group $G_v$ and to each edge $e$ (say connecting $v_1, v_2$) we also associate a group $G_e$, together with a pair of injective homomorphisms $f_1: G_e \rightarrow G_{v_1}$, $f_2: G_e \rightarrow G_{v_2}$. In our context, we should think of this as gluing together a bunch of spaces and take the fundamental group of those spaces, along with their pairwise intersections, as our vertice and edge groups. Just note that we need to have injections from the edge group to vertice groups. For simplicity one may first restrict oneself to the case where all edge groups are trivial (say spaces glued along contractible spaces). There is something called the fundamental group of a graph of groups which is essentially the fundamental group of the resulting space after you glued spaces according to the given graph of groups. Note that the injection associated to edges takes into account how gluing of different pairs interact with each other (that is to say, for example, on a homotopy level it knows about triple intersections, etc.) Let’s look at an application in this paper of Kleiner and Kapovich which I also talked about in an earlier post. Continue from that pervious post, now we know that Theorem: Any hyperbolic group (plus obvious conditions, namely torsion free and does not split over a finite cyclic group) with 1-dimensional boundary has $\partial_\infty G$ homeomorphic to $\mathbb{S}^1$, the Sierpinski carpet or the Menger curve. When $\partial_\infty G = \mathbb{S}^1$, my wonderful advisor Dave Gabai proved that $G$ would act discrete and cocompactly on $\mathbb{H}^2$ by isometries. (i.e. it’s almost the fundamental group of some hyperbolic surface except for possible finite order elements which make the action not properly discontinuous.) Now the next step is of course figuring out when does groups act on $\mathbb{H}^3$, we have: Cannon’s conjecture: If hyperbolic group $G$ has boundary $\mathbb{S}^2$, then $G$ acts discretely and cocompactly on $\mathbb{H}^3$ by isometries. This conjecture was also mentioned another pervious post. Turns our we do not know much about groups with $\mathbb{S}^2$ boundary. However, using graph of groups, they were able to show: Theorem: If Cannon’s conjecture is true, then those hyperbolic groups with Sierpinski carpet boundary are fundamental groups of hyperbolic 3-manifolds with totally geodesic boundary. i.e. the idea is to ‘extend’ the group with Sierpinski carpet boundary to a group having sphere boundary. Of course as sets we can embed the carpet into a sphere and start to ‘reflect it along the boundary of the ‘holes’, continue the process and eventually the union of all copies of the carpets is the entire $\mathbb{S}^2$. The problem is how to ‘reflect’ a group? First, since the boundary is homeomorphic to the carpet, there are countably many well-defined ‘boundary circles’, the group $G$ acts on the set of boundary circles. They showed this action has only finitely many different orbits. (those orbits of boundary circles will eventually correspond to those totally geodesic boundary components of our resulting 3-manifold). We pick one boundary circle from each orbit and denote their stabilizers $H_1, \cdots, H_k$ each $H_i < G$. Define a graph of groups $\mathcal{G}$ with two vertices both labeled $G$, with $k$ edges, all going from one vertex to the other. Let the edge groups be $H_1, \cdots, H_k$. Now we can start to 'unfold' the graph： Let $X_G$ be a 2-complex associated to a set of generators and relations for $G$ and $X_i$ be 2-complexes associated to $H_i$. The inclusion map induces cellular maps $h_i: X_i \rightarrow X_G$. Hence we have $\displaystyle h: \sqcup_{i=1}^n X_i \rightarrow X_G$ Let $X$ be the mapping cylinder of $h$. i.e. $X$ has boundary components $\sqcup_{i=1}^n X_i$ and $X_G$. Let $DX$ be the complex obtained by gluing together two copies of $X$ along $\sqcup_{i=1}^n X_i$, take it’s universal cover $\widetilde{DX}$. Now the fundemental group $\hat{G}$ of $DX$ is, in some sense, the group obtained by doubling $G$ along each $H_i$. i.e. $\hat{G}$ is the fundamental group of the graph graph of groups $\mathcal{G}$. Now by studying the 1-skeleton of the complex $\widetilde{DX}$, one is able to conclude that $\hat{G}$ is Gromov hyperbolic with $\mathbb{S}^2$ boundary, as expected. Hence from groups with Sierpinski carpet boundary we are able to produce groups with sphere boundary. Now if Cannon’s conjecture is true, $\hat{G}$ is fundamental group of some hyperbolic 3-manifold, together with Gabai’s result that $H_i$ are fundamental groups of hyperbolic surfaces, we would have that $G$ is the fundamental group of a hyperbolic 3-manifold with $n$ totally geodesic boundary components. Well, since now we don’t have Cannon’s conjecture, there is still something we can conclude: Definition: A n-dimensional Poincare duality group is a group which has group cohomology satisfying n-dimensional Poincare duality. Those should be thought of as fundamental groups of manifolds in the level of homology. Well, I know nothing about group cohomologies, luckily we have: Theorem: (Bestvina-Mess) $\Gamma$ is a n-dimensional Poincare duality group iff it’s torsion free and $\partial \Gamma$ has integral Cech cohomology of $\mathbb{S}^{n-1}$ Great! In our case $\partial \hat{G}$ IS the sphere! So it’s a 3-dimensional Poincare duality group~ Now we have a splitting of $\hat{G}$ over a bunch of 2-dimensional Poincare duality groups (namely $H_i$) it follows that $(G; H_1, \cdots, H_n)$ is a Poincare duality pair. It is not known whether all such pairs can be realized as fundamental groups of 3-manifolds with boundary. If so, then by Thurston’s geometrization we can also obtain what we derived assuming Cannon’s conjecture. ### A remark on a mini-course by Kleiner in Sullivan’s 70th birthday June 7, 2011 I spent the last week on Long Island for Dennis Sullivan’s birthday conference. The conference is hosted in the brand new Simons center where great food is served everyday in the cafe (I think life-wise it’s a wonderful choice for doing a post-doc). Anyways, aside from getting to know this super-cool person named Dennis, the talks there were interesting~ There are many things I found so exciting and can’t help to not say a few words about, however due to my laziness, I can only select one item to give a little stupid remark on: So Bruce Kleiner gave a 3-lecture mini-course on boundaries of Gromov hyperbolic spaces (see this related post on a piece of his pervious work in the subject) Cannon’s conjecture: Any Gromov hyperbolic group with $\partial_\infty G \approx \mathbb{S}^2$ acts discretely and cocompactly by isometries on $\mathbb{H}^3$. As we all know, in the theory of Gromov hyperbolic spaces, we have the basic theorem that says if a groups acts on a space discretely and cocompactly by isometries, then the group (equipped with any word metric on its Cayley graph) is quasi-isometric to the space it acts on. Since I borrowed professor Sullivan as an excuse for writing this post, let’s also state a partial converse of this theorem (which is more in the line of Cannon’s conjecture): Theorem: (Sullivan, Gromov, Cannon-Swenson) For $G$ finitely generated, if $G$ is quasi-isometric to $\mathbb{H}^n$ for some $n \geq 3$, then $G$ acts on $\mathbb{H}^n$ discretely cocompactly by isometries. This essentially says that due to the strong symmetries and hyperbolicity of $\mathbb{H}^n$, in this case quasi-isometry is enough to guarantee an action. (Such thing is of course not true in general, for example any finite group is quasi-isometric to any compact metric space, there’s no way such action exists.) In some sense being quasi-isometric is a much stronger condition once the spaces has large growth at infinity. In light of the above two theorems we know that Cannon’s conjecture is equivalent to saying that any hyperbolic group with boundary $\mathbb{S}^2$ is quasi-isometric to $\mathbb{H}^3$. At first glance this seems striking since knowing only the topology of the boundary and the fact that it’s hyperbolic, we need to conclude what the whole group looks like geometrically. However, the pervious post on one dimensional boundaries perhaps gives us some hint on the boundary can’t be anything we want. In fact it’s rather rigid due to the large symmetries of our hyperbolic group structure. Having Cannon’s conjecture as a Holy Grail, they developed tools that give raise to some very elegant and inspring proofs of the conjecture in various special cases. For example: Definition: A metric space $M$, is said to be Alfors $\alpha$-regular where $\alpha$ is its Hausdorff dimension, if there exists constant $C$ s.t. for any ball $B(p, R)$ with $R \leq \mbox{Diam}(M)$, we have: $C^{-1}R^\alpha \leq \mu(B(p,R)) \leq C R^\alpha$ This is saying it’s of Hausdorff dimension $\alpha$ in a very strong sense. (i.e. the Hausdorff $\alpha$ measure behaves exactly like the regular Eculidean measure everywhere and in all scales). For two disjoint continua $C_1, C_2$ in $M$, let $\Gamma(C_1, C_2)$ denote the set of rectifiable curves connecting $C_1$ to $C_2$. For any density function $\rho: M \rightarrow \mathbb{R}^+$, we define the $\rho$-distance between $C_1, C_2$ to be $\displaystyle \mbox{dist}_\rho(C_1, C_2) = \inf_{\gamma \in \Gamma(C_1, C_2)} \int_\gamma \rho$. Definition: The $\alpha$-modulus between $C_1, C_2$ is $\mbox{Mod}_\alpha(C_1, C_2) = \inf \{ \int_M \rho^\alpha \ | \ \mbox{dist}_\rho(C_1, C_2) \geq 1 \}$, OK…I know this is a lot of seemingly random definitions to digest, let’s pause a little bit: Given two continua in our favorite $\mathbb{R}^n$, new we are of course Hausdorff dimension $n$, what’s the $n$-modulus between them? This is equivalent to asking for a density function for scaling the metric so that the total n-dimensional volume of $\mathbb{R}^n$ is as small as possible but yet the length of any curve connecting $C_1, \ C_2$ is larger than $1$. So intuitively we want to put large density between the sets whenever they are close together. Since we are integrating the $n$-th power for volume (suppose $n>1$, since our set is path connected it’s dimension is at least 1), we would want the density as ‘spread out’ as possible while keeping the arc-length property. Hence one observation is this modulus depends on the pair of closest points and the diameter of the sets. The relative distance between $C_1, C_2$ is $\displaystyle \Delta (C_1, C_2) = \frac{\inf \{ d(p_1, p_2) \ | \ p_1 \in C_1, \ p_2 \in C_2 \} }{ \min \{ \mbox{Diam}(C_1), \mbox{Diam}(C_2) \} }$ We say $M$ is $\alpha$-Loewner if the $\alpha$ modulus between any two continua is controlled above and below by their relative distance, i.e. there exists increasing functions $\phi, \psi: [0, \infty) \rightarrow [0, \infty)$ s.t. for all $C_1, C_2$, $\phi(\Delta(C_1, C_2)) \leq \mbox{Mod}_\alpha(C_1, C_2) \leq \psi(\Delta(C_1, C_2))$ Those spaces are, in some sense, regular with respect to it’s metric and measure. Theorem: If $\partial_\infty G$ is Alfors 2-regular and 2-Loewner, homeomorphic to $\mathbb{S}^2$, then $G$ acts discrete cocompactly on $\mathbb{H}^3$ by isometries. Most of the material appeared in the talk can be found in their paper. There are many other talks I found very interesting, especially that of Kenneth Bromberg, Mario Bonk and Peter Jones. Unfortunately I had to miss Curt McMullen, Yair Minski and Shishikura… ### 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.

### When k looks and smells like the unknot…

February 14, 2011

Valentine’s day special issue~ ^_^

Professor Gabai decided to ‘do some classical topology before getting into the fancy stuff’ in his course on Heegaard structures on 3-manifolds. So we covered the ‘loop theorem’ by Papakyriakopoulos last week. I find it pretty cool~ (So I started applying it to everything regardless of whether a much simpler argument exists >.<)

Let $M$ be a three dimensional manifold with (non-empty) boundary. In what follows everything is assumed to be in the smooth category.

Theorem: (Papakyriakopoulos, ’58)
If $f: \mathbb{D}^2 \rightarrow M$ extends continuously to $\partial \mathbb{D}$ and the image $f(\partial \mathbb{D}) \subseteq \partial M$ is homotopically non-trivial in $\partial M$. Then in any neighborhood $N(f(\mathbb{D}))$ we can find embedded disc $D \subseteq M$ such that $\partial D$ is still homotopically non-trivial in $\partial M$.

i.e. this means that if we have a loop on $\partial M$ that is non-trivial in $\partial M$ but trivial in $M$, then in any neighborhood of it we can find a simple loop that’s still non-trivial in $\partial M$ and bounds an embedded disc in $M$.

We apply this to the following:

Corollary: If a knot $k \subseteq \mathbb{S}^3$ has $\pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$ then $k$ is the unknot.

Proof: Take tubular neighborhood $N_\varepsilon(k)$, consider $M=\mathbb{S}^3 \backslash \overline{N_\varepsilon(k)}$, boundary of $M$ is a torus.

By assumption we have $\pi_1(M) = \pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}$.

Let $k' \subseteq \partial M$ be a loop homotopic to $k$ in $N_\varepsilon(k)$.

Since $\pi_1(M) = \mathbb{Z}$ and any loop in $M$ is homotopic to a loop in $\partial M = \mathbb{T}^2$. Hence the inclusion map $i: \pi_1(\mathbb{T}^2) \rightarrow \pi_1(M)$ is surjective.

Let $l \subseteq \partial M$ be the little loop winding around $k$.

It’s easy to see that $i(l)$ generates $\pi_1(M)$. Hence there exists $n$ s.t. $k'-n \cdot l = 0$ in $\pi_1(M)$. In other words, after $n$ Dehn twists around $l$, $k'$ is homotopically trivial in $M$ i.e. bounds a disk in $M$. Denote the resulting curve $k''$.

Since $k''$ is simple, there is small neighborhood of $k''$ s.t. any homotopically non-trivial simple curve in the neighborhood is homotopic to $k''$. The loop theorem now implies $k''$ bounds an embedded disc in $M$.

By taking a union with the embedded collar from $k$ to $k''$ in $N_\varepsilon(k)$:

We conclude that $k$ bounds an embedded disc in $\mathbb{S}^3 \backslash k$ hence $k$ is the unknot.

Establishes the claim.

Happy Valentine’s Day, Everyone! ^_^

November 8, 2010

As I was trying to understand the Whitehead manifold and related constructions of non-tame manifolds batter, I guess it makes a cool blog post ^^

The Whitehead manifold is an example of a 3-manifold that’s contractable but not homeomorphic to $\mathbb{R}^3$ i.e. the manifold is homotopically equivalent to a point but not tame. (See the earlier post on tameness for more explanations)

Construction:
Take a solid torus $T_1 \subseteq \mathbb{R}^3$, embed a thinner torus $T_2 \subseteq T_1$ as shown:

Iterate the process: at each step, embed solid torus $T_i$ into $T_{i-1}$ so that $T_i$ “links with itself” inside $T_{i-1}$:

Let the Whitehead continuum be the intersection of the $T_i$s.

i.e. $\displaystyle W=\bigcap_{i=1}^\infty T_i$

As some people know, I have a weird hobby of describing strange continua in terms of Cantor sets…So here comes ‘Conan’s translation’ of the Whitehead continuum:

Take two copies of $C \times [0,1]$ where $C$ is the standard middle-third Cantor set. Bend them into Rainbow-shape with the open ends facing each other:

(I think of this as having a width $1$ ‘brush’ with ink only on the points of the Cantor set, and use the brush to draw two semicircles)

Now we connect the open ends: take a width $1/3$ brush and connect the top pair of ends so that they link with each other, and then a width $1/9$ brush for the highest remaining pair, etc. Take the union of all those connecting sets, union a line segment joining the bottom-most pair of points, we get the Whitehead continuum:

The Whitehead manifold is the complement of $W$ in the three-sphere $\mathbb{S}^3$, equipped with the

What’s the fundamental group of the Whitehead manifold?

Claim: $T_i^c$ is null-homotopic in $T_{i+1}^c$.

pf: $T_1^c$ is contractible to a loop in $T_{i+1}^c$, hence it suffice to homotope the red loop to a point without touching the black loop:

Note that for homotopy, the loop is allowed to pass through itself: (in contrast to isotopy)

The loop can now be easily contracted:

Hence we deduce the Whitehead manifold is null-homotopic. (by collapsing each $T_1$ at some finite time)

In particular, it has trivial fundamental group! (This might seem hard to believe especially when looking at my Cantor-set picture) Infact for this, we can directly see from the picture that all loop can be homotoped to constant:

Since loop is compact, there is a ‘finest gap’ in the Cantor set which the loop passes through, say it’s a gap with width $1/3^i$. Now by performing the operation above, we can homotope all parts of the loop that goes through the $1/3^i$ to segments that goes through $1/3^{i-1}$-gaps, by having the segments crossing themselves once. Now we pass to the $1/3^{i-2}$-gaps, etc. until the loop lie completely outside the thickened disc which the continua lies in. Once it’s outside the disc, the loop can be contracted.

The manifold is not homeomorphic to $\mathbb{R}^3$ as we can easily see that, unlike in $\mathbb{R}^3$, the red loop in picture cannot be isotoped to a trivial loop.

As an alternative point of view, we should note that in fact $T_i^c$ and $T_{i+1}$ form a thickened Whitehead link:

Since the Whitehead link is symmetric, this gives an simpler (but less direct, in my opinion) way of knowing that red loop is homotopically trivial in the complement of the black loop. (As the black loop is obviously homotopically trivial in the complement of the red loop.)

In light of this, one may construct many different non-tame manifolds with finitely generated fundamental group by embedding a handlebody inside another copy of itself and take the complement of the infinite intersection.

Here is an example of embedding a genus $3$ handlebody. The resulting manifold (after taking the complement of the intersection) is a homotopy genus $2$ handlebody. (As in the whitehead case, $T_i^c$ can be homotoped to a genus $2$ handlebody inside $T_{i+1}^c$. The ‘third loop’ can be unknotted by crossing itself once.) But it’s of coruse not homeomorphic to the genus $2$ handlebody.

### Proving the tameness conjecture

October 17, 2010

I have recently went through professor Gabai’s wonderful paper that gives a proof of the tameness conjecture. (This one is a simplified version of the argument given in Gabai and Calegari, where everything is done in the smooth category instead of PL). It’s been a quite exciting reading with many amazing ideas, hence I decided to write a summary from my childish viewpoint (as someone who knew nothing about the subject beforehand).

We say a manifold is tame if it an be embedded in a compact manifold s.t. the closure of the embedding is the whole compact manifold.

To motivate the concept, let’s look at surfaces: Any compact surface is, of course, tame. However, if we “shoot out” a few points of the surface to infinity, as the figure below, it become non-compact but still tame, as we can embed the infinite tube to a disk without a point.

Of course, we can also make a surface non-compact by shooting any closed subset to infinity (e.g. a Cantor set), but such construction will always result in a tame surface. (This can be realized using similar embeddings as above, we may embed the resulting surface into the original surface with image being the original surface subtract the closed set. If the closed set has interior, we further contract each interior components.)

On the other hand, any surface with infinite genus would be non-tame since if there is an embedding into a compact set, the image of ‘genesis’ would have limit points, which will force the compact space fail to be a manifold at that point.

Hence in spirit, being tame means that although the manifold may not be compact itself, but all topology happens in bounded regions (we can think of a complete embedding of the manifold into some $\mathbb{R}^N$ so bounded make sense)

As usual, life gets more complicated for three-manifolds.

Tameness conjecture: Every complete hyperbolic 3-manifold with finitely generated fundamental group is tame.

A bubble chart for capturing the structure of the proof:

A few highlights of the proof: The key idea here is shrinkwrapping, very roughly speaking, to prove an geometrically infinite end is tame one needs to find a sequence of simplicial hyperbolic surfaces exiting at the end. Bonahon’s theorem gives us a sequence of closed geodesics exiting the end. By various pervious results, one is able to produce (topological) surfaces that are ‘in between’ those geodesics. Shrinkwrapping takes the given surface and shrinks it until it’s ‘tightly wrapped’ around the given sequence of geodesics. The fact that each of the curve the surface is wrapping around is a geodesic guarantees the resulting surface simplicial hyperbolic. (think of this as folding a piece of paper along a curve would effect its curvature, but alone a straight line would not; geodesics are like straight lines).

Once we have that, the remaining part would be showing the position of the surfaces are under control so that they would exit the end. Since simplicial hyperbolic surfaces has curvature $\leq -1$, by Gauss-Bonnet they have uniformly bounded area (given our surfaces also has bounded genus). By passing to a subsequence, we may choose the sequence of geodesics to be separated by some uniform constant, which will guarantee the wrapped surfaces are not too thin in the thick parts of the manifold, hence we have control over the diameter of the surface, from which we can conclude that the surfaces must exit the manifold.

Remark: Note that in general, unlike in two dimensions, a three manifold with finitely generated fundamental group does not need to be tame as the Whitehead manifold is homotopic to $\mathbb{R}^3$ (hence trivial fundamental group) but is not tame. On the other hand, if we have infinitely generated fundamental group, then the manifold can never be tame. The theorem says all examples of non-tame manifolds with finitely generated fundamental group does not admit hyperbolic structure.

April 21, 2010

I’ve been going through Thurston’s book ‘The Geometry and Topology of Three-Manifolds‘ in a reading course with Amie Wilkinson. In Chapter 3, p32, when he’s constructing a hyperbolic structure on the Whitehead link complement, there is a picture on how to glue the 2-cells to the knot, to quite Thurston, ‘the attaching map for the two-cells are indicated by the dotted lines.’ However, for me it’s impossible to see where are the dotted lines going. So I reconstruct it here with some more clear pictures. The construction itself was a cool reading that I wish to share.

First, we have the Whitehead link, looking like the first figure below:

We attach three 1-cells (line segments) as in the second figure, note that the ‘x’ in the middle represents a line segment orthogonal to the screen, connecting the top and bottom line in the figure ‘8’ loop.

Now we will start to attach four 2-cells to the 1-complex above: First, we attach a 2-cell spanning the top part of the figure ‘8’ loop, spanning one side of the middle segment and two sides of the top segment (denote this by cell A):

Do the same with the bottom half (cell B). Note that each cell is attached to three edges, hence they are triangles without vertices in the knot complement with three one-cells attached.

For the other two cells, we attach as follows (cells C and D):

Combining the four 2-cells, we get something like the figure showed below. Note that at the top, cell A is under cell C in the left, intersecting the surface spanned by cells C and D at the edge, and comes above cell D to the right of the edge.

It’s easy to see that the complement of the above 2-complex does not separate $\mathbb{R}^3$, hence it’s a 3-cell with eight faces (i.e. it has to go through both sides of each 2-cell in order to fill the 3-space) each of its face has three edges. Hence we may glue an octahedron to the 2-complex after the gluing, pairs of faces of the octahedron will be identified groups of four edges will be identified to single edges. Hence to put a hyperbolic structure on the link complement, it suffice to put an hyperbolic structure to the octahedron with vertices deleted.

Since each edge is glued up by four edges of the octahedron, it suffice to find an octahedron (without vertices) in the hyperbolic 3-space that has all adjacent faces intersect in dihedral angle $2 \pi / 4$ i.e. all adjecent faces are orthogonal in the hyperbolic space. But this is achieved if we inscribe the regular octahedron into the Klein model (also called projective model of hyperbolic 3-space.

The gluing map for the faces are merely rotations and reflections of the ball which are certainly hyperbolic isometries. Hence this gives a hyperbolic structure to the link complement.