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Last week I was at Oberwolfach for a meeting on geometric group theory. My friend and collaborator Koji Fujiwara gave a very nice talk about constructing actions of groups on quasi-trees (i.e. spaces quasi-isometric to trees). The construction is inspired by the famous subsurface projection construction, due to Masur-Minsky, which was a key step in their proof that the complex of curves (a natural simplicial complex on which the mapping class group acts cocompactly) is hyperbolic. Koji’s talk was very stimulating, and shook up my thinking about a few related matters; the purpose of this blog post is therefore for me to put some of my thoughts in order: to describe the Masur-Minsky construction, to point out a connection to certain geometric phenomena like winding numbers of curves on surfaces, and to note that a variation on their construction gives rise directly to certain natural chiral invariants of surface automorphisms (and their generalizations) which should be relevant to 4-manifold topologists.

A few weeks ago, Ian Agol, Vlad Markovic, Ursula Hamenstadt and I organized a “hot topics” workshop at MSRI with the title Surface subgroups and cube complexes. The conference was pretty well attended, and (I believe) was a big success; the organizers clearly deserve a great deal of credit. The talks were excellent, and touched on a wide range of subjects, and to those of us who are mid-career or older it was a bit shocking to see how quickly the landscape of low-dimensional geometry/topology and geometric group theory has been transformed by the recent breakthrough work of (Kahn-Markovic-Haglund-Wise-Groves-Manning-etc.-) Agol. Incidentally, when I first started as a graduate student, I had a vague sense that I had somehow “missed the boat” — all the exciting developments in geometry due to Thurston, Sullivan, Gromov, Freedman, Donaldson, Eliashberg etc. had taken place 10-20 years earlier, and the subject now seemed to be a matter of fleshing out the consequences of these big breakthroughs. 20 years and several revolutions later, I no longer feel this way. (Another slightly shocking aspect of the workshop was for me to realize that I am older or about as old as 75% of the speakers . . .)

The rationale for the workshop (which I had some hand in drafting, and therefore feel comfortable quoting here) was the following:

Recently there has been substantial progress in our understanding of the related questions of which hyperbolic groups are cubulated on the one hand, and which contain a surface subgroup on the other. The most spectacular combination of these two ideas has been in 3-manifold topology, which has seen the resolution of many long-standing conjectures. In turn, the resolution of these conjectures has led to a new point of view in geometric group theory, and the introduction of powerful new tools and structures. The goal of this conference will be to explore the further potential of these new tools and perspectives, and to encourage communication between researchers working in various related fields.

I have blogged a bit about cubulated groups and surface subgroups previously, and I even began this blog (almost 4 years ago now) initially with the idea of chronicling my efforts to attack Gromov’s surface subgroup question. This question asks the following:

Gromov’s Surface Subgroup Question: Does every one-ended hyperbolic group contain a subgroup which is isomorphic to the fundamental group of a closed surface of genus at least 2?

The restriction to one-ended groups is just meant to rule out silly examples, like finite or virtually cyclic groups (i.e. “elementary” hyperbolic groups), or free products of simpler hyperbolic groups. Asking for the genus of the closed surface to be at least 2 rules out the sphere (whose fundamental group is trivial) and the torus (whose fundamental group $\mathbb{Z}^2$ cannot be a subgroup of a hyperbolic group). It is the purpose of this blog post to say that Alden Walker and I have managed to show that Gromov’s question has a positive answer for “most” hyperbolic groups; more precisely, we show that a random group (in the sense of Gromov) contains a surface subgroup (in fact, many surface subgroups) with probability going to 1 as a certain natural parameter (the “length” $n$ of the random relators) goes to infinity. (update April 8: the preprint is available from the arXiv here.)

The purpose of this brief blog post is to advertise that I wrote a little piece of software called wireframe which can be used to quickly and easily produce .eps figures of surface for inclusion in papers. The main use is that one can specify a graph in an ASCII file, and the program will then render a nice 3d picture of a surface obtained as the boundary of a tubular neighborhood of the graph. The software can be downloaded from my github repository at

https://github.com/dannycalegari/wireframe

and then compiled on any unix machine running X-windows (e.g. linux, mac OSX) with “make”.

The program is quite rudimentary, but I believe it should be useful even in its current state. Users are strenuously encouraged to tinker with it, modify it, improve it, etc. If you use the program and find it useful (or not), please let me know.

A couple of examples of output (which can be created in about 5 minutes) are:

and

(added Feb. 20, 2013): I couldn’t resist; here’s another example:

(update April 12, 2013:) Scott Taylor used wireframe to produce a nice figure of a handlebody (in 3-space) having the Kinoshita graph as a spine. He kindly let me post his figure here, as an example. Thanks Scott!

Let $F=\langle a,b\rangle$ be the free group on two generators, and let $\phi:F \to F$ be the endomorphism defined on generators by $\phi(a)=ab$ and $\phi(b)=ba$. We define Sapir’s group $C$ to be the ascending HNN extension

$F*_\phi:=\langle a,b,t\; | \; a^t=ab,b^t=ba\rangle$

This group was studied by Crisp-Sageev-Sapir in the context of their work on right-angled Artin groups, and independently by Feighn (according to Mark Sapir); both sought (unsuccessfully) to determine whether $C$ contains a subgroup isomorphic to the fundamental group of a closed, oriented surface of genus at least 2. Sapir has conjectured in personal communication that $C$ does not contain a surface subgroup, and explicitly posed this question as Problem 8.1 in his problem list.

After three years of thinking about this question on and off, Alden Walker and I have recently succeeded in finding a surface subgroup of $C$, and it is the purpose of this blog post to describe this surface, how it was found, and some related observations. By pushing the technique further, Alden and I managed to prove that for a fixed free group $F$ of finite rank, and for a random endomorphism $\phi$ of length $n$ (i.e. one taking the generators to random words of length $n$), the associated HNN extension contains a closed surface subgroup with probability going to 1 as $n \to \infty$. This result is part of a larger project which we expect to post to the arXiv soon.

I am currently teaching a class at the University of Chicago on hyperbolic groups, and I have just introduced the concept of $\delta$-hyperbolic (geodesic) metric spaces. A geodesic metrix space $(X,d_X)$ is $\delta$-hyperbolic if for any geodesic triangle $abc$, and any $p \in ab$ there is some $q \in ac \cup bc$ with $d_X(p,q)\le \delta$. The quintessential $\delta$-hyperbolic space is the hyperbolic plane, the unique (up to isometry) simply-connected complete Riemannian 2-manifold of constant curvature $-1$. It follows that any simply-connected complete Riemannian manifold of constant curvature $K<0$ is $\delta$-hyperbolic for some $\delta$ depending on $K$; roughly one can take $\delta \sim (-K)^{-1/2}$.

What gives this condition some power is the rich class of examples of spaces which are $\delta$-hyperbolic for some $\delta$. One very important class of examples are simply-connected complete Riemannian manifolds with upper curvature bounds. Such spaces enjoy a very strong comparison property with simply-connected spaces of constant curvature, and are therefore the prime examples of what are known as CAT(K) spaces.

Definition: A geodesic metric space $(X,d_X)$ is said to be $CAT(K)$, if the following holds. If $abc$ is a geodesic triangle in $X$, let $\bar{a}\bar{b}\bar{c}$ be a comparison triangle in a simply connected complete Riemannian manifold $Y$ of constant curvature $K$. Being a comparison triangle means just that the length of $\bar{a}\bar{b}$ is equal to the length of $ab$ and so on. For any $p \in bc$ there is a corresponding point $\bar{p}$ in the comparison edge $\bar{b}\bar{c}$ which is the same distance from $\bar{b}$ and $\bar{c}$ as $p$ is from $b$ and $c$ respectively. The $CAT(K)$ condition says, for all $abc$ as above, and all $p \in bc$, there is an inequality $d_X(a,p) \le d_Y(\bar{a},\bar{p})$.

The term CAT here (coined by Gromov) is an acronym for Cartan-Alexandrov-Toponogov, who all proved significant theorems in Riemannian comparison geometry. From the definition it follows immediately that any $CAT(K)$ space with $K<0$ is $\delta$-hyperbolic for some $\delta$ depending only on $K$. The point of this post is to give a short proof of the following fundamental fact:

CAT(K) Theorem: Let $M$ be a complete simply-connected Riemannian manifold with sectional curvature $\le K_0$ everywhere. Then $M$ with its induced Riemannian (path) metric is $CAT(K_0)$.

I am in Paris attending a workshop at the IHP where Ian Agol has just given the first of three talks outlining his proof of the Virtual Haken Conjecture and Virtual Fibration Conjecture in 3-manifold topology (hat tip to Henry Wilton at the Low Dimensional Topology blog from whom I first learned about Ian’s announcement last week). I think it is no under overstatement to say that this marks the end of an era in 3-manifold topology, since the proof ties up just about every loose end left over on the list of problems in 3-manifold topology from Thurston’s famous Bulletin article (with the exception of problem 23 — to show that volumes of closed hyperbolic 3-manifolds are not rationally related — which is very close to some famous open problems in number theory). The purpose of this blog post is to say what the Virtual Haken Conjecture is, and some of the background that goes into Ian’s argument. I hope to follow this up with more details in another post (after Agol gives talks 2 and 3 this coming Wednesday). Needless to say this post has been written in a bit of a hurry, and I have probably messed up some crucial details; but if that caveat is not enough to dissuade you, then read on.

The purpose of this blog post is to try to give some insight into the “meaning” of the Hall-Witt identity in group theory. This identity can look quite mysterious in its algebraic form, but there are several ways of describing it geometrically which are more natural and easier to understand.

If $G$ is a group, and $a,b$ are elements of $G$, the commutator of $a$ and $b$ (denoted $[a,b]$) is the expression $aba^{-1}b^{-1}$ (note: algebraists tend to use the convention that $[a,b]=a^{-1}b^{-1}ab$ instead). Commutators (as their name suggests) measure the failure of a pair of elements to commute, in the sense that $ab=[a,b]ba$. Since $[a,b]^c = [a^c,b^c]$, the property of being a commutator is invariant under conjugation (here the superscript $c$ means conjugation by $c$; i.e. $a^c:=cac^{-1}$; again, the algebraists use the opposite convention).

In this post, I will cover triangles and area in spaces of constant (nonzero) curvature. We are focused on hyperbolic space, but we will talk about spheres and the Gauss-Bonnet theorem.

1. Triangles in Hyperbolic Space

Suppose we are given 3 points in hyperbolic space ${\mathbb{H}^n}$. A triangle with these points as vertices is a set of three geodesic segments with these three points as endpoints. The fact that there is a unique triangle requires a (brief) proof. Consider the hyperboloid model: three points on the hyperboloid determine a unique 3-dimensional real subspace of ${\mathbb{R}^{n+1}}$ which contains these three points plus the origin. Intersecting this subspace with the hyperboloid gives a copy of ${\mathbb{H}^2}$, so we only have to check there is a unique triangle in ${\mathbb{H}^2}$. For this, consider the Klein model: triangles are euclidean triangles, so there is only one with a given three vertices.

In hyperbolic space, it is still true that knowing enough side lengths and/or angles of a triangles determines it. For example, knowing two side lengths and the angle between them determines the triangle. Similarly, knowing all the angles determines it. However, not every set of angles can be realized (in euclidean space, for example, the angles must add to ${\pi}$), and the inequalities which must be satisfied are more complicated for hyperbolic space.

2. Ideal Triangles and Area Theorems

We can think about moving one (or more) of the points of a hyperbolic triangle off to infinity (the boundary of the disk). An ideal triangle is one with all three “vertices” (the vertices do not exist in hyperbolic space) on the boundary. Using a conformal map of the disk (which is an isometry of hyperbolic space), we can move any three points on the boundary to any other three points, so up to isometry, there is only one ideal triangle. We have fixed our metric, so we can find the area of this triangle. The logically consistent way to find this is with an integral since we will use this fact in our proof sketch of Gauss-Bonnet, but as a remark, suppose we know Gauss-Bonnet. Imagine a triangle very close to ideal. The curvature is ${-1}$, and the euler characteristic is ${1}$. The sum of the exterior angles is just slightly under ${3\pi}$, so using Gauss-Bonnet, the area is very close to ${\pi}$, and goes to ${\pi}$ as we push the vertices off to infinity.

One note is that suppose we know what the geodesics are, and we know what the area of an ideal triangle is (suppose we just defined it to be ${\pi}$ without knowing the curvature). Then by pasting together ideal triangles, as we will see, we could find the area of any triangle. That is, really the key to understanding area is knowing the area of an ideal triangle.

As mentioned above, there is a single triangle, up to isometry, with given angles, so denote the triangle with angles ${\alpha, \beta, \gamma}$ by ${\Delta(\alpha, \beta, \gamma)}$.

2.1. Area

Knowing the area of an ideal triangle allows us to calculate the area of any triangle. In fact:

Theorem 1 (Gauss) ${\mathrm{area}(\Delta(\alpha, \beta, \gamma)) = \pi - (\alpha + \beta + \gamma)}$

This geometric proof relies on the fact that the angles in the Poincare model are the euclidean angles in the model. Consider the generic picture:

We have extended the sides of ${\Delta(\alpha, \beta, \gamma)}$ and drawn the ideal triangle containing these geodesics. Since the angles are what they look like, we know that the area of ${\Delta(\alpha,\beta,\gamma)}$ is the area of the ideal triangle (${\pi}$), minus the sum of the areas of the smaller triangles with two points at infinity:

$\displaystyle \mathrm{area}(\Delta(\alpha, \beta, \gamma)) = \pi - \mathrm{area}(\Delta(\pi-\alpha, 0,0)) - \mathrm{area}(\Delta(\pi-\beta, 0, 0)) - \mathrm{area}(\Delta(\pi-\gamma, 0, 0))$

Thus it suffices to show that ${\mathrm{area}(\Delta(\pi - \alpha, 0, 0)) = \alpha}$.

For this fact, we need another picture:

Define ${f(\alpha) = \mathrm{area}(\Delta(\pi-\alpha, 0, 0))}$. The picture shows that the area of the left triangle (with two vertices at infinity and one near the origin) plus the area of the right triangle is the area of the top triangle plus the area of the (ideal) bottom triangle:

$\displaystyle f(\alpha) + f(\beta) = f(\alpha+\beta-\pi) + \pi$

We also know some boundary conditions on ${f}$: we know ${f(0) = 0}$ (this is a degenerate triangle) and ${f(\pi) = \pi}$ (this is an ideal triangle). We therefore conclude that

$\displaystyle f(\frac{\pi}{2}) + f(\frac{\pi}{2}) = f(0) + \pi \qquad \Rightarrow \qquad f(\frac{\pi}{2}) = \frac{\pi}{2}$

Similarly,

$\displaystyle 2f(\frac{3\pi}{4}) = f(\frac{\pi}{2}) + \pi \qquad \Rightarrow \qquad f(\frac{3\pi}{4}) = \frac{3\pi}{4}$

And we can find ${f(\pi/4) = \pi/4}$ by observing that

$\displaystyle f(\frac{3\pi}{4}) + f(\frac{\pi}{2}) = f(\frac{\pi}{4}) + \pi$

Similarly, if we know ${f(\frac{k\pi}{2^n}) = \frac{k\pi}{2^n}}$, then

$\displaystyle f(\frac{(2^{n+1}-1)\pi}{2^{n+1}}) = \frac{(2^{n+1}-1)\pi}{2^{n+1}}$

And by subtracting ${\pi/2^n}$, we find that ${f(\frac{k\pi}{2^{n+1}}) = \frac{k\pi}{2^{n+1}}}$. By induction, then, ${f(\alpha) =\alpha}$ if ${\alpha}$ is a dyadic rational times ${\pi}$. This is a dense set, so we know ${f(\alpha) = \alpha}$ for all ${\alpha \in [0,\pi]}$ by continuity. This proves the theorem.

3. Triangles On Spheres

We can find a similar formula for triangles on spheres. A lune is a wedge of a sphere:

A lune.

Since the area of a lune is proportional to the angle at the peak, and the lune with angle ${2\pi}$ has area ${4\pi}$, the lune ${L(\alpha)}$ with angle ${\alpha}$ has area ${2\alpha}$. Now consider the following picture:

Notice that each corner of the triangle gives us two lunes (the lunes for ${\alpha}$ are shown) and that there is an identical triangle on the rear of the sphere. If we add up the area of all 6 lunes associated with the corners, we get the total area of the sphere, plus twice the area of both triangles since we have triple-counted them. In other words:

$\displaystyle 4\pi + 4\mathrm{area}(\Delta(\alpha, \beta,\gamma)) = 2L(\alpha) + 2L(\beta) + 2L(\gamma) = 4(\alpha + \beta + \gamma)$

Solving,

$\displaystyle \mathrm{area}(\Delta(\alpha, \beta,\gamma)) = \alpha + \beta + \gamma - \pi$

4. Gauss-Bonnet

If we encouter a triangle ${\Delta}$ of constant curvature ${K(\Delta)}$, then we can scale the problem to one of the two formulas we just computed, so

$\displaystyle \mathrm{area}(\Delta) = \frac{\sum \mathrm{angles} - \pi}{K(\Delta)}$

This formula allows us to give a slightly handwavy, but accurate, proof of the Gauss-Bonnet theorem, which relates topological information (Euler characteristic) to geometric information (area and curvature). The proof will precede the statement, since this is really a discussion.

Suppose we have any closed Riemannian manifold (surface) ${S}$. The surface need not have constant curvature. Suppose for the time being it has no boundary. Triangulate it with very small triangles ${\Delta_i}$ such that ${\mathrm{area}(\Delta_i) \sim \epsilon^2}$ and ${\mathrm{diameter}(\Delta_i) \sim \epsilon}$. Then since the deviation between the curvature and the curvature at the midpoint ${K_\mathrm{midpoint}}$ is ${o(\epsilon^2)}$ times the distance from the midpoint,

$\displaystyle \int_{\Delta_i} K d\mathrm{area} = K_\mathrm{midpoint}\cdot \mathrm{area}(\Delta_i) + o(\epsilon^3)$

For each triangle ${\Delta_i}$, we can form a comparison triangle ${\Delta^c_i}$ with the same edge lengths and constant curvature ${K_\mathrm{midpoint}}$. Using the formula from the beginning of this section, we can rewrite the right hand side of the formula above, so

$\displaystyle \int_{\Delta_i} K d\mathrm{area} = \sum_{\Delta_i^c} \mathrm{angles} - \pi + o(\epsilon^3)$

Now since the curvature deviates by ${o(\epsilon^2)}$ times the distance from the midpoint, the angles in ${\Delta_i}$ deviate from those in ${\Delta_i^c}$ just slightly:

$\displaystyle \sum_{\Delta_i} \mathrm{angles} = \sum_{\Delta_i^c} \mathrm{angles} + o(\epsilon^3)$

So we have

$\displaystyle \int_{\Delta_i} K d\mathrm{area} = \sum_{\Delta_i} \mathrm{angles} - \pi + o(\epsilon^3)$

Therefore, summing over all triangles,

$\displaystyle \int_{S} K d\mathrm{area} = \sum_i \left[ \sum_{\Delta_i} \mathrm{angles} - \pi \right] + o(\epsilon)$

The right hand side is just the total angle sum. Since the angle sum around each vertex in the triangulation is ${2\pi}$,

$\displaystyle \sum_i \left[ \sum_{\Delta_i} \mathrm{angles} - \pi \right] = 2\pi V - \pi T$

Where ${V}$ is the number of vertices, and ${T}$ is the number of triangles. The number of edges, ${E}$, can be calculated from the number of triangles, since there are ${3}$ edges for each triangle, and they are each double counted, so ${E = \frac{3}{2} T}$. Rewriting the equation,

$\displaystyle \int_{S} K d\mathrm{area} = 2\pi (V - \frac{1}{2}T) = 2\pi (V - E + T) = 2\pi\chi(S) + o(\epsilon)$

Taking the mesh size ${\epsilon}$ to zero, we get the Gauss-Bonnet theorem ${\int_S K d\mathrm{area} = 2\pi\chi(S)}$.

4.1. Variants of Gauss-Bonnet

• If ${S}$ is compact with totally geodesic boundary, then the formula still holds, which can be shown by doubling the surface, applying the theorem to the doubled surface, and finding that euler characteristic also doubles.
• If ${S}$ has geodesic boundary with corners, then$\displaystyle \int_S K d\mathrm{area} + \sum_\mathrm{corners} \mathrm{turning angle} = 2\pi\chi(S)$Where the turning angle is the angle you would turn tracing the shape from the outside. That is, it is ${\pi - \alpha}$, where ${\alpha}$ is the interior angle.

• Most generally, if ${S}$ has smooth boundary with corners, then we can approximate the boundary with totally geodesic segments; taking the length of these segments to zero gives us geodesic curvature (${k_g}$):$\displaystyle \int_S K d\mathrm{area} + \sum_\mathrm{corners} \mathrm{turning angle} + \int_{\partial S} k_g d\mathrm{length} = 2\pi\chi(S)$

4.2. Examples

• The Euler characteristic of the round disk in the plane is ${1}$, and the disk has zero curvature, so ${\int_{\partial S} k_g d\mathrm{length} = 2\pi}$. The geodesic curvature is constant, and the circumference is ${2\pi r}$, so ${2\pi r k_g = 2\pi}$, so ${k_g = 1/r}$.
• A polygon in the plane has no curvature nor geodesic curvature, so ${\sum_\mathrm{corners} \pi - \mathrm{angle} = 2\pi}$.

The Gauss-Bonnet theorem constrains the geometry in any space with nonzero curvature. This the “reason” similarities which don’t preserve length and/or area exist in euclidean space; it has curvature zero.