You are currently browsing the category archive for the ‘Hyperbolic geometry’ category.

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 $CAT(K_0)$.

Ian gave his second and third talks this afternoon, completing his (quite detailed) sketch of the proof of the Virtual Haken Theorem. Recall that after work of Kahn-Markovic, Wise, Haglund-Wise and Bergeron-Wise, the proof reduces to showing the following:

Theorem (Agol): Let G be a hyperbolic group acting properly discontinuously and cocompactly on a CAT(0) cube complex X. Then there is a finite index subgroup G’ so that X/G’ is special; in other words, G is virtually special.

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.

Patrick Foulon and Boris Hasselblatt recently posted a preprint entitled “Nonalgebraic contact Anosov flows on 3-manifolds”. These are flows which are at the same time Anosov (i.e. the tangent bundle splits in a flow-invariant way into stable, unstable and flow directions) and contact (i.e. they preserve a contact form — that is, a 1-form $\alpha$ for which $\alpha \wedge d\alpha$ is a volume form). Their preprint gives some very interesting new constructions of such flows, obtained by surgery along a Legendrian knot (one tangent to the kernel of the contact form) which is transverse to the stable/unstable foliations of the Anosov flow.

My student Steven Frankel has just posted his paper Quasigeodesic flows and Mobius-like groups on the arXiv. This heartbreaking work of staggering genius interesting paper makes a deep connection between dynamics, hyperbolic geometry, and group theory, and represents the first significant progress that I know of on a conjectural program I formulated a few years ago.

One of the main results of the paper is to show that every quasigeodesic flow on a closed hyperbolic 3-manifold either has a closed orbit, or the fundamental group of the manifold admits an action on a circle with some very peculiar properties, namely that it is Mobius-like but not Mobius. The problem of giving necessary and sufficient conditions on a vector field on a 3-manifold to guarantee the existence of a closed orbit is a long and interesting one, and the introduction to the paper gives a brief sketch of this history as follows:

1. Mostow Rigidity

For hyperbolic surfaces, Moduli space is quite large and complicated. However, in three dimensions Moduli space is trivial:

Theorem 1 If ${f: M\rightarrow N}$ is a homotopy equivalence of closed hyperbolic ${n}$ manifolds with ${n\ge 3}$, then ${f}$ is homotopic to an isometry.

In other words, Moduli space is a single point.

This post will go through the proof of Mostow rigidity. Unfortunately, the proof just doesn’t work as well on paper as it does in person, especially in the later sections.

1.1. Part 1

First we need a definition familiar to geometric group theorists: a map between metric spaces (not necessarily Riemannian manifolds) ${f: (X, d_X) \rightarrow (Y, d_Y)}$ is a ${(k,\epsilon)}$ quasi-isometry if for all ${p,q \in X}$, we have

$\displaystyle \frac{1}{k} d_X(p,q) - \epsilon \le d_Y(f(p), f(q)) \le k d_X(p,q) + \epsilon$

Without the ${\epsilon}$ term, ${f}$ would be called bilipschitz.

First, we observe that if ${f: M \rightarrow N}$ is a homotopy equivalence, then ${f}$ lifts to a map ${\tilde{f} : \tilde{M} \rightarrow \tilde{N}}$ in the sense that ${\tilde{f}}$ is equivariant with respect to ${\pi_1(M) \cong \pi_1(N)}$ (thought of as the desk groups of ${\tilde{M}}$ and ${\tilde{N}}$, so for all ${\alpha \in \pi_1(M)}$, we have ${\tilde{f} \circ \alpha = f_*(\alpha) \circ \tilde{f}}$.

Now suppose that ${M}$ and ${N}$ are hyperbolic. Then we can lift the Riemannian metric to the covers, so ${\pi_1(M)}$ and ${\pi_1(N)}$ are specific discrete subgroups in ${\mathrm{Isom}(\mathbb{H}^n)}$, and ${\tilde{f}}$ maps ${\mathbb{H}^n \rightarrow \mathbb{H}^n}$ equivariantly with respect to ${\pi_1(M)}$ and ${\pi_1(N)}$.

Lemma 2 ${\tilde{f}}$ is a quasi-isometry.

Proof: Since ${f}$ is a homotopy equivalence, there is a ${g:N \rightarrow M}$ such that ${g\circ f \simeq \mathrm{id}_M}$. Perturbing slightly, we may assume that ${f}$ and ${g}$ are smooth, and as ${M}$ and ${N}$ are compact, there exists a constant ${k}$ such that ${\sup_{x\in M} \Vert \mathrm{d}f \Vert \le k}$ and ${\sup_{x \in N} \Vert \mathrm{d}g \Vert \le k}$. In other words, paths in ${M}$ and ${N}$ are stretched by a factor of at most ${k}$: for any path ${\gamma \in M}$, ${\mathrm{length}(f(\gamma)) \le k \mathrm{length}(\gamma)}$. The same is true for ${g}$ going in the other direction, and because we can lift the metric, the same is true for the universal covers: for any path ${\gamma \in \tilde{M} = \mathbb{H}^n}$, ${\mathrm{length}(\tilde{f}(\gamma)) \le k \mathrm{length}(\gamma)}$, and similarly for ${\tilde{g}}$.

Thus, for any ${p,q}$ in the universal cover ${\mathbb{H}^n}$,

$\displaystyle d(\tilde{f}(p), \tilde{f}(q)) \le k d(p,q).$

and

$\displaystyle d(\tilde{g}(p), \tilde{g}(q)) \le k d(p,q).$

We see, then, that ${\tilde{f}}$ is Lipschitz in one direction. We only need the ${\epsilon}$ for the other side.

Since ${g \circ f \simeq \mathrm{id_{\mathbb{H}^n}}}$, we lift it to get an equivariant lift ${\widetilde{g\circ f} = \tilde{g}\circ \tilde{f} \simeq \mathrm{id}}$ For any point ${p}$, the homotopy between ${\tilde{g}\circ \tilde{f}}$ gives a path between ${p}$ and ${(\tilde{g}\circ \tilde{f})(p)}$. Since this is a lift of the homotopy downstairs, this path must have bounded length, which we will call ${\delta}$. Thus,

$\displaystyle d(\tilde{g}\circ \tilde{f}(p), p) \le \delta$

Putting these facts together, for any ${p,q}$ in ${\mathbb{H}^n}$,

$\displaystyle d(\tilde{g}\circ \tilde{f}(p), \tilde{g}\circ\tilde{f}(q)) \le k d(\tilde{f}(p),\tilde{f}(q)).$

And

$\displaystyle d(\tilde{g}\circ \tilde{f}(p), p) \le \delta, \qquad d(\tilde{g}\circ \tilde{f}(q), q) \le \delta$

By the triangle inequality,

$\displaystyle \frac{1}{k} d(p,q) -\frac{2\delta}{k} \le \frac{1}{k}d(\tilde{g}\circ \tilde{f}(p), \tilde{g}\circ\tilde{f}(q)) \le d(\tilde{f}(p),\tilde{f}(q))$

This is the left half of the quasi-isometry definition, so we have shown that ${\tilde{f}}$ is a quasi-isometry. $\Box$

Notice that the above proof didn’t use anything hyperbolic—all we needed was that ${f}$ and ${g}$ are Lipschitz.

Our next step is to prove that a quasi-isometry of hyperbolic space extends to a continuous map on the boundary. The boundary of hyperbolic space is best thought of as the boundary of the disk in the Poincare model.

Lemma 3 A ${(k,\epsilon)}$ quasi-isometry ${\mathbb{H}^n \rightarrow \mathbb{H}^n}$ extends to a continuous map on the boundary ${\partial f:\mathbb{H}^n \cup \partial S_\infty^{n-1} \rightarrow \mathbb{H}^n \cup S_\infty^{n-1}}$.

The basic idea is that given a geodesic, it maps under ${f}$ to a path that is uniformly close to a geodesic, so we map the endpoints of the first geodesic to the endpoints of the second. We first need a sublemma:

Lemma 4 Take a geodesic and two points ${x}$ and ${y}$ a distance ${t}$ apart on it. Draw two perpendicular geodesic segments of length ${s}$ from ${x}$ and ${y}$. Draw a line ${l}$ between the endpoints of these segments such that ${l}$ has constant distance from the geodesic. Then the length of ${l}$ is linear in ${t}$ and exponential in ${s}$.

Proof: Here is a representative picture:

So we see that ${\frac{d}{ds} \mathrm{area} (R_s) = l_s}$. By Gauss-Bonnet,

$\displaystyle -\mathrm{area}(R_s) + 2\pi + \kappa \cdot l_s = 2\pi$

Where the ${2\pi}$ on the left is the sum of the turning angles, and ${\kappa}$ is the geodesic curvature of the segment ${l_s}$. What is this geodesic curvature ${\kappa}$? If we imagine increasing ${s}$, then the derivative of the length ${l_s}$ with respect to ${s}$ is the geodesic curvature ${\kappa}$ times the length ${l_s}$, i.e.

$\displaystyle \kappa \cdot l_s = \frac{d}{ds} l_s$

So ${\kappa \cdot l_s = \frac{d^s}{ds^2} \mathrm{area}(R_s)}$. Therefore, by the Gauss-Bonnet equality,

$\displaystyle \frac{d^2}{ds^2} \mathrm{area}(R_s) - \mathrm{area}(R_s) = 0$

so ${\mathrm{area}(R_s) = \cosh(s)}$. Therefore, ${l_s = \sinh(s)}$, which proves the lemma

$\Box$

With this lemma in hand, we move on the next sublemma:

Lemma 5 If ${\tilde{f}: \mathbb{H}^n \rightarrow \mathbb{H}^n}$ is a ${(k,\epsilon)}$ quasi-isometry, there is a constant ${C}$ depending only on ${k}$ and ${\epsilon}$ such that for all ${r}$ on the geodesic from ${p}$ to ${q}$ in ${\mathbb{H}^n}$, ${\tilde{f}(r)}$ is distance less than ${C}$ from any geodesic from ${\tilde{f}(p)}$ to ${\tilde{f}(q)}$.

Proof: Fix some ${C}$, and suppose the image ${\tilde{f}(\gamma)}$ of the geodesic ${\gamma}$ from ${p}$ to ${q}$ goes outside a ${C}$ neighborhood of the geodesic ${\beta}$ from ${\tilde{f}(p)}$ to ${\tilde{f}(q)}$. That is, there is some segment ${\sigma}$ on ${\gamma}$ between the points ${r}$ and ${s}$ such that ${\tilde{f}(\sigma)}$ maps completely outside the ${C}$ neighborhood.

Let’s look at the nearest point projection ${\pi}$ from ${\tilde{f}(\sigma)}$ to ${\beta}$. By the above lemma, ${\mathrm{length}(\pi(\tilde{f}(\sigma))) \le e^{-C} \mathrm{length}(\tilde{f}(\sigma))}$. Thus means that

$\displaystyle d(\tilde{f}(r), \tilde{f}(s)) \le 2C + e^{-C} \mathrm{length}(\tilde{f}(\sigma)).$

On the other hand, because ${\tilde{f}}$ is a quasi-isometry,

$\displaystyle \mathrm{length}(\tilde{f}(\sigma)) \le k \mathrm{length}(\sigma) + \epsilon = k d(r,s) + \epsilon$

and

$\displaystyle d(\tilde{f}(r), \tilde{f}(s)) \ge \frac{1}{k} d(r,s) - \epsilon$

So we have

$\displaystyle \frac{1}{k} d(r,s) + \epsilon \le 2C + e^{-C}(k d(r,s) + \epsilon)$

Which implies that

$\displaystyle d(r,s) \le \frac{2Ck + k\epsilon + ke^{-C}\epsilon}{1-k^2e^{-c}}$

That is, the length of the offending path ${\sigma}$ is uniformly bounded. Thus, increase ${C}$ by ${k}$ times this length plus ${\epsilon}$, and every offending path will now be inside the new ${C}$ neighborhood of ${\beta}$. $\Box$

The last lemma says that the image under ${\tilde{f}}$ of a geodesic segment is uniformly close to an actual geodesic. Now suppose that we have an infinite geodesic in ${\mathbb{H}^n}$. Take geodesic segments with endpoints going off to infinity. There is a subsequence of the endpoints converging to a pair on the boundary. This is because the visual distance between successive pairs of endspoints goes to zero. That is, we have extended ${\tilde{f}}$ to a map ${\tilde{f} : S_\infty^{n-1} \times S_\infty^{n-1} / \Delta \rightarrow S_\infty^{n-1} \times S_\infty^{n-1} / \Delta}$, where ${\Delta}$ is the diagonal ${\{(x,x)\}}$. This map is actually continuous, since by the same argument geodesics with endpoints visually close map (uniformly close) to geodesics with visually close endpoints.

1.2. Part 2

Now we know that a quasi-isometry ${\tilde{f} : \mathbb{H}^n \rightarrow \mathbb{H}^n}$ extends continuously to the boundary of hyperbolic space. We will end up showing that ${\partial \tilde{f}}$ is conformal, which will give us the theorem.

We now introduce the Gromov norm. if ${X}$ is a topological space, then singular chain complex ${C_i(X) \otimes \mathbb{R}}$ is a real vector space with basis the continuous maps ${\Delta^i \rightarrow X}$. We define a norm on ${C_i(X)}$ as the ${L^1}$ norm:

$\displaystyle \Vert \sum t_n \sigma_n \Vert = \sum_n | t_n|$

This defines a pseudonorm (the Gromov norm) on ${H_i(X;\mathbb{R})}$ by:

$\displaystyle \Vert \alpha \Vert_{\mathrm{Gromov}} = \inf_{[\sum t_n \sigma_n] = \alpha} \sum_n |t_n|$

This (pseudo) norm has some nice properties:

Lemma 6 If ${f:X\rightarrow Y}$ is continuous, and ${\alpha \in H_n(X;\mathbb{R})}$, then ${\Vert f_*(\alpha) \Vert_Y \le \Vert \alpha \Vert_X}$.

Proof: If ${\sum_n t_n \sigma_n}$ represents ${\alpha}$, then ${\sum_n t_n (f\circ \sigma_n)}$ represents ${f_*(\alpha)}$. $\Box$

Thus, we see that if ${f}$ is a homotopy equivalence, then ${\Vert f_*(\alpha) \Vert = \Vert \alpha \Vert}$.

If ${M}$ is a closed orientable manifold, then we define the Gromov norm of ${M}$ to be the Gromov norm ${\Vert M \Vert = \Vert [M] \Vert}$.

Here is an example: if ${M}$ admits a self map of degree ${d>1}$, then ${\Vert M \Vert = 0}$. This is because we can let ${C}$ represent ${[M]}$, so ${f_*[M] = \deg(f) [M]}$, so ${\frac{1}{\deg(f)} f_*C}$ represents ${[M]}$. Thus ${\Vert M \Vert = \Vert \frac{1}{\deg(f)} f_*C \Vert \le \frac{1}{\deg(f)}\Vert C\Vert}$. Notice that we can repeat the composition with ${f}$ to get that ${\Vert M\Vert}$ is as small as we’d like, so it must be zero.

Theorem 7 (Gromov) Let ${M^n}$ be a closed oriented hyperbolic ${n}$-manifold. Then ${\Vert M \Vert = \frac{\mathrm{vol}(M)}{\nu_n}}$. Where ${\nu_n}$ is a constant depending only on ${n}$.

We now go through the proof of this theorem. First, we need to know how to straighten chains:

Lemma 8 There is a map ${\mathrm{str} : C_n(\mathbb{H}^n) \rightarrow G^g(\mathbb{H}^n)}$ (the second complex is totally geodesic simplices) which is ${\mathrm{Isom}(\mathbb{H}^n)}$-equivariant and ${\mathrm{Isom}^+(\mathbb{H}^n)}$ – equivariantly homotopic to ${\mathrm{id}}$.

Proof: In the hyperboloid model, we imagine a simplex mapping in to ${\mathbb{H}^n}$. In ${\mathbb{R}^{n+1}}$, we can connect its vertices with straight lines, faces, etc. These project to being totally geodesics in the hyperboloid. We can move the original simplex to this straightened one via linear homotopy in ${\mathbb{R}^n}$; now project this homotopy to ${\mathbb{H}^n}$. $\Box$

Now, if ${\sum t_i \sigma_i}$ represents ${[M]}$, then we can straighten the simplices, so ${\sum t_i \sigma_t^g}$ represents ${[M]}$, and ${\Vert \sum t_i \sigma_i\Vert \le \Vert \sum t_i \sigma_t^g \Vert}$, so when finding the Gromov norm ${\Vert M \Vert}$ it suffices to consider geodesic simplices. Notice that every point has finitely many preimages, and total degree is 1, so for any point ${p}$, ${\sum_{q\in \sigma^{-1}(p)} t_i (\pm 1) = 1}$.

Next, we observe:

Lemma 9 If given a chain ${\sum t_i \sigma_i}$, there is a collection ${t_i' \in \mathbb{Q}}$ such that ${|t_i - t_i'| < \epsilon}$ and ${\sum t_i' \sigma_i}$ is a cycle homologous to ${\sum t_i \sigma_i}$.

Proof: We are looking at a real vector space of coefficients, and the equations defining what it means to be a cycle are rational. Rational points are therefore dense in it. $\Box$

By the lemma, there is an integral cycle ${\sum n_i \sigma_i = N[M]}$, where ${N}$ is some constant. We create a simplicial complex by gluing these simplices together, and this complex comes together with a map to ${M}$. Make it smooth. Now by the fact above, ${\sum n_i (\pm 1) = N}$, so ${\sum t_i (\pm 1) = 1}$. Then

$\displaystyle \int_M \sum_{q\in \sigma^{-1}(p)} t_i (\pm 1) dp = \mathrm{vol}(M)$

on the one hand, and on the other hand,

$\displaystyle \int_M \sum_{q\in \sigma^{-1}(p)} t_i (\pm 1) dp = \sum_i t_i \int_{\sigma_i(\Delta)}dp = \sum_i t_i \mathrm{vol}(\sigma_i(\Delta))$

The volume on the right is at most ${\nu_n}$, the volume of an ideal ${n}$ simplex, so we have that

$\displaystyle \sum_i | t_i | \ge \frac{\mathrm{vol}(M)}{\nu_n}$

i.e.

$\displaystyle \Vert M \Vert \ge \frac{\mathrm{vol}(M)}{\nu_n}$

This gives the lower bound in the theorem. To get an upper bound, we need to exhibit a chain representing ${[M]}$ with all the simplices mapping with degree 1, such that the volume of each image simplex is at least ${\nu_n - \epsilon}$.

We now go through the construction of this chain. Set ${L >> 0}$, and fix a fundamental domain ${D}$ for ${M}$, so ${\mathbb{H}^n}$ is tiled by translates of ${D}$. Let ${S_{g_1, \cdot, g_{n+1}}}$ be the set of all simplices with side lengths ${\ge L}$ with vertices in a particular ${(n+1)}$-tuple of fundamental domains ${(g_1D, \cdots g_{n+1}D)}$. Pick ${\Delta_{g_1, \cdot, g_{n+1}}}$ to be a geodesic simplex with vertices ${g_1p, \cdots, g_2p, \cdots g_{n+1}p}$, and let ${\Delta^M(g_1; \cdots; g_{n+1})}$ be the image of ${\Delta_{g_1, \cdot, g_{n+1}}}$ under the projection. This only depends on ${g_1, \cdots, g_{n+1}}$ up to the deck group of ${M}$.

Now define the chain:

$\displaystyle C_L = \sum_{(g_1; \cdots; g_{n+1})} \pm \mu(S_{g_1, \cdot, g_{n+1}}) \Delta^M(g_1; \cdots; g_{n+1})$

With the ${\pm}$ to make it orientation-preserving, and where ${\mu}$ is an ${\mathrm{Isom}(\mathbb{H}^n)}$-invariant measure on the space of regular simplices of side length ${L}$. If the diameter of ${D}$ is ${d}$ every simplex with ${\mu(S_{g_1, \cdot, g_{n+1}}) \ne 0}$ has edge length in ${[L - 2d, L+2d]}$, so:

1. The volume of each simplex is ${\ge \nu_n - \epsilon}$ if ${L}$ is large enough.
2. ${C_L}$ is finite — fix a fundamental domain; then there are only finitely many other fundamental domains in ${[L-2d, L+2d]}$.

Therefore, we just need to know that ${C_L}$ is a cycle representing ${[M]}$: to see this, observe that every for every face of every simplex, there is an equal weight assigned to a collection of simplices on the front and back of the face, so the boundary is zero.

By the equality above, then,

$\displaystyle \Vert M \Vert \le \sum_i t_i = \frac{\mathrm{vol}(M)}{\nu_n - \epsilon}$

Taking ${\epsilon}$ to zero, we get the theorem.

1.3. Part 3 (Finishing the proof of Mostow Rigidity

We know that for all ${\epsilon>0}$, there is a cycle ${C_\epsilon}$ representing ${[M]}$ such that every simplex is geodesic with side lengths in ${[L-2d, L+2d]}$, and the simplices are almost equi-distributed. Now, if ${f:M\rightarrow N}$, and ${C}$ represents ${[M]}$, then ${\mathrm{str}(f(C))}$ represents ${[N]}$, as ${f}$ is a homotopy equivalence.

We know that ${\tilde{f}}$ extends to a map ${\mathbb{H}^n \cup S_{\infty}^{n+1} \rightarrow \mathbb{H}^n \cup S_{\infty}^{n+1}}$. Suppose that there is an ${n+1}$ tuple in ${S_{\infty}^{n+1}}$ which is the vertices of an ideal regular simplex. The map ${\tilde{f}}$ takes (almost) regular simplices arbitrarily close to this regular ideal simplex to other almost regular simplices close to an ideal regular simplex. That is, ${\tilde{f}}$ takes regular ideal simplices to regular ideal simplices. Visualizing in the upper half space model for dimension 3, pick a regular ideal simplex with one vertex at infinity. Its vertices form an equilateral triangle in the plane, and ${\tilde{f}}$ takes this triangle to another equilateral triangle. We can translate this simplex around by the set of reflections in its faces, and this gives us a dense set of equilateral triangles being sent to equilateral triangles. This implies that ${\tilde{f}}$ is conformal on the boundary. This argument works as long as the boundary sphere is at least 2 dimensional, so this works as long as ${M}$ is 3-dimensional.

Now, as ${\tilde{f}}$ is conformal on the boundary, it is a conformal map on the disk, and thus it is an isometry. Translating, this means that the map conjugating the deck group ${\pi_1(M)}$ to ${\pi_1(N)}$ is an isometry of ${\mathbb{H}^n}$, so ${f}$ is actually an isometry, as desired. The proof is now complete.

1. Fenchel-Nielsen Coordinates for Teichmuller Space

Here we discuss a very nice set of coordinates for Teichmuller space. The basic idea is that we cut the surface up into small pieces (pairs of pants); hyperbolic structures on these pieces are easy to parameterize, and we also understand the ways we can put these pieces together.

In order to define these coordinates, we first cut the surface up. A pair of pants is a thrice-punctured sphere.

Another way to specify it is that it is a genus ${0}$ surface with euler characteristic ${-1}$ and three boundary components. We can cut any surface up into pairs of pants with simple closed curves. To see this, we can just exhibit a general cutting: slice with ${3g-3}$ “vertical” simple closed curves.

This is not the only way to cut a surface into pairs of pants. For example, with the once-punctured torus any pair of coprime integers gives us a curve which cuts the surface into a pair of pants. We are going to show that a point in Teichmuller space is determined by the lengths of the ${3g-3}$ curves, plus ${3g-3}$ other coordinates, which record the “twisting” of each gluing curve.

Now, given a choice of ${3g-3}$ disjoint simple closed surves ${\{\alpha_i\}}$, we associate to ${(f, \Sigma) \in \mathrm{Teich}(S)}$ the family of geodesics in ${\Sigma}$ in the homotopy classes of the ${f(\alpha_i)}$. In each class, there is a unique geodesic, but how do we know the geodesics in ${\{f(\alpha_i)\}}$ are pairwise disjoint?

Lemma 1 Suppose ${\{\alpha_i\}}$ is a family of pairwise disjoint simple closed curves in a hyperbolic surface ${\Sigma}$, and ${\{\gamma_i\}}$ are the (unique) geodesic representatives in the homotopy classes of the ${\alpha_i}$.

• The geodesics in ${\{\gamma_i\}}$ are pairwise disjoint simple closed curves.
• As a family, the ${\{\gamma_i\}}$ are ambient isotopic to ${\{\alpha_i\}}$.

Proof: Consider a loop ${\alpha}$ and its geodesic representative ${\gamma}$. Suppose that ${\gamma}$ intersects itself. Now ${\alpha}$ and ${\gamma}$ cobound an annulus, which lifts to the universal cover: in the universal cover we must find the lift of the intersection as an intersection between two lifts ${\tilde{\gamma}}$ and ${\tilde{\gamma}'}$. Because the annulus bounding ${\alpha}$ and ${\gamma}$ lifts to the universal cover, there are two lifts ${\tilde{\alpha}}$ and ${\tilde{\alpha}'}$ of ${\alpha}$ which are uniformly close to ${\tilde{\gamma}}$ and ${\tilde{\gamma}'}$. We therefore find that ${\tilde{\alpha}}$ and ${\tilde{\alpha}'}$ intersect, which means that ${\alpha}$ intersects itself, which is a contradiction. The same idea shows that the geodesic representatives ${\gamma_i}$ are pairwise disjoint.

To see that they are ambient isotopic as a family, it is easiest to lift the picture to the universal cover. At that point, we just need to “wiggle” everything a little to match up the lifts of the ${\alpha_i}$ and ${\gamma_i}$. $\Box$

With the lemma, we see that to a point in Teichmuller space we get ${3g-3}$ pairwise disjoint simple closed geodesics, which gives us ${3g-3}$ positive coordinates, namely, the lengths of these curves. We might wonder: what triples of points can arise as the lengths of the boundary curves in hyperbolic pairs of pants? It turns out that:

Lemma 2 There exists a unique hyperbolic pair of pants with cuff lengths ${(l_1, l_2, l_3)}$, for any ${l_1, l_2, l_3 > 0}$. Cuff lengths here refers to the lengths of the three boundary components.

Proof: We will now prove the lemma, which involves a little discussion. Suppose we are given a hyperbolic pair of pants. We can double it to obtain a genus two surface:

The ${\alpha}$ curves are shown in red, and representatives of the other isotopy class fixed by the involution are in blue.

There is an involution (rotation around a skewer stuck through the surface horizontally) which fixes the (glued up) boundaries of the pairs of pants. This involution also fixes the isotopy classes of three other disjoint simple closed curves, and there is a unique geodesic ${\beta_i}$ in these isotopy classes. Since the ${\beta_i}$ are fixed by the involution, they must intersect the ${\alpha_i}$ at right angles. If we cut along the ${\alpha_i}$ to get (two copies of) our original pair of pants, we have found that there is a unique triple of geodesics ${\beta_i}$ which meet the boundaries at right angles:

Cutting along the ${\beta_i}$, we get two hyperbolic hexagons:

We will prove in a moment that there is a unique hyperbolic right-angled hexagon with three alternating edge lengths specified. In particular, there is a unique hyperbolic right-angled hexagon with alternating edge lengths ${(l_1/2, l_2/2, l_3/2)}$. Since there is a unique way to glue up the hexagons to obtain our original ${(l_1, l_2, l_3)}$ pair of pants, there is a unique hyperbolic pair of pants with specified edge lengths. $\Box$

Lemma 3 There is a unique hyperbolic right-angled hexagon with alternating edge lengths ${(l_1, l_2, l_3)}$.

Proof: Pick some geodesic ${g_1}$ and some point ${x_1}$ on it. We will show the hexagon is now determined, and since we can map a point on a geodesic to any other point on a geodesic, the hexagon will be unique up to isometry. Draw a geodesic segment of length ${l_1}$ at right angles from ${x_1}$. Call the other end of this segment ${x_2}$. There is a unique geodesic ${g_2}$ passing through ${x_2}$ at right angles to the segment. Pick some point ${x_3}$ on ${g_2}$ at length ${y}$ from ${x_2}$ (we will be varying ${y}$). From ${x_3}$ there is a unique geodesic segment of length ${l_2}$ at right angles to ${g_2}$; call its endpoint ${x_4}$. There is a unique geodesic ${g_3}$ through ${x_4}$ at right angles to this segment. Now, there is a unique geodesic segment at right angles to ${g_1}$ and ${g_3}$. Of course, the length ${z}$ of this segment depends on ${y}$.

If we make ${y}$ large, then ${z}$ becomes large, and there is some positive ${y}$ such that ${z}$ goes to ${0}$. Therefore, there is a unique length ${y}$ making ${z = l_3}$. We have now determined the hexagon, and, up to isometry, all of our choices were forced, so there is only one. $\Box$

Since there is a unique hyperbolic pair of pants with specified cuff lengths, when we cut our surface of interest ${S}$ up into pairs of pants, we get a map ${\mathrm{Teich}(S) \rightarrow (\mathbb{R}^+)^{3g-3}}$ which takes a point ${(f, \Sigma)}$ to the ${3g-3}$ lengths of the curves cutting ${S}$ into pairs of pants. This map is not injective: the fiber over a point is all the ways to glue together the pairs of pants.

The issue is that when we want to glue two ${\alpha}$ curves together, we have to decide whether to twist them at all before gluing. Up to isometry, there are ${\mathbb{R}/\mathbb{Z}}$ ways to glue these curves together (all the angles). However, in (marked) Teichmuller space, there are ${\mathbb{R}}$ ways to glue it up. Draw another curve ${\beta}$ (this ${\beta}$ is not the same as the ${\beta_i}$ before). The marking on ${S}$ lets us observe what happens to ${\beta}$ under ${f}$, and we can see that twisting the pairs of pants around ${\alpha}$ results in nontrivial movement in Teichmuller space.

The twist above results in the following new ${\beta}$ curve:

The length of ${\beta}$ determines how twisted the gluing is, since twisting requires increasing its length. That is, given the image of ${\beta}$, there is a unique way to untwist it to get a minimum length. This tells us how twisted the original gluing was.

To understand the twisting around all the ${3g-3}$ curves in ${S}$, we must pick another ${3g-3}$ curves; one simple way is to declare that ${\beta}$ looks like the above pictures if we are gluing two distinct pairs of pants, and like this:

if we are gluing a pair of pants to itself. This construction gives us a global homeomorphism

$\displaystyle \mathrm{Teich}(S) \rightarrow (\mathbb{R}^+)^{3g-3} \times \mathbb{R}^{3g-3} \cong \mathbb{R}^{6g-6}$

Here is an example of a choice of ${\alpha}$ and ${\beta}$ curves. The ${\beta}$ curves get a little messy in the middle: try to fit the pictures above into the context of the one below to see that they are correct.

1.1. A Symplectic Form on Moduli Space

The length and twist coordinates ${l_i}$ and ${t_i}$ are not well-defined on Moduli space, but their derivatives are: define the 2 form on Teichmuller space

$\displaystyle \omega = \sum_i dl_i \wedge dt_i$

It is a theorem of Wolpert that this 2-form is independent of the choice of coordinates, so it descends to a 2-form on Moduli space. It is very usful that Modi space is symplectic.

This post introduces Teichmuller and Moduli space. The upcoming posts will talk about Fenchel-Nielsen coordinates for Teichmuller space; it’s split up because I figured this was a relatively nice break point. Hopefully, I will later add some pictures to this post.

1. Uniformization

This section starts to talk about Teichmuller space and related stuff. First, we recall the uniformization theorem:

If ${S}$ is a closed surface (Riemannian manifold), then there is a unique* metric of constant curvature in its conformal class. The asterisk * refers to the fact that the metric is unique if we require that it has curvature ${\pm 1}$. If ${\chi(S)=0}$, then the metric has curvature zero and it is unique up to euclidean similarities.

2. Teichmuller and Moduli Space of the Torus

Let us see what we can conclude about flat metrics on the torus. We would like to classify them in some way. Choose two straight curves ${\alpha}$ and ${\beta}$ on the torus intersecting once (a longitude and a meridian) and cut along these curves. We obtain a parallelogram which can be glued up along its edges to retrieve the original torus. This parallelogram lives/embeds in ${\mathbb{C}^2}$, and, by composing the embedding with euclidean similarities, we may assume that the bottom left corner is at ${0}$ and the bottom right is ${1}$. The parallelogram is therefore determined by where the upper left hand corner is: some complex number ${z}$ with ${\mathrm{Im}(z) > 0}$. Notice that this is the upper half-plane, which we can think of as hyperbolic space. Therefore, there is a bijection:

{ Torii with two chosen loops up to euclidean similarity } ${\leftrightarrow}$ { ${z \in \mathbb{C} \, | \, \mathrm{Im}(z) > 0}$ }

This set is called the Teichmuller space of the torus. We don’t really care about the loops ${\alpha}$ and ${\beta}$, so we’d like to find a group which takes one choice of loops to another and acts transitively. The quotient of this will be the set of flat metrics on the torus up to isometry, which is known as Moduli space.

We are interested in the mapping class group of the torus, which is defined to be

$\displaystyle \mathrm{MCG}(T^2) = \mathrm{Homeo}^+(T^2) / \mathrm{Homeo}_\circ(T^2)$

Where ${\mathrm{Homeo}_\circ(T^2)}$ denotes the connected component of the identity. That is, the mapping class group is the group of homeomorphisms (homotopy equivalences), up to isotopy (homotopy). The reason for the parentheses is that for surfaces, we may replace homeomorphism and isotopy by homotopy equivalence and homotopy, and we will get the same group (these catagories are equivalent for surfaces).

To find ${\mathrm{MCG}(T^2)}$, think of the torus as the unit square in ${\mathbb{R}^2}$ spanned by the standard unit basis vectors. Then a homeomorphism of ${T^2}$ must send the integer lattice to itself, so the standard basis must go to a basis for this lattice, and the transformation must preserve the area of the torus. Up to isotopy, this is just linear maps of determinant ${1}$ (not ${-1}$ because we want orientation-preserving) preserving the integer lattice, which we care about up to scale, otherwise known as ${\mathrm{PSL}(2,\mathbb{Z})}$.

Using the bijection above, the mapping class group of the torus acts on ${\{ z \in \mathbb{C} \, | \, \mathrm{Im}(z) > 0 \}}$, and this action is

$\displaystyle \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] z = \frac{az + b}{cz+d}$

This action is probably familiar to you from complex analysis.

In summary, the Teichmuller space of the torus is (can be represented as) ${\{ z \in \mathbb{C} \, | \, \mathrm{Im}(z) > 0 \}}$, and the mapping class group ${\mathrm{PSL}(2,\mathbb{Z})}$ acts on this space, and the quotient of this action is the set of flat metrics up to isometry, which is Moduli space. What is the quotient? A fundamental region for the action is the set

$\displaystyle \{ z\in\mathbb{C} \,\, |\,\, |\mathrm{Re}(z)| \le \frac{1}{2}, \, |z| \ge 1\}$

Which is glued to itself by a flip in the ${y}$ axis. The resulting Moduli space is an orbifold: one point is ideal and goes off to infinity, one point looks locally like ${\mathbb{R}^2}$ quotiented by a rotation of ${\frac{2\pi}{3}}$, and the other point looks like ${\mathbb{R}^2}$ quotiented by a rotation of ${\pi}$.

3. Teichmuller Space and Moduli Space for Negatively Curved Surfaces

Now we will go through a similar process for closed, boundaryless, oriented surfaces of negative Euler characteristic. It is possible to do this for surfaces with boundary, etc, but for simplicity, we will stick to multi-holed torii (this what closed, boundaryless, oriented surfaces of negative Euler characteristic are) for now.

We start with a topological surface ${S}$. Topological meaning we do not associate with it a metric. We want to classify the hyperbolic metrics we could give to ${S}$. Define Teichmuller space ${\mathrm{Teich}(S)}$ to be the set of equivalence classes of pairs ${(f, \Sigma)}$ where ${\sigma}$ is a hyperbolic surface and ${f: S \rightarrow \Sigma}$ is a homotopy equivalence. As mentioned earlier, anywhere “homotopy equivalence” appears here, you may replace it with “homeomorphism” as long as you replace “homotopy” with “isotopy.” The equivalence relation on pairs is the following: ${(f, \Sigma_1) \sim (g, \Sigma_2)}$ iff there exists an isometry ${i: \Sigma_1 \rightarrow \Sigma_2}$ such that ${i \circ f}$ is homotopic to ${g}$.

Define the Moduli space ${\mathcal{M}(S)}$ of ${S}$ to be isometry classes of surfaces ${\Sigma}$ which are homotopy equivalent to ${S}$. There is an obvious map ${\mathrm{Teich}(S) \rightarrow \mathcal{M}(S)}$ defined by mapping ${(f, \Sigma) \mapsto \Sigma}$, and this map respects the equivalence relations, because if ${(f, \Sigma_1) \sim (g, \Sigma_2)}$, then ${\Sigma_1}$ is isometric to ${\Sigma_2}$ (since it is isometric by an isometry commuting with ${f}$ and ${g}$).

As with the torus, define the mapping class group ${\mathrm{MCG}(S)}$ to be the group of homotopy equivalences of ${S}$ with itself, up to homotopy. Then ${\mathrm{MCG}(S)}$ acts on ${\mathrm{Teich}(S)}$ by ${\varphi \cdot (f,\Sigma) = (f \circ \varphi, \Sigma)}$. The quotient of ${\mathrm{Teich}(S)}$ by this action is ${\mathcal{M}(S)}$: clearly we never identify surfaces which are not isometric, and if ${i : \Sigma_1 \rightarrow \Sigma_2}$ is an isometry, and ${(f,\Sigma_1)}$, ${(g,\Sigma_2)}$ are points in Teichmuller space with any ${f,g}$, then notice ${f}$ has an inverse (up to homotopy), so if we act on ${(f,\Sigma_1)}$ by ${f^{-1}\circ g}$, we get ${(f\circ f^{-1}\circ g, \Sigma_1)}$, which is the same point in ${\mathrm{Teich}(S)}$ as ${(g,\Sigma_2)}$. We are abusing notation here, because we are thinking of ${\Sigma_1}$, ${\Sigma_2}$ and ${S}$ as the same surface (which they are, topologically). The point is that by acting by ${\mathrm{MCG}(S)}$ we can rearrange ${S}$ so that after mapping by ${f \circ i}$ we are homotopic to ${g}$. The result of this is that

$\displaystyle \mathrm{Teich}(S) / \mathrm{MCG}(S) \cong \mathcal{M}(S)$

A priori, we are interested in hyperbolic metrics on ${S}$ up to isometry — Moduli space. The reason for defining Teichmuller space is that Moduli space is rather complicated. Teichmuller space, on the other hand, will turn out to be as nice as you could want (${\mathbb{R}^{6g-6}}$ for a genus ${g}$ surface). By studying the very nice Teichmuller space plus the less-nice-but-still-understandable mapping class group, we can approach Moduli space.

4. Coordinates for Teichmuller Space

Now we will take a closer look at Teichmuller space and give it coordinates.

4.1. Very Overdetermined (But Easy) Coordinates

One way to give this space coordinates is the following. Let us choose a homotopy class of loop in ${S}$ (this is a conjugacy class in ${\pi_1(S)}$), and we’ll represent this class by the loop ${\gamma : S^1 \rightarrow S}$. Given a point ${(f,\Sigma) \in \mathrm{Teich}(S)}$, there is a unique geodesic representative in the free homotopy class of the loop ${f\circ \gamma}$. Define ${l_\gamma(f,\Sigma) = \mathrm{length}(f\circ \gamma)}$ to be the length of this representative. Let ${C}$ be the set of conjugacy classes in ${\pi_1(S)}$. Then we have defined a map

$\displaystyle l : \mathrm{Teich}(S) \rightarrow \mathbb{R}^C$

by

$\displaystyle (f,\Sigma) \mapsto (l_\gamma(f,\Sigma))_\gamma$

This is nice in the sense that it’s a real vector space, but not nice in that it’s infinite dimensional. We will see that we need a finite number of dimensions.

4.2. Dimension Counting

Method 1

Let’s try to count the dimension of ${\mathrm{Teich}(S)}$. Suppose that ${S}$ has genus ${g}$. We can obtain ${S}$ by gluing the edges of a ${4g}$-gon in pairs (going counterclockwise, the labels read ${a_1}$, ${b_1}$, ${a_1^{-1}}$, ${b_1^{-1}}$, ${a_2}$ …, ${a_g}$, ${b_g}$, ${a_g^{-1}}$, ${b_g^{-1}}$). Since we will be given ${S}$ a hyperbolic metric, let us look at what this tells us about this polygon. We have a hyperbolic polygon; in order to glue it up, we must have

1. The paired sides must have equal length.
2. The corner angles must add to ${2\pi}$.

For a triangle in hyperbolic space, the edges lengths are enough to specify the triangle up to isometry. Similarly, for a hyperbolic 4-gon (square), we need all the exterior edge lengths, plus 1 angle (the angle gives the length of a diagonal). By induction, a ${n}$-gon needs ${n}$ side lengths and ${n-3}$ angles. For our ${4g}$-gon, then, we need to specify ${4g}$ side lengths and ${4g-3}$ angles. This is ${8g-3}$ dimensions. However, we have ${2g}$ pairs, each of which gives a constraint, plus our single constraint about the angle sum. This reduces our dimension to ${6g-4}$. Finally, we made an arbitrary choice about where the vertex of this polygon was in our surface. This is an extra two dimensions that we don’t care about (we disregard those coordinates), so we have ${6g-6}$ dimensions.

Method 2

A marked hyperbolic structure on ${S}$ gives a ${\pi_1(S)}$-equivariant isometry ${\widetilde{\Sigma} \rightarrow \mathbb{H}^2}$. That is, an element of ${\mathrm{Teich}(S)}$ is ${(f,\Sigma)}$, which tells us how to map ${\pi_1(S)}$ isomorphically onto ${\pi_1(\Sigma)}$, which is the same as the deck group of the universal cover ${\widetilde{\Sigma}}$, which is ${\mathbb{H}^2}$. Therefore, to an element of ${\mathrm{Teich}(S)}$ is associated a discrete faithful representation of ${\pi_1(S)}$ to ${\mathrm{PSL}(2,\mathbb{R})}$, the group of isometries of ${\mathbb{H}^2}$, and this representation is unique up to conjugacy (if we conjugate the image of the representation, then the quotient manifold is the same). The dimension of ${\mathrm{Teich}(S)}$ is therefore the dimension of the space of representations of ${\pi_1(S)}$ in ${\mathrm{PSL}(2,\mathbb{R})}$ up to conjugacy.

The fundamental group of ${S}$ has a nice presentation in terms of the polygon we can glue up to make it; the interior of the polygon gives us a single relation:

$\displaystyle \pi_1(S) = \langle a_1, b_1, \cdots, a_g, b_g \,| \, \prod_i [a_i,b_i]\rangle$

So ${\mathrm{Hom}(\pi_1(S), \mathrm{PSL}(2,\mathbb{R}))}$ is the subset of ${\mathrm{Hom}(F_{2g}, \mathrm{PSL}(2,\mathbb{R}))}$ such that ${\prod_i [a_i,b_i] = 1}$ (here ${F_{2g}}$ is the free group on 2 generators, which is what we get if we forget the single relation). Now a representation in ${\mathrm{Hom}(F_{2g}, \mathrm{PSL}(2,\mathbb{R}))}$ is completely free: we can send the generators anywhere we want, so

$\displaystyle \mathrm{Hom}(F_{2g}, \mathrm{PSL}(2,\mathbb{R})) \cong \left( \mathrm{PSL}(2,\mathbb{R}) \right)^{2g}$

Since ${\mathrm{PSL}(2,\mathbb{R})}$ is 3-dimensional, the right hand side is a real manifold of dimension ${6g}$. Insisting that ${\prod_i [a_i,b_i]}$ map to ${1}$ is a 3-dimensional constraint (it gives 4 equations, when you think of it as a matrix equation, but there is an implied equation already taken into account). Therefore we expect that ${\mathrm{Hom}(\pi_1(S), \mathrm{PSL}(2,\mathbb{R}))}$ will be ${6g-3}$ dimensional. However, we are interested in representations up to conjugacy, so this removes another 3 dimensions, giving us the same dimension estimate for ${\mathrm{Teich}(S)}$ as ${6g-6}$ dimensional.