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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.)

There is an old puzzle which starts by asking: what is the next number in the sequence 1,2,4,? We are supposed to recognize the start of the sequence and answer that the next number is surely 8, because the first three numbers are consecutive powers of 2, and so the next number should be the cube of 2 which is 8. The puzzler then explains (contrary to expectations) that the successive terms in the sequence are actually the number of regions into which the plane is divided by a collection of lines in general position (so that any two lines intersect, and no three lines intersect in a single point). Thus:

So the “correct” answer to the puzzle is 7 (and the sequence continues 11, 26, $\cdots (n^2+n+2)/2$). This is somehow meant to illustrate some profound point; I don’t quite see it myself. Anyway, I would like to suggest that there is a natural sense in which the “real” answer should actually be 8 after all, and it’s the point of this short blog post to describe some connections between this puzzle, the theory of cube complexes (which is at the heart of Agol’s recent proof of the Virtual Haken Conjecture), and the location of the missing 8th region.

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.

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.

Today Jason Manning gave a talk on a vital ingredient in the proof of Agol’s theorem, which is a result in geometric group theory. The theorem is a joint project of Agol-Groves-Manning, and generalizes some earlier work they did a few years ago. Jason referred to the main theorem during his talk as the “Goal Theorem” (I guess it was the goal of his lecture), but I’m going to call it the Weak Separation Theorem, since that is a somewhat more descriptive name. The statement of the theorem is as follows.

Weak Separation Theorem (Agol-Groves-Manning): Let G be a hyperbolic group, let H be a subgroup of G which is quasiconvex, and isomorphic to the fundamental group of a virtually special NPC cube complex, and let g be an element of G which is not contained in H. Then there is a surjection $\phi:G \to \bar{G}$ so that

1. $\bar{G}$ is hyperbolic;
2. $\phi(H)$ is finite; and
3. $\phi(g)$ is not contained in $\phi(H)$.

In the remainder of this post I will try to explain the proof of this theorem, to the extent that I understand it. Basically, this amounts to my summarizing Manning’s talk (or the part of it that I managed to get down in my notes); again, any errors, foolishness, silly blog post titles etc. are due to me.

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).

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.