You are currently browsing the category archive for the ‘Groups’ category.

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.

I am Alden, one of Danny’s students. Error/naivete that may (will) be found here is mine. In these posts, I will attempt to give notes from Danny’s class on hyperbolic geometry (157b). This first post covers some models for hyperbolic space.

1. Models

We have a very good natural geometric understanding of ${\mathbb{E}^3}$, i.e. 3-space with the euclidean metric. Pretty much all of our geometric and topological intuition about manifolds (Riemannian or not) comes from finding some reasonable way to embed or immerse them (perhaps locally) in ${\mathbb{E}^3}$. Let us look at some examples of 2-manifolds.

• Example (curvature = 1) ${S^2}$ with its standard metric embeds in ${\mathbb{E}^2}$; moreover, any isometry of ${S^2}$ is the restriction of (exactly one) isometry of the ambient space (this group of isometries being ${SO(3)}$). We could not ask for anything more from an embedding.
• Example (curvature = 0) Planes embed similarly.
• Example (curvature = -1) The pseudosphere gives an example of an isometric embedding of a manifold with constant curvature -1. Consider a person standing in the plane at the origin. The person holds a string attached to a rock at ${(0,1)}$, and they proceed to walk due east dragging the rock behind them. The movement of the rock is always straight towards the person, and its distance is always 1 (the string does not stretch). The line traced out by the rock is a tractrix. Draw a right triangle with hypotenuse the tangent line to the curve and vertical side a vertical line to the ${x}$-axis. The bottom has length ${\sqrt{1-y^2}}$, which shows that the tractrix is the solution to the differential equation$\displaystyle \frac{-y}{\sqrt{1-y^2}} = \frac{dy}{dx}$

The Tractrix

The surface of revolution about the ${x}$-axis is the pseudosphere, an isometric embedding of a surface of constant curvature -1. Like the sphere, there are some isometries of the pseudosphere that we can understand as isometries of ${\mathbb{E}^3}$, namely rotations about the ${x}$-axis. However, there are lots of isometries which do not extend, so this embeddeding does not serve us all that well.

• Example (hyperbolic space) By the Nash embedding theorem, there is a ${\mathcal{C}^1}$ immersion of ${\mathbb{H}^2}$ in ${\mathbb{E}^3}$, but by Hilbert, there is no ${\mathcal{C}^2}$ immersion of any complete hyperbolic surface.That last example is the important one to consider when thinking about hypobolic spaces. Intuitively, manifolds with negative curvature have a hard time fitting in euclidean space because volume grows too fast — there is not enough room for them. The solution is to find (local, or global in the case of ${\mathbb{H}^2}$) models for hyperbolic manfolds such that the geometry is distorted from the usual euclidean geometry, but the isometries of the space are clear.

2. 1-Dimensional Models for Hyperbolic Space

While studying 1-dimensional hyperbolic space might seem simplistic, there are nice models such that higher dimensions are simple generalizations of the 1-dimensional case, and we have such a dimensional advantage that our understanding is relatively easy.

2.1. Hyperboloid Model

Parameterizing ${H}$

Consider the quadratic form ${\langle \cdot, \cdot \rangle_H}$ on ${\mathbb{R}^2}$ defined by ${\langle v, w \rangle_A = \langle v, w \rangle_H = v^TAw}$, where ${A = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]}$. This doesn’t give a norm, since ${A}$ is not positive definite, but we can still ask for the set of points ${v}$ with ${\langle v, v \rangle_H = -1}$. This is (both sheets of) the hyperbola ${x^2-y^2 = -1}$. Let ${H}$ be the upper sheet of the hyperbola. This will be 1-dimensional hyperbolic space.

For any ${n\times n}$ matrix ${B}$, let ${O(B) = \{ M \in \mathrm{Mat}(n,\mathbb{R}) \, | \, \langle v, w \rangle_B = \langle Mv, Mw \rangle_B \}}$. That is, matrices which preserve the form given by ${A}$. The condition is equivalent to requiring that ${M^TBM = B}$. Notice that if we let ${B}$ be the identity matrix, we would get the regular orthogonal group. We define ${O(p,q) = O(B)}$, where ${B}$ has ${p}$ positive eigenvalues and ${q}$ negative eigenvalues. Thus ${O(1,1) = O(A)}$. We similarly define ${SO(1,1)}$ to be matricies of determinant 1 preserving ${A}$, and ${SO_0(1,1)}$ to be the connected component of the identity. ${SO_0(1,1)}$ is then the group of matrices preserving both orientation and the sheets of the hyperbolas.

We can find an explicit form for the elements of ${SO_0(1,1)}$. Consider the matrix ${M = \left[ \begin{array}{cc} a & b \\ c& d \end{array} \right]}$. Writing down the equations ${M^TAM = A}$ and ${\det(M) = 1}$ gives us four equations, which we can solve to get the solutions

$\displaystyle \left[ \begin{array}{cc} \sqrt{b^2+1} & b \\ b & \sqrt{b^2+1} \end{array} \right] \textrm{ and } \left[ \begin{array}{cc} -\sqrt{b^2+1} & b \\ b & -\sqrt{b^2+1} \end{array} \right].$

Since we are interested in the connected component of the identity, we discard the solution on the right. It is useful to do a change of variables ${b = \sinh(t)}$, so we have (recall that ${\cosh^2(t) - \sinh^2(t) = 1}$).

$\displaystyle SO_0(1,1) = \left\{ \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \, | \, t \in \mathbb{R} \right\}$

These matrices take ${\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]}$ to ${\left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$. In other words, ${SO_0(1,1)}$ acts transitively on ${H}$ with trivial stabilizers, and in particular we have parmeterizing maps

$\displaystyle \mathbb{R} \rightarrow SO_0(1,1) \rightarrow H \textrm{ defined by } t \mapsto \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \mapsto \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]$

The first map is actually a Lie group isomorphism (with the group action on ${\mathbb{R}}$ being ${+}$) in addition to a diffeomorphism, since

$\displaystyle \left[ \begin{array}{cc} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{array} \right] \left[ \begin{array}{cc} \cosh(s) & \sinh(s) \\ \sinh(s) & \cosh(s) \end{array} \right] = \left[ \begin{array}{cc} \cosh(t+s) & \sinh(t+s) \\ \sinh(t+s) & \cosh(t+s) \end{array} \right]$

Metric

As mentioned above, ${\langle \cdot, \cdot \rangle_H}$ is not positive definite, but its restriction to the tangent space of ${H}$ is. We can see this in the following way: tangent vectors at a point ${p \in H}$ are characterized by the form ${\langle \cdot, \cdot \rangle_H}$. Specifically, ${v\in T_pH \Leftrightarrow \langle v, p \rangle_H}$, since (by a calculation) ${\frac{d}{dt} \langle p+tv, p+tv \rangle_H = 0 \Leftrightarrow \langle v, p \rangle_H}$. Therefore, ${SO_0(1,1)}$ takes tangent vectors to tangent vectors and preserves the form (and is transitive), so we only need to check that the form is positive definite on one tangent space. This is obvious on the tangent space to the point ${\left[ \begin{array}{c} 0 \\ 1 \end{array} \right]}$. Thus, ${H}$ is a Riemannian manifold, and ${SO_0(1,1)}$ acts by isometries.

Let’s use the parameterization ${\phi: t \mapsto \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$. The unit (in the ${H}$ metric) tangent at ${\phi(t) = \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right]}$ is ${\left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right]}$. The distance between the points ${\phi(s)}$ and ${\phi(t)}$ is

$\displaystyle d_H(\phi(s), \phi(t)) = \left| \int_s^t\sqrt{\langle \left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right], \left[ \begin{array}{c} \cosh(t) \\ \sinh(t) \end{array} \right] \rangle_H dv } \right| = \left|\int_s^tdv \right| = |t-s|$

In other words, ${\phi}$ is an isometry from ${\mathbb{E}^1}$ to ${H}$.

1-dimensional hyperbollic space. The hyperboloid model is shown in blue, and the projective model is shown in red. An example of the projection map identifying ${H}$ with ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is shown.

2.2. Projective Model

Parameterizing

Real projective space ${\mathbb{R}\mathrm{P}^1}$ is the set of lines through the origin in ${\mathbb{R}^2}$. We can think about ${\mathbb{R}\mathrm{P}^1}$ as ${\mathbb{R} \cup \{\infty\}}$, where ${x\in \mathbb{R}}$ is associated with the line (point in ${\mathbb{R}\mathrm{P}^1}$) intersecting ${\{y=1\}}$ in ${x}$, and ${\infty}$ is the horizontal line. There is a natural projection ${\mathbb{R}^2 \setminus \{0\} \rightarrow \mathbb{R}\mathrm{P}^1}$ by projecting a point to the line it is on. Under this projection, ${H}$ maps to ${(-1,1)\subseteq \mathbb{R} \subseteq \mathbb{R}\mathrm{P}^1}$.

Since ${SO_0(1,1)}$ acts on ${\mathbb{R}^2}$ preserving the lines ${y = \pm x}$, it gives a projective action on ${\mathbb{R}\mathrm{P}^1}$ fixing the points ${\pm 1}$. Now suppose we have any projective linear isomorphism of ${\mathbb{R}\mathrm{P}^1}$ fixing ${\pm 1}$. The isomorphism is represented by a matrix ${A \in \mathrm{PGL}(2,\mathbb{R})}$ with eigenvectors ${\left[ \begin{array}{c} 1 \\ \pm 1 \end{array} \right]}$. Since scaling ${A}$ preserves its projective class, we may assume it has determinant 1. Its eigenvalues are thus ${\lambda}$ and ${\lambda^{-1}}$. The determinant equation, plus the fact that

$\displaystyle A \left[ \begin{array}{c} 1 \\ \pm 1 \end{array} \right] = \left[ \begin{array}{c} \lambda^{\pm 1} \\ \pm \lambda^{\pm 1} \end{array} \right]$

Implies that ${A}$ is of the form of a matrix in ${SO_0(1,1)}$. Therefore, the projective linear structure on ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is the “same” (has the same isometry (isomorphism) group) as the hyperbolic (Riemannian) structure on ${H}$.

Metric

Clearly, we’re going to use the pushforward metric under the projection of ${H}$ to ${(-1,1)}$, but it turns out that this metric is a natural choice for other reasons, and it has a nice expression.

The map taking ${H}$ to ${(-1,1) \subseteq \mathbb{R}\mathrm{P}^1}$ is ${\psi: \left[ \begin{array}{c} \sinh(t) \\ \cosh(t) \end{array} \right] \rightarrow \frac{\sinh(t)}{\cosh(T)} = \tanh(t)}$. The hyperbolic distance between ${x}$ and ${y}$ in ${(-1,1)}$ is then ${d_H(x,y) = |\tanh^{-1}(x) - \tanh^{-1}(y)|}$ (by the fact from the previous sections that ${\phi}$ is an isometry).

Recall the fact that ${\tanh(a\pm b) = \frac{\tanh(a) \pm \tanh(b)}{1 \pm \tanh(a)\tanh(b)}}$. Applying this, we get the nice form

$\displaystyle d_H(x,y) = \frac{y-x}{1 - xy}$

We also recall the cross ratio, for which we fix notation as ${ (z_1, z_2; z_3, z_4) := \frac{(z_3 -z_1)(z_4-z_2)}{(z_2-z_1)(z_4-z_3)}}$. Then

$\displaystyle (-1, x;y,1 ) = \frac{(y+1)(1-x)}{(x+1)(1-y)} = \frac{1-xy + (y-x)}{1-xy + (x-y)}$

Call the numerator of that fraction by ${N}$ and the denominator by ${D}$. Then, recalling that ${\tanh(u) = \frac{e^{2u}-1}{e^{2u}+1}}$, we have

$\displaystyle \tanh(\frac{1}{2} \log(-1,x;y,1)) = \frac{\frac{N}{D} -1}{\frac{N}{D} +1} = \frac{N-D}{N+D} = \frac{2(y-x)}{2(1-xy)} = \tanh(d_H(x,y))$

Therefore, ${d_H(x,y) = \frac{1}{2}\log(-1,x;y,-1)}$.

3. Hilbert Metric

Notice that the expression on the right above has nothing, a priori, to do with the hyperbolic projection. In fact, for any open convex body in ${\mathbb{R}\mathrm{P}^n}$, we can define the Hilbert metric on ${C}$ by setting ${d_H(p,q) = \frac{1}{2}\log(a,p,q,b)}$, where ${a}$ and ${b}$ are the intersections of the line through ${a}$ and ${b}$ with the boundary of ${C}$. How is it possible to take the cross ratio, since ${a,p,q,b}$ are not numbers? The line containing all of them is projectively isomorphic to ${\mathbb{R}\mathrm{P}^1}$, which we can parameterize as ${\mathbb{R} \cup \{\infty\}}$. The cross ratio does not depend on the choice of parameterization, so it is well defined. Note that the Hilbert metric is not necessarily a Riemannian metric, but it does make any open convex set into a metric space.

Therefore, we see that any open convex body in ${\mathbb{R}\mathrm{P}^n}$ has a natural metric, and the hyperbolic metric in ${H = (-1,1)}$ agrees with this metric when ${(-1,1)}$ is thought of as a open convex set in ${\mathbb{R}\mathrm{P}^1}$.

4. Higher-Dimensional Hyperbolic Space

4.1. Hyperboloid

The higher dimensional hyperbolic spaces are completely analogous to the 1-dimensional case. Consider ${\mathbb{R}^{n+1}}$ with the basis ${\{e_i\}_{i=1}^n \cup \{e\}}$ and the 2-form ${\langle v, w \rangle_H = \sum_{i=1}^n v_iw_i - v_{n+1}w_{n+1}}$. This is the form defined by the matrix ${J = I \oplus (-1)}$. Define ${\mathbb{H}^n}$ to be the positive (positive in the ${e}$ direction) sheet of the hyperbola ${\langle v,v\rangle_H = -1}$.

Let ${O(n,1)}$ be the linear transformations preserving the form, so ${O(n,1) = \{ A \, | \, A^TJA = J\}}$. This group is generated by ${O(1,1) \subseteq O(n,1)}$ as symmetries of the ${e_1, e}$ plane, together with ${O(n) \subseteq O(n,1)}$ as symmetries of the span of the ${e_i}$ (this subspace is euclidean). The group ${SO_0(n,1)}$ is the set of orientation preserving elements of ${O(n,1)}$ which preserve the positive sheet of the hyperboloid (${\mathbb{H}^n}$). This group acts transitively on ${\mathbb{H}^n}$ with point stabilizers ${SO(n)}$: this is easiest to see by considering the point ${(0,\cdots, 0, 1) \in \mathbb{H}^n}$. Here the stabilizer is clearly ${SO(n)}$, and because ${SO_0(n,1)}$ acts transitively, any stabilizer is a conjugate of this.

As in the 1-dimensional case, the metric on ${\mathbb{H}^n}$ is ${\langle \cdot , \cdot \rangle_H|_{T_p\mathbb{H}^n}}$, which is invariant under ${SO_0(n,1)}$.

Geodesics in ${\mathbb{H}^n}$ can be understood by consdering the fixed point sets of isometries, which are always totally geodesic. Here, reflection in a vertical (containing ${e}$) plane restricts to an (orientation-reversing, but that’s ok) isometry of ${\mathbb{H}^n}$, and the fixed point set is obviously the intersection of this plane with ${\mathbb{H}^n}$. Now ${SO_0(n,1)}$ is transitive on ${\mathbb{H}^n}$, and it sends planes to planes in ${\mathbb{R}^{n+1}}$, so we have a bijection

{Totally geodesic subspaces through ${p}$} ${\leftrightarrow}$ ${\mathbb{H}^n \cap}$ {linear subspaces of ${\mathbb{R}^{n+1}}$ through ${p}$ }

By considering planes through ${e}$, we can see that these totally geodesic subspaces are isometric to lower dimensional hyperbolic spaces.

4.2. Projective

Analogously, we define the projective model as follows: consider the disk ${\{v_{n+1} \,| v_{n+1} = 1, \langle v,v \rangle_H < 0\}}$. I.e. the points in the ${v_{n+1}}$ plane inside the cone ${\langle v,v \rangle_H = 0}$. We can think of ${\mathbb{R}\mathrm{P}^n}$ as ${\mathbb{R}^n \cup \mathbb{R}\mathrm{P}^{n-1}}$, so this disk is ${D^\circ \subseteq \mathbb{R}^n \subseteq \mathbb{R}\mathrm{P}^n}$. There is, as before, the natural projection of ${\mathbb{H}^n}$ to ${D^\circ}$, and the pushforward of the hyperbolic metric agrees with the Hilbert metric on ${D^\circ}$ as an open convex body in ${\mathbb{R}\mathrm{P}^n}$.

Geodesics in the projective model are the intersections of planes in ${\mathbb{R}^{n+1}}$ with ${D^\circ}$; that is, they are geodesics in the euclidean space spanned by the ${e_i}$. One interesting consequence of this is that any theorem which is true in euclidean geometry which does not reply on facts about angles is still true for hyperbolic space. For example, Pappus’ hexagon theorem, the proof of which does not use angles, is true.

4.3. Projective Model in Dimension 2

In the case that ${n=2}$, we can understand the projective isomorphisms of ${\mathbb{H}^2 = D \subseteq \mathbb{R}\mathrm{P}^2}$ by looking at their actions on the boundary ${\partial D}$. The set ${\partial D}$ is projectively isomorphic to ${\mathbb{R}\mathrm{P}^1}$ as an abstract manifold, but it should be noted that ${\partial D}$ is not a straight line in ${\mathbb{R}\mathrm{P}^2}$, which would be the most natural way to find ${\mathbb{R}\mathrm{P}^1}$‘s embedded in ${\mathbb{R}\mathrm{P}^2}$.

In addition, any projective isomorphism of ${\mathbb{R}\mathrm{P}^1 \cong \partial D}$ can be extended to a real projective isomorphism of ${\mathbb{R}\mathrm{P}^2}$. In other words, we can understand isometries of 2-dimensional hyperbolic space by looking at the action on the boundary. Since ${\partial D}$ is not a straight line, the extension is not trivial. We now show how to do this.

The automorphisms of ${\partial D \cong \mathbb{R}\mathrm{P}^1}$ are ${\mathrm{PSL}(2,\mathbb{R}}$. We will consider ${\mathrm{SL}(2,\mathbb{R})}$. For any Lie group ${G}$, there is an Adjoint action ${G \rightarrow \mathrm{Aut}(T_eG)}$ defined by (the derivative of) conjugation. We can similarly define an adjoint action ${\mathrm{ad}}$ by the Lie algebra on itself, as ${\mathrm{ad}(\gamma '(0)) := \left. \frac{d}{dt} \right|_{t=0} \mathrm{Ad}(\gamma(t))}$ for any path ${\gamma}$ with ${\gamma(0) = e}$. If the tangent vectors ${v}$ and ${w}$ are matrices, then ${\mathrm{ad}(v)(w) = [v,w] = vw-wv}$.

We can define the Killing form ${B}$ on the Lie algebra by ${B(v,w) = \mathrm{Tr}(\mathrm{ad}(v)\mathrm{ad}(w))}$. Note that ${\mathrm{ad}(v)}$ is a matrix, so this makes sense, and the Lie group acts on the tangent space (Lie algebra) preserving this form.

Now let’s look at ${\mathrm{SL}(2,\mathbb{R})}$ specifically. A basis for the tangent space (Lie algebra) is ${e_1 = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]}$, ${e_2 = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]}$, and ${e_3 = \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right]}$. We can check that ${[e_1,e_2] = e_3}$, ${[e_1,e_3] = -2e_1}$, and ${[e_2, e_3]=2e_2}$. Using these relations plus the antisymmetry of the Lie bracket, we know

$\displaystyle \mathrm{ad}(e_1) = \left[ \begin{array}{ccc} 0 & 0 & -2 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right] \qquad \mathrm{ad}(e_2) = \left[ \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 2 \\ -1 & 0 & 0 \end{array}\right] \qquad \mathrm{ad}(e_3) = \left[ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right]$

Therefore, the matrix for the Killing form in this basis is

$\displaystyle B_{ij} = B(e_i,e_j) = \mathrm{Tr}(\mathrm{ad}(e_i)\mathrm{ad}(e_j)) = \left[ \begin{array}{ccc} 0 & 4 & 0 \\ 4 & 0 & 0 \\ 0 & 0 & 8 \end{array}\right]$

This matrix has 2 positive eigenvalues and one negative eigenvalue, so its signature is ${(2,1)}$. Since ${\mathrm{SL}(2,\mathbb{R})}$ acts on ${T_e(\mathrm{SL}(2,\mathbb{R}))}$ preserving this form, we have ${\mathrm{SL}(2,\mathbb{R}) \cong O(2,1)}$, otherwise known at the group of isometries of the disk in projective space ${\mathbb{R}\mathrm{P}^2}$, otherwise known as ${\mathbb{H}^2}$.

Any element of ${\mathrm{PSL}(2,\mathbb{R})}$ (which, recall, was acting on the boundary of projective hyperbolic space ${\partial D}$) therefore extends to an element of ${O(2,1)}$, the isometries of hyperbolic space, i.e. we can extend the action over the disk.

This means that we can classify isometries of 2-dimensional hyperbolic space by what they do to the boundary, which is determined generally by their eigevectors (${\mathrm{PSL}(2,\mathbb{R})}$ acts on ${\mathbb{R}\mathrm{P}^1}$ by projecting the action on ${\mathbb{R}^2}$, so an eigenvector of a matrix corresponds to a fixed line in ${\mathbb{R}^2}$, so a fixed point in ${\mathbb{R}\mathrm{P}^1 \cong \partial D}$. For a matrix ${A}$, we have the following:

• ${|\mathrm{Tr}(A)| < 2}$ (elliptic) In this case, there are no real eigenvalues, so no real eigenvectors. The action here is rotation, which extends to a rotation of the entire disk.
• ${|\mathrm{Tr}(A)| = 2}$ (parabolic) There is a single real eigenvector. There is a single fixed point, to which all other points are attracted (in one direction) and repelled from (in the other). For example, the action in projective coordinates sending ${[x:y]}$ to ${[x+1:y]}$: infinity is such a fixed point.
• ${|\mathrm{Tr}(A)| > 2}$ (hyperbolic) There are two fixed point, one attracting and one repelling.
•

5. Complex Hyperbolic Space

We can do a construction analogous to real hyperbolic space over the complexes. Define a Hermitian form ${q}$ on ${\mathbb{C}^{n+1}}$ with coordinates ${\{z_1,\cdots, z_n\} \cup \{w\}}$ by ${q(x_1,\cdots x_n, w) = |z_1|^2 + \cdots + |z_n|^2 - |w|^2}$. We will also refer to ${q}$ as ${\langle \cdot, \cdot \rangle_q}$. The (complex) matrix for this form is ${J = I \oplus (-1)}$, where ${q(v,w) = v^*Jw}$. Complex linear isomorphisms preserving this form are matrices ${A}$ such that ${A^*JA = J}$. This is our definition for ${\mathrm{U}(q) := \mathrm{U}(n,1)}$, and we define ${\mathrm{SU}(n,1)}$ to be those elements of ${\mathrm{U}(n,1)}$ with determinant of norm 1.

The set of points ${z}$ such that ${q(z) = -1}$ is not quite what we are looking for: first it is a ${2n+1}$ real dimensional manifold (not ${2n}$ as we would like for whatever our definition of “complex hyperbolic ${n}$ space” is), but more importantly, ${q}$ does not restrict to a positive definite form on the tangent spaces. Call the set of points ${z}$ where ${q(z) = -1}$ by ${\bar{H}}$. Consider a point ${p}$ in ${\bar{H}}$ and ${v}$ in ${T_p\bar{H}}$. As with the real case, by the fact that ${v}$ is in the tangent space,

$\displaystyle \left. \frac{d}{dt} \right|_{t=0} \langle p + tv, p+tv\rangle_q = 0 \quad \Rightarrow \quad \langle v, p \rangle_q + \langle p,v \rangle_q = 0$

Because ${q}$ is hermitian, the expression on the right does not mean that ${\langle v,p\rangle_q = 0}$, but it does mean that ${\langle v,p \rangle_q}$ is purely imaginary. If ${\langle v,p \rangle_q = ik}$, then ${\langle v,v\rangle_q < 0}$, i.e. ${q}$ is not positive definite on the tangent spaces.

However, we can get rid of this negative definite subspace. ${S^1}$ as the complex numbers of unit length (or ${\mathrm{U}(1)}$, say) acts on ${\mathbb{C}^{n+1}}$ by multiplying coordinates, and this action preserves ${q}$: any phase goes away when we apply the absolute value. The quotient of ${\bar{H}}$ by this action is ${\mathbb{C}\mathbb{H}^n}$. The isometry group of this space is still ${\mathrm{U}(n,1)}$, but now there are point stabilizers because of the action of ${\mathrm{U}(1)}$. We can think of ${\mathrm{U}(1)}$ inside ${\mathrm{U}(n,1)}$ as the diagonal matrices, so we can write

$\displaystyle \mathrm{SU}(n,1) \times \mathrm{U}(1) \cong U(n,1)$

And the projectivized matrices ${\mathrm{PSU}(n,1)}$ is the group of isometries of ${\mathbb{C}\mathbb{H}^n \subseteq \mathbb{C}^n \subseteq \mathbb{C}\mathrm{P}^n}$, where the middle ${\mathbb{C}^n}$ is all vectors in ${\mathbb{C}^{n+1}}$ with ${w=1}$ (which we think of as part of complex projective space). We can also approach this group by projectivizing, since that will get rid of the unwanted point stabilizers too: we have ${\mathrm{PU}(n,1) \cong \mathrm{PSU}(n,1)}$.

5.1. Case ${n=1}$

In the case ${n=1}$, we can actually picture ${\mathbb{C}\mathrm{P}^1}$. We can’t picture the original ${\mathbb{C}^4}$, but we are looking at the set of ${(z,w)}$ such that ${|z|^2 - |w|^2 = -1}$. Notice that ${|w| \ge 1}$. After projectivizing, we may divide by ${w}$, so ${|z/w| - 1 = -1/|w|}$. The set of points ${z/w}$ which satisfy this is the interior of the unit circle, so this is what we think of for ${\mathbb{C}\mathbb{H}^1}$. The group of complex projective isometries of the disk is ${\mathrm{PU}(1,1)}$. The straight horizontal line is a geodesic, and the complex isometries send circles to circles, so the geodesics in ${\mathbb{C}\mathbb{H}^1}$ are circles perpendicular to the boundary of ${S^1}$ in ${\mathbb{C}}$.

Imagine the real projective model as a disk sitting at height one, and the geodesics are the intersections of planes with the disk. Complex hyperbolic space is the upper hemisphere of a sphere of radius one with equator the boundary of real hyperbolic space. To get the geodesics in complex hyperbolic space, intersect a plane with this upper hemisphere and stereographically project it flat. This gives the familiar Poincare disk model.

5.2. Real ${\mathbb{H}^2}$‘s contained in ${\mathbb{C}\mathbb{H}^n}$

${\mathbb{C}\mathbb{H}^2}$ contains 2 kinds of real hyperbolic spaces. The subset of real points in ${\mathbb{C}\mathbb{H}^n}$ is (real) ${\mathbb{H}^n}$, so we have a many ${\mathbb{H}^2 \subseteq \mathbb{H}^n \subseteq \mathbb{C}\mathbb{H}^n}$. In addition, we have copies of ${\mathbb{C}\mathbb{H}^1}$, which, as discussed above, has the same geometry (i.e. has the same isometry group) as real ${\mathbb{H}^2}$. However, these two real hyperbolic spaces are not isometric. the complex hyperbolic space ${\mathbb{C}\mathbb{H}^1}$ has a more negative curvature than the real hyperbolic spaces. If we scale the metric on ${\mathbb{C}\mathbb{H}^n}$ so that the real hyperbolic spaces have curvature ${-1}$, then the copies of ${\mathbb{C}\mathbb{H}^1}$ will have curvature ${-4}$.

In a similar vein, there is a symplectic structure on ${\mathbb{C}\mathbb{H}^n}$ such that the real ${\mathbb{H}^2}$ are lagrangian subspaces (the flattest), and the ${\mathbb{C}\mathbb{H}^1}$ are symplectic, the most negatively curved.

An important thing to mention is that complex hyperbolic space does not have constant curvature(!).

6. Poincare Disk Model and Upper Half Space Model

The projective models that we have been dealing with have many nice properties, especially the fact that geodesics in hyperbolic space are straight lines in projective space. However, the angles are wrong. There are models in which the straight lines are “curved” i.e. curved in the euclidean metric, but the angles between them are accurate. Here we are interested in a group of isometries which preserves angles, so we are looking at a conformal model. Dimension 2 is special, because complex geometry is real conformal geometry, but nevertheless, there is a model of ${\mathbb{R}\mathbb{H}^n}$ in which the isometries of the space are conformal.

Consider the unit disk ${D^n}$ in ${n}$ dimensions. The conformal automorphisms are the maps taking (straight) diameters and arcs of circles perpendicular to the boundary to this same set. This model is abstractly isomorphic to the Klein model in projective space. Imagine the unit disk in a flat plane of height one with an upper hemisphere over it. The geodesics in the Klein model are the intersections of this flat plane with subspaces (so they are straight lines, for example, in dimension 2). Intersecting vertical planes with the upper hemisphere and stereographically projecting it flat give geodesics in the Poincare disk model. The fact that this model is the “same” (up to scaling the metric) as the example above of ${\mathbb{C}\mathbb{H}^1}$ is a (nice) coincidence.

The Klein model is the flat disk inside the sphere, and the Poincare disk model is the sphere. Geodesics in the Klein model are intersections of subspaces (the angled plane) with the flat plane at height 1. Geodesics in the Poincare model are intersections of vertical planes with the upper hemisphere. The two darkened geodesics, one in the Klein model and one in the Poincare, correspond under orthogonal projection. We get the usual Poincare disk model by stereographically projecting the upper hemisphere to the disk. The projection of the geodesic is shown as the curved line inside the disk

The Poincare disk model. A few geodesics are shown.

Now we have the Poincare disk model, where the geodesics are straight diameters and arcs of circles perpendicular to the boundary and the isometries are the conformal automorphisms of the unit disk. There is a conformal map from the disk to an open half space (we typically choose to conformally identify it with the upper half space). Conveniently, the hyperbolic metric on the upper half space ${d_H}$ can be expressed at a point ${(x,t)}$ (euclidean coordinates) as ${d_H = d_E/t}$. I.e. the hyperbolic metric is just a rescaling (at each point) of the euclidean metric.

One of the important things that we wanted in our models was the ability to realize isometries of the model with isometries of the ambient space. In the case of a one-parameter family of isometries of hyperbolic space, this is possible. Suppose that we have a set of elliptic isometries. Then in the disk model, we can move that point to the origin and realize the isometries by rotations. In the upper half space model, we can move the point to infinity, and realize them by translations.

I am (update: was) currently (update: but am no longer) in Brisbane for the “New directions in geometric group theory” conference, which has been an entirely enjoyable and educational experience. I got to eat fish and chips, to watch Australia make 520 for 7 (declared) against the West Indies at the WACA, and to hear Masato Mimura give a very nice talk about his recent results on rigidity of the “universal lattice”.

His talk included a quick and beautiful survey of some geometric aspects of the theory of rigidity for infinite groups, which I will attempt to partially reproduce (despite the limitations of the wordpress format). In this context, rigidity is expressed in terms of isometric affine actions of groups on Banach spaces. This means the following. Suppose $B$ is a Banach space (i.e. a complete, normed vector space) and $G$ is a group. A linear isometric action is a representation $\rho$ from $G$ to the group of linear isometries of $B$ — i.e. linear norm-preserving automorphisms. An affine action is a representation from $G$ to the group of affine isometries of $B$ — i.e. isometries as a metric space that do not necessarily fix the zero element. The group of isometries of a Banach space $B$ is a semi-direct product $B \rtimes U(B)$ where $U(B)$ is the group of linear isometries, and $B$ is the Banach space, thought of as an Abelian group, acting on itself by (isometric) translations. Such an action is usually encoded by a pair $\rho:G \to U(B)$ which records the “linear” part of the action, and a 1-cocycle with coefficients in $\rho$, i.e. a function $c:G \to B$ satisfying $c(gh) = c(g) + \rho(g)c(h)$ for every $g,h \in G$. This formula might look strange if you don’t know where it comes from: it is just the way that factors transform in semi-direct products. The affine action is given by sending $g \in G$ to the transformation that sends each $b \in B$ to $\rho(g)b + c(g)$. Consequently, $gh$ is sent to the transformation that sends $b$ to $\rho(gh)b + c(gh)$ and the fact that this is a group action becomes the formula

$\rho(gh)b + c(gh) = \rho(g)(\rho(h)b + c(h)) + c(g) = \rho(gh)b + \rho(g)c(h) + c(g)$

Equating the left and right hand sides gives the cocycle condition. Given one affine isometric action, one can obtain another in a silly way by conjugating by an isometry $b \to b + b'$ for some $b' \in B$. Under conjugation by such an isometry, a cocycle $c$ transforms by $c(g) \to c(g) + \rho(g)b' - b'$. A function of the form $c(g) = \rho(g)b' - b'$ is called a 1-coboundary, and the quotient of the space of 1-cocycles by the space of 1-coboundaries is the 1 dimensional cohomology of $G$ with coefficients in $\rho:G \to U(B)$. This is usually denoted $H^1(G,\rho)$, where $B$ is suppressed in the notation. In particular, an affine isometric action of $G$ on $B$ with linear part $\rho$ has a global fixed point iff it represents $0$ in $H^1(G,\rho)$. Contrapositively, $G$ admits an affine isometric action on $B$ without a global fixed point iff $H^1(G,\rho) \ne 0$ for some $\rho$.

A group $G$ is said to satisfy Serre’s Property (FH) if every affine isometric action of $G$ on a Hilbert space has a global fixed point. In 2007, Bader-Furman-Gelander-Monod introduced a property (FB) for a group $G$ to mean that every affine isometric action of $G$ on some (out of a class of) Banach space(s) $B$ has a global fixed point. Mimura used the notation property (FL_p) for the case that $B$ is allowed to range over the class of $L_p$ spaces (for some fixed $1 < p < \infty$).

Intimately related is Kazhdan’s Property (T), introduced by Kazhdan in this paper. Let $G$ be a locally compact topological group (for example, a discrete group). The set of irreducible unitary representations of $G$ is called its dual, and denoted $\hat{G}$. This dual is topologized in the following way. Associated to a representation $\rho:G \to U(L)$, a unit vector $X \in L$, a positive number $\epsilon > 0$ and a compact subset $K \subset G$ there is an open neighborhood of $\rho$ consisting of representations $\rho':G \to U(L')$ for which there is a unit vector $Y \in L$ such that $|\langle \rho(g)X,X\rangle - \langle \rho(g')Y, Y\rangle| < \epsilon$ whenever $g \in K$. With this topology (called the Fell topology), one says that a group $G$ has property (T) if the trivial representation is isolated in $\hat{G}$. Note that this topology is very far from being Hausdorff: the trivial representation fails to be isolated exactly when there are a sequence of representations $\rho_i:G \to U(L_i)$, unit vectors $X_i \in L_i$, numbers $\epsilon_i \to 0$ and compact sets $K_i$ exhausting $G$ so that $|\langle\rho_i(g)X_i,X_i\rangle| < \epsilon_i$ for any $g \in K_i$. The vectors $X_i$ are said to be (a sequence of) almost invariant vectors. Hence (informally) a group has property (T) if some compact subset must move some unit vector a definite amount in every irreducible nontrivial unitary representation. If a group fails to have property (T), one can rescale a sequence of irreducible actions near a sequence of almost invariant vectors in such a way that one obtains in the geometric limit a nontrivial isometric action on $L^2$ without a global fixed point. A famous theorem of Delorme-Guichardet says that property (T) and property (FH) are equivalent for (locally compact second countable) groups. Property (T) passes to quotients, and to lattices (i.e. finite covolume discrete subgroups of a topological group). Kazhdan already showed in his paper that $\text{SL}(n,\mathbb{R})$ has property (T) for $n$ at least $3$, and therefore the same is true for lattices in this groups, such as $\text{SL}(n,\mathbb{Z})$, a fact which is not easy to see directly from the definition. One beautiful application, already pointed out by Kazhdan, is that this means that all lattices in $\text{SL}(n,\mathbb{R})$, for instance the groups $\text{SL}(n,\mathbb{Z})$ (and in fact, all discrete groups with property (T)) are finitely generated. Kazhdan’s proof of this is incredibly short: let $G$ be a discrete group and $g_i$ and sequence of elements. For each $i$, let $G_i$ be the subgroup of $G$ generated by $\lbrace g_1,g_2,\cdots,g_i\rbrace$. Notice that $G$ is finitely generated iff $G_i=G$ for all sufficiently large $i$. On the other hand, consider the unitary representations of $G$ induced by the trivial representations on the $G_i$. Every compact subset of $G$ is finite, and therefore eventually fixes a vector in every one of these representations; thus there is a sequence of almost fixed vectors. If $G$ has property (T), this sequence eventually contains a fixed vector, which can only happen if $G/G_i$ is finite, in which case $G$ is finitely generated, as claimed.

Property (FL_p) generalizes (FH) (equivalently (T)) in many significant ways, with interesting applications to dynamics. For example, Navas showed that if $G$ is a group with property (T) then every action of $G$ on a circle which is at least $C^{1+1/2 + \epsilon}$ factors through a finite group. Navas’s argument can be generalized straightforwardly to show that if $G$ has (FL_p) for some $p>2$ then every action of $G$ on a circle which is at least $C^{1+1/p+\epsilon}$ factors through a finite group. The proof rests on a beautiful construction due to Reznikov (although a similar construction can be found in Pressley-Segal) of certain functions on a configuration space of the circle which are not in $L^p$ but have coboundaries which are; this gives rise to nontrivial cohomology with $L^p$ coefficients for groups acting on the circle in a sufficiently interesting way.

(Update: Nicolas Monod points out in an email that the “function on a configuration space” is morally just the derivative. In fact, he made the nice remark that if $D$ is any elliptic operator on an $n$-manifold, then the commutator $[D,g]$ is of Schatten class $(n+1)$ whenever $g$ is a sufficiently smooth function; morally this should give rise to nontrivial cohomology with suitable coefficients for groups acting with enough regularity on any given $n$-manifold, and one would like to use this e.g. to approach Zimmer’s conjecture, but nobody seems to know how to make this work as yet; in fact the work of Monod et. al. on (FL_p) is at least partly motivated by this general picture.)

Mimura discussed a spectrum of rigid behaviour for infinite groups, ranging from most rigid (property (FL_p) for every $p$) to least rigid (amenable) (note: every finite group is both amenable and has property (T), so this only really makes sense for infinite groups; moreover, every reasonable measure of rigidity for infinite groups is usually invariant under passing to subgroups of finite index). Free groups, $\text{SL}(2,\mathbb{Z})$ and so on are very non-rigid. However, it is well-known that certain infinite families of (word) hyperbolic groups, including lattices in groups of isometries of quaternion-hyperbolic symmetric spaces, and “random” groups with relations having density parameter $1/3 < d < 1/2$ (see Zuk or Ollivier) are both hyperbolic and have property (T). Nevertheless, these groups are not as rigid as higher rank lattices like $\text{SL}(n,\mathbb{Z})$ for $n>2$. The latter have property (FL_p) for every $1< p < \infty$, whereas Yu showed that every hyperbolic group admits a proper affine isometric action on $\ell^p$ for some $p$ (the existence of a proper affine isometric action on a Hilbert space is called “a-T-menability” by Gromov, and the “Haagerup property” by some. Groups satisfying this property, or even Yu’s weaker property, are known to satisfy some version of the Baum-Connes conjecture, the subject of a very nice minicourse by Graham Niblo at the same conference).

It is in this context that one can appreciate Mimura’s results. His first main result is that the group $\text{SL}_n(\mathbb{Z}[x_1,x_2,\cdots,x_n])$ (i.e. the “universal lattice”) has property (FL_p) for every $1 provided $n$ is at least 4. Since property (FL_p) (like (T)) passes to quotients, this implies that $\text{SL}_n(R)$ has (FL_p) for every unital, commutative, finitely generated ring $R$.

His second main result concerns a “quasification” of FL_p, to a property called (FFL_p). Without getting too technical, this property concerns “quasi-actions” of a group on a Banach space by affine isometries; algebraically these are encoded by 1-cochains $c:G \to B$ for which there is a universal constant $D$ so that $|c(gh) - c(g) -\rho(g)c(h)| < D$ as measured in the Banach norm on $B$. Any bounded map $c:G \to B$ defines a 1-cochain; such (bounded) 1-cochains corresponds to  quasi-action with a bounded orbit. Associated to $\rho: G \to U(B)$ one defines in a similar way a complex of bounded cochains; quasi-actions modulo bounded quasi-actions are parameterized by the kernel of the comparison map $H^2_b(G,\rho) \to H^2(G,\rho)$ from bounded to ordinary cohomology. Mimura’s second main result is that when $G$ is the universal lattice as above, and $\rho$ has no invariant vectors, the comparison map from bounded to ordinary cohomology in dimension 2 is injective.

The fact that $\rho$ as above is required to have no invariant vectors is a technical necessity of Mimura’s proof. When $\rho$ is trivial, one is studying “ordinary” bounded cohomology, and there is an exact sequence

$0 \to H^1(G) \to Q(G) \to H^2_b(G) \to H^2(G)$

with real coefficients for any $G$ (here $Q(G)$ denotes the vector space of homogeneous quasimorphisms on $G$). In this context, one knows by Bavard duality that $H^2_b \to H^2$ is injective if and only if the stable commutator length is identically zero on $[G,G]$. By quite a different method, Mimura shows that for $n$ at least $6$, and for any Euclidean ring $R$ (i.e. a ring for which one has a Euclidean algorithm; for example, $R = \mathbb{C}[x]$) the group $SL_n(R)$ has vanishing stable commutator length, and therefore one has injectivity of bounded to ordinary cohomology in dimension $2$.

(Update 1/9/2010): Nicholas Monod sent me a nice email commenting on a couple of points in this blog entry, and I have consequently modified the language a bit in a few places. Ta much!

Last Friday, Henry Wilton gave a talk at Caltech about his recent joint work with Sang-hyun Kim on polygonal words in free groups. Their work is motivated by the following well-known question of Gromov:

Question(Gromov): Let $G$ be a one-ended word-hyperbolic group. Does $G$ contain a subgroup isomorphic to the fundamental group of a closed hyperbolic surface?

Let me briefly say what “one-ended” and “word-hyperbolic” mean.

A group is said to be word-hyperbolic if it acts properly and cocompactly by isometries on a proper $\delta$-hyperbolic path metric space — i.e. a path metric space in which there is a constant $\delta$ so that geodesic triangles in the metric space have the property that each side of the triangle is contained in the $\delta$-neighborhood of the union of the other two sides (colloquially, triangles are thin). This condition distills the essence of negative curvature in the large, and was shown by Gromov to be equivalent to several other conditions (eg. that the group satisfies a linear isoperimetric inequality; that every ultralimit of the group is an $\mathbb{R}$-tree). Free groups are hyperbolic; fundamental groups of closed manifolds with negative sectional curvature (eg surfaces with negative Euler characteristic) are word-hyperbolic; “random” groups are hyperbolic — and so on. In fact, it is an open question whether a group $G$ that admits a finite $K(G,1)$ is word hyperbolic if and only if it does not contain a copy of a Baumslag-Solitar group $BS(m,n):=\langle x,y \; | \; x^{-1}y^{m}x = y^n \rangle$ for $m,n \ne 0$ (note that the group $\mathbb{Z}\oplus \mathbb{Z}$ is the special case $m=n=1$); in any case, this is a very good heuristic for identifying the word-hyperbolic groups one typically meets in examples.

If $G$ is a finitely generated group, the ends of $G$ really means the ends (as defined by Freudenthal) of the Cayley graph of $G$ with respect to some finite generating set. Given a proper topological space $X$, the set of compact subsets of $X$ gives rise to an inverse system of inclusions, where $X-K'$ includes into $X-K$ whenever $K$ is a subset of $K'$. This inverse system defines an inverse system of maps of discrete spaces $\pi_0(X-K') \to \pi_0(X-K)$, and the inverse limit of this system is a compact, totally disconnected space $\mathcal{E}(X)$, called the space of ends of $X$. A proper topological space is canonically compactified by its set of ends; in fact, the compactification $X \cup \mathcal{E}(X)$ is the “biggest” compactification of $X$ by a totally disconnected space, in the sense that for any other compactification $X \subset Y$ where $Y-X$ is zero dimensional, there is a continuous map $X \cup \mathcal{E}(X) \to Y$ which is the identity on $X$.

For a word-hyperbolic group $G$, the Cayley graph can be compactified by adding the ideal boundary $\partial_\infty G$, but this is typically not totally disconnected. In this case, the ends of $G$ can be recovered as the components of $\partial_\infty G$.

A group $G$ acts on its own ends $\mathcal{E}(G)$. An elementary argument shows that the cardinality of $\mathcal{E}(G)$ is one of $0,1,2,\infty$ (if a compact set $V$ disconnects $e_1,e_2,e_3$ then infinitely many translates of $V$ converging to $e_1$ separate $e_3$ from infinitely many other ends accumulating on $e_1$). A group has no ends if and only if it is finite. Stallings famously showed that a (finitely generated) group has at least $2$ ends if and only if it admits a nontrivial description as an HNN extension or amalgamated free product over a finite group. One version of the argument proceeds more or less as follows, at least when $G$ is finitely presented. Let $M$ be an $n$-dimensional Riemannian manifold with fundamental group $G$, and let $\tilde{M}$ denote the universal cover. We can identify the ends of $G$ with the ends of $\tilde{M}$. Let $V$ be a least ($n-1$-dimensional) area hypersurface in $\tilde{M}$ amongst all hypersurfaces that separate some end from some other (here the hypothesis that $G$ has at least two ends is used). Then every translate of $V$ by an element of $G$ is either equal to $V$ or disjoint from it, or else one could use the Meeks-Yau “roundoff trick” to find a new $V'$ with strictly lower area than $V$. The translates of $V$ decompose $\tilde{M}$ into pieces, and one can build a tree $T$ whose vertices correspond to to components of $\tilde{M} - G\cdot V$, and whose edges correspond to the translates $G\cdot V$. The group $G$ acts on this tree, with finite edge stabilizers (by the compactness of $V$), exhibiting $G$ either as an HNN extension or an amalgamated product over the edge stabilizers. Note that the special case $|\mathcal{E}(G)|=2$ occurs if and only if $G$ has a finite index subgroup which is isomorphic to $\mathbb{Z}$.

Free groups and virtually free groups do not contain closed surface subgroups; Gromov’s question more or less asks whether these are the only examples of word-hyperbolic groups with this property.

Kim and Wilton study Gromov’s question in a very, very concrete case, namely that case that $G$ is the double of a free group $F$ along a word $w$; i.e. $G = F *_{\langle w \rangle } F$ (hereafter denoted $D(w)$). Such groups are known to be one-ended if and only if $w$ is not contained in a proper free factor of $F$ (it is clear that this condition is necessary), and to be hyperbolic if and only if $w$ is not a proper power, by a result of Bestvina-Feighn. To see that this condition is necessary, observe that the double $\mathbb{Z} *_{p\mathbb{Z}} \mathbb{Z}$ is isomorphic to the fundamental group of a Seifert fiber space, with base space a disk with two orbifold points of order $p$; such a group contains a $\mathbb{Z}\oplus \mathbb{Z}$. One might think that such groups are too simple to give an insight into Gromov’s question. However, these groups (or perhaps the slightly larger class of graphs of free groups with cyclic edge groups) are a critical case for at least two reasons:

1. The “smaller” a group is, the less room there is inside it for a surface group; thus the “simplest” groups should have the best chance of being a counterexample to Gromov’s question.
2. If $G$ is word-hyperbolic and one-ended, one can try to find a surface subgroup by first looking for a graph of free groups $H$ in $G$, and then looking for a surface group in $H$. Since a closed surface group is itself a graph of free groups, one cannot “miss” any surface groups this way.

Not too long ago, I found an interesting construction of surface groups in certain graphs of free groups with cyclic edge groups. In fact, I showed that every nontrivial element of $H_2(G;\mathbb{Q})$ in such a group is virtually represented by a sum of surface subgroups. Such surface subgroups are obtained by finding maps of surface groups into $G$ which minimize the Gromov norm in their (projective) homology class. I think it is useful to extend Gromov’s question by making the following

Conjecture: Let $G$ be a word-hyperbolic group, and let $\alpha \in H_2(G;\mathbb{Q})$ be nonzero. Then some multiple of $\alpha$ is represented by a norm-minimizing surface (which is necessarily $\pi_1$-injective).

Note that this conjecture does not generalize to wider classes of groups. There are even examples of $\text{CAT}(0)$ groups $G$ with nonzero homology classes $\alpha \in H_2(G;\mathbb{Q})$ with positive, rational Gromov norm, for which there are no $\pi_1$-injective surfaces representing a multiple of $\alpha$ at all.

It is time to define polygonal words in free groups.

Definition: Let $F$ be free. Let $X$ be a wedge of circles whose edges are free generators for $F$. A cyclically reduced word $w$ in these generators is polygonal if there exists a van-Kampen graph $\Gamma$ on a surface $S$ such that:

1. every complementary region is a disk whose boundary is a nontrivial (possibly negative) power of $w$;
2. the (labelled) graph $\Gamma$ immerses in $X$ in a label preserving way;
3. the Euler characteristic of $S$ is strictly less than the number of disks.

The last condition rules out trivial examples; for example, the double of a single disk whose boundary is labeled by $w^n$. Notice that it is very important to allow both positive and negative powers of $w$ as boundaries of complementary regions. In fact, if $w$ is not in the commutator subgroup, then the sum of the powers over all complementary regions is necessarily zero (and if $w$ is in the commutator subgroup, then $D(w)$ has nontrivial $H_2$, so one already knows that there is a surface subgroup).

Condition 2. means that at each vertex of $\Gamma$, there is at most one oriented label corresponding to each generator of $F$ or its inverse. This is really the crucial geometric property. If $\Gamma,S$ is a van-Kampen graph as above, then a theorem of Marshall Hall implies that there is a finite cover of $X$ into which $\Gamma$ embeds (in fact, this observation underlies Stallings’s work on foldings of graphs). If we build a $2$-complex $Y$ with $\pi_1(Y)=D(w)$ by attaching two ends of a cylinder to suitable loops in two copies of $X$, then a tubular neighborhood of $\Gamma$ in $S$ (i.e. what is sometimes called a “fatgraph” ) embeds in a finite cover $\tilde{Y}$ of $Y$, and its double — a surface of strictly negative Euler characteristic — embeds as a closed surface in $\tilde{Y}$, and is therefore $\pi_1$-injective. Hence if $w$ is polygonal, $D(w)$ contains a surface subgroup.

Not every word is polygonal. Kim-Wilton discuss some interesting examples in their paper, including:

1. suppose $w$ is a cyclically reduced product of proper powers of the generators or their inverses (e.g a word like $a^3b^7a^{-2}c^{13}$ but not a word like $a^3bc^{-1}$); then $w$ is polygonal;
2. a word of the form $\prod_i a^{p_{2i-1}}(a^{p_{2i}})^b$ is polygonal if $|p_i|>1$ for each $i$;
3. the word $abab^2ab^3$ is not polygonal.

To see 3, suppose there were a van-Kampen diagram with more disks than Euler characteristic. Then there must be some vertex of valence at least $3$. Since $w$ is positive, the complementary regions must have boundaries which alternate between positive and negative powers of $w$, so the degree of the vertex must be even. On the other hand, since $\Gamma$ must immerse in a wedge of two circles, the degree of every vertex must be at most $4$, so there is consequently some vertex of degree exactly $4$. Since each $a$ is isolated, at least $2$ edges must be labelled $b$; hence exactly two. Hence exactly two edges are labelled $a$. But one of these must be incoming and one outgoing, and therefore these are adjacent, contrary to the fact that $w$ does not contain a $a^{\pm 2}$.

1 above is quite striking to me. When $w$ is in the commutator subgroup, one can consider van-Kampen diagrams as above without the injectivity property, but with the property that every power of $w$ on the boundary of a disk is positive; call such a van-Kampen diagram monotone. It turns out that monotone van-Kampen diagrams always exist when $w \in [F,F]$, and in fact that norm-minimizing surfaces representing powers of the generator of $H_2(D(w))$ are associated to certain monotone diagrams. The construction of such surfaces is an important step in the argument that stable commutator length (a kind of relative Gromov norm) is rational in free groups. In my paper scl, sails and surgery I showed that monomorphisms of free groups that send every generator to a power of that generator induce isometries of the $\text{scl}$ norm; in other words, there is a natural correspondence between certain equivalence classes of monotone surfaces for an arbitrary word in $[F,F]$ and for a word of the kind that Kim-Wilton show is polygonal (Note: Henry Wilton tells me that Brady, Forester and Martinez-Pedroza have independently shown that $D(w)$ contains a surface group for such $w$, but I have not seen their preprint (though I would be very grateful to get a copy!)).

In any case, if not every word is polygonal, all is not lost. To show that $D(w)$ contains a surface subgroup is suffices to show that $D(w')$ contains a surface subgroup, where $w$ and $w'$ differ by an automorphism of $F$. Kim-Wilton conjecture that one can always find an automorphism $\phi$ so that $\phi(w)$ is polygonal. In fact, they make the following:

Conjecture (Kim-Wilton; tiling conjecture): A word $w$ not contained in a proper free factor of shortest length (in a given generating set) in its orbit under $\text{Aut}(F)$ is polygonal.

If true, this would give a positive answer to Gromov’s question for groups of the form $D(w)$.

Last week, Michael Brandenbursky from the Technion gave a talk at Caltech on an interesting connection between knot theory and quasimorphisms. Michael’s paper on this subject may be obtained from the arXiv. Recall that given a group $G$, a quasimorphism is a function $\phi:G \to \mathbb{R}$ for which there is some least real number $D(\phi) \ge 0$ (called the defect) such that for all pairs of elements $g,h \in G$ there is an inequality $|\phi(gh) - \phi(g) - \phi(h)| \le D(\phi)$. Bounded functions are quasimorphisms, although in an uninteresting way, so one is usually only interested in quasimorphisms up to the equivalence relation that $\phi \sim \psi$ if the difference $|\phi - \psi|$ is bounded. It turns out that each equivalence class of quasimorphism contains a unique representative which has the extra property that $\phi(g^n) = n\phi(g)$ for all $g\in G$ and $n \in \mathbb{Z}$. Such quasimorphisms are said to be homogeneous. Any quasimorphism may be homogenized by defining $\overline{\phi}(g) = \lim_{n \to \infty} \phi(g^n)/n$ (see e.g. this post for more about quasimorphisms, and their relation to stable commutator length).

Many groups that do not admit many homomorphisms to $\mathbb{R}$ nevertheless admit rich families of homogeneous quasimorphisms. For example, groups that act weakly properly discontinuously on word-hyperbolic spaces admit infinite dimensional families of homogeneous quasimorphisms; see e.g. Bestvina-Fujiwara. This includes hyperbolic groups, but also mapping class groups and braid groups, which act on the complex of curves.

Michael discussed another source of quasimorphisms on braid groups, those coming from knot theory. Let $I$ be a knot invariant. Then one can extend $I$ to an invariant of pure braids on $n$ strands by $I(\alpha) = I(\widehat{\alpha \Delta})$ where $\Delta = \sigma_1 \cdots \sigma_{n-1}$, and the “hat” denotes plat closure. It is an interesting question to ask: under what conditions on $I$ is the resulting function on braid groups a quasimorphism?

In the abstract, such a question is probably very hard to answer, so one should narrow the question by concentrating on knot invariants of a certain kind. Since one wants the resulting invariants to have some relation to the algebraic structure of braid groups, it is natural to look for functions which factor through certain algebraic structures on knots; Michael was interested in certain homomorphisms from the knot concordance group to $\mathbb{R}$. We briefly describe this group, and a natural class of homomorphisms.

Two oriented knots $K_1,K_2$ in the $3$-sphere are said to be concordant if there is a (locally flat) properly embedded annulus $A$ in $S^3 \times [0,1]$ with $A \cap S^3 \times 0 = K_1$ and $A \cap S^3 \times 1 = K_2$. Concordance is an equivalence relation, and the equivalence classes form a group, with connect sum as the group operation, and orientation-reversed mirror image as inverse. The only subtle aspect of this is the existence of inverses, which we briefly explain. Let $K$ be an arbitrary knot, and let $K^!$ denote the mirror image of $K$ with the opposite orientation. Arrange $K \cup K^!$ in space so that they are symmetric with respect to reflection in a dividing plane. There is an immersed annulus $A$ in $S^3$ which connects each point on $K$ to its mirror image on $K^!$, and the self-intersections of this annulus are all disjoint embedded arcs, corresponding to the crossings of $K$ in the projection to the mirror. This annulus is an example of what is called a ribbon surface. Connect summing $K$ to $K^!$ by pushing out a finger of each into an arc in the mirror connects the ribbon annulus to a ribbon disk spanning $K \# K^!$. A ribbon surface (in particular, a ribbon disk) can be pushed into a (smoothly) embedded surface in a $4$-ball bounding $S^3$. Puncturing the $4$-ball at some point on this smooth surface, one obtains a concordance from $K\#K^!$ to the unknot, as claimed.

The resulting group is known as the concordance group $\mathcal{C}$ of knots. Since connect sum is commutative, this group is abelian. Notice as above that a slice knot — i.e. a knot bounding a locally flat disk in the $4$-ball — is concordant to the unknot. Ribbon knots (those bounding ribbon disks) are smoothly slice, and therefore slice, and therefore concordant to the trivial knot. Concordance makes sense for codimension two knots in any dimension. In higher even dimensions, knots are always slice, and in higher odd dimensions, Levine found an algebraic description of the concordance groups in terms of (Witt) equivalence classes of linking pairings on a Seifert surface; (some of) this information is contained in the signature of a knot.

Let $K$ be a knot (in $S^3$ for simplicity) with Seifert surface $\Sigma$ of genus $g$. If $\alpha,\beta$ are loops in $\Sigma$, define $f(\alpha,\beta)$ to be the linking number of $\alpha$ with $\beta^+$, which is obtained from $\beta$ by pushing it to the positive side of $\Sigma$. The function $f$ is a bilinear form on $H_1(\Sigma)$, and after choosing generators, it can be expressed in terms of a matrix $V$ (called the Seifert matrix of $K$). The signature of $K$, denoted $\sigma(K)$, is the signature (in the usual sense) of the symmetric matrix $V + V^T$. Changing the orientation of a knot does not affect the signature, whereas taking mirror image multiplies it by $-1$. Moreover, if $\Sigma_1,\Sigma_2$ are Seifert surfaces for $K_1,K_2$, one can form a Seifert surface $\Sigma$ for $K_1 \# K_2$ for which there is some sphere $S^2 \in S^3$ that intersects $\Sigma$ in a separating arc, so that the pieces on either side of the sphere are isotopic to the $\Sigma_i$, and therefore the Seifert matrix of $K_1 \# K_2$ can be chosen to be block diagonal, with one block for each of the Seifert matrices of the $K_i$; it follows that $\sigma(K_1 \# K_2) = \sigma(K_1) + \sigma(K_2)$. In fact it turns out that $\sigma$ is a homomorphism from $\mathcal{C}$ to $\mathbb{Z}$; equivalently (by the arguments above), it is zero on knots which are topologically slice. To see this, suppose $K$ bounds a locally flat disk $\Delta$ in the $4$-ball. The union $\Sigma':=\Sigma \cup \Delta$ is an embedded bicollared surface in the $4$-ball, which bounds a $3$-dimensional Seifert “surface” $W$ whose interior may be taken to be disjoint from $S^3$. Now, it is a well-known fact that for any oriented $3$-manifold $W$, the inclusion $\partial W \to W$ induces a map $H_1(\partial W) \to H_1(W)$ whose kernel is Lagrangian (with respect to the usual symplectic pairing on $H_1$ of an oriented surface). Geometrically, this means we can find a basis for the homology of $\Sigma'$ (which is equal to the homology of $\Sigma$) for which half of the basis elements bound $2$-chains in $W$. Let $W^+$ be obtained by pushing off $W$ in the positive direction. Then chains in $W$ and chains in $W^+$ are disjoint (since $W$ and $W^+$ are disjoint) and therefore the Seifert matrix $V$ of $K$ has a block form for which the lower right $g \times g$ block is identically zero. It follows that $V+V^T$ also has a zero $g\times g$ lower right block, and therefore its signature is zero.

The Seifert matrix (and therefore the signature), like the Alexander polynomial, is sensitive to the structure of the first homology of the universal abelian cover of $S^3 - K$; equivalently, to the structure of the maximal metabelian quotient of $\pi_1(S^3 - K)$. More sophisticated “twisted” and $L^2$ signatures can be obtained by studying further derived subgroups of $\pi_1(S^3 - K)$ as modules over group rings of certain solvable groups with torsion-free abelian factors (the so-called poly-torsion-free-abelian groups). This was accomplished by Cochran-Orr-Teichner, who used these methods to construct infinitely many new concordance invariants.

The end result of this discussion is the existence of many, many interesting homomorphisms from the knot concordance group to the reals, and by plat closure, many interesting invariants of braids. The connection with quasimorphisms is the following:

Theorem(Brandenbursky): A homomorphism $I:\mathcal{C} \to \mathbb{R}$ gives rise to a quasimorphism on braid groups if there is a constant $C$ so that $|I([K])| \le C\cdot\|K\|_g$, where $\|\cdot\|_g$ denotes $4$-ball genus.

The proof is roughly the following: given pure braids $\alpha,\beta$ one forms the knots $\widehat{\alpha\Delta}$, $\widehat{\beta\Delta}$ and $\widehat{\alpha\beta\Delta}$. It is shown that the connect sum $L:= \widehat{\alpha \Delta} \# \widehat{\beta\Delta} \# \widehat{\alpha\beta\Delta}^!$ bounds a Seifert surface whose genus may be universally bounded in terms of the number of strands in the braid group. Pushing this Seifert surface into the $4$-ball, the hypothesis of the theorem says that $I$ is uniformly bounded on $L$. Properties of $I$ then give an estimate for the defect; qed.

It would be interesting to connect these observations up to other “natural” chiral, homogeneous invariants on mapping class groups. For example, associated to a braid or mapping class $\phi \in \text{MCG}(S)$ one can (usually) form a hyperbolic $3$-manifold $M_\phi$ which fibers over the circle, with fiber $S$ and monodromy $\phi$. The $\eta$-invariant of $M_\phi$ is the signature defect $\eta(M_\phi) = \int_Y p_1/3 - \text{sign}(Y)$ where $Y$ is a $4$-manifold with $\partial Y = M_\phi$ with a product metric near the boundary, and $p_1$ is the first Pontriagin form on $Y$ (expressed in terms of the curvature of the metric). Is $\eta$ a quasimorphism on some subgroup of $\text{MCG}(S)$ (eg on a subgroup consisting entirely of pseudo-Anosov elements)?