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1. Mostow Rigidity

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

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

In other words, Moduli space is a single point.

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

1.1. Part 1

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

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

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

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

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

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

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

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

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

and

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

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

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

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

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

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

And

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

By the triangle inequality,

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

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

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

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

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

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

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

Proof: Here is a representative picture:

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

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

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

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

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

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

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

$\Box$

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

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

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

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

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

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

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

and

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

So we have

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

Which implies that

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

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

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

1.2. Part 2

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

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

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

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

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

This (pseudo) norm has some nice properties:

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

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

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

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

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

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

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

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

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

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

Next, we observe:

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

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

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

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

on the one hand, and on the other hand,

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

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

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

i.e.

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

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

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

Now define the chain:

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

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

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

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

By the equality above, then,

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

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

1.3. Part 3 (Finishing the proof of Mostow Rigidity

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

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

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

1. Fenchel-Nielsen Coordinates for Teichmuller Space

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1.1. A Symplectic Form on Moduli Space

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

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

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

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

1. Uniformization

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

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

2. Teichmuller and Moduli Space of the Torus

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

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

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

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

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

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

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

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

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

This action is probably familiar to you from complex analysis.

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

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

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

3. Teichmuller Space and Moduli Space for Negatively Curved Surfaces

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

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

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

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

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

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

4. Coordinates for Teichmuller Space

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

4.1. Very Overdetermined (But Easy) Coordinates

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

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

by

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

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

4.2. Dimension Counting

Method 1

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

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

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

Method 2

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

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

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

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

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

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

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

1. Triangles in Hyperbolic Space

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

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

2. Ideal Triangles and Area Theorems

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

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

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

2.1. Area

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

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

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

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

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

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

For this fact, we need another picture:

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

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

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

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

Similarly,

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

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

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

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

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

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

3. Triangles On Spheres

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

A lune.

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

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

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

Solving,

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

4. Gauss-Bonnet

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

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

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

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

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

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

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

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

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

So we have

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

Therefore, summing over all triangles,

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

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

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

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

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

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

4.1. Variants of Gauss-Bonnet

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

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

4.2. Examples

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

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

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 recently uploaded a paper to the arXiv entitled Knots with small rational genus, joint with Cameron Gordon. The genesis of this paper was a couple of nice (and related) talks at Caltech by Matthew Hedden and Jake Rasmussen in 2007. They both talked about potential applications of the theory of knot Floer homology to the Berge conjecture. A Berge knot is a (tame) knot $K$ in the 3-sphere which lies on a genus two Heegaard surface, and with the property that on each side of the Heegaard surface there is a meridian disk that the knot intersects exactly once. Equivalently, the inclusion of the knot into each (closed) handlebody sends the generator of $\pi_1(K)$ to a generator of $\pi_1(\text{handlebody})$. Note that since the 3-sphere admits a unique (up to isotopy) Heegaard splitting of any genus, one may think of such a knot as lying on a specific genus 2 surface in $S^3$. Such knots were classified by Berge; they admit (Dehn) surgeries which result in (nontrivial) Lens spaces. The Berge conjecture is the converse; i.e.:

Berge Conjecture: Let $K$ be a knot in $S^3$ which admits a nontrivial Lens space surgery; i.e. there is a Lens space $L$ and a knot $K'$ in $L$ for which $S^3 - K$ is homeomorphic to $L - K'$. Then $K$ is a Berge knot.

An equivalent formulation (of course) is to try to classify knots in Lens spaces which admit an $S^3$ surgery, i.e. to identify the knots $K'$ as in the formulation of the conjecture above. The equivalent formulation says that these knots should be 1-bridge. The strategy of Hedden-Rasmussen (building on work of Ken Baker and Eli Grigsby) to approach the Berge conjecture depends on characterizing such knots by properties which can be detected by topological invariants that behave well under surgery. An example of such a topological invariant is the Casson invariant $\lambda(\cdot)$, a $\mathbb{Z}$-valued invariant of integer homology spheres which satisfies the surgery formula $\lambda(M_{n+1}) - \lambda(M_n) = \text{Arf}(K)$ where $M_i$ denotes the result of $1/i$ surgery on some integral homology sphere $M$ along a fixed knot $K$, and $\text{Arf}(K)$ is the Arf invariant. For more sophisticated invariants like knot Floer homology, the surgery formula is replaced by an exact triangle. One important piece of topological information that is detected by knot Floer homology is the genus of a knot. The approach to the Berge conjecture thus rests on Ken Baker’s impressive paper showing that small genus knots (in a sense to be made precise) in Lens spaces have small bridge number.

Hedden remarked in his talk that his work, and that of his collaborators “gave the first examples of an infinite family of knots that were characterized by their knot Floer homology”. Though technically true, I think this overstates the role of knot Floer homology in this case, since the knots (1-bridge knots in Lens spaces) are entirely characterized (up to isotopy) by their genus (and therefore by any topological invariant which detects genus). My immediate instinct was to think that knots with small genus in any 3-manifold should always be quite special, and that a complete classification might even be feasible. My paper with Cameron confirms this suspicion, and gives such a classification. Let me admit at this point that I am not especially interested in the Berge conjecture per se, although I find it interesting that new ideas in 3-manifold topology are starting to have something meaningful to say about it. In any case, I shall not have anything else to say about it (meaningful or otherwise) in this post.

First I should say that I have been using the word “genus” in a somewhat sloppy manner. For an oriented knot $K$ in $S^3$, a Seifert surface is a compact oriented embedded surface $\Sigma \subset S^3$ whose boundary is $K$. The genus of such a surface is a non-negative integer, and the least such genus over all Seifert surfaces is (said to be) the genus of $K$, denoted $g(K)$. Such a surface represents the generator in the relative homology group $H_2(S^3, K)$ which equals $H_1(K) = \mathbb{Z}$ since $S^3$ has vanishing homology in dimensions 1 and 2. This relative homology group is dual to $H^1(S^3 - K)$, which is parameterized by homotopy classes of maps from $S^3 - K$ to a circle (which is a $K(\mathbb{Z},1)$). The preimage of a regular value under a smooth map dual to the homology class is a smooth proper surface in $S^3 - K$ whose closure is a Seifert surface. It is immediate that $g(K)=0$ if and only if $K$ is an unknot; in other words, the unknot is “characterized” by its genus. There are infinitely many knots of any positive genus in $S^3$; on the other hand, there are only two fibered genus 1 knots — the trefoil and the figure 8 knot (three if you distinguish the left-handed from the right-handed trefoil), and it is worth remarking (from the point of view of the motivation of characterizing knots by topological invariants) that a theorem of Yi Ni says that fiberedness of knots can be detected by knot Floer homology.

For knots in integral homology $3$-spheres, the situation is very similar: every knot admits a Seifert surface, and the least genus of such a surface is the genus of a knot. The unknot is (always) characterized by the fact that it has genus $0$, but there are infinitely many knots of every positive genus. For a knot $K$ in a general $3$-manifold $M$ it is not so easy to define genus. A necessary and sufficient condition for $K$ to bound an embedded surface in its complement is that $[K]=1$ in $H_1(M)$. However, if $[K]$ has finite order, one can find an open properly embedded surface $\Sigma$ in the complement of $K$ whose “boundary” wraps some number of times around $K$. Technically, let $\Sigma$ be a compact oriented surface, and $f:\Sigma \to M$ a map which restricts to an embedding from the interior of $\Sigma$ into $M-K$, and which restricts to an oriented covering map from $\partial \Sigma$ to $K$ (note that we allow $\Sigma$ to have multiple boundary components). If $p$ is the degree of the covering map $\partial \Sigma \to K$, we call $\Sigma$ a $p$-Seifert surface, and define the rational genus of $\Sigma$ to be $-\chi^-(\Sigma)/2p$, where $\chi$ denotes Euler characteristic, and $\chi^-(\Sigma) = \min(0,\chi(\Sigma))$ (for a connected surface $\Sigma$). The reason to use Euler characteristic instead of genus is that Euler characteristic is multiplicative under coverings (unlike genus), and behaves well with respect to “local” operations on surfaces like cut-and-paste. Moreover, (negative) Euler characteristic, unlike genus, is a good measure of complexity for surfaces with possibly many boundary components. The coefficient of $2$ in the denominator reflects the fact that genus is “almost” $-2$ times Euler characteristic. With this definition, we say that the rational genus of $K$, for any knot $K \subset M$ with $[K]$ of finite order in $H_1(M)$, is the infimum of $-\chi^-(\Sigma)/2p$ over all $p$-Seifert surfaces for $K$ and all $p$. The purpose of our paper is to give a complete classification of knots with sufficiently small rational genus, and to show that such knots are always “geometric” — i.e. they can be isotoped into a normal form which is sensitive to the geometric decomposition of the ambient $3$-manifold $M$. Thus the concept of rational genus makes contact between the homological world of the Thurston norm, knot Floer homology and such invariants, and the geometric world of hyperbolic structures, JSJ decompositions and so on.

It is worth pointing out at this point that knots with small rational genus are not special by virtue of being rare: if $K$ is any knot in $S^3$ (for instance) of genus $g(K)$, and $K'$ in $M$ is obtained by $p/q$ Dehn surgery on $K$, then the knot $K'$ has order $p$ in $H_1(M)$, and $\|K'\| \le (g-1/2)/2p$. Since for “most” coprime $p/q$ the integer $p$ is arbitrarily large, it follows that “most” knots obtained in this way have arbitrarily small rational genus.

There is a precise connection between rational genus and the Thurston norm. There is an exact sequence in homology, which contains the fragment $H_2(M,K) \to H_1(K) \to H_1(M)$. Since $H_1(K) = \mathbb{Z}$, the kernel of $H_1(K) \to H_1(M)$ is generated by some class $n[K]$, and one can define the affine subspace $\partial^{-1}(n[K]) \subset H_2(M,K)$. By excision, we identify $H_2(M,K)$ with $H_2(M-\text{int}(N(K)), \partial N(K))$ where $N(K)$ is a tubular neighborhood of $K$. Under this identification, the rational genus of $K$ is equal to $\inf \|[\Sigma]\|_T/2$ where $\|\cdot\|_T$ denotes the (relative) Thurston norm, and the infimum is taken over classes in $H_2(M-\text{int}(N(K)), \partial N(K))$ in the affine subspace corresponding to $\partial^{-1}(n[K])$. Since the Thurston norm is a convex piecewise rational function, this infimum is realized at some rational point. In other words, rational genus of any knot is rational, and is realized by some $p$-Seifert surface, where $n$ as above divides $p$ (note: if $M$ is a rational homology sphere, then necessarily $p=n$, but if the rank of $H_1(M)$ is positive, this is not necessarily true, and $p/n$ might be arbitrarily large). This relationship to the Thurston norm also gives a straightforward algorithm to compute rational genus, since one can compute Thurston norm e.g. by linear programming in normal surface space relative to any triangulation.

The precise statement of results depends on the geometric decomposition of the ambient manifold $M$. By the geometrization theorem (of Perelman), a closed, orientable $3$-manifold is either reducible (i.e. contains an embedded sphere that does not bound a ball), or is a Lens space, or is hyperbolic, or is a small Seifert fiber space, or is toroidal (i.e. contains an essential ($\pi_1$-injective) embedded torus). For the record, the complete “classification” is as follows:

Reducible Theorem: Let ${K}$ be a knot in a reducible manifold ${M}$. Then either

1. ${\|K\| \ge 1/12}$; or
2. there is a decomposition ${M = M' \# M''}$, ${K \subset M'}$ and either
1. ${M'}$ is irreducible, or
2. ${(M',K) = (\mathbb{RP}^3,\mathbb{RP}^1)\#(\mathbb{RP}^3,\mathbb{RP}^1)}$

Lens Theorem: Let ${K}$ be a knot in a lens space ${M}$. Then either

1. ${\|K\| \ge 1/24}$; or
2. ${K}$ lies on a Heegaard torus in ${M}$; or
3. ${M}$ is of the form ${L(4k,2k-1)}$ and ${K}$ lies on a Klein bottle in ${M}$ as a non-separating orientation-preserving curve.

Hyperbolic Theorem: Let ${K}$ be a knot in a closed hyperbolic ${3}$-manifold ${M}$. Then either

1. ${\|K\| \ge 1/402}$; or
2. ${K}$ is trivial; or
3. ${K}$ is isotopic to a cable of the core of a Margulis tube.

Small SFS Theorem: Let ${M}$ be an atoroidal Seifert fiber space over ${S^2}$ with three exceptional fibers and let ${K}$ be a knot in ${M}$. Then either

1. ${\|K\| \ge 1/402}$; or
2. ${K}$ is trivial; or
3. ${K}$ is a cable of an exceptional Seifert fiber of ${M}$; or
4. ${M}$ is a prism manifold and ${K}$ is a fiber in the Seifert fiber structure of ${M}$ over ${\mathbb{RP}^2}$ with at most one exceptional fiber.

Toroidal Theorem: Let ${M}$ be a closed, irreducible, toroidal 3-manifold, and let ${K}$ be a knot in ${M}$. Then either

1. ${\|K\| \ge 1/402}$; or
2. ${K}$ is trivial; or
3. ${K}$ is contained in a hyperbolic piece ${N}$ of the JSJ decomposition of ${M}$ and is isotopic either to a cable of a core of a Margulis tube or into a component of ${\partial N}$; or
4. ${K}$ is contained in a Seifert fiber piece ${N}$ of the JSJ decomposition of ${M}$ and either
1. ${K}$ is isotopic to an ordinary fiber or a cable of an exceptional fiber or into ${\partial N}$, or
2. ${N}$ contains a copy ${Q}$ of the twisted ${S^1}$ bundle over the Möbius band and ${K}$ is contained in ${Q}$ as a fiber in this bundle structure;
5. or

6. ${M}$ is a ${T^2}$-bundle over ${S^1}$ with Anosov monodromy and ${K}$ is contained in a fiber.

The constant $1/402$ is presumably not optimal, but reflects the coarseness of certain geometric estimates at a particular step in the argument. Broadly speaking, there are two cases to consider: when the knot complement $M-K$ is hyperbolic, and when it is not. The complement $M-K$ is hyperbolic unless it contains an essential subsurface of non-negative Euler characteristic.

The case that $M-K$ is hyperbolic is conceptually easiest to analyze. Let $\Sigma$ be a surface, embedded in $M$ and with boundary wrapping some number of times around $K$, realizing the rational genus of $K$. The complete hyperbolic structure on $M-K$ may be deformed, adding back $K$ as a cone geodesic. Just as a cone can be obtained from a wedge of paper by gluing the two edges together, the geometry of a cone geodesic is locally modeled on the quotient space obtained from a (3-dimensional hyperbolic) wedge by gluing the two flat faces together. The thinner the wedge, the smaller the cone angle along the geodesic. For all sufficiently small angles $\theta > 0$, Thurston proved that there exists a unique hyperbolic metric on $M$ which is singular along a cone geodesic, isotopic to $K$, with cone angle $\theta$. Call this metric space $M_\theta$. The cone angle can be increased, deforming the geometry in a family of spaces, until one of the following three things happens:

1. The cone angle is increased all the way to $2\pi$, resulting in the complete hyperbolic structure on $M$, in which $K$ is isotopic to an embedded geodesic; or
2. The volume of the family of manifolds $M_\theta$ goes to zero (and either converges after rescaling to a Euclidean cone manifold, or converges after rescaling to have fixed diameter and injectivity radius going to zero everywhere); or
3. The cone locus bumps into itself (this can only happen for $\theta > \pi$).

As the cone angle along $K$ increases, so does the length of the cone geodesic. Simultaneously, the diameter of an embedded tube about this diameter decreases. While the diameter of the tube is big, the deformation can continue. Hodgson-Kerckhoff analyzed the kinds of degenerations that can occur, and obtained universal geometric control on how fast the tube diameter can shrink, or the length of the cone geodesic grow. They showed that the cone angle can be increased (giving rise to a family of singular hyperbolic structures $M_\theta$) either until $\theta = 2\pi$, or until the product $\theta \cdot \ell$, where $\ell$ is the length of the cone geodesic, is at least $1.019675$, at which point the diameter of an embedded tube about this cone geodesic is at least $0.531$. Since $\theta < 2\pi$ in the latter case, one obtains a lower bound on both the length of the cone geodesic and the diameter of an embedded tube, independent of $K$ or $M$.

Now, one would like to use this big tube to conclude that $\|K\|$ is large. This is accomplished as follows. Geometrically, one constructs a $1$-form $\alpha$ which agrees with the length form on the cone geodesic, which is supported in the tube, and which satisfies $\|d\alpha\|\le C$ pointwise for some (universal) constant $C$. Then one uses this $1$-form to control the topology of $\Sigma$. By Stokes theorem, for any surface $S$ homotopic to $\Sigma$ in $M-K$ one has an estimate

$1.019675/2\pi \le \ell = \int_K \alpha = \frac {1}{p} \int_S d\alpha \le \frac {C}{p} \text{area}(S)$

In particular, the area of $S$ divided by $p$ can’t be too small. However, it turns out that one can find a surface $S$ as above with $\text{area}(S) \le -2\pi\chi(S)$; such an estimate is enough to obtain a universal lower bound on $\|K\|$. Such a surface $S$ can be constructed either by the shrinkwrapping method of Calegari-Gabai, or the (related) PL-wrapping method of Soma. Roughly speaking, one uses the cone geodesic as an “obstacle”, and finds a surface $S$ of least area homotopic to $\Sigma$ (rel. boundary) subject to the constraint that it cannot cross the geodesic. Away from the cone geodesic, $S$ looks like an ordinary minimal surface. In particular, its intrinsic curvature is no more than the extrinsic curvature of hyperbolic space, which is $-1$ everywhere. Along the geodesic, $S$ looks like a bedsheet hanging on a clothesline; in particular, it does not accumulate any corners or atoms of positive curvature along this singularity, so the Gauss-Bonnet theorem gives the desired bound on $\text{area}(S)$.

This leaves the case that $M-K$ is not hyperbolic to analyze. As remarked above, this only occurs when $M-K$ contains an essential surface (which might be closed or proper) of non-negative Euler characteristic, i.e. a sphere, a disk, an annulus or a torus. In this case, one tries to make the intersection of $\Sigma$ with this essential surface as simple as possible; if one arranges this just right, every intersection contributes a definite amount to the topology of $\Sigma$, and one can conclude either that $\Sigma$ is complicated (in which case $\|K\|$ is large), or that the intersection is simple, and therefore draw some topological conclusion.

To actually do this in practice is quite complicated, but fortunately it relies on (largely combinatorial) methods developed at length by Gabai, Scharlemann, Gordon and others over the last 30 years to analyze (so-called) “exceptional surgeries”. Of course, the argument is still complicated, and this analysis takes up most of the length of the paper. It is also worth pointing out that every case provided for by the classification above actually occurs, with examples of arbitrarily small rational genus.

This paper raises several natural questions, the most obvious of which is whether the explicit (but quite small) constants can be improved in any way. The constant $1/402$ in the statement of the Toroidal Theorem is really only there to take care of a knot sitting inside a hyperbolic piece in the decomposition; a knot that interacts in a meaningful way with an essential torus necessarily has rational genus at least $1/24$ (for a precise statement, see the paper). As remarked above, knots of (ordinary) genus $1$ are very plentiful, even in $S^3$, and do not “see” any of the ambient geometry, so the wildest and most optimistic guess might be that there is a chance of classifying knots of rational genus at most $1/4$. There are some (very weak) reasons to think that this fraction is critical, at least in some cases, not least of which is the papers of Hedden and Ni mentioned above. But in the hyperbolic case, it is probably not easy to get a better estimate using purely geometric arguments.

Another approach might be to try to substitute another conclusion (again in the hyperbolic case) than that $K$ be isotopic to the cable of a core of a Margulis tube. For instance, one might ask for $K$ to admit an insulator family (of the kind Gabai used here), or one might merely ask that $K$ be unknotted in the universal cover, or satisfy some other condition. This goes to the heart of a very, very difficult and important question, namely how to identify geometric features of codimension 2 objects in (especially hyperbolic) geometric 3-manifolds from purely topological properties. If I am optimistic, then I can imagine that this paper makes a contribution, however small, to this ongoing project.

Martin Bridgeman gave a nice talk at Caltech recently on his discovery of a beautiful identity concerning orthospectra of hyperbolic surfaces (and manifolds of higher dimension) with totally geodesic boundary. The $2$-dimensional case is (in my opinion) the most beautiful, and I would like to take a post to explain the identity, and give a derivation which is slightly different from the one Martin gives in his paper. There are many other things one could say about this identity, and its relation to other identities that turn up in the theory of hyperbolic manifolds (and elsewhere); I hope to get to this in a later post.

Let $\Sigma$ be a hyperbolic surface with totally geodesic boundary. An orthogeodesic is a geodesic segment properly immersed in $\Sigma$, which is perpendicular to $\partial \Sigma$ at its endpoints. The set of orthogeodesics is countable, and their lengths are proper. Denote these lengths by $l_i$ (with multiplicity). The identity is:

$\sum_i \mathcal{L}(1/\cosh^2{l_i/2}) = -\pi^2\chi(\Sigma)/2$

where $\mathcal{L}$ is the Rogers’ dilogarithm function (to be defined in a minute). Treating this function as a black box for the moment, the identity has the form $\sum_i L(l_i) =$ a term depending only on the topology of $\Sigma$. The proof is very, very short and elegant. By the Gauss-Bonnet theorem, the term on the right is equal to $1/8$ of the volume of the unit tangent bundle of $\Sigma$. Almost every tangent vector on $\Sigma$ can be exponentiated to a geodesic on $\Sigma$ which intersects the boundary in finite forward and backward time (eg. by ergodicity of the geodesic flow on a closed hyperbolic surface obtained by doubling). If $v$ is such a tangent vector, and $\gamma_v$ is the associated geodesic arc, then $\gamma_v$ is homotopic keeping endpoints on $\partial \Sigma$ to a unique orthogeodesic (which is the unique length minimizer in this relative homotopy class). The volume of the set of $v$ associated to a given orthogeodesic $\alpha$ can be computed as follows. Lift $\alpha$ to the universal cover, where it is the crossbar of a letter “H” whose vertical lines are lifts of the geodesics it ends on. Any $\gamma_v$ lifts to a unique geodesic segment in the universal cover with endpoints on the edges of the H. So the volume of the set of such $v$ depends only on $\text{length}(\alpha)$, giving rise to the explicit formula for $L$. qed.

That’s it — that’s the whole proof! . . . modulo some calculations, which we now discuss.

The “ordinary” polylogarithms $\text{Li}_k$ are defined by Taylor series

$\text{Li}_k(z) = \sum_{n=1}^\infty \frac {z^n} {n^k}$

which converges for $|z|<1$, and extends by analytic continuation. Taking derivatives, one sees that they satisfy $\text{Li}_k'(z) = \text{Li}_{k-1}(z)/z$, thereby giving rising to integral formulae. $\text{Li}_0(z)$ is the familiar geometric series $z/(1-z)$, so $\text{Li}_1(z) = -\log(1-z)$ and

$\text{Li}_2(z) = -\int \frac {\log(1-z)} {z} dz$

The Rogers dilogarithm is then given by the formula $\mathcal{L}(z) = \text{Li}_2(z) + \frac 1 2 \log(|z|)\log(1-z)$ for real $z<1$. One sees that the Rogers dilogarithm is obtained by symmetrizing the integrand for the integral expression for $\text{Li}_2$ under the involution $z \to 1-z$:

$\mathcal{L}'(z) = -\frac {1}{2} \left(\frac {\log(1-z)}{z} + \frac {\log(z)}{1-z} \right)$

Martin derives his identity by direct calculation, but in fact this calculation can be simplified a bit by some hyperbolic geometry. Consider an ideal quadrilateral $Q$ (whose unit tangent bundle has area $4\pi^2$) with one pair of opposite sides that are distance $l$ apart. Join opposite vertices in pairs to decompose the quadrilateral into four triangles, each with one non-ideal point:

In the (schematic) picture, suppose the two edges of the H are the left and right side (call them $L$ and $R$) and the other two edges are $U$ and $D$. Similarly, call the four triangles $T_L, T_R, T_U, T_D$ depending on which edge of the quadrilateral they bound. The triangle $T_R$ is colored gray in the figure. We secretly identify this figure with the upper half-plane, in such a way that the ideal vertices are (in circular order) $0,x,1,\infty$, where $\infty,0$ are the ideal vertices of the gray triangle. Call $\alpha$ the (hyperbolic) angle of the gray triangle at its vertex, so $x = (1+\cos(\alpha))/2$. Moreover, it turns out that $x = 1/\cosh^2(l/2)$ where $l$ is the distance between $L$ and $R$. We will compute $L$ implicitly as a function of $x$, and show that it is a multiple of the Rogers dilogarithm function, thus verifying Bridgeman’s identity.

Every vector $v$ in $Q$ exponentiates to a (bi-infinite) geodesic $\gamma_v$, and we want to compute the volume of the set of vectors $v$ for which the corresponding geodesic intersects both $L$ and $R$. The point of the decomposition is that for $v$ in $T_L$ (say), the geodesic $\gamma_v$ intersects $L$ whenever it intersects $R$, so we only need to compute the volume of the $v$ in $T_L$ for which $\gamma_v$ intersects $R$. Similarly, we only need to compute the volume of the $v$ in $T_R$ for which $\gamma_v$ intersects $L$. For $v$ in $T_U$, we compute the volume of the $v$ which do not intersect $U$ (since these are exactly the ones that intersect both $L$ and $R$), and similarly for $T_D$.

These volumes can be expressed in terms of integrals of harmonic functions. Let $\chi_L$ denote the harmonic function on the disk which is $1$ on the arc of the circle bounded by $L$, and $0$ on the rest of the circle. This function at each point is equal to $1/2\pi$ times the visual angle (i.e. the length in the unit tangent circle) subtended by the given arc of the circle, as seen from the given point in the hyperbolic plane. Define $\chi_R,\chi_U,\chi_D$ similarly. Then the total volume we need to compute is equal to

$4\pi \left( (\int_{T_L} 2\chi_R) + (\int_{T_U} 1 - 2\chi_U) \right)$

(here we have identified $\int_{T_L} \chi_R = \int_{T_R} \chi_L$ by symmetry, and similarly for the other pair of terms). Let us approach this a bit more systematically. If $\alpha$ denotes the angle at the nonideal vertex of triangle $T_R$, we denote $\int_{T_R} \chi_R = A(\alpha)$, $\int_{T_R} \chi_U = B(\alpha)$ and $\int_{T_R} \chi_L = C(\alpha)$. The integral we want to evaluate can be expressed easily in terms of explicit rational multiples of $\pi$, and the function $A,B,C$. These functions satisfy obvious identities:

$C(\alpha) = \int_{T_R} 1 - A(\alpha) - 2B(\alpha) = \pi-\alpha - A(\alpha) - 2B(\alpha)$

and

$A(\alpha) + B(\pi - \alpha) = \pi/3$

where the last identity comes by observing that we are integrating a certain function over an ideal triangle, and observing that the average of this function under the symmetries of the ideal triangle is equal to the constant function $1/3$. In particular, we see that we can express everything in terms of $A$. After some elementary reorganization, we see that the contribution $V(\alpha)$ to the volume of the unit tangent bundle of the surface associated to this particular orthogeodesic is

$V(\alpha) = \pi^2(8 - 16/3) - 4\pi\alpha - 8\pi(A(\alpha) - A(\pi - \alpha))$

To compute $A(\alpha)$, it makes sense to move to the upper half-space model, and move the endpoints of the interval to $0$ and $\infty$. The harmonic function is equal to $1$ on the negative real axis, and $0$ on the positive real axis. It takes the value $\theta/\pi$ on the line $\text{arg}(z) = \theta$. The area form in the hyperbolic metric is proportional to the Euclidean area form, with constant $1/\text{Im}(z)^2$. In other words, we want to integrate $\text{arg}(z)/\pi\text{Im}(z)^2$ over the region indicated in the figure, where the nonideal angle is $\alpha$, and the base point is $0$:

If we normalize so that the circular arc is part of the semicircle from $0$ to $1$, then the real projection of the vertical lines in the figure are $0$ and $x$. There is no elementary way to evaluate this integral, so instead we evaluate its derivative as a function of $x$ where as before, $x = (1+\cos(\alpha))/2$. This is the definite integral

$A'(x) = \int_{y = \sqrt{x-x^2}}^\infty (\tan^{-1}(y/x)/\pi y^2) dy$

Integrating by parts gives $(\alpha/\pi\sin{\alpha}) + 1/\pi \int_{y = \sqrt{x-x^2}}^\infty xdy/y(y^2+x^2)$. This evaluates to

$A'(x) = (\alpha/\pi\sin{\alpha}) - 1/\pi ( \log(1-x)/2x)$

Thinking of $V(\alpha)$ as a function of $x$, we get

$V'(x) = -4\pi d\alpha/dx - 8\pi(A'(x) + A'(1-x)) = 8\mathcal{L}'(x)$

Comparing values at $x=0$ we see that $V=8\mathcal{L}$ and the identity is proved.

Well, OK, this is not terribly simple, but a posteriori it gives a way to express the Rogers dilogarithm as a sum of integrals of very simple harmonic functions over hyperbolic triangles, which is a nice geometric way to think of it.

(Update 10/30): This paper by Dupont and Sah relates Rogers dilogarithm to volumes of $\text{SL}(2,\mathbb{R})$-simplices, and discusses some interesting connections to conformal field theory and lattice model calculations. I feel like a bit of a dope, since I read this paper while I was in graduate school more than a dozen years ago, but forgot all about it until I was cleaning out my filing cabinet this morning. They cite an older paper of Dupont for the explicit calculations; these are somewhat tedious and unenlightening; however, he does manage to show that the Rogers dilogarithm is characterized by the Abel identity. In other words,

Lemma A.1 (Dupont): Let $f:(0,1) \to \mathbb{R}$ be a three times differentiable function satisfying

$f(s_1) - f(s_2) + f(\frac{s_2}{s_1}) - f(\frac{1-s_1^{-1}}{1-s_2^{-1}}) + f(\frac{1-s_1}{1-s_2})=0$

for all $0 < s_2 < s_1 < 1$. Then there is a real constant $\kappa$ such that $f(x) = \kappa L(x)$ where $L(x)$ is the Rogers dilogarithm (up to an additive constant).

Nevertheless, they don’t seem to have noticed the formula in terms of integrals of harmonic functions over hyperbolic triangles. Perhaps this is also well-known. Do any readers know?

An amenable group $G$ acting by homeomorphisms on a compact topological space $X$ preserves a probability measure on $X$; in fact, one can given a definition of amenability in such terms. For example, if $G$ is finite, it preserves an atomic measure supported on any orbit. If $G = \mathbb{Z}$, one can take a sequence of almost invariant probability measures, supported on the subset $[-n,n] \cdot p$ (where $p \in X$ is arbitrary), and any weak limit will be invariant. For a general amenable group, in place of the subsets $[-n,n] \subset \mathbb{Z}$, one works with a sequence of Folner sets; i.e. subsets with the property that the ratio of their size to the size of their boundary goes to zero (so to speak).

But if $G$ is not amenable, it is generally not true that there is any probability measure on $X$ invariant under the action of $G$. The best one can expect is a probability measure which is invariant on average. Such a measure is called a harmonic measure (or a stationary measure) for the $G$-action on $X$. To be concrete, suppose $G$ is finitely generated by a symmetric generating set $S$ (symmetric here means that if $s \in S$, then $s^{-1} \in S$). Let $M(X)$ denote the space of probability measures on $X$. One can form an operator $\Delta:M(X) \to M(X)$ defined by the formula

$\Delta(\mu) = \frac {1} {|S|} \sum_{s \in S} s_*\mu$

and then look for a probability measure $\nu$ stationary under $\Delta$, which exists for quite general reasons. This measure $\nu$ is the harmonic measure: the expectation of the $\nu$-measure of $s(A)$ under a randomly chosen $s \in S$ is equal to the $\nu$-measure of $A$. Note for any probability measure $\mu$ that $s_*\mu$ is absolutely continuous with respect to $\Delta(\mu)$; in fact, the Radon-Nikodym derivative satisfies $ds_*\mu/d\Delta(\mu) \le |S|$. Substituting $\nu$ for $\mu$ in this formula, one sees that the measure class of $\nu$ is preserved by $G$, and that for every $g \in G$, we have $dg_*\nu/d\nu \le |S|^{|g|}$, where $|g|$ denotes word length with respect to the given generating set.

The existence of harmonic measure is especially useful when $X$ is one-dimensional, e.g. in the case that $X=S^1$. In one dimension, a measure (at least one of full support without atoms) can be “integrated” to a path metric. Consequently, any finitely generated group of homeomorphisms of the circle is conjugate to a group of bilipschitz homeomorphisms (if the harmonic measure associated to the original action does not have full support, or has atoms, one can “throw in” another random generator to the group; the resulting action can be assumed to have a harmonic measure of full support without atoms, which can be integrated to give a structure with respect to which the group action is bilipschitz). In fact, Deroin-Kleptsyn-Navas showed that any countable group of homeomorphisms of the circle (or interval) is conjugate to a group of bilipschitz homeomorphisms (the hypothesis that $G$ be countable is essential; for example, the group $\mathbb{Z}^{\mathbb{Z}}$ acts in a non-bilipschitz way on the interval — see here).

Suppose now that $G = \pi_1(M)$ for some manifold $M$. The action of $G$ on $S^1$ determines a foliated circle bundle $S^1 \to E \to M$; i.e. a circle bundle, together with a codimension one foliation transverse to the circle fibers. To see this, first form the product $\widetilde{M} \times S^1$ with its product foliation by leaves $\widetilde{M} \times \text{point}$, where $\widetilde{M}$ denotes the universal cover of $M$. The group $G = \pi_1(M)$ acts on $\widetilde{M}$ as the deck group of the covering, and on $S^1$ by the given action; the quotient of this diagonal action on the product is the desired circle bundle $E$. The foliation makes $E$ into a “flat” circle bundle with structure group $\text{Homeo}^+(S^1)$. The foliation allows us to associate to each path $\gamma$ in $M$ a homeomorphism from the fiber over $\gamma(0)$ to the fiber over $\gamma(1)$; integrability (or flatness) implies that this homeomorphism only depends on the relative homotopy class of $\gamma$ in $M$. This identification of fibers is called the holonomy of the foliation along the path $\gamma$. If $M$ is a Riemannian manifold, there is another kind of harmonic measure on the circle bundle; in other words, a probability measure on each circle with the property that the holonomy associated to an infinitesimal random walk on $M$ preserves the expected value of the measure. This is (very closely related to) a special case of a construction due to Lucy Garnett which associates a harmonic transverse measure to any foliation $\mathcal{F}$ of a manifold $N$, by finding a fixed point of the leafwise heat flow on the space of probability measures on $N$, and disintegrating this measure into the product of the leafwise area measure, and a “harmonic” transverse measure.

In any case, we normalize our foliated circle bundle so that each circle has length $2\pi$ in its harmonic measure. Let $X$ be the vector field on the circle bundle that rotates each circle at unit speed, and let $\alpha$ be the $1$-form on $E$ whose kernel is tangent to the leaves of the foliation. We scale $\alpha$ so that $\alpha(X)=1$ everywhere. The integrability condition for a foliation is expressed in terms of the $1$-form as the identity $\alpha \wedge d\alpha = 0$, and we can write $d\alpha = -\beta \wedge \alpha$ where $\beta(X)=0$. More intrinsically, $\beta$ descends to a $1$-form on the leaves of the foliation which measures the logarithm of the rate at which the transverse measure expands under holonomy in a given direction (the leafwise form $\beta$ is sometimes called the Godbillon class, since it is “half” of the Godbillon-Vey class associated to a codimension one foliation; see e.g. Candel-Conlon volume 2, Chapter 7). Identifying the universal cover of each leaf with $\widetilde{M}$ by projection, the fact that our measure is harmonic means that $\beta$ “is” the gradient of the logarithm of a positive harmonic function on $\widetilde{M}$. As observed by Thurston, the geometry of $M$ then puts constraints on the size of $\beta$. The following discussion is taken largely from Thurston’s paper “Three-manifolds, foliations and circles II” (unfortunately this mostly unwritten paper is not publicly available; some details can be found in my foliations book, example 4.6).

An orthogonal connection on $E$ can be obtained by averaging $\alpha$ under the flow of $X$; i.e. if $\phi_t$ is the diffeomorphism of $E$ which rotates each circle through angle $t$, then

$\omega = \frac {1} {2\pi} \int_{0}^{2\pi} \phi_t^* \alpha$

is an $X$-invariant $1$-form on $E$, which therefore descends to a $1$-form on $M$, which can be thought of as a connection form for an $\text{SO}(2)$-structure on the bundle $E$. The curvature of the connection (in the usual sense) is the $2$-form $d\omega$, and we have a formula

$d\omega = \frac {1} {2\pi} \int_{0}^{2\pi} \phi_t^*(d\alpha) = \frac {1} {2\pi} \int_{0}^{2\pi} \phi_t^*(-\beta \wedge \alpha)$

The action of the $1$-parameter group $\phi_t$ trivializes the cotangent bundle to $E$ over each fiber. After choosing such a trivialization, we can think of the values of $\alpha$ at each point on a fiber as sweeping out a circle $\gamma$ in a fixed vector space $V$. The tangent to this circle is found by taking the Lie derivative

$\mathcal{L}_X(\alpha) = \iota_X d\alpha + d\iota_X \alpha = \alpha(X)\beta = \beta$

In other words, $\beta$ is identified with $d\gamma$ under the identification of $\alpha$ with $\gamma$, and $\int \phi_t^*(-\beta \wedge \alpha) = \int \gamma \wedge d\gamma$; i.e. the absolute value of the curvature of the connection is equal to $1/\pi$ times the area enclosed by $\gamma$.

Now suppose $M$ is a hyperbolic $n$-manifold, i.e. a manifold of dimension $n$ with constant curvature $-1$ everywhere. Equivalently, think of $M$ as a quotient of hyperbolic space $\mathbb{H}^n$ by a discrete group of isometries. A positive harmonic function on $\mathbb{H}^n$ has a logarithmic derivative which is bounded pointwise by $(n-1)$; identifying positive harmonic functions on hyperbolic space with distributions on the sphere at infinity, one sees that the  “worst case” is the harmonic extension of an atomic measure concentrated at a single point at infinity, since every other positive harmonic function is the weighted average of such examples. As one moves towards or away from a blob at infinity concentrated near this point, the radius of the blob expands like $e^t$; since the sphere at infinity has dimension $n-1$, the conclusion follows. But this means that the speed of $\gamma$ (i.e. the size of $d\gamma$) is pointwise bounded by $(n-1)$, and the length of the $\gamma$ circle is at most $2\pi(n-1)$. A circle of length $2\pi(n-1)$ can enclose a disk of area at most $\pi (n-1)^2$, so the curvature of the connection has absolute value pointwise bounded by $(n-1)^2$.

One corollary is a new proof of the Milnor-Wood inequality, which says that a foliated circle bundle $E$ over a closed oriented surface $S$ of genus at least $2$ satisfies $|e(E)| \le -\chi(S)$, where $e(E)$ is the Euler number of the bundle (a topological invariant). For, the surface $S$ can be given a hyperbolic metric, and the bundle a harmonic connection whose average is an orthogonal connection with pointwise curvature of absolute value at most $1$. The Euler class of the bundle evaluated on the fundamental class of $S$ is the Euler number $e(E)$; we have

$|e(E)| = \frac {1} {2\pi} |\int_S \omega| \le \text{area}(S)/2\pi = -\chi(S)$

where the first equality is the Chern-Weil formula for the Euler class of a bundle in terms of the curvature of a connection, and the last equality is the Gauss-Bonnet theorem for a hyperbolic surface. Another corollary gives lower bounds on the area of an incompressible surface in a hyperbolic manifold. Suppose $S \to M$ is an immersion which is injective on $\pi_1$. There is a cover $\widehat{M}$ of $M$ for which the immersion lifts to a homotopy equivalence, and we get an action of $\pi_1(\widehat{M})$ on the circle at infinity of $S$, and hence a foliated circle bundle as above with $e(E) = -\chi(S)$. Integrating as above over the image of $S$ in $\widehat{M}$, and using the fact that the curvature of $\omega$ is pointwise bounded by $(n-1)^2$, we deduce that the area of $S$ is at least $-2\pi\chi(S)/(n-1)^2$. If $M$ is a $3$-manifold, we obtain $\text{area}(S) \ge -2\pi\chi(S)/4$.

(A somewhat more subtle argument allows one to get better bounds, e.g. replacing $4$ by $(\pi/2)^2$ for $n=3$, and better estimates for higher $n$.)