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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.