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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 1If is a homotopy equivalence of closed hyperbolic manifolds with , then 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) is a *quasi-isometry* if for all , we have

Without the term, would be called *bilipschitz*.

First, we observe that if is a homotopy equivalence, then lifts to a map in the sense that is equivariant with respect to (thought of as the desk groups of and , so for all , we have .

Now suppose that and are hyperbolic. Then we can lift the Riemannian metric to the covers, so and are specific discrete subgroups in , and maps equivariantly with respect to and .

Lemma 2is a quasi-isometry.

*Proof:* Since is a homotopy equivalence, there is a such that . Perturbing slightly, we may assume that and are smooth, and as and are compact, there exists a constant such that and . In other words, paths in and are stretched by a factor of at most : for any path , . The same is true for going in the other direction, and because we can lift the metric, the same is true for the universal covers: for any path , , and similarly for .

Thus, for any in the universal cover ,

and

We see, then, that is Lipschitz in one direction. We only need the for the other side.

Since , we lift it to get an equivariant lift For any point , the homotopy between gives a path between and . Since this is a lift of the homotopy downstairs, this path must have bounded length, which we will call . Thus,

Putting these facts together, for any in ,

And

By the triangle inequality,

This is the left half of the quasi-isometry definition, so we have shown that is a quasi-isometry.

Notice that the above proof didn’t use anything hyperbolic—all we needed was that and 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 3A quasi-isometry extends to a continuous map on the boundary .

The basic idea is that given a geodesic, it maps under 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 4Take a geodesic and two points and a distance apart on it. Draw two perpendicular geodesic segments of length from and . Draw a line between the endpoints of these segments such that has constant distance from the geodesic. Then the length of is linear in and exponential in .

*Proof:* Here is a representative picture:

So we see that . By Gauss-Bonnet,

Where the on the left is the sum of the turning angles, and is the geodesic curvature of the segment . What is this geodesic curvature ? If we imagine increasing , then the derivative of the length with respect to is the geodesic curvature times the length , i.e.

So . Therefore, by the Gauss-Bonnet equality,

so . Therefore, , which proves the lemma

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

Lemma 5If is a quasi-isometry, there is a constant depending only on and such that for all on the geodesic from to in , is distance less than from any geodesic from to .

*Proof:* Fix some , and suppose the image of the geodesic from to goes outside a neighborhood of the geodesic from to . That is, there is some segment on between the points and such that maps completely outside the neighborhood.

Let’s look at the nearest point projection from to . By the above lemma, . Thus means that

On the other hand, because is a quasi-isometry,

and

So we have

Which implies that

That is, the length of the offending path is uniformly bounded. Thus, increase by times this length plus , and every offending path will now be inside the new neighborhood of .

The last lemma says that the image under of a geodesic segment is uniformly close to an actual geodesic. Now suppose that we have an infinite geodesic in . 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 to a map , where is the diagonal . 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 extends continuously to the boundary of hyperbolic space. We will end up showing that is conformal, which will give us the theorem.

We now introduce the Gromov norm. if is a topological space, then singular chain complex is a real vector space with basis the continuous maps . We define a norm on as the norm:

This defines a pseudonorm (the Gromov norm) on by:

This (pseudo) norm has some nice properties:

Lemma 6If is continuous, and , then .

*Proof:* If represents , then represents .

Thus, we see that if is a homotopy equivalence, then .

If is a closed orientable manifold, then we define the Gromov norm of to be the Gromov norm .

Here is an example: if admits a self map of degree , then . This is because we can let represent , so , so represents . Thus . Notice that we can repeat the composition with to get that is as small as we’d like, so it must be zero.

Theorem 7 (Gromov)Let be a closed oriented hyperbolic -manifold. Then . Where is a constant depending only on .

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

Lemma 8There is a map (the second complex is totally geodesic simplices) which is -equivariant and – equivariantly homotopic to .

*Proof:* In the hyperboloid model, we imagine a simplex mapping in to . In , 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 ; now project this homotopy to .

Now, if represents , then we can straighten the simplices, so represents , and , so when finding the Gromov norm it suffices to consider geodesic simplices. Notice that every point has finitely many preimages, and total degree is 1, so for any point , .

Next, we observe:

Lemma 9If given a chain , there is a collection such that and is a cycle homologous to .

*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.

By the lemma, there is an integral cycle , where is some constant. We create a simplicial complex by gluing these simplices together, and this complex comes together with a map to . Make it smooth. Now by the fact above, , so . Then

on the one hand, and on the other hand,

The volume on the right is at most , the volume of an ideal simplex, so we have that

i.e.

This gives the lower bound in the theorem. To get an upper bound, we need to exhibit a chain representing with all the simplices mapping with degree 1, such that the volume of each image simplex is at least .

We now go through the construction of this chain. Set , and fix a fundamental domain for , so is tiled by translates of . Let be the set of all simplices with side lengths with vertices in a particular -tuple of fundamental domains . Pick to be a geodesic simplex with vertices , and let be the image of under the projection. This only depends on up to the deck group of .

Now define the chain:

With the to make it orientation-preserving, and where is an -invariant measure on the space of regular simplices of side length . If the diameter of is every simplex with has edge length in , so:

- The volume of each simplex is if is large enough.
- is finite — fix a fundamental domain; then there are only finitely many other fundamental domains in .

Therefore, we just need to know that is a cycle representing : 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,

Taking to zero, we get the theorem.

** 1.3. Part 3 (Finishing the proof of Mostow Rigidity **

We know that for all , there is a cycle representing such that every simplex is geodesic with side lengths in , and the simplices are almost equi-distributed. Now, if , and represents , then represents , as is a homotopy equivalence.

We know that extends to a map . Suppose that there is an tuple in which is the vertices of an ideal regular simplex. The map takes (almost) regular simplices arbitrarily close to this regular ideal simplex to other almost regular simplices close to an ideal regular simplex. That is, 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 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 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 is 3-dimensional.

Now, as 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 to is an isometry of , so 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 surface with euler characteristic 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 “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 curves, plus other coordinates, which record the “twisting” of each gluing curve.

Now, given a choice of disjoint simple closed surves , we associate to the family of geodesics in in the homotopy classes of the . In each class, there is a unique geodesic, but how do we know the geodesics in are pairwise disjoint?

Lemma 1Suppose is a family of pairwise disjoint simple closed curves in a hyperbolic surface , and are the (unique) geodesic representatives in the homotopy classes of the .

- The geodesics in are pairwise disjoint simple closed curves.
- As a family, the are ambient isotopic to .

*Proof:* Consider a loop and its geodesic representative . Suppose that intersects itself. Now and 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 and . Because the annulus bounding and lifts to the universal cover, there are two lifts and of which are uniformly close to and . We therefore find that and intersect, which means that intersects itself, which is a contradiction. The same idea shows that the geodesic representatives 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 and .

With the lemma, we see that to a point in Teichmuller space we get pairwise disjoint simple closed geodesics, which gives us 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 2There exists a unique hyperbolic pair of pants with cuff lengths , for any . 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 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 in these isotopy classes. Since the are fixed by the involution, they must intersect the at right angles. If we cut along the to get (two copies of) our original pair of pants, we have found that there is a unique triple of geodesics which meet the boundaries at right angles:

Cutting along the , 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 . Since there is a unique way to glue up the hexagons to obtain our original pair of pants, there is a unique hyperbolic pair of pants with specified edge lengths.

Lemma 3There is a unique hyperbolic right-angled hexagon with alternating edge lengths .

*Proof:* Pick some geodesic and some point 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 at right angles from . Call the other end of this segment . There is a unique geodesic passing through at right angles to the segment. Pick some point on at length from (we will be varying ). From there is a unique geodesic segment of length at right angles to ; call its endpoint . There is a unique geodesic through at right angles to this segment. Now, there is a unique geodesic segment at right angles to and . Of course, the length of this segment depends on .

If we make large, then becomes large, and there is some positive such that goes to . Therefore, there is a unique length making . We have now determined the hexagon, and, up to isometry, all of our choices were forced, so there is only one.

Since there is a unique hyperbolic pair of pants with specified cuff lengths, when we cut our surface of interest up into pairs of pants, we get a map which takes a point to the lengths of the curves cutting 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 curves together, we have to decide whether to twist them at all before gluing. Up to isometry, there are ways to glue these curves together (all the angles). However, in (marked) Teichmuller space, there are ways to glue it up. Draw another curve (this is not the same as the before). The marking on lets us observe what happens to under , and we can see that twisting the pairs of pants around results in nontrivial movement in Teichmuller space.

The twist above results in the following new curve:

The length of determines how twisted the gluing is, since twisting requires increasing its length. That is, given the image of , 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 curves in , we must pick another curves; one simple way is to declare that 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

Here is an example of a choice of and curves. The 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 and are not well-defined on Moduli space, but their derivatives *are*: define the 2 form on Teichmuller space

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 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 . If , 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 and 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 , and, by composing the embedding with euclidean similarities, we may assume that the bottom left corner is at and the bottom right is . The parallelogram is therefore determined by where the upper left hand corner is: some complex number with . 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 } { }

This set is called the *Teichmuller space* of the torus. We don’t really care about the loops and , 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

Where 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 , think of the torus as the unit square in spanned by the standard unit basis vectors. Then a homeomorphism of 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 (not because we want orientation-preserving) preserving the integer lattice, which we care about up to scale, otherwise known as .

Using the bijection above, the mapping class group of the torus acts on , and this action is

This action is probably familiar to you from complex analysis.

In summary, the Teichmuller space of the torus is (can be represented as) , and the mapping class group 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

Which is glued to itself by a flip in the axis. The resulting Moduli space is an orbifold: one point is ideal and goes off to infinity, one point looks locally like quotiented by a rotation of , and the other point looks like quotiented by a rotation of .

**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 . Topological meaning we do not associate with it a metric. We want to classify the hyperbolic metrics we could give to . Define Teichmuller space to be the set of equivalence classes of pairs where is a hyperbolic surface and 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: iff there exists an isometry such that is homotopic to .

Define the Moduli space of to be isometry classes of surfaces which are homotopy equivalent to . There is an obvious map defined by mapping , and this map respects the equivalence relations, because if , then is isometric to (since it is isometric by an isometry commuting with and ).

As with the torus, define the mapping class group to be the group of homotopy equivalences of with itself, up to homotopy. Then acts on by . The quotient of by this action is : clearly we never identify surfaces which are not isometric, and if is an isometry, and , are points in Teichmuller space with any , then notice has an inverse (up to homotopy), so if we act on by , we get , which is the same point in as . We are abusing notation here, because we are thinking of , and as the same surface (which they are, topologically). The point is that by acting by we can rearrange so that after mapping by we are homotopic to . The result of this is that

A priori, we are interested in hyperbolic metrics on 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 ( for a genus 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 (this is a conjugacy class in ), and we’ll represent this class by the loop . Given a point , there is a unique geodesic representative in the free homotopy class of the loop . Define to be the length of this representative. Let be the set of conjugacy classes in . Then we have defined a map

by

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 . Suppose that has genus . We can obtain by gluing the edges of a -gon in pairs (going counterclockwise, the labels read , , , , …, , , , ). Since we will be given 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

- The paired sides must have equal length.
- The corner angles must add to .

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 -gon needs side lengths and angles. For our -gon, then, we need to specify side lengths and angles. This is dimensions. However, we have pairs, each of which gives a constraint, plus our single constraint about the angle sum. This reduces our dimension to . 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 dimensions.

**Method 2**

A marked hyperbolic structure on gives a -equivariant isometry . That is, an element of is , which tells us how to map isomorphically onto , which is the same as the deck group of the universal cover , which is . Therefore, to an element of is associated a discrete faithful representation of to , the group of isometries of , 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 is therefore the dimension of the space of representations of in up to conjugacy.

The fundamental group of 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:

So is the subset of such that (here is the free group on 2 generators, which is what we get if we forget the single relation). Now a representation in is completely free: we can send the generators anywhere we want, so

Since is 3-dimensional, the right hand side is a real manifold of dimension . Insisting that map to 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 will be dimensional. However, we are interested in representations up to conjugacy, so this removes another 3 dimensions, giving us the same dimension estimate for as 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 . 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 which contains these three points plus the origin. Intersecting this subspace with the hyperboloid gives a copy of , so we only have to check there is a unique triangle in . 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 ), 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 , and the euler characteristic is . The sum of the exterior angles is just slightly under , so using Gauss-Bonnet, the area is very close to , and goes to 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 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 by .

** 2.1. Area **

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

Theorem 1 (Gauss)

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 and drawn the ideal triangle containing these geodesics. Since the angles are what they look like, we know that the area of is the area of the ideal triangle (), minus the sum of the areas of the smaller triangles with two points at infinity:

Thus it suffices to show that .

For this fact, we need another picture:

Define . 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:

We also know some boundary conditions on : we know (this is a degenerate triangle) and (this is an ideal triangle). We therefore conclude that

Similarly,

And we can find by observing that

Similarly, if we know , then

And by subtracting , we find that . By induction, then, if is a dyadic rational times . This is a dense set, so we know for all 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 has area , the lune with angle has area . Now consider the following picture:

Notice that each corner of the triangle gives us two lunes (the lunes for 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:

Solving,

**4. Gauss-Bonnet **

If we encouter a triangle of constant curvature , then we can scale the problem to one of the two formulas we just computed, so

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) . The surface need not have constant curvature. Suppose for the time being it has no boundary. Triangulate it with very small triangles such that and . Then since the deviation between the curvature and the curvature at the midpoint is times the distance from the midpoint,

For each triangle , we can form a comparison triangle with the same edge lengths and constant curvature . Using the formula from the beginning of this section, we can rewrite the right hand side of the formula above, so

Now since the curvature deviates by times the distance from the midpoint, the angles in deviate from those in just slightly:

So we have

Therefore, summing over all triangles,

The right hand side is just the total angle sum. Since the angle sum around each vertex in the triangulation is ,

Where is the number of vertices, and is the number of triangles. The number of edges, , can be calculated from the number of triangles, since there are edges for each triangle, and they are each double counted, so . Rewriting the equation,

Taking the mesh size to zero, we get the Gauss-Bonnet theorem .

** 4.1. Variants of Gauss-Bonnet **

- If 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 has geodesic boundary with corners, thenWhere the turning angle is the angle you would turn tracing the shape from the outside. That is, it is , where is the interior angle.
- Most generally, if 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 ():

** 4.2. Examples **

- The Euler characteristic of the round disk in the plane is , and the disk has zero curvature, so . The geodesic curvature is constant, and the circumference is , so , so .
- A polygon in the plane has no curvature nor geodesic curvature, so .

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.

## Hyperbolic Geometry (157b) Notes #1

April 8, 2010 in Commentary, Euclidean Geometry, Groups, Hyperbolic geometry, Lie groups, Overview, Visualization | by aldenwalker | 5 comments

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. ModelsWe have a very good natural geometric understanding of , 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 . Let us look at some examples of 2-manifolds.

The Tractrix

The surface of revolution about the -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 , namely rotations about the -axis. However, there are lots of isometries which do not extend, so this embeddeding does not serve us all that well.

2. 1-Dimensional Models for Hyperbolic SpaceWhile 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 ModelParameterizingConsider the quadratic form on defined by , where . This doesn’t give a norm, since is not positive definite, but we can still ask for the set of points with . This is (both sheets of) the hyperbola . Let be the upper sheet of the hyperbola. This will be 1-dimensional hyperbolic space.

For any matrix , let . That is, matrices which preserve the form given by . The condition is equivalent to requiring that . Notice that if we let be the identity matrix, we would get the regular orthogonal group. We define , where has positive eigenvalues and negative eigenvalues. Thus . We similarly define to be matricies of determinant 1 preserving , and to be the connected component of the identity. 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 . Consider the matrix . Writing down the equations and gives us four equations, which we can solve to get the solutions

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 , so we have (recall that ).

These matrices take to . In other words, acts transitively on with trivial stabilizers, and in particular we have parmeterizing maps

The first map is actually a Lie group isomorphism (with the group action on being ) in addition to a diffeomorphism, since

MetricAs mentioned above, is not positive definite, but its restriction to the tangent space of is. We can see this in the following way: tangent vectors at a point are characterized by the form . Specifically, , since (by a calculation) . Therefore, 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 . Thus, is a Riemannian manifold, and acts by isometries.

Let’s use the parameterization . The unit (in the metric) tangent at is . The distance between the points and is

In other words, is an isometry from to .

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 with is shown.

2.2. Projective ModelParameterizingReal projective space is the set of lines through the origin in . We can think about as , where is associated with the line (point in ) intersecting in , and is the horizontal line. There is a natural projection by projecting a point to the line it is on. Under this projection, maps to .

Since acts on preserving the lines , it gives a projective action on fixing the points . Now suppose we have any projective linear isomorphism of fixing . The isomorphism is represented by a matrix with eigenvectors . Since scaling preserves its projective class, we may assume it has determinant 1. Its eigenvalues are thus and . The determinant equation, plus the fact that

Implies that is of the form of a matrix in . Therefore, the projective linear structure on is the “same” (has the same isometry (isomorphism) group) as the hyperbolic (Riemannian) structure on .

MetricClearly, we’re going to use the pushforward metric under the projection of to , but it turns out that this metric is a natural choice for other reasons, and it has a nice expression.

The map taking to is . The hyperbolic distance between and in is then (by the fact from the previous sections that is an isometry).

Recall the fact that . Applying this, we get the nice form

We also recall the cross ratio, for which we fix notation as . Then

Call the numerator of that fraction by and the denominator by . Then, recalling that , we have

Therefore, .

3. Hilbert MetricNotice 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 , we can define the Hilbert metric on by setting , where and are the intersections of the line through and with the boundary of . How is it possible to take the cross ratio, since are not numbers? The line containing all of them is projectively isomorphic to , which we can parameterize as . 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 has a natural metric, and the hyperbolic metric in agrees with this metric when is thought of as a open convex set in .

4. Higher-Dimensional Hyperbolic Space4.1. HyperboloidThe higher dimensional hyperbolic spaces are completely analogous to the 1-dimensional case. Consider with the basis and the 2-form . This is the form defined by the matrix . Define to be the positive (positive in the direction) sheet of the hyperbola .

Let be the linear transformations preserving the form, so . This group is generated by as symmetries of the plane, together with as symmetries of the span of the (this subspace is euclidean). The group is the set of orientation preserving elements of which preserve the positive sheet of the hyperboloid (). This group acts transitively on with point stabilizers : this is easiest to see by considering the point . Here the stabilizer is clearly , and because acts transitively, any stabilizer is a conjugate of this.

As in the 1-dimensional case, the metric on is , which is invariant under .

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

{Totally geodesic subspaces through } {linear subspaces of through }

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

4.2. ProjectiveAnalogously, we define the projective model as follows: consider the disk . I.e. the points in the plane inside the cone . We can think of as , so this disk is . There is, as before, the natural projection of to , and the pushforward of the hyperbolic metric agrees with the Hilbert metric on as an open convex body in .

Geodesics in the projective model are the intersections of planes in with ; that is, they are geodesics in the euclidean space spanned by the . 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 2In the case that , we can understand the projective isomorphisms of by looking at their actions on the boundary . The set is projectively isomorphic to as an abstract manifold, but it should be noted that is not a straight line in , which would be the most natural way to find ‘s embedded in .

In addition, any projective isomorphism of can be extended to a real projective isomorphism of . In other words, we can understand isometries of 2-dimensional hyperbolic space by looking at the action on the boundary. Since is not a straight line, the extension is not trivial. We now show how to do this.

The automorphisms of are . We will consider . For any Lie group , there is an Adjoint action defined by (the derivative of) conjugation. We can similarly define an adjoint action by the Lie algebra on itself, as for any path with . If the tangent vectors and are matrices, then .

We can define the Killing form on the Lie algebra by . Note that 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 specifically. A basis for the tangent space (Lie algebra) is , , and . We can check that , , and . Using these relations plus the antisymmetry of the Lie bracket, we know

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

This matrix has 2 positive eigenvalues and one negative eigenvalue, so its signature is . Since acts on preserving this form, we have , otherwise known at the group of isometries of the disk in projective space , otherwise known as .

Any element of (which, recall, was acting on the boundary of projective hyperbolic space ) therefore extends to an element of , 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 ( acts on by projecting the action on , so an eigenvector of a matrix corresponds to a fixed line in , so a fixed point in . For a matrix , we have the following:

5. Complex Hyperbolic SpaceWe can do a construction analogous to real hyperbolic space over the complexes. Define a Hermitian form on with coordinates by . We will also refer to as . The (complex) matrix for this form is , where . Complex linear isomorphisms preserving this form are matrices such that . This is our definition for , and we define to be those elements of with determinant of norm 1.

The set of points such that is not quite what we are looking for: first it is a real dimensional manifold (not as we would like for whatever our definition of “complex hyperbolic space” is), but more importantly, does not restrict to a positive definite form on the tangent spaces. Call the set of points where by . Consider a point in and in . As with the real case, by the fact that is in the tangent space,

Because is hermitian, the expression on the right does not mean that , but it does mean that is purely imaginary. If , then , i.e. is not positive definite on the tangent spaces.

However, we can get rid of this negative definite subspace. as the complex numbers of unit length (or , say) acts on by multiplying coordinates, and this action preserves : any phase goes away when we apply the absolute value. The quotient of by this action is . The isometry group of this space is still , but now there are point stabilizers because of the action of . We can think of inside as the diagonal matrices, so we can write

And the projectivized matrices is the group of isometries of , where the middle is all vectors in with (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 .

5.1. CaseIn the case , we can actually picture . We can’t picture the original , but we are looking at the set of such that . Notice that . After projectivizing, we may divide by , so . The set of points which satisfy this is the interior of the unit circle, so this is what we think of for . The group of complex projective isometries of the disk is . The straight horizontal line is a geodesic, and the complex isometries send circles to circles, so the geodesics in are circles perpendicular to the boundary of in .

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 ‘s contained incontains 2 kinds of real hyperbolic spaces. The subset of real points in is (real) , so we have a many . In addition, we have copies of , which, as discussed above, has the same geometry (i.e. has the same isometry group) as real . However, these two real hyperbolic spaces are not isometric. the complex hyperbolic space has a more negative curvature than the real hyperbolic spaces. If we scale the metric on so that the real hyperbolic spaces have curvature , then the copies of will have curvature .

In a similar vein, there is a symplectic structure on such that the real are lagrangian subspaces (the flattest), and the 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 ModelThe 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 in which the isometries of the space are conformal.

Consider the unit disk in 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 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 can be expressed at a point (euclidean coordinates) as . 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.