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Ian gave his second and third talks this afternoon, completing his (quite detailed) sketch of the proof of the Virtual Haken Theorem. Recall that after work of Kahn-Markovic, Wise, Haglund-Wise and Bergeron-Wise, the proof reduces to showing the following:

**Theorem (Agol):** Let G be a hyperbolic group acting properly discontinuously and cocompactly on a CAT(0) cube complex X. Then there is a finite index subgroup G’ so that X/G’ is special; in other words, G is *virtually special*.

I am in Paris attending a workshop at the IHP where Ian Agol has just given the first of three talks outlining his proof of the Virtual Haken Conjecture and Virtual Fibration Conjecture in 3-manifold topology (hat tip to Henry Wilton at the Low Dimensional Topology blog from whom I first learned about Ian’s announcement last week). I think it is no ~~under ~~overstatement to say that this marks the end of an era in 3-manifold topology, since the proof ties up just about every loose end left over on the list of problems in 3-manifold topology from Thurston’s famous Bulletin article (with the exception of problem 23 — to show that volumes of closed hyperbolic 3-manifolds are not rationally related — which is very close to some famous open problems in number theory). The purpose of this blog post is to say what the Virtual Haken Conjecture is, and some of the background that goes into Ian’s argument. I hope to follow this up with more details in another post (after Agol gives talks 2 and 3 this coming Wednesday). Needless to say this post has been written in a bit of a hurry, and I have probably messed up some crucial details; but if that caveat is not enough to dissuade you, then read on.

Patrick Foulon and Boris Hasselblatt recently posted a preprint entitled “Nonalgebraic contact Anosov flows on 3-manifolds”. These are flows which are at the same time Anosov (i.e. the tangent bundle splits in a flow-invariant way into stable, unstable and flow directions) and contact (i.e. they preserve a contact form — that is, a 1-form for which is a volume form). Their preprint gives some very interesting new constructions of such flows, obtained by surgery along a Legendrian knot (one tangent to the kernel of the contact form) which is transverse to the stable/unstable foliations of the Anosov flow.

My student Steven Frankel has just posted his paper *Quasigeodesic flows and Mobius-like groups* on the arXiv. This ~~heartbreaking work of staggering genius~~ interesting paper makes a deep connection between dynamics, hyperbolic geometry, and group theory, and represents the first significant progress that I know of on a conjectural program I formulated a few years ago.

One of the main results of the paper is to show that every quasigeodesic flow on a closed hyperbolic 3-manifold either has a closed orbit, or the fundamental group of the manifold admits an action on a circle with some very peculiar properties, namely that it is *Mobius-like* but not *Mobius*. The problem of giving necessary and sufficient conditions on a vector field on a 3-manifold to guarantee the existence of a closed orbit is a long and interesting one, and the introduction to the paper gives a brief sketch of this history as follows:

**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.

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