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In winter and spring of 2001, Nathan Dunfield and I ran a seminar at Harvard whose purpose was to go through Thurston’s proof of the geometrization theorem for Haken manifolds. This was a very useful and productive exercise, and there was wide participation from faculty and students. As well as talks by Nathan and myself, there were talks by David Dumas, Laura de Marco, Maryam Mirzakhani, Curt McMullen, Dylan Thurston, and John Holt. At the conclusion of the semester, Bill Thurston agreed to come out and lead a discussion on geometrization, in which he ended up talking a bit about what had led him to formulate the conjecture in the first place, what ideas had played into it, how and when he had gone about proving it, his ideas about exposition, and so on.

I had recently bought a video camera, and decided to tape Bill’s talk. I never did anything with it until now (in fact, I don’t think I ever re-watched anything that I taped), but it turned out to be not too difficult to transfer the file from tape to computer. Since this seems like an interesting fragment of intellectual history, I thought it might be worthwhile to post the result to YouTube — the video link is here.

This morning I heard the awful news that Bill Thurston died last night. Many of us knew that Bill was very ill, but we all hoped (or imagined?) that he would still be with us for a while yet, and the suddenness of this is very harsh. As Sarah Koch put it in an email to me, “Although this was not unexpected, it is still shocking.” On the other hand, I am glad to hear that he was surrounded by family, and died peacefully.

I counted Bill as my friend, as well as my mentor, and I have many vivid and happy memories of time I spent with him. I hope that writing down a few of these reminiscences will be cathartic for me, and for others who are coping with this loss.

1. Mostow Rigidity

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

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

In other words, Moduli space is a single point.

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

1.1. Part 1

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

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

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

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

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

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

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

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

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

and

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

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

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

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

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

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

And

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

By the triangle inequality,

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

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

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

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

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

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

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

Proof: Here is a representative picture:

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

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

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

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

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

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

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

$\Box$

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

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

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

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

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

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

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

and

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

So we have

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

Which implies that

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

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

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

1.2. Part 2

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

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

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

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

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

This (pseudo) norm has some nice properties:

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

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

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

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

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

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

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

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

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

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

Next, we observe:

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

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

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

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

on the one hand, and on the other hand,

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

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

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

i.e.

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

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

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

Now define the chain:

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

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

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

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

By the equality above, then,

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

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

1.3. Part 3 (Finishing the proof of Mostow Rigidity

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

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

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

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.