## Hyperbolicity of the curve graph: the proof from The Book

In this post I’ll talk about a lovely paper by Sebastian Hensel, Piotr Przytycki and Richard Webb. They show that all curve graphs are 17-hyperbolic. Hyperbolicity of curve graphs is a very very very useful* property because mapping class groups act nicely on them, so you can prove all sorts of things using this action.
The proofs available before HPW were quite complicated and relied on Teichmüller theory, so I was quite amazed to see that this could be proven in 6 self-contained pages. And, as the authors pointed out to me: “Well, there’s one page of introduction and one page of references.”

This post is an illustrated guide to the proof. I won’t define arc/curve complexes here, but I plan to write a BI soon. The more loudly you complain, the sooner I will actually do.

Ok, let’s start. First of all, as it turns out it’s more convenient to deal with arc graphs instead of curve graphs.

Reduction to the arc graph. (Really, just skip this ) Here’s a very quick sketch of how to deduce hyperbolicity of curve graphs once we know it for arc graphs, which for the moment we assume we do.
If we are dealing with a (complicated) surface with boundary then the idea we can use is that one can construct a path in the curve complex starting from a path $\gamma$ in the arc complex by considering curves disjoint from the arcs appearing along $\gamma$ (it will be clearer later how to exploit this, if you don’t know how to do it already). I’ll be honest, just this one time: There is a subtle issue I’m sweeping under the carpet here, but I would like not to make the post too technical…
Anyway, suppose instead that your surface, $S$, is closed. If you call $S'$ the surface obtained by adding a boundary component, then you have a boundary-forgetting map. This map is Lipschitz, this is immediate, and has a section, constructed as follows. Choose a hyperbolic metric on $S$ and realise all (homotopy classes of) curves as geodesics. Now you can take a puncture outside the union of all such (countably many) geodesics. Using the two properties it’s not hard to see that the curve graph of $S$ must be hyperbolic as well. The details are in section 5 of the paper.

Unicorns! Ok, let’s now take an arc graph. The strategy to show it’s hyperbolic is to prove that, as David puts it, “it smells hyperbolic”**, i.e. that there exists a family of preferred paths with nice properties (you may wish to take a look at this post ). The most important property that this family should satisfy is that triangles of preferred paths should be slim. The other properties you need (in simplified form) are that a preferred path is a geodesic if the endpoints have distance at most 1 and that a subpath of a preferred path is a preferred path.
HPW called the nice paths they constructed unicorn paths (!). They were initially called one-corner paths, but then Piotr realised that one-corner and unicorn are almost the same word in Polish, and that unicorn would be a much more awesome name.
And now, here are the paths. You have to start with two arcs in minimal position, $a$ and $b$, and it’s convenient to fix preferred endpoints, $\alpha$ and $\beta$. We can assume that $a,b$ intersect minimally. The arcs appearing along the path from $a$ to $b$, the unicorn arcs, have a rather simple form: You travel along $a$ until you reach some intersection point, $\pi$, and then you go to $\beta$.

Not all intersection points work, because the procedure described above might not give an embedded arc (look at $\tau$, for example).
Let us also include $a$ and $b$ in the collection of unicorn arcs. So, given $a,b$ you have finitely many arcs, and there is a natural order on them: You write $c if the common subpath of $c$ and $a$ is longer than the common subpath of $c'$ and $a$. The idea is that you are describing a path from $a$ to $b$, so the paths more closely resembling $a$ should appear first.
Now, we have to check that we actually described a path, i.e. that if $c'$ is the successor of $c$ then $c,c'$ can be realised disjointly.
The recipe to construct the successor is simple. Let $\pi$ be the intersection point defining $c$, and let $b'$ be the subpath of $c$ from $\pi$ to $\beta$. Just travel along $a$ after $\pi$ until you hit $b'$ in some point $\pi'$. Then it’s not hard to convince yourself that $\pi'$ defines $c'$.

The picture also illustrates how to realise $c,c'$ disjointly.

Slim unicorns! And here is the cool part. Unicorn triangles are 1-slim, i.e. given $a,b,d$ and a unicorn arc $c$, defined by $\pi\in a\cap b$, on the unicorn path $P(a,b)$ from $a$ to $b$ there is a point on $P(a,d)\cup P(b,d)$ at distance at most 1 from $c$. The proof is very easy, just travel along $d$ until you hit $c$, then go to $\alpha$ or $\beta$, whichever you can reach avoiding $\pi$.

The picture hopefully clearly illustrates that the unicorn arc $c'$ we just constructed can be realised disjointly from $c$.
The authors told me that the initial argument for 1-slimness of triangles was based on their dismantlability, you may wish to take a look at this paper.

The technical bit. You need one more property to show that arc graphs are hyperbolic and unicorn paths are close to geodesics, that is to say that a subpath of a unicorn path is a unicorn path. Well, this is not actually true, sometimes you can have a subpath of a unicorn path of length 2 connecting points at distance 1 in the arc graph. But that’s the only bad thing that can happen, and this is good enough for us.
Stare at the recipe I told you about earlier to construct the successor. Done? It might now seem to you that if you have two unicorn arcs $c_1,c_2$ constructed from $a,b$, the recipe will give you exactly the same successor of $c_1$ regardless of whether you work “between $a$ and $b$” or “between $c_1$ and $c_2$.” So… what’s the problem? The problem is that $c_1,c_2$ might not be in minimal position!
So you have to understand when they don’t intersect minimally (Sublemma 3.6 in the paper). You can reduce to the case $a=c_1$ and $c_2$ the predecessor of $b$, once again because of the recipe for the successor. The predecessor of $b$ is the unicorn arc travelling as little as possible along $a$, so you just have to find the first intersection point $\pi$ of $a,b$ along $a$.

(Yes, there’s a reason why I’ve drawn the picture in that weird way, see below )
Let me recall that two arcs are in minimal position if there are no bigons or half bigons:

Now, there cannot be a bigon between $a$ and $c_2$, for otherwise you can easily convince yourself that there would be one between $a$ and $b$. Similarly, you can only have a half-bigon if $\pi$ appears on it:

So, if there is a half-bigon, then it must look like this:

So, you have a very good control on what can go wrong. I won’t go into details but I hope that you’ll find it reasonable to believe that it’s possible to deal with this problem.

And that’s the end

Let me add the comment that Lemma 3.4, which I didn’t go into, is there just to get a better constant at the end, but without it you can prove, say, 19-hyperbolicity.
The lemma says that given 3 unicorn paths forming a triangle you can find a triple of points on them at pairwise distance at most 1. If you allow yourself to substitute 1 with 3, this follows from what we already know. In fact, suppose you start at $a$ and move towards $b$. At first, you’ll be 1-close to $P(a,c)$, and at some point you’ll be 1-close to $P(b,c)$. When the switch happens, you have the 3 points you’re looking for. Btw, analysing this idea gives you a way of finding your candidate points at pairwise distance 1.

*David Hume was next to me when I wrote this and said “I have a paper that relies on that, so I’m not going to argue.”
** David suggested I add “I would like to thank David for being so quotable.”

## Tracking of random walks with geodesics

In this post I’ll tell you about a property of random walks on hyperbolic groups. To make a random walk on a group $G$ just start from the identity in the Cayley graph, then move to a neighbor with uniform probability and keep going. You have for each $n$ a probability measure $\mu_n$ on $G$ telling you the probability of ending up at any given $g\in G$. $\mu_n$ is the law of a random variable $X_n$, and this sequence of random variables is referred to as (simple) random walk.

Theorem: Let $\{X_n\}$ be a simple random walk on a non-virtually cyclic hyperbolic group $G$. Then the expected Hausdorff distance between $\{X_i\}_{i\leq n}$ and a geodesic connecting $1$ to $X_n$ is $O(\log(n))$.

Surprisingly, I couldn’t find this result in the literature, except for free groups (Ledrappier) [EDIT: I've been pointed out a reference (Corollary 3.9). The proof is quite different, but the overall aims are deeper.]. This post contains an essentially complete proof of the theorem. It’ll be a bit more technical than usual, but I just couldn’t resist blogging the full proof as it’s so short. I’m almost done writing up the same result for relatively hyperbolic groups (as well as a related result for mapping class groups). [EDIT: here it is.]

First, we show that any point on $[1,X_n]$ is close to a point in $\gamma_n=\{X_i\}_{i\leq n}$. This has very little to do with random walks, it’s just a fact about paths connecting the endpoints of a geodesic in a hyperbolic space.

Lemma 1: There exists $C_1\geq 1$ so that if the path $\alpha$ (say of length at leat $2$) connects the endpoints of a geodesic $\gamma$ in a $\delta-$hyperbolic space and $p\in\gamma$, then $d(p,\alpha)\leq C_1\log(length(\alpha))$.

Another way of saying this is: hyperbolic spaces have (at least) exponential divergence.

The proof is not hard, you can find all the details in Bridson-Haefliger’s book, Proposition III.H.1.6.
The procedure is the following: you split $\alpha$ into two subpaths of equal length and consider the triangle as in the picture:

$p$ is close to some $p_1$ on one of the sides. Then you repeat using $p_1$ instead of $p$, and if you keep going in logarithmically many steps you get to $\alpha$.

And now we have to exclude the existence of large detours, something like in the following picture:

The idea is that the first and last point of the detour $X_{i_0},X_{i_1}$ will be close to some point in $[1,X_n]$, and hence close to each other, the details are below. Anyway, this motivates showing that if $|i-j|$ is at least logarithmically large, then so is $d(X_i,X_j)$.

Lemma 2: There exists $C_2$ so that
$\mathbb{P}[\exists i,j\leq n: |i-j|\geq C_2\log(n), d(X_i,X_j)\leq |i-j|/C_2]$  $=o(n^{-2})$.

The “$|i-j|\geq C_2\log(n)$” part is there just to make the proof work, but we’re happy if $|i_1-i_2|\leq C_2\log(n)$ for $i_1,i_2$ as in the picture above, because this gives a logarithmic bound on the length of the detour.

The proof is based on the following classical inequality due (afaik) to Kesten, which holds for nonamenable groups: There exists $K>0$ so that

$\mathbb{P}[d(1,X_n)\leq n/K]\leq Ke^{-n/K}$.

In words, the probability that $X_n$ does not make linear progress decays exponentially.

In order to prove the lemma, we just have to sum these inequalities up (using the fact that $d(X_i,X_j)$ has the same distribution as $d(1,X_{|i-j|})$).
Fixing $i$ first ($n$ possibilities) and letting the difference between $i$ and $j$ vary, we see that the probability in the statement of Lemma 2 is at most

$n\times \sum_{j\geq C_2\log(n)} Ke^{-j/K}\leq K n\cdot n^{-C_2/K} \times \sum_{j\geq 0} e^{-j/K}$,

and this is it, for $C_2$ large.

Ok, the endgame. If there are no points on $\alpha_n$ at distance more than $C_1\log(n)$ from $[1,X_n]$ we are happy. Otherwise, let $i$ be so that $d(X_i,[1,X_n])> C_1\log(n)$. Then each point on $[1,X_n]$ is at distance at most $C_1\log(n)$ from either $\alpha^-_n=\{X_j\}_{j or $\alpha^+_n=\{X_j\}_{j>i}$. As you move from $1$ to $X_n$ along the geodesic you’ll be at first close to $\alpha^-_n$ and then you’ll be close to $\alpha^+_n$, so at some point $p$ you’ll be close to both. In particular, there are $i_0 so that $d(X_i,X_j)\leq 2C_1\log(n)$. By Lemma 2 (we can assume that) we have $|i_0-i_1|\leq 2C_1C_2\log(n)$. Hence, $d(X_i,X_{i_0})\leq |i-i_0|$ is also logarithmically bounded, as well as $d(X_i,[1,X_n])\leq d(X_i,X_{i_0})+d(X_{i_0},[1,X_n])$. The end.

## Just for fun: genus 2 madness in Bonn

Real life model of the L-surface, realised by Mark Pedron with the help of Dawid Kielak.

Yeah, too much light, I know… Clearer one:

Explanation on the blackboard:

The cone point. Behold the negative curvature!

## BI: Finite decomposition complexity (is preserved by relative hyperbolicity)

In this post I’ll define finite decomposition complexity (FDC) for you, tell you what it’s good for (Stable Borel Conjecture!) and point out that an argument by Osin shows that it is preserved by relative hyperbolicity.

First, the motivation. Here is a very cool conjecture.

Borel Conjecture: If two closed aspherical manifolds have isomorphic fundamental groups, then they are homeomorphic.

A manifold $M$ is aspherical if its universal cover is contractible, so that $\pi_1(M)$ determines the homotopy type of $M$. In case you don’t see the point, let me tell you that the Borel Conjecture for $S^1\times S^1\times S^1$ implies the Poincaré Conjecture.
The conjecture has been established for high dimensional manifolds of non-positive curvature by Farrell and Jones, and several people have worked on it since then, most notably showing that the conjecture holds if the fundamental groups involved are hyperbolic or CAT(0).

There are several conjectures related to the Borel Conjecture, for example the Novikov Conjecture, and here is another one.

Stable Borel Conjecture (SBC): If the closed aspherical manifolds $M, M'$ have isomorphic fundamental groups, then there exists $n$ so that $M\times \mathbb{R}^n$ is homeomorphic to $M'\times \mathbb{R}^n$.

Stabilisation procedures appear all over topology. For example, there are many contractible 3-manifolds (without boundary) that are not homeomorphic to $\mathbb{R}^3$. However, when taking the product of such a manifold with $\mathbb{R}$  one actually gets $\mathbb{R}^4$. Just to say that the SBC is a natural weaker version of the full Borel Conjecture. Of course, $M$ is said to satisfy the SBC if the statement holds for the given $M$.

And here’s the motivation for being interested in FDC.

Theorem (Guentner, Tessera, Yu): If $\pi_1(M)$ has finite decomposition complexity, then $M$ satisfies the Stable Borel Conjecture.

(Guoliang Yu told me that one can take $n=3$ in statement of the SBC.)

Time for the definition of FDC. We say that $X=\bigcup X_\alpha$ is an $R$-decomposition of the metric space $X$ if $d(X_\alpha,X_\beta)>R$ for each $\alpha\neq \beta$. Let us consider the following decomposition game. Fix a metric space $X$. Bob’s aim is to find a nice decomposition of $X$, and Alice’s aim is to prevent this.

Step 0: Alice gives Bob a huge real number $R_0$. Bob finds an $R_0$-decomposition $\bigcup X^1_\alpha$ of $X$.

Inductive step: A family of subspaces $X^i_\alpha$, coming from the previous step, is given. Alice gives Bob a real number $R_i$ and Bob finds  an $R_i$-decomposition of each $X^i_\alpha$. $X^{i+1}_\alpha$ is the union of all subspaces appearing in each decomposition.

Definition: $X$ has FDC if Bob has a strategy to end up with a collection of uniformly bounded subspaces of $X$.

Having FDC is a quasi-isometry invariant, and actually a coarse invariant.

If you know what asymptotic dimension is, it’s a nice exercise to show that it implies FDC. Otherwise, try to do it for $\mathbb{R}^n$. Hint: find a nice coloured covering with finitely many subsets of diameter $\leq D$, for each $D$.
In particular, say, hyperbolic groups have FDC.

Not all groups with FDC have finite asymptotic dimension. Indeed, GTY showed that linear groups have FDC, but some of them contain an infinite sum of copies of $\mathbb{Z}$ and hence have infinite asymptotic dimension.

Another nice fact is that the class of groups with FDC is closed under extensions. Indeed, GTY observed that the following holds in the context of metric spaces. In order to state it, notice that it makes sense to play the decomposition game starting from a family of metric spaces, rather than one metric space.

Fibering Theorem: Suppose that $f:X\to Y$ is Lipschitz and that $Y$ as well as $\{f^{-1}(B_i)\}$ have FDC, for each family of uniformly bounded subsets $\{B_i\}$ of $Y$. Then $X$ has FDC.

A similar statement for asymptotic dimension is usually called Hurewicz Theorem (as it is inspired by an actual theorem of Hurewicz about topological dimension).
The proof is simple. Bob first pretends he’s playing the decomposition game on $Y$, and when he as won in $Y$ he’s left with preimages of bounded sets of $Y$ in $X$, where he also knows how to win.

Let me state another permanence result. It is not true that if a metric space is an infinite union of subspaces with FDC then the ambient space has FDC. However, the following result tells us that this is true if the subspaces can be “separated”. You may think of $X$ as obtained gluing metric spaces $\{X_i\}$ to some $Y_0$, and then you can let $Y_R$ be the $R/2$-neighborhood of $Y_0$.

Infinite Union Theorem: Suppose that $X=\bigcup X_i$ and that $\{X_i\}$ has FDC. Also, suppose that for every $R$ there exists $Y_R$ with FDC  so that $\{X_i\backslash Y_R\}$ are pairwise $R$-disjoint. Then $X$ has FDC.

Finally, as promised, I’ll tell you how one can show the following.

Theorem: A relatively hyperbolic group has FDC if and only if the peripherals do.

The only if part follows from the fact that having FDC is stable under taking subgroups.
For the if part, we can use arguments used by Denis Osin to show the analogue statement for finite asymptotic dimension.

What he did is considering the (Lipschitz) map $f$ from the relatively hyperbolic group $G$ to its coned off graph, $\hat{G}$, and showing that the hypothesis of Fibering/Hurewicz Theorem for asymptotic dimension apply to such map. In particular, he showed that $\hat{G}$ has finite asymptotic dimension, and so in particular it has FDC. To show that the Fibering Theorem applies we now have to look at preimages of balls in $\hat{G}$, and proceed by induction. Let $X_n$ be the preimage of the ball of radius $n$ in $\hat{G}$. Then $X_{n+1}$ is obtained as a union of $X_n$ and some cosets of peripheral subgroups (almost true). This is the perfect setting for using the Infinite Union Theorem, with $Y_R$ being suitable neighborhoods of $X_n$, and indeed Osin showed that the result applies. It shouldn’t be too hard to convince yourself of this fact in the case of a free product.

This was the proof outline, and one can actually check that Denis only used the Fibering Theorem and the Infinite Union Theorem (and properties of relatively hyperbolic groups), and hence his proof also gives permanence of FDC.

## BI: Guessing geodesics in hyperbolic spaces

In this post I’ll discuss a cool lemma due to Brian Bowditch (from this paper) which is very useful when you want to show that a space is hyperbolic.

The most direct way of showing that $X$ is hyperbolic is trying to figure out how geodesics in $X$ look like, show that you guessed the geodesics correctly and then show that triangles are thin, right? Well, Bowditch tells you that you can skip the second step…

This might not seem much, but showing that a given path is (close to) a geodesic might be quite annoying… and instead  you get this for free!

More formally, the statement is the following (in this form it’s due to Ursula  Hamenstädt, see Proposition 3.5 here). We denote the $D$-neighborhood by $N_D$.

Guessing Geodesics Lemma: Suppose that we’re given, for each pair of points $x,y$ in the geodesic metric space $X$, a path $\eta(x,y)$ connecting them so that, for each $x,y,z$,

$\eta(x,y)\subseteq N_D(\eta(x,z)\cup \eta(z,y)),$

for some constant $D$. Then $X$ is $K$-hyperbolic and each $\eta(x,y)$ is $K$-Hausdorff-close to a geodesic, where $K=K(D)$.

(The statement is almost true: You also have to require two “coherence conditions”, i.e. that $diam(\eta(x,y))\leq D$ if $d(x,y)\leq 1$ and that for each $x',y'\in \eta(x,y)$ the Hausdorff distance between $\eta(x',y')$ and the subpath of $\eta(x,y)$ between $x'$ and $y'$ is at most $D$.)

This trick has been used many times in the literature: Bowditch used it to show that curve complexes are hyperbolic, Hamenstädt to study convex-cocompact subgroups of mapping class groups, Druţu and Sapir to show that asymptotic tree-graded is the same as relatively hyperbolic, Ilya Kapovich and Rafi to show that the free-factor complex is hyperbolic, etc.

The proof is quite neat. First, one shows a weaker estimate, i.e. that if $\beta$ is any path connecting, say, $x$ to $y$, then the distance from any $p\in\eta(x,y)$ from $\beta$ is bounded, roughly, by $\log(length(\beta))$.

To show this, you just split $\beta$ in two halves of equal length, draw the corresponding $\eta$‘s and notice that $p$ is close to one of them.

Then repeat the procedure logarithmically many times or, more formally, use induction (starting from $diam(x,y)\leq D$ when $d(x,y)\leq 1$).

Ok, now that we have the weaker estimate (for any rectifiable path) we have to improve it (for geodesics).
Let now $\beta$ be a geodesic from $x$ to $y$, let $p\in\eta(x,y)$ be the furthest point from $\beta$, and set $d(p,\beta)=\xi$. Pick $x',y'\in\eta(x,y)$ before and after $p$ at distance $2\xi$ from $p$ (let’s just ignore the case when $d(p,x)<2\xi$ or $d(p,y)<2\xi$, I hope you trust it’s not hard to handle).

As $p$ is the worst point, we have $d(x',\beta),d(y',\beta)\leq \xi$, so that we can draw a red path of length at most $8\xi$ like so:

And now the magic happens. In view of the weak estimate for the red path $\alpha$, we get

$\xi= d(p,\alpha)\leq O(\log(\xi))$,

which gives a bound on $\xi$. (You might have noticed that here we need the second “coherence condition”). That’s it!

Exercise: complete the proof showing that each $q\in \beta$ is close to $\eta(x,y)$.

A relatively hyperbolic version of this trick will be available soon…

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## BI: Teichmüller space, part I

It is a remarkable fact that the group of orientation-preserving isometries of $\mathbb{H}^2$ can be identified with the group of bi-holomorphisms of the unit disk $\mathbb{D}$ in $\mathbb{C}$. [Fun fact: Poincaré said that he discovered (basically) this while stepping on a bus.]

As a consequence, for $S$ an orientable closed surface of genus at least 2 fixed from now on, we have a natural bijection

{complex structures on $S$ } $\leftrightarrow$ {hyperbolic metrics on $S$}

The bijection is given by the fact that a complex structure on $S$ is a way of seeing $S$ as a quotient of $\mathbb{D}$ by a group of biholomophisms, and similarly for hyperbolic metrics.
There’s a more explicit way of comparing a complex structure with its corresponding  hyperbolic metric. Look at the tangent space $T_p(S)$ at some point $p$ of $S$. The hyperbolic metric (as well as any other Riemannian metric) naturally gives  a collection of circles covering $T_p(S)$. The complex structure does as well (consider multiplication by $e^{i\theta}$ for $\theta\in [0,2\pi]$). As it turns out, a complex structure and its corresponding hyperbolic metric give the same collection of circles in every tangent space.
(In other words, they induce the same conformal structure.)

Teichmüller space $Teich(S)$ is a space that parametrizes the complex structures/hyperbolic metrics on $S$. It does not parameterize them up to biholorphism/isometry, but up to biholorphisms/isometries isotopic to the identity. The formal definition is below the informal discussion in the next paragraph.

Suppose you have a homeomorphism $f$ of $S$ supported in a disk and a hyperbolic metric $g$ on $S$. Then one can take the push-forward $f_*(g)$ of $g$ through $f$ and get another hyperbolic metric on $S$. Teichmüller space does not distinguish between these metrics (if it did it would be huge!).
On the other hand, suppose you have a hyperbolic metric $g$ on $S$ with some non-trivial isometry $f$. Then $g$ and $f_*(g)$ actually give distinct points of $Teich(S)$.

And, now, if you really want to see it, here is the formal definition.

$Teich(S)=\{(X,f)| X \mathrm{\ hyperbolic\ surface}, f:S\to X \mathrm{\ homeomorphism} \} /_\sim,$

where “hyperbolic surface” just means surface endowed with a hyperbolic metric and the equivalence relation $\sim$ given by $(X_1,f_1)\sim (X_2,f_2)$ if and only if there exists an isometry $\iota: X_1\to X_2$ so that the diagram

commutes up to isotopy, i.e. $f_2^{-1}\iota\circ f_1:S\to S$ is isotopic to the identity. (One can safely substitute isotopies with homotopy equivalences.)

There is a topology on $Teich(S)$, but I will not discuss it. Here is probably the most important theorem about $Teich(S)$.

Theorem:  $Teich(S)$ if homeomorphic to $\mathbb{R}^{6g-6}$, where $g$ is the genus of $S$.

More precisely, there is a system of coordinates for $Teich(S)$, the so-called Fenchel-Nielsen coordinates. Danny Calegari has a post on this which contains way more details than the discussion below.

Digression: The very nice structure of $Teich(S)$ given by this theorem explains why it is reasonable to consider hyperbolic metrics up to isometry isotopic to the identity, rather than up to isometry, at least as a first step. Then, if one is interested in hyperbolic metrics up to isometry, one can then take the quotient of $Teich(S)$ by the natural action of the mapping class group $MCG(S)$, and hence form the so-called moduli space of $S$. This action is very nice as well, e.g. it is properly discontinuous, so that moduli space has the structure of an orbifold.

Back to the theorem. The proof is quite interesting, and is based on a pants decomposition of $S$, that is to say describing $S$ as a union of pairs of pants, i.e. the following object:

Any maximal collection of disjoint simple closed curves on $S$ gives a pants decomposition, here is an example:

One can endow a pair of paints with a hyperbolic metric so that the boundary is a geodesic in the following way. In $\mathbb{H}^2$ there exist right-angled hexagons, and one can also freely choose the lengths $a,b,c$ of three sides as in the picture.

Gluing two copies of such a hexagon along the blue sides yields a hyperbolic metric on a pair of pants. What is more, one can glue pairs of pants and still obtain a hyperbolic metric as long as corresponding boundary components have the same length. So, given a pair of pants decomposition of $S$, one can assign a length to all the curves appearing in it, consider the corresponding metrics on pairs of pants and glue them all together to get a hyperbolic metric on $S$.

Any maximal collection of disjoint simple closed curves in $S$ contains $3g-3$ curves, not $6g-6$, which means that there are other parameters to consider.
The hyperbolic metrics we constructed on pairs of pants have the feature that for any given pair of boundary components  there is a unique geodesic connecting them and orthogonal to both of them, namely the suitable blue side of one of the hexagons. Hence you see that gluing two pairs of pants in this way (the blue lines represent the geodesics we just described, in adjacent pairs of pants):

or this way:

gives different metrics. So, in order to specify a metric on $S$ we also have to assign a “twist parameter” to each simple closed curve in the decomposition. You might think that these parameters are defined modulo $2\pi$, but it is not the case: If we change one of the twist parameters by $2\pi$ we obtain a new hyperbolic metric which is indeed isometric to the first one, but there’s no isometry isotopic to the identity between them.

To sum up, we choose a maximal collection of disjoint simple closed curves in $S$, and in order to specify a hyperbolic metric on $S$ we assign to each of them a “length parameter” and a “twist parameter”. Some routine and some not-so-routine checks and the theorem is proven.

In another post I’ll talk about the tangent space of $Teich(S)$ and the metrics one can put on $Teich(S)$ using it.

## Just for fun: trefoil knot complement cake

Based on a joint work with Rob Clancy, presented at the Oxford Mathematical Cake Seminar. You might think that this is just a regular cake.

But it’s not! The icing is in the shape of a trefoil knot, so that the cake itself has the shape of a trefoil knot complement (in $S^3$). Here, take a look at the layers before they were put on top of each other:

-its fundamental groups is the braid group on 3 strands and has presentations $\langle x,y | xyx=yxy\rangle=\langle x,y | x^2=y^3\rangle$,
-it contains approximately 500 grams of butter, 500 grams of sugar and 7 eggs,
-it admits finite volume metrics with universal covers $\mathbb{H}^2\times \mathbb{R}$ and $\widetilde{SL_2(\mathbb{R})}$,
-the icing was very buttery,
-it has a Seifert fibration with base orbifold a once-punctured sphere with two cone-points,
-it took an afternoon minus a (spectacular!) rugby match to bake it.

Credits:

Designer: Rob
Manager: Rob
Cook: Rob
Sculptor: Rob
Icing artist: Rob
Dish washing: Alex

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