Posted by: matheuscmss | January 18, 2012

## “Surfaces à petits carreaux (suite)” by Jean-Christophe Yoccoz

From January 11 to March 21, Jean-Christophe Yoccoz delivers (on Wednesdays) his course (of academic year 2011-2012) at Collège de France. As the reader can find in his webpage, he decided to make a continuation of his last course (about square-tiled surfaces) and so he entitled the current series of lectures “Surfaces à petits carreaux (suite)”.

After following the first two lectures, I thought it could be a nice idea to try to make available the notes I’m taking for this course. So, I plan to write a series of posts whose titles will have the form “SPCS x” (where SPCS stands for“Surfaces à petits carreaux (suite)” and “x” stands for the number of the lecture). Of course, this goes without saying that any errors and/or mistakes are surely my sole fault: indeed, since the course is delivered in French, it may happen that I misinterpret some points.

Below the fold the reader will find my first set of notes, i.e., SPCS 1, corresponding to Yoccoz’s lecture on last January 11th.

-Introduction-

We start by recalling some distinct (but completely equivalent) points of view (discussed in last year’s course) on square-tiled surfaces / origamis.

1. Topological point of view

An origami is a finite (usually ramified) covering $\pi:M\to\mathbb{T}^2=\mathbb{R}^2/\mathbb{Z}^2$ where $M$ is a connected surface and $\pi$ is not ramified on $\mathbb{T}^2-\{0\}$.

The squares are the connected components of $\pi^{-1}((0,1)^2)$. We denote by $Sq(M)$ the set of squares.

A morphism $p$ between two origamis $\pi:M\to\mathbb{T}^2$ and $\pi':M'\to\mathbb{T}^2$ is an application $p:M\to M'$ such that $\pi = \pi'\circ p$. An isomorphism is a homeomorphism $p$ such that $\pi = \pi'\circ p$.

2. Geometrical-Analytical point of view

We begin by reviewing the notion of translation structures. Let $M$ be a compact connected orientable surface genus $g\geq 1$, and $\emptyset\neq\Sigma=\{A_1,\dots,A_s\}\subset M$, and $\kappa=(k_1,\dots,k_s)\in\mathbb{N}^s$, $k_i\geq 1$, $\sum\limits_{i=1}^s k_i=2g-2$.

A structure of translation surface on $(M,\Sigma)$ of type $\kappa$ is a maximal atlas $\zeta$ of local charts on $M-\Sigma$ verifying:

• the coordinate changes are locally given by translations;
• for each $1\leq i\leq s$, there are neighborhoods $V_i$ of $A_i\in M$, $W_i$ of $0\in\mathbb{R}^2$ and a ramified covering $\pi_i:(V_i,A_i)\to(W_i,0)$ of degree $k_i$ such that the injective restrictions of $\pi_i$ are local charts of $\zeta$ (in particular, the “total angle” around $A_i$ is $2\pi k_i$).

Equivalently, we can say that a translation surface is a Riemann surface structure on $M$ together with a choice of a non-trivial holomorphic $1$-form $\omega\neq 0$ with a zero of order $(k_i-1)$ at $A_i$. Indeed, this last definition can be connected to the previous one by noticing that the local primitives of $\omega$ provide local charts for an atlas $\zeta$ as above.

Next, we recall that, given a translation surface structure, one has a period map $\Theta:H_1(M,\Sigma,\mathbb{Z})\to\mathbb{R}^2=\mathbb{C}$, $\Theta([\gamma]):=\int_{\gamma}\omega$.

In this setting, an origami is a translation surface whose period map $\Theta$ takes values in $\mathbb{Z}^2$.

Remark (ambiguity on $\Sigma$). In the previous definition, the set $\Sigma$ was not explicitly chosen. In fact, by letting $\Sigma_{\min}=\{\textrm{ramification points of }\pi:M\to\mathbb{T}^2\}$ and $\Sigma_{\max}:=\pi^{-1}(\{0\})$, we will see that any choice of $\Sigma_{\min}\subset\Sigma\subset\Sigma_{\max}$ works.

In order to see that the geometrical-analytical definition of origami is equivalent to the topological one, we take $A_1\in\Sigma$, and we observe that $\pi(z):=\int_{A_1}^z\omega$ (mod $\mathbb{Z}^2$) is a well-defined topological origami and, conversely, a topological origami $\pi:M\to\mathbb{T}^2$ defines a geometrical-analytical origami by taking the inverses of injective restrictions of $\pi$ as the local charts of an atlas $\zeta$ with the desired properties.

3. Algebraic-Combinatorial point of view

An origami is a finite set $\mathcal{O}$ equipped with two permutations $r, u\in S(\mathcal{O})$ generating a transitive action on $\mathcal{O}$. Given a topological origami $\pi:M\to\mathbb{T}^2$, we set $\mathcal{O}=Sq(M)$ and $r(s):=\textrm{square to the right of }s$, $u(s):=\textrm{square on the top of }s$. In this way, the connectedness of $M$ corresponds to the transitivity of the action of $r,u$ on $\mathcal{O}$.

Conversely, given an algebraic-combinatorial origami, we see that $M=(\mathcal{O}\times [0,1]^2)/\sim$, where $\sim$ is the equivalence relation $(s,1,x)\sim (r(s),0,x)$ and $(s,x,1)\sim (u(s),x,0)$, is a topological origami.

A morphism between $(\mathcal{O},r,u)$ and $(\mathcal{O}',r',u')$ is an application $p:\mathcal{O}\to\mathcal{O}'$ such that $p\circ r = r'\circ p$ and $p\circ u = u'\circ p$. An isomorphism is a bijective morphism $p$.

-Monodromy of origamis and description of origamis as quotients of groups-

This algebraic-combinatorial point of view of origamis was studied by D. Zmiaikou in his PhD thesis (under the supervision of J.-C. Yoccoz). In the sequel, we will follow closely some aspects of D. Zmiaikou’s thesis.

Definition (D. Zmiaikou). The monodromy group $Mon(\mathcal{O})$ is the subgroup of the permutation group $S(\mathcal{O})$ generated by $r$ and $u$.

Convention. We will let $S(\mathcal{O})$ acts on the right on $\mathcal{O}$, i.e., $(sg)g' = s(gg')$ (because we want automorphisms of origamis to act on the left).

1. Ramification

The type $\kappa$ is determined by the action of the commutator $c=r^{-1}u^{-1}ru$:

In fact, from the figure above one deduces that $s = s\cdot c^{k(s)}$ where $k(s)$ is the ramification index of the left-down corner of the square $s$. (Warning: the figure above is somewhat misleading as $s$ and $s\cdot c$ are usually not the same square!).

2. Origamis and quotients of groups

Denote by $G=Mon(\mathcal{O})$ and let $g_r, g_u\in G$ be the elements corresponding to $r,u$. Fix $\star\in\mathcal{O}$, and denote by $H$ the stabilizer of $\star$ with respect to the (right-) action of $G$ on $\mathcal{O}$.

We identify $\mathcal{O} = H\backslash G=\{Hg: g\in G\}$, $\star = H$ and $r:Hg\mapsto Hgg_r$, $u: Hg\mapsto Hgg_u$.

Note that $H$ is not an arbitrary subgroup of $G$. Indeed, since the stabilizer of $Hg$ is $g^{-1}Hg$, the transitivity faithfulness of the action implies that the intersection of the conjugates of $H$ is the trivial group $\{1\}$. In other words, $H$ doesn’t contain non-trivial normal subgroups.

Conversely, given a finite group $G$ generated by two elements $g_r,g_u\in G$, and $H\subset G$ a subgroup containing no non-trivial normal subgroup, we obtain an origami $M$ by taking $\mathcal{O}=H\backslash G$ and $r:Hg\mapsto Hgg_r$, $u: Hg\mapsto Hgg_u$.

Of course, in this language, a change of basepoint $\star$ corresponds to replace $H$ by one of its conjugates $g^{-1}Hg$.

3. Automorphisms of an origami

By definition, an automorphism is a bijection $p:\mathcal{O}\to\mathcal{O}$ such that $p\circ r = r\circ p$ and $p\circ u = u\circ p$.

Denote by $n\in G$ an element with $H\cdot n = p(H) (=p(\star)$. For every $g\in G$, one has $Hng=p(Hg)$ (since $p$ is an automorphism). Thus, for every $h\in H$, $Hn=p(H)=Hnh$, i.e., $nhn^{-1}\in H$, i.e., $n$ belongs to the normalizer $N$ of $H$ in $G$. In particular, for every $g\in G$, one has $p(Hg)=Hng=nHg$, i.e., the automorphisms of $M$ are given by the left action of the normalizer $N$ of $H$ in $G$ (see the convention above). So, by putting $n\cdot Hg = nHg=Hng$, we can identify

$Aut(M)\simeq N/H$

Definition (D.Zmiaikou). We say that $M$ is a regular origami if $Aut(M)$ acts transitively on $Sq(M)$, or, equivalently, $H=\{1\}$ (and, a fortiori, $N=G$).

The general plan for the beginning of the course is to explain some results from a work (still in progress) by J.-C. Yoccoz, D. Zmiaikou and myself, where the following topics are studied:

• the action of $Aut(M)$ on $H_1(M,K)$ with $K=\mathbb{Q},\mathbb{R},\mathbb{C}$;
• consequences of the previous topic to the action of the affine group $Aff(M)$ on $H_1(M,K)$ and the Lyapunov exponents of the Kontsevich-Zorich cocycle;
• application of the previous topics to concrete examples.

In particular, the results we’re going to discuss below (and in the next two or three lectures) are part of a forthcoming paper by J-C. Y., D. Z., C. M.

-Homology of a origami $M$ as a $Aut(M)$-module-

Let $K$ be a subfield of $\mathbb{C}$. Given an origami $\pi:M\to\mathbb{T}^2$, take $\Sigma=\Sigma_{\max}=\pi^{-1}(\{0\})$ and consider the first relative homology group

$H_1(M,\Sigma,K)$

Recall that $Sq(M)=H\backslash G$, $G=\langle g_r, g_u\rangle$, and $\bigcap\limits_{g\in G} g^{-1}Hg=\{1\}$. We denote by $K(M)=K^{H\backslash G}$ the $K$-vector space of canonical basis $e_s$, $s\in H\backslash G$. Next, we take two copies $K(M)\oplus K(M)$ of $K(M)$ and we denote by $\sigma_s$, $s\in H\backslash G$, the canonical basis of the first copy, and by $\zeta_s$, $s\in H\backslash G$, the canonical basis of the second copy.

Proposition. We have an exact sequence of $Aut(M)$-modules

$0\to K\stackrel{\varepsilon}{\to} K(M)\stackrel{i}{\to} K(M)\oplus K(M)\stackrel{j}{\to} H_1(M,\Sigma,K)\to 0$

where $\varepsilon(1)=\sum\limits_{s\in H\backslash G} e_s$, $i(e_s)=\square_s:=\sigma_s + \zeta_{sg_r} - \sigma_{s g_u} - \zeta_s$, and the map $j$ is explained in the picture below:

Proof. Let $Sk(M) = M-\bigcup\limits_{s\in H\backslash G}s$ be the skeleton of $M$. We have that $\Sigma\subset Sk(M)\subset M$ gives rise to an exact sequence in homology:

$H_2(Sk(M),\Sigma,K)\to H_2(M,\Sigma,K)\to H_2(M,Sk(M),K)\stackrel{\partial}{\to}$

$H_1(Sk(M),\Sigma,K)\to H_1(M,\Sigma,K)\to H_1(M,Sk(M),K)$

where $\partial$ is the usual boundary operator.

Since $Sk(M)$ is a $1$-dimensional skeleton, $H_2(Sk(M),\Sigma,K)=0$ and $H_1(M,Sk(M),K)=0$, $H_2(M,Sk(M),K)=K(M)$ and $H_1(Sk(M),\Sigma,K)=K(M)\oplus K(M)$. Also, since $M$ is $2$-dimensional, $H_2(M,Sk(M),K)=K(M)$. By plugging this information into the previous exact sequence (and by reinterpreting the operators between the homology groups above), one can check that this exact sequence is precisely the one appearing in the conclusion of the proposition. $\square$

Corollary. One has $H_1(M,\Sigma,K)=K(M)\oplus K$ as $Aut(M)$-module. $\square$

We recall that one can write a decomposition of $Aut(M)$-modules:

$H_1(M,K)=H_1^{st}(M,K)\oplus H_1^{(0)}(M,K)$

where $H_1^{st}(M,K)$ is the $2$-dimensional $Aut(M)$-module generated by $\sum\limits_{s\in H\backslash G}\sigma_s$ and $\sum\limits_{s\in H\backslash G}\zeta_s$ (hence it is isomorphic to $K^2$) and $H_1^{(0)}(M,K)$ is the codimension $2$ (i.e., dimension $2g-2$) given by the kernel of the map

$\pi_*:H_1(M,K)\to H_1(\mathbb{T}^2,K)=K^2,$

that is, $H_1^{(0)}(M,K)$ consists of homology classes projecting to zero in the torus.

On the other hand, the exact sequence in homology associated to $\emptyset\subset\Sigma\subset M$ is

$0=H_1(\Sigma,K)\to H_1(M,K)\to H_1(M,\Sigma,K)\to$

$H_0(\Sigma,K)\to H_0(M,K)=K\to H_0(\emptyset,K)=0$

By combining this with the previous corollary, we get

Corollary. $H_1^{(0)}(M,K)\simeq K(M)\ominus H_0(\Sigma,K)$ as $Aut(M)$-module. $\square$

As we told above, our current goal is to understand the action of $Aut(M)$ on $H_1(M,K)$. Of course, since the action of $Aut(M)$ on $H_1^{st}(M,K)$ is trivial, it suffices to understand the action of $Aut(M)$ on $H_1^{(0)}(M,K)$, and, by the previous corollary, this amounts to study $H_0(\Sigma,K)=K^{\Sigma}$. Therefore, we’ll close today’s post with some preliminaries on the action of $Aut(M)=N/H$ on $\Sigma$.

Denote by $k$ the order of the commutator $c=g_r^{-1}g_u^{-1}g_r g_u$, and let $\langle c\rangle$ be the subgroup generated by it. During the study of ramifications of origamis in terms of the commutator, we saw that the points of $\Sigma$ correspond to orbits of the action of $\langle c\rangle$ on $H\backslash G$, or, equivalently, orbits of the action of $H\times\langle c\rangle$ on $G$ given by $(h,c^m)\cdot g\mapsto hgc^m$.

In this way, for each $g\in G$, we denote by $A_g$ the point of $\Sigma$ corresponding to the action of $g$, and $Stab(g)$ the stabilizer of $A_g$ for the action of $N$ on $\Sigma$.

Lemma. $Stab(g)=N\cap H\cdot\langle gcg^{-1}\rangle = H\cdot (N\cap\langle gcg^{-1}\rangle)$.

Proof. Observe that

$n\in Stab(g)\iff nHg\langle c\rangle=Hg\langle c\rangle\iff Hng\langle c\rangle=Hg\langle c\rangle\iff$

$n\in H\langle gcg^{-1}\rangle\iff n=h \underbrace{gc^mg^{-1}}_{\substack{\in N \\ \textrm{ since } n,h\in N}}\iff n\in H\cdot (N\cap\langle gcg^{-1}\rangle)$

This completes the proof. $\square$

At this point, Jean-Christophe ended his lecture (as he was running out of time), and, so we also stop here for today. Next time, we will continue the study of the action of $Aut(M)$ on $H_1^{(0)}(M,K)$.

## Responses

1. Hi Matheus,

An error in section 2 ? The fact that H does not contain a normal subgroup does not follow from “transitivity” of the action but from “faithfulness”. More precisely, the intersection of g^-1 H g for g in G is exactly the set of elements in G that acts as the identity on H \ G. It is also the maximal normal subgroup of G contained in H.

In section 3. you did not insist on the fact that the automorphism group is defined as a centralizer. Actually, you prove a nice group theoretical statement : the centralizer in Sym(H \ G) of G is isomorphic to N_G(H) / H.

V. D.

2. Hi Vincent,

Thanks for the comments! In fact, you’re right that it is faithfulness that we’re using in section 2, and that the argument showing that the automorphism group is N/H in section 3 translates into the fact that the centralizer of G in Aut(M) is N/H (but I confess that I did not insist on this because it was part of the material of last year’s course)

Best,

Matheus

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