Posted by: matheuscmss | January 23, 2016

## Some remarks on Sarnak’s question about thin Kontsevich-Zorich monodromies

Sometime ago, Alex Eskin and Alex Wright told me about the following question posed to them by Peter Sarnak:

How frequent are thin groups among Kontsevich-Zorich monodromies?

Instead of explaning the meaning of Sarnak’s question in general, we shall restrict ourselves to the case of Kontsevich-Zorich (KZ) monodromies associated to square-tiled surfaces.

More concretely, let ${X=(M,\omega)}$ be a square-tiled surface (also called origamis) of genus ${g\geq 1}$, i.e., ${p:M\rightarrow\mathbb{R}^2/\mathbb{Z}^2}$ is a finite branched covering which is unramified off ${0\in\mathbb{R}^2/\mathbb{Z}^2}$ and ${\omega = p^*(dz)}$ is the pullback of ${dz=dx+i dy}$ on ${\mathbb{R}^2/\mathbb{Z}^2 \simeq \mathbb{C}/\mathbb{Z}\oplus i\mathbb{Z}}$. We have a natural representation

$\displaystyle \rho_X:\textrm{Aff}(X)\rightarrow Sp(H_1^{(0)}(X,\mathbb{Z}))\simeq Sp(2g-2,\mathbb{Z})$

from the group ${\textrm{Aff}(X)}$ of affine homeomorphisms of ${X}$ to the group ${Sp(H_1^{(0)}(X,\mathbb{Z}))}$ of symplectic matrices of the subspace ${H_1^{(0)}(X,\mathbb{Z})}$ of integral homology classes of ${X}$ projecting to zero under ${p}$. In this setting, the Kontsevich-Zorich monodromy ${\Gamma_X}$ (associated to the ${SL(2,\mathbb{R})}$-orbit of ${X}$ in the moduli space of translation surfaces) is the image of ${\rho_X}$, i.e.,

$\displaystyle \Gamma_X := \rho_X(\textrm{Aff}(X))$

(See e.g. these posts here for more background material on square-tiled surfaces.)

By following Sarnak’s terminology, we will say that ${\Gamma_X}$ is a thin group if ${\Gamma_X}$ is an infinite index subgroup of ${Sp(2g-2,\mathbb{Z})}$ whose Zariski closure is

$\displaystyle \overline{\Gamma_X}^{\textrm{Zariski}} = Sp(2g-2,\mathbb{R})$

In the particular case of square-tiled surfaces, Sarnak’s question above is related to the following two problems:

• (a) find examples of square-tiled surfaces ${X}$ whose KZ monodromies ${\Gamma_X}$ are thin;
• (b) decide whether the “majority” of square-tiled surfaces in a given connected component ${\mathcal{C}}$ of a stratum of the moduli spaces of unit area translation surfaces has thin KZ monodromy (here, “majority” could mean “all but finitely many” or “almost full probability as the number of squares/tiles grows”.)

The goal of this post is to record (below the fold) some discussions with Vincent Delecroix and certain participants of MathOverFlow around item (a).

Remark 1 While we will not give answers to items (a) and/or (b) in this post, we decided to write it down anyway with the hope that it might be of interest to some readers of this blog: in fact, by the end of this post, we will show the following conditional statement: if the group generated by the matrices

$\displaystyle A=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & -1 & 0 \end{array}\right), \quad B=\left(\begin{array}{cccc} -1 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right)$

has infinite-index in ${Sp(4,\mathbb{Z})}$, then a certain square-tiled surface of genus ${3}$ answers item (a) affirmatively.

Remark 2 Some “evidence” supporting a positive answer to item (b) is provided by this recent paper of Fuchs-Rivin where it is shown that two “randomly chosen” elements (in ${SL(n,\mathbb{Z})}$) “tend” to generate thin groups.

1. Faithfulness and thinness

Let ${X}$ be a square-tiled surface of genus ${g\geq 3}$. For the sake of simplicity, suppose that ${X}$ has no non-trivial automorphisms. In this case, the group ${\textrm{Aff}(X)}$ is naturally identified with a finite-index subgroup ${SL(X)}$ of ${SL(2,\mathbb{Z})}$ called Veech group, and, therefore, we get a representation

$\displaystyle \rho_X: SL(X)\rightarrow Sp(2g-2,\mathbb{Z})$

The following proposition produces a certificate of thinness for ${\Gamma_X=\rho_X(SL(X))}$.

Proposition 1 Assume that ${SL(X)}$ is virtually free and ${\rho_X}$ is faithful. Then, ${\Gamma_X=\rho_X(SL(X))}$ has infinite index in ${Sp(2g-2,\mathbb{Z})}$.

Proof: If ${SL(X)}$ is virtually free, ${\rho_X}$ is faithful and ${\Gamma_X=\rho_X(SL(X))}$ has finite index in ${Sp(2g-2,\mathbb{Z})}$, then the group ${Sp(2g-2,\mathbb{R})}$ of higher rank ${g-1\geq 2}$ would contain a lattice isomorphic to a free group.

However, this is impossible because higher rank linear groups do not contain lattices isomorphic to free groups: on one hand, a higher rank linear group satisfies Kazhdan’s property (T), so that its lattices also have this property (by Kazhdan’s theorem), and, on the other hand, a free group does not have Kazhdan’s property (T). $\Box$

This proposition says that a positive answer to item (a) would follow from a positive answer to the following problem:

• (c) find a square-tiled surface ${X}$ of genus ${g\geq 3}$ (without non-trivial automorphisms) such that the Veech group ${SL(X)}$ is virtually free, the image ${\Gamma_X}$ of ${\rho_X:SL(X)\rightarrow Sp(2g-2,\mathbb{Z})}$ is Zariski dense in ${Sp(2g-2,\mathbb{R})}$ and ${\rho_X}$ is faithful.

As we are going to see in the next section, it is not hard to construct explicit examples of square-tiled surfaces ${X}$ (of genus ${g\geq 3}$ without non-trivial automorphisms) such that the Veech group ${SL(X)}$ is virtually free. In other words, a positive solution of (c) is somewhat related to a positive answer to the following problem of independent interest (for the experts on translation surfaces):

• (d) find a square-tiled surface ${X}$ of genus ${g\geq 3}$ such that the monodromy representation ${\rho_X:\textrm{Aff}(X)\rightarrow Sp(H_1^{(0)}(X,\mathbb{Z}))}$ is faithful.

Remark 3 If a square-tiled surface ${X}$ has a rational direction of homological dimension one (i.e., the waist curves of maximal cylinders of ${X}$ in a direction of rational slope span a one-dimensional subspace in absolute homology), then ${\rho_X}$ is not faithful: indeed, this happens because Dehn twists in such a direction act by the identity matrix on ${H_1^{(0)}(X,\mathbb{Z})}$ (see, e.g., Lemma 5.3 of this paper here), that is, they induce elements of ${\textrm{Ker}(\rho_X)-\{\textrm{Id}\}}$. In particular, since any square-tiled surface ${X}$ of genus ${2}$ always have rational directions of one-cylinder decompositions (by the works of Hubert-Lelièvre and McMullen), it follows that the answer for the analog of item (d) in genus ${2}$ is always negative.

This remark suggests to try to answer items (c) and (d) by looking at square-tiled surfaces of genus ${g\geq 3}$ with no one-cylinder rational directions. After a computer search (using Sage), Vincent Delecroix found a beautiful example of such a square-tiled surface whose description occupies the next section.

Remark 4 For any prescribed ${k\in\mathbb{N}}$, one can construct infinite families of square-tiled surfaces without ${k}$-cylinders rational directions (via the Hubert-Lelièvre-Kani invariant described in this paper here).

2. An origami without one-cylinder decompositions

Consider the square-tiled surface ${\mathcal{O}_1}$ associated to the pair of permutations

$\displaystyle h_{\mathcal{O}_1} = (1) (2,3,4,5)(6,7,8,9), \quad v_{\mathcal{O}_1} = (1,2,3,6)(4,7,9,8)(5)$

The commutator ${[h_{\mathcal{O}_1}, v_{\mathcal{O}_1}]:=v_{\mathcal{O}_1} h_{\mathcal{O}_1} v_{\mathcal{O}_1}^{-1} h_{\mathcal{O}_1}^{-1}}$ is

$\displaystyle [h_{\mathcal{O}_1}, v_{\mathcal{O}_1}]=(1,9)(2,3)(4,6)(5,8)(7),$

so that ${\mathcal{O}_1\in\mathcal{H}(1,1,1,1)}$ is a genus ${3}$ square-tiled surface.

The ${SL(2,\mathbb{Z})}$-orbit of ${\mathcal{O}_1}$ consists of four elements. Indeed, this fact can be checked as follows. We recall that:

• the generators ${T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)}$ and ${S=\left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right)}$ of ${SL(2,\mathbb{Z})}$ act on pairs of permutations ${(h,v)}$ by the rules ${T(h,v) = (h,vh^{-1})}$ and ${S(h,v)=(hv^{-1}, v)}$;
• the pairs of permutations ${(h,v)}$ and ${(\phi h \phi^{-1}, \phi v \phi^{-1})}$ give rise to the same square-tiled surface.

Therefore, the ${T}$-orbit of ${\mathcal{O}_1}$ is ${\{\mathcal{O}_1, \mathcal{O}_2, \mathcal{O}_3, \mathcal{O}_4\}}$ where ${\mathcal{O}_k:=T^k(\mathcal{O}_1)}$ is given by the pair of permutations ${(h_{\mathcal{O}_1}, v_{\mathcal{O}_k})}$ with

$\displaystyle v_{\mathcal{O}_2} = (1,2,5,7)(3)(4,6,8,9), \quad v_{\mathcal{O}_3} = (1,2,7,8)(3,5,6,4)(9),$

$\displaystyle v_{\mathcal{O}_4} = (1,2,6,9)(3,7,4,5)(8)$

As it turns out, the ${T}$-orbit of ${\mathcal{O}_1}$ accounts for its entire ${SL(2,\mathbb{Z})}$-orbit because

$\displaystyle S(\mathcal{O}_1)=(\phi_4^{-1} h_{\mathcal{O}_1} \phi_4, \phi_4^{-1} v_{\mathcal{O}_4} \phi_4)\simeq \mathcal{O}_4, \quad S^2(\mathcal{O}_1)=(\phi_3^{-1} h_{\mathcal{O}_1} \phi_3, \phi_3^{-1} v_{\mathcal{O}_3} \phi_3)\simeq \mathcal{O}_3,$

$\displaystyle S^3(\mathcal{O}_1)=(\phi_2^{-1} h_{\mathcal{O}_1} \phi_2, \phi_2^{-1} v_{\mathcal{O}_2} \phi_2)\simeq \mathcal{O}_2,$

where

$\displaystyle \phi_4 = (1,6,2,9,4,3)(5,8)(7), \quad \phi_3 = (1,5,9,8)(2,6,3,4)(7)$

and

$\displaystyle \phi_2 = (1,9)(2,4,5,3,6,8)(7).$

Remark 5 For later use, observe that the matrix ${-\textrm{Id}}$ acts on pairs of permutations by ${-\textrm{Id}(h,v)=(h^{-1},v^{-1})}$. In particular, the action of ${-\textrm{Id}}$ on ${SL(2,\mathbb{Z})\cdot\mathcal{O}_1}$ is completely described by the formulas

$\displaystyle -\textrm{Id}(\mathcal{O}_1) = (\psi_{3}^{-1} h_{\mathcal{O}_1} \psi_{3}, \psi_3^{-1} v_{\mathcal{O}_3} \psi_3)\simeq \mathcal{O}_3, \,\,\, -\textrm{Id}(\mathcal{O}_2) = (\psi_{4}^{-1} h_{\mathcal{O}_1} \psi_{4}, \psi_4^{-1} v_{\mathcal{O}_4} \psi_4)\simeq \mathcal{O}_4$

where ${\psi_3 := (1)(2,8,4,6)(3,7,5,9)}$ and ${\psi_4 := (1)(2,9,4,7)(3,8,5,6)}$.

In summary, the ${SL(2,\mathbb{Z})}$-orbit of ${\mathcal{O}_1}$ can be depicted as follows.

Since the cylinder decompositions of ${\mathcal{O}_1}$ in any rational direction described by the horizontal cylinder decomposition of some element of ${SL(2,\mathbb{Z})\cdot\mathcal{O}_1=\{\mathcal{O}_1, \mathcal{O}_2, \mathcal{O}_3, \mathcal{O}_4\}}$, we have that all cylinder decompositions in rational directions of ${\mathcal{O}_1}$ have exactly three cylinders.

Remark 6 It follows from this discussion that ${SL(2,\mathbb{R})\cdot\mathcal{O}_1}$ has a single cusp (i.e., single ${T}$-orbit in ${SL(2,\mathbb{Z})\cdot\mathcal{O}_1}$). Also, the homological dimension of ${SL(2,\mathbb{R})\cdot\mathcal{O}_1}$ in the sense of Forni is three. Thus, by the results in this paper of Forni, the Lyapunov spectrum of the Kontsevich-Zorich cocycle over ${SL(2,\mathbb{R})\cdot\mathcal{O}_1}$ with respect to the Haar measure has the form

$\displaystyle 1=\lambda_1>\lambda_2\geq\lambda_3>0>-\lambda_3\geq-\lambda_2>-\lambda_1=-1$

Moreover, the Eskin-Kontsevich-Zorich formula for the sum of non-negative Lyapunov exponents of the Kontsevich-Zorich cocycle imply that ${1+\lambda_2+\lambda_3 = 2}$, i.e.,

$\displaystyle \lambda_2+\lambda_3=1$

Some numerical experiments (with Sage) indicate that ${\lambda_2\simeq 0.57...}$ and ${\lambda_3\simeq 0.43...}$

3. The affine homeomorphisms of ${\mathcal{O}_1}$

The group ${\textrm{Aff}(\mathcal{O}_1)}$ of affine homeomorphisms of ${\mathcal{O}_1}$ is the stabilizer of ${\mathcal{C}:=SL(2,\mathbb{R})\cdot\mathcal{O}_1}$ in the moduli space of translation surfaces.

It is not hard to see that the subgroup ${\textrm{Aut}(\mathcal{O}_1)\subset \textrm{Aff}(\mathcal{O}_1)}$ of automorphisms of ${\mathcal{O}_1}$ is trivial. It follows that the elements of ${\textrm{Aff}(\mathcal{O}_1)}$ are determined by their linear parts in ${SL(2,\mathbb{R})}$, that is, the natural map

$\displaystyle \textrm{Aff}(\mathcal{O}_1)\rightarrow SL(2,\mathbb{R})$

is injective. Hence, ${\textrm{Aff}(\mathcal{O}_1)}$ is isomorphic to its image ${SL(\mathcal{O}_1)}$ under this map.

The group ${SL(\mathcal{O}_1)}$ is the finite-index subgroup of ${SL(2,\mathbb{Z})}$ consisting of all elements of ${SL(2,\mathbb{R})}$ stabilizing ${\mathcal{O}_1}$: in the literature, ${SL(\mathcal{O}_1)}$ is called the Veech group of ${\mathcal{O}_1}$.

We saw in the previous section that ${\# SL(2,\mathbb{Z})\cdot\mathcal{O}_1 = 4}$. Thus, ${SL(\mathcal{O}_1)}$ is an index four subgroup of ${SL(2,\mathbb{Z})}$. Furthermore, ${SL(\mathcal{O}_1)}$ is a congruence subgroup of level ${4}$, and the Teichmüller curve ${\mathcal{C} =SL(2,\mathbb{R})/SL(\mathcal{O}_1)}$ has genus zero. Thus, ${SL(\mathcal{O}_1)}$ is generated by elliptic and parabolic elements: indeed, one can check that ${SL(\mathcal{O}_1)}$ is generated by the following two elliptic matrices

$\displaystyle a:=\left(\begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array}\right), \quad b:=\left(\begin{array}{cc} 1 & -3 \\ 1 & -2\end{array}\right)$

of orders ${3}$.

The group structure of ${SL(\mathcal{O}_1)}$ is provided by the following lemma:

Lemma 2 ${SL(\mathcal{O}_1)}$ is the free product

$\displaystyle SL(\mathcal{O}_1)=\langle a\rangle\ast\langle b\rangle \simeq \mathbb{Z}/3\mathbb{Z}\ast \mathbb{Z}/3\mathbb{Z}$

Proof: Consider the twelve cones ${C_k\subset\mathbb{R}^2-\{(0,0)\}}$ defined by the following properties:

• ${C_{6+l} = -C_l}$ for each ${l=1,\dots, 6}$;
• each ${C_l}$, ${l=1,\dots, 6}$, consists of the convex combinations of positive multiples of the vectors ${v_l}$ and ${v_{l+1}}$, where ${v_1:=(1,0)}$, ${v_2:=(2,1)}$, ${v_3:=(1,1)}$, ${v_4:=(1,2)}$, ${v_5:=(0,1)}$, ${v_6:=(-1,1)}$ and ${v_7:=(-1,0)}$.

A simple calculation shows that

• ${a(v_l)=v_{l+4}}$ for each ${k=1,\dots, 6}$;
• ${b(v_1)=v_3}$, ${b(v_2)=v_7}$, ${b(v_3)=-v_2}$, ${b(v_4)=(-5,-3)\in C_8}$, ${b(v_5)=(-3,-2)\in C_8}$, ${b(v_6)=(-4,-3)\in C_8}$ and ${b(v_7)=-v_3}$.

It follows that ${\{a, a^2\}=\langle a\rangle-\{\textrm{Id}\}}$ and ${\{b, b^2\}=\langle b\rangle-\{\textrm{Id}\}}$ play ping-pong with the tables

$\displaystyle X:=(C_1\cup C_2)\cup (C_7\cup C_8)$

and

$\displaystyle Y:=C_3\cup C_4\cup C_5\cup C_6\cup C_9\cup C_{10}\cup C_{11}\cup C_{12}$

in the sense that ${X}$ and ${Y}$ are disjoint subsets of ${\mathbb{R}^2}$ such that

• ${a(X)=(C_5\cup C_6)\cup(C_{11}\cup C_{12})\subset Y}$, ${a^2(X) = (C_9\cup C_{10})\cup (C_3\cup C_4)\subset Y}$;
• ${b(Y)\subset C_2\cup C_8\subset X}$, ${b^2(Y)\subset C_1\cup C_7\subset X}$.

By the ping-pong lemma, we conclude that ${SL(\mathcal{O}_1) = \langle a\rangle\ast\langle b\rangle}$. $\Box$

Remark 7 The construction of the cones ${C_k}$ above is similar to the cones studied in Section 2 of this paper of Brav-Thomas.

4. The Kontsevich-Zorich monodromy of ${\mathcal{O}_1}$

The natural representation ${\alpha:\textrm{Aff}(\mathcal{O}_1)\rightarrow \textrm{Sp}(H_1(\mathcal{O}_1,\mathbb{Z}))}$ is called Kontsevich-Zorich cocycle over the arithmetic Teichmüller curve ${SL(2,\mathbb{R})\cdot\mathcal{O}_1\simeq \mathcal{C}}$. In the sequel, we will compute the image under ${\alpha}$ of the generators ${a}$ and ${b}$ of ${SL(\mathcal{O}_1)\simeq\textrm{Aff}(\mathcal{O}_1)}$.

4.1. The relative homology groups of ${\mathcal{O}_k}$, ${k=1,\dots, 4}$

Given ${\mathcal{O}_k\in SL(2,\mathbb{Z})\cdot\mathcal{O}_1}$, ${k=1,\dots, 4}$, let us denote by ${\sigma_g^{(k)}}$, resp., ${\zeta_g^{(k)}}$ the relative cycles on ${\mathcal{O}_k}$ consisting of the bottommost horizontal and leftmost vertical sides of the square numbered ${g\in\{1,\dots,9\}}$.

Note that each square ${g}$ of ${\mathcal{O}_k}$ gives a relation ${\sigma_g^{(k)} + \zeta_{h_{O_1}(g)}^{(k)} = \zeta_g^{(k)} + \sigma_{v_{\mathcal{O}_k}(g)}^{(k)}}$, that is,

• ${\sigma_1^{(1)} = \sigma_2^{(1)}}$, ${\sigma_2^{(1)}+\zeta_3^{(1)} = \zeta_2^{(1)} + \sigma_3^{(1)}}$, ${\sigma_3^{(1)}+\zeta_4^{(1)} = \zeta_3^{(1)} + \sigma_6^{(1)}}$, ${\sigma_4^{(1)}+\zeta_5^{(1)} = \zeta_4^{(1)} + \sigma_7^{(1)}}$, ${\zeta_2^{(1)} = \zeta_5^{(1)}}$, ${\sigma_6^{(1)}+\zeta_7^{(1)} = \zeta_6^{(1)} + \sigma_1^{(1)}}$, ${\sigma_7^{(1)}+\zeta_8^{(1)} = \zeta_7^{(1)} + \sigma_9^{(1)}}$, ${\sigma_8^{(1)}+\zeta_9^{(1)} = \zeta_8^{(1)} + \sigma_4^{(1)}}$, ${\sigma_9^{(1)}+\zeta_6^{(1)} = \zeta_9^{(1)} + \sigma_8^{(1)}}$;
• ${\sigma_1^{(2)} = \sigma_2^{(2)}}$, ${\sigma_2^{(2)}+\zeta_3^{(2)} = \zeta_2^{(2)} + \sigma_5^{(2)}}$, ${\zeta_3^{(2)} = \zeta_4^{(2)}}$, ${\sigma_4^{(2)}+\zeta_5^{(2)} = \zeta_4^{(2)} + \sigma_6^{(2)}}$, ${\sigma_5^{(2)}+\zeta_2^{(2)} = \zeta_5^{(2)} + \sigma_7^{(2)}}$, ${\sigma_6^{(2)}+\zeta_7^{(2)} = \zeta_6^{(2)}+ \sigma_8^{(2)}}$, ${\sigma_7^{(2)}+\zeta_8^{(2)} = \zeta_7^{(2)} + \sigma_1^{(2)}}$, ${\sigma_8^{(2)}+\zeta_9^{(2)} = \zeta_8^{(2)} + \sigma_9^{(2)}}$, ${\sigma_9^{(2)}+\zeta_6^{(2)} = \zeta_9^{(2)} + \sigma_4^{(2)}}$;
• ${\sigma_1^{(3)} = \sigma_2^{(3)}}$, ${\sigma_2^{(3)}+\zeta_3^{(3)} = \zeta_2^{(3)} + \sigma_7^{(3)}}$, ${\sigma_3^{(3)}+\zeta_4^{(3)} = \zeta_3^{(3)} + \sigma_5^{(3)}}$, ${\sigma_4^{(3)}+\zeta_5^{(3)} = \zeta_4^{(3)} + \sigma_3^{(3)}}$, ${\sigma_5^{(3)}+\zeta_2^{(3)} = \zeta_5^{(3)} + \sigma_6^{(3)}}$, ${\sigma_6^{(3)}+\zeta_7^{(3)} = \zeta_6^{(3)} + \sigma_4^{(3)}}$, ${\sigma_7^{(3)}+\zeta_8^{(3)} = \zeta_7^{(3)} + \sigma_8^{(3)}}$, ${\sigma_8^{(3)}+\zeta_9^{(3)} = \zeta_8^{(3)} + \sigma_1^{(3)}}$, ${\zeta_6^{(3)} = \zeta_9^{(3)}}$;
• ${\sigma_1^{(4)} = \sigma_2^{(4)}}$, ${\sigma_2^{(4)}+\zeta_3^{(4)} = \zeta_2^{(4)} + \sigma_6^{(4)}}$, ${\sigma_3^{(4)}+\zeta_4^{(4)} = \zeta_3^{(4)} + \sigma_7^{(4)}}$, ${\sigma_4^{(4)}+\zeta_5^{(4)} = \zeta_4^{(4)} + \sigma_5^{(4)}}$, ${\sigma_5^{(4)}+\zeta_2^{(4)} = \zeta_5^{(4)} + \sigma_3^{(4)}}$, ${\sigma_6^{(4)}+\zeta_7^{(4)} = \zeta_6^{(4)} + \sigma_9^{(4)}}$, ${\sigma_7^{(4)}+\zeta_8^{(4)} = \zeta_7^{(4)} + \sigma_4^{(4)}}$, ${\zeta_8^{(4)} = \zeta_9^{(4)}}$, ${\sigma_9^{(4)}+\zeta_6^{(4)} = \zeta_9^{(4)} + \sigma_1^{(4)}}$.

4.2. The action of ${SL(2,\mathbb{Z})}$ on the relative homology groups

The matrix ${T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)}$ takes ${\mathcal{O}_k}$ to ${\mathcal{O}_{k+1}}$, and it acts on the corresponding relative homology groups by the matrix ${T_{k,k+1}}$ such that

$\displaystyle T_{k,k+1}(\sigma_g^{(k)}) = \sigma_g^{(k+1)}, \quad T_{k,k+1}(\zeta_g^{(k)}) = \zeta_g^{(k+1)} + \sigma_{h_{\mathcal{O}_1}(g)}^{(k+1)},$

Similarly, the matrix ${S=\left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right)}$ takes ${\mathcal{O}_{k}}$ to ${\mathcal{O}_{k-1}}$, and it acts on the corresponding relative homology groups by the matrix ${S_{k+1,k}}$ such that

$\displaystyle S_{k+1,k}(\sigma_g^{(k+1)}) = \zeta_{\phi_k(g)}^{(k)} + \sigma_{v_{\mathcal{O}_k}(\phi_k(g))}^{(k)}, \quad S_{k+1,k}(\zeta_g^{(k+1)}) = \zeta_{\phi_k(g)}^{(k)}$

Finally, ${-\textrm{Id}}$ exchange ${\mathcal{O}_1}$ and ${\mathcal{O}_3}$, resp. ${\mathcal{O}_2}$ and ${\mathcal{O}_4}$, and it acts on the corresponding relative homology groups by the matrices ${(-\textrm{Id})_{1,3}=(-\textrm{Id})_{3,1}^{-1}}$ and ${(-\textrm{Id})_{2,4}=(-\textrm{Id})_{4,2}^{-1}}$ such that

$\displaystyle (-\textrm{Id})_{1,3}(\sigma_g^{(1)}) = -\sigma_{v_{\mathcal{O}_3(\psi_3(g))}}^{(3)}, \quad (-\textrm{Id})_{1,3}(\zeta_g^{(1)}) = -\zeta_{h_{\mathcal{O}_3(\psi_3(g))}}^{(3)},$

and

$\displaystyle (-\textrm{Id})_{2,4}(\sigma_g^{(2)}) = -\sigma_{v_{\mathcal{O}_4}(\psi_4(g))}^{(4)}, \quad (-\textrm{Id})_{2,4}(\zeta_g^{(2)}) = -\zeta_{h_{\mathcal{O}_4}(\psi_4(g))}^{(4)}$

4.3. The absolute homology groups of ${\mathcal{O}_k}$, ${k=1,\dots, 4}$

The absolute homology group ${H_1(\mathcal{O}_1,\mathbb{Q})}$ has a basis ${\mathcal{B}_k:=\{\Sigma_0^{(k)}, Z_0^{(k)}, \Sigma_1^{(k)}, \Sigma_2^{(k)}, Z_1^{(k)}, Z_2^{(k)}\}}$ where

$\displaystyle \Sigma_0^{(k)} := \sum\limits_{g=1}^9 \sigma_g^{(k)}, \quad Z_0^{(k)} := \sum\limits_{g=1}^9 \zeta_g^{(k)},$

$\displaystyle \Sigma_1^{(k)} := \sum\limits_{j=1}^4\sigma_{h_{\mathcal{O}_1}^j(2)}^{(k)} - 4\sigma_1^{(k)}, \Sigma_2^{(k)} := \sum\limits_{j=1}^4\sigma_{h_{\mathcal{O}_1}^j(6)}^{(k)} - 4\sigma_1^{(k)},$

and

$\displaystyle Z_1^{(1)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(1)}^{(1)} - 4\zeta_5^{(1)}, \quad Z_2^{(1)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(4)}^{(1)} - 4\zeta_5^{(1)},$

$\displaystyle Z_1^{(2)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(1)}^{(2)} - 4\zeta_3^{(2)}, \quad Z_2^{(2)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(4)}^{(1)} - 4\zeta_3^{(1)},$

$\displaystyle Z_1^{(3)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(1)}^{(1)} - 4\zeta_9^{(1)}, \quad Z_2^{(3)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(4)}^{(1)} - 4\zeta_9^{(1)},$

$\displaystyle Z_1^{(4)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(1)}^{(1)} - 4\zeta_8^{(1)}, \quad Z_2^{(4)} := \sum\limits_{j=1}^4 \zeta_{v_{\mathcal{O}_1}^j(4)}^{(1)} - 4\zeta_8^{(1)}.$

Note that this basis is adapted to the decomposition ${H_1(\mathcal{O}_k,\mathbb{Q}) = H_1^{st}(\mathcal{O}_k,\mathbb{Q}) \oplus H_1^{(0)}(\mathcal{O}_k,\mathbb{Q})}$ in the sense that this decomposition corresponds to the partition ${\mathcal{B}_k=\mathcal{B}_k^{st}\cup \mathcal{B}_k^{(0)}}$ where ${\mathcal{B}_k^{st}=\{\Sigma_0^{(k)}, Z_0^{(k)}\}}$ and ${\mathcal{B}_k^{(0)}=\mathcal{B}_k - \mathcal{B}_k^{st}}$, i.e.,

$\displaystyle H_1^{st}(\mathcal{O}_k,\mathbb{Q}) = \mathbb{Q}\Sigma_0^{(k)}\oplus\mathbb{Q} Z_0^{(k)}$

and

$\displaystyle H_1^{(0)}(\mathcal{O}_k,\mathbb{Q}) = \mathbb{Q}\Sigma_1^{(k)}\oplus\mathbb{Q} Z_1^{(k)}\oplus \mathbb{Q}\Sigma_2^{(k)}\oplus\mathbb{Q} Z_2^{(k)}.$

Moreover, it is worth to point out that, in the basis ${\mathcal{B}_1^{(0)}}$, the matrix of the restriction to ${H_1^{(0)}(\mathcal{O}_1,\mathbb{Z})}$ of the intersection form ${\Omega}$ is

$\displaystyle \Omega = \left(\begin{array}{cccc} 0 & 0 & -6 & -3 \\ 0 & 0 & -3 & 3 \\ 6 & 3 & 0 & 0 \\ 3 & -3 & 0 & 0 \end{array}\right)$

4.4. The action of ${\textrm{Aff}(\mathcal{O}_1)}$ on the absolute homology group

The matrices of ${T_{k,k+1}}$, ${S_{k+1,k}}$ and ${-(\textrm{Id})_{k,k+2}}$ with respect to the basis ${\mathcal{B}_l}$ are

$\displaystyle T_{1,2}=\left(\begin{array}{cccccc} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1\end{array}\right), \quad T_{2,3}=\left(\begin{array}{cccccc} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & -1 & -1 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & -1\end{array}\right),$

$\displaystyle T_{3,4}=\left(\begin{array}{cccccc} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1\end{array}\right), \quad T_{4,1}=\left(\begin{array}{cccccc} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & -1 & -1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & -1 & -1\end{array}\right),$

$\displaystyle S_{1,4}=\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1\end{array}\right), \quad S_{4,3}=\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & -1 & -1 & 1 & 0\end{array}\right),$

$\displaystyle S_{3,2}=\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1\end{array}\right), \quad S_{2,1}=\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & -1 & -1 & 1 & 0\end{array}\right),$

$\displaystyle (-\textrm{Id})_{1,3} = \left(\begin{array}{cccccc} -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1\end{array}\right) = (-\textrm{Id})_{2,4}$

This allows us to compute the images ${\alpha(a)}$ and ${\alpha(b)}$ of the generators ${a}$ and ${b}$ of ${SL(\mathcal{O}_1)\simeq\textrm{Aff}(\mathcal{O}_1)}$ under the KZ cocycle ${\alpha:\textrm{Aff}(\mathcal{O}_1)\rightarrow \textrm{Sp}(H_1(\mathcal{O}_1,\mathbb{Z}))}$. Indeed,

$\displaystyle a=\left(\begin{array}{cc} 0 & -1 \\ 1 & -1 \end{array}\right) = (-\textrm{Id}) T S^{-1}, \quad b=\left(\begin{array}{cc} 1 & -3 \\ 1 & -2\end{array}\right) = ST^{-3},$

so that

$\displaystyle \alpha(a) = (-\textrm{Id})_{3,1} T_{2,3} S_{2,1}^{-1}, \quad \alpha(b) = S_{2,1}T_{2,3}^{-1}T_{3,4}^{-1}T_{4,1}^{-1}$

For later use, we observe that these formulas give that the non-tautological subrepresentation ${\rho:=\rho_{\mathcal{O}_1}:\textrm{Aff}(\mathcal{O}_1)\rightarrow \textrm{Sp}(H_1^{(0)}(\mathcal{O}_1,\mathbb{Z}))}$ of ${\alpha}$ takes values

$\displaystyle \rho(a) = \left(\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 1 \end{array}\right), \quad \rho(b) = \left(\begin{array}{cccc} 1 & 0 & 3 & 3 \\ -1 & -1 & -2 & -1 \\ 0 & 1 & -1 & -1 \\ -1 & -1 & -1 & -1\end{array}\right)$

(with respect to the basis ${\mathcal{B}_1^{(0)}}$ of ${H_1^{(0)}(\mathcal{O}_1,\mathbb{Z})}$) at the two generators ${a}$ and ${b}$ of ${SL(\mathcal{O}_1)}$. Moreover, if we denote by ${p_1 = ST^{-4}S^{-1}T^4}$, ${p_2 = ST^{-4}ST^6}$, then the characteristic polynomials ${\chi_{p_1}(x)}$ and ${\chi_{p_2}(x)}$ of the matrices ${\rho(p_1)}$ and ${\rho(p_2)}$ are

$\displaystyle \chi_{p_1}(x) = x^4 - 11x^3 + 29x^2 - 11x + 1$

and

$\displaystyle \chi_{p_2}(x) = x^4 -2x^3 -16x^2 -2x +1$

5. Zariski density of ${\rho(\text{Aff}(\mathcal{O}_1))}$ in ${\text{Sp}(H_1^{(0)}(\mathcal{O}_1,\mathbb{R}))}$

The matrices ${\rho(p_1)}$ and ${\rho(p_2)}$ are Galois-pinching, i.e., all roots of their characteristic polynomials ${\chi_{p_1}}$ and ${\chi_{p_2}}$ are real and simple, and the Galois groups of ${\chi_{p_1}}$ and ${\chi_{p_2}}$ have order ${8}$ (that is, the largest possible for symplectic ${4\times 4}$matrices). Furthermore, the splitting fields of their characteristic polynomials are disjoint.

Indeed, these facts follow from the analysis of the following discriminants

$\displaystyle \Delta_1(\chi_{p_1}) = (-11)^2 - 4\times (29-2) = 13, \quad \Delta_1(\chi_{p_2}) = (-2)^2 - 4\times (-16-2) = 2^2\times 19$

and

$\displaystyle \Delta_2(\chi_{p_1}) = (29+2)^2 - 4\times (-11)^2 = 3^2\times 53, \,\, \Delta_2(\chi_{p_2}) = (-16+2)^2 - 4\times(-2)^2 = 6^2\times 5$

related to the quadratic subfields of the splitting fields of ${\chi_{p_1}}$ and ${\chi_{p_2}}$: see Proposition 6.14, Remark 6.15 and Proposition 6.16 in this paper here for more details.

By the Zariski density criterion of Prasad-Rapinchuk (see also page 3 of Rivin’s paper or this blog post here), we deduce that:

Proposition 3 The Kontsevich-Zorich monodromy ${\Gamma_{\mathcal{O}_1}:=\rho(\textrm{Aff}(\mathcal{O}_1))}$ is Zariski-dense in ${\textrm{Sp}(H_1^{(0)}(\mathcal{O}_1,\mathbb{R}))\simeq Sp(4,\mathbb{R})}$.

Remark 8 By the main result in this paper here, the Zariski-density of ${\Gamma_{\mathcal{O}_1}}$ in ${Sp(4,\mathbb{R})}$ implies that the Lyapunov spectrum in Remark 6 is simple, i.e.,

$\displaystyle 1=\lambda_1>\lambda_2>\lambda_3>-\lambda_3>-\lambda_2>-\lambda_1=-1$

6. Non-faithfulness of the representation ${\rho}$

Once we have Proposition 3 in our toolkit, it is natural to investigate the thinness of ${\Gamma_{O_1}}$. Here, it is tempting to try to use the thinness certificate discussed in Proposition 1 from Section 1 in order to give a positive (partial) answer to Sarnak’s question.

By definition, the faithfulness of ${\rho}$ would amount to show that ${\Gamma_{\mathcal{O}_1}}$ has the same group structure of the Veech group ${SL(\mathcal{O}_1)}$ (described in Lemma 2), i.e., ${\rho(\textrm{Aff}(\mathcal{O}_1))\simeq\langle\rho(a)\rangle\ast\langle\rho(b)\rangle \simeq \mathbb{Z}/3\mathbb{Z}\ast \mathbb{Z}/3\mathbb{Z}}$.

After some numerical experiments with all non-trivial words of length ${\leq 23}$ on ${\rho(a)}$ and ${\rho(b)}$ and some non-trivial words of length ${24}$ on ${\rho(a)}$ and ${\rho(b)}$, I thought that ${\rho}$ could be faithful.

As it turns out, after I asked about the faithfulness of ${\rho}$ on MathOverflow, Stefan Kohl noticed that ${\rho}$ is not faithful because (a computer-assisted calculation shows that) the kernel of ${\rho}$ contains a certain non-trivial word of length ${24}$ in ${\rho(a)}$ and ${\rho(b)}$.

Proposition 4 (S. Kohl) The representation ${\rho:\textrm{Aff}(\mathcal{O}_1)\rightarrow \textrm{Sp}(H_1^{(0)}(\mathcal{O}_1),\mathbb{Z})}$ is not faithful because

$\displaystyle \rho\left(\begin{array}{cc} -24587 & 42408 \\ 15048 & -25955\end{array}\right) = \rho((aba^{-1}ba^{-1}bab^{-1})^3) = \textrm{Id}_{4\times 4}$

Remark 9 The matrices ${A}$ and ${B}$ appearing in my MathOverflow question generate a conjugate of ${\langle \rho(a), \rho(b)\rangle}$: indeed,

$\displaystyle A:= \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & -1 & 0 \end{array}\right) = \Theta^{-1}\rho(a)\Theta$

and

$\displaystyle B:= \left(\begin{array}{cccc} -1 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right) = \Theta^{-1}\rho(b)\Theta$

where

$\displaystyle \Theta := \left(\begin{array}{cccc} 1 & 1 & 1 & -1 \\ -1 & 0 & 0 & 1 \\ -1 & -1 & 0 & -1 \\ 0 & 1 & -1 & 1 \end{array}\right)$