Zorich conjecture for hyperelliptic Rauzy–Veech groups

Abstract

We describe the structure of hyperelliptic Rauzy diagrams and hyperelliptic Rauzy–Veech groups. In particular, this provides a solution of the hyperelliptic cases of a conjecture of Zorich on the Zariski closure of Rauzy–Veech groups.

Appendix A: A pinching and twisting group with small Zariski closure

Let $\rho$ be the third symmetric power of the standard representation of $SL(2,\mathbb {R})$. In concrete terms, $\rho$ is constructed as follows. Consider the basis $\mathcal {B} = \{X^3, X^2Y, XY^2, Y^3\}$ of the space V of homogenous polynomials of degree 3 on two variables X and Y. By letting $g=\left( \begin{array}{ll} a &{} b \\ c &{} d \end{array}\right) \in SL(2,\mathbb {R})$ act on X and Y as $g(X)=aX+cY$ and $g(Y)=bX+dY$, we obtain an induced action $\rho (g)$ on V whose matrix in the basis $\mathcal {B}$ is

$$\begin{aligned} \rho \left( \begin{array}{ll} a &{} \quad b \\ c &{} \quad d \end{array}\right) = \left( \begin{array}{cccc} a^3 &{} \quad a^2 b &{} \quad a b^2 &{} \quad b^3 \\ 3 a^2 c &{} \quad a^2 d + 2 a b c &{} \quad b^2 c + 2 a b d &{} \quad 3 b^2 d \\ 3 a c^2 &{} \quad b c^2 + 2 a c d &{} \quad a d^2 + 2 b c d &{} \quad 3 b d^2 \\ c^3 &{} \quad c^2 d &{} \quad c d^2 &{} \quad d^3 \end{array}\right) \end{aligned}$$

Note that the faithful representation $\rho$ is the unique irreducible four-dimensional representation of $SL(2,\mathbb {R})$. Furthermore, the matrices $\rho (g)$ preserve the symplectic structure on V associated to the matrix

$$\begin{aligned} J = \left( \begin{array}{llll} 0 &{} \quad 0 &{} \quad 0 &{} \quad -1 \\ 0 &{} \quad 0 &{} \quad 1/3 &{} \quad 0 \\ 0 &{} \quad -1/3 &{} \quad 0 &{} \quad 0 \\ 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \end{array}\right) \end{aligned}$$

Indeed, a direct calculation shows that if $g=\left( \begin{array}{ll} a &{} b \\ c &{} d \end{array}\right)$, then

$$\begin{aligned} {}^t\rho (g)\cdot J\cdot \rho (g) = \left( \begin{array}{llll} 0 &{} \quad 0 &{} \quad 0 &{} \quad -(a d - b c)^3 \\ 0 &{} \quad 0 &{} \quad \frac{(a d - b c)^3}{3} &{} \quad 0 \\ 0 &{} \quad - \frac{(a d - b c)^3}{3} &{} \quad 0 &{} \quad 0 \\ (a d - b c)^3 &{} \quad 0 &{} \quad 0 &{} \quad 0 \end{array}\right) \end{aligned}$$

where ${}^t\rho (g)$ stands for the transpose of $\rho (g)$.

Denote by $\mathcal {M}$ the group generated by the matrices

$$\begin{aligned} A = \rho \left( \begin{array}{ll} 1 &{} \quad 1 \\ 0 &{} \quad 1 \end{array}\right) = \left( \begin{array}{llll} 1 &{} \quad 1 &{} \quad 1 &{} \quad 1 \\ 0 &{} \quad 1 &{} \quad 2 &{} \quad 3 \\ 0 &{} \quad 0 &{} \quad 1 &{} \quad 3 \\ 0 &{} \quad 0 &{} \quad 0 &{} \quad 1\end{array}\right) \quad \text {and} \quad B = \rho \left( \begin{array}{ll} 1 &{} \quad 0 \\ 1 &{} \quad 1 \end{array}\right) = \left( \begin{array}{llll} 1 &{} \quad 0 &{} \quad 0 &{} \quad 0 \\ 3 &{} \quad 1 &{} \quad 0 &{} \quad 0 \\ 3 &{} \quad 2 &{} \quad 1 &{} \quad 0 \\ 1 &{} \quad 1 &{} \quad 1 &{} \quad 1\end{array}\right) \end{aligned}$$

On one hand, the group $\mathcal {M}$ has small Zariski closure.

Proposition A.1

The group $\mathcal {M}$ is not Zariski dense in Sp(V).

Proof

Note that $\rho$ is a polynomial ⁹ homomorphism from $SL(2,\mathbb {R})$ to Sp(V). In particular, it follows that the image $H=\rho (SL(2,\mathbb {R}))$ of $\rho$ is Zariski closed in Sp(V): see, e.g., Corollary 4.6.5 in Witte-Morrris book [12]. Since $SL(2,\mathbb {Z})$ is a Zariski dense subgroup of $SL(2,\mathbb {R})$ generated by $\left( \begin{array}{ll} 1 &{} 1 \\ 0 &{} 1 \end{array}\right)$ and $\left( \begin{array}{ll} 1 &{} 0 \\ 1 &{} 1 \end{array}\right)$, we see that the Zariski closure of the group $\mathcal {M}$ is $H=\rho (SL(2,\mathbb {R}))$ ($\simeq SL(2,\mathbb {R})$ because $\rho$ is faithful). This proves the proposition (since H is a proper linear algebraic subgroup of Sp(V)). $\square$

On the other hand, the group $\mathcal {M}$ is pinching and twisting in the sense of Avila–Viana (see [3] and [9, Sect. 2]):

Proposition A.2

The matrix $A.B\in \mathcal {M}$ is a pinching element¹⁰ and the matrix $A\in \mathcal {M}$ is twisting¹¹ with respect to the pinching element $A.B\in \mathcal {M}$.

Proof

The first assertion follows from the fact that

$$\begin{aligned} 9+4\sqrt{5}> \frac{3+\sqrt{5}}{2}> \frac{3-\sqrt{5}}{2} > \frac{1}{9+4\sqrt{5}} \end{aligned}$$

are the eigenvalues of $A.B\in \mathcal {M}$.

The second assertion is established by the following reasoning. The columns of

$$\begin{aligned} M = \left( \begin{array}{llll} -\frac{1}{4} + \frac{(9 + 4 \sqrt{5})}{4} &{} \quad 1 - \frac{(3 + \sqrt{5})}{2} &{} \quad 1 - \frac{(3 - \sqrt{5})}{2} &{} \quad -\frac{1}{4} + \frac{(9 - 4 \sqrt{5})}{4} \\ \frac{9}{8} + \frac{3(9 + 4 \sqrt{5})}{8} &{} \quad -2 + \frac{(3 + \sqrt{5})}{2} &{} \quad -2 + \frac{(3 - \sqrt{5})}{2} &{} \quad \frac{9}{8} + \frac{3(9 - 4 \sqrt{5})}{8} \\ -\frac{15}{8} + \frac{3(9 + 4 \sqrt{5})}{8} &{} \quad \frac{(3 + \sqrt{5})}{2} &{} \quad \frac{(3 - \sqrt{5})}{2} &{} \quad -\frac{15}{8} + \frac{3(9 - 4 \sqrt{5})}{8} \\ 1 &{} \quad 1 &{} \quad 1 &{} \quad 1 \end{array}\right) \end{aligned}$$

consist of eigenvectors of A.B. Thus, $T=M^{-1}\cdot A\cdot M$ is the matrix of A in the corresponding basis of eigenvectors of A.B. By definition, A is twisting with respect to A.B when all entries of T and all of its $2\times 2$ minors associated to Lagrangian planes are non-zero. As it turns out, this last property holds because a direct computation reveals that T and the matrix $T^{\wedge 2}$ of $2\times 2$ minors are given by:

$$\begin{aligned} T = \left( \begin{array}{llll} \frac{8(5 + 2 \sqrt{5})}{25} &{} \quad \frac{2(5 + 3 \sqrt{5})}{25} &{} \quad \frac{(5 + \sqrt{5})}{25} &{} \quad \frac{1}{( 5 \sqrt{5})} \\ -\frac{6(5 + 3 \sqrt{5})}{25} &{} \quad \frac{2(5 + \sqrt{5})}{25} &{} \quad \frac{7}{5 \sqrt{5}} &{} \quad -\frac{3(-5 + \sqrt{5})}{25} \\ \frac{3(5 + \sqrt{5})}{25} &{} \quad -\frac{7}{5 \sqrt{5}} &{} \quad -\frac{2(-5 + \sqrt{5})}{25} &{} \quad \frac{6(-5 + 3 \sqrt{5})}{25} \\ -\frac{1}{5 \sqrt{5}} &{} \quad \frac{(5 - \sqrt{5})}{25} &{} \quad \frac{2}{5} - \frac{6}{5 \sqrt{5}} &{} \quad -\frac{(8(-5 + 2 \sqrt{5})}{25} \end{array}\right) \end{aligned}$$

and

$$\begin{aligned} T^{\wedge 2} = \left( \begin{array}{llllll} \frac{56}{25} + \frac{24}{5 \sqrt{5}} &{} \quad \frac{32}{25} + \frac{16}{5 \sqrt{5}} &{} \quad \frac{18}{25} + \frac{6}{5 \sqrt{5}} &{} \quad \frac{6}{25} + \frac{2}{5 \sqrt{5}} &{} \quad \frac{2}{25} + \frac{2}{5 \sqrt{5}} &{} \quad \frac{1}{25} \\ -\frac{32}{25} - \frac{16}{5 \sqrt{5}} &{} \quad \frac{6}{25} + \frac{2}{5 \sqrt{5}} &{} \quad \frac{9}{25} + \frac{9}{5 \sqrt{5}} &{} \quad \frac{3}{25} + \frac{3}{5 \sqrt{5}} &{} \quad \frac{11}{25} &{} \quad -\frac{2}{25} + \frac{2}{5 \sqrt{5}} \\ \frac{6}{25} + \frac{2}{5 \sqrt{5}} &{} \quad -\frac{3}{25} - \frac{3}{5 \sqrt{5}} &{} \quad \frac{13}{25} &{} \quad -\frac{4}{25} &{} \quad -\frac{3}{25} + \frac{3}{5 \sqrt{5}} &{} \quad \frac{6}{25} - \frac{2}{5 \sqrt{5}} \\ \frac{18}{25} + \frac{6}{5 \sqrt{5}} &{} \quad -\frac{9}{25} - \frac{9}{5 \sqrt{5}} &{} \quad -\frac{36}{25} &{} \quad \frac{13}{25} &{} \quad -\frac{9}{25} + \frac{9}{5 \sqrt{5}} &{} \quad \frac{18}{25} - \frac{6}{5 \sqrt{5}} \\ -\frac{2}{25} - \frac{2}{5 \sqrt{5}} &{} \quad \frac{11}{25} &{} \quad \frac{9}{25} - \frac{9}{5 \sqrt{5}} &{} \quad \frac{3}{25} - \frac{3}{5 \sqrt{5}} &{} \quad \frac{6}{25} - \frac{2}{5 \sqrt{5}} &{} \quad -\frac{32}{25} + \frac{16}{5 \sqrt{5}} \\ \frac{1}{25} &{} \quad \frac{2}{25} - \frac{2}{5 \sqrt{5}} &{} \quad \frac{18}{25} - \frac{6}{5 \sqrt{5}} &{} \quad \frac{6}{25} - \frac{2}{5 \sqrt{5}} &{} \quad \frac{32}{25} - \frac{16}{5 \sqrt{5}} &{} \quad \frac{56}{25} - \frac{24}{5 \sqrt{5}} \end{array}\right) \end{aligned}$$

$\square$

In summary, these propositions say that $\mathcal {M}$ is the desired group: it is pinching and twisting, but not Zariski dense in Sp(V).

Remark A.3

Observe that the group $\rho (SL(2,\mathbb {Z}))$ does not contain Galois-pinching¹² elements of Sp(V) in the sense of [9] because $H=\rho (SL(2,\mathbb {R}))$ has rank 1. Alternatively, this fact can be shown as follows. A straightforward computation reveals that the characteristic polynomial of $\rho (g)$ is

$$\begin{aligned} (x^2-\text {tr}(g)\det (g)x+\det (g)^3)\cdot (x^2 - \text {tr}(g)(\text {tr}(g)^2 - 3\det (g))x + \det (g)^3) \end{aligned}$$

and, consequently, the eigenvalues of $\rho (g)$ are

$$\begin{aligned} \frac{1}{2}\det (g)\left( \text {tr}(g)\pm \sqrt{\text {tr}(g)^2 - 4 \det (g)}\right) , \end{aligned}$$

and

$$\begin{aligned} \frac{1}{2}\left( \text {tr}(g)(\text {tr}(g)^2 - 3\det (g)) \pm (\text {tr}(g)^2 - \det (g)) \sqrt{\text {tr}(g)^2 - 4 \det (g)}\right) . \end{aligned}$$

Therefore, the Galois group of the characteristic polynomial $\rho (g)$, $g\in SL(2,\mathbb {Z})$, is not the largest possible among reciprocal polynomials of degree four.

Remark A.4

It seems unlikely to find pinching and twisting monoids of symplectic matrices which are not Zariski dense in the context of the Kontsevich–Zorich cocycle. Indeed, Filip’s classification theorem [7] says that, modulo finite-index and compact factors, a Kontsevich–Zorich monodromy $\mathcal {M}_V$ has Zariski closure $\text {Sp}(V)$, $\text {SU}(p,q)$, $\text {SO}^*(2n)$, $\wedge ^k \text {SU}(p,1)$ or some spin groups. Thus, all matrices in $\mathcal {M}_V$ have two eigenvalues with the same modulus unless the Zariski closure of $\mathcal {M}_V$ is isomorphic to $\text {Sp}(V)$ modulo finite-index and compact factors. It follows that if $\mathcal {M}_V$ is pinching, then $\mathcal {M}_V$ is Zariski dense in $\text {Sp}(V)$ modulo finite-index and compact factors.