Posted by: matheuscmss | July 17, 2015

## A pinching and twisting monoid of symplectic matrices which is not Zariski dense in the full symplectic group

Consider a random product of two symplectic matrices ${A_0, A_1 \in Sp(V)}$ on a real symplectic vector space ${V}$ of dimension ${\textrm{dim}(V)=2d}$, that is, the (symplectic) linear cocycle

$\displaystyle F: \{0,1\}^{\mathbb{Z}}\times V\rightarrow \{0,1\}^{\mathbb{Z}}\times V$

given by

$\displaystyle F(x,v) = (\sigma(x), A_{x_0}v)$

where ${x=(x_n)_{n\in\mathbb{Z}}}$ and ${\sigma(x):=(x_{n+1})_{n\in\mathbb{Z}}}$ is the shift map equipped with the Bernoulli measure ${\mathbb{P}=(\frac{1}{2}\delta_0 + \frac{1}{2}\delta_1)^{\mathbb{Z}}}$.

By Oseledets multiplicative ergodic theorem, the Lyapunov exponents of the random product of ${A_0}$ and ${A_1}$ (i.e., the linear cocycle ${F}$) are well-defined quantities ${\lambda_1\geq\dots\geq\lambda_{2d}}$ (depending only on ${A_0}$ and ${A_1}$) describing the exponential growth of the singular values of the random products

$\displaystyle A_{x_{m}}\dots A_{x_0}, \quad m\in\mathbb{N}$

for any ${\mathbb{P}}$-typical choice of ${x=(x_n)_{n\in\mathbb{Z}}}$.

Moreover, the fact that ${A_0}$ and ${A_1}$ are symplectic matrices implies that the Lyapunov exponents are symmetric with respect to the origin, i.e., ${\lambda_{2d-k-1} = -\lambda_k}$ for each ${k=1,\dots,d}$. In other words, the Lyapunov exponents of the symplectic linear cocycle ${F}$ have the form:

$\displaystyle \lambda_1\geq\dots\geq\lambda_d\geq-\lambda_d\geq\dots\geq-\lambda_1$

In fact, this structure of the Lyapunov exponents of a symplectic linear cocycle reflects the fact that if ${\theta}$ is an eigenvalue of a symplectic matrix ${B}$, then ${\theta^{-1}}$ is also an eigenvalue of ${B}$.

A natural qualitative question about Lyapunov exponents concerns their simplicity in the sense that there are no repeated numbers in the list above (i.e., ${\lambda_j > \lambda_{j+1}}$ for all ${k=1,\dots,d}$).

The simplicity property for Lyapunov exponents is the subject of several papers in the literature: see, e.g., the works of Furstenberg, Goldsheid-Margulis, Guivarch-Raugi, and Avila-Viana (among many others).

Very roughly speaking, the basic philosophy behind these papers is that the simplicity property holds whenever the monoid ${\mathcal{M}}$ generated by ${A_0}$ and ${A_1}$ is rich. Of course, there are several ways to formalize the meaning of the word “rich”, for example:

• Goldsheid-Margulis and Guivarch-Raugi asked ${\mathcal{M}}$ to be Zariski-dense in ${Sp(V)}$;
• Avila-Viana required ${\mathcal{M}}$ to be
• pinching: there exists ${C\in\mathcal{M}}$ whose eigenvalues are all real with distinct moduli; such a ${C}$ is called a pinching matrix;
• twisting: there exists a pinching matrix ${C\in\mathcal{M}}$ and a twisting matrix ${D\in\mathcal{M}}$ with respect to ${C}$ in the sense that ${D(F)\cap F'=\{0\}}$ for all isotropic ${C}$-invariant subspaces ${F}$ and all coisotropic ${C}$-invariant subspaces ${F'}$ with ${\textrm{dim}(F) + \textrm{dim}(F')=2d}$.

Of course, these notions of “richness” of a monoid ${\mathcal{M}}$ are “close” to each other, but they differ in a subtle detail: while the Zariski-density condition on ${\mathcal{M}}$ is an algebraic requirement, the pinching and twisting condition on ${\mathcal{M}}$ makes no reference to the algebraic structure of the linear group ${Sp(V)}$.

In particular, this leads us to the main point of this post:

How the Zariski-density and pinching and twisting conditions relate to each other?

The first half of this question has a positive answer: a Zariski-dense monoid ${\mathcal{M}}$ is also pinching and twisting. Indeed:

• (a) a modification of the arguments in this blog post here (in Spanish) permits to prove that any Zariski-dense monoid ${\mathcal{M}}$ contains a pinching matrix ${C}$, and
• (b) the twisting condition on a matrix ${D}$ with respect to a pinching matrix ${C}$ can be phrased in terms of the non-vanishing of certain (isotropic) minors of the matrix of ${D}$ written in a basis of eigenvectors of ${C}$; thus, a Zariski-dense monoid ${\mathcal{M}}$ contains a twisting matrix with respect to any given pinching matrix.

On the other hand, the second half of this question has a negative answer: we exhibit below a pinching and twisting monoid ${\mathcal{M}}$ which is not Zariski dense.

Remark 1 The existence of such examples of monoids is certainly known among experts. Nevertheless, I’m recording it here because it partly “justifies” a forthcoming article joint with Artur Avila and Jean-Christophe Yoccoz in the following sense.

The celebrated paper of Avila-Viana quoted above (on Kontsevich-Zorich conjecture) shows that the so-called “Rauzy monoids” are pinching and twisting (and this is sufficient for their purposes of proving simplicity of Lyapunov exponents of the Kontsevich-Zorich cocycle for Masur-Veech measures).

On the other hand, since a pinching and twisting monoid is not necessarily Zariski dense (as we are going to see below), the results of Avila-Viana (per se) can not answer a question of Zorich (see also Remark 6.12 in Avila-Viana paper) about the Zariski density of Rauzy monoids.

In this direction, Artur, Jean-Christophe and I solve (in an article still in preparation) Zorich’s question about Zariski density of Rauzy monoids in the special case of hyperelliptic Rauzy diagrams, and the main example of this post (which will be included in our forthcoming article with Artur and Jean-Christophe) serves to indicate that the results obtained by Artur, Jean-Christophe and myself can not be deduced as “abstract consequences” of the arguments in Avila-Viana paper.

Remark 2 The main example of this post also shows that (a version of) Prasad-Rapinchuk’s criterion for Zariski density (cf. Theorem 9.10 of Prasad-Rapinchuk paper or Theorem 1.5 in Rivin’s paper) based on Galois-pinching (in the sense of this paper here) and twisting properties is “sharp”: indeed, an important feature of the main example of this post is the failure of the Galois-pinching property (cf. Remark 4 below for more comments).

1. A monoid of 4×4 symplectic matrices

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 ${g=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\in SL(2,\mathbb{R})}$ act on ${X}$ and ${Y}$ as ${g(X)=aX+cY}$ and ${g(Y)=bX+dY}$, we get a linear map ${\rho(g)}$ on ${V}$ whose matrix in the basis ${\mathcal{B}}$ is

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

Remark 3 The faithful representation ${\rho}$ is the unique irreducible four-dimensional representation of ${SL(2,\mathbb{R})}$.

The matrices ${\rho(g)}$ preserve the symplectic structure on ${V}$ associated to the matrix

$\displaystyle J = \left(\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & 1/3 & 0 \\ 0 & -1/3 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right)$

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

$\displaystyle (\rho(g))^T\cdot J\cdot \rho(g) = \left(\begin{array}{cccc} 0 & 0 & 0 & -(a d - b c)^3 \\ 0 & 0 & \frac{(a d - b c)^3}{3} & 0 \\ 0 & - \frac{(a d - b c)^3}{3} & 0 & 0 \\ (a d - b c)^3 & 0 & 0 & 0 \end{array}\right)$

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

Therefore, the image ${H=\rho(SL(2,\mathbb{R}))}$ is a linear algebraic subgroup of the symplectic group ${Sp(V)}$, and the Zariski closure of the monoid ${\mathcal{M}}$ generated by the matrices

$\displaystyle A = \rho\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1\end{array}\right)$

and

$\displaystyle B = \rho\left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array}\right) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 \\ 1 & 1 & 1 & 1\end{array}\right)$

is precisely ${\overline{\mathcal{M}}^{Zariski}=H}$.

Remark 4 Coming back to Remark 2, observe that ${H}$ does not contain Galois-pinching elements of ${Sp(V)}$ in the sense of this paper here (i.e., pinching elements of ${Sp(V)}$ with integral entries whose characteristic polynomial has the largest possible Galois group for a reciprocal polynomial [namely, the hyperoctahedral group]) because its rank is ${1}$. Alternatively, a straightforward computation reveals that the characteristic polynomial of ${\rho(g)}$ is

$\displaystyle (x^2-\textrm{tr}(g)\det(g)x+\det(g)^3)\cdot (x^2 - \textrm{tr}(g)(\textrm{tr}(g)^2 - 3\det(g))x + \det(g)^3)$

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

$\displaystyle \frac{1}{2}\det(g)\left(\textrm{tr}(g)\pm\sqrt{\textrm{tr}(g)^2 - 4 \det(g)}\right),$

and

$\displaystyle \frac{1}{2}\left(\textrm{tr}(g)(\textrm{tr}(g)^2 - 3\det(g)) \pm (\textrm{tr}(g)^2 - \det(g)) \sqrt{\textrm{tr}(g)^2 - 4 \det(g)}\right)$

In particular, since the characteristic polynomial of ${\rho(g)}$ always splits, it is never the case that ${\rho(g)}$ is Galois-pinching.

On the other hand, the element ${A.B\in\mathcal{M}}$ is pinching because its eigenvalues are

$\displaystyle 9+4\sqrt{5} > \frac{3+\sqrt{5}}{2} > \frac{3-\sqrt{5}}{2} > \frac{1}{9+4\sqrt{5}}$

Also, the matrix ${A\in\mathcal{M}}$ is twisting with respect to ${A.B}$. Indeed, the columns of the matrix

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

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 ${A.B}$. Moreover, ${A}$ is twisting with respect to ${A.B}$ if and only if all entries of ${T}$ and all of its ${2\times 2}$ minors associated to isotropic planes are non-zero (cf. Lemma 4.8 in this paper here). Finally, this last fact is a consequence of the following exact calculation (see also the numerical approximations) for ${T}$ and its matrix of ${2\times 2}$ minors:

$\displaystyle \begin{array}{rcl} T&=& \left(\begin{array}{cccc} \frac{8(5 + 2 \sqrt{5})}{25} & \frac{2(5 + 3 \sqrt{5})}{25} & \frac{(5 + \sqrt{5})}{25} & \frac{1}{( 5 \sqrt{5})} \\ -\frac{6(5 + 3 \sqrt{5})}{25} & \frac{2(5 + \sqrt{5})}{25} & \frac{7}{5 \sqrt{5}} & -\frac{3(-5 + \sqrt{5})}{25} \\ \frac{3(5 + \sqrt{5})}{25} & -\frac{7}{5 \sqrt{5}} & -\frac{2(-5 + \sqrt{5})}{25} & \frac{6(-5 + 3 \sqrt{5})}{25} \\ -\frac{1}{5 \sqrt{5}} & \frac{(5 - \sqrt{5})}{25} & \frac{2}{5} - \frac{6}{5 \sqrt{5}} & -\frac{(8(-5 + 2 \sqrt{5})}{25} \end{array}\right) \\ &=&\left(\begin{array}{cccc} 3.03108 & 0.936656 & 0.289443 & 0.0894427 \\ -2.80997 & 0.578885 & 0.626099 & 0.331672 \\ 0.868328 & -0.626099 & 0.221115 & 0.409969 \\ -0.0894427 & 0.110557 & -0.136656 & 0.168916 \end{array}\right) \end{array}$

and

$\displaystyle \begin{array}{rcl} & & 2\times 2 \textrm{ minors of } T = \\ & & \left(\begin{array}{cccccc} \frac{56}{25} + \frac{24}{5 \sqrt{5}} & \frac{32}{25} + \frac{16}{5 \sqrt{5}} & \frac{18}{25} + \frac{6}{5 \sqrt{5}} & \frac{6}{25} + \frac{2}{5 \sqrt{5}} & \frac{2}{25} + \frac{2}{5 \sqrt{5}} & \frac{1}{25} \\ -\frac{32}{25} - \frac{16}{5 \sqrt{5}} & \frac{6}{25} + \frac{2}{5 \sqrt{5}} & \frac{9}{25} + \frac{9}{5 \sqrt{5}} & \frac{3}{25} + \frac{3}{5 \sqrt{5}} & \frac{11}{25} & -\frac{2}{25} + \frac{2}{5 \sqrt{5}} \\ \frac{6}{25} + \frac{2}{5 \sqrt{5}} & -\frac{3}{25} - \frac{3}{5 \sqrt{5}} & \frac{13}{25} & -\frac{4}{25} & -\frac{3}{25} + \frac{3}{5 \sqrt{5}} & \frac{6}{25} - \frac{2}{5 \sqrt{5}} \\ \frac{18}{25} + \frac{6}{5 \sqrt{5}} & -\frac{9}{25} - \frac{9}{5 \sqrt{5}} & -\frac{36}{25} & \frac{13}{25} & -\frac{9}{25} + \frac{9}{5 \sqrt{5}} & \frac{18}{25} - \frac{6}{5 \sqrt{5}} \\ -\frac{2}{25} - \frac{2}{5 \sqrt{5}} & \frac{11}{25} & \frac{9}{25} - \frac{9}{5 \sqrt{5}} & \frac{3}{25} - \frac{3}{5 \sqrt{5}} & \frac{6}{25} - \frac{2}{5 \sqrt{5}} & -\frac{32}{25} + \frac{16}{5 \sqrt{5}} \\ \frac{1}{25} & \frac{2}{25} - \frac{2}{5 \sqrt{5}} & \frac{18}{25} - \frac{6}{5 \sqrt{5}} & \frac{6}{25} - \frac{2}{5 \sqrt{5}} & \frac{32}{25} - \frac{16}{5 \sqrt{5}} & \frac{56}{25} - \frac{24}{5 \sqrt{5}} \end{array}\right) = \\ & & \left(\begin{array}{cccccc} 4.38663 & 2.71108 & 1.25666 & 0.418885 & 0.258885 & 0.04 \\ -2.71108 & 0.418885 & 1.16498 & 0.388328 & 0.44 & 0.0988854 \\ 0.418885 & -0.388328 & 0.52 & -0.16 & 0.148328 & 0.0611146 \\ 1.25666 & -1.16498 & -1.44 & 0.52 & 0.444984 & 0.183344 \\ -0.258885 & 0.44 & -0.444984 & -0.148328 & 0.0611146 & 0.151084 \\ 0.04 & -0.0988854 & 0.183344 & 0.0611146 & -0.151084 & 0.0933747 \end{array}\right) \end{array}$

In summary, the monoid ${\mathcal{M}}$ is pinching and twisting, but not Zariski dense in ${Sp(V)}$.

## Responses

1. Dear Professor Matheus,
I have a question about unique invarinat measure in base map. Is it true if cocycle be pinching and twisting, then the measure is a unique on base map?
Best