A criterion for the simplicity of the Lyapunov spectrum of square-tiled surfaces

Abstract

We present a Galois-theoretical criterion for the simplicity of the Lyapunov spectrum of the Kontsevich–Zorich cocycle over the Teichmüller flow on the ${\mathrm {SL}}_2(\mathbb {R})$-orbit of a square-tiled surface. The simplicity of the Lyapunov spectrum has been proved by A. Avila and M. Viana with respect to the so-called Masur–Veech measures associated to connected components of moduli spaces of translation surfaces, but is not always true for square-tiled surfaces of genus ${\geqslant }3$. We apply our criterion to square-tiled surfaces of genus 3 with one single zero. Conditionally to a conjecture of Delecroix and Lelièvre, we prove with the aid of Siegel’s theorem (on integral points on algebraic curves of genus ${>}0$) that all but finitely many such square-tiled surfaces have simple Lyapunov spectrum.

Appendices

Appendix A: A version of Avila–Viana simplicity criterion

In this appendix we will present a streamlined proof of Theorem 2.13. Here, we will use notations and definitions introduced in Sect. 2.5 without further comments.

We begin by noticing that one may consider the cocycle $A$ over the invertible dynamics $\hat{f}:\hat{\Sigma }\rightarrow \hat{\Sigma }$ because the Lyapunov spectrum is not affected by this procedure.

A crucial fact from (bi)linear algebra allows to adapt the proof for $\mathbb {G}= {\mathrm {GL}}(d,\mathbb {R})$ to other matrix groups $\mathbb {G}$ (symplectic or unitary). Let $k$ be an admissible integer, $A \in \mathbb {G}$; denote by $\sigma _1(A)\geqslant \cdots \geqslant \sigma _d(A)$ the singular values of $A$, i.e the lengths of the semi-major axes of the ellipsoid $A(\{v:\Vert v\Vert =1\})$.¹² Then, the subspace $\xi _A$ spanned by the $k$ largest semi-major axes belongs to $G(k)$.¹³ This fact is a simple consequence of the polar decomposition of matrices in $\mathbb {G}$.

In the statement below, we use the following notations:

• $\underline{\ell }(x,n)$ is the terminal word of $x\in \Sigma _-$ of length $n$;

• For $\underline{\ell } \in \Omega$, we write $\xi _{\underline{\ell }}$ for $\xi _{A^{\underline{\ell }}}$.

The main result towards the proof of Theorem 2.13 above is:

Theorem 9.1

(A. Avila and M. Viana) For every admissible integer $k,$ there exists a map $\Sigma _-\rightarrow G(k),x\mapsto \xi (x)$ verifying the properties:

• Invariance: the map $\hat{\xi } = \xi \circ p^-$ satisfies $A(x)\hat{\xi }(x) = \hat{\xi }(\hat{f}(x));$

• for $\mu _-$-almost every $x\in \Sigma _-,$ $\frac{\sigma _k(A^{\underline{\ell }(x,n)})}{\sigma _{k+1}(A^{\underline{\ell }(x,n)})}\rightarrow +\infty$ and $\xi _{\underline{\ell }(x,n)}\rightarrow \xi (x)$ as $n\rightarrow +\infty ;$

• for all $F'\in G(d-k),$ we have $\xi (x)\cap F'=\{0\}$ for a set of positive $\mu _-$-measure.

This result corresponds to [4, Theorem A.1]. As the reader can check (see Subsection A.6 “Proof of Theorem 7.1” of [4]), it is not hard to deduce Theorem 2.13 from Theorem 9.1.

Sketch of proof of Theorem 2.13 assuming Theorem 9.1. For each admissible integer $k$ and each $x\in \Sigma _-$, Theorem 9.1 provides us, with a subspace $\xi (x)\in G(k)$ verifying the properties above. By using the same theorem with the time “reversed”, one gets for $y\in \Sigma _+$ a subspace $\xi _*(y) \in G(d-k)$ verifying similar properties. From the third property in the theorem, one deduces the transversality property $\xi (x)\cap \xi _*(y)=\{0\}$ for almost every $(x,y)\in \hat{\Sigma }$. The second property in the theorem implies that $\xi (x)$ is associated with the $k$ largest exponents and $\xi _*(y)$ is associated with the $d-k$ smallest exponents. Then the transversality property permits to show that, for any admissible integer $k$, the $k$th Lyapunov exponent is strictly larger than the $(k+1)$th exponent. This shows that the Lyapunov spectrum of $A$ is simple in the sense defined in Sect. 2.5.

$\square$

This reduces our considerations to the discussion of Theorem 9.1. Let $k$ be an admissible integer. We denote by $p', p''$ the natural projections from $\hat{\Sigma }\times G(k)$ onto $\hat{\Sigma }$ and $G(k)$ respectively.

Definition 9.2

A u-state is a probability measure $\hat{m}$ on $\hat{\Sigma }\times G(k)$ such that $p'_*(\hat{m})=\hat{\mu }$ and there exists a constant $C(\hat{m})$ with

$$\begin{aligned} \frac{\hat{m}(\Sigma _-(\underline{\ell }^0)\times \Sigma (\underline{\ell })\times X)}{\mu (\Sigma (\underline{\ell }))} \leqslant C(\hat{m})\frac{\hat{m}(\Sigma _-(\underline{\ell }^0)\times \Sigma (\underline{\ell }')\times X)}{\mu (\Sigma (\underline{\ell }'))} \end{aligned}$$

for any Borelian $X\subset G(k)$, $\underline{\ell }^0,\underline{\ell }, \underline{\ell }'\in \Omega$.

Roughly speaking, the previous condition says that u-states are almost product measures.

Example 9.3

Given any probability measure $\nu$ on $G(k)$, $\hat{m}:=\hat{\mu }\times \nu$ is a u-state with $C(\hat{m})=C(\mu )^2$, where $C(\mu )$ is the constant appearing in the bounded distortion property (see Definition 2.6).

Proposition 9.4

There exists a u-state invariant under $(\hat{f},A)$.

Proof

The argument is very classical and we will only sketch its main steps. Even though the space $\hat{\Sigma }$ may not be compact (in the case of an alphabet $\Lambda$ with countably many symbols), the space of probability measures on $\hat{\Sigma }\times G(k)$ projecting to $\hat{\mu }$ is compact in the weak-* topology. In particular, for each $C>0$, it follows that the space of u-states $\hat{m}$ with $C(\hat{\mu })\leqslant C$ is a convex compact set.

A short direct computation [4, Lemma A.2] shows that, for any u-state $\hat{m}_0$ and any $n>0$, $\hat{m}(n):=(\hat{f},A)_*^n\hat{m}_0$ is a u-state with $C(\hat{m}(n))\leqslant C(\hat{m})C(\mu )^2$. Then the standard Krylov–Bogolyubov argument completes the proof: any accumulation point of the Cesaro averages of $\hat{m}(n)$ is a u-state invariant under $(\hat{f},A)$. $\square$

The following result is a simple application of the martingale convergence theorem (see [4, Lemma A.4] for a short proof). We recall that $\underline{\ell }(x,n)$ denotes the terminal word of $x\in \Sigma _-$ of length $n$.

Proposition 9.5

Let $\hat{m}$ be a probability measure on $\hat{\Sigma }\times G(k)$ with $p'_*\hat{m} = \hat{\mu }$. For any $x\in \Sigma _-$, and any Borelian subset $X\subset G(k),$ let

$$\begin{aligned} \hat{m}_n(x)(X):=\frac{\hat{m}(\Sigma _-(\underline{\ell }(x,n))\times \Sigma \times X)}{\hat{m}(\Sigma _-(\underline{\ell }(x,n))\times \Sigma \times G(k))} \end{aligned}$$

Then$,$ for $\mu _-$-almost every $x\in \Sigma _-,\hat{m}_n(x)$ converges in the weak$*$ topology to some $\hat{m}(x).$

Let $\hat{m}$ be a $(\hat{f},A)$-invariant u-state given by Proposition 9.4. Define $\hat{m}_n (x)$ as in Proposition 9.5. Let also $\nu =p''_*\hat{m}$. For any $x\in \Sigma _-$, define a sequence of probability measures on $G(k)$ by

$$\begin{aligned} \nu _n(x):=A^{\underline{\ell }(x,n)}_*\nu . \end{aligned}$$

Let $X$ be a Borelian subset of $G(k)$. As $\hat{m}$ is $(\hat{f},A)$-invariant, we have

$$\begin{aligned} \hat{m}_n(x)(X)&= \frac{\hat{m}(\Sigma _-(\underline{\ell }(x,n))\times \Sigma \times X)}{\hat{m}(\Sigma _-(\underline{\ell }(x,n))\times \Sigma \times G(k))}\\&= \frac{\hat{m}(\Sigma _-\times \Sigma (\underline{\ell }(x,n))\times A^{-\underline{\ell }(x,n)}(X))}{\hat{m}(\Sigma _-\times \Sigma (\underline{\ell }(x,n))\times G(k))} \end{aligned}$$

On the other hand, by definition, we have

$$\begin{aligned} \nu _n(x)(X) = \nu (A^{-\underline{\ell }(x,n)}(X)) = \hat{m}(\hat{\Sigma }\times A^{-\underline{\ell }(x,n)}(X)). \end{aligned}$$

Since $\hat{m}$ is a u-state, we obtain

$$\begin{aligned} C(\hat{m})^{-2}\leqslant \hat{m}_n(x)(X)/\nu _n(x)(X)\leqslant C(\hat{m})^2. \end{aligned}$$

In particular,

Corollary 9.6

For $\mu _-$-almost every $x\in \Sigma _-,$ the probability measure $\hat{m}(x)=\lim \hat{m}_n(x)$ is equivalent to any accumulation point of the sequence $\nu _n(x)$.

The crucial step in the proof of Theorem 9.1 is given by the

Proposition 9.7

For $\mu _-$-almost every $x\in \Sigma _-,$ there exists a subsequence $\nu _{n_k}(x),n_k=n_k(x)\rightarrow \infty ,$ converging to a Dirac mass.

Sketch of proof of Theorem 9.1 assuming Proposition 9.7. (see end of Subsection A.4 of [4]). By Corollary 9.6 and Proposition 9.7, $\hat{m}(x)$ is a Dirac mass $\delta _{\xi (x)}$ for $\mu _-$-almost every $x\in \Sigma _-$.

Then $x\mapsto \xi (x)$ has the desired properties: the invariance property (first item of Theorem 9.1) follows from the $(\hat{f},A)$-invariance of $\hat{m}$; the other two items are a consequence of the pinching and twisting assumptions on the cocycle $A$. $\square$

Proof of Proposition 9.7. Let $\underline{\ell }^* \in \Omega$ be a word such that the matrix $A^{\underline{\ell }^*}$ is pinching. As there are only finitely many $A^{\underline{\ell }^*}$-invariant subspaces in $G(k)$ (see Remark 2.15), one can use the twisting hypothesis to choose $\underline{\ell }^0\in \Omega$ such that, for each admissible integer $k$ and for every pair of $A^{\underline{\ell }^*}$-invariant subspaces $F\in G(k), F'\in G(d-k)$, one has $A^{\underline{\ell }^0}(F)\cap F'=\{0\}$.

We claim that there exists $m\geqslant 1$, $\underline{\ell }_1,\ldots ,\underline{\ell }_m\in \Omega$ and $\delta >0$ such that, for each admissible integer $k$ and for every $F'\in G(d-k)$, there exists $\underline{\ell }_i$ with $A^{\underline{\ell }_i}(F_+(A^{\underline{\ell }^*}))\cap F'=\{0\}$ and the angle between $A^{\underline{\ell }_i}(F_+(A^{\underline{\ell }^*}))$ and $F'$ is $\geqslant \delta$. Here, $F_+(A^{\underline{\ell }^*})$ is the subspace associated to $k$ largest exponents of $A^{\underline{\ell }^*}$. Indeed, it is sufficient to prove this for a given admissible integer $k$, in which case it follows from the twisting assumption and the compactness of $G(d-k)$.

Lemma 9.8

(Lemma A.6 of [4]) Let $\varepsilon >0$ and $\rho$ a probability measure on $G(k).$ There exists $n_0=n_0(\rho ,\varepsilon )$ and$,$ for each $\widetilde{\underline{\ell }}\in \Omega ,$ there exists $i=i(\widetilde{\underline{\ell }})\in \{1,\ldots ,m\}$ such that$,$ for $n\geqslant n_0,$ we have

$$\begin{aligned} A^{\underline{\ell }}_*(\rho )(B)>1-\varepsilon \end{aligned}$$

where $\underline{\ell }:=(\underline{\ell }^*)^n\underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i\widetilde{\underline{\ell }}$ and $B$ is the ball of radius $\varepsilon >0$ centered at $\xi _{\underline{\ell }}$.

This lemma is harder to state than to explain: geometrically, it says that, although the word $\widetilde{\underline{\ell }}$ may be very long, we can choose an appropriate “start” ($(\underline{\ell }^*)^n\underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i$) so that the word $\underline{\ell }$ obtained by the concatenation of $(\underline{\ell }^*)^n\underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i$ and $\widetilde{\underline{\ell }}$ has the property that $A^{\underline{\ell }}$ concentrates most of the mass of any probability measure $\rho$ on $G(k)$ (given in advance) in a tiny ball $B$.

We defer the proof of the lemma to the end of the appendix and first end the proof of Proposition 9.7. The details are slightly different from [4].

We will apply Lemma 9.8 with $\rho = \nu$. As $\nu _n(x):=A^{\underline{\ell }(x,n)}_*\nu$, the conclusion of the lemma will imply the conclusion of Proposition 9.7 if we can show that, for any $n\geqslant 0$ and $\mu _-$-almost every $x$, there are infinitely many integers $N$ such that $(\underline{\ell }^*)^n \underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i\underline{\ell }(x,N)$, with $i=i(\underline{\ell }(x,N))$, is a terminal word of $x$.

Assume that this is not true. Then there exist integers $n, N_0$ and a positive measure set $E \subset \Sigma _-$ such that, for any $x \in E$, $N \geqslant N_0$, the word $(\underline{\ell }^*)^n \underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i \underline{\ell }(x,N)$, with $i=i(\underline{\ell }(x,N))$, is not a terminal word of $x$.

By the bounded distortion property, there exists $c>0$ such that

$$\begin{aligned} \mu _{-} (\Sigma _{-} ((\underline{\ell }^{*})^{n} \underline{\ell }^{0} (\underline{\ell }^{*})^{n} \underline{\ell }_{i}\widetilde{\underline{\ell }}))\geqslant c \mu _{-}(\Sigma _{-} (\widetilde{\underline{\ell }})) \end{aligned}$$

(9.1)

for every $1\leqslant i\leqslant m$ and $\widetilde{\underline{\ell }}\in \Omega$.

Let $x_0$ be a density point of $E$. There exists $N \geqslant N_0$ such that

$$\begin{aligned} \mu _-(\Sigma _-(\underline{\ell }(x_0,N))\cap E^c) < \frac{c}{2} \mu _-(\Sigma _-(\underline{\ell }(x_0,N))), \end{aligned}$$

where $E^c$ is the complement of $E$.

Taking $\widetilde{\underline{\ell }} = \underline{\ell }(x_0,N)$, $i = i(\widetilde{\underline{\ell }})$ in (9.1) above, we find a point in $E$ with terminal word $(\underline{\ell }^*)^n \underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i \underline{\ell }(x_0,N)$. This contradiction to the definition of $E$ proves the claim and ends the proof of Proposition 9.7. $\square$

Proof of Lemma 9.8. An elementary calculation shows that, as $A^{\underline{\ell }^*}$ is pinching, the sequence $(A^{\underline{\ell }^*})^n(\xi )$ converges for every $\xi \in G(k)$. The limit is one of the finitely many $A^{\underline{\ell }^*}$-invariant subspaces in $G(k)$. Moreover, the limit is the subspace $F_+(A^{\underline{\ell }^*})$ associated to the $k$ largest exponents whenever $\xi$ is transverse to every $A^{\underline{\ell }^*}$-invariant subspace in $G(d-k)$.

For any probability measure on $G(k)$, the sequence $(A^{\underline{\ell }^*})_*^n(\rho )$ converges, as $n$ goes to $+ \infty$, to a limit which is a convex combination of Dirac masses at the $A^{\underline{\ell }^*}$-invariant subspaces in $G(k)$. By definition of $\underline{\ell }^0$, the images under $A^{\underline{\ell }^0}$ of these invariant subspaces are transverse to every $A^{\underline{\ell }^*}$-invariant subspace in $G(d-k)$. We conclude that $(A^{\underline{\ell }^*})_*^n (A^{\underline{\ell }^0})_* (A^{\underline{\ell }^*})_*^n (\rho )$ converges to the Dirac mass at $F_+(A^{\underline{\ell }^*})$.

Let $\widetilde{\underline{\ell }} \in \Omega$ be given. Denote by $\xi ^*_{\widetilde{\underline{\ell }}}$ the $(d-k)$-dimensional subspace which is least dilated by $A^{\widetilde{\underline{\ell }}}$, i.e whose image is spanned by the $(d-k)$ shortest semi-major axes of the ellipsoid $A^{\widetilde{\underline{\ell }}} (\{||v|| =1\})$. Taking $F' = \xi ^*_{\widetilde{\underline{\ell }}}$ in the defining property of $\ell _1,\ldots ,\ell _m$, we find $i$ such that $A^{\underline{\ell }_i} ( F_+(A^{\underline{\ell }^*}))$ is transverse to $\xi ^*_{\widetilde{\underline{\ell }}}$, the angle between these subspaces being $\geqslant \delta$. From the claim below, we conclude that for large $n$ (independently of $\widetilde{\underline{\ell }}$), most of the mass of the probability measure

$$\begin{aligned} (A^{\widetilde{\underline{\ell }}})_* (A^{\underline{\ell }_i})_* (A^{\underline{\ell }^*})_*^n(A^{\underline{\ell }^0})_*(A^{\underline{\ell }^*})_*^n (\rho ) \end{aligned}$$

is concentrated in a small ball in $G(k)$ around $A^{\widetilde{\underline{\ell }}} A^{\underline{\ell }_i} ( F_+(A^{\underline{\ell }^*}))$. Considering the case where $\rho$ is the Lebesgue measure on $G(k)$, we conclude that this small ball is contained in a small ball around $\xi _{\underline{\ell }}$, where $\underline{\ell }:=(\underline{\ell }^*)^n\underline{\ell }^0(\underline{\ell }^*)^n\underline{\ell }_i\widetilde{\underline{\ell }}$. $\square$

Claim

Let $A \in {\mathrm {GL}}_d(\mathbb {K})$ act on the Grassmannian of $k$-dimensional subspaces. Denote by $\xi ^*_A$ the $(d-k)$ dimensional subspace which is least dilated by $A$, and by $K_{\delta }(A)$ the set of $k$-dimensional subspaces which form an angle $\geqslant \delta$ with $\xi ^*_A$. Then the modulus of continuity of the restriction of $A$ to $K_{\delta }(A)$ is controlled by $\delta$ only$,$ independently of $A$.

Proof

This is an elementary computation: write any subspace in $K_{\delta }(A)$ as the graph of a linear map from $(\xi ^*_A)^{\perp }$ to $\xi ^*_A$, whose norm is bounded in terms of $\delta$. After composing if necessary by an isometry, the action of $A$ on the matrix of this linear map is given by the multiplication of each coefficient by a number $\in (0, 1)$ (the ratio of two singular values of $A$). $\square$

Appendix B: Twisting properties

In Appendix A above, we studied a version of Avila–Viana simplicity criterion in the context of locally constant cocycles with values on

$$\begin{aligned} \mathbb {G} = {\mathrm {GL}}(d, {\mathbb {R}}), {\mathrm {Sp}}(d, {\mathbb {R}}), U_{\mathbb {K}}(p,q), \mathbb {K}= \mathbb {R},\,\mathbb {C}, \text { or } \mathbb {H}\end{aligned}$$

(10.1)

over shifts on at most countably many symbols.

In this way, based on the setting of Sect. 3 above, we can already get a simplicity criterion for the Kontsevich–Zorich cocycle over ${\mathrm {SL}}_2(\mathbb {R})$-orbits of square-tiled surfaces based on pinching and a strong form of twisting. However, such a simplicity criterion is not easy to apply directly, so it is desirable to replace the strong form of twisting by the relative form, with respect to some pinching matrix. This lead us to the statement of Proposition 2.16 whose proof is the main purpose of this appendix. But, before explaining the proof of Proposition 2.16, it is convenient to revisit a little bit the features of Noetherian topological spaces.

1.1 Noetherian spaces

Let $\mathbb {G}$ be as in (10.1). We use the notations and definitions introduced in Sect. 2.5. Let $k$ be an admissible integer. For each $F'\in G(d-k)$, we define an hyperplane section as

$$\begin{aligned} \{F\in G(k): F\cap F'\ne \{0\}\} \end{aligned}$$

We consider then the coarsest topology on $G(k)$ such that the hyperplane sections are closed. The closed sets are the (arbitrary) intersections of finite unions of hyperplane sections. For sake of convenience, we will refer to this “pseudo-Zariski” topology as the Schubert topology.

Notice that the Schubert topology is coarser than the Zariski topology: hyperplane sections are defined by degree one (linear) equations while Zariski topology involves taking equations of arbitrary degree. In particular, this topology is not Hausdorff as the same is true for the Zariski topology.

Definition 10.1

A topological space $X$ is Noetherian if one of the following equivalent conditions is satisfied:

• (i) any decreasing sequence $F_1\supset F_2\supset \cdots$ of closed sets is stationary (in the sense that there exists $m\in \mathbb {N}$ such that $F_i=F_m$ for all $i\geqslant m$).

• (ii) any increasing sequence of open sets is stationary.

• (iii) every intersection of a family $(F_{\alpha })$ of closed sets is the intersection of a finite subfamily $F_1,\ldots , F_m$.

• (iv) every union of a family $(U_{\alpha })$ of open sets is the union of a finite subfamily $U_1,\ldots ,U_m$.

Observe that any subspace of a Noetherian space is also Noetherian. A topology which is coarser than a Noetherian topology is also Noetherian.

Example 10.2

It is a classical fact that the Zariski topology is Noetherian. Therefore the Schubert topology is also Noetherian.

Definition 10.3

A Noetherian (topological) space $X$ is irreducible if $X$ is not the union of two proper closed sets.

The Grassmannian $G(k)$, equipped with the Zariski topology, is irreducible. It is a fortiori irreducible when it is equipped with the coarser Schubert topology.

We will need the following properties of Noetherian spaces.

Proposition 10.4

(Proposition 1.5 in [26]) A Noetherian space $X$ can be written as a finite union $X=X_1\cup \cdots \cup X_m$ of irreducible closed subsets $X_i,$ $1\leqslant i\leqslant m$. Moreover$,$ this decomposition is unique $($up to a permutation of the $X_i$’s$)$ if we ask that $X_i\not \subset X_j$ for $i\ne j$.

Proposition 10.5

Let $X_1, \ldots , X_n$ be Noetherian spaces.

• (i) The product space $X= X_1 \times \cdots \times X_n$ is Noetherian.

• (ii) It is irreducible iff each $X_i$ is irreducible.

• (iii) Open subsets of $X$ are exactly the finite unions of products of open subsets of the $X_i$.

• (iv) Closed subsets of $X$ are exactly the finite unions of products of closed subsets of the $X_i$.

• (v) A closed subset of $X$ is irreducible iff it is the product of closed irreducible subsets of the $X_i$.

Proof

The first assertion is Exercise 8, p. 142 of [8]. Item (iii) is an immediate consequence of item (i) and Definition 10.1, item (iv). Then item (iv) follows from some Boolean manipulations. From item (iv), it follows that a closed irreducible subset of $X$ is the product of closed subsets of the $X_i$. It is also clear that if $X$ is irreducible, then each $X_i$ must be irreducible. Finally we show that a product of irreducible spaces is irreducible. Let

$$\begin{aligned} X=F_1\cup \cdots \cup F_m \end{aligned}$$

be the minimal decomposition of $X$ into irreducible subsets. Each $F_j$ is a product

$$\begin{aligned} F_j=F_j^{(1)}\times \cdots \times F_j^{(n)} \end{aligned}$$

where each $F_j^{(i)}$ is an irreducible closed subset of $X_i$. For each $1 \leqslant i \leqslant n$, define

$$\begin{aligned} F^{(i)} := \bigcup \limits _{F_j^{(i)}\ne X} F_j^{(i)}. \end{aligned}$$

As $X_i$ is irreducible, one has $F^{(i)} = X_i$ iff $F_j^{(i)} = X_i$ for all $1 \leqslant j \leqslant m$. For $1 \leqslant i \leqslant n$, choose $x_i \in X_i - F^{(i)}$ if $F^{(i)} \ne X_i$ and $x_i \in X_i$ otherwise. Let $j$ be an index such that $x:=(x_1, \ldots ,x_n) \in F_j$. One must have $F_j^{(i)} = X_i$ for all $1 \leqslant i \leqslant n$, hence $F_j = X$. $\square$

1.2 Twisting monoids

Let $\mathcal {M}$ be a monoid acting on a Noetherian space $X$ by homeomorphisms. Here, of course, our main example is:

Example 10.6

Given a countable family of matrices $A_{\ell }\in \mathbb {G}$, $\ell \in \Lambda$, we consider the natural action of the monoid $\mathcal {M}$ generated by $A_{\ell }$ acting on the Grassmannian $X_k=G(k)$ equipped with the Schubert topology.

Proposition 10.7

If $g\in \mathcal {M},F\subset X$ is closed and $gF\subset F,$ then $gF=F.$

Proof

Otherwise, $(g^nF)_{n\geqslant 0}$ would be a strictly decreasing infinite sequence of closed subsets of the Noetherian space $X$. $\square$

Proposition 10.8

Let $g\in \mathcal {M},$ let $F=F_1\cup \cdots \cup F_n$ be the $($minimal$)$ decomposition of the closed subset $F\subset X$ into irreducible closed subsets $F_i\subset X.$ If $gF=F,$ then $g$ permutes the irreducible pieces $F_i$.

Proof

This follows from the uniqueness part of Proposition 10.4. $\square$

Proposition 10.9

Assume that $X$ is irreducible. Then$,$ the following properties are equivalent$:$

• (i) there exists no proper closed $\mathcal {M}$-invariant subset of $X;$

• (ii) for every $x\in X$ and every non empty open subset $U\subset X,$ there exists $g\in \mathcal {M}$ such that $gx\in U;$

• (iii) for every $N\geqslant 1,x_1,\ldots , x_N\in X$ and every non empty open subsets $U_1,\ldots ,U_N\subset X,$ there exists $g\in \mathcal {M}$ such that $gx_i\in U_i$ for all $1\leqslant i\leqslant N$.

Proof

It is clear that $(\hbox {iii}) \implies (\hbox {ii})\implies (\hbox {i})$. We will prove by contradiction that $(\hbox {i})$ implies $(\hbox {iii})$. We let $\mathcal {M}$ act diagonally on $X^N$, for any $N \geqslant 1$.

Suppose that there exist $N \geqslant 1$, $x= (x_1,\ldots ,x_N)\in X^N$ and non empty open subsets $U_1,\ldots ,U_N\subset X$ such that

$$\begin{aligned} g.x \notin U^{(N)} := U_1\times \cdots \times U_N \end{aligned}$$

for all $g\in \mathcal {M}$.

Consider the closed set

$$\begin{aligned} F:= \bigcap \limits _{g\in \mathcal {M}}g^{-1}(X^N - U^{(N)}). \end{aligned}$$

It is distinct from $X^N$ and non empty because it contains $x$. It satisfies $gF\subset F$ for all $g\in \mathcal {M}$, hence $gF = F$ for all $g\in \mathcal {M}$ (Proposition 10.7).

Let $F=F_1\cup \cdots \cup F_m$ be the decomposition of $F$ into irreducible closed sets $F_i\subset X^N$. By Proposition 10.5, one has

$$\begin{aligned} F_i=F_i^{(1)}\times \cdots \times F_i^{(N)} \end{aligned}$$

where each $F_i^{(l)}$ is an irreducible closed subset of $X$.

Since $gF=F$ for all $g\in \mathcal {M}$, Proposition 10.8 implies that every $g\in \mathcal {M}$ permutes the subsets $F_i^{(l)}$.

Define the closed subset

$$\begin{aligned} F^*:=\bigcup \limits _{F_i^{(l)}\ne X} F_i^{(l)}. \end{aligned}$$

As $\emptyset \ne F\ne X^N$, the subset $F^*$ is not empty. As $X$ is irreducible, one has $F^* \ne X$. As every $g\in \mathcal {M}$ permutes the subsets $F_i^{(l)}$, one has $gF^*=F^*$ for every $g\in \mathcal {M}$, so item (i) does not hold. $\square$

In view of Example 10.6 and the discussion in Sect. 2.5 (related to Avila–Viana simplicity criterion), it is natural to call (iii) a (“strong form” of) twisting condition for an abstract monoid $\mathcal {M}$ acting by homeomorphisms on a Noetherian space $X$.

Remark 10.10

The equivalent conditions of the proposition are satisfied by the monoid $\mathcal {M}$ if and only if they are satisfied by the group $\mathcal {G}=\langle g, g^{-1}:g\in \mathcal {M}\rangle$ generated by $\mathcal {M}$: this follows immediately from the statement of item (i).

1.3 Twisting with respect to pinching matrices

In the context of Example 10.6 and aiming to the proof of Proposition 2.16, consider a word $\underline{\ell }^*\in \Omega =\bigcup \limits _{n\geqslant 0}\Lambda ^n$ such that $A_*:=A^{\underline{\ell }^*}$ has simple spectrum (i.e., the cocycle is pinching). In this notation, Proposition 2.16 can be restated as:

Proposition 10.11

The $($strong form of the$)$ twisting condition is realized for the cocycle $A$ if and only if$,$ for each admissible integer $k,$ there exists a word $\underline{\ell }(k)\in \Omega$ such that the matrix $B_k:=A^{\underline{\ell }(k)}$ satisfies

$$\begin{aligned} B_k(F)\cap F'=\{0\} \end{aligned}$$

for every $A_*$-invariant subspaces $F\in G(k)$ and $F'\in G(d-k)$.

Proof

The condition is clearly necessary.

Conversely, assume that the condition in the proposition is satisfied. Let $\mathcal {M}$ denote the monoid generated by the matrices $A_\ell$, $\ell \in \Lambda$. Recall that each $G(k)$ is irreducible.

Lemma 10.12

For each admissible integer $k,$ the action of $\mathcal {M}$ on $G(k)$ satisfies the equivalent conditions of Proposition 10.9.

Proof of Lemma. We check that item (ii) in Proposition 10.9 is satisfied. It is sufficient to show that, given $F \in G(k)$ and $F'_1, \ldots , F'_m \in G(d-k)$, there exists $C \in \mathcal M$ such that $C(F)$ is transverse to each $F'_i$. We claim that $C = A_*^n B_k A_*^n$ is an appropriate choice if $n$ is large enough. Indeed, when $n$ goes to $+\infty$, the sequence $(A_*^n(F))$ converges to some $A_*$-invariant subspace in $G(k)$, and each sequence $(A_*^{-n}(F'_i))$ converges to some $A_*$-invariant subspace in $G(d-k)$. As transversality is an open property, the claim follows from the property of $B_k$. $\square$

We now finish the proof of the proposition. Consider the diagonal action of $\mathcal {M}$ on the irreducible Noetherian space $X=\prod \limits _{k \text { admissible}}G(k)$. The strong form of the twisting condition will be satisfied if the action of $\mathcal {M}$ on $X$ satisfies item (iii) in Proposition 10.9. We check the equivalent item (i). Let $F$ be a non-empty closed subset $F \subset X$ invariant under $\mathcal {M}$. Let $F=F_1\cup \cdots \cup F_m$ be the minimal decomposition of $F$ into irreducible closed sets $F_i\subset X$. By Proposition 10.5, one has

$$\begin{aligned} F_i= \prod \limits _{k \text { admissible}} F_i^{(k)} \end{aligned}$$

where each $F_i^{(k)}$ is an irreducible closed subset of $G(k)$. Define, for each admissible integer $k$,

$$\begin{aligned} F^{(k)}:=\bigcup \limits _{F_i^{(k)}\ne G(k)} F_i^{(k)}. \end{aligned}$$

By Proposition 10.8, the closed subset $F^{(k)}$ is invariant under $\mathcal M$. From Lemma 10.12, $F^{(k)}$ must be either empty or equal to $G(k)$ for each admissible integer $k$. The second case cannot occur (since $G(k)$ is irreducible), and the first means that $F_i^{(k)} = G(k)$ for all $i$. We conclude that $F = X$, and the proof of the proposition is complete. $\square$

Appendix C: Completely periodic configurations in $\mathcal {H}(4)$ by Samuel Lelièvre

A surface in the stratum $\mathcal {H}(4)$ has one singularity of angle $10\pi$. At this singularity, 5 outgoing separatrices start and 5 incoming separatrices end (see Fig. 6). We label the outgoing separatrices from 1 to 5 (see Fig. 6).

Fig. 6

Outgoing and incoming separatrices, and a numbering of outgoing separatrices

In a completely periodic direction, outgoing separatrices pair with incoming separatrices. Define a permutation $\sigma$ of $\{ 1, 2, 3, 4, 5 \}$ by setting $\sigma (i) = j$ if the outgoing separatrix $i$ comes back between the outgoing separatrices $j$ and $j+1 \hbox {mod} 5$. We can enumerate permutations and draw the corresponding separatrix diagrams. Since the separatrix at which we start the labelling is arbitrary, we only need to enumerate permutations up to conjugation by cyclic permutations.

A diagram makes sense if the ribbons which follow unions of separatrices above or below form compatible bottoms and tops of cylinders. So, given a permutation, we look for the cycles of $\sigma$ and of $\sigma '$ defined by $\sigma '(i) = \sigma (i) + 1 \hbox {mod} 5$ and then look for all the ways to match them.

A first example: $\sigma = \text {id}$ gives the diagram on Fig. 7. In this example, $\sigma$ has cycles $(1)(2)(3)(4)(5)$, while $\sigma '$ has only one cycle $(12345)$, therefore there is no possible pairing. We can’t glue any cylinders on this separatrix diagram, there would need to be five bottoms of cylinders and only one top of cylinder.

Another example: $\sigma = (354)$ gives the diagram on Fig. 8. In this example, $\sigma$ has cycles $(1)(2)(354)$, while $\sigma '$ has cycles $(123)(4)(5)$, and two pairings are possible: cylinders can fit on this separatrix diagram in two different ways.

Fig. 7

There is no way to glue cylinders using this pairing of separatrices

Fig. 8

With this pairing of separatrices there are two ways to glue cylinders

As checked with Sage, there are exactly 16 permutations $\sigma$ of $\{1, 2, 3, 4, 5\}$ (up to cyclic permutation) for which $\sigma$ and $\sigma '$ have the same number of cycles. The corresponding pairs $(\sigma ,\sigma ')$ are listed below, expressed as products of nontrivial cycles. It turns out that each of them gives one, two or three cylinder diagrams, and we get 22 cylinder diagrams in all.

$$\begin{aligned} \begin{array}{cccc} (354), (123) &{}\quad (23)(45), (124) &{}\quad (2345), (124)(35) &{}\quad (2354), (1243)\\ (2453), (1254) &{}\quad (2435), (1253) &{}\quad (253), (12)(45) &{}\quad (2534), (12)(354)\\ (25)(34), (12)(35) &{}\quad (12)(345), (1354) &{}\quad (12345), (13524) &{}\quad (12453), (13254)\\ (124)(35), (13)(254) &{}\quad (13524), (14253) &{}\quad (13)(254), (1432) &{}\quad (14253), (15432) \end{array} \end{aligned}$$

Below, we list next to each of these 16 pairs $(\sigma ,\sigma ')$ the associated cylinder diagrams and, furthermore, for each cylinder diagram, we put the letter $H$, resp. $O$, when the corresponding translation surfaces belong the hyperelliptic, resp., odd, connected component of $\mathcal {H}(4)$.