Posted by: matheuscmss | March 1, 2012

## SPCS 8

In SPCS6 and SPCS7, we studied a version of Avila-Viana symplicity criterion in the context of locally constant cocycles with values on

$\displaystyle G=GL(d,\mathbb{R}), Sp(d,\mathbb{R}), U_{\mathbb{C}}(p,q) \textrm{ or } U_{\mathbb{H}}(p,q) \ \ \ \ \ (1)$

over shifts on countably many symbols, and the codification of the Kontsevich-Zorich cocycle over ${SL(2,\mathbb{R})}$-orbits of origamis by a locally constant cocycle over a shift on countably many symbols.

In this way, we can get a simplicity criterion for the Kontsevich-Zorich cocycle over ${SL(2,\mathbb{R})}$-orbits of origamis based on a “pinching” property and a “twisting” property (see this post). However, since the twisting property was stated in a “strong form” (in SPCS6), before extracting consequences from SPCS6 and SPCS7, we will spend the first half of today’s post with the study of several versions of the twisting properties.

1. Twisting properties

For the discussion of twisting properties, it is convenient to revisit a little bit the features of Noetherian topological spaces.

1.1. Noetherian spaces

Given ${G}$ as in (1), we recall the following notations (from SPCS6). For each ${k}$ admissible, ${G(k)}$ denotes the Grassmanian of ${k}$-planes (isotropic or coisotropic depending on the group ${G}$).

Then, for each ${F'\in G(d-k)}$, we can define an hyperplane section as

$\displaystyle \{F\in G(k): F\cap F'\neq\{0\}\}$

Using these hyperplane sections, we have a (“pseudo-Zariski”) topology on ${G(k)}$ by considering the coarsest topology such that the hyperplane sections are closed. Equivalently, we consider the topology on ${G(k)}$ whose closed sets are the (arbitrary) intersections of finite unions of hyperplane sections.

Notice that this topology is coarser than the Zariski topology: for instance, hyperplane sections are defined by degree one (linear) equations while Zariski topology involves taking equations of arbitrary degree (in other words, this topology is a sort of “Zariski topology for dummies” … by the way, I would be happy to know if this topology has a “standard” name in the literature…). In particular, this topology is not Hausdorff as the same is true for the Zariski topology. As we’re going to see in a moment (see Example 1 below), this topology has the following property described in the definition below:

Definition 1 A topological space ${X}$ is Noetherianif one of the following equivalent conditions is satisfied:

• (i) any decreasing sequence ${F_1\supset F_2\supset\dots}$ of closed sets is stationary (in the sense that there exists ${m\in\mathbb{N}}$ such that ${F_i=F_m}$ for all ${i\geq m}$).
• (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 ${F_1,\dots, F_m}$.
• (iv) every union of a family ${(U_{\alpha})}$ of open sets is the union of a finite subfamily ${U_1,\dots,U_m}$.

Observe that any subspace of a Noetherian space is also Noetherian.

Example 1 It is a classical fact that the Zariski topology is Noetherian. In particular, since the topology on ${G(k)}$ associated to hyperplane sections is coarser than the Zariski topology, it is also Noetherian.

Definition 2 A Noetherian (topological) space ${X}$ is irreducible if ${X}$ is not the union of two non-trivial closed sets.

Proposition 3 A Noetherian space ${X}$ can be written (decomposed) as a finite union ${X=X_1\cup\dots\cup X_m}$ of irreducible closed sets ${X_i}$, ${1\leq i\leq m}$. Moreover, this decomposition is unique (up to permutation of ${X_i}$‘s) with the property ${X_i\not\subset X_j}$ for ${i\neq j}$.

Proof: We will show only the existence part of this statement (leaving the uniqueness as an exercise). We argue by contradiction: otherwise, since ${X}$ is Noetherian, we could find ${\overline{X}\subset X}$ closed set that is not the finite union of irreducible sets and minimal (with respect to the inclusion) for this property. Of course, by construction, ${\overline{X}}$ is not irreducible. Hence, we can write ${\overline{X}=X'\cup X''}$ with ${X',X''\neq \overline{X}}$. By minimality, ${X'=\bigcup\limits_{i=1}^m X_i'}$ and ${X''=\bigcup\limits_{j=1}^n X_j''}$ where ${X_i'}$ and ${X_j''}$ are irreducible closed sets. So,

$\displaystyle \overline{X} = X'\cup X''=(\bigcup\limits_{i=1}^m X_i)\cup(\bigcup\limits_{j=1}^n X_j'')$

is also a finite union of irreducible closed sets, a contradiction. $\Box$

Proposition 4 A finite product of Noetherian spaces is Noetherian. An open set of the product space is the finite union of products of open sets.

Remark 1 This proposition is not true if we replace “Noetherian space” by “Zariski topology”, i.e., the product of Zariski topologies is not the Zariski topology on the product. In particular, this is the reason why one does not insist on the product of Zariski topologies in Algebraic Geometry.

Proof: We illustrate the arguments in the case of the product of two Noetherian spaces ${X}$, ${Y}$ as the general case is similar.

It is not hard to see that our task is to prove that every union of products ${U_{\alpha}\times V_{\alpha}}$ is the union of a finite subfamily. Suppose that this is not true. Then, for each ${i\in\mathbb{N}}$, ${x_i, y_i, U_i, V_i}$ such that ${x_i\in U_i}$, ${y_i\in V_i}$ for all ${i\in\mathbb{N}}$ and ${(x_n,y_n)\notin U_i\times V_i}$ for all ${n>i}$.

Define ${A_i=\{n>i:x_n\notin U_i\}}$ and ${B_i=\{n>i: y_n\notin V_i\}}$. By construction, ${A_i\cup B_i=\{n\in\mathbb{N}: n>i\}}$. We claim that:

• (a) either there exists an infinite sequence ${i_0 with ${i_k\in A_{i_0}\cap\dots\cap A_{i_{k-1}}}$ for all ${k\geq 1}$
• (b) or there exists an infinite sequence ${j_0 with ${i_k\in B_{j_0}\cap\dots\cap B_{j_{k-1}}}$ for all ${k\geq 1}$.

Here, we note that these possibilities are notmutually exclusive, i.e., it may happen that both of them occur. In any event, assuming momentarily that this claim is true, say (a) holds, it follows that ${U_{i_0}, U_{i_1}, \dots}$ is an infinite sequence of open sets such that ${x_{i_k}\in U_k}$ but ${x_{i_k}\notin U_{i_0}\cup\dots\cup U_{i_{k-1}}}$, a contradiction with the fact that ${X}$ is Noetherian. In other words, we reduced the proof of the proposition to showing the our claim above.

We prove the claim by recurrence. Put ${A_{i_0}:=\{n>1\}}$ and assume that the sets ${A_{i_0},\dots, A_{i_{k-1}}}$ are already constructed with the properties that ${i_{l}\in A_{i_0}\cap \dots\cap A_{i_{l-1}}}$ for all ${l, and ${A^*_k:=A_{i_0}\cap\dots\cap A_{i_{k-1}}}$ is infinite.

Consider the sets ${A_i\cap A_{i_0}\cap\dots\cap A_{i_{k-1}}=A_i\cap A_k^*}$ for ${i\in A_k^*}$. We have two possible scenarios:

• there exists ${i\in A_k^*}$ such that ${A_i\cap A_k^*}$ is infinite: in this case, we can define ${i_k=i}$ and advance one step in our argument by recurrence;
• for all ${i\in A_k^*}$ the set ${A_i\cap A_k^*}$ is finite: in this case, since ${A_i\cup B_i=\{n>i\}}$, one has that, for all ${i\in A^*_k}$, the set ${B_i\cap A^*_k}$ contains all but finitely many elements of ${A_k^*}$; then, we can pick ${j_0\in A_k^*}$ and we can define ${j_1\in B_{j_0}\cap A_k^*, \dots, j_k\in B_{j_{k-1}}\cap A_k^*,\dots}$ and it is not hard to see that ${j_0 satisfies item (b) of our claim.

In resume, the recursive argument above says that either we can advance indefinitely in the construction of the sequence ${i_0 fitting item (a) of our claim, or we get “blocked” at some stage ${k}$ but then we can produce in “one shot” the sequence ${j_0 meeting the requirements of item (b) of our claim. Of course, this proves the claim and thus the proof of the proposition is complete. $\Box$

Proposition 5 A closed irreducible set of ${X_1\times\dots\times X_n}$ has the form

$\displaystyle F_1\times\dots\times F_n$

where ${F_i\subset X_i}$ are closed irreducible sets.

Proof: By the previous proposition, one can check that closed sets of ${X_1\times\dots\times X_n}$ are finite unions of products of closed sets of ${X_i}$‘s. Thus, the irreducible closed sets of ${X_1\times\dots\times X_n}$ are precisely the products ${F_1\times\dots\times F_n}$ of irreducible closed sets ${F_i\subset X}$. $\Box$

After this little digression on Noetherian spaces, we can pass to twisting properties.

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 2 Given a countable family of matrices ${A_{\ell}\in G}$, ${\ell\in\Lambda}$, with ${G}$ as in (1), we can consider the natural action of the monoid ${\mathcal{M}}$ generated by ${A_{\ell}}$ acting on the Grassmanian ${X=G(k)}$ equipped with the Noetherian topology associated to hyperplane sections.

Proposition 6 If ${g\in\mathcal{M}}$, ${F\subset X}$ is closed and ${gF\subset F}$, then ${gF=F}$.

Proof: Otherwise, ${(g^nF)_{n\geq 0}}$ would be a strictly decreasing infinite sequence of closed sets of the Noetherian space ${X}$. $\Box$

Proposition 7 If ${g\in\mathcal{M}}$, ${F=F_1\cup\dots\cup F_n}$ is the (minimal) decomposition of the closed set ${F\subset X}$ into irreducible closed sets ${F_i\subset X}$, and ${gF=F}$, then ${g}$ permutes the irreducible pieces ${F_i}$.

Proof: This follows from the uniqueness part of Proposition 3. $\Box$

Proposition 8 Let ${X}$ be irreducible. Then, the following properties are equivalent:

• (i) there exists no non-trivial closed invariant set by ${\mathcal{M}}$
• (ii) for every ${x\in X}$ and ${\emptyset\neq U\subset X}$ open set, there exists ${g\in\mathcal{M}}$ such that ${gx\in U}$
• (iii) for every ${N\geq 1}$, ${x_1,\dots, x_N\in X}$ and ${\emptyset\neq U\subset X}$ open set, there exists ${g\in\mathcal{M}}$ such that ${gx_i\in U}$ for all ${1\leq i\leq N}$
• (iii)’ for every ${N\geq 1}$, ${x_1,\dots, x_N\in X}$ and ${\emptyset\neq U_1,\dots,U_N\subset X}$ open sets, there exists ${g\in\mathcal{M}}$ such that ${gx_i\in U_i}$ for all ${1\leq i\leq N}$

Proof: The equivalence between ${(iii)}$ and ${(iii)'}$ follows from the fact that the monoid ${\mathcal{M}}$ also acts (diagonally) on the Noetherian space ${X^N}$ (see Proposition 4).

Also, it is clear that ${(iii) \implies (ii)\implies (i)}$, so that it remains only to prove that ${(i)}$ implies ${(iii)}$.

Suppose that ${(i)}$ holds but ${(iii)}$ is not true. Then, we can find ${N\in\mathbb{N}}$, a point ${(x_1,\dots,x_N)\in X^N}$ and ${U^N=U\times\dots\times U}$ with ${U}$ a non-trivial open set of ${X}$ such that the (diagonal) action of ${\mathcal{M}}$ on ${X^N}$ satisfies:

$\displaystyle g(x_1,\dots,x_N)\notin U^N$

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

Observe that, by Proposition 4, ${U^N}$ is a (non-trivial) open set of ${X^N}$. So, ${F_0:=X^N-U^N}$ is a closed set such that

$\displaystyle (x_1,\dots,x_N)\in\bigcap\limits_{g\in\mathcal{M}}g^{-1}F_0:=F$

In particular, ${\emptyset\neq F\neq X}$ is a closed set such that ${gF\subset F}$ for all ${g\in\mathcal{M}}$. By Proposition 6, it follows that ${gF=F}$ for all ${g\in\mathcal{M}}$.

Now, we write the decomposition ${F=F_1\cup\dots\cup F_m}$ of ${F}$ into irreducible closed sets ${F_i\subset X^N}$. By Proposition 5,

$\displaystyle F_i=F_i^{(1)}\times\dots\times F_i^{(N)}$

where ${F_i^{(l)}}$ are irreducible closed subsets of ${X}$. Since ${gF=F}$ for all ${g\in\mathcal{M}}$, we have that, by Proposition 7, every ${g\in\mathcal{M}}$ permutes the irreducible closed sets ${F_i^{l}}$.

Define

$\displaystyle \emptyset\neq F^*:=\bigcup\limits_{F_i^{(l)}\neq X} F_i^{(l)}$

Here, we used that ${\emptyset\neq F\neq X}$ to ensure ${\emptyset\neq F^*}$. Because ${\mathcal{M}}$ acts by homeomorphisms and ${\mathcal{M}}$ permutes the irreducible closed subsets ${F_i^{(l)}}$, any ${F_i^{(l)}\neq X}$ is sent by ${g\in\mathcal{M}}$ into some ${F_j^{(k)}\neq X}$, and hence

$\displaystyle gF^*=F^*$

for every ${g\in\mathcal{M}}$. On the other hand, since ${X}$ is irreducible (by hypothesis), ${F^*\neq X}$, a contradiction with the statement of item ${(i)}$. $\Box$

In view of Example 2 and the discussion in SPCS6 (related to Avila-Viana simplicity criterion), it is natural to call ${(iii)}$ a (“strong form” of) twisting condition.

Remark 2 The twisting condition (i.e., ${(i), (ii), (iii)}$ or ${(iii)'}$) is satisfied for the monoid ${\mathcal{M}}$ if and only if it is satisfied for the group ${\mathcal{G}=\langle g, g^{-1}:g\in\mathcal{M}\rangle}$ generated by ${\mathcal{M}}$ (e.g., this follows immediately from the statement of item ${(i)}$).

1.3. Twisting with respect to pinching matrices

Coming back to the context of Example 2 (and using the notations from SPCS6), assume that we have a word ${\underline{\ell}\in\Omega=\bigcup\limits_{n\geq 0}\Lambda^n}$ such that ${A=A_{\underline{\ell}}}$ is simple (i.e., pinching).

Proposition 9 The twisting condition is realized if and only if for every admissible ${k}$ there exists ${\underline{\ell}(k)}$ such that the matrix ${B_k=A_{\underline{\ell}(k)}}$ verifies

$\displaystyle B_kF\cap F'=\{0\}$

for every ${A}$-invariant ${F\in G(k)}$ and ${F'\in G(d-k)}$ (that is, ${A(F)=F}$ and ${A(F')=F'}$).

For later use, we will abbreviate the second statement in this proposition by saying that ${B_k}$ is twisting with respect to ${A}$.

Remark 3 It can be checked that ${F\in G(k)}$ isotropic is ${A}$-invariant (i.e., ${A(F)=F}$) if and only if ${F}$ is generated by the sum of eigenvectors ${v_1,\dots,v_k}$ of ${A}$ with eigenvalues ${A(v_i)=\lambda_i v_i}$ such that ${|\lambda_i|\neq 1}$ and ${\lambda_i\lambda_j\neq 1}$ for every ${i\neq j}$. In particular, there are only finitely many ${F\in G(k)}$ with ${A(F)=F}$, and hence the proposition above says that, in the presence of a pinching matrix ${A}$, the (strong form of) twisting condition (an “infinitary” statement about all ${F\in G(k)}$ and ${F'\in G(d-k)}$) can be reduced to the property of twisting with respect to ${A}$ (a “finitary” statement as it concerns only the finitely many [isotropic and coisotropic] ${A}$-invariant subspaces).

Proof: Of course, “twisting with respect to ${A}$” is a necessary condition. Conversely, given ${F\in G(k)}$ and ${F_1',\dots, F_m'\in G(d-k)}$, we are searching for a matrix ${C}$ such that ${C(F)\cap F_i'=\{0\}}$ for all ${1\leq i\leq m}$. We affirm that it suffices to take ${C = C_n := A_{\underline{\ell}_0}^n B_k A_{\underline{\ell}_0}^n}$ for ${n}$ sufficiently large: indeed, as ${n\rightarrow\infty}$, we have

• ${A_{\underline{\ell}_0}^n(F)}$ converges (in the usual topology) to some ${F_0\in G(k)}$ with ${A_{\underline{\ell}_0}(F_0)=F_0}$;
• ${A_{\underline{\ell}_0}^{-n}(F_i')}$ converges (also in the usual topology) to ${F_{i,0}'\in G(d-k)}$ with ${A_{\underline{\ell}_0}(F_{i,0}')=F_{i,0}'}$.

It follows that, by assumption, ${B_k(F_0)\cap F_{i,0}'=\{0\}}$ for ${1\leq i\leq m}$ (as ${F_0}$ and ${F_{i,0}'}$ are ${A}$-invariant). Since these transversality conditions ${B_k(F_0)\cap F_{i,0}'=\{0\}}$ are open, one gets that ${B_kA_{\underline{\ell}_0}^n(F)\cap A_{\underline{\ell}_0}^{-n}(F_i') = \{0\}}$ for ${n}$ sufficiently large (depending on ${F}$ and ${F_i'}$), that is, ${C_n(F)\cap F_i'=\{0\}}$ for ${n}$ sufficiently large. $\Box$

At this stage, we are ready to state our main result towards the verification (based on Galois theory) of the pinching and twisting conditions for the Kontsevich-Zorich cocycle over the ${SL(2,\mathbb{R})}$-orbits of square-tiled surfaces.

2. Sufficient conditions for pinching and twisting in the case of origamis

Theorem 10 (M., Möller, Yoccoz) Suppose ${A_{\ell}\in Sp(2d,\mathbb{Z})}$. Assume also that there exists ${A=A_{\underline{\ell}_0}}$ and ${B=A_{\underline{\ell}_1}}$ such that

• (i) the eigenvalues of ${A}$ are real
• (ii) the Galois group ${G}$ of the characteristic polynomial ${P}$ of ${A}$ is the largest possible, that is, ${G\simeq S_d\times (\mathbb{Z}/2\mathbb{Z})^d}$
• (iii) ${B^2}$ and ${A}$ don’t share a common non-trivial invariant subspace

Then, the pinching and twisting conditions are satisfied.

Remark 4 Concerning the Galois group ${G}$ (see item ${(ii)}$), since ${A}$ is symplectic, ${P}$ is a reciprocal polynomial (i.e., ${P(1/x) = x^{-2d} P(x)}$), so that its roots have the form ${\lambda_1, \lambda_1^{-1}, \dots, \lambda_d, \lambda_d^{-1}}$. Thus, by saying that ${G}$ is the largest possible, we want to be able make all permutations of roots of ${P}$ respecting the symplectic constraint/symmetry, that is, we want ${G}$ to permute all pairs ${\{\lambda_i,\lambda_i^{-1}\}}$, ${1\leq i\leq d}$ (this gives the ${S_d}$ factor of ${G}$), and for each given pair ${\{\lambda_i, \lambda_i^{-1}\}}$ we want ${G}$ to permute ${\lambda_i}$ and ${\lambda_i^{-1}}$ without touching the other pairs (this gives the ${(\mathbb{Z}/2\mathbb{Z})^d}$ factor of ${G}$).

Remark 5 The proof of this result is somewhat long (as it involves considering several cases), and hence the general plan for the next two lectures (March 7 and 14) by J.-C. Yoccoz is the following: for the next lecture (March 7), we will see some ideas of the proof of the theorem above and for the final lecture (March 14) we will see some applications of this result to the case of origamis in the stratum ${\mathcal{H}(4)}$ (of genus ${3}$ square-tiled surfaces with a single zero of order ${4}$) where we will take ${d=g-1=2}$ and we will consider matrices on ${Sp(H_1^{(0)})\simeq Sp(4,\mathbb{Z})}$.

The rest of this post concerns some preparatory facts towards the proof of this theorem.

We begin by noticing that the matrix ${A}$ verifies the pinching condition: indeed, since the Galois group ${G}$ of ${P}$ is the largest possible, we have that ${P}$ is irreducible, and thus its roots are simple. By the assumption ${(i)}$, all roots ${\lambda_i, \lambda_i^{-1}}$, ${1\leq i\leq d}$, of ${P}$ are real, so that the pinching condition is violated by ${A}$ precisely when there are ${i\neq j}$ such that ${\lambda_i=-\lambda_j^{\pm}}$. However, this is impossible because ${G}$ is the largest possible: for instance, since ${i\neq j}$, we have an element of ${G}$ fixing ${\lambda_i}$ and exchanging ${\lambda_j}$ and ${\lambda_j^{-1}}$; applying this element to the relation ${\lambda_i=-\lambda_j^{\pm}}$, we would get that ${\lambda_i=-\lambda_j}$ and ${\lambda_i=-\lambda_j^{-1}}$, so that ${\lambda_j=\pm1}$ a contradiction with the fact that ${P}$ is irreducible.

Remark 6 Concerning the applications of this theorem to the case of origamis, we observe that item ${(iii)}$ is satisfied whenever the splitting fields ${\mathbb{Q}(P_B)}$ and ${\mathbb{Q}(P)}$ of the characteristic polynomials of ${B}$ and ${A}$ are disjointas extensions of ${\mathbb{Q}}$, i.e., ${\mathbb{Q}(P_B)\cap\mathbb{Q}(P)=\mathbb{Q}}$. Indeed, if ${E\subset\mathbb{R}^{2d}}$ is invariant by ${A}$ and ${B^2}$, one has that

• ${E}$ is generated by eigenvectors of ${A}$ (as ${A}$ is pinching, i.e., ${A}$ has simple spectrum), so that ${E}$ is defined over ${\mathbb{Q}(P)}$, and
• ${E}$ is invariant by ${B^2}$, so that ${E}$ is also defined over ${\mathbb{Q}(P_B)}$.

Since ${\mathbb{Q}(P)}$ and ${\mathbb{Q}(P_B)}$ are disjoint, it follows that ${E}$ is defined over ${\mathbb{Q}}$. But this is impossible as ${A}$ doesn’t have rational invariant subspaces (by ${(i)}$ and ${(ii)}$).

Once we know that the matrix ${A}$ satisfies the pinching condition, the proof of Theorem 10 is reduced to checking the twisting condition, and, by Proposition 9, it suffices to check the condition of twisting with respect to the pinching matrix ${A}$. Keeping this goal in mind, we introduced the following notations.

We denote by ${\widetilde{R}}$ the set of roots of the polynomial ${P}$ (so that ${\#\widetilde{R}=2d}$), for each ${\lambda\in\widetilde{R}}$, we put ${p(\lambda)=\lambda+\lambda^{-1}}$, and we define ${R=p(\widetilde{R})}$ (so that ${\# R=d}$).

Given ${1\leq k\leq d}$, let ${\widetilde{R}_k}$, resp. ${R_k}$ be the set whose elements are subsets ${\underline{\lambda}}$ of ${\widetilde{R}}$, resp. ${R}$ with ${k}$ elements, and let ${\widehat{R}_k}$ be the set whose elements are subsets ${\underline{\lambda}}$ of ${\widetilde{R}}$ with ${k}$ elements such that ${p|_{\underline{\lambda}}}$ is injective. In other words, ${\widehat{R}_k}$ consist of those ${\underline{\lambda}\in\widetilde{R}_k}$ such that if ${\lambda\in\underline{\lambda}}$, then ${\lambda^{-1}\notin\underline{\lambda}}$.

Next, we make a choice of basis of ${\mathbb{R}^{2d}}$ as follows. For each ${\lambda\in\widetilde{R}}$, we select an eigenvector ${v_{\lambda}}$ of ${A}$ associated to ${\lambda}$, i.e., ${Av_\lambda=\lambda v_{\lambda}}$. In particular, ${v_{\lambda}}$ is defined over ${\mathbb{Q}(\lambda)\subset \mathbb{Q}(P)}$). Then, we assume that the choices of ${v_{\lambda}}$‘s are coherent with the action of the Galois group ${G}$, i.e., ${v_{g\lambda}=gv_{\lambda}}$ (and thus ${A(v_{g\lambda}) = (g\lambda) v_{g\lambda}}$) for each ${g\in G}$. In this way, for each ${\underline{\lambda}\in\widetilde{R}_k}$, we can associated a multivector ${v_{\underline{\lambda}} = v_{\lambda_1}\wedge\dots\wedge v_{\lambda_k}\in\bigwedge^k\mathbb{R}^{2d}}$ (using the natural order of the elements of ${\underline{\lambda}=\{\lambda_1<\dots<\lambda_k\}}$).

By definition, ${(\bigwedge^k A)(v_{\underline{\lambda}}) = N(\underline{\lambda})v_{\underline{\lambda}}}$ where ${N(\underline{\lambda}):=\prod\lambda_i}$.

From our assumptions ${(i)}$ and ${(ii)}$ on ${A}$, we have that:

• ${v_{\underline{\lambda}}}$, ${\underline{\lambda}\in\widetilde{R}_k}$, is a basis of ${\bigwedge^k\mathbb{R}^{2d}}$
• the subspace generated by ${v_{\lambda_1}, \dots, v_{\lambda_k}}$ is isotropic if and only if ${\underline{\lambda}=\{\lambda_1,\dots,\lambda_k\}\in\widehat{R}_k}$

Also, by an elementary (linear algebra) computation, it is not hard to check that a matrix ${C}$ is twisting with respect to ${A}$ if and only if

$\displaystyle C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq 0 \ \ \ \ \ (2)$

for all ${\underline{\lambda}, \underline{\lambda}'\in\widehat{R}_k}$, where ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}}$ are the coefficients of the matrix ${\bigwedge^k C}$ in the basis ${v_{\underline{\lambda}}}$.

In order to organize our discussions, we observe that the condition (2) can be used to define an oriented graph ${\Gamma_k(C)}$ as follows. Its vertices ${\textrm{Vert}(\Gamma_k(C))}$ are ${\textrm{Vert}(\Gamma_k(C))=\widehat{R}_k}$, and we have an arrow from ${\underline{\lambda}_0}$ to ${\underline{\lambda}_1}$ if and only if ${C^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_1}\neq 0}$. In this language, (2) corresponds to the fact that ${\Gamma_k(C)}$ is a complete graph.

Of course, the verification of the completeness of ${\Gamma_k(C)}$ is not simple in general, and hence it could be interesting to look for more soft properties of ${\Gamma_k(C)}$ ensuring completeness of ${\Gamma_k(D)}$ for some matrix ${D}$ constructed as a product of powers of ${C}$ and ${A}$. Here, we take our inspiration from Dynamical Systems and we introduce the following classical notion:

Definition 11 ${\Gamma_k(C)}$ is mixing if there exists ${m\geq 1}$ such that for all ${\underline{\lambda}_0,\underline{\lambda}_1\in\widehat{R}_k}$ we can find an oriented path in ${\Gamma_k(C)}$ of length ${m}$ going from ${\underline{\lambda}_0}$ to ${\underline{\lambda}_1}$.

Here, we note that it is important in this definition that we can connect two arbitrary vertices by a path of length exactly ${m}$ (and not of length ${\leq m}$). For instance, the figure below shows a connected graph that is not mixing because all paths connecting ${A}$ to ${B}$ have odd length while all paths connecting ${A}$ to ${D}$ have even length.

As the reader can guess by now, mixing is a soft property ensuring completeness of a “related” graph. This is the content of the following proposition:

Proposition 12 Suppose that ${\Gamma_k(C)}$ is mixing with respect to an integer ${m\geq 1}$. Then, there is a finite family of hyperplanes ${V_1,\dots,V_t}$ of ${\mathbb{R}^{m-1}}$ such that, for any ${\underline{\ell}=(\ell_1,\dots,\ell_{m-1})\in \mathbb{Z}^{m-1}-(V_1\cup\dots\cup V_{m-1})}$, the matrix

$\displaystyle D(n):=C A^{n\ell_1}\dots C A^{n\ell_{m-1}} C$

satisfies (2) for all sufficiently large ${n}$.

We close this post with the proof of this proposition.

Proof: To alleviate the notation, let’s put ${D=D(n)}$. By definition,

$\displaystyle \begin{array}{rcl} D^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_m} &=& \sum_{ \substack{\gamma \textrm{ path of length }m \\ \textrm{ in } \Gamma_k(C) \textrm{ from } \underline{\lambda}_0 \textrm{ to }\underline{\lambda}_m}} C^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_1}N(\underline{\lambda}_1)^{n\ell_1} C^{(k)}_{\underline{\lambda}_1,\underline{\lambda}_2}\dots N(\underline{\lambda}_{m-1})^{n\ell_{m-1}} C^{(k)}_{\underline{\lambda}_{m-1},\underline{\lambda}_m} \\ &=:& \sum\limits_{\gamma} c_{\gamma} \exp(n L_{\gamma}(\underline{\ell})) \end{array}$

where ${c_{\gamma}\neq 0}$ and ${L_\gamma(\underline{\ell}) = \sum\limits_{i=1}^{n-1} \ell_i\left(\sum\limits_{\lambda\in\underline{\lambda}_i} \log|\lambda|\right)}$.

Our goal is to prove that ${D^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_m}\neq0}$. Of course, even though ${D^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_m}}$ was expressed as a sum of exponentials with non-vanishing coefficients, there is a risk of getting non-trivial cancelations so that the resulting expression vanishes. The idea is to show that ${\underline{\ell}}$ can be chosen suitably to avoid such cancelations, and the heart of this argument is the observation that, for ${\gamma\neq\gamma'}$, the linear forms ${L_{\gamma}}$ and ${L_{\gamma'}}$ are distinct. Indeed, given ${\underline{\lambda}\in\widehat{R}_k}$ and ${\underline{\lambda}'\in\widetilde{R}_k}$, ${\underline{\lambda}'\neq\underline{\lambda}}$, we claim that the following coefficients of ${L_{\gamma}}$ and ${L_{\gamma'}}$ differ:

$\displaystyle \sum\limits_{\lambda\in\underline{\lambda}} \log|\lambda| \neq \sum\limits_{\lambda'\in\underline{\lambda}'} \log|\lambda'|$

Otherwise, we would have a relation

$\displaystyle \prod\limits_{\lambda\in\underline{\lambda}}\lambda = \pm\prod\limits_{\lambda'\in\underline{\lambda}'}\lambda' :=\phi$

But, since ${\underline{\lambda}\in\widehat{R}_k}$, we have that if ${\lambda\in\underline{\lambda}}$ then ${\lambda^{-1}\notin\underline{\lambda}}$. In particular, by taking an element ${\lambda(0)\in\underline{\lambda}-\underline{\lambda}'}$, and by considering an element ${g}$ of the Galois group ${G}$ with ${g(\lambda(0))=\lambda(0)^{-1}}$ and ${g(\lambda)=\lambda}$ otherwise, one would get on one hand that

$\displaystyle g\phi = \prod\limits_{\lambda\in\underline{\lambda}}g\lambda = \lambda(0)^{-2}\phi$

but, on the other hand,

$\displaystyle g\phi=\pm\prod\limits_{\lambda'\in\underline{\lambda}'}g\lambda'=\pm\left\{\begin{array}{cc}\lambda(0)^2\phi & \textrm{ if } \lambda(0)^{-1}\in\underline{\lambda}' \\ \phi & \textrm{ otherwise }\end{array}\right.$

so that ${\lambda(0)^{-2}\phi=\pm\lambda(0)^2\phi}$ or ${\pm\phi}$, a contradiction in any event (as ${\lambda(0)}$ is real and ${\lambda(0)\neq\pm1}$ by the pinching conditions on ${A}$).

Now, we define ${V(\gamma,\gamma')=\{\underline{\ell}: L_{\gamma}(\underline{\ell}) = L_{\gamma'}(\underline{\ell})\}}$. Since ${L_{\gamma}}$ and ${L_{\gamma'}}$ are distinct linear forms for ${\gamma\neq\gamma'}$, it follows that ${V(\gamma,\gamma')}$ is a hyperplane. Because there are only finitely many paths ${\gamma,\gamma'}$ of length ${m}$ on ${\Gamma_k(C)}$, the collection of ${V(\gamma,\gamma')}$ corresponds to a finite family of hyperplanes ${V_1,\dots,V_t}$.

Finally, we complete the proof by noticing that if ${\underline{\ell}\notin V_1\cup\dots\cup V_t}$, then

$\displaystyle D^{(k)}_{\underline{\lambda}_0,\underline{\lambda}_m}=\sum\limits_{\gamma} c_{\gamma} \exp(n L_{\gamma}(\underline{\ell}))\neq 0$

for ${n\rightarrow\infty}$ sufficiently large because the coefficients ${L_{\gamma}(\underline{\ell})}$ are mutually distinct. $\Box$