Posted by: matheuscmss | March 11, 2012

## SPCS9

We begin with a quick revision of the second part of SPCS8. Let ${A\in Sp(2d,\mathbb{Z})}$ be a symplectic ${2d\times 2d}$ matrix with integer coefficients. Denote by ${P}$ the characteristic polynomial of ${A}$ and let ${\widetilde{R}\subset\mathbb{R}}$ be the set of roots of ${P}$. Since ${A}$ is symplectic, ${P}$ is a reciprocal polynomial, i.e., ${x^{2d}P(1/x) = P(x)}$, so that ${\lambda^{-1}\in\widetilde{R}}$ whenever ${\lambda\in\widetilde{R}}$. Then, we put ${p(\lambda):=\lambda+\lambda^{-1}}$, and we define ${R:=p(\widetilde{R})}$. Also, for each ${1\leq k\leq d}$, we let:

• ${\widetilde{R}_k}$ the set of subsets of ${\widetilde{R}}$ of cardinality ${k}$,
• ${R_k}$ the set of subsets of ${R}$ of cardinality ${k}$, and
• ${\widehat{R}_k=\{\underline{\lambda}\in\widetilde{R}_k:\#p(\underline{\lambda})=k\}}$.

Next, we suppose that the Galois group ${G}$ of ${P}$ is the largest possible, i.e., ${G\simeq S_d\times (\mathbb{Z}/2\mathbb{Z})^d}$. In other words, by dividing ${\widetilde{R}}$ into ${d}$ pairs ${\{\lambda,\lambda^{-1}\}}$, we have that ${G}$ contains all permutations of these ${d}$ pairs (and this is the ${S_d}$ factor), and for each pair ${\{\lambda,\lambda^{-1}\}}$, we may or may not permute ${\lambda}$ and ${\lambda^{-1}}$ independently of other pairs (and this is the ${(\mathbb{Z}/2\mathbb{Z})^d}$).

Now, assuming that ${G}$ is the largest possible, we fix a choice of basis ${v_{\lambda}\in\mathbb{R}^{2d}}$ consisting of eigenvectors of ${A}$ that is coherent with the action of the Galois group ${G}$, i.e., ${Av_{\lambda}=\lambda v_{\lambda}}$ and ${v_{g\lambda}=gv_{\lambda}}$ for all ${g\in G}$.

Then, we consider a matrix ${B\in Sp(2d,\mathbb{Z})}$ such that ${B^2}$ and ${A}$ don’t share a common invariant subspace, and we want to prove that

Theorem 1 Under the assumptions above, adequate products of powers of ${A}$ and ${B}$ are twisting with respect to the pinching matrix ${A}$.

Indeed, this is a main step towards the application (in the next and final post of this series) of the simplicity criterion for the Kontsevich-Zorich cocycle over ${SL(2,\mathbb{R})}$-orbits of origamis to concrete cases of genus ${3}$ origamis because it permits to reduce the verification of the twisting condition to an algebraic condition on Galois groups of characteristic polynomials of matrices in ${Sp(2d,\mathbb{Z})}$.

Moreover, by the end of SPCS8, we saw that given ${C\in Sp(2d,\mathbb{Z})}$, it is twisting with respect to ${A}$ if and only if ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq 0}$ for all ${\underline{\lambda}, \underline{\lambda}'\in\widehat{R}_k}$ for ${1\leq k\leq d}$, where ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}}$ is the coefficient of the matrix ${\wedge^k C}$ acting on ${\wedge^k\mathbb{R}^{2d}}$. Here, we used that ${\wedge^k\mathbb{R}^{2d}}$ is generated by ${v_{\underline{\lambda}}:=v_{\lambda_1}\wedge\dots\wedge v_{\lambda_k}}$, ${\underline{\lambda}=\{\lambda_1<\dots<\lambda_k\}\in\widetilde{R}_k}$, so that morally ${\widehat{R}_k}$ corresponds to an “isotropy condition”. Furthermore, we constructed a graph ${\Gamma_k(C)}$ whose set of vertices is ${\widehat{R}_k}$ and whose set of arrows is ${\{(\underline{\lambda}, \underline{\lambda}'): C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq 0\}}$. Thus, in this language ${C}$ is twisting with respect to ${A}$ if and only if ${\Gamma_k(C)}$ is a complete graph. Finally, we mentioned that checking directly that ${\Gamma_k(C)}$ is complete is not obvious, and we proved that:

Proposition 2 If ${\Gamma_k(C)}$ is mixing, then an adequate product ${D}$ of powers of ${A}$ and ${C}$ is ${k}$-twisting with respect to ${A}$ (in the sense that ${D}$ twists ${k}$-dimensional isotropic ${A}$-invariant subspaces).

So, after this brief revision of SPCS8, we will spend today’s post with the presentation of some elements of the proof of Theorem 1.

1. Strategy of proof of Theorem 1

The outline of the proof of Theorem 1 is:

• Step 0: We will show that the graphs ${\Gamma_k(C)}$ are always non-trivial, i.e., there is at least one arrow starting at each of its vertices.
• Step 1: Starting from ${A}$ and ${B}$ as above, we will show that ${\Gamma_1(B)}$ is mixing and hence, by Proposition 2, there exists ${C}$ twisting ${1}$-dimensional (isotropic) ${A}$-invariant subspaces.
• By Step 1, the treatment of the case ${d=1}$ is complete, so that we have to consider ${d\geq 2}$. Unfortunately, there is no “unified” argument to deal with all cases and we are obliged to separate the case ${d=2}$ from ${d\geq 3}$.
• Step 2: In the case ${d\geq 3}$, we will show that ${\Gamma_k(C)}$ (with ${C}$ as in Step 1) is mixing for all ${1\leq k. Hence, by Proposition 2, we can find ${D}$ twisting ${k}$-dimensional isotropic ${A}$-invariant subspaces for all ${1\leq k. Then, we will prove that ${\Gamma_d(D)}$ is mixing and, by Proposition 2, we have ${E}$ twisting with respect to ${A}$, so that this completes the argument in this case.
• Step 3: In the special case ${d=2}$, we will show that either ${\Gamma_2(C)}$ or a closely related graph ${\Gamma_2^*(C)}$ are mixing and we will see that this is sufficient to construct ${D}$ twisting ${2}$-dimensional isotropic ${A}$-invariant subspaces.

In the sequel, the following easy remarks will be repeatedly used:

Remark 1 If ${C\in Sp(2d,\mathbb{Z})}$, then the graph ${\Gamma_k(C)}$ is invariant under the action of Galois group ${G}$ on the set ${\widehat{R}_k\times \widehat{R}_k}$ (parametrizing all possible arrows of ${\Gamma_k(C)}$). In particular, since the Galois group ${G}$ is the largest possible, whenever an arrow ${\underline{\lambda}\rightarrow\underline{\lambda}'}$ belongs to ${\Gamma_k(C)}$, the inverse arrow ${\underline{\lambda}'\rightarrow\underline{\lambda}}$ also belongs to ${\Gamma_k(C)}$. Consequently, ${\Gamma_k(C)}$ always contains loop of even length.

Remark 2 A connected graph ${\Gamma}$ is not mixing if and only if there exists an integer ${m\geq 2}$ such that the lengths of all of its loops are multiples of ${m}$.

2. Step 0: ${\Gamma_k(C)}$ is nontrivial

Lemma 3 Let ${C\in Sp(2d,\mathbb{R})}$. Then, each ${\underline{\lambda}\in\widehat{R}_k}$ is the start of at least one arrow of ${\Gamma_k(C)}$.

Remark 3 Notice that we allow symplectic matrices with real (not necessarily integer) coefficients in this lemma. However, the fact that ${C}$ is symplectic is important here and the analogous lemma for general invertible (i.e., ${GL}$) matrices is false.

Proof: For ${k=1}$, since every ${1}$-dimensional subspace is isotropic, ${\widehat{R}_1=\widetilde{R}}$ and the lemma follows in this case from the fact that ${C}$ is invertible. So, let’s assume that ${k\geq 2}$ (and, in particular, ${\widehat{R}_k}$ is a proper subset of ${\widetilde{R}_k}$). Since ${C}$ is invertible, for each ${\underline{\lambda}\in\widehat{R}_k}$, there exists ${\underline{\lambda}'\in\widetilde{R}_k}$ with

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

Of course, one may have a priori that ${\underline{\lambda}'\in\widetilde{R}_k - \widehat{R}_k}$, i.e., ${\# p(\underline{\lambda}), and, in this case, our task is to “convert” ${\underline{\lambda}'}$ into some ${\underline{\lambda}''\in\widehat{R}_k}$ with ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}''}\neq 0}$.

Evidently, in order to accomplish this task it suffices to show that if ${\#p(\underline{\lambda}') and ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq 0}$, then there exists ${\underline{\lambda}''}$ with ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}''}\neq 0}$ and ${\#p(\underline{\lambda}'')=\#p(\underline{\lambda}')+1}$. Keeping this goal in mind, we observe that ${\underline{\lambda}'\notin\widehat{R}_k}$ implies that we can write ${\underline{\lambda}'=\{\lambda_1',\lambda_2',\dots,\lambda_k'\}}$ with ${\lambda_1'\cdot \lambda_2'=1}$. Also, the fact that ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq0}$ is equivalent to say that the ${k\times k}$ minor of ${C}$ associated to ${\underline{\lambda}}$ and ${\underline{\lambda}'}$ is invertible, and hence, by writing ${\underline{\lambda}=\{\lambda_1,\lambda_2,\dots,\lambda_k\}}$, we can find ${w_1,\dots, w_k\in\mathbb{R}^{2d}}$ such that ${\textrm{span}\{w_1,\dots,w_k\}=\textrm{span}\{v_{\lambda_1},\dots, v_{\lambda_k}\}}$ and

$\displaystyle C(w_i)=v_{\lambda_i'}+\sum\limits_{\lambda\notin\underline{\lambda}'}C_{i\lambda}^* v_{\lambda}.$

In other words, we can make a change of basis to convert the invertible minor of ${C}$ into the ${k\times k}$ identity matrix.

Now, denoting by ${\{,.\}}$ the symplectic form, we observe that ${\{w_1, w_2\}=0}$ because ${\underline{\lambda}\in\widehat{R}_k}$, i.e., the span of ${v_{\lambda_i}}$ is an isotropic subspace, and ${w_1,w_2\in\textrm{span}\{v_{\lambda_1},v_{\lambda_2},\dots, v_{\lambda_k}\}}$. On the other hand, since ${C}$ is symplectic, we get that

$\displaystyle 0=\{w_1,w_2\}=\{C(w_1), C(w_2)\} = \{v_{\lambda_1'}, v_{\lambda_2'}\} + \sum_{\substack{\lambda', \lambda''\notin \underline{\lambda}' \\ \lambda'\cdot\lambda''=1}} C_{1\lambda'}^* C_{2\lambda''}^* \{v_{\lambda'}, v_{\lambda''}\}$

Since ${\{v_{\lambda_1'}, v_{\lambda_2'}\}\neq 0}$ (as ${\lambda_1'\cdot\lambda_2'=1}$), it follows that there exists ${\lambda', \lambda''\notin \underline{\lambda}'}$ with ${C^*_{1\lambda'}\neq 0}$ and ${C_{2\lambda''}^*\neq 0}$.

Then we define ${\underline{\lambda}'':=(\underline{\lambda}'-\{\lambda_1'\})\cup\{\lambda'\}}$. We have that ${\#p(\underline{\lambda}'')=\#p(\underline{\lambda}')+1}$. Furthermore, the minor ${C[\underline{\lambda}, \underline{\lambda}'']}$ of ${C}$ associated to ${\underline{\lambda}}$ and ${\underline{\lambda}''}$ is obtained from the minor ${C[\underline{\lambda}, \underline{\lambda}']}$ of ${C}$ associated to ${\underline{\lambda}}$ and ${\underline{\lambda}'}$ by removing the line associated to ${v_{\lambda_1'}}$ and replacing it by the line associated to ${v_{\lambda'}}$. By looking in the basis ${w_1,\dots, w_k}$, this means that the minor ${C[\underline{\lambda}, \underline{\lambda}'']}$ differs from the identity minor ${C[\underline{\lambda}, \underline{\lambda}']}$ by the fact that the line associated to ${v_{\lambda_1'}}$ was replaced by the line associated to ${v_{\lambda'}}$. In other words, in the basis ${w_1, \dots, w_k}$, one of the entries ${1}$ of ${C[\underline{\lambda}, \underline{\lambda}']}$ was replaced by the coefficient ${C^*_{1\lambda'}\neq 0}$. Thus, we conclude that the determinant ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}''}}$ of the the minor ${C[\underline{\lambda}, \underline{\lambda}'']}$ is

$\displaystyle C^{(k)}_{\underline{\lambda}, \underline{\lambda}''} = C^*_{1\lambda'}\neq 0$

Therefore, ${\underline{\lambda}''}$ satisfies the desired properties and the argument is complete. $\Box$

3. Step 1: ${\Gamma_1(B)}$ is mixing

For ${d=1}$, the set ${\widehat{R}_1}$ consists of exactly one pair ${=\{\lambda,\lambda^{-1}\}}$, so that the possible Galois invariant graphs are:

In the first case, by definition, we have that ${B(\mathbb{R}v_{\lambda})=\mathbb{R}v_{\lambda}}$ (and ${B(\mathbb{R}v_{\lambda^{-1}}) = \mathbb{R}v_{\lambda^{-1}}}$), so that ${B}$ and ${A}$ share a common subspace, a contradiction with our hypothesis in Theorem 1.

In the second case, by definition, we have that ${B(\mathbb{R}v_{\lambda})=\mathbb{R}v_{\lambda^{-1}}}$ and ${B(\mathbb{R}v_{\lambda^{-1}}) = \mathbb{R}v_{\lambda}}$, so that ${B^2(\mathbb{R}v_{\lambda})=\mathbb{R}v_{\lambda}}$ and thus ${B^2}$ and ${A}$ share a common subspace, a contradiction with our assumptions in Theorem 1.

Finally, in the third case, we have that the graph ${\Gamma_1(B)}$ is complete, and hence ${B}$ is ${1}$-twisting with respect to ${A}$.

Now, after this “warm up”, we pass to the general case ${d\geq 2}$. Firstly, suppose that the sole arrows in ${\Gamma_1(B)}$ are of the form ${\lambda\rightarrow\lambda^{\pm1}}$. Then, ${B(\mathbb{R}v_{\lambda}\oplus\mathbb{R}v_{\lambda^{-1}}) = \mathbb{R}v_{\lambda}\oplus\mathbb{R}v_{\lambda^{-1}}}$, and, since ${d\geq 2}$, the subspace ${\mathbb{R}v_{\lambda}\oplus\mathbb{R}v_{\lambda^{-1}}}$ is non-trivial. In particular, in this case, ${B}$ and ${A}$ share a common non-trivial subspace, a contradiction. Of course, this arguments breaks up for ${d=1}$ (and this is why we had a separate argument for this case).

Therefore, we may assume that ${\Gamma_1(B)}$ has some arrow ${\lambda\rightarrow\lambda'}$ with ${\lambda'\neq\lambda^{\pm1}}$. Because the Galois group ${G}$ is the largest possible and ${\Gamma_1(B)}$ is invariant under the action of ${G}$ (see Remark 1), we have that all arrows of this type belong to ${\Gamma_1(B)}$. In view of Remarks 1 and 2, it suffices to construct a loop of odd length in ${\Gamma_1(B)}$.

Since we dispose of all arrows ${\lambda\rightarrow\lambda'}$ with ${\lambda'\neq\lambda^{\pm1}}$, if ${d\geq 3}$, we can easily construct a loop of length ${3}$:

On the other hand, for ${d=2}$, we have two possibilities. If ${\Gamma_1(B)}$ is the non-mixing graph invariant under the Galois group:

we get that ${B(\mathbb{R}v_{\lambda_1}\oplus \mathbb{R}v_{\lambda_1}^{-1}) = \mathbb{R}v_{\lambda_2}\oplus \mathbb{R}v_{\lambda_2}^{-1}}$ and ${B(\mathbb{R}v_{\lambda_2}\oplus \mathbb{R}v_{\lambda_2}^{-1}) = \mathbb{R}v_{\lambda_1}\oplus \mathbb{R}v_{\lambda_1}^{-1}}$, so that ${B^2}$ and ${A}$ share a common invariant subspace, a contradiction.

So, we have some extra arrow in the previous picture, say:

In this case, it is not hard to see that the addition of any extra arrow allows to build up loops of lenght ${3}$, so that, by Remarks 1 and 2, the argument is complete.

Therefore, in any event, we proved that ${\Gamma_1(B)}$ is mixing.

4. Step 2: For ${d\geq 3}$, ${\Gamma_k(C)}$ is mixing for ${2\leq k, and ${\Gamma_d(D)}$ is mixing

Given ${C\in Sp(2d,\mathbb{Z})}$ twisting ${1}$-dimensional ${A}$-invariant subspaces, we wish to prove that ${\Gamma_k(C)}$ is mixing for all ${2\leq k whenever ${d\geq 3}$. Since ${\Gamma_k(C)}$ is invariant under the Galois group ${G}$ (see Remark 1), we start by considering the orbits of the action of ${G}$ on ${\widehat{R}_k\times \widehat{R}_k}$.

Proposition 4 The orbits of the action of ${G}$ on ${\widehat{R}_k\times \widehat{R}_k}$ are

$\displaystyle \mathcal{O}_{\widetilde{\ell}, \ell} = \{(\underline{\lambda}, \underline{\lambda}')\in \widehat{R}_k\times \widehat{R}_k: \#(\underline{\lambda}\cap \underline{\lambda}')=\widetilde{\ell}, \#(p(\underline{\lambda}) \cap p(\underline{\lambda}'))=\ell\}$

where

$\displaystyle 0\leq\widetilde{\ell}\leq\ell\leq k, \ell\geq 2k-d \ \ \ \ \ (1)$

We leave the proof of this proposition as an exercise to the reader. This proposition says that the orbits of the action ${G}$ on ${\widehat{R}_k\times \widehat{R}_k}$ are naturalized parametrized by

$\displaystyle \widetilde{I}=\{(\widetilde{\ell}, \ell) \textrm{ satisfying } (1)\}$

In particular, since ${\Gamma_k(C)}$ is ${G}$-invariant, we can write ${\Gamma_k(C)=\Gamma_k(\widetilde{J})}$ for some ${\widetilde{J}:=\widetilde{J}(C)\subset\widetilde{I}}$, where ${\Gamma_k(J)}$ is the graph whose vertices are ${\widehat{R}_k}$ and whose arrows are

$\displaystyle \bigcup\limits_{(\widetilde{\ell}, \ell)\in\widetilde{J}} \mathcal{O}_{\widetilde{\ell}, \ell}$

Proposition 5 The graph ${\Gamma_k(\widetilde{J})}$ is not mixing if and only if

• either ${k\neq d/2}$ and ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq \widetilde{\ell}\leq k\}}$
• or ${k=d/2}$ and ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq \widetilde{\ell}\leq k\}\cup\{(0,0)\}}$

Proof: Let ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq\widetilde{\ell}\leq k\}}$ for ${k\neq d/2}$ or ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq\widetilde{\ell}\leq k\}\cup\{(0,0)\}}$ for ${k=d/2}$. Then, one can see that, since ${k, ${\Gamma_k(\widetilde{J})}$ is not mixing simply because it is not connected! For the proof of the converse statement, due to the usual space-time limitations, we’re going only to say a few words on (referring to the forthcoming article by M. Möller, J.-C. Yoccoz and C. M. for formal arguments). Essentially, one starts by converting pairs ${\{\lambda,\lambda^{-1}\}}$ into a single point ${p(\lambda)=p(\lambda^{-1})}$, so that ${\Gamma_k(\widetilde{J})}$ becomes a new graph ${\overline{\Gamma}_k(\widetilde{J})}$. Then, one proves that, if ${\widetilde{J}\not\subset \{(\widetilde{\ell},k): 0\leq \widetilde{\ell}\leq k\}\cup\{(0,0)\}}$, then ${\overline{\Gamma}_k(\widetilde{J})}$ is connected. Using that ${\overline{\Gamma}_k(\widetilde{J})}$ is connected, it is possible to prove that ${\Gamma_k(\widetilde{J})}$ is connected and from this one can construct loops of odd length, thus getting the mixing property. $\Box$

Coming back to the study of ${\Gamma_k(C)}$, ${1\leq k, ${d\geq 3}$, we set ${\widetilde{J}:=\widetilde{J}(C)}$. By the previous proposition, if ${\Gamma_k(C)}$ is not mixing, then ${\widetilde{J}(C)\subset \{(\widetilde{\ell}, k): 0\leq \widetilde{\ell}\leq k\}}$ for ${k\neq d/2}$ or ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq\widetilde{\ell}\leq k\}\cup\{(0,0)\}}$ for ${k=d/2}$. For sake of concreteness, we will deal “only” with the case ${\widetilde{J}\subset\{(\widetilde{\ell},k): 0\leq\widetilde{\ell}\leq k\}\cup\{(0,0)\}}$ (leaving the particular case ${\widetilde{J}=\{(0,0)\}}$ when ${k=d/2}$ as an exercise to the reader). In this situation, we have an arrow ${\{\lambda_1, \dots, \lambda_k\}=\underline{\lambda}\rightarrow\underline{\lambda}'=\{\lambda_1', \dots, \lambda_k'\}}$ of ${\Gamma_k(C)}$ with ${p(\underline{\lambda})=p(\underline{\lambda}')}$. This means that ${C^{(k)}_{\underline{\lambda}, \underline{\lambda}'}\neq 0}$, and hence we can find ${w_1, \dots, w_k}$ such that ${\textrm{span}\{w_1,\dots,w_k\} = \textrm{span}\{v_{\lambda_1}, \dots, v_{\lambda_k}\}}$ and

$\displaystyle C(w_i)=v_{\lambda_i'}+\sum\limits_{\lambda\notin\underline{\lambda}'} C_{i\lambda}^* v_{\lambda}$

In other words, as we also did in Step 0, we can use ${w_1,\dots, w_k}$ to “convert” the minor of ${C}$ associated to ${\underline{\lambda}, \underline{\lambda}'}$ into the identity.

We claim that if ${\lambda, \lambda^{-1}\notin \underline{\lambda}'}$, then ${C_{i\lambda}^*=0}$ for all ${i=1,\dots,k}$. Indeed, by the same discussion around minors and replacement of lines, if this were not true, say ${C_{i\lambda}^*\neq 0}$, we could find an arrow from ${\underline{\lambda}}$ to ${\underline{\lambda}''=(\underline{\lambda}'-\{\lambda_i'\})\cup\{\lambda\}}$. Since ${p(\underline{\lambda})=p(\underline{\lambda}')}$, we have that ${\#(p(\underline{\lambda})\cap p(\underline{\lambda}''))=k-1}$, so that, for some ${\widetilde{\ell}_0}$, one has ${(\widetilde{\ell}_0, k-1)\in \widetilde{J}\subset \{(\widetilde{\ell}, k): 0\leq \widetilde{\ell}\leq k\}}$, a contradiction showing that the claim is true.

From the claim above we deduce that e.g. ${C(v_{\lambda_1})}$ is a linear combination of ${v_{\lambda_i'}}$, ${i=1,\dots, k}$, a contradiction with the fact that ${C}$ twists ${1}$-dimensional ${A}$-invariant subspaces. In other words, we proved that ${\Gamma_k(C)}$ is mixing for each ${2\leq k whenever ${C}$ twists ${1}$-dimensional ${A}$-invariant subspaces.

By Proposition 2, it follows that we can construct a matrix ${D}$ twisting ${k}$-dimensional isotropic ${A}$-invariant subspaces for ${1\leq k, and we wish to show that ${\Gamma_d(D)}$ is mixing. In this direction, we consider the orbits of the action of the Galois group ${G}$ on ${\widehat{R}_d\times\widehat{R}_d}$. By Proposition 4, the orbits are

$\displaystyle \mathcal{O}_{\widetilde{\ell},\ell} = \{(\underline{\lambda}, \underline{\lambda}')\in \widehat{R}_k\times \widehat{R}_k: \#(\underline{\lambda}\cap \underline{\lambda}')=\widetilde{\ell}, \#(p(\underline{\lambda}) \cap p(\underline{\lambda}'))=\ell\}$

with ${\ell\leq k}$, ${\ell\geq 2k-d}$ and ${k=d}$. In particular, ${\ell=d}$ in this case, and the orbits are parametrized by the set

$\displaystyle I=\{0\leq\widetilde{\ell}\leq d\}$

For sake of simplicity, we will denote the orbits of ${G}$ on ${\widehat{R}_d\times\widehat{R}_d}$ by

$\displaystyle \mathcal{O}(\widetilde{\ell})=\{(\underline{\lambda}, \underline{\lambda}')\in \widehat{R}_d\times \widehat{R}_d: \#(\underline{\lambda}\cap \underline{\lambda}')=\widetilde{\ell}\}$

and we write

$\displaystyle \Gamma_d(D)=\Gamma_d(J)=\bigcup\limits_{\widetilde{\ell}\in J}\mathcal{O}(\widetilde{\ell})$

where ${J=J(D)\subset I=\{0\leq\widetilde{\ell}\leq d\}}$.

It is possible to show (again by the arguments with “minors” we saw above) that if ${D}$ is ${k}$-twisting with respect to ${A}$, then ${J}$ contains two consecutive integers say ${\widetilde{\ell}, \widetilde{\ell}+1}$.

We claim that ${\Gamma_d(D)}$ is mixing whenever ${J}$ contains two consecutive integers.

Indeed, we start by showing that ${\Gamma_d(J)}$ is connected. Notice that it suffices to connect two vertices ${\underline{\lambda}_0}$ and ${\underline{\lambda}_1}$ with ${\#(\underline{\lambda}_0\cap\underline{\lambda}_1)=d-1}$ (as the general case of two general vertices ${\underline{\lambda}}$ and ${\underline{\lambda}'}$ follows by producing a series of vertices ${\underline{\lambda}=\underline{\lambda}_0}$, ${\underline{\lambda}_1}$, ${\dots}$, ${\underline{\lambda}_a=\underline{\lambda}'}$ with ${\#(\underline{\lambda}_i\cap \underline{\lambda}_{i+1})=d-1}$, ${i=0,\dots, a-1}$). Given ${\underline{\lambda}_0}$ and ${\underline{\lambda}_1}$ with ${\#(\underline{\lambda}_0\cap\underline{\lambda}_1)=d-1}$, we select ${\underline{\lambda}'\subset \underline{\lambda}_0\cap \underline{\lambda}_1}$ with ${\#\underline{\lambda}'=d-\widetilde{\ell}-1}$. Then, we consider ${\underline{\lambda}''}$ obtained from ${\underline{\lambda}_0}$ by replacing the elements of ${\underline{\lambda}'}$ by their inverses. By definition, ${\#(\underline{\lambda}''\cap\underline{\lambda}_0) = \widetilde{\ell}+1}$ and ${\#(\underline{\lambda}''\cap\underline{\lambda}_1)=\widetilde{\ell}}$ (because ${\#(\underline{\lambda}_0\cap\underline{\lambda}_1)=d-1}$). By assumption, ${J}$ contains ${\widetilde{\ell}+1}$ and ${\widetilde{\ell}}$, so that we have the arrows ${\underline{\lambda}_0\rightarrow\underline{\lambda}''}$ and ${\underline{\lambda}''\rightarrow\underline{\lambda}_1}$ in ${\Gamma_d(J)}$. Thus, the connectedness of ${\Gamma_d(J)}$ follows.

Next, we show that ${\Gamma_d(J)}$ is mixing. Since ${\Gamma_d(J)}$ is invariant under the Galois group, it contains loops of length ${2}$ (see Remark 1). By Remark 2, it suffices to construct some loop of odd length in ${\Gamma_d(J)}$. We fix some arrow ${\underline{\lambda}\rightarrow\underline{\lambda}'\in\mathcal{O}(\widetilde{\ell})}$ of ${\Gamma_d(J)}$. By the construction “${\underline{\lambda}_0\rightarrow\underline{\lambda}''\rightarrow\underline{\lambda}_1}$” when ${\#(\underline{\lambda}_0\cap\underline{\lambda}_1)=d-1}$ performed in the proof of the connectedness of ${\Gamma_d(J)}$, we can connect ${\underline{\lambda}'}$ to ${\underline{\lambda}}$ by a path of length ${2\widetilde{\ell}}$ in ${\Gamma_d(J)}$. In this way, we have a loop (based on ${\underline{\lambda}_0}$) in ${\Gamma_d(J)}$ of length ${2\widetilde{\ell}+1}$.

5. Step 3: Special case ${d=2}$

We consider the symplectic form ${\{.,.\}:\wedge^2\mathbb{R}^4\rightarrow\mathbb{R}}$. Since ${\wedge^2\mathbb{R}^4}$ has dimension ${6}$ and ${\{.,.\}}$ is non-degenerate, ${K:=\textrm{Ker}\{.,.\}}$ has dimension ${5}$.

By denoting by ${\lambda_1> \lambda_2>\lambda_2^{-1}> \lambda_1^{-1}}$ the eigenvalues of ${A}$, we have the following basis of ${K}$

• ${v_{\lambda_1}\wedge v_{\lambda_2}}$, ${v_{\lambda_1}\wedge v_{\lambda_2^{-1}}}$, ${v_{\lambda_1^{-1}}\wedge v_{\lambda_2}}$, ${v_{\lambda_1^{-1}}\wedge v_{\lambda_2^{-1}}}$;
• ${v_*=\frac{v_{\lambda_1}\wedge v_{\lambda_1^{-1}}}{\omega_1} - \frac{v_{\lambda_2}\wedge v_{\lambda_2^{-1}}}{\omega_2}}$ where ${\omega_i = \{v_{\lambda_i}, v_{\lambda_i^{-1}}\}\neq 0}$.

In general, given ${C\in Sp(4,\mathbb{Z})}$, we can use ${\wedge^2C|_K}$ to construct a graph ${\Gamma_2^*(C)}$ whose vertices are ${\widehat{R}_2\simeq \{ v_{\lambda_1}\wedge v_{\lambda_2}, \dots, v_{\lambda_1^{-1}}\wedge v_{\lambda_2^{-1}}\}}$ and ${v_*}$, and whose arrows connect vertices associated to non-zero entries of ${\wedge^2C|_K}$.

By definition, ${\wedge^2A(v_*)=v_*}$, so that ${1}$ is an eigenvalue of ${\wedge^2A|_K}$. In principle, this poses a problem to apply Proposition 12 of SPCS8 (to deduce ${2}$-twisting properties of ${C}$ from ${\Gamma_2^*(C)}$ is mixing), but, as it turns out, the fact that the eigenvalue ${1}$ of ${\wedge^2A|_K}$ is simple can be exploited to rework the proof of Proposition 12 of SPCS8 to check that ${\Gamma_2^*(C)}$ is mixing implies the existence of adequate products ${D}$ of powers of ${C}$ and ${A}$ satisfying the ${2}$-twisting condition (i.e., ${D}$ twists ${2}$-dimensional ${A}$-invariant isotropic subspaces).

Therefore, it “remains” to show that either ${\Gamma_2(C)}$ or ${\Gamma_2^*(C)}$ is mixing to complete the considerations of this section (and today’s post).

We write ${\Gamma_2(C)=\Gamma_2(J)}$ with ${J\subset\{0, 1, 2\}}$.

• If ${J}$ contains ${2}$ two consecutive integers, then one can check that the arguments of the end of the previous sections work and ${\Gamma_2(C)}$ is mixing.
• Otherwise, since ${J\neq\emptyset}$ (see Step 0), we have ${J=\{0\}}$, ${\{2\}, \{1\}}$ or ${\{0,2\}}$. As it turns out, the cases ${J=\{0\}, \{2\}}$ are “symmetric”, as well as the cases ${J=\{1\}, \{0,2\}}$.

For sake of concreteness, we will consider the cases ${J=\{2\}}$ and ${J=\{1\}}$ (leaving the treatment of their “symmetric” as an exercise). We will show that the case ${J=\{2\}}$ is impossible while the case ${J=\{1\}}$ implies that ${\Gamma_2^*(C)}$ is mixing.

We begin by ${J=\{2\}}$. This implies that we have an arrow ${\underline{\lambda}\rightarrow\underline{\lambda}}$ with ${\underline{\lambda}= \{\lambda_1,\lambda_2\}}$. Hence, we can find ${w_1, w_2}$ with ${\textrm{span}\{w_1,w_2\}=\textrm{span}\{v_{\lambda_1}, v_{\lambda_2}\}}$ and

$\displaystyle C(w_1) = v_{\lambda_1}+ C^*_{11} v_{\lambda_1^{-1}} + C^*_{12} v_{\lambda_2^{-1}},$

$\displaystyle C(w_2) = v_{\lambda_2}+ C^*_{21} v_{\lambda_1^{-1}} + C^*_{22} v_{\lambda_2^{-1}}.$

Since ${J=\{2\}}$, the arrows ${\underline{\lambda}\rightarrow\{\lambda_1^{-1}, \lambda_2\}}$, ${\underline{\lambda}\rightarrow\{\lambda_1, \lambda_2^{-1}\}}$, and ${\underline{\lambda}\rightarrow\{\lambda_1^{-1}, \lambda_2^{-1}\}}$ do not belong ${\Gamma_2(J)}$. Thus, ${C_{11}^*=C_{22}^*=0=C_{12}^*C_{21}^*}$. On the other hand, because ${C}$ is symplectic, ${\omega_1 C_{21}^* - \omega_2 C_{12}^* = 0}$ (with ${\omega_1,\omega_2\neq 0}$). It follows that ${C^*_{ij}=0}$ for all ${1\leq i,j\leq 2}$, that is, ${C}$ preserves the ${A}$-subspace spanned by ${v_{\lambda_1}}$ and ${v_{\lambda_2}}$, a contradiction with the fact that ${C}$ is ${1}$-twisting with respect to ${A}$.

Now we consider the case ${J=\{1\}}$ and we wish to show that ${\Gamma_2^*(C)}$. We claim that, in this situation, it suffices to construct arrows from the vertex ${v_*}$ to ${\widehat{R}_2}$ and vice-versa. Notice that the action of the Galois group can’t be used to revert arrows of ${\Gamma_2^*(C)}$ involving the vertex ${v_*}$, so that the two previous statements are “independent”. Assuming the claim holds, we can use the Galois action to see that once ${\Gamma_2^*(C)}$ contains some arrows from ${v_*}$ and some arrows to ${v_*}$, it contains all such arrows. In other words, if the claim is true, we have the following situation:

Thus, we have loops of length ${2}$ (in ${\widehat{R}_2}$), and also loops of length ${3}$ (based on ${v_*}$), so that ${\Gamma_2^*(C)}$ is mixing. In particular, our task is reduced to show the claim above.

The fact that there are arrows from ${\widehat{R}_2}$ to ${v_*}$ follows from the same kind of arguments involving “minors” (i.e., selecting ${w_1, w_2}$ as above, etc.) and we will not repeat it here.

Instead, we will focus on showing that there are arrows from ${v_*}$ to ${\widehat{R}_2}$. The proof is by contradiction: otherwise, one has ${\wedge^2C(v_*)\in\mathbb{R}v_*}$. Then, we invoke the following elementary lemma (whose proof is a straightforward computation):

Lemma 6 Let ${H\subset\mathbb{R}^4}$ be a symplectic ${2}$-plane. Given ${e, f}$ a basis of ${H}$ with ${\{e,f\}=1}$, we define

$\displaystyle i(H):=e\wedge f$

The bi-vector ${i(H)}$ is independent of the choice of ${e, f}$ as above. Denote by ${H^\perp}$ the symplectic orthogonal of ${H}$ and put

$\displaystyle v(H):=i(H)-i(H^\perp)\in K$

Then, ${v(H)}$ is collinear to ${v(H')}$ if and only if ${H'=H}$ or ${H'=H^\perp}$.

Since ${v_*:=i(\textrm{span}(v_{\lambda_1}, v_{\lambda_1^{-1}}))}$, from this lemma we obtain that ${\wedge^2C(v_*)\in\mathbb{R}v_*}$ implies ${C(\textrm{span}(v_{\lambda_1}, v_{\lambda_1^{-1}})) = \textrm{span}(v_{\lambda_1}, v_{\lambda_1^{-1}})}$ or ${\textrm{span}(v_{\lambda_2}, v_{\lambda_2^{-1}})}$, a contradiction with the fact that ${C}$ is ${1}$-twisting with respect to ${A}$.