Posted by: matheuscmss | March 15, 2012

## SPCS10

For the final post of this series, we will study some applications of the simplicity criterion for the Kontsevich-Zorich cocycle over the ${SL(2,\mathbb{R})}$-orbits of square-tiled surfaces (discussed in SPCS 6, 7, 8 and 9).

We recall the notation ${H(k_1-1,\dots, k_s-1)}$ for the stratum of the moduli space of Abelian differentials with zeros of orders ${k_1-1,\dots, k_s-1}$, or, equivalently, the stratum of the moduli space of translation structures with conical singularities of total angle ${2\pi k_i}$. Today, we will impose the restriction ${k_1,\dots,k_s\geq 2}$. Since the genus ${g}$ of the Riemann surfaces supporting translation structures in ${H(k_1-1,\dots,k_s-1)}$ satisfies

$\displaystyle \sum\limits_{i=1}^s (k_i-1)=2g-2,$

our restriction forces ${g\geq 2}$.

In general ${H(k_1-1,\dots,k_s-1)}$ is a complex orbifold of complex dimension

$\displaystyle d=\textrm{dim}_{\mathbb{C}}H(k_1-1,\dots,k_s-1)=2g+s-1$

The list of strata (in increasing order of “complexity”) is thus:

• ${H(2)}$ corresponding to ${d=4}$, ${g=2}$, ${s=1}$, ${k_1=3}$: after the works of P. Hubert and S. Lelievre, C. McMullen and K. Calta, our current knowledge is satisfactory.
• ${H(1,1)}$ corresponding to ${d=5}$, ${g=2}$, ${s=2}$, ${k_1=k_2=2}$: by the works of C. McMullen and K. Calta, we dispose of partial information about this stratum.
• ${H(4)}$ corresponding to ${d=6}$, ${g=3}$, ${s=1}$, ${k_1=5}$: by the end of this post we will see a conjectural picture (due to V. Delecroix and S. Lelievre) about the classification of square-tiled surfaces in this stratum, and we will indicate how to apply the simplicity criterion for a
• ${d=7}$, ${g=3}$, ${s=2}$: for ${k_1=4}$, ${k_2=1}$ we have ${H(3,1)}$, and for ${k_1=k_2=3}$ we have ${H(2,2)}$.
• etc.

In the sequel we will need the following lemma about the “minimal” strata:

Lemma 1 In a stratum ${H(2g-2)}$, a translation surface has no non-trivial automorphisms.

Proof: Let ${(M,\omega)\in H(2g-2)}$ and denote by ${p_M\in M}$ the unique zero of the Abelian differential ${\omega}$. Any automorphism ${\phi}$ of the translation structure ${(M,\omega)}$ satisfies ${\phi(p_M)=p_M}$. Suppose that there exists ${\phi}$ a non-trivial automorphism of ${(M,\omega)}$. We have that ${\phi}$ has finite order, say ${\phi^{\kappa}=\textrm{id}}$ where ${\kappa\geq 2}$ is the order of ${\phi}$. Since ${\phi}$ fixes ${p_M}$ and ${\phi}$ is non-trivial, ${p_M}$ is the sole fixed point of ${\phi}$. Hence, the quotient ${N}$ of ${M}$ by the cyclic group ${\langle\phi\rangle\simeq\mathbb{Z}/\kappa\mathbb{Z}}$ can be viewed as a normal covering ${\pi:(M,p_M)\rightarrow (N,p_N)}$ of degree ${\kappa}$ such that ${\pi}$ is not ramified outside ${p_N:=\pi(p_M)}$, ${\pi^{-1}(p_N)=\{p_M\}}$ and ${\pi}$ is ramified of index ${\kappa}$ at ${p_N}$.

Now let us fix ${\ast\in N-\{p_N\}}$ a base point, so that the covering ${\pi}$ is given by a homomorphism ${h:\pi_1(N-\{p_N\},\ast)\rightarrow\mathbb{Z}/\kappa\mathbb{Z}}$. In this language, we see that a loop ${\gamma\in \pi_1(N-\{p_N\}, \ast)}$ around ${p_N}$ based at ${\ast}$ is a product of commutators in ${\pi_1(N-\{p_N\},\ast)}$, so that ${h(\gamma)=1}$ (as ${\mathbb{Z}/\kappa\mathbb{Z}}$), a contradiction with the fact that ${\pi}$ is ramified of order ${\kappa\geq 2}$ at ${p_N}$. $\Box$

Remark 1 In a certain sense this lemma implies that origamis of minimal strata ${H(2g-2)}$ are not concerned by the discussions of the first 5 posts of this series: indeed, there the idea was to use representation theory of the automorphism group of the origami to detect multiple (i.e., non-simple) and/or zero exponents; in particular, we needed rich groups of automorphisms and hence, from this point of view, origamis without non-trivial automorphisms are “uninteristing”.

Next, we recall that M. Kontsevich and A. Zorich classified all connected components of strata ${H(k_1-1,\dots, k_s-1)}$, and, as an outcome of this classification, any stratum has at most 3 connected components. We will come back to this latter when discussing the stratum ${H(4)}$. For now, let us mention that the two strata ${H(2)}$ and ${H(1,1)}$ of genus ${2}$ translation structures are connected.

Once we fix a connected component ${\mathcal{C}}$ of a stratum, we can ask about the classification of ${SL(2,\mathbb{Z})}$-orbits of (reduced/primitive) origamis in ${\mathcal{C}}$. In this direction, it is important to dispose of invariants to distinguish ${SL(2,\mathbb{Z})}$-orbits. As it turns out, the so-called monodromy group of an origami is such an invariant, and the following result (from the PhD thesis) of D. Zmiaikou shows that this invariant can take only two values when the number of squares of the origami is sufficiently large:

Theorem 2 (D. Zmiaikou) Given an stratum ${H(k_1-1,\dots, k_s-1)}$, there exists an integer ${N_0=N_0(k_1,\dots,k_s)}$ such that any primitive origami of ${H(k_1-1,\dots,k_s-1)}$ with ${N\geq N_0}$ squares has monodromy group isomorphic to ${A_N}$ or ${S_N}$.

Remark 2 The integer ${N_0=N_0(k_1,\dots,k_s)}$ has explicit upper bounds (as it was shown by D. Zmiaikou), but we did not include it in the previous statement because it is believed that the current upper bounds are not sharp.

After this (brief) general discussion, in the next section we’ll specialize to the case of genus 2 origamis.

1. Classification of ${SL(2,\mathbb{Z})}$-orbits of reduced origamis in ${H(2)}$

Denote by ${N\geq 3}$ the number of squares of a reduced origami in ${H(2)}$. By the results of P. Hubert and S. Lelievre, and C. McMullen quoted above, it is possible to show that the ${SL(2,\mathbb{Z})}$-orbits of origamis are organized as follows

• if ${N\geq 4}$ is even, then there is exactly ${1}$ ${SL(2,\mathbb{Z})}$-orbit and the monodromy group is ${S_N}$.
• if ${N\geq 5}$ is odd, then there are exactly ${2}$ ${SL(2,\mathbb{Z})}$-orbits distinguished by the monodromy group being ${A_N}$ or ${S_N}$.
• if ${N=3}$, there is exactly ${1}$ ${SL(2,\mathbb{Z})}$-orbit and the monodromy group is ${S_3}$.

Concerning the Lyapunov exponents of the Kontsevich-Zorich (KZ) cocycle, it was proved by M. Bainbridge, and A. Eskin, M. Kontsevich and A. Zorich (EKZ for short) that the second (non-negative) exponent ${\theta_2}$ is always ${\theta_2=1/3}$ in ${H(2)}$ (independently of the ${SL(2,\mathbb{R})}$-invariant probability considered).

Remark 3 Recall that, in general, the KZ cocycle is symplectic its Lyapunov spectrum has the form ${\theta_1\geq\theta_2\geq\dots\geq\theta_g\geq-\theta_g\geq\dots\geq-\theta_1}$ (where ${g}$ is the genus), and, by general geometric reasons, ${\theta_1=1}$ and ${\theta_1>\theta_2}$, i.e., the top exponent ${\theta_1=1}$ is always simple.

By this general remark above, the results of Bainbridge and EKZ imply that the Lyapunov spectrum of KZ cocycle is always simple in ${H(2)}$. Of course, they say much more as the explicit value of ${\theta_2}$ is given, while the simplicity in ${H(2)}$ amounts only to say that ${\theta_2>0(>-\theta_2)}$. But, as it turns out, the arguments employed by Bainbridge and EKZ are sophisticated (involving the Deligne-Mumford compactification of the moduli space of curves, etc.) and long (both of them has ${\geq100}$ pages). So, partly motivated by this, we will discuss in this section the application of the simplicity criterion of SPCS 6 — 9 to the case of origamis in ${H(2)}$: evidently, this gives only the fact that ${\theta_2>0}$ but it has the advantage of relying on the elementary methods developed in previous posts.

We begin by selecting the following ${L}$-shaped origami ${L(m,n)}$:

In terms of the parameters ${m,n\geq 2}$, the total number of squares of ${L(m,n)}$ is ${N=n+m-1\geq 3}$. Notice that the horizontal permutation ${r}$ associated ${L(m,n)}$ is a ${m}$-cycle while the vertical permutation ${u}$ is a ${n}$-cycle. Thus, by our discussion so far above, one has that:

• if ${m+n}$ is odd, the monodromy group is ${S_N}$,
• if ${m}$ and ${n}$ are odd, the monodromy group is ${A_N}$, and
• if ${m}$ and ${n}$ are even, the monodromy group is ${S_N}$.

As we saw in the first post of this series, given an origami ${p:M\rightarrow\mathbb{T}^2}$, we have a decomposition

$\displaystyle H_1(M,\mathbb{Q}) = H_1^{st}(M,\mathbb{Q})\oplus H_1^{(0)}(M,\mathbb{Q})$

such that the standardpart ${H_1^{st}(M,\mathbb{Q})}$ has dimension ${2}$ and associated Lyapunov exponents ${\theta_1=1}$ and ${-\theta_1=-1}$, while the part ${H_1^{(0)}(M,\mathbb{Q})}$ consisting of homology classes projecting to ${0}$ under ${p}$ has dimension ${2g-2}$ and associated Lyapunov exponents ${\theta_2,\dots,\theta_g}$ and their opposites.

In the particular case of ${L(m,n)}$, in terms of the homology cycles ${\sigma_1, \sigma_2, \zeta_1, \zeta_2}$ showed in the previous picture, we can construct the following basis of ${H_1^{(0)}(M,\mathbb{Q})}$:

$\displaystyle \sigma:=\sigma_1-m\sigma_2, \quad\zeta:=\zeta_1-n\zeta_2.$

Now, we choose the elements ${S=\left(\begin{array}{cc}1 & m \\ 0 & 1\end{array}\right)}$ and ${T=\left(\begin{array}{cc}1 & 0 \\ n & 1\end{array}\right)}$ of the Veech group of ${L(m,n)}$. Since ${L(m,n)\in H(2)}$ has no automorphisms (see Lemma 1), we can also think of ${S}$ and ${T}$ as elements of the affine group of ${L(m,n)}$.

As the reader can check in the figures, the action of ${S}$ on the homology cycles ${\sigma_1, \sigma_2, \zeta_1, \zeta_2}$ is

• ${S(\sigma_i)=\sigma_i}$ for ${i=1, 2}$;
• ${S(\zeta_1)=\zeta_1+\sigma_1+(n-1)m\sigma_2}$
• ${S(\zeta_2)=\zeta_2+\sigma_1}$

Therefore, ${S(\sigma)=\sigma}$ and ${S(\zeta)=\zeta+(n-1)\sigma}$. Actually, the same computation with ${S}$ replaced by any power ${S^a}$, ${a\in\mathbb{N}}$, gives the “same” result

$\displaystyle S^a(\sigma)=\sigma, \quad S^a(\zeta)=\zeta+a(1-n)\sigma$

By the natural symmetry between the horizontal and vertical directions in ${L(m,n)}$, there is no need to compute twice to get the corresponding formulas for ${T}$ and/or ${T^b}$:

$\displaystyle T^b(\sigma)=\sigma+b(1-m)\zeta, \quad T^b(\zeta)=\zeta$

By combining ${S}$ and ${T}$, we can find pinching elements acting on ${H_1^{(0)}(M,\mathbb{Q})}$: for instance,

$\displaystyle A:=ST=\left(\begin{array}{cc}1+(m-1)(n-1) & (1-n) \\ (1-m) & 1 \end{array}\right)$

has eigenvalues given by the roots of ${\lambda^2-t\lambda+1=0}$ where ${t=t_A=2+(m-1)(n-1)}$ is the trace of ${A}$; since ${|t_A|>2}$ (as ${m,n\geq 2}$) and ${t_A^2 -4 = (m-1)(n-1)[(m-1)(n-1)+4]}$ is not a square, ${A}$ has two real irrational eigenvalues, and hence ${A}$ is a pinchingmatrix whose eigenspaces are defined over ${\mathbb{Q}(\sqrt{t_A})}$.

Thus, we can apply the simplicity criterion as soon as one has a twisting matrix ${B}$ with respect to ${A}$. We take

$\displaystyle B:=S^2T^2$

By applying the previous formulas for the powers ${S^a}$ and ${T^b}$ of ${S}$ and ${T}$ (with ${a=b=2}$), we find that the trace of ${B}$ is

$\displaystyle t_B=2+4(m-1)(n-1)$

so that ${t_B^2-4 = 16(m-1)(n-1)[(m-1)(n-1)+1]}$ is also not a square. Hence, ${B}$ is also a pinching matrix whose eigenspaces are defined over ${\mathbb{Q}(\sqrt{t_B})}$. Furthermore, these formulas for ${t_A}$ and ${t_B}$ above show that the quadratic fields ${\mathbb{Q}(\sqrt{t_A})}$ and ${\mathbb{Q}(\sqrt{t_B})}$ are disjointin the sense that ${\mathbb{Q}(\sqrt{t_A})\cap\mathbb{Q}(\sqrt{t_B})=\mathbb{Q}}$. So, ${B}$ is twisting with respect to ${A}$, and therefore ${1=\theta_1>\theta_2>0}$.

Remark 4 For the “other” stratum of genus ${2}$, namely ${H(1,1)}$, the results of Bainbridge and EKZ show that ${\theta_2=1/2}$. In principle, the (weaker) fact that ${\theta_2>0}$ in this situation can be derived for particular origamis in ${H(1,1)}$ along the lines sketched above for ${H(2)}$, but it is hard to treat such origamis in a systematic way because currently there is only a conjectural classification of ${SL(2,\mathbb{Z})}$-orbits.

This closes our (preliminary) discussions of the application of the simplicity criterion in the (well-established) case of ${H(2)}$. Now, we pass to the case of the stratum ${H(4)}$.

2. ${SL(2,\mathbb{Z})}$-orbits of origamis in ${H(4)}$

2.1. Connected components of the stratum ${H(4)}$

By the work of M. Kontsevich and A. Zorich, we know that ${H(4)}$ has two connected components, a hyperelliptic connected component ${H(4)^{hyp}}$, and a odd spin connected component ${H(4)^{odd}}$. In this particular case, they can be distinguished as follows:

• ${-\textrm{id}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1 \end{array}\right)}$ belongs to the Veech group of all translation surfaces in the connected component ${H(4)^{hyp}}$.
• ${-\textrm{id}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1 \end{array}\right)}$ doesn’t belong to the Veech group of the generic translation surface of the connected component ${H(4)^{odd}}$.

2.2. Weierstrass points and ${SL(2,\mathbb{Z})}$-orbits of “symmetric” origamis: an invariant of E. Kani, and P. Hubert and S. Lelievre

In general, we have the exact sequence

$\displaystyle 1\rightarrow Aut(M)\rightarrow Aff(M)\rightarrow SL(M)\rightarrow 1$

If the origami ${M}$ belongs to a mininal stratum ${H(2g-2)}$ (e.g., ${H(4)}$), by Lemma 1, one has ${Aut(M)=1}$ and therefore ${Aff(M)\simeq SL(M)}$. In other words, given ${g\in SL(M)}$, there exists an uniqueaffine diffeomorphism of ${M}$ with derivative ${g}$.

Suppose now that ${p:M\rightarrow\mathbb{T}^2}$ is a reduced origami such that ${-\textrm{id}=\left(\begin{array}{cc}-1 & 0 \\ 0 & -1 \end{array}\right)}$ belongs to the Veech group ${SL(M)}$ of ${M}$. Thus, if ${M\in H(2g-2)}$, there exists an unique affine diffeomorphism ${\phi}$ of ${M}$ with derivative ${-\textrm{id}}$. Of course, ${\phi}$ is an involution and it corresponds to a lift under ${p}$ of the involution ${x\mapsto -x}$ of ${\mathbb{T}^2}$. It follows that the fixed points of ${\phi}$ project to the points of ${\mathbb{T}^2}$ of order ${1}$ and ${2}$. (Actually, it is also possible to show that the fixed points of ${\phi}$ are Weierstrass points of ${M}$, but we will not use this here). In the figure below we indicated the 4 points of ${\mathbb{T}^2}$ of orders ${1}$ and ${2}$, and we denoted by

• ${\ell_0}$ the number of fixed points of ${\phi}$ over the integer point ${0=(0,0)\in\mathbb{T}^2}$,
• ${\ell_1}$ the number of fixed points of ${\phi}$ over the half-integer point ${(0,1/2)\in\mathbb{T}^2}$,
• ${\ell_2}$ the number of fixed points of ${\phi}$ over the half-integer point ${(1/2,1/2)\in\mathbb{T}^2}$,
• ${\ell_3}$ the number of fixed points of ${\phi}$ over the half-integer point ${(1/2,0)\in\mathbb{T}^2}$

Furthermore, the action of ${SL(2,\mathbb{Z})}$ conjugates the involutions of the origamis in the ${SL(2,\mathbb{Z})}$-orbit of ${M}$ (by an element of ${SL(2,\mathbb{Z})}$). Since ${SL(2,\mathbb{Z})}$ fixes the integer point ${0}$ while it permutes (arbitrarily) the 3 points of order ${2}$, we have that the list

$\displaystyle [\ell_0, \{\ell_1,\ell_2,\ell_3\}]$

is an invariant of the ${SL(2,\mathbb{Z})}$-orbit of ${M}$ if one considers the sublist ${\{\ell_1,\ell_2,\ell_3\}}$ up to permutations.

Remark 5 The sole zero of ${M\in H(2g-2)}$ is fixed under the involution ${\phi}$, so that ${\ell_0\geq 1}$.

Remark 6 For ${i>0}$, ${\ell_i\equiv N (\textrm{mod }2)}$: for instance, the involution permute the squares forming ${M}$ and, e.g. ${\ell_2}$ is the number of squares fixed by ${\phi}$; since ${\phi}$ is an involution, the ${\phi}$-orbits of squares have size ${1}$ or ${2}$, so that ${\ell_2\equiv N (\textrm{mod } 2)}$; finally, since ${SL(2,\mathbb{Z})}$ acts transitively on the set of half-integer points of ${\mathbb{T}^2}$, we can “replace” ${\ell_2}$ by ${\ell_i}$, ${i>0}$, to get the same conclusion for all ${i>0}$.

The invariant ${[\ell_0, \{\ell_1,\ell_2,\ell_3\}]}$ was introduced by E. Kani, and P. Hubert and S. Lelievre to study ${SL(2,\mathbb{Z})}$-orbits of origamis of ${H(2)}$ (historically speaking this invariant came before the monodromy invariant used in the previous section of this post). Since any genus ${2}$ Riemann surface is hyperelliptic, ${-\textrm{id}}$ belongs to the Veech group of any translation surface ${M\in H(2)}$, and the quotient of ${M}$ by involution ${\phi}$ has genus ${0}$. In particular, by Riemann-Hurwitz formula, the involution ${\phi}$ has ${6}$ fixed points (the ${6}$ Weierstrass points of ${M}$). For the origami ${L(m,n)}$ with ${N=n+m-1}$ squares shown in the second figure above, one can compute the invariant ${[\ell_0, \{\ell_1,\ell_2,\ell_3\}]}$:

$\displaystyle \ell_2=\left\{\begin{array}{cc}1, & \textrm{if } n,m \textrm{ are odd} \\ 2, & \textrm{if } n+m \textrm{ is odd} \\ 3, & \textrm{if } n,m \textrm{ are even}\end{array}\right.$

$\displaystyle \ell_3 (\textrm{resp. } \ell_1)=\left\{\begin{array}{cc}0 (\textrm{resp. } 2), & \textrm{if } m \textrm{ is odd and } n \textrm{ is even} \\ 1, & \textrm{if } n+m \textrm{ is even} \\ 2 (\textrm{resp. } 0), & \textrm{if } m \textrm{ is even and } n \textrm{ is odd}\end{array}\right.$

In resume, we get that

• if ${N=n+m-1}$ is even, there is exactly one ${SL(2,\mathbb{Z})}$-orbit of origamis of ${H(2)}$ with invariant ${[2,\{2,2,0\}]}$.
• if ${N=n+m-1}$ is odd, there are exactly two ${SL(2,\mathbb{Z})}$-orbit of origamis of ${H(2)}$: \subitem – if ${m,n}$ are odd, the monodromy group is ${A_N}$ and the invariant is ${[3,\{1,1,1\}]}$; \subitem – if ${m,n}$ are even, the monodromy group is ${S_N}$ and the invariant is ${[1,\{3,1,1\}]}$.

Coming back to ${H(4)}$, after some numerical experiments of V. Delecroix and S. Lelievre (using SAGE), currently we dispose of a conjectural classification of ${SL(2,\mathbb{Z})}$ orbits of origamis in ${H(4)}$.

3. Conjecture of Delecroix and Lelievre in ${H(4)}$

• For the hyperelliptic component ${H(4)^{hyp}}$, the quotient of the involution ${\phi}$ of any translation surface ${M\in H(4)^{hyp}}$ has genus ${0}$ and (by Riemann-Hurwitz formula) ${8}$ fixed points. The (experimentally observed) Kani-Hubert-Lelievre invariants for the ${SL(2,\mathbb{Z})}$-orbits of reduced origamis in ${H(4)}$ with ${N}$ squares are:
• For ${N\geq 9}$ odd, ${[5,\{1,1,1\}]}$, ${[3,\{3,1,1\}]}$, ${[1,\{5,1,1\}]}$ and ${[1,\{3,3,1\}]}$;
• For ${N\geq 10}$ even, ${[4, \{2,2,0\}]}$, ${[2,\{4,2,0\}]}$, ${[2,\{2,2,2\}]}$
• For the “odd spin” connected component ${H(4)^{odd}}$ and ${N\geq 9}$:
• For the generic origami ${M\in H(4)^{odd}}$ (i.e., ${-\textrm{id}\notin SL(M)}$), there are exactly two ${SL(2,\mathbb{Z})}$-orbits distinguished by the monodromy group ${A_N}$ or ${S_N}$;
• For the so-called “Prym” case, i.e., ${M\in H(4)^{odd}}$ and ${-\textrm{id}\in SL(M)}$, the quotient of the involution ${\phi}$ has genus ${1}$, and hence, by Riemann-Hurwitz formula, ${\phi}$ has ${4}$ fixed points. In this situation, the ${SL(2,\mathbb{Z})}$-orbits were classified by a recent theorem of E. Lanneau and D.-Manh Nguyen whose statement we recall below.

Theorem 3 (E. Lanneau and D.-Mahn Nguyen) In the Prym case in ${H(4)^{odd}}$:

• if ${N}$ is odd, there exists precisely one ${SL(2,\mathbb{Z})}$-orbit whose associated Kani-Hubert-Lelievre invariant is ${[1,\{1,1,1\}]}$;
• if ${N\equiv 0 (\textrm{mod }4)}$, there exists precisely one ${SL(2,\mathbb{Z})}$-orbit whose associated Kani-Hubert-Lelievre invariant is ${[2,\{2,0,0\}]}$;
• if ${N\equiv 2 (\textrm{mod }4)}$, there are precisely two ${SL(2,\mathbb{Z})}$-orbits whose associated Kani-Hubert-Lelievre invariant are ${[2, \{2,0,0\}]}$ and ${[4,\{0,0,0\}]}$.

4. Simplicity of Lyapunov spectrum of origamis in ${H(4)}$

Concerning Lyapunov exponents of KZ cocycle, it is possible to prove (see, e.g., this previous post here) that the Prym case of ${H(4)^{odd}}$ always lead to Lyapunov exponents ${\theta_2=2/5}$ and ${\theta_3=1/5}$. In particular, the Lyapunov spectrum of KZ cocycle in the Prym case is simple. As a side remark, we would like to observe that, interestingly enough, the simplicity criterion developed here can’t be used in the Prym case: indeed, the presence of the involution permits to decompose or, more precisely, diagonalize by blocks, the KZ cocycle so that the twisting property is never satisfied! Also, some numerical experiments by A. Zorich and V. Delecroix indicate that the second Lyapunov exponent ${\theta_2}$ of generic origamis (as well as generic translation surfaces) of the stratum ${H(4)^{odd}}$ is ${\theta_2=0.41...>2/5}$ where the second decimal here seems “reliable” (so that the second Lyapunov exponent of the generic translation surface of ${H(4)^{odd}}$ is strictly bigger than the Prym case).

In any event, in the forthcoming article by C.M., M. Möller and J.-C. Yoccoz we will apply the simplicity criterion to all (non-Prym) ${SL(2,\mathbb{Z})}$-orbits in ${H(4)}$ described in Delecroix-Lelievre conjecture. The outcome is that for such ${SL(2,\mathbb{Z})}$-orbits, the Lyapunov spectrum of the KZ cocycle is simple except (maybe) for finitely many ${SL(2,\mathbb{Z})}$-orbits (corresponding to “small” values of ${N}$, the number of squares).

In other words, if the conjecture of Delecroix-Lelievre is true, the simplicity criterion from previous posts in this series proves that all but (maybe) finitely many ${SL(2,\mathbb{Z})}$-orbits of origamis of ${H(4)}$ have simple Lyapunov spectrum.

Unfortunately, since there are several ${SL(2,\mathbb{R})}$-orbits in Delecroix-Lelievre conjecture above, we will not be able to discuss all of them today. Moreover, as we’re going to see in a moment, the proof of the simplicity statement above involves long computer-assisted calculations, so that our plan for the rest of this post will be the following: we will select a family of ${SL(2,\mathbb{Z})}$-orbits and we will start some hand-made computations towards the simplicity statement; then, we will see why we run in “trouble” with the hand-made calculation (essentially it is “too naive”) and why we need some computer assistance to complete the argument; however, we will not present the computer assisted computations as we prefer to postpone them to the forthcoming article, and we will end this series with this “incomplete” argument as it already contains the main ideas and it is not as technical as the complete argument.

In the case of an origami ${M\in H(4)}$, we have that ${\textrm{dim}(H_1^{(0)}(M,\mathbb{R}))=4}$, so that the KZ cocycle takes its values in ${Sp(4,\mathbb{Z})}$. Given ${A, B\in Sp(4,\mathbb{Z})}$, denote by

$\displaystyle P_A(x)=x^4+ax^3+bx^2+ax+1$

and

$\displaystyle P_B(x)=x^4+a'x^3+b'x^2+a'x+1$

their characteristic polynomials.

Recall that (see SPCS9) ${A}$ is pinching if the roots of ${P_A}$ are all real, and the Galois group is ${G}$ is the largest possible, i.e., ${G\simeq S_2\times(\mathbb{Z}/2\mathbb{Z})^2}$.

Following the usual trick to determine the roots of ${P_A}$, we pose ${y=x+1/x}$, so that we get an equation

$\displaystyle Q_A(y)=y^2+ay+(b-2)$

with ${a,b\in\mathbb{Z}}$. The roots of ${Q_A}$ are

$\displaystyle y_{\pm}=\frac{-a\pm\sqrt{\Delta_1}}{2}$

where ${\Delta_1:=a^2-4(b-2)}$. Then, we recover the roots of ${P_A}$ by solving the equation

$\displaystyle x^2-y_{\pm}x+1=0$

of discriminant ${\Delta_{\pm}:=y_{\pm}^2-4}$.

A simple calculation shows that

$\displaystyle \Delta_2:=\Delta_+\Delta_-=(b+2)^2-4a^2 = (b+2+2a)(b+2-2a)$

For later use, we denote by ${K_A}$, resp. ${K_B}$, the splitting field of ${P_A}$, resp. ${P_B}$.

We leave the following proposition as an exercise in Galois theory for the reader:

Proposition 4 It holds:

• (a) If ${\Delta_1}$ is a square (of an integer number), then ${Q_A}$ and ${P_A}$ are not irreducible.

In the sequel, we will always assume that ${\Delta_1}$ is not a square.

• (b) If ${\Delta_2}$ is a square, then ${[K_A:\mathbb{Q}]=4}$ and the Galois group ${G}$ is ${(\mathbb{Z}/2\mathbb{Z})\times(\mathbb{Z}/2\mathbb{Z})}$.
• (c) If ${\Delta_1\Delta_2}$ is a square, then ${[K_A:\mathbb{Q}]=4}$ and the Galois group is ${\mathbb{Z}/4\mathbb{Z}}$.
• (d) If (${\Delta_1}$,) ${\Delta_2}$ and ${\Delta_1\Delta_2}$ are not squares, then ${[K_A:\mathbb{Q}]=8}$, the Galois group is ${G\simeq D_4\simeq S_2\times (\mathbb{Z}/2\mathbb{Z})^2}$, ${K_A}$ contains exactly ${3}$ quadratic fields ${\mathbb{Q}(\sqrt{\Delta_1}), \mathbb{Q}(\sqrt{\Delta_2})}$ and ${\mathbb{Q}(\sqrt{\Delta_1\Delta_2})}$, and each intermediate field ${\mathbb{Q}\subset L\subset K_A}$ contains one of these quadratic fields.
• (e) If (${\Delta_1}$,) ${\Delta_2}$ and ${\Delta_1\Delta_2}$ are not squares, and ${\Delta_1\Delta_2>0}$, then the roots of ${P_A}$ are real and the Galois group of ${P_A}$ is the largest possible.

This proposition establishes a sufficient criterion for ${A, B\in Sp(4,\mathbb{Z})}$ to fit the pinching and twisting conditions. Indeed, we can use item (e) to get pinching matrices ${A}$, and we can apply item (d) to produce twisting matrices ${B}$ with respect to ${A}$ as follows: recall that (see SPCS9) the twisting condition is true if the splitting fields ${K_A}$ and ${K_B}$ are “disjoint” (i.e., ${K_A\cap K_B=\mathbb{Q}}$), and, by item (d), this disjointness can be checked by computing the quantities ${\Delta_1, \Delta_2, \Delta_1\Delta_2}$ associated to ${A}$, and ${\Delta_1', \Delta_2', \Delta_1'\Delta_2'}$ associated to ${B}$, and verifying that they generated distinct quadratic fields.

Now, we consider the following specific family ${M=M_n}$, ${n\geq 5}$, of origamis in ${H(4)^{odd}}$:

Here, the total number of squares is ${N=n+4}$ and ${-\textrm{id}}$ doesn’t belong to the Veech group (i.e., this origami is “generic” in ${H(4)^{odd}}$). The horizontal permutation is the product of a ${n}$-cycle with a ${2}$-cycle (so its parity equals the parity of ${n}$), and the vertical permutation is the product of a ${4}$-cycle with a ${2}$-cycle (so that its parity is even). In particular, here the monodromy group is ${S_N}$ if ${N}$ is odd, and ${A_N}$ if ${N}$ is even.

In terms of the homology cycles ${\sigma_1, \sigma_2, \sigma_3, \zeta_1, \zeta_2, \zeta_3}$ showed in the figure above, we can select the following basis of ${H_1^{(0)}(M,\mathbb{Q})}$:

• ${\Sigma_1=\sigma_1-n\sigma_3}$, ${\Sigma_2=\sigma_2-2\sigma_3}$,
• ${Z_1=\zeta_1-4\zeta_3}$, ${Z_2=\zeta_2-2\zeta_3}$.

Then, we take the following elements of the Veech group of ${M}$

$\displaystyle S=\left(\begin{array}{cc}1 & 2n \\ 0 & 1\end{array}\right), \quad T=\left(\begin{array}{cc}1 & 0 \\ 4 & 1\end{array}\right)$

Notice that when ${n}$ is even we can replace ${2n}$ by ${n}$ in the definition of ${S}$, but we prefer to consider ${S}$ directly (to avoid performing to similar discussions depending on the parity of ${n}$).

By direct inspection of the figure above we deduce that:

• ${S(\sigma_i)=\sigma_i}$, for ${i=1, 2, 3}$,
• ${S(\zeta_1)=\zeta_1+2\sigma_1+n\sigma_2+4n\sigma_3}$,
• ${S(\zeta_2)=\zeta_1+2\sigma_1+n\sigma_2}$,
• ${S(\zeta_3)=\zeta_1+2\sigma_1}$.

Hence,

• ${S(\Sigma_i)=\Sigma_i}$, for ${i=1, 2}$,
• ${S(Z_1)=Z_1-6\Sigma_1+n\Sigma_2}$,
• ${S(Z_2)=\zeta_1-2\Sigma_1+n\sigma_2}$,

Thus, in terms of the basis ${\{\Sigma_1,\Sigma_2, Z_1, Z_2\}}$ of ${H_1^{(0)}(M,\mathbb{Q})}$, the matrix of ${S|_{H_1^{(0)}}}$ has the following decomposition into ${2\times 2}$ blocks:

$\displaystyle S|_{H_1^{(0)}}=\left(\begin{array}{cc}1 & \overline{S} \\ 0 & 1\end{array}\right), \quad \textrm{where } \overline{S}=\left(\begin{array}{cc}-6 & n \\ -2 & n\end{array}\right)$

By symmetry, the matrix of ${T|_{H_1^{(0)}}}$ is:

$\displaystyle T|_{H_1^{(0)}}=\left(\begin{array}{cc}1 & 0 \\ \overline{T} & 1\end{array}\right), \quad \textrm{where } \overline{T}=\left(\begin{array}{cc}1-n & -1 \\ 2 & 2\end{array}\right)$

In particular,

$\displaystyle (ST)|_{H_1^{(0)}}=\left(\begin{array}{cc}1+\overline{S}\overline{T} & \overline{S} \\ \overline{T} & 1\end{array}\right)$

Of course one can extract the characteristic polynomial ${x^4+ax^3+bx^2+ax+1}$ of ${(ST)|_{H_1^{(0)}}}$ directly from these formulas, but in our particular we can take the following “shortcut”: observe that ${ST(v_1,v_2)=\lambda(v_1,v_2)}$ (where ${v_1,v_2\in\mathbb{R}^2}$) if and only if ${\overline{S}\overline{T}(v_1) = (\lambda+\lambda^{-1}-2)v_1=(y_{\pm}-2)v_1}$. By computing with ${\overline{S}}$ and ${\overline{T}}$, and using that ${y_{\pm}}$ are the solutions of ${y^2-ay+(b-2)=0}$, one finds ${a=-7n+6}$, ${b=8n^2-2n-14}$ and

$\displaystyle \Delta_1=17n^2-76n+100$

By Proposition 4, we wish to know e.g. how frequently ${\Delta_1}$ is a square. Evidently, one can maybe produce some ad hoc method here, but since we want to verify simplicity conditions in a systematic way, it would be nice to this type of question in “general”.

Here, the idea is very simple: the fact that ${\Delta_1=17n^2-76n+100}$ is a square is equivalent to get integer and/or rational solutions to

$\displaystyle z^2=17t^2-76t+100,$

that is, we need to understand integer/rational points in this curve.

This hints the following solution to our problem: if we can replace ${A=ST}$ by more complicated products of (powers of) ${S}$ and ${T}$ chosen more or less by random, it happens that ${\Delta_1}$ becomes a polynomial ${P(n)}$ of degree ${\geq 5}$ without square factors. In this situation, the problem of knowing whether ${\Delta_1=P(n)}$ is a square of an integer number becomes the problem of finding integer/rational points in the curve

$\displaystyle z^2=P(n).$

Since this a non-singular curve (as ${P(n)}$ has no square factors) of genus ${g\geq 2}$ (as ${\textrm{deg}(P)\geq 5}$), we know by Faltings’ theorem (previously known as Mordell’s conjecture) that the number of rational solutions is finite.

Remark 7 In the case we get polynomials ${\widetilde{P}(n)}$of degree ${3}$ or ${4}$ after removing square factors of ${P(n)}$, we still can apply (and do [apply in the forthcoming article]) Siegel’s theorem saying that the genus ${1}$ curve ${z^2=\widetilde{P}(n)}$ has finite many integer points, but we preferred to mention the case of higher degree polynomials because it is the “generic situation” of the argument (in some sense).

In other words, by the end of the day, we have that ${\Delta_1=P(n)}$ is not the square of an integer number for all but finitely many values of ${n}$.

Of course, at this point, the general idea to get the pinching and twisting conditions (and hence simplicity) for the origamis ${M=M_n}$ for all but finitely many ${n\in\mathbb{N}}$ is clear: one produces “complicated” products of ${S}$, ${T}$ (and also a third auxiliary parabolic matrix ${U}$) leading to elements ${A, B\in Sp(4,\mathbb{Z})}$ such that the quantities ${\Delta_i=\Delta_i(n), i=1,2}$, ${\Delta_3=\Delta_1\Delta_2(n)}$ associated to ${A}$, ${\Delta_i'=\Delta_i'(n), \Delta_3'=\Delta_1'\Delta_2'(n)}$ associated to ${B}$ (and also the “mixed products” ${\Delta_i\Delta_j'}$ for ${i,j=1,2,3}$) are polynomials of the variable ${n}$ of high degree (and without square factors if possible), so that these quantities don’t take square of integers as their values for all but finitely many ${n\in\mathbb{N}}$.

Unfortunately, it seems that there is no “systematic way” of choosing these “complicated” products of ${S}$, ${T}$, etc. and, as a matter of fact, the calculations are too long and that’s why we use computer-assisted calculations at this stage. Again, as mentioned by the beginning of this section, these computer-assisted computations are not particularly inspiring (i.e., there are no mathematical ideas behind them), so we will close this post by giving an idea of what kind of complicated coefficient ${b=b(n)}$ of the characteristic polynomial of ${A}$ we get in our treatment of the family ${M=M_n}$:

$\displaystyle b(n)=3840 n^5 - 17376 n^4 + 14736 n^3 + 25384n^2 -28512n-7066$