Posted by: matheuscmss | November 1, 2011

## Unramified double covers of origamis in H(2) are Lagrangian

In a recent article (quickly discussed in a previous post of this blog), Giovanni Forni gave a criterion for the non-uniform hyperbolicity of the Kontsevich-Zorich cocycle over the support of certain ${SL(2,\mathbb{R})}$-invariant measures, and, after a kind invitation by Giovanni, I ended up contributing with an Appendix to Giovanni’s paper presenting some examples related to his main theorems.

As it turns out, while I was preparing the Appendix, some “scenes” were deleted from the final version of the article due to the usual limitations of space. Indeed, the following two remarks were omitted:

• (a) all ${SL(2,\mathbb{R})}$-invariant measures supported on the stratum ${\mathcal{H}(1,1,1,1)}$ coming from (unramified) double covers of Veech surfaces of ${\mathcal{H}(1,1)}$ have non-simple Lyapunov spectrum;
• (b) all ${SL(2,\mathbb{R})}$-invariant measures supported on the stratum ${\mathcal{H}(2,2)}$ coming from (unramified) double covers of square-tiled surfaces of ${\mathcal{H}(2)}$ are Lagrangian (in the sense of Giovanni’s article).

Of course, since the proofs of these statements are (very) straightforward, they are not publication-quality, except (maybe) in this blog. So, I will use today’s post to present the (easy) proofs of (a) and (b).

Proof of (a)

Let ${(S,\omega_S)\in\mathcal{H}(1,1)}$ be a Veech surface (of genus 2), and consider ${(R,\omega_R)\in\mathcal{H}(1,1,1,1)}$ be a (unramified) double cover of ${(S,\omega_S)}$, i.e., we have ${p:R\rightarrow S}$ a (unramified) double cover and ${\omega_R=p^*(\omega_S)}$. In this situation, by Lemma A.1 in Giovanni’s article, we have that the Lyapunov spectrum of (the Kontsevich-Zorich cocycle over the ${SL(2,\mathbb{R})}$-orbit of) ${(S,\omega_S)}$ is contained in the Lyapunov spectrum of ${(R,\omega_R)}$. On the other hand, by a result of Matthew Bainbridge, we have that the (two) Lyapunov exponents of the (genus 2) Abelian differential ${(S,\omega_S)}$ is ${\{1,1/2\}}$. It follows that the Lyapunov spectrum of the genus ${3}$ Abelian differential ${(R,\omega_R)}$ has the form ${\{1,1/2,\lambda\}}$ for some ${0\leq\lambda<1}$.

Next, we observe that ${(R,\omega_R)}$ lives in the hyperelliptic locus of ${\mathcal{H}(1,1,1,1)}$: for instance, it is not hard to show that the sheet exchange map (induced by the unramified double cover ${p:R\rightarrow S}$) and a lift of the hyperelliptic involution of the genus ${2}$ (and hence hyperelliptic) surface ${S}$ can be lifted to ${R}$ generates a copy of Klein four group ${\mathbb{Z}/2\times\mathbb{Z}/2}$ inside the automorphism group of ${R}$ (and this is sufficient to conclude hyperellipticity of ${R}$). In any case, see page 268 of H. Farkas and I. Kra’s book for two detailed proofs of the hyperellipticity of ${(R,\omega_R)}$.

Once we know that (the ${SL(2,\mathbb{R})}$-orbit of) ${(R,\omega_R)}$ lies in the hyperelliptic locus, we can apply Theorem 1.1 of this recent paper of Dawei Chen and Martin Möller to obtain that the sum of Lyapunov exponents of ${(R,\omega_R)}$ is ${2}$, i.e., ${1+1/2+\lambda=2}$, that is, ${\lambda=1/2}$. Therefore, the Lyapunov spectrum of ${(R,\omega_R)}$ is ${\{1,1/2,1/2\}}$, i.e., ${1/2}$ is a double Lyapunov exponent. This proves (a).

Proof of (b)

Recall that an unramified covering of a Riemann surface with group of deck transformations ${G}$ corresponds to a surjective morphism from the fundamental group of the Riemann surface to ${G}$. In particular, the number of unramified double covers of a genus ${g}$ Riemann surface is the number of epimorphisms from its fundamental group to ${\mathbb{Z}_2}$ (and, as ${\mathbb{Z}_2}$ is Abelian, this is the same as the number of epimorphisms from its first homology group to ${\mathbb{Z}_2}$), that is, ${2^{2g}-1}$ (and, moreover, it is possible to write algebraic equations for each of them, see this article of Y. Fuertes and G. González-Diez). In fact, if we denote by ${\alpha_i,\beta_i}$, ${i=1,\dots,2g}$, a “canonical” basis of the fundamental group of our Riemann surface, an epimorphism from the fundamental group to ${\mathbb{Z}_2=\{-1,+1\}}$ is determined by an assignment ${\alpha_i\mapsto (\pm1)_{\alpha_i}}$, ${\beta_i\mapsto (\pm1)_{\beta_i}}$ such that ${-1}$ is assigned to some ${\alpha_i}$ or ${\beta_i}$. Geometrically, such an epimorphism corresponds to taking two copies of our Riemann surface and declaring that the lift of a curve ${\alpha_i}$ or ${\beta_i}$ corresponds to two disjoint copies (one in each copy of our Riemann surface) of the curve if we’ve assigned ${+1}$ to it or one closed curve (with “twice the length of the original curve”) if we’ve assigned ${-1}$ to it. In this language, it is easy to see why ${2^{2g}-1}$ is the number of unramified double coverings: we can assign ${\pm1}$ to ${\alpha_i}$, ${\beta_i}$, ${i=1,\dots,2g}$ in ${2^{2g}}$ ways, but we have to exclude the trivial assignment (${\alpha_i,\beta_i\mapsto+1}$ for all ${i}$) because it corresponds to two disjoint copies of our initial Riemann surface (and hence it gives a non-connected cover [a case that we want to rule out]).

In the setting of item (b), we are looking at unramified double covers of a genus 2 Riemann surface ${S}$ (since ${(S,\omega_S)\in\mathcal{H}(2)}$), that is, we have to deal with ${15}$ coverings of ${(S,\omega_S)}$.

Remark 1 Just to “double-check” the counting of unramified double coverings of ${(S,\omega_S)\in\mathcal{H}(2)}$, we recall that all such coverings are hyperelliptic genus ${3}$ surfaces (see the proof of item (a)). On the other hand, by a general result of H. Farkas, among the ${2^{2g}-1}$ unramified double coverings of a genus ${g}$ Riemann surfaces, ${\binom{2g+2}{2}}$ of them are hyperelliptic. In the case ${(S,\omega_S)\in\mathcal{H}(2)}$, ${g=2}$ and ${\binom{2g+2}{2}=15}$, so that the global scenario is coherent.

In the sequel, these ${15}$ coverings will be rendered explicit.

Let ${(S,\omega_S)\in\mathcal{H}(2)}$ be a square-tiled surface. By a celebrated classification result of P. Hubert and S. Lelièvre (when ${(S,\omega_S)}$ is tiled by a prime number of squares) and C. McMullen (in general), it follows that the ${SL(2,\mathbb{Z})}$-orbit of any ${(S,\omega_S)}$ contains a ${L}$shaped representative:

In particular, given that item (b) concerns entire ${SL(2,\mathbb{R})}$-orbits, the classification result above says that we can assume (without loss of generality) that ${(S,\omega_S)}$ is ${L}$-shaped. In this case, the ${15}$ unramified double covers of ${(S,\omega_S)}$ are very easy to describe. In the ${L}$-shaped origami ${(S,\omega_S)}$ in Figure 1 above, we have 4 pairs of sides identified by translations: as a representative of each pair, we have two bottom horizontal sides (${\sigma_I}$ and ${\sigma_{II}}$) connecting the double zero of ${\omega_S}$, and two leftmost vertical sides (${\zeta_I}$ and ${\zeta_{II}}$) also connecting the double zero of ${\omega_S}$. To each such pair, we assign ${\pm 1}$ (i.e., we consider an epimorphism to ${\mathbb{Z}_2}$). Now, we take two copies of ${(S,\omega_S)}$ and we glue the sides in a given pair with its “patner” in the same copy of ${(S,\omega_S)}$ if we’ve assigned ${+1}$ to the pair, and with its “patner” in the other copy of ${(S,\omega_S)}$ if we’ve assigned ${-1}$ to the pair. To aleviate the notation, given an unramified double cover ${(R,\omega_R)}$ of ${(S,\omega_S)}$ presented as two copies of ${(S,\omega_S)}$ with certain identifications of sides (determined by an surjective map ${\{\sigma_I,\sigma_II,\zeta_I,\zeta_{II}\}\rightarrow\mathbb{Z}_2=\{-1,+1\}}$), we denote by ${\sigma_I}$ and ${\sigma_{II}}$ the two bottom horizontal sides in the first copy of ${(S,\omega_S)}$, and by ${\sigma_{III}}$ and ${\sigma_{IV}}$ the two bottom horizontal sides in the second copy of ${(S,\omega_S)}$. Similarly, we introduce ${\zeta_I}$, ${\zeta_{II}}$, ${\zeta_{III}}$ and ${\zeta_{IV}}$. For instance, the figure below shows the location of these sides in the unramified double cover ${(R,\omega_R)}$ of ${(S,\omega_S)}$ determined by the map ${\sigma_I\rightarrow-1}$, ${\sigma_{II}\rightarrow+1}$, ${\zeta_I\rightarrow+1}$, and ${\zeta_{II}\rightarrow+1}$.

At this stage, the proof of the statement of item (b) will be complete as soon as we show that all ${15}$ possible unramified double coverings of ${(S,\omega_S)}$ are Lagrangian in the sense of Giovanni Forni, that is, for each such covering we can find a (periodic, i.e., rational) direction such that the genus ${3}$ Riemann surface (upstairs) decomposes into cylinders whose waist curves span a ${3}$-dimensional subspace in homology. To simplify a little more our notation, we represent a map ${\{\sigma_{I},\sigma_{II},\zeta_{I},\zeta_{II}\}\rightarrow\{\pm1\}}$ by the list ${((\pm1)_{\sigma_{I}}, (\pm1)_{\sigma_{II}}, (\pm1)_{\zeta_{I}}, (\pm1)_{\zeta_{II}})}$ of images of ${\sigma_{I},\sigma_{II},\zeta_{I},\zeta_{II}}$ in this order.

Keeping this notation in mind, we begin by treating the cases where the horizontal direction of the unramified double covering is Lagrangian.

Remark 2 We strongly encourage the reader to draw the coverings to check the geometrical facts claimed in the items below.

• ${(-1,+1,+1,+1)}$: the waist curves of the 4 horizontal cylinders are homologous to ${\sigma_{I} + \sigma_{II}}$, ${\sigma_{III} + \sigma_{IV}}$, ${\sigma_I}$ and ${\sigma_{III}}$ (see Figure … above); by direct inspection, we see that ${\sigma_{I}=\sigma_{III}}$, and, moreover, the (non-trivial) homology classes ${\sigma_{II}}$ and ${\sigma_{IV}}$ are distinct (for instance, ${\sigma_{II}}$ has intersection with the closed vertical curve connecting the middle of the ${\sigma_{II}}$ sides in the first copy of ${(S,\omega_S)}$, while ${\sigma_{IV}}$ doesn’t intersect this curve); therefore, ${\sigma_I=\sigma_{III}}$, ${\sigma_I+\sigma_{II}}$ and ${\sigma_{III}+\sigma_{IV}}$ span a ${3}$-dimensional subspace in homology, that is, we have the Lagrangian property.
• ${(+1,-1,+1,+1)}$: the waist curves of the 4 horizontal cylinders are homologous to ${\sigma_{I} + \sigma_{II}}$, ${\sigma_{III} + \sigma_{IV}}$, ${\sigma_{I}}$ and ${\sigma_{III}}$, but this time ${\sigma_{II}=\sigma_{IV}}$ (as one can see that ${\sigma_{I}+\sigma_{II}=\sigma_{I}+\sigma_{IV}}$ from the first copy), and ${\sigma_{I}\neq\sigma_{III}}$ (as an intersection argument [similar to the previous item] with an adequate vertical path shows); thus, ${\sigma_{I} + \sigma_{II}}$, ${\sigma_{III} + \sigma_{IV}}$, ${\sigma_{I}}$ and ${\sigma_{III}}$ span a ${3}$-dimensional space in homology, and the Lagrangian property follows.
• ${(-1,-1,+1,+1)}$: the waist curves of the 4 horizontal cylinders are homologous to ${\sigma_{I} + \sigma_{II}}$, ${\sigma_{III} + \sigma_{IV}}$, ${\sigma_{I}}$ and ${\sigma_{III}}$, but this time ${\sigma_{I} + \sigma_{II}=\sigma_{III} + \sigma_{IV}}$ (as one can read from the first copy of ${(S,\omega_S)}$), and ${\sigma_{I} + \sigma_{II}}$, ${\sigma_{I}}$ and ${\sigma_{III}}$ are linearly independent in homology (as an intersection argument shows); hence, this covering is also Lagrangian.
• ${(+1,+1,-1,+1)}$, ${(-1,+1,-1,+1)}$, ${(+1,-1,-1,+1)}$ and ${(-1,-1,-1,+1)}$: the waist curves of the 3 horizontal cylinders are homologous to ${\sigma_{I}+\sigma_{II}+\sigma_{III}+\sigma_{IV}}$, ${\sigma_{I}}$ and ${\sigma_{III}}$, and they are linearly independent in homology by appropriate intersection arguments in each case; thus, these covering are Lagrangian as well.
• ${(+1,+1,+1,-1)}$, ${(-1,+1,+1,-1)}$, ${(+1,-1,+1,-1)}$ and ${(-1,-1,+1,-1)}$: the waist curves of the 3 horizontal cylinders are homologous to ${\sigma_{I}+\sigma_{II}}$, ${\sigma_{III}+\sigma_{IV}}$ and ${\sigma_{I}+\sigma_{III}}$, and they are linearly independent in homology by appropriate intersection arguments in each case; thus, these covering are Lagrangian.

Next, we analyze the following three cases:

• ${(+1,+1,-1,-1)}$, ${(+1,-1,-1,-1)}$ and ${(-1,+1,-1,-1)}$: by looking the vertical direction of these coverings, or equivalently, by rotating by ${\pi/2}$ the associated square-tiled surfaces and looking the horizontal direction of the resulting origamis, we come back to the cases ${(-1,-1,+1,+1)}$, ${(-1,-1,+1,-1)}$ and ${(-1,-1,-1,+1)}$ (resp.) treated above. So, the Lagrangian property for these coverings follows.

In resume, we showed that ${14}$ out of the ${15}$ unramified double coverings of ${(S,\omega_S)}$ are Lagrangian. However, the remaining case ${(-1,-1,-1,-1)}$ has only 2 horizontal cylinders (so that the span in homology of its waist curves has dimension ${\leq 2}$) and the situation doesn’t change after rotation by ${\pi/2}$, that is, we still have a covering of type ${(-1,-1,-1,-1)}$ and hence ${2}$ cylinders only (in some sense this last case is “self-dual”).

To overcome this technical difficulty, the idea is to apply the parabolic element ${T=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)}$ an appropriate number of times (i.e., apply ${T^i}$ for a convenient ${i}$) to the covering ${(-1,-1,-1,-1)}$ in order to change its “type”: since ${T^i}$ “twists” horizontal cylinders (it is a Dehn twist), one can hope to get back to one of the previous cases already treated after a certain amount of “twists”.

More precisely, we begin by noticing that the classification result of P. Hubert, S. Lelièvre and C. McMullen mentioned above implies that:

• any square-tiled surface ${(S,\omega_S)\in\mathcal{H}(2)}$ tiled by an even number of squares belongs to the ${SL(2,\mathbb{Z})}$-orbit of the ${L}$-shaped origami in Figure 1 above with parameter ${a=2}$;
• any square-tiled surface ${(S,\omega_S)\in\mathcal{H}(2)}$ tiled by an odd number of squares belongs to the ${SL(2,\mathbb{Z})}$-orbit of the ${L}$-shaped origami in Figure 1 above with parameter ${a=2}$ or ${a=3}$;

In particular, we can assume that ${(R,\omega_R)}$ is an unramified double covering of type ${(-1,-1,-1,-1)}$ of a ${L}$-shaped origami ${(S,\omega_S)}$ as above with parameter ${a=2}$ or ${3}$. In these cases, we can number the squares of ${(R,\omega_R)}$ as follows:

In other words, by using a pair of permutations to code ${(R,\omega_R)}$ (one permutation giving the neighbor to the right of each square and another permutation giving the neighbor on the top of each square), one gets

$\displaystyle h_{(2)}=(1,2,\dots,2n)(2n+1,2n+2),$

$\displaystyle v_{(2)}=(1,2n+1,n+1,2n+2)(2,n+2)(3,n+3)\dots(n,2n)$

in the case of parameter ${a=2}$, and

$\displaystyle h_{(3)}=(1,2,\dots,2n)(2n+1,2n+2)(2n+3,2n+4),$

$\displaystyle v_{(3)}=(1,2n+1,2n+3)(n+1,2n+2,2n+4)(2,n+2)\dots (n,2n)$

in the case of parameter ${a=3}$. In terms of a pair of permutations ${(h,v)}$, the action of the parabolic element ${T=\left(\begin{array}{cc}1&1 \\ 0&1\end{array}\right)}$ is ${T(h,v)=(h,vh^{-1})}$.

Therefore, in the case ${a=2}$, we get ${T(h_{(2)},v_{(2)}) = (h_{(2)}),v_{(2)}h_{(2)}^{-1})}$, where

$\displaystyle v_{(2)}h_{(2)}^{-1}(x)=\left\{\begin{array}{rl} n+x-1, & \textrm{if } x\notin\{2,n+2,2n+1,2n+2\} \\ 2n+1, & \textrm{if } x=2 \\ 2n+2, & \textrm{if } x=n+2 \\ 1, & \textrm{if } x=2n+1 \\ n+1, & \textrm{if } x=2n+2 \end{array}\right.,$

${T^2(h_{(2)},v_{(2)}) = (h_{(2)}, v_{(2)}h_{(2)}^{-2})}$, where

$\displaystyle v_{(2)}h_{(2)}^{-2}(x)=\left\{\begin{array}{rl} n+x-2, & \textrm{if } x\notin\{3,n+3,2n+1,2n+2\} \\ 2n+1, & \textrm{if } x=3 \\ 2n+2, & \textrm{if } x=n+3 \\ 1, & \textrm{if } x=2n+2 \\ n+1, & \textrm{if } x=2n+1 \end{array}\right.,$

and, by induction, ${T^n(h_{(2)},v_{(2)})=(h_{(2)},v_{(2)}h_{(2)}^{-n})}$, where

$\displaystyle v_{(2)}h_{(2)}^{-n}=\left\{\begin{array}{rl} n+x-n = x, & \textrm{if } x\notin\{n+1,1,2n+1,2n+2\} \\ 2n+1, & \textrm{if } x=n+1 \\ 2n+2, & \textrm{if } x=1 \\ 1, & \textrm{if } x=2n+ \left\{\begin{array}{rl}1 & \textrm{if } n \textrm{ is odd} \\ 2 & \textrm{if } n \textrm{ is even}\end{array}\right. \\ n+1, & \textrm{if } x=2n+\left\{\begin{array}{rl}2 & \textrm{if } n \textrm{ is odd} \\ 1 & \textrm{if } n \textrm{ is even}\end{array}\right. \end{array}\right.$

Hence, one has ${T^n(h_{(2)},v_{(2)})}$ is an unramified double covering of ${(S,\omega_S)}$ of type ${((\pm1)_n, +1, -1,-1)}$ with

$\displaystyle (\pm1)_n=\left\{\begin{array}{rl} +1, & \textrm{if } n \textrm{ is odd} \\ -1, & \textrm{if } n \textrm{ is even }\end{array}\right.$

so that we come back to a previous considered case.

Similarly, in the case ${a=3}$, we get ${T(h_{(3)},v_{(3)}) = (h_{(3)},v_{(3)}h_{(3)}^{-1})}$, where

$\displaystyle v_{(3)}h_{(3)}^{-1}=\left\{\begin{array}{rl} n+x-1, & \textrm{if } x\notin\{2,n+2,2n+1,2n+2,2n+3,2n+4\} \\ 2n+1, & \textrm{if } x=2 \\ 2n+2, & \textrm{if } x=n+2 \\ 2n+4, & \textrm{if } x=2n+1 \\ 2n+3, & \textrm{if } x=2n+2 \\ 1, & \textrm{if } x=2n+3 \\ n+1, & \textrm{if } x=2n+4 \end{array}\right.,$

${T^2(h_{(3)},v_{(3)}) = (h_{(3)},v_{(3)}h_{(3)}^{-2})}$, where

$\displaystyle v_{(3)}h_{(3)}^{-2}=\left\{\begin{array}{rl} n+x-2, & \textrm{if } x\notin\{3,n+3,2n+1,2n+2,2n+3,2n+4\} \\ 2n+1, & \textrm{if } x=3 \\ 2n+2, & \textrm{if } x=n+3 \\ 2n+4, & \textrm{if } x=2n+2 \\ 2n+3, & \textrm{if } x=2n+1 \\ 1, & \textrm{if } x=2n+4 \\ n+1, & \textrm{if } x=2n+3 \end{array}\right.,$

and, by induction, ${T^n(h_{(3)},v_{(3)})=(h_{(3)},v_{(3)}h_{(3)}^{-n})}$, where

$\displaystyle v_{(3)}h_{(3)}^{-n}=\left\{\begin{array}{rl} n+x-n = x, & \textrm{if } x\notin\{n+1,1,2n+1,2n+2,2n+3,2n+4\} \\ 2n+1, & \textrm{if } x=n+1 \\ 2n+2, & \textrm{if } x=1 \\ 2n+4, & \textrm{if } x=2n+ \left\{\begin{array}{rl}1 & \textrm{if } n \textrm{ is odd} \\ 2 & \textrm{if } n \textrm{ is even}\end{array}\right. \\ 2n+3, & \textrm{if } x=2n+\left\{\begin{array}{rl}2 & \textrm{if } n \textrm{ is odd} \\ 1 & \textrm{if } n \textrm{ is even}\end{array}\right. \\ 1, & \textrm{if } x=2n+2+ \left\{\begin{array}{rl}1 & \textrm{if } n \textrm{ is odd} \\ 2 & \textrm{if } n \textrm{ is even}\end{array}\right. \\ n+1, & \textrm{if } x=2n+2+\left\{\begin{array}{rl}2 & \textrm{if } n \textrm{ is odd} \\ 1 & \textrm{if } n \textrm{ is even}\end{array}\right. \end{array}\right.$

Hence, one has ${T^n(h_{(3)},v_{(3)})}$ is an unramified double covering of ${(S,\omega_S)}$ of type ${((\pm1)_n, +1, -1,-1)}$ with

$\displaystyle (\pm1)_n=\left\{\begin{array}{rl} +1, & \textrm{if } n \textrm{ is odd} \\ -1, & \textrm{if } n \textrm{ is even }\end{array}\right.$

so that we come back to a previous considered case. This completes the proof of item (b).

Remark 3 Actually, the same kind of computation works with any parameter ${a}$, but the notation becomes less transparent. So that’s why I decided to treat the cases ${a=2}$ and ${a=3}$ separately (since the classification result says that it suffices to treat these two cases).