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).

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