Posted by: matheuscmss | August 6, 2013

## Hodge-Teichmüller planes and finiteness results for Teichmüller curves

Alex Wright and I have just upload to ArXiv our paper Hodge-Teichmüller planes and finiteness results for Teichmüller curves.

As the title of the paper indicates, this article concerns finiteness results for certain classes of Teichmüller curves (i.e., closed ${SL(2,\mathbb{R})}$-orbits in the moduli spaces of Abelian differentials). For example, one of the results in this paper is the finiteness of algebraically primitive Teichmüller curves in minimal strata of prime genus $\geq 3$.

In fact, some parts of this paper were previously discussed in this blog here, and, for this reason, we will not make further comments on the contents of this article today.

Instead, let us take the opportunity to briefly discuss a “deleted scene” of this paper.

More concretely, at some stage of the paper, Alex and I need to know that there are Abelian differentials/translation surfaces with “rich monodromy” (whatever this means) in each connected component of every stratum of genus ${g\geq 3}$.

In this direction, we ensure the existence of such translation surfaces by an inductive argument going as follows.

Starting with an hypothetical connected component ${\mathcal{H}}$ of genus ${g\geq 3}$ containing only translation surfaces with “poor monodromy” (${g-1}$ orthogonal Hodge-Teichmüller planes in the notation of the paper), we follow and adpat some arguments of Kontsevich-Zorich paper on the classification of connected components of strata (cf. Section 5 of our paper) and we work a little bit with the Deligne-Mumford compactification of moduli spaces (cf. Section 6 of our paper) to show that the “poorness of monodromy” property passes down to all translations surfaces in both connected components ${\mathcal{H}(4)^{hyp}}$ and ${\mathcal{H}(4)^{odd}}$. Here, very roughly speaking, the basic idea is that, if we “degenerate” (e.g., pinch off a curve) in a careful way the translation surfaces in ${\mathcal{H}}$, we can decrease the genus without loosing the “poorness of monodromy” property, so that, if we keep degenerating, then we will find ourselves with plenty of genus ${3}$ translation surfaces with poor monodromy.

However, it is easy to see that the property that all translation surfaces in ${\mathcal{H}(4)^{hyp}}$ and ${\mathcal{H}(4)^{odd}}$ have “poor monodromy” is false: indeed, in Section 4 of our paper, we exhibit explicitely two translation surfaces ${M_{\ast}\in \mathcal{H}(4)^{odd}}$ and ${M_{\ast\ast}\in\mathcal{H}(4)^{hyp}}$ with “rich monodromy”. Thus, the hypothetical connected component ${\mathcal{H}}$ can not exist.

As it turns out, before thinking about this argument based on Kontsevich-Zorich classification of connected components and the features of Deligne-Mumford compactification, Alex and I thought of using the following simple-minded argument. One can check that the “richness of monodromy” property passes through finite branched coverings. Hence, it suffices to produce for each connected component ${\mathcal{H}}$ an explicit finite branched cover of one of the translation surfaces ${M_{\ast}}$ or ${M_{\ast\ast}}$ lying in ${\mathcal{H}}$ to deduce that all connected components of all strata possess translation surfaces with “rich monodromy”.

Remark 1 The strategy in the previous paragraph is very natural: for instance, after reading a preliminary version of our paper, Giovanni Forni asked us if we could not make the arguments in Sections 5 and 6 of our paper (related to Kontsevich-Zorich classification of connected components of strata and Deligne-Mumford compactification) more elementary by taking finite branched covers to produce explicit translation surfaces with rich monodromy on each connected component of every stratum. As we will see below, this does not quite work to fully recover the statements in Section 5 of our paper, but it allows to obtain at least part of our statements.

In particular, around November/December 2012, Alex and I started looking at finite branched covers of ${M_{\ast}}$ and ${M_{\ast\ast}}$ “hitting” all connected components of all strata. Unfortunately, this strategy does not work as well as one could imagine: firstly, there are restrictions on the genera of strata we can reach using finite branched covers (thanks to Riemann-Hurwitz formula), and, secondly, it is not so easy to figure out in what connected component our finite branched cover lives.

Nevertheless, this elementary strategy permits to deduce the following proposition (allowing to deduce partial versions of the statements in Section 5 of our paper). Let ${M_{\ast}\in \mathcal{H}(4)^{odd}}$ and ${M_{\ast\ast}\in\mathcal{H}(4)^{hyp}}$ be the square-tiled surfaces with “rich monodromy” constructed in our paper (see also this post), i.e., the square-tiled surfaces below

$M_{\ast}\in\mathcal{H}(4)^{odd}$

$M_{\ast\ast}\in\mathcal{H}(4)^{hyp}$

associated to the pairs of permutations ${h_{\ast}=(1)(2,3)(4,5,6)}$ and ${v_{\ast}=(1,4,2)(3,5)(6)}$, and ${h_{\ast\ast}=(1)(2,3)(4,5,6)}$ and ${v_{\ast\ast}=(1,2)(3,4)(5)}$.

Remark 2 Here, as it is usual in this theory, we are constructing square-tiled surfaces from a pair of permutations ${h, v\in S_N}$ on ${N}$ elements by taking ${N}$ unit squares and gluing them (by translations) so that ${h(i)}$ is the square to the right of the square ${i}$ and ${v(i)}$ is the square on the top of the square ${i}$.

Proposition 1 For each ${d\geq 3}$ odd, there exists ${M_{\ast}(d)}$ a finite branched cover of ${M_{\ast}}$ of degree ${d}$ in the odd connected component ${\mathcal{H}(5d-1)^{odd}}$ of the minimal stratum ${\mathcal{H}(5d-1)}$ (of Abelian differentials with a single zero of order ${5d-1}$).

Also, there exists ${M_{\ast\ast}(3)}$ a finite branched cover of ${M_{\ast\ast}}$ of degree ${3}$ in the hyperelliptic connected component ${\mathcal{H}(14)^{hyp}}$ of the minimal stratum ${\mathcal{H}(14)}$ (of Abelian differentials with a single zero of order ${14}$).

Finally, for each ${d\geq 5}$, there exists ${M_{\ast\ast}(d)}$ a finite branched cover of ${M_{\ast\ast}}$ of degree ${d}$ in the even connected component ${\mathcal{H}(5d-1)^{even}}$ of the minimal stratum ${\mathcal{H}(5d-1)}$ (of Abelian differentials with a single zero of order ${5d-1}$).

Remark 3 In this statement, the nomenclature “hyperelliptic”, “even” and “odd” refers to the invariants introduced by Kontsevich and Zorich to distinguish between the connected components of strata. We will briefly review these notions below (as we will need them to prove Proposition 1.

We will dedicate the remainder of this post to outline the construction of the translation surfaces ${M_{\ast}(d)}$ and ${M_{\ast\ast}(d)}$ satisfying the conclusions of Proposition 1. In particular, we will divide the discussion into three sections: in the next one, we will quickly review the invariants introduced by Kontsevich-Zorich to classify connected components of strata, and in the last two sections we will construct ${M_{\ast}(d)}$ and ${M_{\ast\ast}(d)}$ respectively.

1. Parity of spin structure and components of strata

The results of Kontsevich-Zorich imply that any stratum ${\mathcal{H}(k_1,\dots,k_{\sigma})}$ (of Abelian differentials with zeroes of prescribed orders ${k_1,\dots k_{\sigma}}$) has at most three connected components

$\displaystyle \mathcal{H}(k_1,\dots,k_{\sigma})^{hyp}, \mathcal{H}(k_1,\dots,k_{\sigma})^{even} \textrm{ and } \mathcal{H}(k_1,\dots,k_{\sigma})^{odd}$

characterized by the following properties:

• ${\mathcal{H}(k_1,\dots,k_{\sigma})^{hyp}}$ consists entirely of hyperelliptic translation surfaces;
• ${\mathcal{H}(k_1,\dots,k_{\sigma})^{even}}$, resp. ${\mathcal{H}(k_1,\dots,k_{\sigma})^{odd}}$, consists of translation surfaces whose parity of the spin structure is even, resp. odd, (and they contain some non-hyperelliptic translation surface).

For convenience of the reader, we recall that the parity of the spin structure ${\Phi(M)}$ of a translation surface ${M=(X,\omega)}$ of genus ${g\geq 1}$ can be computed as follows.

Firstly, one chooses a canonical (symplectic) basis ${\{a_i, b_i\}_{i=1}^g}$ of ${H_1(M,\mathbb{R})}$ represented by paths avoiding the singularities of ${M}$(i.e., zeroes of ${\omega}$).

Secondly, we define the index ${\textrm{ind}(\gamma)}$ of a simple closed curved avoiding the singularities of ${M}$ as the degree of the Gauss map associated to ${\gamma}$ (i.e., the degree of the map measuring the angle the tangents to ${\gamma}$ make with the horizontal direction on ${M}$ [a notion that is well-defined outside the singularities]).

Finally, we form the quantity ${\Phi(\gamma):=\textrm{ind}(\gamma)+1 \,(\textrm{mod } 2)}$ and we define the parity ${\phi(M)}$ of the spin structure of ${M}$ as

$\displaystyle \phi(M)=\sum\limits_{i=1}^g\Phi(a_i)\Phi(b_i)\in\mathbb{Z}/2\mathbb{Z}$

Remark 4 Of course, it is implicit in this definition that ${\Phi(M)}$ does not depend on the choice of the canonical basis ${\{a_i, b_i\}}$. This fact is true and the reader is invited to consult Kontsevich-Zorich paper for more discussion on this (as well as a motivation for the nomenclature “parity of spin structure” for ${\phi(M)}$).

In this notation, we say that ${M}$ has even, resp. odd, parity of spin structure if and only if ${\phi(M)=0}$, resp. ${1}$.

2. Explicit covers of ${M_{\ast}\in\mathcal{H}(4)^{odd}}$

Consider again the square-tiled surface ${M_{\ast}\in \mathcal{H}(4)^{\textrm{odd}}}$ given by the pair of permutations ${h_{\ast}=(1)(2,3)(4,5,6)}$ and ${v_{\ast}=(1,4,2)(3,5)(6)}$.

For technical reason, it is desirable to “simplify” the geometry of ${M_{\ast}}$ in order to produce finite covers that are “simple” to handle.

More concretely, we will replace ${M_{\ast}}$ by the square-tiled surface ${\overline{M}_{\ast}}$ given by the pair of permutations ${\overline{h}_{\ast}=(1,2,3,4,5,6)}$, ${\overline{v}_{\ast}=(1)(2,5,4)(3,6)}$ in the ${SL(2,\mathbb{Z})}$orbit of ${M_{\ast}}$: indeed, ${\overline{M}_{\ast}=J\cdot T^2(M_{\ast})}$ where ${J=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)}$ and ${T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)}$.

Depiction of $\overline{M}_{\ast}$

Given ${d\geq 3}$ an odd integer, let ${\overline{M}_{\ast}(d)}$ be the square-tiled surface associated to the pair of permutations

$\displaystyle \overline{h}_{\ast}(d):=(1_1,2_1,3_1,4_1,5_1,6_1)(1_2,2_2,3_2,4_2,5_2,6_2,1_3,2_3,3_3,4_3,5_3,6_3)\dots$

$\displaystyle (1_{d-1},2_{d-1},3_{d-1},4_{d-1},5_{d-1},6_{d-1},1_d,2_d,3_d,4_d,5_d,6_d),$

$\displaystyle \overline{v}_{\ast}(d):=(1_1,1_2,\dots,1_d)(2_1,5_1,4_1)(3_1,6_1)\dots(2_d,5_d,4_d)(3_d,6_d)$

By defintion, ${\overline{M}_{\ast}(d)}$ is a degree ${d}$ cover of ${\overline{M}_{\ast}}$ belonging to the stratum ${\mathcal{H}(5d-1)}$.

Remark 5 The “shape” of the covering ${\overline{M}_{\ast}(d)}$ was “guessed” with the help of SAGE: in fact, we tried a few simple-minded finite coverings of ${\overline{M}_{\ast}}$ (including ${\overline{M}_{\ast}(d)}$ for ${d=3, 5, \dots, 13}$) and we asked SAGE to determine their connected components; then, once we got the “correct” connected components (in minimal strata), we looked at the permutations corresponding these square-tiled surfaces and we found the “partner” leading to the expressions for the permutations ${\overline{h}_{\ast}(d)}$ and ${\overline{v}_{\ast}(d)}$.

Let us now determine, for ${d\geq 3}$ odd, the connected component of ${\mathcal{H}(5d-1)}$ containing ${\overline{M}_{\ast}(d)}$.

We think of it as ${(d+1)/2}$ horizontal cylinders determined by the permutation ${\overline{h}_{\ast}(d)}$ whose top and bottom boundaries are glued accordingly to the permutation ${\overline{v}_{\ast}(d)}$. Using this geometrical representation, we can define the following cycles in ${H_1(\overline{M}_{\ast}(d),\mathbb{Z})}$.

• for each ${i=1,\dots,d}$, let ${c_{23}^{(i)}}$, ${c_4^{(i)}}$, ${c_5^{(i)}}$ and ${c_6^{(i)}}$ be the homology classes of the vertical cycles within the horizontal cylinders connecting the middle of the bottom side of the squares ${5_i}$, ${2_i}$, ${4_i}$ and ${3_i}$ (resp.) to the top side of the squares ${2_i}$, ${4_i}$, ${5_i}$ and ${6_i}$ (resp.);
• for each ${j=2l}$, ${l=1,\dots, (d-1)/2}$, let ${c_1^{(j)}}$ be the homology classes of the vertical cycles within the horizontal cylinders connecting the middles of the bottom side of the square ${1_{j+1}}$ to the top side of the squares ${1_{j}}$;
• let ${c_1^{(1)}}$ be the homology class of the concatenation of the vertical cycles within the horizontal cylinders connecting the middles of the bottom side of the square ${1_{2l}}$ to the top side of the square ${1_{2l+1}}$ for ${l=1,\dots, (d-1)/2}$, the bottom side of the square ${1_2}$ to the left side of the square ${1_2}$, and the right side of the square ${6_d}$ to the top side of the square ${1_d}$;
• ${\sigma_1}$ is the horizontal cycle connecting the left vertical side of the square ${1_1}$ to the right vertical side of the square ${6_1}$, and, for ${j=2l}$, ${l=1,\dots, (d-1)/2}$, ${\sigma_{j}}$ is the horizontal cycle connecting the left vertical side of the square ${1_{j}}$ to the right vertical side of the square ${6_{j+1}}$.

Depiction of $\overline{M}_{\ast}(d)$ and some of the cycles $\sigma_i$ and $c_{a}^{(j)}$

It is not hard to check that these ${5d+1}$ cycles form a basis of ${H_1(\overline{M}_{\ast}(d),\mathbb{Z})}$. From this basis, we will compute the parity of the spin of ${\overline{M}_{\ast}(d)}$ using the orthogonalization procedure described in Appendix C of this paper of Zorich (a Gram-Schmidt orthogonalization process to produce canonical basis of homology modulo ${2}$). More precisely, we start with the cycles ${a_1:=\sigma_2}$ and ${b_1:=c_1^{(2)}}$. Then, by induction, we use the cycles ${a_i}$, ${b_i}$ to successively render the cycles

$\displaystyle c_{23}^{(2)}, c_4^{(2)}, c_5^{(2)}, c_6^{(2)}, \dots, \sigma_1, c_1^{(1)}, c_{23}^{(1)}, c_4^{(1)}, c_5^{(1)}, c_6^{(1)}$

into a canonical basis ${a_1, b_1, a_2, b_2, \dots, a_g, b_g}$ (where ${g=(5d+1)/2}$). After performing this procedure, one has that the parity ${\phi(\overline{M}_{\ast}(d))\in\mathbb{Z}_2}$ of the spin structure of ${\overline{M}_{\ast}(d)}$ is

$\displaystyle \phi(\overline{M}_{\ast}(d))=\sum\limits_{i=1}^{(5d+1)/2} \Phi(a_i)\Phi(b_i) \textrm{ (mod } 2),$

where ${\Phi:H_1(\overline{M}_{\ast}(d),\mathbb{Z}_2)\rightarrow\mathbb{Z}_2}$ satisfies the following properties:

• ${\Phi(\gamma)=\textrm{ind}(\gamma)+1}$ whenever ${\gamma}$ is a simple smooth closed curve in ${\overline{M}_{\ast}(d)}$ not passing through singular points whose Gauss map has degree ${\textrm{ind}(\gamma)}$;
• ${\Phi}$ is a quadratic form representing the intersection form ${(.,.)}$ in the sense that ${\Phi(\alpha+\beta)=\Phi(\alpha)+\Phi(\beta)+(\alpha,\beta)}$.

We affirm that ${\phi(\overline{M}_{\ast}(d+2))=\phi(\overline{M}_{\ast}(d))}$ for each ${d\geq 1}$ odd. Indeed, we note that the ${10}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_{23}^{(d+1)}, c_4^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_{23}^{(d+2)}, c_4^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)}$

of ${\overline{M}_{\ast}(d+2)}$ do not intersect the other ${5d-1}$ cycles of ${\overline{M}_{\ast}(d+2)}$ defined above except for ${c_1^{(1)}}$, and, moreover, they have trivial intersection with the cycle

$\displaystyle \begin{array}{rcl} \widetilde{c}_{1}^{(1)}&:=& c_1^{(1)}-c_1^{(d+1)}+c_{23}^{(d+1)}-c_4^{(d+1)}+2c_5^{(d+1)}-2c_6^{(d+1)} \\ &+& c_4^{(d+2)}-c_{23}^{(d+2)}+c_6^{(d+2)}-c_5^{(d+2)} \end{array}$

satisfying ${\Phi(\widetilde{c}_{1}^{(1)})=\Phi(c_1^{(1)})=1}$. Thus, by replacing ${c_1^{(1)}}$ by ${\widetilde{c}_1^{(1)}}$ and by applying the orthogonalization procedure to these two sets of ${10}$ and ${5d-1}$ cycles in an independent way, we deduce that ${\phi(\overline{M}_{\ast}(d+2))}$ differs from ${\phi(\overline{M}_{\ast}(d))}$ by a term of the form

$\displaystyle \phi(\overline{M}_{\ast}(d+2))-\phi(\overline{M}_{\ast}(d))=\sum\limits_{i=1}^{5}\Phi(a_i^{(d+1)})\Phi(b_i^{(d+1)})$

where ${a_i^{(d+1)}, b_i^{(d+1)}}$, ${i=1,\dots, 5}$ is an orthogonalization of the ${10}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_{23}^{(d+1)}, c_4^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_{23}^{(d+2)}, c_4^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)}.$

On the other hand, by a direct computation, one can check that

• ${a_1^{(d+1)}=\sigma_{d+1}}$, ${b_1^{(d+1)}=c_{1}^{(d+1)}}$,
• ${a_2^{(d+1)}=c_4^{(d+1)}-a_1^{(d+1)}-b_1^{(d+1)}}$, ${b_2^{(d+1)} = c_{23}^{(d+1)}-a_1^{(d+1)}-b_1^{(d+1)}}$,
• ${a_3^{(d+1)}=c_6^{(d+1)}-a_1^{(d+1)}-b_1^{(d+1)}+a_2^{(d+1)}}$, ${b_3^{(d+1)}=c_5^{(d+1)}-a_1^{(d+1)}-b_1^{(d+1)}+a_2^{(d+1)}}$,
• ${a_4^{(d+1)}=c_4^{(d+2)}-b_1^{(d+1)}+b_2^{(d+1)}-a_2^{(d+1)}+2b_3^{(d+1)}-2a_3^{(d+1)}}$, ${b_4^{(d+1)}=c_{23}^{(d+2)}-b_1^{(d+1)}+b_2^{(d+1)}-a_2^{(d+1)}+2b_3^{(d+1)}-2a_3^{(d+1)}}$,
• ${a_5^{(d+1)}=c_5^{(d+2)}-c_4^{(d+2)}}$, ${b_5^{(d+1)}=c_6^{(d+2)}-c_4^{(d+2)}}$

is an orthogonalization of the ${10}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_{23}^{(d+1)}, c_4^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_{23}^{(d+2)}, c_4^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)}.$

Moreover, ${\Phi(a_i^{(d+1)})=\Phi(b_i^{(d+1)})=1}$ for ${i=1,4}$, and ${\Phi(a_j^{(d+1)})=\Phi(b_j^{(d+1)})=0}$ for ${j=2,3,5}$, so that

$\displaystyle \phi(\overline{M}_{\ast}(d+2))-\phi(\overline{M}_{\ast}(d))=\sum\limits_{i=1}^{5}\Phi(a_i^{(d+1)})\Phi(b_i^{(d+1)}) = 0,$

as it was claimed.

It follows that ${\phi(\overline{M}_{\ast}(d))=\phi(\overline{M}_{\ast}(1)):=\phi(\overline{M}_{\ast})=1\in\mathbb{Z}_2}$. Since the square-tiled surfaces ${\overline{M}_{\ast}(d)}$ are not hyperelliptic, we deduce that:

Proposition 2 For each ${d\geq 3}$ odd, let ${\overline{M}_{\ast}(d)}$ be the square-tiled surface (defined above) covering the square-tiled ${\overline{M}_{\ast}}$ in the ${SL(2,\mathbb{Z})}$-orbit of the square-tiled surface ${M_{\ast}}$. Then, ${\overline{M}_{\ast}(d)}$ belongs to ${\mathcal{H}(5d-1)^{\textrm{odd}}}$.

3. Explicit covers of ${M_{\ast\ast}\in\mathcal{H}(4)^{hyp}}$

Consider now the square-tiled surface ${M_{\ast\ast}\in \mathcal{H}(4)^{\textrm{hyp}}}$ given by the pair of permutations ${h_{\ast}=(1)(2,3)(4,5,6)}$ and ${v_{\ast}=(1,2)(3,4)(5)(6)}$.

Once more, let us “simplify the geometry” by taking the square-tiled surface ${\overline{M}_{\ast\ast}}$ given by the pair of permutations ${\overline{h}_{\ast\ast}=(1,2,3,4,5,6)}$, ${\overline{v}_{\ast\ast}=(1)(2,6)(3,4,5)}$ belongs to the ${SL(2,\mathbb{Z})}$-orbit of ${M_{\ast\ast}}$: indeed, ${\overline{M}_{\ast\ast}=J\cdot T(M_{\ast\ast})}$ where ${J=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right)}$ and ${T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)}$.

Depiction of $\overline{M}_{\ast\ast}$

Given ${d\geq 3}$ an odd integer, let ${\overline{M}_{\ast\ast}(d)}$ be the square-tiled surface associated to the pair of permutations

$\displaystyle \overline{h}_{\ast\ast}(d):=(1_1,2_1,3_1,4_1,5_1,6_1)(1_2,2_2,3_2,4_2,5_2,6_2,1_3,2_3,3_3,4_3,5_3,6_3) \dots$

$\displaystyle (1_{d-1},2_{d-1},3_{d-1},4_{d-1},5_{d-1},6_{d-1},1_d,2_d,3_d,4_d,5_d,6_d)$

$\displaystyle \overline{v}_{\ast\ast}(d):=(1_1,1_2,\dots,1_d)(2_1,6_1)(3_1,4_1,5_1)\dots(2_d,6_d)(3_d,4_d,5_d)$

By defintion, ${\overline{M}_{\ast\ast}(d)}$ is a degree ${d}$ cover of ${\overline{M}_{\ast\ast}}$ belonging to the stratum ${\mathcal{H}(5d-1)}$.

Remark 6 Similarly to Remark 5, we “guessed” the shape of ${\overline{M}_{\ast\ast}(d)}$ with the help of SAGE.

Now, let us determine, for ${d\geq 3}$ odd, the connected component of ${\mathcal{H}(5d-1)}$ containing ${\overline{M}_{\ast\ast}(d)}$. Again, we think of it as ${(d+1)/2}$ horizontal cylinders determined by the permutation ${\overline{h}_{\ast\ast}(d)}$ whose top and bottom boundaries are glued accordingly to the permutation ${\overline{v}_{\ast\ast}(d)}$. Using this geometrical representation, we can define the following cycles in ${H_1(\overline{M}_{\ast\ast}(d),\mathbb{Z})}$.

• for each ${i=1,\dots,d}$, let ${c_{2}^{(i)}}$, ${c_{34}^{(i)}}$, ${c_5^{(i)}}$ and ${c_6^{(i)}}$ be the homology classes of the vertical cycles within the horizontal cylinders connecting the middle of the bottom side of the squares ${6_i}$, ${4_i}$, ${3_i}$ and ${2_i}$ (resp.) to the top side of the squares ${2_i}$, ${3_i}$, ${5_i}$ and ${6_i}$ (resp.);
• for each ${j=2l}$, ${l=1,\dots, (d-1)/2}$, let ${c_1^{(j)}}$ be the homology classes of the vertical cycles within the horizontal cylinders connecting the middles of the bottom side of the square ${1_{j+1}}$ to the top side of the squares ${1_{j}}$;
• let ${c_1^{(1)}}$ be the homology class of the concatenation of the vertical cycles within the horizontal cylinders connecting the middles of the bottom side of the square ${1_{2l}}$ to the top side of the square ${1_{2l+1}}$ for ${l=1,\dots, (d-1)/2}$, the bottom side of the square ${1_2}$ to the left side of the square ${1_2}$, and the right side of the square ${6_d}$ to the top side of the square ${1_d}$;
• ${\sigma_1}$ is the horizontal cycle connecting the left vertical side of the square ${1_1}$ to the right vertical side of the square ${6_1}$, and, for ${j=2l}$, ${l=1,\dots, (d-1)/2}$, ${\sigma_{j}}$ is the horizontal cycle connecting the left vertical side of the square ${1_{j}}$ to the right vertical side of the square ${6_{j+1}}$.

Depiction of $\overline{M}_{\ast\ast}(d)$ and some of the cycles $\sigma_i$ and $c_{a}^{(j)}$

It is not hard to check that these ${5d+1}$ cycles form a basis of ${H_1(\overline{M}_{\ast\ast}(d),\mathbb{Z})}$. We affirm that ${\phi(\overline{M}_{\ast\ast}(d+4))=\phi(\overline{M}_{\ast\ast}(d))}$ for each ${d\geq 1}$ odd. Indeed, we note that the ${20}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_2^{(d+1)}, c_{34}^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_2^{(d+2)}, c_{34}^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)},$

$\displaystyle \sigma_{d+3}, c_1^{(d+3)}, c_2^{(d+3)}, c_{34}^{(d+3)}, c_5^{(d+3)}, c_6^{(d+3)}, c_2^{(d+4)}, c_{34}^{(d+4)}, c_{5}^{(d+4)}, c_6^{(d+4)}$

of ${\overline{M}_{\ast\ast}(d+4)}$ do not intersect the other ${5d-1}$ cycles of ${\overline{M}_{\ast\ast}(d+4)}$ defined above except for ${c_1^{(1)}}$, and, moreover, they have trivial intersection with the cycle

$\displaystyle \begin{array}{rcl} \widetilde{\widetilde{c}}_{1}^{(1)}&:=& c_1^{(1)}-c_1^{(d+1)}+c_{2}^{(d+1)}-c_{34}^{(d+1)}+c_5^{(d+1)}-c_6^{(d+1)} \\ &+& c_{34}^{(d+2)}-c_{2}^{(d+2)}+c_6^{(d+2)}-c_5^{(d+2)} \\ &-&c_1^{(d+3)}+c_{2}^{(d+3)}-c_{34}^{(d+3)}+c_5^{(d+3)}-c_6^{(d+3)} \\ &+& c_{34}^{(d+4)}-c_{2}^{(d+4)}+c_6^{(d+4)}-c_5^{(d+4)} \end{array}$

satisfying ${\Phi(\widetilde{\widetilde{c}}_1^{(1)})=\Phi(c_1^{(1)})=1}$. Thus, by replacing ${c_1^{(1)}}$ by ${\widetilde{\widetilde{c}}_1^{(1)}}$ and by applying the orthogonalization procedure to these two sets of ${20}$ and ${5d-1}$ cycles in an independent way, we deduce that ${\phi(\overline{M}_{\ast\ast}(d+4))}$ differs from ${\phi(\overline{M}_{\ast\ast}(d))}$ by a term of the form

$\displaystyle \phi(\overline{M}_{\ast\ast}(d+4))-\phi(\overline{M}_{\ast\ast}(d))=\sum\limits_{k\in\{d+1, d+3\}}\sum\limits_{i=1}^{5}\Phi(a_i^{(k)})\Phi(b_i^{(k)})$

where ${a_i^{(d+1)}, b_i^{(d+1)}, a_i^{(d+3)}, b_i^{(d+3)}}$, ${i=1,\dots, 5}$ is an orthogonalization of the ${20}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_2^{(d+1)}, c_{34}^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_2^{(d+2)}, c_{34}^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)},$

$\displaystyle \sigma_{d+3}, c_1^{(d+3)}, c_2^{(d+3)}, c_{34}^{(d+3)}, c_5^{(d+3)}, c_6^{(d+3)}, c_2^{(d+4)}, c_{34}^{(d+4)}, c_{5}^{(d+4)}, c_6^{(d+4)}.$

On the other hand, by a direct computation, one can check that, for ${k=d+1,d+3}$,

• ${a_1^{(k)}=\sigma_{k}}$, ${b_1^{(k)}=c_{1}^{(k)}}$,
• ${a_2^{(k)}=c_{34}^{(k)}-a_1^{(k)}-b_1^{(k)}}$, ${b_2^{(k)} = c_{2}^{(k)}-a_1^{(k)}-b_1^{(k)}}$,
• ${a_3^{(k)}=c_6^{(k)}-a_1^{(k)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}}$, ${b_3^{(k)}=c_5^{(k)}-a_1^{(k)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}}$,
• ${a_4^{(k)}=c_{34}^{(k+1)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}+b_3^{(k)}-a_3^{(k)}}$, ${b_4^{(k)}=c_{2}^{(k+1)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}+b_3^{(k)}-a_3^{(k)}}$,
• ${a_5^{(k)}=c_6^{(k+1)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}+b_3^{(k)}-a_3^{(k)}+b_4^{(k)}-a_4^{(k)}}$, ${b_5^{(d+1)}=c_5^{(k+1)}-b_1^{(k)}+b_2^{(k)}-a_2^{(k)}+b_3^{(k)}-a_3^{(k)}+b_4^{(k)}-a_4^{(k)}}$

is an orthogonalization of the ${20}$ cycles

$\displaystyle \sigma_{d+1}, c_1^{(d+1)}, c_2^{(d+1)}, c_{34}^{(d+1)}, c_5^{(d+1)}, c_6^{(d+1)}, c_2^{(d+2)}, c_{34}^{(d+2)}, c_{5}^{(d+2)}, c_6^{(d+2)},$

$\displaystyle \sigma_{d+3}, c_1^{(d+3)}, c_2^{(d+3)}, c_{34}^{(d+3)}, c_5^{(d+3)}, c_6^{(d+3)}, c_2^{(d+4)}, c_{34}^{(d+4)}, c_{5}^{(d+4)}, c_6^{(d+4)}.$

Moreover, for ${k=d+1, d+3}$, one has ${\Phi(a_i^{(k)})=\Phi(b_i^{(k)})=1}$ for ${i=1, 3, 5}$, and ${\Phi(a_j^{(k)})=\Phi(b_j^{(k)})=0}$ for ${j=2,4}$, so that

$\displaystyle \phi(\overline{M}_{\ast\ast}(d+4))-\phi(\overline{M}_{\ast\ast}(d))=\sum\limits_{k\in\{d+1, d+3\}}\sum\limits_{i=1}^{5}\Phi(a_i^{(k)})\Phi(b_i^{(k)}) = 0,$

as it was claimed.

Remark 7 Contrary to the last section, we compared ${\phi(\overline{M}_{\ast\ast}(d))}$ to ${\phi(\overline{M}_{\ast\ast}(d+4))}$ instead of ${\phi(\overline{M}_{\ast\ast}(d+2))}$ because the 20 cycles indicated above have a better geometrical behavior (for the orthogonalization process) than the corresponding 10 cycles (as a direct computation shows).

It follows that ${\phi(\overline{M}_{\ast\ast}(d))=\phi(\overline{M}_{\ast\ast}(1))}$ for ${d\equiv 1 \, (\textrm{mod } 4)}$, and ${\phi(\overline{M}_{\ast\ast}(d))=\phi(\overline{M}_{\ast\ast}(3))}$ for ${d\equiv 3 \, (\textrm{mod } 4)}$. Since ${\overline{M}_{\ast\ast}:=\overline{M}_{\ast\ast}(1)}$ and ${\overline{M}_{\ast\ast}(3)}$ are hyperelliptic square-tiled surfaces of genus ${3}$ and ${8}$ resp., i.e., ${\overline{M}_{\ast\ast}(1)\in \mathcal{H}(4)^{\textrm{hyp}}}$ and ${\overline{M}_{\ast\ast}(3)\in \mathcal{H}(14)^{\textrm{hyp}}}$, we deduce that ${\phi(\overline{M}_{\ast\ast}(1))=\phi(\overline{M}_{\ast\ast}(3))=0\in\mathbb{Z}_2}$: this last claim follows directly from the fact that the parity of the spin of a translation surface in ${\mathcal{H}(2g-2)^{\textrm{hyp}}}$ is ${[(g+1)/2] (\textrm{mod }2)}$ (where ${[x]}$ is the integer part of ${x}$), or by a direct calculation.

Because ${\overline{M}_{\ast\ast}(d)}$ is not hyperelliptic for ${d\geq 5}$ odd, we conclude that:

Proposition 3 For each ${d\geq 3}$ odd, let ${\overline{M}_{\ast\ast}(d)}$ be the square-tiled surface (defined above) covering the square-tiled ${\overline{M}_{\ast\ast}}$ in the ${SL(2,\mathbb{Z})}$-orbit of the square-tiled surface ${M_{\ast\ast}}$. Then, ${\overline{M}_{\ast\ast}(3)\in \mathcal{H}(14)^{\textrm{hyp}}}$ and ${\overline{M}_{\ast\ast}(d)\in \mathcal{H}(5d-1)^{\textrm{even}}}$.

This completes the proof of Proposition 1 (as this proposition follows from Propositions 2 and 3 above).