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Zapponi-orientable dessins d'enfants

Almost two decades ago Zapponi introduced a notion of orientability of a clean dessin d'enfant, based on an orientation of the embedded bipartite graph. We extend this concept, which we call Z-orientability to distinguish it from the traditional topological definition, to the wider context of all dessins, and we use it to define a concept of twist orientability, which also takes account of the Z-orientability properties of those dessins obtained by permuting the roles of white and black vertices and face-centres. We observe that these properties are Galois-invariant, and we study the extent to which they are determined by the standard invariants such as the passport and the monodromy and automorphism groups. We find that in general they are independent of these invariants, but in the case of regular dessins they are determined by the monodromy group.


Introduction
According to a famous theorem of Belyi [1], a compact Riemann surface S, regarded as a complex projective algebraic curve, can be defined over the field Q of algebraic numbers if and only if it admits a non-constant meromorphic function β : S → C = C ∪ {∞} ≃ P 1 (C) which is branched over at most three points (which we can take to be 0, 1 and ∞). Such a function β is called a Belyi function, and (S, β) is known as a Belyi pair. The 'certificate' for possessing such a function is the existence of a map embedding a bipartite graph in S, with the fibres β −1 (0) and β −1 (1) providing the white and black vertices, and the unit interval I = [0, 1] ⊂ R lifting to the edges (the complement of the embedded graph is a disjoint union of discs, called the faces of the map). Grothendieck [6] called such a map D a dessin d'enfant, or simply a dessin, and noted that the natural action of the absolute Galois group Gal(Q/Q) on Belyi pairs induces a faithful action on (isomorphism classes of) dessins.
One of the major problems in this area is that of determining Galois orbits, or more specifically of determining whether or not two given dessins are in the same orbit of Gal(Q/Q). If they differ in some Galois-invariant property, such as their genus, type, monodromy group, automorphism group or passport, then we know that they are in different orbits. However, if they agree in these properties then nothing can be concluded, and we need to try further tests. For this reason it is useful to know and to be able to determine as many Galois invariants as possible, in order to have a more powerful toolkit in tackling this problem. Of course, a new invariant needs to be independent of the others (for instance, the passport determines the genus and the type, but not the monodromy or automorphism group), or to be more easily computed, in order to be of any use.
In [10,11] Zapponi introduced a new Galois invariant, which he called 'orientability' (motivated by the graph-theoretic concept of an orientation of the embedded graph). Of course, the term 'orientability' has a traditional meaning in topology, and in that sense all dessins are orientable since they lie on Riemann surfaces, so to avoid confusion we will use the term 'Zapponi-orientability' for his concept. By abbreviating it to 'Z-orientability' we will commemorate both his name and the special role played by the third standard generator Z of a triangle group in its definition. Thus a dessin is said to be Z-orientable if its faces can be assigned labels + or − so that each edge is incident with faces having different labels. (In graph theory this is known as a 2-face colouring, so that the dual map is bipartite.) We will show that this property can be reinterpreted in a number of different ways, so that it can be studied in terms of function theory, group theory or combinatorics. One of our main results is as follows, where M denotes the monodromy group of D (that is, of the covering β : S → C) and σ 0 , σ 1 and σ ∞ , with σ 0 σ 1 σ ∞ = 1, are its standard generators, giving the monodromy permutations of the edges (equivalently, the sheets of the covering) at the critical values.
Theorem 1. Each of the following properties of a dessin D is equivalent to its Z-orientability: (1) The function β(1 − β) is the square of a meromorphic function on S.
(2) The edges of D can be labelled + or − so that successive edges around each vertex have different signs. Part (1) of Theorem 1 will be proved in Section 3, whereas the remaining parts will be proved in Section 4. It follows immediately from part (1) of the above theorem that the Z-orientability or otherwise of a dessin D is a Galois invariant. As we will show by example, it is independent of the more obvious and widely used Galois invariants such as the monodromy group, automorphism group and passport, so it represents an additional tool which can be used in trying to distinguish Galois orbits.
The traditional choice of using the fibres over 0 and 1, but not ∞, as vertices of the dessin is often convenient, but it has the disadvantage of disguising the complete symmetry of the situation. To rectify this, we will also consider the dessins D ′ and D ′′ which can be obtained from D by cyclically permuting the roles of the three critical values of β (of course, simply transposing vertex colours has no effect on Zorientability). There are obvious analogues of Theorem 1 for these 'twisted' dessins, representing a further refinement of Z-orientability as a useful Galois invariant. We will introduce a new Galois invariant, the twist-invariant orientability type tot(D), which is the number of Z-orientable dessins among the triple {D, D ′ , D ′′ }, and we will show in Proposition 15 that it can take the values 0, 1 or 3, but never 2.
In order to show that the Z-orientability properties of a dessin are independent of the more traditional Galois invariants mentioned above, in our final section we will construct, for each integer n ≥ 8, four uniform dessins (see Section 2 for the definition of uniform dessin) D i (i = 0, 1, 2, 3) with the same degree N = (n!/4!) 2 , the same type, the same monodromy group M = S n × S n (as an abstract group) and the same passport; these dessins also have isomorphic (but non-conjugate) edge-stabilisers and isomorphic automorphism groups. However, we will also show that tot(D 0 ) = 0, tot(D 1 ) = tot(D 2 ) = 1, tot(D 3 ) = 3, and that D 2 is Z-orientable whereas D 1 is not, so the four dessins D i lie in distinct Galois orbits. Thus Zorientability properties can distinguish between these orbits, whereas the other Galois invariants listed above can not. In particular, this example proves the following: Theorem 2. The Galois invariant tot of a dessin is independent of its monodromy group, point stabilisers and automorphism group (as abstract groups) and of its passport.
However, in Section 5 we will prove the following (see Corollary 19): Theorem 3. For regular dessins the invariant tot is determined by the monodromy group (as an abstract group).

Preliminaries and notations
Recall that a dessin d'enfant is a connected bipartite graph D embedded in a closed oriented surface X in such a way that the connected components of X D, which are called faces, are homeomorphic to open discs. The vertices of a dessin can be coloured white or black in such a way that the two vertices of any edge do not have the same colour. We say that a face has degree r if it is topologically a polygon of 2r sides, while the degree of a vertex is simply the number of edges incident with it. Two dessins d'enfants are called equivalent if there is an orientation-preserving homeomorphism between the closed orientable surfaces inducing a colour-preserving isomorphism of the corresponding bipartite graphs.
The passport of a dessin D ⊂ X is the tuple (a 1 , . . . , a α ; b 1 , . . . , b β ; c 1 , . . . , c γ ), where 1 ≤ a 1 ≤ · · · ≤ a α , 1 ≤ b 1 ≤ · · · ≤ b β , 1 ≤ c 1 ≤ · · · ≤ c γ and a i , b j and c k are the degrees of the white vertices, black vertices and faces, respectively. If the least common multiples of the degrees of the white vertices, black vertices and faces are a, b and c respectively, then we say that D has type (a, b, c). If D has n edges, then n = a 1 + · · · + a α = b 1 + · · · + b β = c 1 + · · · + c γ , and, as a consequence of Euler's formula, the genus of X is g = 1 + 1 2 (n − α − β − γ). After labelling the edges of D from 1 to n, the orientation of X induces permutations σ 0 , σ 1 encoding the cyclic ordering of the different edges around the white or black vertices respectively. The pair σ 0 , σ 1 is called the permutation representation of D, and the group M = σ 0 , σ 1 , which is transitive since D is connected, is called the monodromy group of the dessin.
Conversely, any pair of permutations σ 0 , σ 1 generating a transitive subgroup of the symmetric group S n arises as the permutation representation of a dessin d'enfant D with n edges. The disjoint cycles of σ 0 , σ 1 and σ ∞ := (σ 0 σ 1 ) −1 are in bijective correspondence with the white vertices, black vertices and faces of D, and their lengths encode its passport.
A Belyi pair (S, β) consists of a compact Riemann surface S (called a Belyi surface) and a non-constant meromorphic function β on S (called a Belyi function) with at most three branching values which can be assumed to be 0, 1 and ∞. Two Belyi pairs (S 1 , β 1 ) and (S 2 , β 2 ) are called equivalent if there is a bi-holomorphism ψ : S 1 → S 2 such that β 2 • ψ = β 1 . There is a well-known bijective correspondence between the following classes of objects (see e.g. [3], [9]): (1) Equivalence classes of dessins.
(3) Conjugacy classes of finite index subgroups of triangle groups ∆(a, b, c).
(4) Conjugacy classes of finite index subgroups of the free group F 2 of rank two. (5) Isomorphism classes of two-generator transitive finite permutation groups. The link between these classes of objects is made as follows.
(1) ↔ (2) : Given a Belyi pair (S, β) one gets a dessin d'enfant by setting X = S, D = β −1 (I), where I stands for the unit interval [0, 1] in C, and declaring white (resp. black) vertices to be the points in the fibre over 0 (resp. 1). Conversely, given a dessin D ⊂ X one can define a Riemann surface structure on the topological surface X and a Belyi function β on it such that D = f −1 (I), with poles corresponding to the centres of faces.
(2) ↔ (4) : Given a Belyi pair (S, β), we may consider the smooth covering β : S * → C \ {0, 1, ∞}, where S * is the complement in S of the β-preimages of 0, 1 and ∞. By standard covering space theory, these coverings are in bijective correspondence with the finite index subgroups of the fundamental group of C \ {0, 1, ∞}, which is isomorphic to the free group of rank two.
(1) ↔ (3) : The permutation representation of a dessin D ⊂ X of type (a, b, c) and n edges induces a group homomorphism ω, called the monodromy homomorphism of the dessin, ω : ∆(a, b, c) = x, y, z | x a = y b = z c = xyz = 1 −→ S n , determined by ω(x) = σ 0 , ω(y) = σ 1 and ω(z) = σ ∞ , whose image is the monodromy group M . Denoting by Stab M (1) the stabiliser subgroup of 1 in M and by Γ its preimage in the triangle group ∆ = ∆(a, b, c), it can be easily checked that the inclusion Γ ≤ ∆ induces a Belyi map β : D/Γ → D/∆ on S = D/Γ, where D is the unit disc; the corresponding dessin is precisely D and the corresponding monodromy homomorphism ω β : π 1 ( C {0, 1, ∞}) −→ S n is obtained from ω by pre-composition with the obvious homomorphism (4) ↔ (5) : Given a subgroup of finite index in F 2 , we obtain a two-generator transitive finite permutation group by letting F 2 act on its cosets by right multiplication, and factoring out the kernel of the action. Conversely, any such permutation group is a quotient of F 2 , acting on the cosets of a subgroup of finite index, which is unique up to conjugacy.
It is often useful to reinterpret (5) in the following way, in order to distinguish between the monodromy group of a dessin as an abstract group, and as a permutation group. We The group Γ is torsion-free exactly when a i = a, b j = b and c k = c for all i, j and k respectively. Such dessins are called uniform dessins. Regular dessins are a subclass of uniform dessins characterized by any of the following equivalent conditions: i) Stab M (1) is trivial, ii) Γ is normal in ∆, iii) β is a Galois (or normal or regular) covering, and iv) the group of orientation-preserving automorphisms of D as an embedded bicolored graph acts transitively on the set of edges of the dessin.
Clearly, the roles played by the points 0, 1 and ∞ in the above discussion can be interchanged in order to produce dessins related to each other in a natural way. For instance, if β is a Belyi function with dessin D then 1 − β is a Belyi function whose dessin is obtained from D by interchanging the colours of the vertices. More generally, something similar happens by post-composition of β with any element of the order 6 group of Möbius transformations preserving the set {0, 1, ∞}. We will say that dessins obtained in this way are twist-related to each other. Note that twist-related dessins have the same monodromy group, and that their passports are related by permutations of the branching indices.
For example, let D ′ be the dessin obtained from D by replacing β with the Belyi function (β − 1)/β, so that the white and black vertices and face centres of D become the face centres and white and black vertices of D ′ ; this operation cyclically permutes their roles by replacing the triple (σ 0 , σ 1 , σ ∞ ) with the triple (σ 1 , σ ∞ , σ 0 ). Repeating (or inverting) this operation gives a dessin D ′′ corresponding to the Belyi function 1/(1 − β) and the triple (σ ∞ , σ 0 , σ 1 ); the white and black vertices and face centres of D are the black vertices, face centres and white vertices of D ′′ .
A famous result by Belyi [1] shows that a Riemann surface admits an algebraic model with coefficients in Q if and only if it admits a Belyi function, which in turn can also be defined over the algebraic numbers. The action of the absolute Galois group Gal(Q/Q) on the coeficients of Belyi pairs induces an action on the collection of dessins d'enfants. If D is a dessin and τ ∈ Gal(Q/Q), then D τ denotes the image of D under the action of τ . Some basic invariants of the Galois action are the passport, the genus, the number of edges, and the isomorphism classes of the monodromy and automorphism groups [3,8,9].

Z-orientable dessins d'enfants
Following Zapponi [10,11] we shall say that a dessin d'enfant is Zapponi-orientable (abbreviated here to Z-orientable) if its faces can be labelled with symbols + and − in such a way that each edge meets two faces of different sign. Z-orientability imposes the obvious necessary condition that white and black vertices must have even degree. Note also that according to this definition a dessin with monofacial edges, meaning edges meeting the same face on both sides (e.g. a dessin with only one face) is never Z-orientable. Proof. We begin with the observation that β(1 − β) is the square of a meromorphic function if and only if the quadratic differential is the square of a meromorphic 1-form. This is because any meromorphic 1-form can be written as dβ/h for a suitable meromorphic function h (see e.g. [3], Proposition 1.36).
So, suppose that Φ = η ⊗2 for some meromorphic 1-form η. Let F be a face of D and p ∈ S the pole of β lying inside F . Suppose that the multiplicity of p is m so that the number of the edges e k of F is 2m, where monofacial edges are counted twice. An easy calculation in local coordinates shows that the residue of η at p, res p η, has to be either im or −im. Now we can write this residue ±im as • γ ⊂ F stands for any Jordan curve encircling p traversed in the same direction as ∂F (the oriented boundary of F ), • the edges e k of F are assumed to be endowed with the orientation induced by ∂F so in particular each monofacial edge occurs twice, once for each orientation (so that its contributions to the above sum cancel out), • the first equality holds because ∂F and γ are homology equivalent loops in the punctured Riemann surface S \ {poles of η}, and • the last equality holds because We notice that the identity ±im = 1 2πi 2m k=1 πε k implies that the integrals e k η = ± must all take the same value, namely π when res p η = −im and −π when res p η = im. In particular monofacial edges do not occur when β(1 − β) is a square. We can now Z-orient D by declaring F positive or negative according to whether res p η = −im or im. By the previous observation this is equivalent to saying that the sign of F is positive (resp. negative) when the integral of η along any of its (oriented) edges equals π (resp. −π). Of course when an edge is oriented according to the adjacent face the integral takes the opposite value, hence our rule assigns opposite signs to adjacent faces, as it should.
Conversely, let us now assume that D is Z-oriented. We want to show that a choice of β(1 − β) can be made around each point so that the expression yields a globally defined meromorphic 1-form. As noted at the beginning of this section the zeroes as well as the preimages of 1 under β all have even degree. This implies that the zeroes and poles of β(1 − β) are also of even degree and therefore there is a well-defined square root of this function (hence, exactly two) on any simply connected domain U .
1) (Choice of β(1 − β) on the interior of a face) Let F be a face of D, p ∈ S the pole of β in F and m its multiplicity. Let U be the interior of F . Since U is an open cell there are two well-defined holomorphic square roots of β(1 − β) on it. If F has positive (resp. negative) sign we choose the one for which the residue of the 1-form on U defined by the expression (3.1) is −im (resp. im). We shall denote this 1-form by η.
2) (Choice of β(1 − β) around the edges) We now need to make a choice of β(1 − β) on a small simply connected neighbourhood U k of each of the edges e k of a given face F . This is equivalent to making a choice on each U k of the 1-form defined by (3.1). If F has positive (resp. negative) sign we choose the 1-form η k that satisfies e k η k = π (resp. −π) with respect to the orientation on e k induced by the orientation of ∂F . Notice that this choice is independent of which of the two faces incident to e k we look at because if e k is oriented according to the other face G the integral e k η k will take the opposite value. Since the sign of G is the opposite to the sign of F we see that η k is also our choice from the perspective of the adjacent face G. We recall that, as was observed in the introduction, being Z-orientable implies the absence of monofacial edges. It remains to see that η = η k on each intersection U ∩ U k . Suppose, for instance, that F is positively oriented. In this case the choice of η we have made is the one fulfilling the identity where γ is any Jordan curve around the pole p lying inside the interior U ⊂ F . Since D does not have monofacial edges, the number of edges of F is exactly 2m. Now let γ approach the boundary ∂F by means of Jordan curves consisting of 2m segments γ k , each of which approaches the corresponding edge e k . Then, for each k the integral γ k η will get as close to one of the values ± = ±π as we wish, the sign depending on whether η = η k or η = −η k on U ∩ U k . But then one sees that in order for the identity to hold each integral γ k η must be close to π and so η must agree with η k on U ∩ U k .
Since being a square is preserved under the Galois action we deduce the following: It is well known that the action of the absolute Galois group on Belyi surfaces of each genus g ≥ 1 is faithful, that is, for each non-identity τ ∈ Gal(Q/Q) there is a Belyi surface S (in each genus g ≥ 1, see [4]) so that S τ is not isomorphic to S. In fact, the Galois action is faithful even when restricted to quasiplatonic Belyi surfaces (those admitting a regular Belyi function), see [5]. Proposition 6. Every Belyi surface admits a Z-orientable dessin. In particular, the absolute Galois group acts faithfully on the class of Z-orientable dessins.
Proof. Let β be a Belyi function on S, and consider the multiplicative inverse of the Klein j-modular function t(x) = 27x 2 (x − 1) 2 /4(x 2 − x + 1) 3 . Both t and t • β are Belyi functions [3] and one has the identity So, by Theorem 4, the Belyi function t • β on S defines a Z-orientable dessin.
We next present a useful characterization of Z-orientability of dessins in terms of their permutation representation. This is a criterion established by Zapponi in [11] for the so-called clean dessins, but one can easily see that it works for general dessins.
Proposition 7 (Zapponi's criterion). Let D be a dessin with permutation representation (σ 0 , σ 1 ) ∈ S n × S n . Then D is Z-orientable if and only if the rule such that the stabiliser subgroup Stab M (1) is contained in ker(ρ).
The fiber product enjoys the following universal property, written for our situation (see Example 2.25 in [7, page 30]): given two morphisms φ 1 , φ 2 : S → C such that Q 1 •φ 1 = Q 2 •φ 2 , there exists a morphism ψ : S → C such that φ j = π j •ψ. So, by taking φ 1 = β and φ 2 = h, this asserts the existence of a morphism P : S → C making the following diagram commutative where ω is the monodromy map and θ is defined by θ(x) = θ(y) = −1. It follows that Γ = ω −1 (stab M (1)) ⊂ ω −1 (ker ρ) = ker ρ • ω = ker θ = K and we have a commutative diagram of morphisms of Riemann surfaces The vertical arrow is a degree two regular covering of the Riemann sphere which ramifies over only two points, namely the quotient by the group G = ∆/K. After suitable identifications D/∆ ≃ C and D/K ≃ C, the projection β becomes the Belyi function associated with D, the group G corresponds to the group generated by the involution J(x) = 1/x, and the function π corresponds to π 1 (z) = (1 + z) 2 /4z. It follows that β = π 1 • P = (1 + P ) 2 /4P and a direct computation gives Remark 8. Concerning the use of fiber products in the above proof the following facts should be mentioned.
(1) The Z-orientability of the dessin induced by the Belyi pair (S, β) is equivalent to the reducibility of the fiber product In the case when it is irreducible the pair (S × (β,π1) C, Q 1 • β) is a Belyi pair corresponding to a Z-orientable dessin covering the original non-Zorientable dessin with degree two. Proposition 7 has an obvious immediate consequence: Corollary 10. Let D be a dessin with monodromy group M . If M does not contain an index two subgroup, then D is not Z-orientable. In particular, a dessin whose monodromy group is a simple group of order greater than two is never Z-orientable.
As the monodromy group M of a dessin is a two-generator group, the group M/M 2 , where M 2 is the subgroup generated by the squares of its elements, is a two-generator group of exponent dividing 2 and therefore isomorphic to 1, Z 2 or Z 2 × Z 2 . (All three cases arise, for instance when M has odd order, or when M is a dihedral group of order 2n for n odd or even, respectively.) This fact will be used in Section 5. Example 12. An interesting property of Z-orientability is that, unlike the monodromy group and other more sophisticated invariants, it can often be checked with a simple glance at the dessin. For instance the two dessins in Figure 1 have the same passport (4 2 ; 4 2 ; 4 2 ) and isomorphic monodromy groups (as can be checked using GAP [2]; see below) but they are not Galois conjugate because while the left-hand dessin D is not Z-orientable (edges 1, 3, 6 and 8 are monofacial; see the first paragraph of Section 3), the right-hand dessin D is, as the two faces miraculously share each of the eight edges. If (S, β) is the Belyi pair corresponding to the dessin D of this example, whose permutation representation is given by By Theorem 4, Z-orientability of dessins of genus 0 is encoded in the passport, as the necessary condition of possessing only even degree vertices is also sufficient to ensure Z-orientability in this case (when all the vertices have even degree all the zeroes and poles of the rational function β(1 − β) have even order, hence β(1 − β) is a square). On the other hand, as we have seen in the previous example, the Z-orientability of a dessin cannot be determined in general by its passport and its monodromy group.

An alternative point of view
Here we outline an alternative approach to the Z-orientability properties of dessins, relying more on the concepts and language of group theory and combinatorics. As explained in Section 2, a dessin D may be identified with a 5-tuple of the form (M, H, m 0 , m 1 , m ∞ ), where M is a finite group with generators m 0 , m 1 and m ∞ satisfying m 0 m 1 m ∞ = 1, and H is a subgroup of M with trivial core.
We will now give some alternative characterizations of Z-orientability, implicitly using this correspondence to translate concepts about dessins from one language to another. The following result completes the proof of Theorem 1. Recall that a permutation group is imprimitive if it preserves a non-trivial equivalence relation, in which case it permutes the equivalence classes (called blocks of imprimitivity, or simply blocks).
Proposition 13. For any dessin D, the following are equivalent (with the usual notation): (1) D is Z-orientable.
(2) The faces of D can be labelled + or − so that consecutive faces around each vertex have different labels.  Proof. The proof that each statement is equivalent to the next is straightforward, so we will just outline the basic ideas. For instance, (2) is essentially a restatement of the definition of Z-orientability. To show that (2) implies (3), one can transfer labels from faces to edges by following the orientation around the white vertices (or equivalently following the reverse orientation around the black vertices), and giving the sign on each face to the next edge; the converse is similar. For the equivalence of (3) and (4), having the same label is the required relation on edges. The subgroup N in (5) is the kernel of the action on blocks, and the two blocks in (4) are its orbits. The inclusion H ≤ N in (5) is equivalent to the covering in (6).
The subgroups H ≤ N ≤ M appearing here are the images in M of the subgroups Γ ≤ K ≤ ∆ under the epimorphism ω : ∆ → M defined earlier. As an application of this result, Proposition 7 is an immediate consequence of (1) being equivalent to (5). Similarly, the Galois invariance results in Section 3 follow immediately from the equivalence of (1) and (6), together with the fact that the absolute Galois group preserves degree, type and coverings. The two blocks for the Z-orientable dessin D in Example 12 are the sets {1, 2, 3, 4} and {5, 6, 7, 8}. The dessin in Figure 2 corresponds to the subgroup of the free group of rank 2 consisting of the words of even length in the standard generators. If we place this dessin in C with its white and black vertices at −1 and 1, and its face-centres at 0 and ∞, it corresponds to the Belyi function π 1 : C → C given by z → (z + 1) 2 /4z in the proof of Proposition 7. A quarter turn, with white and black vertices moved to −i and i, gives the Belyi function π 2 : z → (z 2 − 1)/4iz in the same proof.

The twist-invariant orientability type
One might want to avoid issues related to twists (see Example 12) by slightly modifying the notion of Z-orientability in the following way.
Definition 14. The twist-invariant orientability type of a dessin D is the number of dessins in the twist-related triple {D, D ′ , D ′′ } that are Z-orientable. We denote it by tot(D). Note that it is sufficient to restrict attention to D and the dessins D ′ and D ′′ obtained from it by cyclic permutation of the generators σ 0 , σ 1 and σ ∞ : the other three twist-related dessins are obtained from these by transposing vertex colours, which has no effect on Z-orientability.
Proposition 15. The twist-invariant orientability type is a Galois invariant, and tot(D) = 2 for every dessin D.
Proof. Using Theorem 4, one can easily show that if two of the three Belyi functions β, (β − 1)/β, 1/(1 − β) correspond to a Z-orientable dessin then the third one also does. Note that the three functions have product equal to 1, so if two of them are squares of meromorphic functions then so is the third. Obviously, Theorem 4 also shows that tot is a Galois invariant.
Remark 16. Note that for a dessin D with at least one face of odd degree, tot is exactly the same invariant as plain Z-orientability. In fact, if one face has odd degree, the dessins D ′ and D ′′ will each have a vertex of odd degree, so they cannot be Z-orientable. It is an immediate consequence of Corollary 10 that for regular dessins tot = 0 if and only if the monodromy group does not contain a subgroup of index 2. Also, we may observe that tot = 3 is equivalent to the condition that β and 1 − β are the squares of meromorphic functions (by Theorem 4 and the proof of Proposition 15), which can be restated as the existence of two meromorphic functions f and g on S such that f 2 + g 2 = 1 and β = f 2 .
In the following, we provide necessary and sufficient conditions, in terms of the monodromy group, for the Z-orientability properties of a dessin D and its twistrelated dessins D ′ and D ′′ .   Proof. Part (1) is just Corollary 11. In Case (2) there is a unique surjective homomorphism ρ : M → {±1} whose kernel is M 2 . There is exactly one pair (a, b) ∈ {(σ 0 , σ 1 ), (σ 1 , σ ∞ ), (σ ∞ , σ 0 )} with the property that ρ(a) = ρ(b) = −1. Let D ∈ {D, D ′ , D ′′ } be the dessin with permutation representation pair (a, b). Then, by Zapponi's criterion, the other two dessins are not Z-orientable, and the dessin D is Z-orientable if and only if H ≤ M 2 .
Example 18. Let D and D be the dessins in Example 12 and let M be their common monodromy group. Since M is a non-cyclic 2-group we have M/M 2 ∼ = Z 2 × Z 2 , so Case (3) of Theorem 17 applies. We saw earlier that D is not Zorientable, whereas D is. Since D ′ differs from D only by transposition of vertexcolours, it is also Z-orientable, so 1 ≤ tot(D) < 3. It follows from Proposition 15 that tot(D) = 1, so D ′′ is not Z-orientable. This is confirmed by checking that H ≤ M 2 , σ j only for j = 0. (These results, obtained by hand, were verified by using GAP.) As a consequence of Theorem 17, we obtain the following. In particular, for regular dessins the tot invariant is completely determined by the monodromy group.    (2), consider the non-Z-orientable uniform dessin D in Example 12. In this case H, isomorphic to Z 2 × Z 2 , is not a subgroup of M 2 , and M/M 2 ∼ = Z 2 × Z 2 . As the dessin D ′ differs from D in that example only by transposition of the vertex colours, and as it is Z-orientable, we have tot(D) = tot( D) = 1. As in this case H 0 = H ∩ M 2 ∼ = Z 2 has index two in H, the dessins D 0 and D 0 give two-fold (unbranched) coverings of the dessins D and D, respectively. These are uniform but non-regular dessins of type (4,4,4), passport (4 4 ; 4 4 ; 4 4 ) and genus 3. The dessin D 0 , with permutation representation is shown in Figure 4; the covering D 0 → D is given by the mapping e, e ′ → e of edge-labels for e = 1, . . . , 8. By Theorem 1(3), the partition of edges into those labelled e or e ′ shows that D 0 is Z-orientable.  In this last section we will prove Theorem 2 by constructing an infinite set of instances of four uniform but non-regular dessins with the same monodromy group and generating triple (as an abstract group), with isomorphic point stabilisers and automorphism groups and with the same passport, but with different Z-orientability properties, so that no two of them are conjugate under the absolute Galois group. This shows that in some circumstances Z-orientability can be used to distinguish dessins in separate Galois orbits when these other properties are insufficient for this purpose. We will describe all our dessins as tuples (M, H, m 0 , m 1 , m ∞ ), as explained in Section 2.
The construction starts by taking M = S n × S n for some n ≥ 3. As A n has index 2 in S n we have S 2 n ≤ A n , and as A n is generated by 3-cycles these subgroups are equal, so We will now construct four subgroups H i (i = 0, 1, 2, 3) of M , each one with trivial core, and will apply this lemma to the dessins D i = (M, H i , m 0 , m 1 , m ∞ ). The construction proceeds in several steps, as follows.
The rotation group Q of a cube is isomorphic to S 4 , through its faithful action on the four main diagonals, and its isometry group G has the form Q × I, where I ( ∼ = S 2 ) is generated by the antipodal isometry i. The faithful action of G on the faces of the cube gives an embedding of G in S 6 , with Q acting transitively. In addition to Q, there is another subgroup P ∼ = S 4 in G, also acting transitively on the faces: this is the subgroup Q ′ ∪ (Q \ Q ′ )i preserving the two tetrahedra inscribed in the cube, and acting as the isometry group of each. (Here Q ′ ( ∼ = A 4 ) is the commutator subgroup of Q.) Although P and Q are isomorphic subgroups of S 6 , they are not conjugate in S 6 : for instance, their elements of order 4 have cycle structures 2, 4 and 1, 1, 4 respectively. Having index 2 in G, both P and Q are normal in G, so their normalisers in S 6 contain G. They are not normal in S 6 , and G is a maximal subgroup of S 6 (it has index 15, and S 6 has no subgroups of index 3 or 5), so they have the same normaliser N S6 (P ) = N S6 (Q) = G. (Note that P and Q are not equivalent under the non-trivial outer automorphism of S 6 , which instead transposes them with intransitive subgroups of S 6 .) For any n ≥ 6 let A and B be subgroups of S n isomorphic to S 4 , acting as P or Q on a set of six points and fixing the remaining n − 6 points. Thus A and B are isomorphic but non-conjugate subgroups of index n!/4! in S n . It is easy to check that A ≤ A n , whereas B ≤ A n : for example, an element of order 4 in A or B is respectively even or odd.
is not. Thus these four dessins have different Z-orientability properties, so they lie in distinct Galois orbits.
We now consider the monodromy and automorphism groups of these dessins. For each i = 0, 1, 2, 3, the natural action of M on the cosets of the subgroup H i yields an embedding ϕ i : M → Symm(M/H i ) ∼ = S N which allows us to regard M as a transitive subgroup of S N . For each j = 0, 1, ∞ define σ j to be the permutation ϕ i (m j ) ∈ S N induced by m j , so that the triple (σ 0 , σ 1 , σ ∞ ) determines a dessin for each i.
By construction the four dessins D i all have the same monodromy group and generating triple, and have isomorphic point stabilisers and automorphism groups, so it remains only to show that they have the same passport. This is an immediate consequence of the following: Lemma 25. For each n ≥ 8 the dessins D i (i = 0, . . . , 3) are all uniform, having type (n − 1, n(n − 1), 2n) or (2(n − 1), n(n − 1), n) as n is odd or even.
Proof. It is sufficient to show that no non-identity power of any generator m j (j = 0, 1, ∞) of M is conjugate to an element of any subgroup H i , for in that case the dessins D i will all be uniform, with the stated type. This can be seen as follows: if in the cycle decomposition of one of the generators σ j = ϕ i (m j ) ∈ S N we had a cycle of length k less than its order, then σ k j would have to fix some point in {1, 2, . . . , N }, and by the transitivity of M would have to lie in a conjugate of the corresponding subgroup H i . Now by inspection of the cube each non-identity element of P or Q fixes either two or none of the six points it permutes, so each non-identity element of A or B fixes either n − 4 or n − 6 points of the n points permuted. On the other hand, a non-identity power of x, y or z fixes none, n − 2, or 1, so for n ≥ 8 a non-identity power of m j cannot be conjugate to an element of any H i . It follows that for odd n ≥ 9 and for each dessin D i , σ 0 has N/(n − 1) cycles of length (n − 1), while σ 1 has N/n(n − 1) cycles of length n(n − 1) and σ ∞ has N/2n cycles of length 2n, with a similar result for even n ≥ 8. In particular, these four dessins all have the same passport, as claimed. (And hence, of course, they have the same genus.) This completes the proof of Theorem 2.
In this case the dessins have type (14, 56, 8) and genus 1, 108, 801. The authors regret that they are unable to provide drawings of these dessins.
Remark 28. For each n ≥ 6, the four dessins D i all have the same minimal regular cover, namely the regular dessin with monodromy and automorphism group M = S n × S n and generating triple (m 0 , m 1 , m ∞ ). This has degree (n!) 2 , and is a regular 576-sheeted covering, with covering group H i ∼ = S 4 × S 4 , of each D i . These coverings are smooth for n ≥ 8, but branched for n = 6 and 7. This regular dessin may seem rather complicated, but it is simply the minimal common regular cover of the pair of planar dessins of degree n with monodromy group S n , acting naturally, and generating triples (y, z, x) and (z, x, y) given by the first and second coordinates of the triple (m j ) generating M . These twist-related dessins are shown on the left and right in Figure 5; the labelling of edges is obvious, and therefore omitted.