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ALMOST ABELIAN REGULAR DESSINS D’ENFANTS
RUBEN A. HIDALGO
Abstract. A regular dessin d’enfant, in this paper, will be a pair (S, β), where
Sis a closed Riemann surface and β:S→b
Cis a regular branched cover
whose branch values are contained in the set {∞,0,1}. Let Aut(S, β) be the
group of automorphisms of (S, β), that is, the deck group of β. If Aut(S, β)
is Abelian, then it is known that (S, β) can be defined over the field of ra-
tional numbers Q. In this paper we prove that, if Ais an Abelian group and
Aut(S, β)A⋊ Z2, then (S, β) is also definable over Q. Moreover, if AZn,
then we provide explicitly these dessins over Q.
1. Introduction and statement of results
A dessin d’enfant (or just a dessin), as defined by Grothedieck [G], corre-
sponds to a pair (X,D), where Xis a closed orientable surface and D⊂Xis
a bipartite graph (vertices are colored black or white and adjacent vertices have
different colors) such that X−Dconsists of a finite collection of topological discs
(called the faces of the dessin). The genus of (X,D) is by definition the genus of
X. Two dessins d’enfants, say (X1,D1) and (X2,D2), are said to be equivalent if
there exists an orientation preserving homeomorphism H:X1→X2inducing
an isomorphism, as bipartite graphs, between D1and D2(i.e., an isomorphism of
the graphs sending black (white) vertices to black (white) vertices).
A Belyi pair is a pair (S, β), where Sis a closed Riemann surface, called a
Belyi curve, and β:S→b
Cis a non-constant meromorphic map whose branch
values are contained in the set {∞,0,1}, called a Belyi map. The genus of (S, β) is
the genus of S. Two Belyi pairs, say (S1, β1) and (S2, β2), are said to be equivalent
if there is a biholomorphism F:S1→S2such that β2◦F=β1. If S1=S2=S
and β1=β2=β, then the above provides the definition of an automorphism
of (S, β). We denote by Aut(S) the full group of conformal automorphisms of
Sand by Aut(S, β) its subgroup of automorphisms of (S, β). We say that the
Belyi pair (S, β) is regular if βis a regular branch cover; in which case its deck
group is Aut(S, β). The group of M¨obius transformations keeping invariant the
set {∞,0,1}is the symmetric group on three letters S3generated by A(z)=1/z
and B(z)=1/(1 −z). If we have a (regular) Belyi pair (S, β) and M∈S3, then
2010 Mathematics Subject Classification. 14H57, 30F10, 05C65.
Key words and phrases. Dessin d’enfant, Riemann surface, Algebraic curve, Field of defini-
tion, Field of moduli. 1
2 RUBEN A. HIDALGO
(S,M◦β) is again a (regular) Belyi pair and, moreover, Aut(S, β)=Aut(S,M◦β).
In this paper we are interested in regular dessins d’enfants.
As defined above, dessins d’enfants are combinatorial (2-dimensional) ob-
jects and Belyi pairs are analytic objects. By the uniformization theorem, a dessin
d’enfant (X,D) determines a Belyi pair (S, β) (unique up to equivalence) so that
D=β−1([0,1]) (the black vertices being the preimages of 0 and the white ver-
tices being the preimages of 1). Conversely, a Belyi pair (S, β) defines a dessin
d’enfant as just described above. This provides a bijection between equivalence
classes of dessins d’enfants and equivalence classes of Belyi pairs; so we may
work indistinctly with dessins d’enfants and Belyi pairs. In this paper we con-
sider Belyi pairs as the objects under study.
A field of definition of a Belyi pair (S, β) is a subfield Kof Cfor which there
is an equivalent Belyi pair (C, η), where Cis a smooth complex algebraic curve
and ηis a rational map, both defined over K; we also say that (S, β) is definable
over K. Belyi’s theorem [B] asserts that every Belyi pair is definable over the
field of algebraic numbers Q. The field of moduli of (S, β) is the intersection of
all its fields of definition [K].
If a regular Belyi pair (S, β) has genus zero, then, up to conformal equiva-
lence, S=b
C(which is defined over Q) and Aut(S, β) is any of the finite groups
of M¨obius transformations (finite cyclic groups, dihedral groups, the alternating
groups A4,A5and the symmetric group S4). Explicit regular branched cover
maps, for each of these cases, are provided in [Ho]. It can be checked that all of
these are definable over Q.
Let us assume, from now on, that the regular Belyi pair (S, β) has genus g≥1.
In this case, the branch values of βare 0, 1 and ∞, say with branch orders k1,
k2and k3; we say that S/Hhas signature (0; k1,k2,k3) and that the associated
dessin d’enfant has type (k1,k2,k3). By the Riemann-Hurwitz formula, the con-
dition g≥1 is equivalent to have k−1
1+k−1
2+k−1
3≤1 (strictly less than 1 if and
only if g≥2). As a consequence of the uniformization theorem, there is a surjec-
tive homomorphism θ:Γ→Aut(S, β), with a torsion free kernel ker(θ), where
Γ = hx,y:xk1=yk2=(yx)k3=1iis a triangular Kleinian group uniformizing the
orbifold S/Aut(S, β) and ker(θ) uniformizing S(the converse holds). In particu-
lar, Aut(S, β) is generated by two elements, say aand b, with aof order k1,bof
order k2and c=(ab)−1of order k3. This, and the fact that any two generators of
a dihedral group are either both of order two or one of order two with product of
order two, ensures that Aut(S, β) cannot be isomorphic to a dihedral group.
In [W] Wolfart proved that (S, β) and Scan both be defined over their corre-
sponding fields of moduli (this fact can also be obtained from D`ebes-Emsalem’s
ALMOST ABELIAN REGULAR DESSINS D’ENFANTS 3
results in [DE]). In [Hi] it was noticed that if Aut(S, β) is Abelian, then these
two fields are equal to the field of rational numbers Q. We may wonder for other
cases ensuring a regular Belyi pair (S, β) to be definable over Q. A class of groups
which are near to be Abelian groups (in some rough sense) are the semi-direct
products A⋊B, where Aand Bare both Abelian groups. In [SW] Streit-Wolfart
studied the family of those regular Belyi pairs with Aut(S, β)=ha,b:ap=bq=
1,bab−1=amiZp⋊ Zq(i.e. AZpand BZq), where p>3 and q>3
are primes and mq≡1 mod p, and they exhibit explicit curves and the corre-
sponding fields of moduli (which result to be different from Q). In this paper we
consider the case Aut(S, β)A⋊ Z2, where Ais an Abelian group. The case
Aut(S, β)A⋊ Z3will be considered elsewhere (see also Remark 2.2). Our first
result is the following.
Theorem 1.1. Let (S, β)be a regular Belyi pair of genus g ≥1, with Aut(S, β)
A⋊ Z2. If A is an Abelian group, then (S, β)is definable over Q.
If in Theorem 1.1 AZn, that is, Aut(S, β)=ha,b:an=b2=baba−m=
1iZn⋊ Z2, where n∈ {2,3, ...},m∈ {1,2, ..., n−1},m2≡1 mod nand
gcd(n,m)=1, then next result describes explicit models over Q. As already
noticed above, the case m=n−1 (the dihedral case) is not possible.
Theorem 1.2. Let (S, β)be a regular Belyi pair of genus g ≥1, with Aut(S, β)=
ha,b:an=b2=baba−m=1i, where n ∈ {2,3, ...}, m ∈ {1,2, ..., n−2},
m2≡1 mod n and gcd(n,m)=1. Then the following hold.
(1) There exist integers α, ρ, γ ∈ {1, ..., n−1}and non-negative integers
ϑ1, ϑ2, ϑ3, satisfying
(1.1) gcd(n, α, ρ, γ)=1;
(1.2) 1 +γ−ρ=m;
(1.3) (α+ρ+γ)(2 +γ−ρ)≡0mod n;
(1.4) α(γ−ρ)=nϑ1;
(1.5) (ρ−1)(γ−ρ)=nϑ2;
(1.6) (γ+1)(γ−ρ)=nϑ3; and
(1.7) (−1)(α+ρ+γ)(2+γ−ρ)/n=(−1)ϑ1+ϑ2+ϑ3,
so that (S, β)is equivalent to the regular Belyi pair (C, η), where
C:yn=xα(x−1)ρ(x+1)γ,
η:C→b
C: (x,y)7→ x2,
and Aut(C, η)=ha,biwith
a(x,y)=(x, ωy),b(x,y)= −x,δy1+γ−ρ
xϑ1(x−1)ϑ2(x+1)ϑ3!, δ =(−1)(α+ρ+γ)/n, ω =e2πi/n.
4 RUBEN A. HIDALGO
(2) If ξ∈Zis so that (ξ−1)n< α +ρ+γ≤ξn and we set η=ξn−α−ρ−γ
(setting gcd(n,0) :=n), then
g=1+n−(1/2) gcd(n, α)+gcd(n, ρ)+gcd(n, γ)+gcd(n, η).
(3) If α+ρ+γ≡0 mod n, then S/β has signature (0; 2,p,2q), where
p=n/gcd(n, ρ)=n/gcd(n, γ)and q =n/gcd(n, α).
(4) If α+ρ+γ.0 mod n, then S/β has signature (0; p,2q,2u), where p
and q are as above and u =n/gcd(n, η).
In the particular case m=1 (the Abelian situation), Theorem 1.2 can be
written as follows.
Corollary 1.3. Let (S, β)be a regular Belyi pair of genus g ≥1, with deck
group Aut(S, β)Zn×Z2. Then there exist integers α, ρ ∈ {1, ..., n−1}, with
gcd(n, α, ρ)=1and α+2ρ≡0 mod n, so that (S, β)is equivalent to (C, η),
where
C:yn=xα(x2−1)ρ,
η:C→b
C: (x,y)7→ x2.
Moreover, Aut(C, η)=ha,biwith
a(x,y)=(x, ωy),b(x,y)=(−x, δy), δ =(−1)(α+2ρ)/n, ω =e2πi/n.
Proof. This follows from Theorem 1.2 taking m=1; so γ−ρ=0, ϑ1=ϑ2=
ϑ3=0, gcd(n, α, ρ)=1 and α+2ρ≡0 mod n.
2. Proof of Theorems 1.1 and 1.2
Let us fix a regular Belyi pair (S, β), of genus g≥1, with Aut(S, β)A⋊ Z2,
where Ais an Abelian group. Let b∈Aut(S, β) be the conformal automorphism
that generates the Z2component.
The quotient orbifold S/Ahas a signature (h;n1, ..., nr), that is, its underlying
Riemann surface is a closed Riemann surface, say R, of genus hand it has ex-
actly rcone points of orders n1,..., nr, respectively. Let P:S→Rbe a regular
branched cover with Aas its deck group.
As Ais a normal subgroup of Aut(S, β), there is a conformal automorphism
bof R, of order two, such that b◦P=P◦b. The involution bpermutes the cone
points of S/Aand it respects their orders. Let Q:R→b
Cbe a regular branched
cover with deck group hbiand so that Q◦P=β.
Since S/Aut(S, β) is the Riemann sphere b
Cwith cone points at ∞, 0 and
1, it follows that R/hbiis b
Cand that its cone points (that is, the branch values
of Q) are contained in the set {∞,0,1}. So, (R,Q) is a regular Belyi pair with
Aut(R,Q)=hbi.
ALMOST ABELIAN REGULAR DESSINS D’ENFANTS 5
By the Riemann-Hurwitz formula, the number of fixed points of the involu-
tion bis even; say 2s.
Claim 2.1. s=1, h =0and r ∈ {3,4}.
Proof. We first prove that s=1. In fact, if s=0, then necessarily h≥1 (since on
the Riemann sphere every involution has two fixed points). Now, the Riemann-
Hurwitz formula ensures that R/hbihas positive genus, a contradiction. If s≥2,
then Qwill have 2s≥4 branch values, again a contradiction.
Now, as bhas exactly two fixed points and R/hbihas genus zero, it follows
from the Riemann-Hurwitz formula that h=0.
The above states that the signature of S/Ais of the form (0; n1, ..., nr). Since
the genus of Sis at least one, it again follows from the Riemann-Hurwitz formula
that r≥3. Now, as the cone points of S/Aare permuted by the involution band
S/Aut(S, β) has exactly three cone points, it follows that r∈ {3,4}.
The above claim ensures that R=b
Cand that bis a M¨obius transformation of
order 2. So, up to composition at the left of Pby a suitable M¨obius transforma-
tion, we may assume that b(x)=−x; so Q(x)=x2.
If r=3, then one of the cone points is a fixed point of band the other two are
permuted by b.
If r=4, then two of the cone points are fixed by band the other two are
permuted by it.
Up to composition at the left of Pby a M¨obius transformation of the form
T(x)=dx, for a suitable d∈C− {0}, we may also assume that the cone points
of S/Aare ±1 (the ones which are permuted by b), 0 (and ∞for r=4).
2.1. Proof of Theorem 1.1.
2.1.1. If r=3, then the branch values of P:S→b
Care given by the points ±1
and 0. If M(x)=(1 −x)/(1 +x), then PM=M◦Pis a Belyi map with Aas its
deck group. By the results in [Hi], we may assume both Sand PMto be defined
over Q. The induced involution by b, under PM, is b
b(x)=M◦b◦M−1(x)=1/x.
The two-folded branch cover b
Q(x)=(1 −x)2/(1 +x)2has its deck group hb
biand
Q=b
Q◦M. It follows that β=b
Q◦PMis defined over Q.
2.1.2. If r=4, then we may proceed as follows (see [Hi]). Let µ≥2 be the least
common multiple between all the orders of the four cone points (∞, 0, 1 and −1)
of S/A. Let us consider the generalized Fermat curve [GHL] (a closed Riemann
6 RUBEN A. HIDALGO
surface of genus gC=(µ−1)(µ2+µ−1) ≥5)
C=
xµ
1+xµ
2+xµ
3=0
−xµ
1+xµ
2+xµ
4=0
⊂P3
C.
The group K=ha1,a2,a3iZ3
µ, where ajis multiplication at the xj-coordinate
by e2πi/µ, is a group of conformal automorphisms of C. If L:C→b
Cis defined
by L([x1:x2:x3:x4]) =−(x2/x1)µ, then Lis a regular branched cover with K
as its deck group and whose branch values are ±1, 0 and ∞, each one with order
µ.
If Γ = hy1,y2,y3,y4:yµ
1=yµ
2=yµ
3=yµ
4=y1y2y3y4=1iis a Fuchsian group
acting on the hyperbolic plane H2so that H2/Γ = C/K, then H2/Γ′=C, where
Γ′is the derived subgroup of Γ.
It follows that there is a normal subgroup ΓSof Γ(containing Γ′) whose uni-
formized orbifold H2/ΓShas underlying Riemann surface structure isomorphic
to S(and A= Γ/ΓS). In particular, there is a subgroup K0= ΓS/Γ′of Kso that
the underlying Riemann surface structure of C/K0is isomorphic to S.
In [Hi, Section 6] we described (using geometric invariant theory) how to
compute a curve model Efor C/K0(we don’t need the explicit form). By [Hi,
Lemma 5.1.] such a curve E, and the regular branched cover U:C→E(with
deck group K0), are both defined over Q. As L=P◦Uwe may see that Pis also
defined over Qand, in particular, that β=Q◦Pis defined over Q.
Remark 2.2. If Aut(S, β)A⋊ Z3, then one may try to follow the same ideas
as in the above proof. We will have the conformal automorphism b, of order 3,
of the quotient orbifold S/Ainduced by the Z3component of Aut(S, β). As b
permutes the cone points of S/Aand (S/A)/hbi=S/Aut(S, β), we necessarily
have that S/Ais either of genus zero or of genus one. If S/Ahas genus zero,
then we may assume that b(z)=e2πi/3zand that the cone points are inside the set
{∞,0,1,e2πi/3,e4πi/3}. In this case, it is not clear to ensure that Scan be defined
over Q. If the quotient S/Ahas genus one, then bmust have exactly three fixed
points; in other words, the Riemann surface structure of S/Ais given by the
curve C:y3=x(x−1) and cone points being (0,0), (1,0) and ∞. Let us consider
Q:C→b
Cis defined by Q(x,y)=x(a regular branched cover with hbias deck
group) and a regular branched cover map P:S→C, with deck group A, then
β=Q◦P:S→b
C, up to post-composition with a M¨obius transformation in S3.
In order to see if the result in Theorem 1.1 holds or not for this case, we need
to check if it is possible to find an algebraic curve for Sand a regular branched
cover P:S→Cboth defined over Q.
ALMOST ABELIAN REGULAR DESSINS D’ENFANTS 7
2.2. Proof of Theorem 1.2. Let a∈Aut(S, β) be a conformal automorphism
that generates the cyclic group A=Zn, that is, Aut(S, β)=ha,bi.
If r=3, then (as the involution bpermutes two of the cone points and pre-
serves the orders) the signature of S/haimust of the form (0; p,p,q) (where p
and qare divisors of n) and the signature of S/Aut(S, β) is (0; 2,p,2q). The two
cone points ±1 of S/Ahave order pand the other cone point 0 has order q.
If r=4, then the signature of S/haimust be of the form (0; p,p,q,u) (where
p,qand uare divisors of n) and the signature of S/Aut(S, β) is (0; p,2q,2u). The
cone points ±1 have order p, the cone point 0 has order qand the cone point ∞
has order u.
It follows from [BW] that Scan be described by a cyclic n-gonal curve of the
form
C:yn=xα(x−1)ρ(x+1)γ,
where α, ρ, γ ∈ {1, ..., n−1}and gcd(n, α, ρ, γ)=1.
We should note that r=3 if and only if α+ρ+γ≡0 mod n and that
r=4 otherwise. Moreover, also from [BW], p=n/gcd(n, ρ)=n/gcd(n, γ) and
q=n/gcd(n, α) and, if r=4, then u=n/gcd(n, η), where η=ξn−α−ρ−γand
ξ∈Zis so that (ξ−1)n< α +ρ+γ≤ξn(setting gcd(n,0) :=n).
In this model, we have P(x,y)=x,a(x,y)=(x, ωy), where ω=e2πi/n, and
(as Q(x)=x2)β(x,y)=x2.
As b(x)=−xand P◦b=b◦P, it follows that the involution bmust be of the
form
b(x,y)=
−x, δy x−1
x+1!γ−ρ
n
, δn=(−1)α+ρ+γ.
We will distinguish the cases (i) ρ=γand (ii) ρ,γ.
2.3. If γ=ρ, then ϑ1=ϑ2=ϑ3=0,
C:yn=xα(x−1)ρ(x+1)ρ,
and b(x,y)=(−x, δy). As b2is the identity, it follows that δ2=1, that is,
(−1)2(α+2ρ)/n=1, from which we see that nnecessarily divides α+2ρ. In this
case m=1 and Aut(S, β)Zn×Z2.
2.4. If γ,ρ, then we may assume without loss of generality that ρ≤γ. As
x−1=yn
xα(x−1)ρ−1(x+1)γ
we have
b(x,y)= −x,δy1+γ−ρ
xα(γ−ρ)/n(x−1)(ρ−1)(γ−ρ)/n(x+1)(γ+1)(γ−ρ)/n!.
8 RUBEN A. HIDALGO
It follows from the above the existence of non-negative integers ϑ1, ϑ2, ϑ3so
that
α(γ−ρ)=nϑ1; (ρ−1)(γ−ρ)=nϑ2,(γ+1)(γ−ρ)=nϑ3.
In this way
b(x,y)= −x,δy1+γ−ρ
xϑ1(x−1)ϑ2(x+1)ϑ3!
As b2is the identity, the equality
(x,y)=b2(x,y)=
x,δ2+γ−ρ(−1)ϑ1+ϑ2+ϑ3yy(1+γ−ρ)2−1
xϑ1(2+γ−ρ)(x−1)ϑ3+ϑ2(2+γ−ρ)(x+1)ϑ3+ϑ2(2+γ−ρ)
,
ensures that
y(1+γ−ρ)2−1=xϑ1(2+γ−ρ)(x−1)ϑ3+ϑ2(2+γ−ρ)(x+1)ϑ3+ϑ2(2+γ−ρ)
and
δ2+γ−ρ=(−1)ϑ1+ϑ2+ϑ3.
In particular, nnecessarily divides (α+ρ+γ)(2 +γ−ρ) and bab =a1+γ−ρ,
that is, m=1+γ−ρ.
The formula for gis just a consequence of the Riemann-Hurwitz formula (see
also [BW]).
Remark 2.3. As already noted in the Introduction, there is not a regular Belyi
pair (S, β) of genus at least one with Aut(S, β) isomorphic to a dihedral group.
This also follows directly from the first part of the proof of Theorem 1.2. In fact,
let us assume there is a regular Belyi pair (S, β), with Aut(S, β)=ha,b:an=
b2=1,bab =a−1i. The involution bhas as one of its fixed points a cone point
of S/hai. This means that there is some ak(where k∈ {1, ..., n−1}) and some alb
(where l∈ {0,1, ..., n−1}) so that both of them have a common fixed point. This
asserts that hak,albishould be a cyclic group (the stabilizer of any point of Sin
Aut(S) is known to be a cyclic group), a contradiction.
3. AnExample in genus two
Let us describe those regular Belyi pairs (S, β) with
Aut(S, β)=ha,b:a8=b2=1,bab =a3iZ8⋊ Z2.
By Theorem 1.2, taking n=8 and m=3, we know that there are integers
α, ρ, γ ∈ {1, ..., 7}and non-negative integers ϑ1, ϑ2, ϑ3so that gcd(8, α, ρ, γ)=1,
ρ≤γ,γ−ρ=m−1=2, α+ρ+γis even, α=4ϑ1,ρ−1=4ϑ2and γ+1=4ϑ3,
such that (S, β) is equivalent to (C, η), where η(x,y)=x2and
C:y8=xα(x−1)ρ(x+1)γ.
ALMOST ABELIAN REGULAR DESSINS D’ENFANTS 9
By checking at all possibilities, we only obtain the following two cases
(α, ρ, γ)∈ {(4,1,3),(4,5,7)},
that is, Cmust be one of the following two curves of genus 2
C1:y8=x4(x−1)(x+1)3,(α, ρ, γ)=(4,1,3),
C2:y8=x4(x−1)5(x+1)7,(α, ρ, γ)=(4,5,7).
The group Aut(Cj, η) is generated by
a(x,y)=(x, ωy) (ω=eπi/4)
and
b(x,y)=
−x,−y3
x(x+1)!for C1
−x,y3
x(x−1)(x+1)2!for C2
In both cases, r=3, p=8, q=2, Cj/haihas signature (0; 2,8,8) and the
regular Belyi pair (S, β) has type (0; 2,4,8).
If we multiply the tuple (4,1,3) by 5 (and taking congruence module 8) we
obtain the tuple (4,5,7), that is, C1and C2are isomorphic. In fact, this also
follows from the fact that, up to isomorphisms, there s a unique Riemann surface
of genus 2 whose reduced group of automorphisms contains a group of order
8 (the quotient of Aut(S, β) by the cyclic group generated by the hyperelliptic
involution) [Ig]. That surface has as full reduced group the symmetric group S4
and it is described by the hyperelliptic curve
E:w2=u(u4−1).
The Belyi pair (S, β) is equivalent to (E, θ), where
θ(u,w)=(u8−2u4+1)/(−4u4)
and Aut(E, θ) is generated by the element of order 8
A(u,w)=(iu,√i w)
and the involution
B(u,w)=(i/u,i√i w/u3).
Acknowledgements. The author is deeply grateful to the referee for the various
comments and suggestions to the first draft of this article. This research was
partly supported by project Fondecyt 1110001 and UTFSM 12.13.01.
10 RUBEN A. HIDALGO
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Verlag Stuttgart, Stuttgart, 2006.
Departamento de Matem ´atica, Universidad T´ecnica Federico Santa Mar´ıa. Casilla 110-V,
Valpara´ıso, Chile
E-mail address:ruben.hidalgo@usm.cl