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QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m
ARE DEFINABLE OVER Q
RUB ´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
Abstract. It is well known that every closed Riemann surface Sof genus g2, admitting a group Gof conformal
automorphisms so that S/Ghas triangular signature, can be defined over a finite extension of Q. It is interesting to
know, in terms of the algebraic structure of G, if Scan in fact be defined over Q. This is the situation if Gis either
abelian or isomorphic to AoZ
2, where Ais an abelian group. On the other hand, as shown by Streit and Wolfart,
if GZpoZ
qwhere p,q>3 are prime integers, then Sis not necessarily definable over Q. In this paper, we
observe that if GZ2
2oZ
mwith m3, then Scan be defined over Q. Moreover, we describe explicit models
for S, the corresponding groups of automorphisms and an isogenous decomposition of their Jacobian varieties as
product of Jacobians of hyperelliptic Riemann surfaces.
1. Introducci´
on
As it was defined by Grothendick in [8], a dessin d’enfant of genus gis a bipartite map on a closed
orientable surface of genus g. The dessin d’enfant induces a unique, up to isomorphism, Riemann surface
structure Stogether with a non-constant meromorphic map :S!b
Cwhose branch values are contained in
the set {1,0,1};Sis called a Belyi curve, a Belyi map and (S,) a Belyi pair. Conversely, as a consequence
of the uniformization theorem, each Belyi pair (S,) induces a dessin d’enfant (the bipartite map is provided
by the preimage under of the closed interval [0,1]).
A Belyi pair (S,) (and the corresponding dessin d’enfant) is called regular (in which case Sis called a
quasiplatonic curve) if is a regular branched cover, that is, if there is a group of conformal automorphisms
of Sbeing the deck group of (see, for instance, [4,21] for more details). It is well known that a finite group
Gcan be seen as the deck group of a Belyi pair (we say that the action has triangular signature) if and only if
it can be generated by two elements [21].
Two Belyi pairs (S1,
1) and (S2,
2) are called isomorphic if there is an isomorphism (a biholomorphic
map) h:S1!S2such that 1=2h. Let us note that if (S1,
1) and (S2,
2) are regular Belyi pairs with
respective deck groups G1and G2, then the isomorphism hconjugates G1onto G2.
As a consequence of Belyi’s theorem [1], each Belyi pair can be defined over the field of algebraic numbers
Q,that is, there is an isomorphic Belyi pair (C,), where C(as an irreducible algebraic curve) and (as a
rational map) are defined over Q. This fact permits to define an action of the absolute Galois group Gal(Q/Q)
on Belyi pairs (or dessins d’enfants), as follows. Let P1,...,Prbe polynomials (with coefficients in Q)
defining C, that is, Cis the set of common zeroes of these polynomials. Each 2Gal(Q/Q) provides
new polynomials P
1,...,P
r(where P
jis obtained from Pjby applying to its coefficients). These new
polynomials define a new irreducible algebraic curve C. Similarly, we may apply to the coefficients of
and at the end we obtain a new Belyi pair (C,
). It is well known that the absolute Galois group acts
faithfully. Recently, Gonz´
alez-Diez and Jaikin-Zapirain [7] proved that the absolute group acts faithfully on
regular Bely pairs (even at the level of quasiplatonic curves).
2000 Mathematics Subject Classification. 14H37, 14H55, 14H25, 14H30, 30F10.
Key words and phrases. Riemann surfaces, regular Belyi pairs, dessins d’enfants, Galois action.
Partially supported by Fondecyt Project 1150003, Postdoctoral Fondecyt Projects 3160002 and 3140050, Beca Chile Fellowship for
Postdoctoral Research and Project Anillo ACT 1415 PIA-CONICYT.
1
2 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
The fixed points of the absolute group action are provided by those Belyi pairs which can be defined over
Q. It is a difficult task to decide if a given Belyi pair (or dessin d’enfant) can or cannot be definable over
Q. In the case of regular ones, some answers are known in terms of the corresponding deck group G. For
instance, if either Gis an abelian group or a semidirect product AoZ
2, where Ais abelian group, then the
corresponding regular Belyi pair can be defined over Q(see [9] and [10]). On the other hand, in [20] it was
noted that if G=ha,b:ap=bq=1,bab1=aniZpoZ
q, where p,q>3 are prime integers and nq⌘1
mod p, then the regular Belyi pair is not necessarily definable over Q.
In this paper we consider regular Belyi pairs (S,) with deck group GZ2
2oZ
mwhere m2. As
previously noted, for the case m=2 these are definable over Q. So we only need to take care of the case
m3. Also, as the abelian situation is also definable over Q, we assume Gto be non-abelian. Theorem 1
asserts that for these left cases such regular Belyi pairs can be defined over Q.
We are also able to construct explicit rational models of these pairs, their full groups of conformal au-
tomorphisms and isogenous decompositions of their Jacobian varieties as a product of Jacobian varieties of
hyperelliptic Riemann surfaces.
Acknowledgments. The authors are very grateful to Professor Anita Rojas for sharing her MAGMA routines
with us; they were very useful for the calculations in the last section of this paper. The authors would also
like to thank the referee for his/her valuable suggestions.
2. Main results
We consider regular Belyi pairs (S,G), where GZ2
2oZ
m,m3, and Gnon-abelian.
2.1. Signatures. Before we proceed to our main result, we first describe the possible signatures for the
quotient orbifold S/G.
Proposition 1. Let (S,)be a regular Belyi pair of genus g 2admitting G Z2
2oZ
m, where m 3and G
non-abelian, as its deck group. Then the possible signatures for the quotient S/G are:
(1) (0; 2,2q,4q)if m =2q and q 3is odd, or
(2) (0; 2,m,m)if m 6is either divisible by 3or by 4.
Proof. It is not difficult to see that we only have two cases to consider (up to automorphisms).
(i) G=ha,b,t:a2=b2=(ab)2=tm=1,tat1=a,tbt1=abi,
(ii) G=ha,b,t:a2=b2=(ab)2=tm=1,tat1=b,tbt1=abi.
In case (i), as tat1=a,tbt1=ab and tabt1=b, the integer mmust be even. Moreover, in this situation
G=hb,t:b2=tm=[t,b]2=1,(tb)2=(bt)2i,
where [x,y]=xyx1y1. So it can be seen that bt has order mif mis divisible by 4 and order 2motherwise;
in particular, S/Ghas signature (0;2,m,m) if mis divisible by 4 and (0; 2,m,2m) if m=2qwith q3 odd.
In case (ii), as tat1=b,tbt1=ab and tabt1=a, the integer mmust be divisible by 3. Moreover, in this
situation
G=ha,t:a2=tm=[a,t]2=1,t3=(at)3i,
so it can be seen that at has order mand that S/Ghas signature (0; 2,m,m). ⇤
2.2. Main theorem. Our main result provides the explicit algebraic descriptions of S, its full group of con-
formal automorphisms Aut(S) and an isogenous decomposition of its Jacobian variety JS as product of
Jacobian varieties of hyperelliptic curves.
Theorem 1. Let m 3and !m=exp(2⇡i/m).Let (S,)be a regular Belyi pair of genus g 2admitting a
non-abelian semidirect product G =ha,biohtiZ2
2oZ
mas its deck group. Then (S,)is definable over Q
and the following holds.
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m3
(1) If S/G has signature (0; 2,2q,4q), where m =2q and q 3is odd, then g =2(q1) and the Belyi
pair (S,)is unique up to isomorphisms. Moreover,
(a) S can be described by the algebraic curve
(y2=xq1
z2=xm1)⇢C3,
the Belyi map corresponds to (x,y,z)=xmand
a(x,y,z)=(x,y,z),b(x,y,z)=(x,y,z),t(x,y,z)=(!mx,(iz)/y,z).
(b) The group G is the full group of conformal automorphisms of S .
(c) The Jacobian variety JS is isogenous to (JS b)4, where
Sb:y2=xq1.
(2) If S/G has signature (0; 2,m,m), where m 6is either divisible by 3or by 4, then g =m3and the
following holds.
(a) If m =3l, l 2, is not divisible by 12, then (S,)is unique up to isomorphisms. Moreover,
(i) S is described by the algebraic curve
(y2=(xl1)(xl!2
3)=x2l+!3xl+!2
3
z2=(xl!3)(xl!2
3)=x2l+xl+1)⇢C3,
the Belyi map corresponds to (x,y,z)=xmand
a(x,y,z)=(x,y,z),b(x,y,z)=(x,y,z),t(x,y,z)=0
B
B
B
B
@!mx,!3z,!3yz
xl!2
31
C
C
C
C
A.
(ii) The Riemann surface S has the extra automorphism
u(x,y,z)=0
B
B
B
B
@1
x,!3yz
xl(xl!2
3),z
xl1
C
C
C
C
A.
(iii) The group of conformal automorphisms Aut(S)is generated by a,b,t and u. In fact,
Aut(S)=ht,u:u4=tm=(ut)2=1,t3=(u2t)3,([t1,u]u1)2=1i,
has order 8m and S/Aut(S)has signature (0; 2,4,m).
(iv) The Jacobian variety JS is isogenous to (JS a)3, where
Sa:z2=(xl!3)(xl!2
3)=x2l+xl+1.
In fact,
JS ⇠(JSa,1)3⇥(JS a,2)3,
where
Sa,1:w2
1=(1 v1)l+2
l
X
j=0 2l
2j!vj
1,
and
Sa,2:w2
2=v20
B
B
B
B
B
B
@(1 v2)l+2
l
X
j=0 2l
2j!vj
21
C
C
C
C
C
C
A.
(b) If m =4l, l 2, is not divisible by 12, then (S,)is unique up to isomorphisms. Moreover,
4 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
(i) S is described by the algebraic curve
(y2=x2l1
z2=xm1
in C3,the Belyi map corresponds to (x,y,z)=xmand
a(x,y,z)=(x,y,z),b(x,y,z)=(x,y,z),t(x,y,z)= !mx,iz
y,z!.
(ii) The Riemann surface S has the extra automorphism
u(x,y,z)= 1
x,iy
xl,iz
x2l!.
(iii) The group of conformal automorphisms Aut(S)is generated by a,b,t and u. In fact,
Aut(S)=ht,u:u4=tm=(tu)2=[u2,t]2=[u2,tut1]=1,(tu2)2=(u2t)2i
has order 8m and S/Aut(S)has signature (0; 2,4,m).
(iv) The Jacobian variety JS is isogenous to the product of JS a,2⇥(JSb)3, where
Sa,2:w2
2=v2(v2l
21),Sb:y2=x2l1.
(c) If m is divisible by 12, then there are exactly two non-isomorphic pairs (S,); they are alge-
braically represented as in (2a) and (2b) above.
(d) In any of the above cases (2a),(2b)and (2c), the subgroup ha,biis the unique subgroup (so a
normal subgroup) of Aut(S)isomorphic to Z2
2,Aut(S)/ha,biDm, and S/ha,biis the Riemann
sphere with exactly m cone points, each one of order two.
3. Some remarks concerning Theorem 1
3.1. Equations over Q.The provided curves in Theorem 1, with the only exception of the case (2a), are
defined over Q. In the left case the provided curve is defined over a degree two extension of Q; namely
Q(!3).However, the uniqueness property asserts that it is definable over Q. In this case, we may follow the
computational method presented in [11] to find a rational model (this is done in Section 5).
3.2. Projective models. Projective models associated to the affine ones provided in Theorem 1, are the
following:
(1) In case (1),given by the curve
ˆ
C:(y2wq2=xqwq
z2w2q2=x2qw2q)⇢P3
with Belyi map ([x:y:z:w]) =(x/w)m. Moreover,
(a) If (p)2Cis not a 2q-root of unity, then ˆ
Cis smooth at p.
(b) If (p)2Cis a q-root of unity, then pis a node of ˆ
C(that is, locally looks like two cones glued
at a common vertex).
(c) If (p)2Cis a 2q-root of unity but not a q-root of unity, then pis a singular point of ˆ
Cof cone
type (i.e., locally looks as a topological disc).
(d) Above 1we have two points on ˆ
C; each one being a singular point of cone type (similarly as
above).
In particular, a smooth model of ˆ
Cis obtained by separating at its qnodes by a blowing-up process.
(2a) In case (2a),given by the curve
ˆ
C:(y2w2l2=(xlwl)(xl!2
3wl)=x2l+!3xlwl+!2
3w2l
z2w2l2=(xl!3wl)(xl!2
3wl)=x2l+xlwl+w2l)⇢P3
and the Belyi map corresponds to ([x:y:z:w]) =(x/w)m. Moreover,
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m5
(a) Observe that ˆ
Cis reducible as it contains a projective line Lat infinity. The surface Scorre-
sponds to the union of Cwith certain 4 points of L.
(b) If (p)2Cis not a 3l-root of unity, then ˆ
Cis smooth at p.
(c) If (p)2Cis a l-root of unity or a l-root of !2
3, then pis a node of ˆ
C(that is, locally looks like
two cones glued at a common vertex).
(d) If (p)2Cis a l-root of unity or a l-root of !3, then pis a singular point of ˆ
Cof cone type (i.e.,
locally looks as a topological disc).
The di↵erence of the two equations of Cpermits to obtain xl:
(!31)xly2wl2+z2wl2(2 +!3)wl=0
Now, using this value of xlin the first equation of Cpermits to obtain x2l:
(1 !3)x2ly2w2l2+!3z2w2l2!3(1 !3)w2l=0
and this first equation is the same as
y4+z42y2z23z2w2+3(1 +!3)y2w2=0
In this way, the above affine curve can be written as follows:
C:8
>
>
>
<
>
>
>
:
y4+z42y2z23z2+3(1 +!3)y2=0
(1 !3)x2ly2+!3z2!3(1 !3)=0
(!31)xly2+z2(2 +!3)=0
9
>
>
>
=
>
>
>
;⇢C3
and its projectivization as
ˆ
C:8
>
>
>
<
>
>
>
:
y4+z42y2z23z2w2+3(1 +r)y2w2=0
(1 !3)x2ly2w2l2+!3z2w2l2!3(1 !3)w2l=0
(!31)xly2wl2+z2wl2(2 +!3)wl=0
9
>
>
>
=
>
>
>
;⇢P3
which is now irreducible. This has two singular points at infinity, given by the two points [0 : 1 :
±1 : 0], each one, after desingularization, produces two smooth points (i.e., we obtain the four points
at infinity).
(2b) In case (2b),given by the curve
ˆ
C:(y2w2l2=x2lw2l
z2w4l2=x4lw4l)⇢P3
with Belyi map ([x:y:z:w]) =(x/w)m. Moreover,
(a) If (p)2Cis not a 4l-root of unity, then ˆ
Cis smooth at p.
(b) If (p)2Cis a 2l-root of unity, then pis a node of ˆ
C(that is, locally looks like two cones glued
at a common vertex).
(c) If (p)2Cis a 4l-root of unity but not a 2l-root of unity, then pis a singular point of ˆ
Cof cone
type (i.e., locally looks like a topological disc).
(d) Above 1we have four points on ˆ
C; each one being a smooth point.
3.3. Fiber product. In any of the cases in Theorem 1, the surface Sis just the fiber product of (Sa,⇡
a(x,z)=
x) and (Sb,⇡
b(x,y)=x).
3.4. Hyperelliptic cases. If we take m=6 in case (2a), then ht,uiZ2⇥S4, where the Z2component
is generated by an element of order two with exactly 8 fixed points (that is, the hyperelliptic involution).
It follows that Sis the only hyperelliptic Riemann surface of genus three admitting as group of conformal
automorphisms S4. In fact, this is the only hyperelliptic situation appearing in Theorem 1(see Proposition 2
below).
Proposition 2. The only hyperelliptic situation in Theorem 1is for m =6in case (2a).
6 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
Proof. Let us consider the group K=ha,biZ2
2as above. By Theorem 1, we may identify S/Kwith the
Riemann sphere b
C. Let us consider a regular branched covering P:S!b
Cwith Kas its deck group. As the
number of fixed points of a,band ab is less than 2g+2, it follows that ◆<K, in particular, there is an order
two M¨
obius transformation ⌧so that P◆=⌧P. As ◆cannot have a common fixed point with a,band ab,
it follows that ⌧cannot fix any of the branch values of P. It follows that (as the fixed points of ◆are projected
by Pto the two fixed points of ⌧)2g+28, i.e., g3. As the cases g=1,2 are not possible, we must have
g=3 (this is only possible for m=6). ⇤
3.5. Completely decomposable Jacobians. Isogenous decompositions of Jacobian varieties with group ac-
tion (see sections 4.3 and 6). have been extensively studied from di↵erent points of view; see for example
[3], [12], [13], [14] and [17]. In particular, completely decomposable Jacobians (having only elliptic factors)
are a big subfield of study; see for example [5] and [16]. In our case, in the Theorem 1we have
•In case (1) for q=3, Shas genus g=4 and its Jacobian variety is isogenous to E4
1, where E1is the
elliptic curve y2=x31.
•In case (2(a)i) for l=2, Shas genus g=3 and its Jacobian variety is isogenous to E3
2, where E2is
the elliptic curve y2=x(3x2+10x+3)
3.6. Fuchsian uniformizations. In each case as in Theorem 1one may provide the corresponding Fuchsian
uniformizations.
3.6.1. Case (1): m =2q, q 3odd. Let us consider the triangular group
=hx,y:x2m=ym=(xy)2=1i
and the surjective homomorphism
⇥:!ht:tm=1i
⇥(x)=t,⇥(y)=t1.
The kernel of ⇥is the subgroup
K=h↵1,...,↵
m+1:↵2
1=··· =↵2
m+1=↵1···↵m+1=1i,
where
↵j=xjyx1j,j=1,...,m
↵m+1=xm.
Let us now consider the surjective homomorphism
⌘:K!ha,b:a2=b2=(ab)2=1i
where
⌘(↵j)=8
>
>
>
<
>
>
>
:
b,j⌘1 mod 2,j,m+1
ab,j⌘0 mod 2
a,j=m+1
and let be its kernel. Then S=H2/,S/ha,bi=H2/Kand S/G=H2/.
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m7
3.6.2. Case (2): m 2{3l,4l},m4.Let us consider the triangular group
0=hx,z:xm=z2=(xz)4=1i
the index two subgroup (y=zxz)
=hx,y:xm=ym=(xy)2=1i
and the surjective homomorphism
⇥:!ht:tm=1i
⇥(x)=t,⇥(y)=t1.
The kernel of ⇥is the subgroup
K=h↵1,...,↵
m:↵2
1=··· =↵2
m=↵1···↵m=1i,
where
↵j=xjyx1j,j=1,...,m.
Let us now consider the surjective homomorphism
⌘:K!ha,b:a2=b2=(ab)2=1i
where
⌘(↵j)=8
>
>
>
<
>
>
>
:
a,j⌘1 mod 3
b,j⌘2 mod 3
ab,j⌘0 mod 3
if m=3l
⌘(↵j)=(b,j⌘1 mod 2
ab,j⌘0 mod 2 if m=4l
and let be its kernel. Then S=H2/,S/ha,bi=H2/K,S/G=H2/and S/Aut(S)=H2/0.
4. Proof of Theorem 1
4.1. It can be checked that the algebraic curves Sand the groups Gas described in the theorem are such
that S/Ghas signature as required. This provides the existence for the values of mas desired.
4.1.1. For the curve described in (1), the quotient S/Ghas signature (0; 2,m,2m), where m=2qand q3 is
odd. Let us assume that Aut(S),G.Then, by the lists in Singerman’s paper [19], the signature of S/Aut(S)
must be (0; 2,3,2m). Following the same article (see pp. 37), for the surjective homomorphism
✓:=hx,y:x3=y2m=(xy)2=1i!S3
✓(y)=(1,2),✓(x)=(1,2,3),✓(xy)=(1,3)
the group =✓1(h(1,2)i) is the Fuchsian group uniformizing the orbifold S/Gand is uniformizing
S/Aut(S). If u=yand v=x1yx1, then =hu,v:u2m=v2=(uv)m=1i.
If we set xj=uj1vu1j, where j=1,...,m, and xm+1=um, then the subgroup 0generated by these
elements has the presentation 0=hx1,...,xm+1:x2
1=··· =x2
m+1=x1x2···xm+1=1iand it uniformizes
the orbifold S/ha,bi.
The group uniformizing Sis the kernel Kof the surjective homomorphism
⌘:0:!ha,bi
⌘(x2j1)=b,⌘(x2j)=ab,j=1,...,q,
⌘(xm+1)=a.
In order to get a contradiction, we only need to check that Kis not a normal subgroup of . If it is a
normal subgroup, then, as x1x32K, we must have that xx1x3x12K. Since
xx1x3x1=yx1y2x1yx1y2x1
8 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
and we are assuming Knormal in , we also must have that
x1yx1y2x1yx1y22K.
Since x1yx12K, the above asserts that y2x1yx1y22K, which (again by assuming the normality)
asserts that x1=x1yx12K, a contradiction.
4.1.2. For the curves described in (2), the signature of the quotient S/ha,b,t,ui(for any of the two cases)
has signature (0; 2,4,m).
If m6 and m,8, then the signature (0; 2,4,m) is maximal [19]. In particular,
Aut(S)=ha,b,t,ui,if m5 and m,8.
If m=8, then Sis the Riemann surface os genus g=5 described by the algebraic curve
(y2=x4+1
z2=x81.
In this case, the group b
G=ha,b,t,uihas order 8m=64. It follows that the order of Aut(S) is of the form
64d, some integer d1. By Singerman’s list [19], either d=1 (in which case, Aut(S)=b
G) or d=3 (in
which case S/Aut(S) must have signature (0; 2,3,8)). Let us assume d=3. Following Singerman’s paper
(see pp. 37), for the surjective homomorphism
✓:=hx,y:x3=y8=(xy)2=1i!S3
✓(xy)=(1,3),✓(x)=(1,2,3),✓(y)=(1,2)
the group =✓1(h(1,2)i) is the Fuchsian group uniformizing the orbifold S/b
Gand is uniformizing
S/Aut(S). If u=yand v=x1yx1, then =hu,v:u8=v2=(uv)4=1i. If we set xj=uj1vu1j, where
j=1,...,8, then the subgroup 0generated by these elements has the presentation 0=hx1,...,x8:x2
1=
··· =x2
8=x1x2···x8=1iand uniformizes the orbifold S/b
G. The derived subgroup 0
0of 0uniformizes S.
Since v20and upermutes the generators x1,...,x8, we may see that 0
0is also normal subgroup of as
supossed to be. Since y2, we may see that ynormalizes 0
0. In our assumption (d=3) it must happen that
xalso must normalize 0
0. But, xx1x1=yx satisfies that ✓(yx)=(1)(2,3), that is, xx1x1does not belong to
, in particular, it cannot belong to 0
0; we get a contradiction.
4.2. Next, we will see that the cases shown in part (2) are the only situations (the case of part (1) uses similar
arguments and it is left to the interested reader to make the suitable modifications).
Let us assume that Sis a closed Riemann surface admitting a group of conformal automorphisms G
Z2
2oZ
mso that the quotient S/Ghas signature (0; 2,m,m). Let us denote by Aand Bthe generators of the
normal factor Z2
2and by b
Tthe one of the cyclic factor Zm.
The quotient orbifold O=S/hA,Bihas a signature of the form (;2,r
···,2) and it admits a conformal
automorphism e
Tof order minduced by b
Twhich permutes the cone points. Moreover, since O/he
Ti=S/G
has signature of the form (0; 2,m,m), the automorphism e
Tmust have two fixed points and must permute the
rpoints in one orbit (so r=m). This in particular asserts that =0 and, by the Riemann-Hurwitz formula
applied to the branched regular cover induced by hA,Bi, we obtain that g=m3.
We may assume that Ois given by the Riemann sphere b
Cand e
Tis a M¨
obius transformation of order m.
Up to a M¨
obius transformation we may assume that e
T(z)=!z, where !=e2⇡i/mand that the mcone points
are given by the m-roots of unity. If we now consider the M¨
obius transformation
M(z)= !+1
!!z!
z1
and we set
j=M(!2+j)=(!+1)(!1+j1)
!2+j1,j=1,...,m3,
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m9
we might assume that the mcone points are 1,0,1,
1,...,
m3and
e
T(z)=(1 +!)2
(1 +!)2!z.
Let us now consider the following generalized Fermat curve
C:8
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
:
x2
1+x2
2+x2
3=0
1x2
1+x2
2+x2
4=0
.
.
..
.
..
.
..
.
.
m3x2
1+x2
2+x2
m=0
9
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
;
⇢Pm1
which is a closed Riemann surface of genus gC=1+2m3(m4) (for details, see [2,6]).
The curve Cadmits the linear automorphisms
aj([x1:··· :xm]) =[x1:··· :xj1:xj:xj+1:··· :xm],j=1,...,m1.
Set am=a1a2···am1(multiplication by 1 the coordinate xm). So, the curve Cadmits the following
abelian group of conformal automorphisms
Zm1
2F=ha1,...,am1i.
The map
⇡:C!b
C;⇡([x1:··· :xm]) =(x2/x1)2
is a regular branched cover with Fas its deck group and whose branch values are 1,0,1,
1,...,
m3.
As consequence of the results in [9], there must be a subgroup HZm3
2of Facting freely on Cso that
S=C/H.
Observe that
e
T(1)=0,e
T(0) =1,e
T(1) =1,e
T(1)=2,...,
e
T(m4)=m3,T(m3)=1.
If L(z)=1/zand
e
U(z)=MLM1(z)=z+(1 +!)2/!
then
e
U(1)=1,e
U(0) =m3,e
U(1) =m4,e
U(j)=m4j,j=1,...,m5.
Note that for modd none of the values jis fixed by e
U. If mis even, then e
Uonly fixes (m4)/2=1 and
none of the others. Moreover, he
U,e
TiDm.
As a consequence of the results in [9], there exist linear automorphisms T,U2Aut(C) (each one normal-
izing F) so that ⇡T=e
T⇡and ⇡U=e
U⇡. In fact, by [6], we have that
T([x1:··· :xn+1]) =[xm:↵1x1:↵2x2:··· :↵m2xm2:↵m1xm1]
↵1=pn2,↵
2=1,↵
j+2=ipj,j=1,...,m3.
As Tinduces b
T, the subgroup His normalized by T. Let us observe that
Taj=aj+1T,j=1,...,m1,
Tam=a1T,
hT,Fi=FohTiZm1
2oZ
m.
Again from [6],
U([x1:··· :xm]) =[x1:xm:ixm1:ixm2:···:ix4:ix3:x2].
10 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
Note that
U2([x1:··· :xm]) =[x1:x2:x3:x4:··· :xm2:xm1:xm]2F
As a consequence of all the above, the subgroup Fis normal in hF,T,Uiand
hF,T,Ui/FDm.
The quotient orbifold C/hF,Tiis equal to the quotient orbifold O/he
Ti=S/Gwhose of signature is
(0; 2,m,m).We notice that C/hF,T,Uihas signature (0; 2,4,m).
If m5 and m,8, then the signature (0; 2,4,m) is maximal [19]; so
Aut(C)=hF,T,Ui
and C/Aut(C) has signature (0; 2,4,m).
Next we proceed to see that there exist subgroups Has above only in the cases that mis either divisible by
3 or by 4.
Lemma 1. Let H <F so that H Zm3
2acts freely on C and such that THT 1=H. Then one of the
following holds.
(1) m⌘0 mod 4 and
H=ha1a3,a2a4,a3a5,...,am1a1,ama2i.
(2) m⌘0 mod 3 and
H=ha1a2a3,a2a3a4,a3a4a5,...,am2am1am,am1ama1,ama1a2i.
Proof. Let us consider a surjective homomorphism :F!Z2
2so that
(i) H=ker() does not contains the elements a1,..., am(these are the only elements of Facting with
fixed points on C; see [6]); and
(ii) THT1=H.
If we denote by T⇤the automorphism of Fgiven by conjugation by T, then there is an automorphism ⇢of
Z2
2so that
⇢=T⇤.
As a1<Hwe must have that (a1),id. Set a:=(a1). Now, as T⇤(aj)=aj+1and is surjective, it
should happen that ⇢(a),a. Set b:=⇢(a). Then, Z2
2=ha,bi.
There are only two possibilities for ⇢; these being the following ones:
(1) ⇢(a)=b,⇢(b)=a,⇢(ab)=ab.
(2) ⇢(a)=b,⇢(b)=ab,⇢(ab)=a.
In case (1) it holds that (a2j1)=aand (a2j)=b. In this situation, we must have that mis divisible by
4, and
H=ker(⇢)=ha1a3,a2a4,a3a5,...,am1a1,ama2i
In case (2) it holds that (a1+3j)=a,(a2+3j)=band (a3+3j)=ab. In this situation we must now have
that mis divisible by 3, and
H=ker(⇢)=ha1a2a3,a2a3a4,a3a4a5,...,am2am1am,am1ama1,ama1a2i.
⇤
The Riemann surface defined by Cis the highest abelian branched cover of the orbifold O=S/hA,Bi; thus
it is uniquely determined up to isomorphisms. If 2Gal(Q/Q), then Cis also a highest abelian branched
cover of the orbifold O; let us denote by F0the associated deck group. By the uniqueness property, there
exists an isomorphism ':C!C.We observe that
F0=hb1,...,bm1i
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m11
where bi='ai'1.We can also consider the regular branched cover S!O; we shall denote by H0its deck
group.
Lemma 1asserts that the only possibility for the existence of a closed Riemann surface Sadmitting a
group GZ2
2oZ
m,m3, of conformal automorphisms with S/Gof signature (0; 2,m,m) is that mis either
divisible by 3 or by 4.
4.2.1. If mis divisible by 4 and not by 3, then the normal subgroup hA,BiZ2
2has two non-trivial elements
acting with fixed points (each one having exactly mfixed points) whose product acts freely on S. Without
loss of generality, we can suppose that the branch values µ1=1,µ
2=0,µ
3=1,µ
4=1,...,µ
m=m3of
⇡are the mroots of the unity and that induces the permutation µi7! µi+2sfor some 1 sm/21.We
can see that the above implies that bi=ai+2sand that
H0=hbibi+2=ai+2sbi+2+2s:1imi=H.
It follows that SS.
4.2.2. If mis divisible by 3 and not by 4, then the three non-trivial elements of the normal subgroup hA,Bi
Z2
2acts with fixed points (each one having exactly 2m/3 fixed points). Without loss of generality, we can
suppose that the branch values µ1=1,µ
2=0,µ
3=1,µ
4=1,...,µ
m=m3of ⇡are the mroots of the
unity and that induces the permutation µi7! µi+3sfor some 1 sm/31.We can see that the above
implies that bi=ai+3sand that
H0=hbibi+1bi+2=ai+3sai+1+3sai+2+3s:1imi=H.
It follows that SS.
In both cases {:SS}=Gal(Q/Q). Now, from a result of Wolfart on minimal field of definition
of regular Belyi curves [21], we are in position to conclude that Sis definable over Q.These curves are
described by the algebraic curves of parts (2a) and (2b) of the theorem.
4.2.3. If mis divisible by 3 and 4, then there are two possible actions. But, as observed above, in one case
the three elements of order two of the normal subgroup Z2
2act with fixed points and in the other case this is
not the case. In particular, these two pairs are non-isomorphic and, for any of the two cases, the surface Sis
definable over Q.
4.3. Isogenous decomposition of the Jacobian variety. In this section we prove the isogenous decomposi-
tion of the Jacobian variety JS for the quasiplatonic curves described in Theorem 1. We will use Kani-Rosen’s
decomposition theorem [13].
Corollary 1 ([13]).Let S be a closed Riemann surface of genus g 1and let H1,...,Ht<Aut(S)such
that:
(i) HiHj=HjHi, for all i,j=1,...,t;
(ii) g(S/HiHj)=0, for 1i<jt
(iii) g=Pt
j=1g(S/Hj).
Then
JS ⇠
t
Y
j=1
J(S/Hj).
In each of our cases we use the following three order two cyclic groups
H1=hai,H2=hbi,H3=habi.
If k2{a,b,ab}, we denote by Skthe Riemann surface structure subjacent of the orbifold S/hkiand its
genus by gk.
It is clear that conditions (i) and (ii) of Corollary 1are satisfied and we only need to check the condition
(iii).
12 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
4.3.1. In the case (1) of our theorem, that is, m=2q(qodd), we have that Sa,Sband Sab are, respectively,
the following hyperelliptic curves
z2=xm1y2=xq1w2=xq+1
which have respective genera equal to ga=(m2)/2, gb=(m2)/4 and gab =(m2)/4; so ga+gb+gab =m2
and condition (iii) is then satisfied. Moreover, we may see that Sband Sab are isomorphic; so we obtain that
JS ⇠JSa⇥(JS b)2.
Also, let us observe that the hyperelliptic curve Saadmits an extra conformal involution d, induced by tq,
with two fixed points. The quotient Sa,1=Sa/hdihas equation w2
1=vq
11 and Sa,2=Sa/h◆adi(where
◆adenotes the hyperelliptic involution), has equation w2
2=v2(vq
21). Clearly, Sa,1and Sa,2are isomorphic
curves. Again, applying Kani-Rosen result, we obtain that JS a⇠(JS a,1)2=(JS b)2.
4.3.2. In the case (2a) of our theorem (m=3l), we have that Sa,Sband Sab are, respectively, the following
hyperelliptic curves
z2=(xl!3)(xl!2
3)=x2l+xl+1y2=(xl1)(xl!2
3)w2=(xl1)(xl!3)
of genera ga=gb=gab =l1; so ga+gb+gc=m3 and condition (iii) is then satisfied. Moreover, we
may see that Sa,Sband Sab are isomorphic, that is, JS ⇠(JS a)3.
But in this case, the order 4 automorphism uof Sinduces the automorphism d(x,z)=(1/x,z/xl) of order
two of Sa(acting with two fixed points for lodd and four fixed points if lis even). We may apply Kani-
Rosen’s result using the groups K1=hdiand K2=hjadi, where ja(x,z)=(x,z) is the hyperelliptic
involution of Sato obtain that JSa⇠JS u⇥JS bu , where Suis the subjacent Riemann surface of the quotient
Sa/K1=S/huiand Sbu is the subjacent Riemann surface of the quotient Sa/K2=S/hbui. To obtain explicit
equations we first observe that Sais isomorphic to
w2=(v+1)2l+(1 v2)l+(1 v)2l
by the isomorphism
(x,z)7! (v,w)=0
B
B
B
B
B
@x1
x+1,z 2
x+1!l1
C
C
C
C
C
A=(1 v2)l+2
l
X
j=0 2l
2j!v2j.
In this new model, the automorphism dis given as d(v,w)=(v,w) and jadis (v,w)7! (v,w). Then
an equation for Suis given by
w2
1=(1 v1)l+2
l
X
j=0 2l
2j!vj
1
and an equation for Sbu is given by
w2
2=v20
B
B
B
B
B
B
@(1 v2)l+2
l
X
j=0 2l
2j!vj
21
C
C
C
C
C
C
A.
4.3.3. In the case (2b) of our theorem (m=4l), we have that Sa,Sband Sab are, respectively, the following
hyperelliptic curves
z2=xm1y2=x2l1w2=x2l+1
of genera ga=2l1, gb=gab =l1; so ga+gb+gab =m3 and condition (iii) is then satisfied. Moreover,
we may see that Sband Sab are isomorphic; so JS ⇠JS a⇥(JS b)2.
Also, let us observe that the hyperelliptic curve Saadmits an extra conformal involution d, induced by t2l,
with four fixed points. The quotient Sa,1=Sa/hdihas equation w2
1=v2l
11 and Sa,2=Sa/h◆adi(where ◆a
denotes the hyperelliptic involution), has equation w2
2=v2(v2l
21). Clearly, Sa,1and Sa,2are not isomorphic
curves as they have di↵erent genera. Applying Kani-Rosen result, we obtain that JS a⇠JS a,1⇥JS a,2=
JSb⇥JS a,2; so JS ⇠(JS b)3⇥JS a,2.
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m13
5. Curves over Qfor the case mdivisible by 3
Theorem 1asserts that if Sadmits a group of conformal automorphisms HZ2
2oZ
mso that S/Hhas
triangular signature, then Sis definable over Q. In the same theorem explicit curves are provided, all of them
defined over Qwith the exception of one case: m=3land S/Hof triangular signature (0; 2,m,m). In this
last case, there is provided a curve over the degree two extension Q(!3)
C:(y2=x2l+!3xl+!2
3
z2=x2l+xl+1
In this section we use the computation algorithm presented in [11] in order to indicate how to find another
curve representation of Sdefined over Q.
Let =Gal(Q(!3)/Q)=hiZ2, where (!3)=!2
3. In this way the Galois orbit of Cconsists of Cand
the curve
C:(y2=x2l+!2
3xl+!3
z2=x2l+xl+1
The map
f(x,y,z)=0
B
B
B
B
@1
x,!2
3y
xl,z
xl1
C
C
C
C
A
provides an isomorphism f:C!C.
We may observe that f
fis the identity map; so the set {I,f}defines a Weil datum for Cwith respect
to the Galois extension Q(!3)/Q.
The map
:C!(C)⇢C6:(x,y,z)7! 0
B
B
B
B
@x,y,z,1
x,!2
3y
xl,z
xl1
C
C
C
C
A
defines an isomorphism between Cand (C) (its inverse is just projection on the first three coordinates).
We consider the permutation action ⇥:!GL(6,C) given by
⇥()(x1,x2,x3,x4,x5,x6)=(x4,x5,x6,x1,x2,x3).
A set of generators of the algebra of -invariant polynomials C[x1,x2,x3,x4,x5,x6]is given by
t1=x1+x4,t2=x2+x5,t3=x3+x6,
t4=x1x4,t5=x2x5,t6=x3x6,
t7=x1x2+x4x5,t8=x1x3+x4x6,t9=x2x3+x5x6.
Remark 1. Observe that if (x1,...,x6)2(C), then t4=1.
Let us consider the branched cover
:C6!C9:(x1,...,x6)7! (t1,...,t9).
The results in [11] asserts that Cis isomorphic to D= ((C)) and that Dis defined over Q. In order to
find the equations for Dwe proceed as follows.
It is possible to observe that
(C6)=8
>
>
>
<
>
>
>
:
t2
1t5t1t2t7+t2
2t44t4t5+t2
7=0
t2
1t6t1t3t8+t2
3t44t4t6+t2
8=0
t2
2t6t2t3t9+t2
3t54t5t6+t2
9=0
9
>
>
>
=
>
>
>
;.
We also have the equalities
(1) y=t7t1t2+t2x
2xt1
(2) z=t8t1t3+t3x
2xt1
14 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
(3) x2=t1xt4
Now, (1) and (3) above asserts that the equality
(4) y2=x2l+!2
3xl+!3=(xl1)(xl!3)
can be written as
(t7t1t2)2t2
2t4+t2(2t7t1)x=(xl1)(xl!3)(2xt1)2.
Equality (3) asserts that
(2xt1)2=t2
14t4
and that there are polynomials P,Q2Q(!3)[t1,t4] so that
(xl1)(xl!3)=P(t1,t4)x+Q(t1,t4).
Remark 2. If l=2, then P(t1,t4)=t1(t2
1+2t4!2
3) and Q(t1,t4)=(1 +t4)(!3+t4)t2
1t4.
All the above asserts that (4) is equivalent to
(t7t1t2)2t2
2t4+t2(2t7t1)x=(P(t1,t4)x+Q(t1,t4))(t2
14t4),
from which we obtain
x=R(t1,t2,t4,t7)=(t7t1t2)2t2
2t4(t2
14t4)Q(t1,t4)
(t2
14t4)P(t1,t4)t2(2t7t1).
Now, using this expression for x, we use (1) and (2) to obtain rational expressions for yand zas follows:
y=P(t1,t4)(t3
1t2t2
1t74t1t2t4+4t4t7)+Q(t1,t4)(t2
1t24t2t4)t2
1t3
2+t2
1t2
2t1t2t7+t3
2t4+t2t2
7
P(t1,t4)(t3
14t1t4)+2Q(t1,t4)(t2
14t4)2t2
1t2
2+t2
1t2+2t1t2t7+2t2
2t42t2
7
z=P(t1,t4)(t3
1t3t2
1t84t1t3t4+4t4t8)+Q(t1,t4)(t2
1t34t3t4)t2
1t2
2t3+t2
1t2t3t1t2t8+t2
2t3t4+2t2t7t8t3t2
7
P(t1,t4)(t3
14t1t4)+2Q(t1,t4)(t2
14t4)2t2
1t2
2+t2
1t2+2t1t2t7+2t2
2t42t2
7
In this way, equation
y2=(xl1)(xl!3)
can be written as an equation
E(t1,...,t9)=0
and the equation
z2=(xl!3)(xl!2
3)
can be written as an equation
F(t1,...,t9)=0,
where E,F2Q(!3)[t1,...,t9].
All the above asserts that Dis defined as the common zeroes of
D:8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
t2
1t5t1t2t7+t2
2t44t4t5+t2
7=0
t2
1t6t1t3t8+t2
3t44t4t6+t2
8=0
t2
2t6t2t3t9+t2
3t54t5t6+t2
9=0
E(t1,...,t9)=0
F(t1,...,t9)=0
9
>
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
>
;
.
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m15
The first three equations are given by polynomials with coefficients in Q. The last two, Eand F, may still
have coefficients on Q(!3). In this case we may change them by the following traces (which are defined over
Qas desired), so equations for Dover Qare:
D:
8
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
:
t2
1t5t1t2t7+t2
2t44t4t5+t2
7=0
t2
1t6t1t3t8+t2
3t44t4t6+t2
8=0
t2
2t6t2t3t9+t2
3t54t5t6+t2
9=0
E(t1,...,t9)+E(t1,...,t9)=0
!3E(t1,...,t9)+!2
3E(t1,...,t9)=0
F(t1,...,t9)+F(t1,...,t9)=0
!3F(t1,...,t9)+!2
3F(t1,...,t9)=0
9
>
>
>
>
>
>
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
>
>
>
>
>
>
;
.
6. A remark on the decomposition of JS for the case m=6. Group algebra point of view.
The isogenous decomposition obtained in Theorem 1for the Jacobian variety of Swas obtained by a
simple application of Kani-Rosen’s decomposition result (Corollary 1). In this section we shall show how
the methods of Lange-Recillas [14], Carocca-Rodr´
ıguez [3], Rojas [17] and Jim´
enez [12] about decomposi-
tions of abelian varieties, using the rational algebra of finite groups, can also be applied to obtain the same
decomposition of Theorem 1. We only describe it for m=6 in case (2a) as the general case follows the same
ideas.
6.1. The group algebra decomposition for abelian varieties with non-trivial automorphisms. We start
by recalling some definitions and results about the isotypical decomposition of any abelian variety Awith a
non-trivial (finite) group of automorphisms in terms of the complex and rational irreducible representations
of G.
Let Vbe an irreducible representation of Gover C. If we denote by Fits field of definition and by Kthe
field obtained by extending Qby the values of the character V, then Fis a finite extension of Kand the
extension degree mV=[F:K] is called the Schur index of V. For details see [18].
The action of Gon Ainduced a Qalgebra homomorphism ⇢:Q[G]!EndQ(A). For any element
↵2Q[G] we define an abelian subvariety B↵:=Im(↵)=⇢(l↵)(A)⇢A, where lis some positive integer such
that l↵2Z[G].
The semi-simple algebra Q[G] decomposes into a product Q0⇥···⇥Qrof simple Qalgebras; the simple
algebras Qiare in bijective correspondence with the rational irreducible representations of G. That is, for any
rational irreducible representation Wiof Gthere is a uniquely determined central idempotent ei. This idem-
potent defines an abelian subvariety of A, namely Bi=Bei. These varieties, called isotypical components, are
uniquely determined by the representation Wi. Moreover, the decomposition of every Qi=L1⇥···⇥Lniinto
a product of minimal left ideals (all isomorphic) gives a further decomposition of A. More precisely, there
are idempotents fi1,..., fini2Qisuch that ei=fi1+··· +finiwhere ni=dimVi/mVi, with Vithe complex
irreducible representation associated to Wi. These idempotents provide subvarieties Bij :=Bfij ⇠Bi, for all
j. Then we have the following theorem
Theorem 2 ([14], [3]).Let G be a finite group acting on an abelian variety A.Let W1,...,Wrdenote the
irreducible rational representations of G. Then there are abelian subvarieties B1,...,Brof A and an isogeny
(1) A⇠Bn1
1⇥···⇥Bnr
r.
The above isogenous decomposition of the abelian variety Ais called the group algebra decomposition of
A.
16 RUB´
EN A. HIDALGO, LESLIE JIM´
ENEZ, SA ´
UL QUISPE, AND SEBASTI ´
AN REYES-CAROCCA
6.2. The group algebra decomposition for the Jacobian variety of Riemann surfaces. We next assume
that A=JS, where Sis a closed Riemann surface and Gis a (finite) group of conformal automorphisms of
it. In this particular case, in Theorem 2we always have that one of the factors Bjis isogenous to the Jacobian
variety JSG, where SGis the subjacent Riemann surface structure associated to the Riemann orbifold S/G.
In this situation, the isotypical decomposition can be made more explicitly as follows.
Let Hbe a subgroup of G.We denote by ⇡H:S!S/Hthe associated regular covering map and by
⇢Hthe representation of Ginduced by the trivial representation of H. If Uand Vare representations of G,
then hU,Videnotes the usual inner product of the corresponding characters. By the Frobenius Reciprocity
Theorem h⇢H,Vi=dimCVH, where VHis the subspace of Vfixed by H. Define pH=1
|H|Ph2Hhas the
central idempotent in Q[H]; corresponding to the trivial representation of H. Also, we define fi
Has pHei, an
idempotent element in Q[G]ei.
With the previous notations, the corresponding group algebra decomposition of JS His given as follows
[3, Proposition 5.2]:
(2) JS H⇠JSG⇥B
dimVH
1
m1
1⇥···⇥B
dimVH
r
mr
r,
whit mi=mVi. Moreover,
(3) Im(pH)=⇡⇤
H(JH)
where ⇡⇤
H(JS H) is the pullback of JS Hby ⇡H. If dimVH
i,0 then
(4) Im( fi
H)=B
dimVH
i
mi
i.
We should notice that the previous results do not depend on the action of G.The next result related to the
dimension of the factors in (1) involves the way the group Gacts.
Theorem 3. [17]Let G be a finite group acting on a compact Riemann surface S with geometric signature
given by (;[m1,C1],...,[mr,Cr]). Then the dimension of factor Biassociated to a non trivial rational
irreducible representation Wiin (1) is given by
dim Bi=ki(dim Vi(1) +1
2
r
X
k=1
(dim Vidim(VGk
i))
where Gkis a representative of the conjugacy class of Ckand ki=mi|Gal(Ki/Q)|.
The following lemma gives us conditions under which a factor in the group algebra decomposition can
be described as the image of a concrete idempotent, in particular, when it corresponds to a Jacobian of an
intermediate quotient.
Lemma 2. [12]Let S be a Riemann surface with an action of a finite group G such that the genus of S /G
is equal to zero. Assume that V1,...,V
qare the non-isomorphic complex irreducible representations of G.
Let us consider the group algebra decomposition of JS given by (1). Let H be a subgroup of G so that
dimCVH
i=mi, for some fixed index i. Then
(i) Im( fi
H)=Bi;
(ii) if, moreover, dimCVH
l=0for all l, l ,i, such that dimCBl,0then
JS H⇠Im(pH)=Bi.
On this way, in order to obtain factors isogenous to Jacobian varieties at the group algebra decomposition
of JS, we need to look for subgroups Hof Gsatisfying dimCVH
i=h⇢H,Vii=miand dimCVH
l=0 for all l,
l,i, such that dimCBl,0.
QUASIPLATONIC CURVES WITH SYMMETRY GROUP Z2
2oZ
m17
6.3. Our examples. In our case, the group
G=ha,t:a2=t6=[a,t]2=1,t3=(at)3i
has eight complex irreducible representations V1,...,V8, as shown in the following character table.
Conj. class id at3a t3t2t5t4t
V11 1 1 1 1 1 1 1
V21 -1 1 -1 1 -1 1 -1
V31 -1 1 -1 ⇠2⇠2⇠⇠
V41 -1 1 -1 ⇠⇠ ⇠2⇠2
V51 1 1 1 ⇠2⇠2⇠ ⇠
V61 1 1 1 ⇠ ⇠ ⇠2⇠2
V73 1 -1 -3 0 0 0 0
V83 -1 -1 3 0 0 0 0
where ⇠=exp(2⇡i/3).
It is not difficult to see that the rational irreducible representations of Gare
W1:=V1,W2:=V2,W3:=V3V4,W4:=V5V6,W5:=V7,W6:=V8.
By applying Theorem 2we obtain
JS ⇠B1
1⇥B1
2⇥B1
3⇥B1
4⇥B3
5⇥B3
6.
Moreover, as B1⇠JSG(and S/Ghas genus zero), B1=0.Finally, with the help of a computational
program such as MAGMA [15] we can obtain that dim(B2)=dim(B3)=dim(B4)=dim(B6)=0.Combining
this fact with the previous isogenies, we are in position to conclude that JS ⇠B3
5.
Then now we are looking for every (conjugacy class of) subgroup Hof Gsatisfying dimCVH
7=h⇢H,V7i=
1. Remember that W5=V7. Hence, again using MAGMA we obtain the table of induced representation by
any H✓G. Therefore, the class of subgroups Hsatisfying this is which given by H=hai. Thus
JS ⇠B3
5⇠(JShai)3.
We obtain that dim(B5)=1; thus JS in this case is completely decomposable (see also Section 3.5).
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AN REYES-CAROCCA
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E-mail address:ruben.hidalgo@ufrontera.cl
Departamento de Matem´
atica y Estad´
ıstica,Universidad de LaFrontera,Temuco,Chile.
E-mail address:leslie.jimenez@liu.se
Matematiska Institutionen,Link¨
opings Universitet,Link ¨
oping,Sweden
E-mail address:saul.quispe@ufrontera.cl
Departamento de Matem´
atica y Estad´
ıstica,Universidad de LaFrontera,Temuco,Chile.
E-mail address:sebastian.reyes@ufrontera.cl
Departamento de Matem´
atica y Estad´
ıstica,Universidad de LaFrontera,Temuco,Chile.