Hyperelliptic plane curves of type (d,d-2)

0.2 MB·English

by  Fumio Sakai

Hyperelliptic plane curves of type (d,d 2)
−
Fumio SAKAI, Mohammad SALEEM∗ and Keita TONO
8
0 Abstract
0
In [7], we classified and constructed all rational plane curves of type
2
(d,d−2). In this paper, we generalize these results to irreducible plane

n curvesof type(d,d−2) with positive genus.
a
J

2 1 Introduction

G] Let C P2 =P2(C) be a plane curve of degree d. We call C a plane curve of
⊂
type (d,ν) if the maximal multiplicity of singular points on C is equal to ν. A
A
unibranchedsingularityis calleda cusp. Rationalcuspidal plane curves oftype
.
h (d,d 2) and (d,d 3) were classified by Flenner–Zaidenberg [5, 6] (See also
− −
t [4,8]forsomecases). In[7],weclassifiedrationalplanecurvesoftype(d,d 2)
a −
m with arbitrary singularities. In order to describe a multibranched singularity
P, we introducedthe notionof the system of the multiplicity sequences m (C)
[ P
(See Sect.2). We denote by Data(C), the collection of such systems of multi-

1 plicity sequences. The purpose of this paper is to complete the classification of
v
irreducibleplanecurvesoftype(d,d 2)withpositivegenusg. Weremarkthat
9 −
if g 2, then C is a hyperelliptic curve, for the projection from the singular
3
≥
3 point of multiplicity d 2 induces a double covering of C over P1.
−
0
. Theorem 1. Let C be a plane curve of type (d,d 2) with genus g. Let Q C
1 − ∈
be the singular point of multiplicity d 2. Then, we have
0 −
8
(i) Data(C)= m (C), 1 ,..., 1 ,(2 ),...,(2 ) , where
0 Q 1 b1 1 bn bn+1 bn+n′
v: h (cid:0) (cid:1) (cid:0) (cid:1) i
i k1 1
X
ar mQ(C)= kkk...s1s′′  111...!!aa1s 
ks+1 2as+1
and the following conditionks...Naresat2...iasNfied:
2000Mathematics SubjectClassification: 14H50, 14E07
∗PartiallysupportedbyPostDoctoralFellowshipforForeignResearchers,1705292,JSPS.
1
N s N n+n′
′
(1) kh+ kh′ =d−2and ai+ bj =d−g−2,whereai ≥0
h=1 h′=1 i=1 j=1
X X X X
(a >0 for i=1,...,s), b >0,
i j
(2) we have n,n′,s 0 and n′+s′ 2g+2, where s′ =# j a >0 ,
s+j
≥ ≤ { | }
(3) for i = 1,2,...,s, if k′ = k , then a k and if k′ > k , then
i i i ≥ i i i
a =k ,
i i
(4) for i = s+1,...,N, if a > 0, then either k is even and a k /2
i i i i
≥
or k is odd and a =(k 1)/2.
i i i
−
Note that the N is the number of the different tangent lines to C at Q.
(ii) Data(C) can be derived from Degtyarev’s 2–formula T(C) defined for the
defining equation of C (See Proposition 9 for details).
Corollary. Let C be an irreducible plane curve of type (d,d 2) with genus g.
−
(i) If C has only cusps, then C has the following data (b >0,k >0,j 0):
i
≥
Class Data(C)
(a) [(k),(2 ),...,(2 )] (k =g+ n′ b )
b1 bn′ i=1 i
(n′ 2g+P2)
≤
(b) [(2k+1,2 ),(2 ),...,(2 )] (k+1=g+ n′ b )
k b1 bn′ i=1 i
(n′ 2g+1)P
≤
(c) [(2k,2 ),(2 ),...,(2 )] (k =g+j+ n′ b )
k+j b1 bn′ i=1 i
(n′ 2g+1)P
≤
(ii) If C has only bibranched singularities, then C has the following data (b >
i
0,k>0,r>0,j 0,l 0):
≥ ≥
Class Data(C)
(e) k 1 , 1 ,..., 1 (k =g+j+ n b )
k 1 k+j 1 b1 1 bn i=1 i
(f) h(cid:0) (cid:1)k(cid:0) (cid:1)1 ,(cid:0)1(cid:1) ,...,(cid:0)1(cid:1) i (k+r=g+Pn b )
k+r 1 k 1 b1 1 bn i=1 i
(aa) h(cid:0)k ,(cid:1)1(cid:0) (cid:1),.(cid:0)..(cid:1), 1 (cid:0) (cid:1) i (k+r=g+Pn b )
r 1 b1 1 bn i=1 i
(ab) h(cid:0) (cid:1)2k(cid:0)+1(cid:1) 2k , 1(cid:0) (cid:1),.i.., 1 (k+r+1=gP+ n b )
r 1 b1 1 bn i=1 i
(ac) h(cid:8)(cid:0)2k 2(cid:1)k+j (cid:9), (cid:0)1(cid:1) ,..., (cid:0)1(cid:1) i (k+r=g+j+Pn b )
r 1 b1 1 bn i=1 i
(bb) h(cid:8)(cid:0)2k+(cid:1)1 2k(cid:9),(cid:0)1(cid:1) ,...,(cid:0)1(cid:1) i (k+r+2=g+Pn b )
2r+1 2r 1 b1 1 bn i=1 i
(bc) h(cid:8)(cid:0)2k+1(cid:1)2k(cid:9) (cid:0),(cid:1)1 ,...(cid:0),(cid:1)1 i (k+r+1=g+Pl+ n b )
2r 2r+l 1 b1 1 bn i=1 i
(cc) h(cid:8)(cid:0)2k 2(cid:1)k+j ,(cid:9)1(cid:0) (cid:1),..., 1(cid:0) (cid:1) i (k+r=g+j+l+Pn b )
2r 2r+l 1 b1 1 bn i=1 i
h(cid:8)(cid:0) (cid:1) (cid:9) (cid:0) (cid:1) (cid:0) (cid:1) i P
2
Theorem 2 (Cf. Coble [1], Coolidge [2]). Let C be an irreducible plane curve
of type (d,d 2) with genus g. Then, there exists a Cremona transformation
−
which transforms C into a plane curve:
2g+2
Γ:y2 = (x λ ),
i
−
i=1
Y
with some distinct λ ’s.
i
Conversely, given a plane curve Γ as above and a collection of systems of
multiplicity sequences M satisfying the conditions (1)–(4) in Theorem 1, (i) for
d g+2, then we can find an irreducible plane curve C of type (d,d 2) such
≥ −
that
(a) Data(C)=M,
(b) C is Cremona birational to Γ.
In Sect.2, we review the systemof the multiplicity sequences,the 2–formula
andquadratic Cremonatransformations. InSect.3 (resp. Sect.4), we will prove
Theorem 1 (resp. Theorem 2). In Sect.5, we discuss the defining equations for
those curves given in Corollary.
2 Preliminaries
AcuspP canbe describedbyits multiplicity sequencem =(m ,m ,m ,...).
P 0 1 2
For a multibranched singular point P on C, we introduced the system of the
multiplicity sequences of P.
Definition 3 ([7]). Let P C be a multibranched singular point, having r lo-
∈
cal branchesγ ,...,γ . Let m(γ )=(m ,m ,m ,...) denote the multiplicity
1 r i i0 i1 i2
sequences of the branches γ , respectively. We define the system of the multi-
i
plicity sequences, which will be denoted by the same symbol m (C), to be the
P
combination of m(γ ) with brackets indicating the coincidence of the centers of
i
the infinitely near points of the branchesγ . For instance, for the case in which
i
r =3, we write it in the following form:
m1,0 m1,ρ m1,ρ+1 ... m1,ρ′ m1,ρ′+1,...,m1,s1
m2,0...m2,ρ m2,ρ+1! m2,ρ!′ m2,ρ′+1,...,m2,s2.
m3,0 m3,ρ m3,ρ+1,...,m3,s3 
   
We alsouse some simplifications such as 
a
a
1 1 1
(2 )=(2,...,2,1,1), (2 )=(1), = ... .
a 0
(cid:18)1(cid:19)a (cid:18)z1(cid:19)}|(cid:18)1(cid:19){
z }| {
Example 4. We examine our notations for ADE singularities.
P A2n A2n−1 D2n−1 D2n E6 E7 E8
1 1
mP(C) (2n) 11 n 21 2n−3 1 1 n−2 (3) 21 11 (3,2)
( 1!(cid:16) (cid:17) )
(cid:0) (cid:1) (cid:8)(cid:0) (cid:1) (cid:9) (cid:0) (cid:1)(cid:0) (cid:1)
3
Example 5. The hyperelliptic curvey2 = 2g+2(x λ )has onesingularityQ
on the line at infinity with m = g 1 . i=1 − i
Q g 1 g Q
Let C be an irreducible plane(cid:0) (cid:1)c(cid:0)ur(cid:1)ve of type (d,d 2). Let Q C be
− ∈
the singular point with multiplicity d 2. Choosing homogeneous coordinates
−
(x,y,z) so that Q=(0,0,1), the curve C is defined by an equation:
F(x,y)z2+2G(x,y)z+H(x,y)=0,
where F, G and H are homogeneous polynomials of degree d 2, d 1 and d,
− −
respectively. Set ∆ = G2 FH. Let t ,...,t P1 be all the distinct roots
1 l
− ∈
of the equation F(t)∆(t) = 0. For each i, let (p ,q ) = (ord (F),ord (∆)),
i i ti ti
whereord (F)(resp. ord (∆))isthe multiplicity ofthe roott ofthe equation
ti ti i
F(t) = 0 (resp. ∆(t ) = 0). Set T(C) = (p ,q ),...,(p ,q) . This unordered
i 1 1 l l
{ }
l–tupleT(C)iscalledthe2–formulaofC (Degtyarev[3]). WeremarkthatT(C)
does not depend on the choice of the coordinates (x,y,z) with Q=(0,0,1).
Lemma 6. The 2–formula T(C) satisfies the following properties:
l l
(i) q =2 p +2,
i i
i=1 i=1
X X
(ii) p =q or min p ,q is even for each i,
i i i i
{ }
(iii) there exists a pair (p ,q ) such that q is an odd number.
i i i
Proof. (i) By definition, l p = d 2 and l q = 2d 2. (ii) Suppose
i=1 i − i=1 i −
that p = q . We may assume t = (0,1). We can write F, G and ∆ as
i i i
6 P P
F = xpiF0, ∆ = xqi∆0 and G = xmG0, respectively, where m = ordti(G). If
pi > qi, then x2mG20 = xqi(∆0 +xpi−qiF0H). Thus qi = 2m. If 0 < pi < qi,
then we get x2mG20 = xpi(xqi−pi∆0 +F0H), which implies that 2m ≥ pi > 0.
We have x H, since C is irreducible. Hence p = 2m. (iii) Suppose that all
i
6|
q ’s are even. Then we can write as ∆ = ∆2. We have F(Fz2+2Gz+H) =
i 0
(Fz+G+∆ )(Fz+G ∆ ). Since deg(Fz+G ∆ )=d 1, we infer that
0 0 0
− ± −
Fz2+2Gz+H is reducible. This is a contradiction.
Remark 7. We note that P(x,y,z) = Fz2 + 2Gz + H is irreducible if (a)
GCD(F,G,H) = 1, and if (b) the property (iii) holds. Indeed, under the as-
sumption (a), if P is reducible, then P =(Az+B)(Cz+D) with A,B,C,D
∈
C[x,y]. But, in this case, 4∆ = (AD BC)2, which contradicts the prop-
−
erty (iii).
Example8. LetC bethequarticcurvex2y2+y2z2+z2x2 2xyz(x+y+z)=0.
−
We have T(C)= (2,0),(0,3),(0,3) .
{ }
The (degenerate) quadratic Cremona transformation
ϕ :(x,y,z) (xy,y2,x(z cx)) (c C)
c
−→ − ∈
playedanimportantrolein[6,7]. Wefindthatϕ−1(x,y,z)=(x2,xy,yz+cx2).
c
We use the notations:
l:x=0, t:y =0, O =(0,0,1),A=(1,0,c),B =(0,1,0).
4
Note that ϕ (l O )=B and ϕ (t O,A )=O.
c c
\{ } \{ }
Let C be an irreducible plane curve of type (d,d 2) with d 4. Suppose
− ≥
the singular point Q C of multiplicity d 2 has coordinates O. We have
seen in [7] that the st∈rict transform C′ = ϕ−(C) is an irreducible plane curve
c
of type (d′,d′ 2) for some d′. In [7, 8], by analyzing how a local branch γ at
P Sing(C) is−transformed by ϕ , we described Data[C′] from Data[C].
c
∈
3 Proof of Theorem 1
(i) We easily see that P Sing(C) Q is a double point, because LC =
(d 2)Q+2P, where L is∈the line pa\ss{ing}through P,Q. Let π : P˜2 P2 be
− →
theblowing–upatQ. LetE denotetheexceptionalcurve. TakealineLpassing
through Q. Let C′ (resp. L′) be the strict transform of C (resp. L). We have
C′L′ =2. ItfollowsthatP Sing(C′) Eisalsoadoublepoint. Thus,Data(C)
has the shape as in Theore∈m 1. Clea∩rly, k + k′ = mult (C) = d 2.
h h′ Q −
The second part of the condition (1) follows from the genus formula. The
P P
condition (2) follows from the Hurwitz formula applied to the double covering
C˜ P1, which corresponds to the projection of C from Q, where the C˜ is the
→
non–singular model of C. For the proof of the conditions (3), (4), we refer to
[7]. We will give an alternative, direct proof in Proposition 9.
(ii) Let F(x,y)z2+2G(x,y)z+H(x,y)=0 be the defining equation of C as in
Sect. 2. Let T(C) be the 2–formula of C. Setting
′
T (C)= (p,q) T(C) p>0 or q 2 ,
{ ∈ | ≥ }
we renumber the pairs (p ,q ) T′(C) in the following way:
i i
∈
(1) p >0, q >0 and q is even for i=1,...s,
i i i
(2) either p >0, q >0 and q is odd, or p >0,q =0 for i=s+1,...,N,
i i i i i
(3) p =0, q >0 and q is even, for i=N +1,...,N +n,
i i i
(4) p =0, q 3 and q is odd, for i=N +n+1,...,N +n+n′.
i i i
≥
Proposition 9. Set
(1) for i=1,...s,
′
k =k =p /2,a =q /2 if p q ,
i i i i i i ≤ i
(ki =qi/2,ki′ =pi−qi/2,ai =qi/2 if pi >qi,
(2) for i=s+1,...N,
k =p ,a =(q 1)/2 if q >0,
i i i i i
−
(ki =pi,ai =0 if qi =0,
(3) b =q /2, for j =1,...,n,
j N+j
′
(4) b =(q 1)/2. for j =n+1,...,n+n.
j N+j−
5
Then Data(C) is given as in Theorem 1, (i).
Proof. Take (p ,q ) T′(C). Write t as t = (α ,β ). Let L be the line
i i i i i i i
∈
β x=α y. Byarrangingthecoordinates,wemayassume(α ,β )=(0,1). Write
i i i i
F, G and ∆ as F =xpiF0, G=xmG0 and ∆=xqi∆0, where m=ordti(G).
We first consider the case in which p =0. Since ∆(t )=0, we have
i i
F(t )z2+2G(t )z+H(t )=F(t )(z+G(t )/F(t ))2.
i i i i i i
It follows that CL =(d 2)Q+2P, where P =(0,1, G(t )/F(t )). Let U be
i i i
− −
a neighbourhood of P such that y =0 and F(x,y)=0 for all (x,y,z) U. We
6 6 ∈
use the affine coordinates (x,z)=(x/y,z/y). We have
F(x,y)(F(x,y)z2+2G(x,y)z+H(x,y))
=y2d−2((F(x,1)z+G(x,1))2 ∆(x,1)).
−
ThusC isdefinedbytheequation(F(x,1)z+G(x,1))2 =∆(x,1)onU. Letting
u = F(x,1)z +G(x,1) and v = (qi ∆0(x,1))x, C is defined by the equation
u2 =vqi around P. Thus P ∈Sing(pC)\{Q} if qi ≥2. In this case, we have
1 if q is even,
m (C)= 1 qi/2 i
P
(((cid:0)2(cid:1)(qi−1)/2) if qi is odd,
which gives the assertions (3), (4).
Conversely,takeP Sing(C) Q . LetLbethelinepassingthroughP,Q.
∈ \{ }
Write L : βx = αy. Since CL = (d 2)Q+2P, we have F(α,β) = 0 and
− 6
∆(α,β) = 0. For (α,β) P1, we find a pair (0,q) T(C). We see from the
∈ ∈
above argumentthat C is defined by the equation u2 =vq near P. Thus q 2.
We now consider the case in which p > 0. Let π : P˜2 P2 be≥the
i
→
blowing–upatQandE theexceptionalcurveofπ. Weusetheaffinecoordinates
(x,y) = (x/z,y/z) of U := (x,y,z) P2 z = 0 . Put V = π−1(U). There
{ ∈ | 6 }
exist an open cover V = V V (V = C2) with standard coordinates (u ,v )
1∪ 2 j ∼ j j
of V such that π : V (u ,v ) (u v ,u ) and π : V (u ,v )
(u ,ju v ). Note t|hVa1t E1is∋defi1ned1by7→u =1 10 on1 V . Th|eV2stric2t ∋trans2for2m L7→′
2 2 2 j j i
of L is defined by v = 0 on V . Let P be the unique point E L′. We
haveiP =(0,0) on V .1The strict1transformC′ of C is defined by the∩equiation:
1
F(v ,1)+2G(v ,1)u +H(v ,1)u2 =0onV . Bythedefinitionofp andm,the
1 1 1 1 1 1 i
curveC′ isdefinedbytheequation: F vpi+2G vmu +Hu2 =0. Inparticular,
0 1 0 1 1 1
we have (C′E) = p . If q = 0, then we must have m = 0 (See the proof of
P i i
Lemma 6). Hence C′ is smooth at P. If q > 0, then we have m > 0 (Cf. the
i
proof of Lemma 6). Since C is irreducible, we see that H(t ) = 0. We have
i
H(F +2G vmu +Hu2)=(Hu +G vm)2 ∆. This means tha6t C′ is defined
0 1 1 1 1 0 1 −
by the equation:
(H(v ,1)u +G (v ,1)vm)2 ∆(v ,1)=0
1 1 0 1 1 − 1
in a neighborhood of P. Letting u = H(v ,1)u + G (v ,1)vm and v =
1 1 0 1 1
(qi ∆0(v1,1))v1, C′ is defined by the equation u2 =vqi around P. We have
p 1 if q is even,
′ 1 qi/2 i
mP(C )=((cid:0)2(cid:1)(qi−1)/2) if qi is odd,
(1) if q =0,
i

6
which gives the values of a in (1), (2). We prove the remaining assertions in
i
(1). If qi is even, then C′ has two branches γ+,γ− at P defined by
H(v ,1)u +G (v ,1)vm vqi/2 ∆ (v ,1)=0.
1 1 0 1 1 ± 1 0 1
In case p > q , we have m = q /2 (See the propof of Lemma 6). We infer that
i i i
one of the intersection numbers (Eγ+)P and (Eγ−)P is equal to qi/2. The
other one must be equal to p q /2, because (EC′) = p . In case p q ,
i i P i i i
− ≤
we have m pi/2 (Cf. the proof of Lemma 6). Thus (Eγ±)P pi/2, hence
(Eγ±)P =p≥i/2. Consequently, we obtain the pair (ki,ki′). ≥
Conversely, take P C′ E. We assume P V . Write the coordinates of
1
∈ ∩ ∈
P as P = (0,β). The equation F(β,1)+2G(β,1)u +H(β,1)u2 = 0 has the
1 1
solution u =0 as C′ passes through P. Thus F(β,1)=0. For (β,1) P1, we
1
∈
find a pair (p,q) T(C) with p>0.
∈
Remark 10. For i = s + 1,...,N, if (k ,a ) = (1,0), then we have either
i i
(p ,q ) = (1,1) or (1,0). The case (p ,q ) = (1,0) occurs if and only if the line
i i i i
L is a flex–tangent line to the corresponding branch at Q.
i
4 Proof of Theorem 2
Let C be given by the equation (See Sect. 2):
F(x,y)z2+2G(x,y)z+H(x,y)=0.
Put ∆ = G2 FH. Via linear coordinates change of x and y, we may assume
−
that y F∆. We then define a Cremona transformation (Cf. [2], Book II,
6|
Chap.V):
Φ(x,y,z)=(xyd−2,yd−1,Fz+G),
We find that Φ−1(x,y,z) = (xF,yF,yd−2z G). We see easily that the strict
transform C′ =Φ(C) is defined by the equa−tion:
y2(d−2)z2 =∆.
Write ∆= ki=1(x−λiy)qi, where the λi’s are distinct. Renumber qi’s so that
q ’s are odd for i = 1,...,l and q ’s are even for i = l + 1,...,k. Letting
i i
si =[qi/2] fQor i=1,...,k, we put S = ki=1(x−λiy)si and s= ki=1si. Note
that 2d 2= k q =2s+l. We next define a Cremona transformation:
− i=1 i Q P
P Ψ(x,y,z)=(xS,yS,ysz),
We find that Ψ−1(x,y,z)=(xys,ys+1,Sz). We see that Γ′ = Ψ(C′) is defined
by the equation:
l
y2(d−2)z2 = (x λ y)
i
−
i=1
Y
We see that l = 2g+2 and g = d s 2. Take a projective transformation:
ι:(x,y,z) (x,z,y). Finally, the i−mag−e Γ=ι(Γ′) has the affine equation:
→
2g+2
y2 = (x λ ).
i
−
i=1
Y
7
We now prove the latter half of Theorem 2. We start with the curve Γ and
a collection of systems of multiplicity sequences:
1 1
M = m, ,..., ,(2 ),...,(2 ) ,
1 1 bn+1 bn+n′
" (cid:18) (cid:19)b1 (cid:18) (cid:19)bn #
wherethemisthesystemofthemultiplicitysequencesofthesingularpointwith
multiplicity d 2. Letr(M), N(M)denotethe numberofthe branchesandthe
−
number ofthe differenttangent lines ofm. We haveto constructan irreducible
plane curveoftype (d,d 2) with Data(C)=M. In[7], we consideredthe case
−
in which g =0. We here assume that g 1. We follow the arguments in [7].
≥
First we deal with the cuspidal case given in Corollary of Theorem 1. See
also Proposition13.
Case(a): M =[(k),(2 ),...,(2 )],wherek =g+ b . Weusetheinduction
on n′. (i) M =[(g)]. Ibn1terchangbinn′g coordinates, we starit with the curve:
P
2g+2
Γ :x2gz2 = (y λ x).
0 i
−
i=1
Y
After a linear change of coordinates,we may assume that c= 2g+2( λ )=0.
Letting c = √c, we have Γ t = (2g)O+A +A′, where A =i(=11,0,−c )i,A6′ =
1 0 1 1 1 Q 1 1
(1,0, c ). LetΓ bethestricttransformofΓ viaϕ . UsingLemma1,(a)and
− 1 1 0 c1
Lemma 2, (e)* in [7], we see that Γ t = (2g 1)O+A . Write A = (1,0,c ).
1 2 2 2
−
LetΓ be the stricttransformofΓ viaϕ . Inthisway,we successivelychoose
2 1 c2
c ,...,c . It turns outthat Data(Γ )=[(g)]. (ii) Suppose we haveconstructed
1 g g
C with Data(C ) = [(k ),(2 ),...,(2 )], where k = g+ n′−1b . After
0 0 0 b1 bn′−1 0 i=1 i
a suitable change of coordinates, we may assume C l = k O+2B and C t =
0 0 1 0
(k +1)O+A . Note that the double covering C˜ P1 definPed through the
0 1
→ ′
projectionfromOtoaline,musthave2g+2branchpoints. Sincen 1<2g+2,
−
we see that a line passing through O is tangent to C at a smooth point B .
0 1
Write A = (1,0,c ). Let C be the strict transform of C via ϕ . We have
1 1 1 0 c1
C l = (k +1)O+2B and C t = (k +2)O+A . Write A = (1,0,c ). Let
1 0 1 0 2 2 2
C be the strict transform of C via ϕ . We have againC t=(k +3)O+A .
2 1 c2 2 0 3
Repeatinginthisway,wesuccessivelychoosec ,...,c anddefineC ,...,C .
1 bn′ 1 bn′
Then, the curve C =C has the desired property.
bn′
Case (b): M = [(2k+1,2 ),(2 ),...,(2 )], where k+1 = g+ b . As in
k b1 bn′ i
Case(a),wecansimilarlyprovethiscase. Forthefirststep: M =[(2g 1,2g−1)],
P−
it suffices to arrange coordinates so that Γ t = gO+2A with A = (1,0,0).
0 1 1
Put c = 0 and choose c ,...,c arbitrarily. Then we obtain Data(Γ ) = M
1 2 g g
(Cf. Lemma 1, (b) and Lemma 2, (e)* in [7]).
Case(c): M =[(2k,2 ),(2 ),...,(2 )],wherek =g+j+ b . Wealsouse
the induction on n′ aks+ijn Cabs1e (a). Fobrn′the first step: M = [(2(gi+j),2 )],
g+2j
P
westartwithacurveC withData(C )=[(g+j),(2 )]constructedinCase(a).
0 0 j
WeagainarrangecoordinatessothatC t=(g+j)O+2R,wherem (C )=(2 )
0 R 0 j
and R = (1,0,a). Choose c = a and c ,...,c arbitrarily. Then we have
1 2 g+j
6
Data(C )=M (Cf. Lemma 1, (a)*, (c) in [7]).
g+j
Starting with the cuspidal case, we can prove the general case in a simi-
lar manner to that in [7]. We have three subcases: I. N(M) = r(M) = 1,
8
II. N(M) = 1,r(M) = 2, III. N(M) 2. Here, we only give a proof for
M =[(k), 1 ,..., 1 ,(2 ),...,(2≥ )], where k =g+ n+n′b , which
1 b1 1 bn bn+1 bn+n′ j=1 j
isoneoftheremainingcasesinI.Weusetheinductiononn. (i)Weconstructed
(cid:0) (cid:1) (cid:0) (cid:1) P
a cuspidal curve C with Data(C) = [(k),(2 ),...,(2 )]. (ii) Suppose we
bn+1 bn+n′
have already constructed C with
0
1 1
Data(C )=[(k ), ,..., ,(2 ),...,(2 )],
0 0 1 1 bn+1 bn+n′
(cid:18) (cid:19)b1 (cid:18) (cid:19)bn−1
where k = g + n−1b + n+n′ b . By arranging coordinates, we have
0 j=1 j j=n+1 j
C l = k O + B + B′ and C t = (k + 1)O +A . Letting A = (1,0,c ),
0 0 1P 1 P0 0 1 1 1
the strict transform C of C via ϕ has the property C t = (k +2)O+A .
1 0 c1 0 0 2
Write A = (1,0,c ). We successively choose c ,...,c in this way. Then
2 2 2 bn
the strict transform C of C via ϕ ϕ has the desired property (Cf.
0 cbn ◦···◦ c1
Lemma 1, (d) and Lemma 2, (tn) in [7]). In particular, C contains a tacnode
1 at B =(0,1,0).
1 bn
R(cid:0) (cid:1)emark 11. Note that in Coolidge [2] (Book II, Chap.V), the cases in which
M =[(g)] and =[ g 1 ] were discussed.
g 1 g
(cid:0) (cid:1)(cid:0) (cid:1)
5 Defining equations
We now describe the defining equations for those curves listed in Corollary. In
[6, 7, 8], the defining equations were computed step by step by using quadratic
Cremona transformations. But, for some cases, we encountered a difficulty to
evaluate points in some special positions. We here employ the method used by
Degtyarev in [3].
Lemma 12. Consider two polynomials
d 2d
g(t)= c ti, δ(t)= d ti C[t].
i i
∈
i=0 i=0
X X
Suppose δ(0)=d =0. For k d, we have tk (g2 δ) if and only if
0
6 ≤ | −
(1) c = √d ,
0 0
±
(2) cj =(dj − ij=−11cicj−i)/(2c0) for j =1,...,k−1.
Proof. Write g(t)P2 = j=0bjtj. We see that bj = ji=0cicj−i for j ≤d.
Proposition 13. ThPe defining equations of irredPucible plane curves of type
(d,d 2) with genus g having only cusps are the following (up to projective
−
equivalence, the λ ’s are distinct).
i
(a) ykz2+2Gz+ G2 ∆ /yk =0, where
−
n o
n′ 2g+2
∆(x,y)= (x λ y)2bi+1 (x λ y).
i i
− −
i=1 i=n′+1
Y Y
Letting G(x,y) = kh+=10chxk+1−hyh, the coefficients c0,...,ck−1 are de-
termined by the condition yk (G(1,y)2 ∆(1,y)) (See Lemma 12).
P | −
9
k+1 n′ 2g+1
(b) (ykz+ c xk+1−hyh)2y (x λ y)2bi+1 (x λ y)=0.
h i i
− − −
h=0 i=1 i=n′+1
X Y Y
k+1 n′ 2g+1
(c) (ykz+ c xk+1−hyh)2 y2j+1 (x λ y)2bi+1 (x λ y)=0,
h i i
− − −
h=0 i=1 i=n′+1
X Y Y
where c =0.
0
6
Proof. Class (a). In this case, in view of the argument in the proof of Theo-
rem1,(ii),wehaveT(C)= (k,0),(0,2b1+1),...,(0,2bn′+1),(0,1),...,(0,1) .
{ }
Thus, we can write F = yk and ∆ as above. We must have yk (G2 ∆). In
| −
view of Lemma 12, the coefficients c0,...,ck−1 are uniquely determined. In
particular, c = 1. So by Remark 7, the defining equation is irreducible.
0
±
Class (b): We have T(C) = (2k +1,2k+1),(0,2b1 +1),...,(0,2bn′+1),
{
(0,1),...,(0,1) . We can arrange coordinates as
}
n′ 2g+1
F =y2k+1, ∆=y2k+1 (x λ y)2bi+1 (x λ y).
i i
− −
i=1 i=n′+1
Y Y
We infer that G=yk+1G for some G .
0 0
Class (c): We have T(C) = (2k,2k+2j +1),(0,2b1+1),...,(0,2bn′+1),
{
(0,1),...,(0,1) . We can arrange coordinates as
}
n′ 2g+1
F =y2k, ∆=y2k+2j+1 (x λ y)2bi+1 (x λ y).
i i
− −
i=1 i=n′+1
Y Y
It follows that G = ykG for some G . If we write G = k+1c xk+1−hyh,
0 0 0 h=0 h
thenwemusthavec =0,forotherwisethedefiningequationbecomesreducible
0
6 P
(See Remark 7).
Example 14. We give the defining equation of a cuspidal septic curve C with
Data(C) = [(5),(2),(2),(2),(2)] which are birational to the elliptic curve y2 =
(x2 1)(x2 λ2), (λ= 1,0).
− − 6 ±
3
y5z2+ 2x4 3(λ2+1)x2y2+ (λ4+6λ2+1)y4 x2z
− 4
1 (cid:8) 3 (cid:9)
(λ2+1)(λ4 10λ2+1)x6y+ 3λ8 28λ6 78λ4 28λ2+3 x4y3
− 8 − 64 − − −
(cid:8) +3λ4(λ2+1)x2y5 λ6y(cid:9)7 =0
−
Proposition 15. The defining equations of irreducible plane curves of type
(d,d 2) with genus g having only bibranched singularities are the following (up
−
to projective equivalence, the λ ’s are distinct).
i
k+1 n n+2g+2
(e) (ykz+ c xk+1−hyh)2 y2j (x λ y)2bi (x λ y)=0,
h i i
− − −
h=0 i=1 i=n+1
X Y Y
where c =0.
0
6
10