Saitama Math. J. 49 Vol. 26 (2009), 49–65 2-Weierstrass points of certain plane curves of genus three Kamel Alwaleed and Masumi Kawasaki (Received October 28, 2009; Revised November 30, 2009) Abstract In this paper, we completely determine the 2-gap sequences of the 2-Weierstrass points on cyclic coverings of genus 3 with four branch points in the projective line. 1. Introduction Let Cn,m1,m2,m3,λ be the algebraic curves of genus g = 3 defined by the equation: n m1 − m2 − m3 ≥ ∈ C\{ } Cn,m1,m2,m3,λ : y = x (x 1) (x λ) ,n4,λ 0, 1 , ≤ ≤ − such that 1 mi n 1, Σimi and n are relatively prime. Then, Cn,m1,m2,m3,λ is isomorphic to one of the following plane curves [6]: 4 C1,a : y = x(x − 1)(x − a), 6 3 2 2 C2,a : y = x (x − 1) (x − a) , 4 3 C3,a : y = x (x − 1)(x − a), 6 3 3 C4,a : y = x (x − 1) (x − a). The 1-Weierstrass points of C1,a and C2,a are classified as follows ([9] and [6]). Proposition 1. We can classify the 1-Weierstrass points of C1,a as follows: ordinary flex hyperflex a = −1, 2, 1/2 0 12 otherwise 16 4 Proposition 2. We can classify the 1-Weierstrass points of C2,a as follows: ordinary flex hyperflex a = −1 16 4 P (a)=0 10 7 otherwise 22 1 2000 Mathematics Subject Classification. Primary 14H55; Secondary 14H10 Key words and phrases. Weierstrass points, quartic curves, sextactic points. 50 K. Alwaleed and M. Kawasaki where P (a)=11a4 − 1036a3 + 1794a2 − 1036a +11. Remark 1. The curves C3,a and C4,a are hyperelliptic (see subsection 2.3 below). So they have eight 1-Weierstrass points whose 1-gap sequences are {1, 3, 5}. In this paper, we compute the 2-gap sequences of the 2-Weierstrass points on Ci,a,i=1, ··· , 4. We note that C1,a is a smooth plane quartic and C2,a is iso- morphic to the smooth plane quartic curve C2,b which is defined by the equation (see subsection 2.3 below) 3 4 2 2 C2,b : y = x − bx − 1,b+4=0 . Our main results on C1,a and C2,b are stated as follows: Theorem 1. We can classify the 2-Weierstrass points of C1,a as follows: ordinary flex hyperflex 1-sextactic 2-sextactic 3-sextactic a = −1, 2, 1/2 0 12 48 0 0 P (a)=0 16 4 40 16 0 Q(a)=0 16 4 48 0 8 otherwise 16 4 72 0 0 where P (a)=(a2 + a +1)(a2 − 3a + 3)(3a2 − 3a +1) and Q(a)=(a2 − 6a +1)(a2 +4a − 4)(4a2 − 4a − 1). Theorem 2. We can classify the 2-Weierstrass points of C2,b as follows: ordinary flex hyperflex 1-sextactic 2-sextactic 3-sextactic b =0 16 4 72 0 0 P (b)=0 10 7 63 0 0 Q(b)=0 22 1 69 6 0 R(b)=0 22 1 72 0 3 otherwise 22 1 81 0 0 where P (b)=11b4 + 1080b2 + 3888,R(b)=b4 +18b +54 and Q(b) = 11953207059991b48 − 1170934255940539104b46 + ··· + 8494372341823291115301085441425408000000000000. Our main results on C3,a and C4,a are stated as follows: Theorem 3. We can classify the 2-Weierstrass points of C3,a as follows: 2-gap sequence {1, 2, 3, 4, 5, 7} {1, 2, 3, 4, 5, 8} {1, 2, 3, 4, 5, 9} {1, 2, 3, 5, 7, 9} a =3/4, 4/3 24 0 12 8 P (a)=0 16 16 4 8 otherwise 48 0 4 8 2-WEIERSTRASS POINTS OF CERTAIN PLANE CURVES 51 where P (a)=16a2 − 17a +16. Theorem 4. We can classify the 2-Weierstrass points of C4,a as follows: 2-gap sequence {1, 2, 3, 4, 5, 7} {1, 2, 3, 4, 5, 9} {1, 2, 3, 5, 7, 9} a =1/9, 8/9 24 12 8 P (a)=0 42 6 8 otherwise 60 0 8 where P (a) = 5103a4 − 10206a3 + 33183a2 − 28080a − 64. 2. Preliminaries Let C be a non-singular projective curve of genus g ≥ 2. Let f(x, y)=0be the defining equation of C. Take a divisor qK, where K is a canonical divisor and q =1, 2. Let dim |qK| = r ≥ 0. We denote by L(qK)theC-vector space of all meromorphic functions f such that div(f)+qK ≥ 0andby(qK) the dimension of L(qK)overC. For a point P on C,ifn is a positive integer such that (qK − (n − 1)P ) > (qK −nP ), we call this integer n a “ q-gap” at P .Thereareexactlyr +1 q-gaps and the sequence of q-gaps {n1,n2, ··· ,nr+1} such that n1 <n2 < ···<nr+1 is called the q-gap sequence at P. Assume that {f1, ··· ,fr+1} is a basis for L(qK). The Wronskian W (f1, ··· ,fr+1)of{f1, ··· ,fr+1} is given by f1(x) f2(x) ··· fr+1(x) f (x) f (x) ··· f (x) ··· 1 2 r+1 W (f1, ,fr+1)= ··· ··· ··· ··· , (r) (r) (r) f1 (x) f2 (x) ··· fr+1(x) here all the derivatives have taken with respect to x. Consider the divisor E : r(r +1) E =(r +1)qK +div(W (f1, ··· ,fr+1)) + div(dx). 2 r+1 Then the multiplicity of E at a point P can be computed as i=1 (ni − i)(see Miranda [10]). This integer is called q-weight at P and denoted by w(q)(P ). If w(q)(P ) > 0, we call the point P a q-Weierstrass point. Let Ω(q) (C)betheC-vector space of holomorphic q-differentials of C.Itis (q) ∼ known that Ω (C) = L(qK), therefore we have (q) g, q =1 dimCΩ (C)= 3g − 3,q=2 and the number of q-Weierstrass points N (q)(C) counted according to their q- weight is given by 52 K. Alwaleed and M. Kawasaki g(g2 − 1),q=1 N (q)(C)= 9g(g − 1)2,q=2 Lemma 1. An integer n is contained in q-gap sequence at P if and only if there (q) is a holomorphic q-differential ω ∈ Ω (C) such that ordP (ω)=n − 1. Lemma 2. Let P beapointinaplanecurveC of genus 3. Then we can choose (2) abasis{ω1, ··· ,ω6} of Ω (C) in such a way that: 0=ordP (ω1) <ordP (ω2) < ···<ordP (ω6) < 9. Therefore we see that the 2-gap sequence at P is {1,ordP (ω2)+1,ordP (ω3)+1, ··· ,ordP (ω6)+1}. Lemma 3 (Duma [3]). Let σ be an involution of C. If the number of fixed points of σ is ≥ 3, then every fixed point is a q-Weierstrass point (q ≥ 2). Let Wq(C)bethesetofallq-Weierstrass points on a curve C.Wedenoteby G(q)(P )theq-gap sequence at the point P ∈ C. Lemma 4. Let Φ:C −→ C be a birational transformation between the non- singular algebraic curves C and C. Then we have (q) (q) Φ(Wq(C)) = Wq(C ) and G (Φ(P )) = G (P ). Remark 2. We have the following facts: (i) Let C be a plane curve of genus 3. Then for any P ∈ C we have w(2)(P ) ≤ 6. Furthermore, equality occurs if and only if C is hyperelliptic and P is a 1-Weierstrass point [5]. (ii) Let C be a plane curve of genus 3. Let P be a point on C such that P is a 2-Weierstrss point and P is not a 1-Weierstrss point. Then we obtain w(2)(P ) ≤ 4[3]. Using Remark 2, we obtain the following lemma. Lemma 5. The 2-gap sequences of the 2-Weierstrass points of a plane curve of genus three are as follows: 2-weight 2-gap sequence 1 {1, 2, 3, 4, 5, 7} 2 {1, 2, 3, 4, 5, 8} {1, 2, 3, 4, 6, 7} 3 {1, 2, 3, 4, 5, 9} {1, 2, 3, 5, 6, 7} 4 {1, 2, 3, 4, 6, 9} 6 {1, 2, 3, 5, 7, 9} 2-WEIERSTRASS POINTS OF CERTAIN PLANE CURVES 53 We use the following notation to describe the repeated roots of a polynomial. + Notation. Let f(x)beapolynomial.WewriteT (f)=(nα,mβ, ···),n,m∈ Z , if f(x)hasα roots of multiplicities n, β roots of multiplicities m, and so on. For instance the polynomial f(x)=x3(x − 1)2(x +1)2(x3 − 2) is of type T (f)=(3, 22, 13). 2.1 Subresultant Method To determine the multiplicities of the repeated roots of a polynomial with a parameter, we use the subresultant method [6]. We denote by R(k)(f(x),g(x); x)tothesubresultant of degree k for the poly- nomials f(x)andg(x). Lemma 6. The polynomials f(x) and g(x) have a non-constant common factor of multiplicity at least k if and only if R(i)(f(x),g(x); x)=0,i=1, 2, ··· ,k. Definition. For a polynomial f(x), we define s := s(f), if the subresultant of de- gree i, R(i)(f(x),f(x); x)=0, for all i =1, ··· ,s and R(s+1)(f(x),f(x); x) =0 . k ni Lemma 7. Take a polynomial f (x)=c i=1 (x − ai) , where ai = ai if i = j k and c is a complex number. Then s (f)= i=1 (ni − 1) .