The gonality of singular plane curves revisited

(ii) We denote by m 1 , . . . , m n the multiplicity sequence of all singular pointson C (the infinitely near points are also included). We write asData(C) = [m 1 , . . . , m n ].We simply write [3 α , 2 β ] for [3, . . . , 3, 2, . . . , 2]. We also need the invariants:} {{ } } {{ }α β• δ =• V =n∑i=1n∑i=1m i (m i − 1),2m i (m i − 2).4(iii) Let ν i denote the i-th largest multiplicity among {m i }.ν 1 = ν. For i > n, we write ν i = 1. We setIn particular,σ = ν 2ν + ν 3ν + ν 4ν .(iv) Let q be a non-negative integer satisfying the property:2 ≤ d − ν − q ≤ d − ν.(v) Define⎧⎨ max {[d/ν], 3}k 0 =⎩max {[d/ν], 2}if d/ν ≥ σ − q/ν,otherwise.Theorem 1. Let C be a plane curve of type (d, ν) (d ≥ 6). Let q be as above.We have the inequality:G ≥ d − ν − qif the following two conditions are satisfied.(A) δ ≤ Q([d/2]) − (d − ν − q),(B) δ ≤ Q(k 0 ) − (d − ν − q) + V .Remark 1. We make some remarks.(1) In [4], we employed the following assumption, which is stronger than bothConditions (A) and (B) (for the case in which q = 0 and [d/ν] ≥ 2):(C)δ ≤ Q([d/ν]) − (d − ν).(2) In case ν = 2, since Condition (A)= Condition (B), we obtain the result:δ ≤ Q([d/2]) − (d − 2 − q) ⇒ G ≥ d − 2 − qIn particular, if q = 0, then we recover the criteria in Coppens-Kato [1].2

(3) Furthermore, if ν = 2, then the above estimate from below is best possible.For those curvesk∏C k : y 2k+1 = (x − a i ) 2 ,we have d = 2k + 1, Data(C) = [2 k 2], δ = k 2 , Q([d/2]) = k(k + 1). We haveδ = Q([d/2])−(d−2−q) for q = k −1. By Theorem 1, we conclude that G ≥ k.On the other hand, C k is birational to the curve y 2k+1 = ∏ ki=1 (x − a i) and wesee easily that G ≤ k. Hence, the equality G = k holds.(4) In view of the proof, it seems that Condition (A) cannot be improved. Thereremains some possibility to weaken Condition (B). If ν ≥ 6 and if k 0 = [d/ν],then Condition (B) implies Condition (A). For the curveC : y 30 =i=12∏24∏(x − a i ) 6 (x − b j ),i=1we have ν = 6 and k 0 = 6. So Condition (B) is satisfied for q = 0. We obtainG = 30. Note that Data(C) = [6 16 ], δ = 240 and V = 96.For the case in which ν = 3, we succeeded in excluding Condition (B).Theorem 2. Let C be a plane curve of type (d, 3) (d ≥ 6). Let q be as above.We have the inequality:G ≥ d − 3 − qj=1if Condition (A) with ν = 3 in Theorem 1 is satisfied.Remark 2. In case ν ≥ 5 and q = 0, Condition (A) alone does not give theresult G = d − ν. There are counterexamples.For those cases in which ν = 2, 3, we state the results in terms of the genusg of the curve C.Proposition 1. Let C be a plane curve of type (d, 2) with d ≥ 6. If{(d + 1)(d − 3) 1/4 if d is even,g ≥ −40 if d is odd,then the equality: G = d − 2 holds.Proposition 2. Let C be a plane curve of type (d, 3) with d ≥ 6. If{(d + 2)(d − 4) 0 if d is even,g ≥ +41/4 if d is odd,then the equality: G = d − 3 holds.3

Remark 3. For ν = 2, the above result is best possible (Coppens-Kato[1]). Forν = 3, at least for d = 8, the result is best possible. For d = 8, we infer fromProposition 2 that if g ≥ 10, then G = 5. Now, the octic curveC : y(x 2 − 1) 3 (x − 2) − (y 2 − 1) 3 (y − 2) = 0has four ordinary triple points, hence Data(C) = [3 4 ]. Let φ be the rationalfunction on C induced by Φ = (x 2 − 1)/(y 2 − 1). We see that deg(φ) = 4.On the the hand, by Theorem 2, we have G ≥ 4. We conclude therefore that(g, G) = (9, 4).3 Rational functionsLet C : F = 0 be a plane curve of type (d, ν). A rational function φ on C isinduced by a rational function Φ = G/H on the projective plane P 2 as φ = Φ |C ,where the G, H are relatively prime homogeneous polynomials having the samedegrees and F ̸ | H. Letting k = deg(G) = deg(H), we say that Φ is a rationalfunction of degree k on P 2 . The degree r = deg(φ) can be computed as follows.By successive blowing ups over P 2 , we simultaneously resolve the base pointsof the rational map Φ : P 2 → P 1 and the singular points of C. Let ˜Φ : X → P 1denote the resulting morphism.X s = X ⊃ ˜C↓.↓X 1↓X 0 = P 2 ⊃ C˜Φ↘Φ P 1For each i, let π i : X i → X i−1 be the blowing up at a point P i . Let m i denotethe multiplicity of C at P i . Note that m i ≥ 0. Let ˜C be the strict transform ofC on X. Then the degree r is nothing but the intersection number of ˜C and afibre f of the map ˜Φ. By construction,˜C ∼ π ∗ O(d) −f ∼ π ∗ O(k) −s∑m i E i ,i=1s∑a i E ii=1(a i ’s are non-negative integers).where the E i is the total transform of the exceptional curve of π i .4

Lemma 1. We have the formulae:(i) k 2 =s∑a 2 i ,i=1(ii) r = dk −s∑a i m i .Proposition 3. We now come to the inequality stated in Introduction. Namely,we haver + δ − Q(k) ≥ V.Proof. See [4].i=1Example 1. We examine the case in Remark 3.C : y(x 2 − 1) 3 (x − 2) − (y 2 − 1) 3 (y − 2) = 0Φ = x2 − 1y 2 − 1 , φ = Φ |CWe blow up 4 points P 1 = (1, 1), P 2 = (−1, 1), P 3 = (−1, −1), P 4 = (1, −1). Wehave a i = 1, m i = 3 for i = 1, . . . , 4. Thus we see that r = 8 · 2 − (1 · 3) × 4 = 4.˜CfE 1E 2E 3E 4˜ΦP 1Proposition 4. If r < d − ν − q, then we have k ≥ k 0 .Proof. Suppose r < d − ν − q. It is easy to show that k ≥ 2. In case k = 2, wemust have d/ν < σ − q/ν (Cf., [4]). Now we prove the inequality: k ≥ [d/ν].By Lemma 1 (ii), we infer that r ≥ dk − ( ∑ a i )ν. Since ∑ a i ≤ ∑ a 2 i = k2 ,we see that r ≥ k(d − kν). By using the assumption r < d − ν, we have(k 2 − 1)ν > (k − 1)d, from which follows that k + 1 > d/ν. Hence we havek ≥ [d/ν].5

4 ProofsProof of Theorem 1. (1) Assume to the contrary that G < d − ν − q. Thereexists a rational function φ on C such that r = deg(φ) < d − ν − q.(2) According to the following key Lemma, if Condition (A) is satisfied, then,we can find a rational function Φ on P 2 with degree k = deg(Φ) < [d/2], whichinduces φ.Lemma 2 ([4, 5]cf., [1, 2]). Let φ be a rational function on a plane curve Cof degree d. Set r = deg(φ). If there is a positive integer l < d, which satisfiesthe inequality:r + δ < Q(l + 1),then there exists a rational function Φ on P 2 with k = deg(Φ) ≤ l, which inducesφ.Indeed, under Condition (A), we have the inequality:r + δ < Q([d/2]).Therefore by letting l = [d/2] − 1 in Lemma 2, we can find a rational functionΦ on P 2 with degree k = deg(Φ) < [d/2], which induces φ.(3) By Proposition 4, we have k ≥ k 0 . Since k 0 ≤ k < [d/2], we haveQ(k 0 ) ≤ Q(k). By Proposition 3, we haveWe then obtain the inequality:which contradicts Condition (B).r + δ ≥ Q(k) + V.δ > Q(k 0 ) − (d − ν − q) + V,Proof of Theorem 2. In case ν = 3, we can write Data(C) = [3 α , 2 β ]. In asimilar manner to that in the proof of Theorem 1, we may assume the existenceof a rational function φ on C of r = deg(φ) < d − 3 − q, and a rational functionΦ on P 2 with k = deg(Φ) < [d/2], which induces φ.Lemma 3. For ν = 3, we have the following bound:∑ai m i ≤ 2k 2 + α.Proof. We argue as follows:∑ ∑k 2ai m i ≤ ν i ≤ 2k 2 + α.i=16

We first prove the second inequality. We have three subcases: (a) α + β ≤ k 2 ,(b) α < k 2 ≤ α + β, (c) k 2 < α.⎧∑k 2⎨ 3α + 2β + (k 2 − α − β) = k 2 + 2α + β ≤ 2k 2 + α for (a)ν i ≤ 3α + 2(k 2 − α) = 2k 2 + αfor (b)⎩i=1 3k 2 < 2k 2 + αfor (c)To check the first inequality, we may assume that a 1 ≥ . . . ≥ a b ≥ 1 and a j =0 for j > b. Note that b ≤ k 2 and ∑ a i m i ≤ ∑ a i ν i . For the sake of simplicity,we restrict ourselves to the case in which (1)α + β ≤ b. In this case, we have∑ν i −k 2i=1b∑a i ν i = 3α + 2β + (k 2 − α − β) −i=1=⎛b∑α∑a 2 i + 2α + β − ⎝3 a i + 2i=1=i=1α∑(a i − 1)(a i − 2) +i=1α+β∑i=α+1b∑a i ν i ,i=1α+β∑i=α+1a i +(a i − 1) 2 +b∑i=α+β+1b∑i=α+β+1a i⎞⎠ ,a i (a i − 1) ≥ 0.We can similarly deal with the other subcases (2) α ≤ b < α + β, (3) b < α ≤α + β, (4) α ≤ b, (5) b < α. .Using this inequality, by Lemma 1, (ii), we haveOr equivalently, we haved − 3 − q > dk − (2k 2 + α).2(k 2 − 1) > d(k − 1) − (α − 1 − q).By dividing by k − 1 (Note that k ≥ 2), we obtaink > d 2 − α − 1 − q2(k − 1) − 1.Since k < [d/2], we must have α − 1 − q > 0. Using k ≥ [d/3], we obtaink > d (2 − λ − 1 λ =α − 1 − q ).2([d/3] − 1)Lettingk 1 =[ d2 − λ ],we have the inequality: k ≥ k 1 .Lemma 4. We have the inequality: V ≥ Q([d/2]) − Q(k 1 ).7

Now, we complete the proof of Theorem 2. By Proposition 3, we haveHence, we haveBy Lemma 4, we then havewhich contradicts Condition (A).Proof of Lemma 4. Setr + δ ≥ Q(k) + V ≥ Q(k 1 ) + V.δ > Q(k 1 ) + V − (d − 3 − q).δ > Q([d/2]) − (d − 3 − q),J = V − {Q([d/2]) − Q(k 1 )} .We have to prove that J ≥ 0. By definition, V = (3α)/4.Here, we only consider the case in which d is even and λ ∉ N. See [5], fordetails. We have k 1 = [d/2] − t, where t = [λ] + 1. We see thatQ([d/2]) − Q(k 1 ) = t 2 .Since k 1 < [d/2], we must have t ≥ 1. Hence, we have λ > 0. Since δ = 3α + β,Condition (A) gives the inequality:3α ≤ (d − 2) 2 /4 + 2 + q. (1)Noting that [d/3] ≥ (d − 2)/3, the definition of λ gives the inequality:3α ≥ 2λ(d − 5) + 3(1 + q). (2)Step 1. Combining (1), (2), we havewhich gives the inequality:d ≥ 8λ − 1 + 8q − 5d − 5 ,d ≥ 8λ − 7/2 (unless t = 1). (3)Indeed, the above inequality holds except if d = 6 and q = 0. If d = 6, q = 0,then by (1), we have α ≤ 2, hence λ = 1/2. So t = 1.Step 2. Using (2), (3), we haveJ ≥3(12 + q)4(+ 3t t − 49 ).12Consequently, we have J ≥ 0 if t ≥ 3 unless t = 3, q = 0.8

Step 3. We check the cases in which t ≤ 3. Assume to the contrary that J < 0(or, equivalently, 3α/4 < t 2 ). We have to consider the following three cases:(i) t = 1, α = 1, (ii) t = 2, α ≤ 5, (iii) t = 3, q = 0, α ≤ 11.(i) We have λ = 0. This is not the subcase which we consider. (ii) By thedefinition of λ, we have α = 5, d = 6, 8, q ≤ 1, or α = 4, d = 6, q = 0. But, by(1), if d = 8, q ≤ 1, then we have α ≤ 4. If d = 6, q ≤ 1, then we have α ≤ 2.So we arrived at a contradiction. (iii) By (2), we have 10 ≤ α ≤ 11, d ≤ 10, or6 ≤ α ≤ 9, d ≦ 8. By (1) again, if d = 10, q = 0, then α ≤ 6 and if d ≤ 8, q = 0,then α ≤ 2. We arrived at a contradiction.References[1] M.Coppens and T.Kato, The gonality of smooth curves with plane models,Manuscripta Math. 70, 5–25 (1990).[2] M.Coppens and T.Kato, Correction to the gonality of smooth curves withplane models, Manuscripta Math. 71, 337–338 (1991).[3] M.Namba, Families of meromorphic functions on compact Riemann surfaces,Lecture Notes in Math. 767, Springer-Verlag, 1979.[4] M.Ohkouchi and F.Sakai, The gonality of singular plane curves, Tokyo J.Math. 27, 137 – 147 (2004).[5] F.Sakai, The gonality of singular plane curves II, To appear in Proceedingsof the Conference on Affine Algebraic Geometry in Osaka, March, 2011.Fumio SAKAIDepartment of MathematicsGraduate School of Science and EngineeringSaitama UniversityShimo–Okubo 255Sakura–ku, Saitama 338–8570, JapanE–mail: fsakai@rimath.saitama-u.ac.jp9