In 1984, Yoshihara conjectured that if two plane irreducible curves have isomorphic complements, they are projectively equivalent, and proved the conjecture for a special family of unicuspidal curves. Recently, Blanc gave counterexamples of degree 39 to this conjecture, but none of these is unicuspidal. In this text, we give a new family of counterexamples to the conjecture, all of them being unicuspidal, of degree 4m+1 for any m≥2. In particular, we have counterexamples of degree 9, which seems to be the lowest possible degre

Costa, Paolo

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Math. Z. (2012) 271:1185–1191DOI 10.1007/s00209-011-0909-4 Mathematische ZeitschriftNew distinct curves having the same complementin the projective planePaolo CostaReceived: 18 December 2010 / Accepted: 4 May 2011 / Published online: 21 July 2011© Springer-Verlag 2011Abstract In 1984, Yoshihara conjectured that if two plane irreducible curves haveisomorphic complements, they are projectively equivalent, and proved the conjecture fora special family of unicuspidal curves. Recently, Blanc gave counterexamples of degree 39to this conjecture, but none of these is unicuspidal. In this text, we give a new family ofcounterexamples to the conjecture, all of them being unicuspidal, of degree 4m + 1 for anym ≥ 2. In particular, we have counterexamples of degree 9, which seems to be the lowestpossible degree.1 The conjectureIn the sequel, we will work with algebraic varieties over a fixed ground field K, which canbe arbitrary.Conjecture 1.1 ([2]) Suppose that the ground field is algebraically closed of characteristiczero. Let C ⊂ P2 be an irreducible curve. Suppose that P2\C is isomorphic to P2\D forsome curve D. Then C and D are projectively equivalent, i.e. there is an automorphismα : P2 → P2 such that α(C) = D.This conjecture leads to several alternatives. Let ψ : P2\C → P2\D be an isomorphism.If the conjecture holds, then:• either ψ extends to an automorphism of P2 and we can choose α := ψ .• or ψ extends to a strict birational map ψ : P2  P2. In this case, there is an automor-phism α : P2 → P2 such that α(C) = D.P. Costa (B)UniGe, Genève, Suissee-mail: costapa5@etu.unige.ch1231186 P. CostaOtherwise, if ψ gives a counterexample to the conjecture, then:• either C and D are not isomorphic.• or C and D are isomorphic, but not by an automorphism of P2.In this text, we are going to study the conjecture in the case of curves of type I.Definition 1.2 We say that a curve C ⊂ P2 is of type I if there is a point p ∈ C such thatC\p is isomorphic to A1.We say that a curve C ⊂ P2 is of type II if there is a line L ⊂ P2 such that C\L is isomorphicto A1.All curves of type II are of type I, but the converse is false in general. Moreover, a curveof type I is a line, a conic, or a unicuspidal curve (a curve with one singularity of cuspidaltype).In the case of curves of type II, Yoshihara [2] showed that the conjecture is true, but ingeneral the conjecture does not hold. Some counterexamples are given in [1], but these curvesare not of type I.In this article, we give a new family of counterexamples, of degree 4m +1 for any m ≥ 2.These are all of type I, and some of them have degree 9, which seems to be the lowest pos-sible degree (see the end of the article for more details). In Sect. 2 we give a general way toconstruct examples, that we precise in Sect. 3. The last section is the conclusion.2 The constructionWe begin with giving a general construction, which provides isomorphisms of the formP2\C → P2\D where C, D are curves in P2. We start with the following definition:Definition 2.1 We say that a morphism π : S → P2 is a (−1)—tower resolution of a curveC if:(1) π = π1 ◦ · · · ◦ πm where πi is the blow-up of a point pi ,(2) πi (pi+1) = pi for i = 1, . . . , m − 1,(3) the strict transform of C in S is a smooth curve, isomorphic to P1, and has self-intersection −1.The isomorphisms of the form P2\C → P2\D are closely related to (−1)−tower resolu-tions of C and D because of the following Lemma:Lemma 2.2 ([1]) Let C ⊂ P2 be an irreducible algebraic curve and ψ : P2\C → P2\Dan isomorphism. Then, either ψ extends to an automorphism of P2, or it extends to a strictbirational map φ : P2  P2.Consider the second case. Let χ : X → P2 a minimal resolution of the indeterminaciesof φ, call E˜1, . . . , E˜m and C˜ the strict transforms of its exceptional curves and C in X andset  := φ ◦ χ . Then:(1) χ is a (−1)−tower resolution of C(2)  collapses C˜, E˜1, . . . , E˜m−1 and (E˜m) = D,(3)  is a (−1)−tower resolution of D.Remark 2.3 This lemma shows that if C does not admit a (−1)−tower resolution, then everyisomorphism P2\C → P2\D extends to an automorphism of P2. So counterexamples willbe given by rational curves with only one singularity.123New distinct curves having the same complement in the projective plane 1187We start with a smooth conic Q ⊂ P2 and φ ∈ Aut(P2\Q) which extends to a strictbirational map φ : P2  P2. Call p1, . . . , pm the indeterminacies points of φ; accordingto Lemma 2.2, we can order the points so that p1 is a point of P2 and pi is infinitely near topi−1 for i = 2, . . . , n. Consider χ : X → P2, a minimal resolution of the indeterminaciesof φ and set  := φ ◦ χ . Lemma 2.2 says that:(1) χ is a (−1)−tower resolution of Q,(2)  collapses Q˜, E˜1, . . . , E˜m−1 and (E˜m) = Q,(3)  is a (−1)−tower resolution of Q.Now, consider a line L ⊂ P2, which is tangent to Q at p = p1. Since φ contracts Q, thenC := φ(L) is a curve with a unique singular point which is φ(Q). Since L ∩ (P2\Q) 	 A1,we have C ∩ (P2\Q) 	 A1, which means that C is of type I.Consider now a birational map f ∈ Aut(P2\L) which extends to a strict birational mapP2  P2 and satisfies:(1) f (Q) = Q,(2) f (p1) = p1.Now, we are going to get a new birational map φ′ : P2  P2 which restricts to anautomorphism of P2\Q using the pi ’s and f . Set:p′i := f (pi ).Note that p′i is a well-defined point infinitely near to p′i−1 for i > 1.Let’s call χ ′ : X ′ → P2 the blow-up of the p′i ’s and E˜ ′1, . . . , E˜ ′m and Q˜′ the strict transformsof the exceptional curves of χ ′ and of Q in X ′.Since f (Q) = Q and f is an isomorphism in a neighbourhood of p1, the intersectionsbetween E˜1, . . . , E˜m and Q˜′ are the same as those between E˜1, . . . , E˜m and Q˜. Then thereis a morphism ′ : X ′ → P2 which contracts E˜ ′1, . . . , E˜ ′m−1 and Q˜′. Moreover, ′(E˜ ′m) is aconic, and up to composing by an automorphism of P2, we can suppose that ′(E˜ ′m) = Q.By construction, the birational map φ′ restricts to an automorphism of P2\Q. In fact, none ofthe p′i ’s belongs to L (as proper or infinitely near point), so φ′(L) is well defined. Moreover,φ′ collapses Q, so D := φ′(L) is a curve with a unique singular point which is φ′(Q).Set then ψ := φ′ ◦ f ◦ φ−1. We have the following commutative diagram:X ′′χ ′ P2φ′ ~~~~~~~~P2XχP2φ ~~~~~~~~fP2ψLemma 2.4 The map ψ : P2\C → P2\D induced by the birational map defined above isan isomorphism.Proof Since φ, φ′ ∈ Aut(P2\Q) and f ∈ Aut(P2\L), we only have to check thatψ(Q) = Q.Let χ : X → P2 (resp. χ ′ : X ′ → P2) be a minimal resolution of the indetermina-cies of φ (resp. φ′) and write  := φ ◦ χ (resp. ′ := φ′ ◦ χ ′). Call E˜1, . . . , E˜m (resp.E˜ ′1, . . . , E˜ ′m) the strict transforms of the exceptional curves of χ (resp. χ ′) in X (resp. X ′).1231188 P. CostaIt follows from Lemma 2.2 that (E˜m) = Q (resp. ′(E˜ ′m) = Q). Then factorising ψ we getψ(Q) = Q. unionsqNow we study the automorphisms α ∈ Aut(P2) such that α(C) = D.Lemma 2.5 If α ∈ Aut(P2) sends C onto D, then a := (φ′)−1 ◦ α ◦ φ is an automorphismof P2 and satisfies:(1) a(L) = L,(2) a(Q) = Q,(3) a(pi ) = p′i for i = 1, . . . , m.X ′′χ ′ P2φ′ ~~~~~~~~P2XχP2φ ~~~~~~~~aP2αProof Call q1, . . . , qm (resp. q ′1, …, q ′m) the points blown-up by  (resp. ′). Then these pointsare the singular points of C (resp. D). Since α is an automorphism such that α(C) = D, thenα sends qi on q ′i for i = 1, . . . , m, and lifts to an isomorphism X → X ′ which sends E˜i onE˜ ′i for i = 1, . . . , m − 1 and Q˜ on Q˜′.Since Q is the conic through q1, . . . , q5, then α(Q) = Q, and the isomorphism X → X ′sends E˜m on E˜ ′m . So χ and χ ′ contract the curves in X and X ′ which correspond by meanof this isomorphism, and we deduce that a ∈ Aut(P2).It follows then that a sends pi on p′i , a(Q) = Q and that a(L) = L . unionsq3 The counterexampleIn this section, we describe more explicitly the construction given in the previous section, bygiving more concrete examples.We choose n ≥ 1 and will define  : X → P2 which is the blow-up of some pointsp1, . . . , p4+2n , such that p1 ∈ P2, and for i ≥ 2 the point pi is infinitely near to pi−1.We call Ei the exceptional curve associated to pi and E˜i its strict transform in X . The pointswill be chosen so that:• pi belongs to Q (as proper or infinitely near points) if and only if i ∈ {1, . . . , 4},• pi belongs (as a proper or infinitely near point) to E4 if and only if i ∈ {5, . . . , 4 + n},• pi ∈ Ei−1\Ei−2 if i ∈ {5 + n, . . . , 4 + 2n}.Note that p1, . . . , p4+n are fixed by these conditions, and that p5+n, . . . , p4+2n dependon parameters. On the surface X , we obtain the following dual graph of curves (see Fig. 1).The symmetry of the graph implies the existence of a birational morphism  : X → P2which contracts the curves E˜1, . . . , E˜3+2n, Q˜, and which sends E4+2n on a conic. We maychoose that this conic is Q, so that φ =  ◦ −1 restricts to an automorphism of P2\Q.Calculating auto-intersection, the image by φ of a line of the plane which does not passthrough p1 has degree 4n + 1.123New distinct curves having the same complement in the projective plane 11893.1 Choosing the pointsNow we are going to choose the birational maps f and the points which define φ in order toget two curves which give a counterexample to the conjecture of Yoshihara.We choose that L is the line of equation z = 0, Q is the conic of equation xz = y2 andp1 = (0 : 0 : 1).We define the birational map f : P2  P2 by:f (x : y : z) = (μ2(λxz + (1 − λ)y2) : μyz : z2) with λ,μ ∈ K∗ and λ = 1.The map f preserves Q, and is an isomorphism at a local neighbourhood of p1. In conse-quence, f sends respectively p1, . . . , p4+2n on some points p′1, . . . , p′4+2n which will define′ : X → P2, ′ : X ′ → P2 and φ′ = ′ ◦ (′)−1 in the same way as φ was constructed.We describe now the points pi and p′i in local coordinates.Since f preserves Q and fixes p1, we have p′i = pi for i = 1, . . . , 4. Locally, the blow-upof p1, . . . , p4 corresponds to:φ4 : A2 → P2, φ4(x, y) = (xy4 + y2 : y : 1).The curve E4 corresponds to y = 0, and the conic Q˜ to x = 0. The lift of f in thesecoordinates is:(x, y) → (λμ2x, μy).The blow-up of the points p5, . . . , p4+n (which are equal to p′5, . . . , p′4+n) now corre-sponds to:φ4+n : A2 → A2, φ4+n(x, y) = (x, xn y).So the lift of f corresponds to:(x, y) →(λμ2x,yλnμ2n−1).We set p4+n+i = (0, ai ) for i ∈ {1, . . . , n} with an = 0. The blow-up of p5+n, . . . ,p4+n+i now corresponds to:φ4+n+i : A2 → A2, φ4+n+i (x, y) =(x, xi y + Pi (x))where Pi (x) = a1xi−1 + · · · + ai .Since f sends pi on p′i , we can set p′4+n+i = (0, bi ) for i ∈ {1, . . . , n} with bn = 0. Theblow-up of p′5+n, . . . , p′4+n+i then corresponds to:φ′4+n+i : A2 → A2, φ′4+n+i (x, y) =(x, xi y+Qi (x))where Qi (x) = b1xi−1+· · · + bi .So the lift of f corresponds to:(x, y) →(λμ2x,xi y + Pi (x) − λiμ2i−1 Qi (λμ2x)λiμ2i−1xi).The curves E4+n+i and E ′4+n+i correspond to x = 0 in both local charts. Since f is alocal isomorphism which sends pi on p′i for each i , it has to be defined on the line x = 0.Because Pi and Qi have both degree i − 1, this implies that:Pi (x) = λiμ2i−1 Qi (λμ2x) for i = 1, . . . , n.In particular, the coefficients satisfy:ai = λiμ2i−1bi for i = 1, . . . , n.1231190 P. CostaFig. 1 The dual graph of the curves E˜1, . . . , E˜3+2n , E4+2n , Q˜. Two curves have an edge between them ifand only they intersect, and their self-intersection is written in brackets, if and only if it is not −23.2 The counterexampleNow to get a counter example, we must show that any automorphism a : P2 → P2 suchthat a(L) = L , a(Q) = Q and a(p1) = p1 does not send pi on p′i for at least one i ∈{5 + n, . . . , 4 + 2n}. Let’s start with the following Lemma:Lemma 3.1 Let a : P2 → P2 be an automorphism such that a(L) = L , a(Q) = Q anda(p1) = p1. Then a is of the form:a(x : y : z) = (k2x : ky : z) where k ∈ K∗.Proof Follows from a direct calculation. unionsqTheorem 3.2 If n ≥ 2, the curves C and D obtained from the construction of the previoussection give a counter example to the conjecture.Proof Choose an = an−1 = 1.Since a is an automorphism, it lifts to an automorphism which sends E4+n+i on E ′4+n+i . Putλ = 1 and μ = k in the formula for f . Then this lift corresponds to:(x, y) →(k2x,xi y + Pi (x) − k2i−1 Qi (k2x)k2i−1xi)where Pi and Qi are the polynomials defined above.Since E4+n+i and E ′4+n+i both correspond to x = 0 in local charts, this lift has to be welldefined on x = 0. So since Pi and Qi both have degree i − 1, we get:Pi (x) = k2i−1 Qi (k2x) for i = 1, . . . , nand the constant terms satisfy ai = k2i−1bi for i = 1, . . . , n.Since an, an−1 = 0, then bn, bn−1 = 0. As explained in the previous section, a sends pi onp′i , so we get:λiμ2i−1bi = k2i−1bi for i = 1, . . . , n.This formula for i = n and i = n − 1 gives λ = 1 or μ = 0, which leads to acontradiction. unionsq123New distinct curves having the same complement in the projective plane 11914 ConclusionWe conclude by observing that the curves C and D of the previous construction have degree4n + 1 (using Fig. 1) and are of type I. In particular, we get a counterexample with a curveof degree 9 when n = 2. One can check by direct computation that the conjecture holdsfor irreducible curves of type I up to degree 5, because there is only one curve of degree 5which is of type I and not of type II, up to automorphism of P2. One can also check that allirreducible curves of type I of degree 6, 7 and 8 are of type II. So the curves of degree 9given by this construction leads to a counterexample of minimal degree among the curves oftype I.If we consider the conjecture for all rational curves, the counterexamples in [1] are ofdegree 39 (and not of type I). So we have new counterexamples with curves of lower degree.It seems that the curves of degree 9 give counterexamples of minimal degree among therational curves, but it hasn’t been shown yet.Acknowledgments I would like to thank J. Blanc for asking me the question and for his help during thepreparation of this article. I also thank T. Vust for interesting discussions on the result.References1. Blanc, J.: The correspondance between a plane curve and its complement. J. Reine Angew. Math. 633,1–10 (2009)2. Yoshihara, H.: On open algebraic surfaces P2 − C . Math. Ann. 268, 43–57 (1984)3. Yoshihara, H.: Rational curves with one cusp II. Proc. Am. Math. Soc. 100, 405–406 (1987)123

New distinct curves having the same complement in the projective plane

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New distinct curves having the same complement in the projective plane

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