Using the same method we provide negative answers to the following questions:
Is it possible to find real equations for complex polynomials in two variables
up to topological equivalence (Lee Rudolph) ? Can two topologically equivalent
polynomials be connected by a continuous family of topologically equivalent
polynomials ?Comment: 6 pages, to appear in C. R. Acad. Sci. Pari

Bodin, Arnaud

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Arnaud  Bodin

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Non-reality and non-connectivity of complex polynomials

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C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042Géométrie algébrique/Algebraic GeometryNon-reality and non-connectivity of complexpolynomialsArnaud BodinLaboratoire Agat, UFR de mathématiques, Université Lille 1, 59655 Villeneuve d’Ascq cedex, FranceReceived 8 October 2002; accepted 15 October 2002Note presented by Étienne Ghys.Abstract Using the same method we provide negative answers to the following questions: is itpossible to find real equations for complex polynomials in two variables up to topologicalequivalence (Lee Rudolph)? Can two topologically equivalent polynomials be connected bya continuous family of topologically equivalent polynomials? To cite this article: A. Bodin,C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042. 2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SASNon réalité et non connectivité des polynômes complexesRésumé Pour les polynômes de deux variables complexes, nous construisons des contre-exemplesaux questions suivantes : à équivalence topologique près, peut-on toujours trouver une équa-tion réelle à un polynôme complexe (Lee Rudolph) ? Deux polynômes topologiquementéquivalents peuvent-ils être reliés par une famille de polynômes topologiquement équiva-lents ? Pour citer cet article : A. Bodin, C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042. 2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS1. IntroductionTwo polynomials f,g ∈ C[x, y] are topologically equivalent, and we will denote f ≈ g, if there existhomeomorphisms : C2 → C2 and : C → C such that g◦= ◦f . They are algebraically equivalent,and we will denote f ∼ g, if we have  ∈ AutC2 and  = id.It is always possible to find real equations for germs of plane curves up to topological equivalence.In fact the proof is as follows: the topological type of a germ of plane curve (C,0) is determined bythe characteristic pairs of the Puiseux expansions of the irreducible branches and by the intersectionmultiplicities between these branches. Then we can choose the coefficients of the Puiseux expansions in R(even in Z). Now it is possible (see [7], appendix to Chapter 1) to find a polynomial f ∈ R[x, y] (even inZ[x, y]) such that the germ (f = 0,0) is equivalent to the germ (C,0).This property has been widely used by N. A’Campo and others (see [1] for example) in the theory ofdivides. Lee Rudolph asked the question whether it is true for polynomials [10]. We give a negative answer:THEOREM A. – Up to topological equivalence it is not always possible to find real equations for complexpolynomials.2. We now deal with another problem. In [5] we proved that a family of polynomials with someconstant numerical data are all topologically equivalent. More precisely for a polynomial let m =(µ,#Baff, λ,#B∞,#B) be the multi-integer respectively composed of the affine Milnor number, the numberE-mail address: Arnaud.Bodin@agat.univ-lille1.fr (A. Bodin). 2002 Académie des sciences/Éditions scientifiques et médicales Elsevier SAS. Tous droits réservésS1631-073X(02)02597-9/FLA 1039A. Bodin / C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042of affine critical values, the Milnor number at infinity, the number of critical values at infinity, the numberof critical values (with B = Baff ∪ B∞). Then we have a global version of the Lê-Ramanujam µ-constanttheorem:THEOREM ([5]). – Let (ft )t∈[0,1] be a family of complex polynomials in two variables whose coefficientsare polynomials in t . Suppose that the multi-integer m(t) and the degree degft do not depend on t ∈ [0,1].Then the polynomials f0 and f1 are topologically equivalent.It is true that two topologically equivalent polynomials have the same multi-integers m. A naturalquestion is: can two topologically equivalent polynomials be connected by a continuous family oftopologically equivalent polynomials?THEOREM B. – There exist two topologically equivalent polynomials f0, f1 that cannot be connectedby a family of equivalent polynomials. That means that for each continuous family (ft )t∈[0,1] there exists aτ ∈ ]0,1[ such that fτ is not topologically equivalent to f0.It can be noticed that the answer is positive for algebraic equivalence. Two algebraically equivalentpolynomials can be connected by algebraically equivalent polynomials since AutC2 is connected by Jung’stheorem.Such kinds of problems have been studied by V. Kharlamov and V. Kulikov in [9] for cuspidal projectivecurves. They give two complex conjugate projective curves that are not isotopic. The example with lowestdegree has degree 825. In [2], Artal, Carmona and Cogolludo give examples of projective curves C, C′ ofdegree 6 that have conjugate equations in Q(√2) but the pairs (P2,C) and (P2,C′) are not homeomorphicby an orientation-preserving homeomorphism.3. The method used in this note is based on the relationship between topological and algebraic equivalence:we set a family (fs)s∈C of polynomials such that (fs = 0) is a line arrangement in C2. One of the linedepends on a parameter s ∈ C. There are enough lines in order that each polynomial is algebraicallyessentially unique. Moreover every polynomial topologically equivalent to fs is algebraic equivalent toa fs ′ , where s′ may be different from s.For generic parameters the polynomials are topologically equivalent all together and the function fs is aMorse function on C2 \f−1s (0). We choose our counter-examples with non-generic parameters, for such anexample fk is not a Morse function on C2 \ f−1k (0). The fact that non-generic parameters are finite enablesus to prove the requested properties.4. Non-realityLetfs(x, y)= xy(x − y)(y − 1)(x − sy).Let k, k̄ be the roots of s2 − s + 1.THEOREM A. – There does not exist a polynomial g with real coefficients such that g ≈ fk .Let C = {0,1, k, k̄}. Then for s ∈ C \ C , fs verifies µ = 14, #Baff = 3 and B∞ = ∅. By the connectivityof C \ C and the global version of the µ-constant theorem, two polynomials fs and fs ′ , with s, s′ /∈ C , aretopologically equivalent.The polynomials fk and fk̄ verify µ = 14, but #Baff = 2. Then such a polynomial is not topologicallyequivalent to a generic one fs , s /∈ C . In fact for s /∈ C there are two non-zero critical fibers with one doublepoint for each one. For s = k or s = k̄, there is only one non-zero critical fiber with an ordinary cusp.LEMMA 1. – Let s, s′ ∈ C. The polynomials fs and fs ′ are algebraically equivalent if and only if s = s′or s = 1 − s′.1040To cite this article: A. Bodin, C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042In particular the polynomials fk and fk̄ are algebraically equivalent.Proof. – Let us suppose that fs and fs ′ are algebraically equivalent. Then we can suppose that there exists ∈ AutC2 such that fs ′ = fs ◦. Such a  must send the lines (x = 0), (y = 0) to two lines, then  islinear:(x,y)= (ax+ by, cx+ dy). A calculus proves that (x,y)= (x, y) or (x,y)= (y− x, y) thatis to say s = s′ or s = 1 − s′. ✷LEMMA 2. – Fix s ∈ C and let f be a polynomial such that f ≈ fs . There exists s′ such that f ∼ fs ′ .Then Lemma 1 implies that there are only two choices for s′, but s′ can be different from s.Proof. – The curve f−1s (0) contains the simply connected curve xy(x−y)(x−sy), then the curve f−1(0)contains also a simply connected curve (with 4 components), by the generalization of Zaı̆denberg–Lintheorem (see [4]) this simply connected curve is algebraically equivalent to xy(x − y)(x − s′y). Then thepolynomial f is algebraically equivalent to xy(x− y)(x− s′y)P (x, y). The curve C defined by (P = 0) ishomeomorphic to C and admits a polynomial parameterization (α(t), β(t)) with α,β ∈ C[t]. Since C doesnot intersect the axe (y = 0), β is a constant polynomial; since C intersects the axe (x = 0) at one point αis monomial. An equation of P is now P(x, y)= yn−λ. By the irreducibility of C and up to an homothetywe get P(x, y)= y − 1. That is to say f is algebraically equivalent to fs ′ . ✷5. Let g ∈ C[x, y], if g(x, y) = ∑ai,j xiyj then we denote by ḡ the polynomial defined by ḡ(x, y) =∑āi,j xiyj . Then g = ḡ if and only if all the coefficients of g are real.We prove Theorem A. Let suppose that there exists a polynomial g such that g = ḡ and g ≈ fk . Thereexists s ∈ C such that g ∼ fs . Since fk has only two critical values, g and fs have two critical values. Thens = k or s = k̄ (s = 0 or s = 1 gives a polynomial with non-isolated singularities). As fk ∼ fk̄ we canchoose s = k. As a consequence we have  ∈ AutC2 such that g = fk ◦.Let  be = (p, q). Then g = pq(p− q)(q − 1)(p− kq). As g = ḡ we have:{p,q,p− q, q − 1,p− kq} = {p̄, q̄, p̄− q̄, q̄ − 1, p̄− k̄q̄}.Moreover by the configuration of the lines we have that q − 1 = q̄ − 1. So q = q̄ . Hence q ∈ R[x, y]. So{p,p− q,p− kq} = {p̄, p̄− q̄, p̄− k̄q̄}.Let suppose that p = p̄. Then p = p̄ − q or p = p̄ − k̄q . So p − p̄ equals −q or −k̄q . But p − p̄ hascoefficients in iR, which is not the case of q ∈ R[x, y] nor of k̄q . Then p = p̄. We have proved that = (p, q) has real coefficients. From g = fk ◦ we get ḡ = f̄k ◦ ̄. So g = fk̄ ◦ . On the one handfk = g ◦−1 and on the other hand fk̄ = g ◦−1. So fk = fk̄ , then k = k̄ which is false. It ends the proof.We could have end in the following way:  = (p, q) is in AutC2 with real coefficients, then ,considered as a real map, is in AutR2 (see [3, Theorem 2.1] for example). Then fk = g ◦−1 with g,−1with real coefficients, then fk has real coefficients which provides the contradiction.6. Non-connectivityLetfs(x, y)= xy(y − 1)(x + y − 1)(x − sy).Let C be the roots ofs(s − 1)(s + 1)(256s4 + 736s3 + 825s2 + 736s + 256)(256s4 + 448s3 + 789s2 + 448s + 256).Then for s ∈ C \ C , fs verifies µ= 14, #Baff = 4 and B∞ = ∅. For s, s′ /∈ C , fs and fs ′ are topologicallyequivalent. The roots of 256s4 + 448s3 + 789s2 + 448s + 256 are of the form {k, k̄,1/k,1/k̄}. Thepolynomials fk and fk̄ verifyµ= 14, but #Baff = 3. Then such a polynomial is not topologically equivalentto a generic one fs , s /∈ C .1041A. Bodin / C. R. Acad. Sci. Paris, Ser. I 335 (2002) 1039–1042THEOREM B. – The polynomials fk and fk̄ are topologically equivalent and it is not possible to find acontinuous family (gt )t∈[0,1] such that g0 = fk , g1 = fk̄ and gt ≈ fk for all t ∈ [0,1].The polynomials fk and fk̄ are topologically equivalent since we have the formula fk̄(x̄, ȳ)= fk(x, y).The two following lemmas are similar to Lemmas 1 and 2.LEMMA 3. – The polynomials fs and fs ′ are algebraically equivalent if and only if s = s′ or s = 1/s′.LEMMA 4. – Fix s and let f be a polynomial such that f ≈ fs . Then there exists s′ such that f ∼ fs ′ .7. We now prove Theorem B. Let us suppose that such a family (gt ) does exist. Then by Lemma 4 for eacht ∈ [0,1] there exists s(t) ∈ C such that gt is algebraically equivalent to fs(t) (in fact there are two choicesfor s(t)). We can suppose that there exists t ∈ AutC2 such that fs(t) = gt ◦t .We now prove that the map t →t can be chosen continuous, that is to say the coefficients of the definingpolynomials are continuous functions of t . We write gt = AtBtGt such that A0(x, y) = x , B0(x, y)= yand the maps t →At , t →Bt are continuous. So the automorphism−1t is defined by−1t (x, y)=(At(x, y),Bt (x, y)).By the inverse local theorem with parameter t , we have that t →t is a continuous function. Then the mapt → fs(t) is a continuous function, as the composition of two continuous functions. As s(t) is a coefficientof the polynomial fs(t), the map t → s(t) is a continuous function.As a conclusion we have a map t → s(t) which is continuous and such that s(0)= k and s(1)= k̄. It im-plies that there exists τ ∈ ]0,1[ such that s(τ ) /∈ C . On the one hand gτ is algebraically, hence topologically,equivalent to fs(τ); on the other hand gτ is topologically equivalent to fk (by hypothesis). As s(τ ) /∈ C ,fs(τ) and fk are not topologically equivalent (because #Baff are different), it provides a contradiction.8. I would like to thank Lee Rudolph for the question which initiated this work. The calculus have beendone with the help of SINGULAR, [8], and especially with author’s library critic described in [6]. Thisresearch has been done at the Centre de Recerca Matemàtica of Barcelona and was supported by a MarieCurie Individual Fellowship of the European Community (HPMF-CT-2001-01246).References[1] N. A’Campo, Real deformations and complex topology of plane curve singularities, Ann. Fac. Sci. ToulouseMath. 8 (1999) 5–23.[2] E. Artal, J. Carmona, J. Cogolludo, Effective invariants of braid monodromy and topology of plane curves,Preprint.[3] H. Bass, E. Connell, D. Wright, The Jacobian conjecture: reduction of degree and formal expansion of the inverse,Bull. Amer. Math. Soc. 7 (1982) 287–330.[4] A. Bodin, Classification of polynomials from C2 to C with one critical value, Math. Z. 242 (2002) 303–322.[5] A. Bodin, Invariance of Milnor numbers and topology of complex polynomials, Comment. Math. Helv., to appear.[6] A. Bodin, Computation of Milnor numbers and critical values in affine space and at infinity, Preprint.[7] D. Eisenbud, W. Neumann, Three-Dimensional Link Theory and Invariants of Plane Curve Singularities, in: Ann.Math. Stud., Vol. 110, Princeton University Press, 1985.[8] G.-M. Greuel, G. Pfister, H. Schönemann, SINGULAR 2.0: a computer algebra system for polynomial computa-tions. Centre for Computer Algebra, University of Kaiserslautern, 2001.[9] V. Kharlamov, V. Kulikov, Diffeomorphisms, isotopies, and braid monodromy factorizations of plane cuspidalcurves, C. R. Acad. Sci. Paris, Série I 333 (2001) 855–859.[10] L. Rudolph, Private communication, Geneva, 1998.1042

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