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A new bound on the number of special fibers in a pencil of curves

In the previous paper by Pereira and the author, it was proved that any pencil of plane curves of degree greater than one with irreducible generic fiber can have at most five completely reducible fibers although no examples with five such fibers were ever found. Recently Janis Stipins has proved that if any two fibers of a pencil intersect transversally then it cannot have five completely reducible fibers. In this paper we generalize the Stipins result to arbitrary pencils. We also include into consideration more general special fibers that are the unions of lines and non-reduced curves. These fibers are important for characteristic varieties of line complements.


Introduction
The main object of this note is a 1-dimensional system of homogeneous polynomials of three variables of some degree d > 1 or the respective projective system (pencil) of curves in P 2 . We identify these systems and use the symbol P for both. If the generic member (fiber) of P is irreducible then the set of reducible fibers is finite. There are various known upper bounds on the size of this set that depend on d, see, for instance, [11,1].
We concerned mainly with completely reducible fibers, i.e., unions of lines not necessarily reduced. Certain results for these fibers with some extra conditions have appeared in classical mathematical literature (see [6,7]). New interest to completely reducible fibers was ignited by papers [8,4] where the union of these fibers was characterized by topological property of its complement and by combinatorics of intersection of lines. The best upper bound on the number k of the completely reducible fibers was found in [9] where the inequality k ≤ 5 was proved without any extra conditions on P. There were however no examples known with k = 5.
Then a better bound appeared in dissertation [10] where the inequality k ≤ 4 was proved assuming that any two different completely reducible fibers intersect transversally. The condition is equivalent to all the lines of those fibers forming a (k, d)-net (see [12]).
The goal of this note is to generalize the method from [10] in order to prove that k ≤ 4 without extra conditions. Also a more general kind of fibers appeared in studies by A.Dimca of characteristic varieties for line arrangement complements (see [3]). These are fibers that are unions of lines and non-reduced curves. We called these fibers special and include them into considerations.
Let us remark that there are many examples of pencils with k = 3 (for instance, see [4,9,10]) but the only one (up to isomorphism) with k = 4. This is the well-known Hesse pencil generated by a smooth plane cubic and its Hessian. We consider the case k = 4 in Theorem 3.1 and Problem at the end of the note.
Since the dissertation [10] is not published we do not assume that its proofs are available to the reader. As the result this note has proofs of three different types. Some of them, like Lemma 2.1 (attributed by Stipins to Dolgachev) and Proposition 2.12, are essentially taken from [10], some, like Theorem 2.7, made shorter and fit the more general situation, others, like Theorem 3.1, are new.
The author is grateful to J.Stipins for providing his dissertation and to I.Dolgachev for useful discussions.

Hessians of a pencil of curves
Recall from Introduction that as a rule we do not distinguish between a plane curve and its defining polynomial. Let P be a pencil of curves of degree d (d > 1) whose generic fiber is irreducible. We assume immediately that P has at least four special fibers F i , i = 1, 2, 3, 4, three of which, say F i , i = 1, 2, 3 are completely reduclible. Clearly any two fibers cannot have a common component whence the base B of P (i.e., the mutual intersection of all the fibers) is finite. Also it follows from the conditions on P that |B| > 1, i.e., P does not lie in a pencil of lines. Choosing the fibers F 1 and F 2 as generators we have F i = λ i F 1 + µ i F 2 , i = 3, 4 for some [λ i : µ i ] ∈ P 1 . For symmetry we put (λ 1 , µ 1 ) = (1, 0) and (λ 2 , µ 2 ) = (0, 1). We also put F (λ, µ) = λF 1 + µF 2 for all λ and µ.
Now we study the family parametrized by [λ : µ] ∈ P 1 of the Hessians of curves F (λ, µ). Let us recall that the Hessian of a plane curve (or a homogeneous polynomial of three variables x, y, z) is the polynomial H(F ) that is the determinant of the 3 × 3-matrix of the second derivatives of F . The points of F ∩ H(F ) are precisely the points of F that are either multiple or flexes (see, for instance [5]). In particular H(F ) contains F if and only if F is a special fiber. Also H(F ) is identically zero if and only if F is the union of concurrent lines. We put H(λ, µ) = H(F (λ, µ)). Notice that H(λ, µ) is a homogeneous polynomial of degree 3d − 6 in x, y, z whose coefficients are cubic forms in λ and µ. Proof. For a fixed point P ∈ B the polynomial H(λ, µ) evaluated at P is a cubic form in λ and µ that has at least 4 zeros at (λ i , µ i ). Thus this form is identically 0 which implies the statement. Lemma 2.1 is false for a pencil with just three completely reducible fibers.
The pencil has three completely reducible fibers -the generators and F (1, In fact we need much stronger statement than Lemma 2.1 -we need to check that the triple (F 1 , F 2 , H(λ, µ)) satisfies at any P ∈ B the conditions of the Max Noether Fundamental Theorem ( [5], p. 120). Then we will be able to apply that theorem. For that we need to discuss the multiplicities of the base points at the fibers of the pencil.
Fix P ∈ B and without any loss assume that P = [0 : 0 : 1]. Then the local ring O P (P 2 ) can be identified with C (x,y) [x, y]. Also for any homogeneous polynomial F of variables x, y, z put F * = F (x, y, 1) and write F * = i≥m F (i) where F (i) is the homogeneous component of degree i of F * and F (m) = 0. Here the non-negative integer m = m P (F ) = mult P (F ) is the multiplicity of P on F . The curve F (m) in P 2 is the union of concurrent lines with some multiplicities and each of them is a tangent line at P to F .
In the rest of the paper any time we fix a point P ∈ B we assume that P = [0 : 0 : 1]. Lemma 2.3. (i) The multiplicity of any P ∈ B is the same on any fiber.
(ii) No two fibers have a common tangent line at any P ∈ B.
Proof. (i) Fix P ∈ B and suppose there are fibers J and K such that m = m P (J) < m P (K). Then since J and K linearly generate all fibers we have m p (F ) = m for every fiber F different from K. Since the number of the completely reducible fibers is at least 3 there exist two fibers, say F 1 and F 2 , among the completely reducible ones different from K . Since for a completely reducible fiber F the polynomial F (m) is the product of all factors (with their multiplicities) of F that pass through P we conclude that F are linearly independent over C. Hence F (m) = 0 for every fiber F which contradicts the assumption about K.
(ii) Suppose there exist fibers J and K and a linear form α such that α divides J (m) and K (m) where m = m P (F ) for every fiber F . Then α divides F (m) for every fiber F , in particular for completely reducible fibers F 1 and F 2 . This implies that α divides F 1 and F 2 which contradicts the assumption about B.
Remark 2.4. Notice that the proof of Lemma 2.3 uses only that the pencil has three completely reducible fibers. In a different form it was proved in [4]. Now we study the multiplicity of the Hessians at the base points. First we record a simple general lemma. Thus the curve G is the union of non-concurrent lines whence it divides its Hessian which is not identically zero. Both statements follow.
Remark 2.6. Notice that the previous lemma can be applied to every fiber F of P and every P ∈ B with mult P (F ) > 1. Indeed if F were the union of lines passing through P then P would have been the only point in B which is impossible by conditions on P.
Proof. Fix [λ : µ] ∈ P 1 and consider several cases. (iii) m P (F i ) = 2, i = 1, 2. By Lemma 2.5 m P (H(λ, µ)) = 2 also. Since no ready criterion applicable to this case is known to me we need to work with Noether's conditions directly.
Remark 2.8. In this remark we briefly discuss the simple case d = 2. First of all every point of the base must have the multiplicity equal one on any fiber. Indeed if a point P has multiplicity two (the maximal possible) on F 1 then P is the only point of B which creates a contradiction (cf. the previous remark). Thus |B| = 4 which means that all the special fibers are completely reducible and the union of all completely reducible fibers form a (k, 2)-net for some k > 2 (e.g., see [4]). It is an easy property of nets that the only (k, 2)-net (up to linear isomorphism) is the (3, 2)-net closely related to the Coxeter group of type A 3 (for instance, see [12], p.1617). Because of that we will always assume in the rest of the paper that d ≥ 3.
Now we are ready to prove the main result about the family of the Hessians. Recall that for each special fiber F the Hessian H(F ) is divisible by F and put H(F i ) = C i F i for i = 1, 2, 3, 4 where each C i is a homogeneous polynomial of degree 2d − 6. Theorem 2.9. There exist homogeneous polynomials A and B of degree 2d − 6 in x, y, z whose coefficients are homogeneous cubic polynomials in λ and µ such that H(λ, µ) = AF 1 + BF 2 for all [λ : µ] ∈ P 1 . Moreover these polynomials can be chosen so that Proof. First of all Lemma 2.7 allows us to apply the Max Noether Fundamental Theorem which implies the following. For every (λ, µ) ∈ C 2 there exist homogeneous polynomials A = A(λ, µ), B = B(λ, µ) ∈ C[x, y, z] of degree 2d − 6 such that H(λ, µ) = AF 1 + BF 2 . Now we prove the statement, namely that A and B can be chosen depending polynomially on λ and µ and satisfying the conditions (*). For that we use elementary linear algebra over a polynomial ring. Indeed the coefficients of A and B are solutions of a finite system Σ of linear equations whose coefficients are complex numbers (in fact coefficients of polynomials F i , i = 1, 2) and the right-hand sides are values of cubic forms in λ and µ (coefficients of polynomials H(λ, µ)). It follows from the previous paragraph that this system is consistent for each values of λ and µ. This implies that reducing the coefficient matrix of Σ to a row echelon form one can skip the zero rows and moving the free-variables columns to the right-hand side obtain an equivalent system Σ ′ with a square non-degenerate matrtix of scalar coefficients.
Now we take care about the free variables. To get a solution polynomially depending on λ and µ it is enough to choose the free variables depending on parameters as arbitrary homogeneous cubic polynomials. We want however to find a special solution satisfying the 'boundary' conditions (*) at the special values of parameters. Since the conditions (*) give a solution of the system Σ ′ for these special values of parameters there are values of the free variables corresponding to them. Thus for each free variable there exists a (unique) cubic form admitting each of these special values at the respective special value of parameters. Now combining cubic forms in the right-hand side and using Cramer's rule we obtain a solution with the required properties.
Remark 2.10. If d ≤ 5 then the solution of the system Σ is unique whence it is automatically polynomial and satisfies the conditions (*). In this case there are no free variables. If however d > 5 then one can easily find non-polynomial solutions by choosing free variables non-polynomially depneding on parameters. where D and E are forms of degree 2d − 6 in x, y, z whose coefficients are quadratic forms in λ and µ. Moreover D(λ, 0) = λ 2 C 1 , E(0, µ) = µ 2 C 2 , and D(λ i , µ i ) = E(λ i , µ i ) = C i for i = 3, 4.
In turns out that we can learn more about D and E. The proof of the following proposition is taken from [10].
Proof. We prove that the polynomial D(0, 1) is proportional to E(0, 1). Then the first formula will follow from Corollary 2.11 and the second one can be proved similarly.
Corollary 2.13. The polynomials D and E can be expressed in terms of C 1 and C 2 as follows.
where X and Y are some forms in x, y, z of degree 2d − 6 . Now we are ready to prove the strongest property of the pencil of the Hessians. Proposition 2.14. There exists a form C of degree 2d − 6 in x, y, z and quadratic forms p and q in λ and µ such that Proof. Assume without any loss that C 1 and C 2 are not identically zero. Recalling that D(λ i , µ i ) = E(λ i .µ i ) for i = 3, 4, and using Corollary 2.13 we obtain the following two equalities: where i = 3, 4. Since λ i µ i = 0 for i = 3, 4 and also λ 3 µ 4 − λ 4 µ 3 = 0 we can eliminate Y − X and obtain The equalities v = u = 1 would imply that X = Y whence D = E and F (λ, µ) divides H(λ, µ) for all [λ : µ]. Since this is impossible for generic fibers (5) implies that C 1 and C 2 are proportional. Since we used about F 1 and F 2 only that they are special we can change the basis of the pencil and extend this statement to an arbitrary pair (C i , C j ), i = j . Choosing C = C 1 we see that every C i is proportional to C. Now consider again the equalities (1) and (2). Plugging in λ = λ 3 and µ = µ 3 yields (together with Corollary 2.11) that X and Y are proportional to C. Now factoring out C implies (3).

The main result
In this section we proof the following main result of the note. (ii) if it has three completely reducible fibers and another special fiber then this special fiber is completely reducible and every two completely reducible fibers intersect transversally (i.e., |B| = d 2 ).
Proof. (i) The first statement follows immediately from Proposition 2.14. Indeed for each special fiber its Hessian is divisible by the fiber. Thus the respective values of λ and µ form a root of the polynomial r(λ, µ) = p(λ, µ)λµ − q(λ, µ)λµ of degree four. If there are five such fibers then r is identically zero whence H(λ, µ) = p(λ, µ)CF (λ, µ) which is impossible for generic (λ, µ).
(ii) Notice that the first part of the statement (ii) follows from the second one since non-reduced curves cannot have intersection with d 2 points. Thus it suffices to prove the second part. Now we assume that P has three completely reducible fibers and another special fiber. First suppose that d = 3, i.e., C = ∅ where C is the curve from (3). Then the equality (3) implies that for each cubic which is a fiber of the pencil its Hessian is also a fiber. On the other hand, it is well-known that the Hesse pencil has any plane cubic as a fiber. Thus taking a generic fiber of our pencil and its Hessian we obtain two generators of our pencil and of the Hesse pencil whence these two pencils coincide. The statement (ii) for the Hasse pencil is well-known. Now suppose that d > 3, i.e., C = ∅, and there exists P ∈ B at which the multiplicity m of any fiber is greater than 1. We assume as before P = [0 : 0 : 1] and let α be a linear factor of F (m) (i.e., a tangent line to F at P ) where F is a generic fiber of P. According to Lemma 2.5, F (m) divides H(F ) (3m−4) whence so does α, i.e., α is a tangent line to H(F ) at P . Since F is generic we have H(F ) = CG where G is a fiber different from F . Since α is tangent to H(F ) but not to G (by Lemma 2.3) it is tangent to C. In other words we have proved that every tangent line to every generic fiber at P is tangent to C whence there exist fibers with common tngent lines at P . This contradicts Lemma 2.3.