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Obstructions on fundamental groups of plane curve complements

We survey various Alexander-type invariants of plane curve complements, with an emphasis on obstructions on the type of groups that can arise as fundamental groups of complements to complex plane curves. Also included are some new computations of higher-order degrees of curves, which are invariants defined in a previous paper of the authors.


Introduction
This paper is an attempt to give partial answers to the following question posed by Serre: what restrictions are imposed on a group by the fact that it can appear as fundamental group of a smooth algebraic variety? There are characteristic zero and respectively finite characteristic aspects of this problem, but we will restrict ourselves to the zero characteristic case. More precisely, our ground field will be C. In what follows we only treat the very special case of open varieties which are complements to hypersurfaces in C n (note that complements to closed varieties of complex codimension at least two are simply-connected).
By a Zariski theorem of Lefschetz type (see [5], Thm. 1.6.5), for a generic plane E relative to a given hypersurface V ⊂ C n , the natural map is an isomorphism. Therefore, possible fundamental groups of complements to hypersurfaces in C n are precisely the fundamental groups of plane affine curve complements. Thus, it suffices to restrict ourselves to the case of complements to curves in C 2 .
In view of the above, we can ask now the following refinement of Serre's question: what groups can be realized as fundamental groups of plane curve complements? what obstructions are there?
In the next sections we will discuss invariants of the fundamental group of an affine plane curve complement that are obtained by studying certain covering spaces of the complement: the Alexander polynomial is an invariant of the total linking number infinite cyclic cover, characteristic varieties (in particular, the support) are derived from studying the universal abelian cover, and the higher-order degrees are numerical invariants obtained by studying certain solvable covers associated to terms of the rational derived series of the group. We will see that these invariants obstruct many knot groups from being realized as fundamental groups of plane curve complements.
In the last section we include some examples of explicit calculations of the higher-order degrees associated to some curve complements. We will also find some examples of groups that cannot be realized as the fundamental group of a curve complement (in general position at infinity) because the higher-order degrees obstruct this.

Plane curve complements
Throughout this paper, we consider the following setting: Let G be a group, and assume that there is a reduced curve C = {f (x, y) = 0} in C 2 of degree d, with s irreducible components, such that G = π 1 (C 2 \C). For simplicity, we assume that C is in general position at infinity, that is, its projective completion is transverse to the line at infinity, though many results remain valid without this restriction on the behavior at infinity.
We will perform the dual task of studying topological properties of the curve by studying the fundamental group of its complement, while at the same time deriving obstructions on a group imposed by the fact that it is the fundamental group of an affine plane curve complement. For more comprehensive surveys on the topology of plane curves and a list of open problems, the interested reader may also consult the papers [20,21,28].
First note that H 1 (G) = H 1 (C 2 \ C) = G/G ′ = Z s , generated by meridians about the smooth parts of irreducible components of C.
Although in geometric problems fundamental groups of complements to projective curves play a central role, by switching to the affine setting (i.e., by also removing a generic line) no essential information is lost. Indeed, ifC ⊂ CP 2 is the projective completion of C, the two groups are related by the central extension Moreover, by [28], Lemma 2, the commutator subgroups of the affine and respectively projective complements coincide: 2.1. The linking number infinite cyclic cover of the complement. We begin with a brief survey of results on the Alexander polynomial of the curve C.
Let lk : G = π 1 (C 2 − C) → Z be the total linking number epimorphism, i.e. α → lk#(α, C). Note that lk factors through H 1 (G), sending the basis vectors of Z s to 1. Let U c be the covering of U corresponding to Ker(lk). U c will be called the total linking number infinite cyclic cover of the complement.
The group of deck transformations of U c is Z, and it acts on H 1 (U c ; C) by a generating transformation, thus making H 1 (U c ; C) into a module over C . This module is called the infinite cyclic Alexander module of the curve complement. As for some m ∈ Z and polynomials λ i (t) defined up to a unit of C[t, t −1 ]. In fact, the following result holds: Therefore, it does make sense to associate to C a polynomial, namely the order of H 1 (U c ; C) (cf. [25]). This is a global invariant of C (or of G) defined as follows: It is easy to see that the exponent of (t − 1) in ∆ C (t) is s − 1, where s is the number of irreducible components of C (e.g., see [28]). In particular, if the curve C is irreducible, the Alexander polynomial ∆ C (t) can be normalized so that ∆ C (1) = 1 (cf. [13]).
With each singular point x ∈ C there is an associated local Alexander polynomial, ∆ x (t), defined as the characteristic polynomial of the monodromy of local Milnor fibration at x (cf. [24]). Then: [13,16]) Up to a power of (t−1), the Alexander polynomial ∆ C (t) of a plane curve in general position at infinity divides the product x∈Sing(C) ∆ x (t) of the local Alexander polynomials at the singular points of C. Therefore the local type of singularities has an effect on the topology of C.
Zariski also showed that the position of singularities has an influence on the topology of C. Moreover, as Libgober observed, the Alexander polynomial is sensitive to the position of singularities ( [13]). The classical example of Zariski's sextics with six cusps will be discussed in section 3. Theorem 2.3 remains true without any assumption on the behavior of C at infinity, but one has to take into account the contribution of singularities at infinity. As a corollary of this fact, we have that Corollary 2.4. ∆ C (t) is cyclotomic. Moreover, for a curve C in general position at infinity, the zeros of ∆ C (t) are roots of unity of order d = deg(C).
It follows that many knot groups, e.g. that of figure eight knot (whose Alexander polynomial is t 2 − 3t + 1), cannot be of the form π 1 (C 2 −C). However, the class of possible fundamental groups of plane curve complements includes braid groups, or groups of torus knots of type (p, q) (see [12] §5, and the references therein).
Remark 2.5. The above divisibility result has been generalized to higher dimensions by Libgober ( [14,15]), who considered complements to affine hypersurfaces with only isolated singularities, and also by Maxim ([23]), who in his thesis treated the case of hypersurfaces with non-isolated singularities. From a divisibility result in [23], it follows that in Libgober's divisibility result it suffices to consider only the contribution of local Alexander polynomials at singular points contained in some fixed irreducible component of the hypersurface. In particular, this shows that the Alexander polynomial does not provide enough information about the topology of reducible curves (hypersurfaces). For example, if C is a union of two curves that intersect transversally, then ∆ C (t) = (t − 1) s−1 (see [28]). To overcome this problem, we study higher coverings of the complement.
2.2. The universal abelian cover of the complement. In this section, following [18] we define invariants associated to the universal abelian cover of the complement.
Let U ab be the universal abelian cover of U, i.e., the covering associated to the subgroup G ′ . Under the action of the covering transformation group, the universal abelian module H 1 (U ab ; Similarly, the i-th (algebraic) characteristic variety is defined by the i-th elementary ideal of A. Away from the trivial character, characteristic varieties of A coincide with jumping loci of homology of rank-one local systems on the complement (cf. [19]), defined as where L λ is the rank-one local system associated to the character λ.
In [6], these jumping loci are called topological characteristic varieties. By a result of Arapura ([1]), each V t i (G) is a union of subtori of the character torus, possibly translated by unitary characters. This fact imposes strong obstructions on the group G. Characteristic varieties, both algebraic and topological, give very precise information about the homology of (finite) abelian covers of U (e.g., see [18]).
is the zero-set of the multivariable Alexander polynomial of the link.
Remark 2.8. A multivariable Alexander polynomial of C could be defined as the greatest common divisor of all elements of the order ideal E 0 (A). However, if codim T s Supp(A) > 1, then this polynomial is trivial, so it doesn't contain any interesting information about the topology of C.
The support of the universal abelian module is restricted by the following result ( [18], [6]): Theorem 2.9. (Libgober) If C is a curve in general position at infinity, then where d i is the degree of the i-th irreducible component of C.
In [6], there is a similar characterization of supports of universal abelian invariants associated to complements of hypersurfaces in C n+1 , with any type of singularities. The supports are also shown to depend on the local type of singularities.

2.3.
Higher-order coverings of the complement. In this section we study covers of the curve complement that are associated to terms in the rational derived series of the fundamental group. The invariants arising in this way were originally used in the study of knots and respectively 3-manifolds, e.g. to show that certain groups cannot be realized as the fundamental group of the complement of a knot, or as the fundamental group of a 3-manifold. Some very useful background material is presented in [2,10]. Let For n ≥ 1, we define the n th term of the rational derived series of G inductively by: It is easy to see that G . We use rational derived series as opposed to the usual derived series in order to avoid zero-divisors in the group ring ZΓ n .
The successive quotients of the rational derived series are torsion-free abelian groups. Indeed (cf. [10]), . By construction, it follows that Γ n is a poly-torsion-free-abelian group, in short PTFA ([10]), i.e., it admits a normal series of subgroups such that each of the successive quotients of the series is torsion-free abelian. Then ZΓ n is a right and left Ore domain, so it embeds in its classical right ring of quotients K n , a skew-field.
Definition 2.10. The n-th order Alexander modules of C are where U Γn is the covering of U corresponding to the subgroup G ] as a right ZΓ n -module. The n th order rank of (the complement of) C is: In [12], the authors showed that this is also true for the reducible case (at least for curves in general position at infinity). (See Theorem 2.16.) Example 2.13. (1) If C is non-singular and in general position at infinity, then G = Z.
(2) If C has only nodal singular points (locally defined by x 2 −y 2 = 0), then G is abelian. In both cases above it follows that A Z 0 (C) = 0, and therefore A Z n (C) = 0 for all n (cf. [12], Remark 3.4).
We associate to any curve C (or equivalently, to its group G) a sequence of non-negative integers δ n (C) as follows (it is more convenient to work over a PID, so we look for a "convenient" one): Let ψ ∈ H 1 (G; Z) be the primitive class representing the linking number homomorphism G ψ → Z, α → lk(α, C). Since G ′ is in the kernel of ψ, we have a well-defined induced epimorphismψ : Γ n → Z. Let Γ n = Kerψ. ThenΓ n is a PTFA group, so ZΓ n has a right ring of quotients K n = (ZΓ n )S −1 n , for S n = ZΓ n − 0. Set R n = (ZΓ n )S −1 n . Then R n is a flat left ZΓ n -module.
Crucially, R n is a PID, isomorphic to the ring of skew-Laurent polynomials K n [t ±1 ]. Indeed, by choosing a t ∈ Γ n such thatψ(t) = 1, we get a splitting φ ofψ, and the embedding ZΓ n ⊂ K n extends to an isomorphism R n ∼ = K n [t ±1 ]. However this isomorphism depends in general on the choice of splitting ofψ ! Definition 2.14. (1) The n th -order localized Alexander module of the curve C is defined to be A n (C) = H 1 (U; R n ), viewed as a right R nmodule. If we choose a splitting φ to identify R n with K n [t ±1 ], we define A φ n (C) = H 1 (U; K n [t ±1 ]). (2) The n th -order degree of C is defined to be: δ n (C) = rk Kn A n (C) = rk Kn A φ n (C). Remark 2.15. Note that δ n (C) < ∞ if and only if rk Kn H 1 (U; K n ) = 0, i.e. A n (C) is a torsion module.
The degrees δ n (C) are integral invariants of the fundamental group G of the complement. Indeed, by [11] §1, we have: Since the isomorphism between R n and K n [t ±1 ] depends on the choice of splitting, we cannot define in a meaningful way a "higher-order Alexander polynomial", as we did in the infinite cyclic case. However, for any choice of splitting, the degree of the associated higher-order Alexander polynomial is the same. Therefore although a higher-order Alexander polynomial is not well-defined in general, the degree of the polynomial associated to a choice of a splitting yields a well-defined invariant of G. This is exactly the higher-order degree δ n defined above.
The higher-order degrees of C may be computed by means of Fox free calculus by using a presentation of π 1 (C 2 − C). The latter can be obtained by means of Moishezon's braid monodromy [26]. In general, these steps are difficult to achieve. However in section 3, some examples are explicitly computed.
The obstructions on G obtained from analyzing the higher-order degrees of a plane curve complement are contained in the following theorem of [12]: Theorem 2.16. (Leidy-Maxim [12]) If G = π 1 (C 2 −C) for some plane curve C in general position at infinity, then the higher-order degrees δ n (C) are finite. More precisely: (1) there exists a uniform upper bound in terms of the degree of C: δ n (C) ≤ d(d − 2), for all n. (2) for each n, there is an upper bound in terms of local invariants at singular points of C δ n (C) ≤ Σ l k=1 (µ(C, c k ) + 2n k ) + 2g + d − l where c k , 1 ≤ k ≤ l, are the singularities of C, n k is the number of branches through the singularity c k , µ(C, c k ) is the Milnor number of the singularity germ (C, c k ), and g is the genus of the normalized curve.
We have the following important corollary that provides an obstruction to a group being the fundamental group of the complement of a curve in general position at infinity. This can be combined with the central extension (2.1) in order to obtain obstructions on the fundamental groups of projective plane curve complements. (In the last section of this paper we will use this corollary to find such examples.) Corollary 2.17. If C is a plane curve in general position at infinity, then A Z n (C) is a torsion ZΓ n -module.

Examples
In this section, we will present some explicit calculations of the higher-order degrees of various curve complements. Although computing higher-order degrees can be difficult, we hope that these examples will aide the reader in understanding how a general calculation can be carried out.
Before presenting the calculations, we recall some results from [12].
• If C is either non-singular or has only nodal singular points (and is in general position at infinity), it follows from Example 2.13 that δ n (C) = 0 for all n ≥ 0. • If C is defined by a weighted homogeneous polynomial f (x, y) = 0, then either: -if either n > 0 or β 1 (U) > 1, then δ n (C) = µ(C, 0) − 1.
if β 1 (U) = 1, then δ 0 (C) = µ(C, 0), where µ(C, 0) is the Milnor number of the singularity germ at the origin. • If C is an irreducible affine curve, then δ 0 (C) = deg∆ C (t), where ∆ C (t) denotes the Alexander polynomial of the curve complement. If, moreover, the Alexander polynomial is trivial then all higher-order degrees vanish. In the next three examples, we consider an irreducible curveC ⊂ CP 2 and a generic line (at infinity) H, then set C =C − H.
Example 3.1. LetC ⊂ CP 2 be a degree d curve having only nodes and cusps as its only singularities. If d ≡ 0 (mod 6), then all higherorder degrees of C vanish. (this follows from the divisibility results on ∆ C (t), which imply that ∆ C (t) = 1).
Remark 3.4. From the above example we see that the higher-order degrees of a curve, at any level n, are also sensitive to the position of singular points. An interesting open problem is to find Zariski pairs that are distinguished by some δ k , but not distinguished by any δ n for n < k.
3.1. Line Arrangements. Since we are assuming that our curves are in generic position at infinity, the arrangements that we will consider do not have parallel lines. If we have an arrangement with two intersecting lines, the only singularity is a node, and therefore δ n is trivial for all n. Similarly, δ n = 0 if we have three lines arranged so that the singularities are each nodes. Hence the first interesting case to consider is the arrangement of three lines intersecting in a triple point. Using the techniques of [3] we can find a presentation for the fundamental group of U, the complement of the three lines in C 2 : Here σ 1 , σ 2 , and σ 3 correspond to the meridians of the lines. In particular, they each map to a different generator of of H 1 (U) ∼ = Z 3 and they are all mapped to the same generator of Z under the total linking number homomorphism π 1 (U) → H 1 (U) → Z. It is easier to work with a presentation for π 1 (U) where only one generator maps to the generator of Z under the total linking number homomorphism. Hence we choose new generators: a = σ 1 , b = σ 2 σ −1 1 , and c = σ 3 σ −1 1 . With these new generators we have the following presentation: Using Fox calculus [7], [8], we can obtain a presentation matrix for H 1 (U, u 0 ; Zπ 1 (U)), the homology of the universal cover of U relative to a basepoint u 0 as a left Zπ 1 (U)-module. The 1-chains for the universal cover of U are generated as a Zπ 1 (U)-module by α, β, and γ, where α, β, and γ each represent a single lift of the 1-chains of U corresponding to a, b, and c, respectively. Since we are computing the homology relative to a basepoint, α, β, and γ are in fact 1-cycles in H 1 (U, u 0 ; Zπ 1 (U)). It remains to consider the 2-chains of the universal cover of U. First, the 2-chain of U corresponding to the relation abaca −1 c −1 a −1 b −1 in π 1 (U) lifts to a 2-chain of the universal cover of U whose boundary is: Using the relation abaca −1 c −1 a −1 b −1 = 1 in π 1 (U), we can rewrite this boundary as: Similarly, the 2-chain of U corresponding to the relation bacab −1 a −2 c −1 in π 1 (U) lifts to a 2-chain of the universal cover of U whose boundary is: Using the relation bacab −1 a −2 c −1 = 1 in π 1 (U), we can rewrite this boundary as: We can collect this information to write a presentation matrix: Here the columns correspond to the generators, α, β, and γ, respectively, and the rows correspond to relations. If we allow elements of π 1 (U) (n+1) r to be set equal to 1 in Zπ 1 (U), we can also consider the above as a presentation matrix for H 1 (U, u 0 ; ZΓ n ). Furthermore, since R n is a flat ZΓ n -module, we can also consider it to be a presentation matrix for H 1 (U, u 0 ; R n ). If we think of the matrix in this way, any non-zero element in ZΓ n has an inverse. (Recall that Γ n is the kernel of the mapψ : Γ n → Z, induced by the total linking number homomorphism.) If we choose a splitting ofψ, there is an isomorphism between R n and K n [t ±1 ]. For our example, we choose the splitting that maps t to a. To obtain a presentation for H 1 (U, u 0 ; K n [t ±1 ]) we must replace each entry in the above matrix with its image under the isomorphism R n → K n [t ±1 ]. This results in the following presentation matrix for H 1 (U, u 0 ; K n [t ±1 ]): ] is a skew Laurent polynomial ring, we must be careful when writing elements where t is not originally on the right. For example, tb = aba −1 t in K n [t ±1 ]. The next step in finding δ n is diagonalizing this matrix, which is possible since K n [t ±1 ] is a PID. Since c = 1 in π 1 (U)/π 1 (U) ′ , it follows that c / ∈ π 1 (U) (n) r for all n ≥ 1. Therefore c = 1 in Γ n for all n ≥ 0. Hence 1 − c = 0 in ZΓ n and is therefore invertible in K n [t ±1 ]. This allows us to multiply the last column in our presentation matrix by the unit 1 − c. Since our matrix is a presentation of a left module and since columns correspond to generators, we multiply columns on the right. The result of multiplying the last column (on the right) by the unit 1 − c is the following: Next we add the first column times 1 − t and the second column times 1 − b to the last column. The result is the following: This means that we have a free generator, which is expected since we are computing the homology relative to a basepoint. Next we multiply the first row by ct + c and add it to the second. Since our matrix is a presentation of a left module and since rows correspond to relations, we multiply rows on the left. The result of multiplying the first row (on the left) by ct + c and adding it to the second is: Now we multiply the second row (on the left) by the unit (1 − c) −1 . Then we multiply the second row by 1 − t and add it to the first. This results in the following matrix: Notice that we can now eliminate the second column and row. Hence we have shown that To find H 1 (U; K n [t ±1 ]), we consider the long exact sequence of a pair: Therefore, for the arrangement of three lines intersecting in a triple point, δ n = 3 for all n ≥ 0.
If we add an additional line to this arrangement that intersects previous three lines in nodes (as in the wiring diagram below), δ n = 0 for all n ≥ 0. In fact, for any line arrangement that contains a line whose only intersections are nodes, δ n = 0 for all n ≥ 0.
If instead we add an additional line to this arrangement so that all lines intersect in a single point, δ n = 8 for all n ≥ 0. The arrangement of five lines intersecting in a single point has δ n = 15 for all n ≥ 0. Each of these calculations can be done in the same fashion as the one above. We conjecture that for m lines intersecting in a single point, δ n = m(m − 2) for all n ≥ 0.

3.2.
Artin groups of spherical-type. Deligne [4] showed that each Artin group of spherical-type appears as the fundamental group of the complement of a complex hyperplane arrangement. Mulholland and Rolfsen [27] showed that the commutator subgroups of the following Artin groups are perfect (i.e. G ′ = G ′′ ): A n , n ≥ 4; B n , n ≥ 5; D n , n ≥ 5; E n , n = 6, 7, 8; H n , n = 3, 4. It follows that all higher-order degrees are trivial for curves whose complements have the above fundamental groups. We will explicitly compute the higher-order degrees of the Artin group of type A 3 .
The Artin group of type A 3 is the braid group on four strands, B 4 . A standard presentation for the braid group on four strands is: We give a new presentation by choosing new generators: x = σ 1 , y = σ 2 σ −1 1 , and z = σ 3 σ −1 1 .
B 4 = x, y, z|xz = zx, xyx = yx 2 y, yxzxy = zxyxz Notice that the abelianization of B 4 is Z, and that under the abelianization map, x maps to a generator of Z, while y and z are mapped to 0. If U is a curve complement with π 1 (U) ∼ = B 4 , we can use Fox calculus to obtain the following presentation for H 1 (U, u 0 ; Zπ 1 (U)), as a left Zπ 1 (U)-module: Here the columns correspond to generators and the rows correspond to relations.
To obtain a presentation for H 1 (U, u 0 ; K n [t ±1 ]), we choose the splitting that maps t to x. Then we have the following presentation for We remind the reader that x and z commute in B 4 and therefore in K n [t ±1 ], we have tz = zt.
It follows from Thm 3.6 of [27] that B ′ 4 /B ′′ 4 ∼ = Z 2 , generated by y and xyx −1 . In particular, y / ∈ (B 4 ) (n) r for n ≥ 2. Therefore, 1 − y = 0 in ZΓ n for n ≥ 1. Hence 1 − y is invertible in K n for n ≥ 1. We first consider the case when n = 0 and then continue the calculation for n ≥ 1.
If n = 0, then we set y = z = 1 in the above matrix to obtain: After adding the second row along with t times the first row to the last row, we are able to eliminate the last row and column. Therefore, (Note that this is simply the computation of the degree of the classical Alexander polynomial. ) We now assume that n ≥ 1, and therefore can use the fact that 1 − y is a unit in K n [t ±1 ]. We begin the process of diagonalizing the matrix by multiplying the second column (on the right) by 1 − y. The result is: 1−z 0 t−1 (xyx −1 −y)t+1−y (xyx −1 −y)t 2 +(1−xyx −1 )t+(y−1) 0 (yz−zxyx −1 )t+y−z (yz−zxyx −1 z)t 2 +(zxyx −1 −z)t+(1−y) −zxyx −1 t 2 +yt−1 Next we add the first column times 1 − t and the last column times 1 − z to the second column. This gives us our expected free generator: Now we subtract the first row from the third, add the second row to the third, and then multiply the third row (on the left) by t −1 . The result is:   1 − z 0 t − 1 (xyx −1 − y)t + 1 − y 0 0 x −1 yxz − zy + y − x −1 yx 0 −zyt + x −1 yx − 1   can be shown that z / ∈ (B 4 ) (ω) r , then there is an integer m ≥ 2 such that δ n = 0 for all n ≥ m.
The same kind of calculations can be carried out for the other Artin groups of spherical-type. We summarize these results without providing the explicit calculations. For the Artin group of type A 2 , δ 0 = 2 and δ n = 1 for n ≥ 1. For the Artin group of type B 2 , δ n = 2 for n ≥ 0. For the Artin group of type B 3 , δ 0 = 4 and δ n = 3 for n ≥ 1. For the Artin groups of types B 4 and F 4 , δ 0 = 1 and δ n = 0 for n ≥ 1. For the Artin group of type D 4 , δ 0 = 2, δ 1 = 1, and δ n = 0 for n ≥ 2. For the Artin group of type I 2 (m), where m is odd, δ 0 = m − 1 and δ n = m − 2 for n ≥ 1. For the Artin group of type I 2 (m), where m is even, δ n = m − 2 for n ≥ 0.

3.3.
Obstructions on fundamental groups of plane curve complements. It follows from Theorem 2.16, that if C is a plane curve in general position at infinity, then δ n < ∞ for all n ≥ 0. This is not true for a free group with at least three generators. Therefore, such a group cannot be the fundamental group of a plane curve complement in general position at infinity. Also Harvey [10] has shown that δ 0 = ∞ for the fundamental group of a boundary link complement. (Recall that a boundary link is a link whose components bound mutually disjoint Seifert surfaces.) An example of such a group (which is Ex. Therefore such a group cannot be the fundamental group of a plane curve complement in general position at infinity. These facts can be combined with the sequence (2.1) in order to obtain classes of groups that cannot be the fundamental group of a projective plane curve complement.