LOWER CENTRAL SERIES AND FREE RESOLUTIONS OF HYPERPLANE ARRANGEMENTS

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 54, Number 9, Pages S ( Article electronically published on May 8, 00 LOWER CENTRAL SERIES AND FREE RESOLUTIONS OF HYPERPLANE ARRANGEMENTS HENRY K SCHENCK AND ALEXANDER I SUCIU Abstract If M is the complement of a hyperplane arrangement, and A = H (M,k is the cohomology ring of M over a field k of characteristic 0, then the ranks, φ k, of the lower central series quotients of π 1 (M can be computed from the Betti numbers, b ii =dimtor A i (k, k i,ofthelinearstrandinaminimal free resolution of k over A We use the Cartan-Eilenberg change of rings spectral sequence to relate these numbers to the graded Betti numbers, b ij = dim Tor E i (A, k j, of a minimal resolution of A over the exterior algebra E From this analysis, we recover a formula of Falk for φ,andobtainanew formula for φ 4 The exact sequence of low-degree terms in the spectral sequence allows us to answer a question of Falk on graphic arrangements, and also shows that for these arrangements, the algebra A is Koszul if and only if the arrangement is supersolvable We also give combinatorial lower bounds on the Betti numbers, b i,i+1,ofthe linear strand of the free resolution of A over E; if the lower bound is attained for i =, then it is attained for all i For such arrangements, we compute the entire linear strand of the resolution, and we prove that all components of the first resonance variety of A are local For graphic arrangements (which do not attain the lower bound, unless they have no braid subarrangements, we show that b i,i+1 is determined by the number of triangles and K 4 subgraphs in the graph 1 Introduction 11 LCS formulas The lower central series (LCS of a finitely-generated group G is a chain of normal subgroups, G = G 1 G G, defined inductively by G k =[G k 1,G] Fix a field k of characteristic 0 Associated to G is a graded Lie algebra, gr(g k:= k=1 G k/g k+1 k, with Lie bracket induced by the commutator map Let φ k = φ k (G denote the rank of the k-th quotient Onecaseinwhichthenumbersφ k have aroused great interest occurs when M is the complement of a complex hyperplane arrangement and G = π 1 (M is its fundamental group In certain situations, the celebrated lower central series formula holds: (11 (1 t k φ k = P (M, t, k=1 Received by the editors August, 001 and, in revised form, September 19, Mathematics Subject Classification Primary 16E05, 0F14, 5C5; Secondary 16S7 Key words and phrases Lower central series, free resolution, hyperplane arrangement, change of rings spectral sequence, Koszul algebra, linear strand, graphic arrangement The first author was partially supported by an NSF postdoctoral research fellowship The second author was partially supported by NSF grant DMS c 00 American Mathematical Society

2 410 HENRY K SCHENCK AND ALEXANDER I SUCIU where P (M,t = b i t i is the Poincaré polynomial of the complement This formula was first proved by Kohno [1] for braid arrangements, and then generalized by Falk and Randell [15] to supersolvable arrangements In [9], Shelton and Yuzvinsky gave a beautiful interpretation of the LCS formula in terms of Koszul duality The LCS formula holds for rational K(π, 1 spaces (see Papadima and Yuzvinsky [6], and an analogue holds for hypersolvable arrangements (see Jambu and Papadima [0] All these LCS formulas ultimately hinge on the assumption that the cohomology algebra (or its quadratic closure is Koszul The goal of this paper is to explore formulas of type (11 for arrangements where such assumptions fail In complete generality, there is no good substitute for the polynomial P (X, t on the right side, as illustrated by the non-fano plane (see Peeva [7] But, as conjectured in [1], and as verified here in certain situations (at least in low degrees, there do exist large classes of arrangements for which a modified LCS formula holds, with the Poincaré polynomial replaced by another (combinatorially determined polynomial 1 LCS ranks and Betti numbers of the linear strand Our starting point is a formula expressing the LCS ranks of an arrangement group in terms of the Betti numbers, b ii =dim k Ext i A(k, k i =dim k Tor A i (k, k i, of the linear strand of the free resolution of the residue field k, viewed as a module over the Orlik-Solomon algebra, A = H (M,k: (1 (1 t k φ k = b ii t i k=1 A bit of history: In [1], Kohno showed that the left-hand quantity is equal to the Hilbert series of the universal enveloping algebra, U = U(g, of the holonomy Lie algebra g of the complement M (This follows from the Poincaré-Birkhoff-Witt theorem, together with the isomorphism gr(g k = g, which is a consequence of the formality of M, in the sense of Sullivan In [9], Shelton and Yuzvinsky proved that U = A!, the Koszul dual of the quadratic closure of the Orlik-Solomon algebra Finally, results of Priddy [8] and Löfwall [] relate the Koszul dual of a quadratic algebra to the linear strand in the Yoneda Ext-algebra: A! (1 = Ext i A (k, k i i Since obviously Ext i A (k, k i =Ext i A(k, k i, formula (1 follows at once (compare [7, Theorem 6] If A is a Koszul algebra (ie, Ext i A (k, k j =0,fori j, then A = A, andthe Koszul duality formula, Hilb(A!,t Hilb(A, t = 1, yields the LCS formula (11 This computation applies to supersolvable arrangements, which do have Koszul OS-algebras, cf [9] The more general formula (1 was first exploited by Peeva in [7] to obtain bounds on the LCS ranks φ k 1 The Orlik-Solomon algebra and free resolutions A minimal free resolution of a graded module N over a graded k-algebra R is simply a graded exact sequence i=0 (14 P : j R bj ( j j R b1j ( j j R b0j ( j N 0,

3 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 411 where the exponents are b ij (N =dim k Tor R i (N,k j We are especially interested in two cases: N = A = E/I, the Orlik-Solomon algebra of an arrangement of n hyperplanes, and R = E is the exterior algebra on n generators, in which case we will write b ij =dim k Tor E i (A, k j N =k and R = A, in which case we will write b ij =dim k Tor A i (k, k j We study the relationship between these two free resolutions by means of the change of rings spectral sequence associated to the composition E A k: (15 Tor A ( i Tor E j (A, k, k = Tor E i+j (k, k The goal is to translate the right-hand side of formula (1 into an expression involving the resolution of A over E, which is a much smaller (in the sense that the ranks of the free modules appearing in the resolution are much smaller object than the resolution of k over A AresolutionP as in (14 is linear if each free module F i is generated in a single degree α i and α i = α i 1 1 By convention, when N = R/J and J is generated in degree, the linear strand is the complex with terms F i = R bi,i+1 ( i 1, i 1 Most of the results we obtain from the change of rings spectral sequence stem from the happy fact that the resolution of k as a module over E is linear (with dim k Tor E i (k, k i = ( n+i 1 i In [10], Eisenbud, Popescu and Yuzvinsky prove that A = H (M,k, viewed as an E-module via the cap product, has a linear free resolution; we had hoped that this would make one of the change of rings spectral sequences particularly simple, but it does not 14 LCS ranks of arrangement groups Let a i denote the number of minimal generators of degree i in the Orlik-Solomon ideal I It is elementary to see that φ 1 = b 1 and φ = a In [11], Falk gave a formula for φ of an arbitrary arrangement, which can be interpreted as φ = b, but no general formula for φ 4 was known (besides the one implicit in (1 Using the above spectral sequence, we obtain a formula for φ 4 which depends solely on the resolution of A over E: ( a (16 φ 4 = + b 4 δ 4, where δ 4 denotes the number of minimal quadratic syzygies on the degree generators of I which are Koszul syzygies For many arrangements, this expression yields an explicit combinatorial formula for φ 4 15 Minimal linear strand arrangements For arrangements with ( µ(x+1 (17 b =, we have (18 φ 4 = X L (A X L (A µ(x (µ(x 1 4 Arrangements for which (17 holds have b i,i+1 as small as possible, for all i (given the combinatorics; in other words, the linear strand in the free resolution of A over E is minimal In view of this, we call such arrangements minimal linear strand (MLS arrangements Examples include Kohno s arrangements X and X, the non-pappus arrangement, and graphic arrangements with no K 4 subgraphs

4 41 HENRY K SCHENCK AND ALEXANDER I SUCIU For MLS arrangements in C, we compute the whole Betti diagram of the minimal free resolution of A over E, as well as the differentials in the linear strand As a generalization of formula (18, we conjecture that, for all MLS arrangements and all k, (19 φ k = 1 k µ(dµ(x k d, d k X L (A where µ stands for both the Möbius function of the integers, and the Möbius function of the intersection lattice of the arrangement Equivalently, (110 (1 t k φ k =(1 t b1 b (1 µ(x t k=1 X L (A The proof of the conjecture hinges upon computing the image of a d mapinthe change of rings spectral sequence 16 Graphic arrangements As another application our methods, we study the free resolutions and LCS quotients of a graphic arrangement Let G be a (simple graph on l vertices, with edge-set E, andleta G = {z i z j =0 (i, j E} be the corresponding arrangement in C l Using a well-known result of Stanley, we show that the OS-algebra A is Koszul if and only if the arrangement A G is supersolvable As for the LCS ranks of the fundamental group of the complement of A G,we conjecture that they are given by (111 φ k = 1 k k d k j=1 s=j k ( s ( 1 s j κ s µ(dj k d, j where κ s is the number of complete subgraphs on s + 1 vertices; equivalently, (11 ( 1 t k l 1 φ k = (1 jt k=1 j=1 l 1 ( 1 s j( s j κs s=j We verify this conjecture for k In particular, we find φ = (κ + κ, thereby answering a question of Falk [14] For k = 4, we verify the inequality φ 4 (κ +κ +κ 4, and show that equality holds if κ =0 Acknowledgment The computations for this work were done primarily with the algebraic geometry and commutative algebra system Macaulay, by Grayson and Stillman [19] Additional computations were done with Mathematica 40, from Wolfram Research Resolution of the OS-algebra over the exterior algebra We start by analyzing the minimal free resolution of the Orlik-Solomon algebra over the exterior algebra For minimal linear strand arrangements, the graded Betti numbers of this resolution can be computed explicitly from the Möbius function of the intersection lattice

5 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 41 1 Orlik-Solomon algebra Let A be a (central arrangement of complex hyperplanes in C l, with complement M(A =C l \ H A HLetL(A ={ H B H B A}be the intersection lattice of all flats of the arrangement, with order given by reverse inclusion, and rank given by codimension The cohomology ring of M(A with coefficients in a field k is isomorphic to a certain k-algebra, A = A(A, which can be described solely in terms of L(A, as follows (see the book by Orlik and Terao [5] as a general reference Let E be the exterior algebra on generators e 1,,e n in degree 1, corresponding to the hyperplanes This is a differential graded algebra over k, with grading E r =span k {e J = e i1 e ir J = {i 1,,i r } {1,,n}}, and differential : E i E i 1 given by (1 = 0, (e i = 1, together with the graded Leibnitz rule The Orlik-Solomon algebra, A, is the quotient of E by the (homogeneous ideal I = ( e B codim H< B The algebra A = E/I inherits a grading from E in the obvious fashion We denote the Betti numbers of the arrangement by b i =dima i Asiswellknown, (1 b i = ( 1 i µ(x, X L i(a where µ: L(A Z is the Möbius function of the lattice Let a j be the number of minimal generators of I having degree exactly j, and let I[j] I be the ideal of E generated by those a j elements We have H B ( a j =dim k (I[j] j =dim k Tor E 1 (A, k j From the exact sequence 0 I E A 0, we get a j + b j ( n j Furthermore, since I isgeneratedindegree, we have a 1 =0and ( a = ( n b = X L (A ( µ(x If A is quadratic (ie, I = I[], then a j =0forj> No explicit combinatorial formula for the numbers a j (j> is known in general Free resolution of the OS-algebra By general results on modules over the exterior algebra, the Orlik-Solomon algebra A admits a resolution over E by finitely-generated free modules, (4 F F 1 E A 0 Furthermore, there exist minimal free resolutions, in the sense that dim k F i = dim k Tor E i (A, k, for all i 1 Such resolutions are unique up to chain-equivalence Our goal in this section is to obtain a better understanding of the graded Betti numbers (5 We start with some simple observations b ij =dim k Tor E i (A, k j

6 414 HENRY K SCHENCK AND ALEXANDER I SUCIU Lemma Let A be an essential, central arrangement of rank l If j i + l, then Tor E i (A, k j =0 Proof As in the proof of Theorem 11 of [10], to study the free E-resolution of A, we may pass to a decone of A Let A = E /I be the OS-algebra of the decone Then A j =0,forj l The free E -resolution of k is linear, so after tensoring this resolution with A, we obtain a complex of modules with i-th term A ( i, which vanishes in degrees i + l and higher Lemma 4 Let A be an arrangement of n hyperplanes in C l The Betti numbers of the minimal free resolution of A over E are related to the Betti numbers of A, as follows: ( l 1 l (6 ( 1 i b i,i+jt i+j (1 + t n = b i t i i=0 j=0 Proof From (4, we get ( i ( 1i Hilb(F i,t P (E,t =P (A, t 5 Local syzygies We now analyze in more detail a certain type of linear syzygy in the resolution of A over E These local syzygies come from flats in L (A := {X L (A µ(x > 1} We start with the simplest situation Example 6 (Pencil of lines Let A = {H 0,H 1,H } be an arrangement of lines through the origin of C The Orlik-Solomon ideal is generated by e 01 = (e 1 e (e 0 e It is readily seen that the minimal free resolution of A over E is ( ( e 1 e e 0 e 0 e01 e1 e e 0 e 0 e 1 e e 0 e 0 A E F 1 F F e 1 e e 0 e e 1 e e 0 e e 1 e e 0 e F i F i+1 with F i = E i ( i Thus, b i,i+1 = i for i 1, and b i,i+r =0forr>1 Example 7 (Pencils of n +1hyperplanes More generally, let A = {H 0,, H n } be a central arrangement of n+1 hyperplanes whose common intersection is of codimension two The Orlik-Solomon ideal is generated by all elements e ijk = (e j e i (e k e i with0 i<j<k n Infact,I = I[], with minimal set of generators (e j e 0 (e k e 0, for 1 j<k n The minimal resolution of A over E has a form similar to the one above The graded Betti numbers are given by ( n + i 1 (7 b i,i+1 = i, for all i 1, i +1 and b i,i+r =0,forr>1 To see this, note that, after a suitable linear change of variables, the ideal I becomes identified with the monomial ideal m, where m =(e 1,,e n Formula (7 then follows from general results of [1], or directly, from the exact sequence 0 Tor E i+1(k, k Tor E i (m/m ( 1, k Tor E i (E /m, k 0, i=0

7 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 415 where E = E/(e 0, together with the isomorphism m/m =k n Going back to an arbitrary arrangement, we observe that for each flat X L (A with µ(x there are linear syzygies of the type discussed in the examples above, which we will call local linear syzygies In fact, these syzygies persist in the free resolution of A over E, as the next lemma shows Lemma 8 Let A be an arrangement Then b i,i+1 i ( µ(x+i 1 (8 i +1 X L (A Moreover, if equality holds for i =, then it holds for all i Proof Sets of local linear syzygies corresponding to different elements of L (A are linearly independent Indeed, they are supported only on generators coming from the same element of L (A 9 MLS arrangements In view of the preceding lemma, we single out a class of arrangements for which the linear strand of a minimal free resolution of A over E is completely determined by the Möbius function of L (A Definition 10 An arrangement is called minimal linear strand (MLS if b = ( µ(x+1 (9 X L (A In [10], Eisenbud, Popescu, and Yuzvinsky remark that it is an interesting problem to determine explicitly a minimal free resolution of the OS-algebra over the exterior algebra In the case of MLS arrangements in C, Lemmas 4 and 8 permit us to compute the whole Betti diagram of the minimal resolution of A over E, as well as the differentials in the linear strand Theorem 11 Let A be an MLS arrangement of n hyperplanes in C Then, the graded Betti numbers of a minimal free resolution of the OS-algebra over the exterior algebra are given by ( µ(x+i 1 (10 b i,i+1 = i, i +1 (11 X L (A b i,i+ = b i+1,i+ + ( i +1 ( n + i i + ( n + i i X L (A ( µ(x Proof The formula for b i,i+1 is given by Lemma 8 The formula for b i,i+ follows from (6, which in the case l = boils down to 1 a t + ( 1 i (b i,i+ b i+1,i+ ti+ = 1+(n 1t + (( n 1 a t (1 + t n 1 i=1 The differentials in the linear strand are obtained in the obvious fashion from the differentials in Examples 6 and 7 Example 1 Let A be the X arrangement, with defining polynomial Q = xyz(x z(y + z(y +x; see Figure 1 We have L = {X 1,X,X },withµ(x i =

8 416 HENRY K SCHENCK AND ALEXANDER I SUCIU Figure 1 The arrangement X and its associated matroid, and so a = Moreover,a = 1, and so condition (9 is satisfied From Theorem 11, we compute: b i,i+1 =i, b i,i+ = i(i +1(i +5i 8 More generally, for each n 6, consider the arrangement with defining polynomial Q = xyz(x z(y + z(y +x (y +(n 4x We then have L = {X 1,X,X },withµ(x 1 =n 4, µ(x =µ(x =, and a = n 5, and so condition (9 is satisfied We leave the computation of the Betti numbers b ij as an exercise 1 A criterion for quadraticity From the proof of Theorem 11, we see that ( b a b1 (1 = b + b 1 a Applying Lemma 8, we get ( b1 (1 a b b 1 a + X L (A ( µ(x+1 This inequality, when coupled with the definition of a, provides a necessary combinatorial criterion for quadraticity (and hence, Koszulness of the OS-algebra Theorem 14 If ( b 1 b b 1 a + ( µ(x+1 X L (A > 0, then A is not quadratic This criterion shows that none of the arrangements in Example 1 has a quadratic OS-algebra Resolution of the residue field over the OS-algebra We now turn to the minimal free resolution of the residue field k over the Orlik-Solomon algebra A, and to the corresponding graded Betti numbers b ij = dim k Tor A i (k, k j A 5-term exact sequence argument relates b j to the number of minimal generators in degree j of the Orlik-Solomon ideal, a j =dim k (I[j] j This computes the first three LCS ranks φ k

9 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS Relating various Betti numbers Since the Hilbert series of the residue field is simply 1, there are simple numerical constraints on the free resolution of k over A Suppose A is a central, essential arrangement in C l Then (1 1= ( 1 b 11 t + b j t j b j t j + (1 + b 1 t + b t + + b l t l j j Thisgivesusawaytosolveforb rr,ifweknowallb ir with i<r: b 11 = b 1, b = b 1 b, b b = b 1 b 1b + b, b 44 b 4 + b 4 = b 4 1 b 1b + b +b 1 b b 4, b 55 b 45 + b 5 b 5 = b 5 1 4b 1 b +b 1 b +b 1 b b b b 1 b 4 + b 5, and, in general, ( r ( 1 i b ir = i= j 1+j + +rj r=r (j j r! ( b 1 j1 ( b r jr j 1! j r! Relation with the OS-algebra We now discuss the relation between the Betti numbers of the resolution of k over A and the numbers a j =dim k Tor E 1 (A, k j Lemma We have b = ( b a,and ( b j = a j, for j> Proof Clearly, b = b 1 b = b 1 ( b 1 + a = ( b 1+1 (4 + a From the change of rings spectral sequence associated to the composition of ring maps E A k (see 4, we obtain a 5-term exact sequence (5 Tor E (k, k TorA (k, k TorE 1 (A, k TorE 1 (k, k TorA 1 (k, k 0 Nowtakefreeresolutions: E (b 1 +1 ( E b 1 ( 1 E k 0, j Abj ( j A b1 ( 1 A k 0, j ( j E A 0 Eaj The Tor s are graded, and we have the following ranks for the respective graded pieces: degree Tor E (k, k TorA (k, k TorE 1 (A, k TorE 1 (k, k TorA 1 (k, k b 1 b ( 1 b1+1 b a b a 0 0 l 0 b l a l 0 0 The result follows at once (Notice that since b = ( b a, the sequence (5 is also exact on the left

10 418 HENRY K SCHENCK AND ALEXANDER I SUCIU 4 Witt s formula Recall that the lower central series of a finitely-generated group G is the chain of normal subgroups G = G 1 G G, with G k equal to the commutator [G k 1,G]fork>1(see [] as a basic reference The successive quotients of the series are finitely-generated abelian groups Let φ k (G =rankg k /G k+1 denote the rank of the k-th LCS quotient The LCS ranks of a finitely-generated free group F n are given by Witt s formula: (6 φ k (F n = 1 k µ(dn k d, or, equivalently, (7 For example, (8 φ 1 (F n =n, d k (1 t k φ k =1 nt k=1 φ (F n = ( n, φ (F n = ( n+1, φ4 (F n = n (n The first three φ k s We now compute the ranks of the first three LCS quotients of an arrangement group, in terms of concrete combinatorial data The results of the previous section allow us to recover a formula of Falk [11], but with a slightly new perspective This approach illustrates that high degree syzygies which appear near the beginning of the resolution can provide information about low degree syzygies further out Contrast this to the approach of Peeva [7], where the Orlik-Solomon algebra is truncated in degree three Corollary 6 The ranks of the first three LCS quotients of an arrangement group are given by: (9 (10 (11 φ 1 = b 1 = hyperplanes, ( b1 φ = b = a, φ = b k=1 ( b1 + b 1 (( b1 b + a = b Proof As explained in 1, we have ( (1 (1 t k φk Hilb Ext i A (k, k i,t =1 i ( b1 + b 1 a + a Expand both terms, using ( and (: (1 t k φ k =1 φ 1 t + (( φ 1 φ t (( φ 1 φ1 φ + φ t i k=1 + (( φ φ1 φ ( φ 1 φ + ( φ φ4 t 4 +, ( Hilb Ext i A (k, k i,t =1+b 1 t +(b 1 b t +(b 1 b 1b + b + a t +, and solve

11 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 419 Formula (10 can be rewritten in terms of the Möbius function of the intersection lattice as φ = X L (A φ (F µ(x Formulas (11 and (1 show that φ equals the number of linear syzygies on the degree two generators of I: (1 φ = b This expression for φ (essentially the same as the one obtained by Falk in [11], depends on more subtle combinatorial data than just the Möbius function of L(A Nevertheless, it was shown by Falk in [1] that, for all k, (14 φ k φ k (F µ(x X L (A For k =, this inequality also follows from (8, together with (1 and (8 4 The change of rings spectral sequence In this section, we examine the change of rings spectral sequence more deeply This spectral sequence is due to Cartan and Eilenberg [4]; a good reference for the material presented here is Eisenbud [9], appendix A (see also McCleary [4], p 508 Our goal is to compute the Betti numbers b ii =dim k Tor A i (k, k i, which will allow us to solve for φ k We concentrate on the case k = 4, since it illustrates well the general situation The Betti number b 44 can be obtained from b 4 and a Hilbert series computation as above, or directly from the spectral sequence We first tackle b 4, since it corresponds to earlier terms in the spectral sequence 41 Change of rings Suppose we have a composition of ring maps: E A k BytakingafreeresolutionQ for k over E and a free resolution P for k over A and tensoring with k, we obtain a first quadrant double complex, which yields a spectral sequence (41 E ij = Tor A i ( Tor E j (A, k, k = Tor E i+j(k, k Since 1 hor Ei,j = 0 unless i =0, (4 hore i,j = hore i,j In our situation (with E A the canonical projection of the exterior algebra onto the OS-algebra, and A k the projection onto the residue field, we have (4 hor E0,j =Tor E j (k, k, and we know how to compute this term from the Koszul complex In particular, Tor E j (k, k is nonzero only in degree j, and we have ( dim k Tor E j (k, k b1 + j 1 (44 j = j We already used this fact to show that the sequence (5 of low degree terms was actually left exact To keep things easy to follow, we display the verte i,j terms:

12 40 HENRY K SCHENCK AND ALEXANDER I SUCIU Tor E (A, k Tor A 1 (Tor E (A, k, k Tor A (Tor E (A, k, k Tor A (Tor E (A, k, k d,1 Tor E 1 (A, k Tor A 1 (Tor E 1 (A, k, k Tor A (Tor E 1 (A, k, k Tor A (Tor E 1 (A, k, k d,0 d,0 d,0 k Tor A 1 (k, k Tor A (k, k Tor A (k, k From the diagram, we compute verte,0 =kerd,0, vert E1,1 =cokerd,0, verte 0, =(cokerd,1 /d,0 (ker d,0 Let tot be the total complex associated with the spectral sequence We then have gr(h k (tot = verte i,j, i+j=k and thus (45 gr(h k (tot r =0, for r>k The exactness of the sequence (5 of terms of low degree implies that both vert E1,1 and vert E0, vanish Hence im d,0 =Tor A 1 (k, Tor E 1 (A, k, coker d,1 = d,0 (ker d,0 Tracing through the spectral sequence and using the equalities above, we obtain exact sequences 0 verte,0 0 ker d,0 Tor A (k, k d,0 Tor A 1 (k, TorE 1 (A, k 0 d,0 coker d,1 0

13 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 41 Now recall that gr(h (tot vanishes in degree greater than three, and so, in particular, ( vert E,0 = 0 We thus have proved the following theorem 4 Theorem 4 The following sequence is exact: (46 0 (coker d,1 4 Tor A (k, k 4 Tor A 1 (k, Tor E 1 (A, k 4 0 Since the dimension of Tor A 1 (k, TorE 1 (A, k 4 is b 1 a, this contribution to b 4 = dim k Tor A (k, k 4 is easy to understand We shall analyze the contribution from dim k (coker d,1 4 next 4 Quadratic syzygies and a Koszulness test We have already seen that if a > 0, then A is not Koszul The sequence (46 gives another necessary condition for Koszulness Indeed, if (47 dim k Tor E (A, k 4 > dim k d,1 (TorA (k, TorE 1 (A, k 4, then A is not a Koszul algebra In order to apply this Koszulness test, we need a way to determine (im d,1 4 This is the content of the following lemma First, recall some definitions Given elements f 1,,f k E, thesyzygy module is a submodule of a free module on ɛ 1,,ɛ k, consisting of E-linear relations among the f i s An (exterior Koszul syzygy is a relation of the form f i ɛ j ± f j ɛ i See [9] for more details Lemma 44 The image of d,1 in degree 4 equals K 4, the space of minimal quadratic syzygies on the generators of I which are Koszul Proof Take (minimal free resolutions: P : 0 k A A b1 ( 1 A (b 1 +1 ( A a (, Q : 0 k E E b1 ( 1 E (b 1 +1 (, and form the double complex P 0 (A Q δ P 1 (A Q P (A Q d d δ P 0 (A Q 1 P 1 (A Q 1 P (A Q 1 P 0 (A Q 0 P 1 (A Q 0 P (A Q 0 The d differential is just the differential from the snake lemma Since all the generators of I may be written as products of linear forms, the result follows from a diagram chase In view of this lemma, we may rephrase the non-koszul criterion (47, as follows Corollary 45 If the minimal resolution of A over E has minimal quadratic non- Koszul syzygies, then A is not a Koszul algebra d

14 4 HENRY K SCHENCK AND ALEXANDER I SUCIU The next lemma gives an upper bound on the number of (minimal quadratic Koszul syzygies, solely in terms of the Möbius function of the intersection lattice of the arrangement A Lemma 46 Let K 4 be the the space of quadratic Koszul syzygies on the generators of I Then (48 dim k K 4 (X,Y ( L (A ( ( µ(x µ(y where ( S denotes the set of unordered pairs of distinct elements of a set S Moreover, if A is MLS, then equality holds in (48 Proof Quadratic Koszul syzygies between elements of I which come from the same dependent set are consequences of linear syzygies (because the free resolution for pencils is linear Any two distinct elements of L, say X and Y, give rise to ( µ(x ( µ(y quadratic Koszul syzygies, whence the bound In the MLS case, all linear syzygies come from pencils; hence any of the above Koszul syzygies is not a consequence of the linear syzygies 47 Computing b ii The basic idea is to divide the problem up, using the exact sequence, (49 0 ker d i,0 Tor A i (k, k i im d i,0 0 We start by analyzing the kernel of d i,0, illustrating the method in the i =4 case Using the fact that the differentials in the spectral sequence are graded, we obtain the following diagram: Tor E (A, k 4 d Tor A 1 (k, Tor E 4 (A, k 4 d Tor A (k, Tor E 1 (A, k 4 d Tor A 4 (k, k 4 Now recall from (45 that gr(h 4 (tot r =0ifr>4 An analysis of the terms in i+j=4 verte i,j shows that verte 4,0 =kerd 4 is the only nonzero term, and ( dim k vert E 4,0 =dim k Tor E b1 + 4 (k, k 4 = 4

15 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 4 From the diagram 0 0 ker d 4 ker d Tor E (A, k 4 0 ker d Tor A 1 (k, TorE (A, k 4 0 we find that dim k ker d 4,0 = ( b b 4 + b 1 b In the general case, we obtain the following ( Lemma 48 dim k ker d i,0 b1 + i 1 i = + b jj b i j 1,i j i j=0 Next, we analyze the image of d i,0 From the diagram 0 ( im d i,0 i we obtain the following 0 ( vert E i,1 i d Tor A i (k, Tor E 1 (A, k i ( coker d i,0 i 0 ( im d i,1 i ( Theorem 49 Let δ i = dim k d Tor A i (k, Tor E 1 (A, k i +dim ( vert k E i,1 i Then ( b1 + i 1 i b ii + δ i = + b jj b i i j 1,i j As a consequence, we can compute the Betti numbers b ii inductively, provided we know all the numbers b j,j+1 and δ j An easy check shows that ( vert E,1 4 =0 j=0 0

16 44 HENRY K SCHENCK AND ALEXANDER I SUCIU Thus, δ 4 =dim k im ( d,1 4 From Lemma 44, we know this equals dim k K 4 We thus obtain the following Corollary 410 Let δ 4 be the number of quadratic Koszul syzygies on the generators of I Then ( b1 + (410 b 44 = 4 + (( b a a + b 4 + b 1b δ Computing φ 4 From formula (1, and the formulas from Corollary (6, we find the following expression for the rank of the fourth LCS quotient: (411 φ 4 = b 1(b 1 6b 1 b 1 8 b 1(b 1 +1 a ( a +1 b 1 (a + b +b 44 Combining (410 and (411, we obtain the following Theorem 41 The rank of the fourth LCS quotient of an arrangement group G is given by ( a (41 φ 4 (G = + b 4 δ 4 Using the bounds for b 4 and δ 4 from Lemmas 8 and 46, together with formulas ( and (8, we recover Falk s lower bound (14 for φ k, in the particular case k =4 5 Minimal linear strand arrangements We now apply the machinery developed in the previous sections to the class of MLS arrangements, introduced in We start with a connection with the resonance varieties 51 Local resonance To an arbitrary arrangement A, with Orlik-Solomon algebra A = H (M(A, k, Falk associated in [1] a sequence of resonance varieties, defined as R p (A ={a E 1 H p (A, a 0} Best understood is the variety R 1 (A, which depends only on L (A To each flat X L (A there corresponds a local component L X of dimension µ(x But, in general, there are other components in R 1 (A, as illustrated by the braid arrangements The next result shows that non-trivial resonance cannot happen for MLS arrangements Theorem 5 If A is MLS, then all components of R 1 (A are local Proof Suppose we have non-local resonance, ie, there exist nonzero a, b E 1 with 0 a b I Say, I =(f 1,,f k, so a b = c i f i Then 0 = a a b = a c i f i = c i a f i and Hence, c 1 a (f 1 f k =0, c k a and so we have a (linear syzygy Since the syzygy involves f 1 =(e r e s (e t e s and, by assumption, all linear syzygies are local, the only variables that can appear in a correspond to hyperplanes incident on the flat X L (A containing {r, s, t} Now pick any f i corresponding to a flat Y L (A withy X Since X Y 1, the element a must be equal to some e u, which is impossible

17 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 45 Remark 5 Another, more indirect proof of this theorem can be assembled from results of [5], [6] Indeed, it can be checked that the condition (9 for an arrangement to be MLS coincides with the condition given in [5, Theorem 79] for the completion of the Alexander invariant to decompose as a direct sum of local Alexander invariants From this decomposition, one infers that the variety R 1 (A decomposes into local components, see [6, Theorem 9] Remark 54 The converse to Theorem 5 does not hold Indeed, if A is a complex realization of the MacLane matroid ML 8, then the Alexander invariant of A does not decompose (and so A is not MLS, even though all the components of R 1 (A are local, see [5, Example 86] and [1, Example 107] 55 LCS formula for MLS arrangements Recall that, for any arrangement, we have φ 1 = b 1, φ = a,andφ = b The assumption that A is MLS means that φ attains the lower bound (14 predicted by the local contributions, ie, (51 φ = = φ (F µ(x X L (A ( µ(x+1 X L (A The next result shows that, under this assumption, φ 4 is given by an analogous formula Theorem 56 If A is an MLS arrangement with group G, then (5 φ 4 (G = φ 4 (F µ(x X L (A Proof By Lemmas 8 and 46, we have b i,i+1 = i ( µ(x+i 1 (5, i +1 X L (A ( ( µ(x µ(y (54 δ 4 = From formula (41, we obtain ( a (55 φ 4 = + Plugging in a = X L (A (56 X L (A φ 4 = Witt s formula ends the proof (X,Y ( L (A ( µ(x+ 4 ( µ(x, we find that X L (A (X,Y ( L (A µ(x (µ(x 1 4 ( ( µ(x µ(y Much of the above argument works for arbitrary b ii and φ k, though the spectral sequence chase that went into proving Corollary 410 becomes much more difficult We state the expected end result as a conjecture about the LCS ranks of an MLS arrangement (the simplest case of the more general Resonance LCS Conjecture from [1], leaving the verification for the future

18 46 HENRY K SCHENCK AND ALEXANDER I SUCIU Conjecture 57 (MLS LCS formula If G is the group of a minimal linear strand arrangement, then, for all k, (57 φ k (G = φ k (F µ(x In other words, (58 X L (A (1 t k φ k =(1 t b1 k=1 X L (A 1 µ(x t 1 t It follows from [1, Prop 1] that the conjecture holds for product -arrangements Yet there are many MLS arrangements which are not products, as illustrated by the examples in A simple situation We now look at what happens in a very simple situation, namely, when A is MLS, and the Möbius function on L (A has values only 1 or This occurs precisely when φ =φ, or, equivalently, ( b1 (59 a + b = (b 1 a In this situation, b = ( b 1+ +(b1 +a Moreover, (510 As a consequence, we get b i,i+1 = ia, and δ 4 = Corollary 59 If φ =a,thenb 44 = ( b 1+ 4 φ 4 =a ( a + a ( b1+ ( + a+1,andso Moreover, in this case Conjecture 57 takes the form φ k (G =a φ k (F, or (511 (1 t k φ k =(1 t b1 a (1 t a k=1 510 Examples We conclude this section with a few examples, illustrating the above formulas and conjectures Example 511 (X arrangement The simplest example of an arrangement for which the OS-algebra is not quadratic is the X arrangement from Example 1 We readily compute φ =a = 6; thus, the arrangement satisfies condition (59 In this case, formula (511 predicts that k=1 (1 tk φ k =(1 t Example 51 (Fan arrangements Let A and A be the pair of arrangements considered by Fan in [17] (see Figure It is easy to see that both arrangements are MLS (in fact, they satisfy condition (59 The arrangement A has group G = F 1 F F F, and thus, by Witt s formula, k=1 (1 tk φ k(g =(1 t(1 t The arrangement A has group G = F 1 G 0,whereG 0 is the group of the X arrangement Conjecture 57 predicts that k=1 (1 tk φ k(g =(1 t(1 t In other words, the two arrangements should have the same φ k s (and thus, the same b ii s, even though the respective matroids are different Nevertheless, the combinatorial difference is picked up by the graded Betti numbers of the free

19 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 47 Figure The Fan matroids Figure The matroid of X resolutions of the respective quadratic OS-algebras Indeed, dim k Tor E (A, k 4 =, whereas dim k Tor E (A, k 4 =4 Example 51 (Kohno arrangement An example of a quadratic OS-algebra that is not Koszul is given by Kohno s X arrangement (the corresponding matroid is depicted in Figure It is readily seen that P (t =(1+t(1+6t+10t anda =0; in particular, A is quadratic Corollary 6 gives φ = 6; hence, the arrangement satisfies condition (59 Conjecture 57 predicts that k=1 (1 tk φ k = (1 t5 (1 t From (510, we compute b 4 =15andδ 4 = 10 Thus, b 4 =5,andsothealgebra A is not Koszul Example 514 (9 Configurations Let A and A be realizations of the classical (9 1 and (9 configurations The two arrangements have the same Möbius function, and thus, the same Poincaré polynomial: P (t =(1+t(1 + 8t +19t Even so, A is MLS (a =,yeta is non-mls (a =4 6 Graphic arrangements There is a particularly nice interpretation of our results and formulas in the case of graphic arrangements 61 Betti numbers and generators of the OS-ideal Let G be a simple graph on l vertices, and let A G be the corresponding graphic arrangement in C l,with complement M The Orlik-Solomon algebra A = H (M,k is a quotient of the exterior algebra on generators e i corresponding to the edges of G The Poincaré polynomial is given by (61 P (M,t =( t l χ G ( t 1, where χ G (t isthechromatic polynomial of the graph, see [5] A direct algorithm for computing the Betti numbers b i (and explicit formulas for the first four may be

20 48 HENRY K SCHENCK AND ALEXANDER I SUCIU found in Farrell [18] As for the number of minimal generators of the Orlik-Solomon ideal, we have the following Lemma 6 For all j>, (6 a j = chordless (j +1-cycles Proof We know the m-cycles are dependent sets, and give rise to degree m 1 elements of the Orlik-Solomon ideal The correspondence is one-to-one because a dependent set e 1 e m gives a relation ( 1 i e 1 ê i e m i But the lead monomial of some lower degree element of I divides e e m iff that monomial corresponds to a chord of (e 1 e m, a contradiction This result has been independently obtained by Cordovil and Forge [7] Notice that the lemma does not generalize directly to arbitrary arrangements For example, for a line configuration with four lines through a point, there are four dependent triples But one of the OS relations can be written as a sum of the other three This situation does not occur with graphic arrangements 6 Koszulness and supersolvability Stanley proved in [0] that a graphic arrangement A G is free iff it is supersolvable iff the graph G is chordal (ie, every circuit in G has a chord For a nice exposition we refer the reader to Edelman and Reiner [8] Using Lemma and Lemma 6, we obtain Theorem 64 A graphic arrangement is supersolvable if and only if its OS-algebra is Koszul (in fact, quadratic The forward implication holds for all arrangements (cf [9, Theorem 46], but the converse is not known in general (see [, Problem 671] Previously, the converse was only known to hold for hypersolvable arrangements (see [0] 65 The first three LCS ranks Recall from 61 that a i equals the number of chordless (i + 1-cycles in G, fori> For each integer r 1, define (6 κ r 1 := {G G G = Kr } to be the number of complete subgraphs on r vertices, so that κ 0 = vertices = l, κ 1 = edges = b 1, κ = triangles = a, etc In other words, (κ 0,κ 1,,κ l 1 is the f-vector of the clique complex associated to G Using this notation, we now answer Problem 14 from Falk s recent survey [14], which asks for a combinatorial formula for φ in the graphic setting Corollary 66 For a graphic arrangement, the first three LCS ranks are given by φ 1 = b 1 = κ 1, ( φ = b1 b = κ, φ = b + b 1 a + a ( b 1 = (κ + κ Proof For a graphic arrangement, the first three Betti numbers are given by b 1 = κ 1, ( b = b1 a, b = ( b 1 b1 a a +(κ + κ Now use Corollary 6

21 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS Resolution of a graphic OS-algebra To proceed further, we need to analyze in more detail the resolution of the Orlik-Solomon algebra over the exterior algebra, in the case of a graphic arrangement We start with an important example e e 0 e e 5 e 4 e 1 Figure 4 ThecompletegraphK 4 Example 68 The braid arrangement (in C 4 is the graphic arrangement associated to the complete graph K 4 (see Figure 4 The free resolution of the Orlik- Solomon algebra A as a module over the exterior algebra E = (e 0,,e 5 begins 0 A E 1 E 4 ( E 10 (, where 1 = ( e 145 e 5 e 04 e 01,and = e 1 e 4 e 1 e e e 0 e e e e e e e 0 e 1 e 0 e e 0 e e 0 e e 1 e 5 e e e 0 e 1 e 0 e e e 5 e 4 e 5 To see how this goes, write the generators of I as f i = α i β i (i =1,,4; for example, f 1 = e 145 =(e 1 e 4 (e 1 e 5 Each f i generates two local linear syzygies, α i and β i, which appear in columns i 1andiof the matrix Let η 1 = e 0 e 1 e + e 5 and η = e 0 e e 4 + e 5 Note that η 1 η = f 1 f + f + f 4 I Thus,η 1 and η belong to the resonance variety R 1 (Infact,η 1 and η span an essential, -dimensional component of R 1, corresponding to the neighborly partition Π = (05 1 4, see [1] We now get two new linear syzygies, expressing the fact that η i (η 1 η =0inE Reducing modulo the local syzygies, we obtain the last two columns of Itiseasytosee that the columns of are linearly independent From Corollary 66, we know that b = 10 Hence, the columns of form a complete set of linear syzygies on I Clearly, each of these 5 pairs of linear syzygies generates i + 1 (independent linear i-th syzygies For example, the pair (α i,β i yields α i α i = β i β i = α i β i + β i α i = 0 as linear second syzygies Hence, b i,i+1 =5i, for all i Finally, note that there are ( 4 = 6 quadratic Koszul syzygies on the generators of I, but that they are all consequences of the linear syzygies on I For example, e 04 e 145 e 145 e 04 =(e 0 e 1 ((e e 0 e 145 +(e 0 e 1 e 5 + (e 1 e 5 e 04 +(e e 5 e 01 +(e0 e (e 1 e 4 e (e 1 e 5 (e 0 e 4 e 04 +(e e 5 (e 0 e 1 e 01

22 40 HENRY K SCHENCK AND ALEXANDER I SUCIU It is easy to check that there are no other quadratic syzygies on the generators of I, and thus b 4 =0 Now let A = A G be an arbitrary graphic arrangement Lemma 69 For i, b i,i+1 = i(κ + κ Proof From Corollary 66, we know there are b = φ =(κ + κ linear first syzygies on the generators of I From the discussion in Example 68, we know that κ of those syzygies are local Each non-local resonance component associated to a K 4 subgraph generates an additional pair of linear first syzygies, as in the example above We claim these κ linear syzygies are linearly independent Indeed, for there to be a linear dependence between a set of such syzygies, some pair would have to overlap at a position Those two syzygies must involve a common element, f, ofi By assumption, the syzygies come from distinct sub-k 4 s, so these two K 4 s share a common triangle corresponding to f Now on each triangle of a K 4, the resonant syzygies do not involve the variables of that triangle (see the above example So the two syzygies on f involving distinct K 4 s involve different sets of variables, hence are independent This contradiction proves the claim Thus, since we know that the number of linear first syzygies is (κ +κ, we have identified them all Each pair of local linear first syzygies associated to a triangle generates i + 1 linear i-th syzygies This is also the case for the linear syzygy pairs associated to the K 4 subgraphs, as described in the example above Such syzygies will be independent, since they are supported only on the linear syzygy pair coming from a K 4, so we are done Remark 610 In general there can be interplay between the syzygies arising from resonance components, and this will be reflected in the free resolution (this happens for example in the case of the non-fano arrangement But as noted above, such a phenomenon does not occur for graphic arrangements Lemma 611 δ 4 ( κ 6(κ + κ 4 Proof There are at most ( κ quadratic Koszul syzygies As seen in Example 68, the linear syzygies of a K 4 -subgraph kill the 6 Koszul syzygies corresponding to that subgraph A similar computation shows that the linear syzygies of a K 5 -subgraph kill 6 Koszul syzygies Of those, 0 are killed by the 5 sub-k 4 s of the K 5,and the remaining 6 are killed by the K 5 itself, whence the bound 61 The fourth LCS rank We now return to the computation of the LCS ranks of an arrangement group Theorem 61 For a graphic arrangement, Furthermore, if κ =0,thenφ 4 =κ φ 4 κ +9κ +6κ 4 Proof The inequality follows from Lemmas 69 and 611, together with Theorem 41 If κ = 0, equality follows from Corollaries 59 and 66

23 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 41 Figure 5 A non-hypersolvable graph Example 614 Let G be the graph in Figure 5 The arrangement A G is the simplest example (in terms of number of vertices of a graphic arrangement which is not hypersolvable In particular, no general method could be applied to obtain avalueforφ 4 On the other hand, Theorem 61 yields φ 4 = Graphic LCS formula Extensive computations suggest that equality holds in Lemma 611, and hence in Theorem 61 In fact, this seems to be part of a more general pattern, which leads us to formulate the following conjecture Conjecture 616 (Graphic LCS formula If G is a graph with l vertices, then (64 ( 1 t k l 1 φ k = (1 jt k=1 j=1 l 1 ( 1 s j( s j κs s=j Expanding both sides, (64 becomes equivalent to the following sequence of equalities: φ 1 = κ 1, φ = κ, φ =(κ + κ, φ 4 =(κ +κ +κ 4, φ 5 =6(κ +5κ +8κ 4 +4κ 5, φ 6 =9κ +89κ + 60κ κ κ 6, φ k = k j=1 s=j k ( s ( 1 s j κ s φ k (F j j By Corollary 66, these equalities hold up to k = By Theorem 61, equality also holds for k =4,providedκ = 0 (otherwise, we only know there is an inequality in one direction Remark 617 For graphic arrangements, the resonance formula for the ranks of the Chen groups, θ k (G =rankgr k (G/G (conjectured in [1] gives (65 θ k =(k 1(κ + κ, for k

24 4 HENRY K SCHENCK AND ALEXANDER I SUCIU The resonance LCS formula (also conjectured in [1] is based upon the assumption that φ 4 = θ 4 In the case of graphic arrangements, formula (64 would imply (66 φ =φ φ 4 = θ 4 κ =0 Hence, if G contains no K 4 subgraphs, then all the above conjectures reduce to (67 θ k = κ θ k (F and φ k = κ φ k (F, for all k Remark 618 The graphic LCS conjecture is true for chordal graphs Indeed, if G is a chordal graph, then its chromatic polynomial is given by κ 0 1 ( ( χ G (t =t κ0 1 jt 1 κ 0 1 s s=j ( 1s j j κ s (68, j=1 see [] Since in this case A G is supersolvable, formula (64 follows at once from (61 and the Falk-Randell LCS formula (11 We warmly thank Yuri Volvovski for bringing formula (68 to our attention References [1]AAramova,LAvramov,JHerzog,Resolutions of monomial ideals and cohomology over exterior algebras, Trans Amer Math Soc 5 (000, MR 000c:101 [] A Beilinson, V Ginzburg, W Soergel, Koszul duality patterns in representation theory, J Amer Math Soc 9 (1996, MR 96k:17010 [] K Braun, M Kretz, B Walter, M Walter, Die chromatischen Polynome unterringfreier Graphen, Manuscripta Math 14 (1974, 4 MR 50:6906 [4] H Cartan, S Eilenberg, Homological algebra, Princeton Univ Press, Princeton, NJ, 1956 MR 17:1040e [5] D Cohen, A Suciu, Alexander invariants of complex hyperplane arrangements, Trans Amer Math Soc 51 (1999, MR 99m:5019 [6] Characteristic varieties of arrangements, Math Proc Cambridge Philos Soc 17 (1999, 5 MR 000m:06 [7] R Cordovil, D Forge, Quadratic Orlik-Solomon algebras of graphic matroids, [8] P Edelman, V Reiner, Free hyperplane arrangements between A n 1 and B n,mathz15 (1994, MR 95b:501 [9] D Eisenbud, Commutative algebra with a view towards algebraic geometry, Graduate Texts in Math, vol 150, Springer-Verlag, Berlin-Heidelberg-New York, 1995 MR 97a:1001 [10] D Eisenbud, S Popescu, S Yuzvinsky, Hyperplane arrangement cohomology and monomials in the exterior algebra, [11] M Falk, The minimal model of the complement of an arrangement of hyperplanes, Trans Amer Math Soc 09 (1988, MR 89d:04 [1], The cohomology and fundamental group of a hyperplane complement, in: Singularities (Iowa City, IA, 1986, Contemporary Math, vol 90, Amer Math Soc, Providence, RI, 1989, pp 55 7 MR 90h:06 [1], Arrangements and cohomology, Ann Combin 1 (1997, MR 99g:5017 [14], Combinatorial and algebraic structure in Orlik-Solomon algebras, European J Combinatorics (001, [15] M Falk, R Randell, The lower central series of a fiber-type arrangement, Invent Math 8 (1985, MR 87c:015b [16], On the homotopy theory of arrangements, II, in: Arrangements Tokyo 1998, Adv Stud Pure Math, vol 7, Math Soc Japan, Kinokuniya, Tokyo, 000, pp 9 15 MR 00b:044 [17] K-M Fan, Position of singularities and fundamental group of the complement of a union of lines, Proc Amer Math Soc 14 (1996, 99 0 MR 97e:140 [18] E J Farrell, On chromatic coefficients, DiscreteMath 9 (1980, MR 81d:0509

25 LOWER CENTRAL SERIES AND HYPERPLANE ARRANGEMENTS 4 [19] D Grayson, M Stillman, Macaulay : a software system for algebraic geometry and commutative algebra; available at [0] M Jambu, S Papadima, A generalization of fiber-type arrangements and a new deformation method, Topology 7 (1998, MR 99g:5019 [1] T Kohno, Série de Poincaré-Koszul associée aux groupes de tresses pures, Invent Math 8 (1985, MR 87c:015a [] C Löfwall, On the subalgebra generated by the one dimensional elements in the Yoneda Ext-algebra, in: Algebra, Algebraic Topology and Their Interactions, Lecture Notes in Math, vol 118, Springer-Verlag, Berlin-Heidelberg-New York, 1986, pp 91 8 MR 88f:1600 [] W Magnus, A Karrass, D Solitar, Combinatorial group theory (nd ed, Dover, New York, 1976 MR 54:104 [4] J McCleary, A user s guide to spectral sequences, nd edition, Cambridge Univ Press, Cambridge, 001 MR 00c:5507 [5] P Orlik, H Terao, Arrangements of hyperplanes, Grundlehren Math Wiss, Bd 00, Springer-Verlag, Berlin-Heidelberg-New York, 199 MR 94e:5014 [6] S Papadima, S Yuzvinsky, On rational K[π,1] spaces and Koszul algebras, J Pure Appl Alg 144 (1999, MR 000k:55017 [7] I Peeva, Hyperplane arrangements and linear strands in resolutions, preprint, 1997 [8] S Priddy, Koszul resolutions, Trans Amer Math Soc 15 (1970, 9 60 MR 4:46 [9] B Shelton, S Yuzvinsky, Koszul algebras from graphs and hyperplane arrangements, J London Math Soc 56 (1997, MR 99c:16044 [0] R Stanley, Supersolvable lattices, Algebra Universalis (197, MR 46:890 [1] A Suciu, Fundamental groups of line arrangements: Enumerative aspects, in: Advances in algebraic geometry motivated by physics, Contemporary Math, vol 76, Amer Math Soc, Providence, RI, 001, pp 4 79 [] S Yuzvinsky, Orlik-Solomon algebras in algebra and topology, Russian Math Surveys 56 (001, 9 64 Department of Mathematics, Harvard University, Cambridge, Massachusetts 018 Current address: Department of Mathematics, Texas A&M University, College Station, Texas address: schenck@mathtamuedu URL: Department of Mathematics, Northeastern University, Boston, Massachusetts address: alexsuciu@neuedu URL: