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On the homotopy Lie algebra of an arrangement



ON THE HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT
GRAHAM DENHAM1 AND ALEXANDER I. SUCIU2

arXiv:math/0502417v2 [math.AT] 22 Jul 2005

Abstract. Let A be a graded-commutative, connected k-algebra generated in degree 1. The homotopy Lie algebra gA is de?ned to be the Lie algebra of primitives of the Yoneda algebra, ExtA (k, k). Under certain homological assumptions on A and its quadratic closure, we express gA as a semi-direct product of the well-understood holonomy Lie algebra hA with a certain hA -module. This allows us to compute the homotopy Lie algebra associated to the cohomology ring of the complement of a complex hyperplane arrangement, provided some combinatorial assumptions are satis?ed. As an application, we give examples of hyperplane arrangements whose complements have the same Poincar? polynomial, the same fundamental group, and the same holonomy e Lie algebra, yet di?erent homotopy Lie algebras.

1. Definitions and statements of results 1.1. Holonomy and homotopy Lie algebras. Fix a ?eld k of characteristic 0. Let A be a graded, graded-commutative algebra over k, with graded piece Ak , k ≥ 0. We will assume throughout that A is locally ?nite, connected, and generated in degree 1. In other words, A = T (V )/I, where V is a ?nite-dimensional k-vector space, T (V ) = k≥0 V ?k is the tensor algebra on V , and I is a two-sided ideal, generated in degrees 2 and higher. To such an algebra A, one naturally associates two graded Lie algebras over k (see for instance [3], [14]). De?nition 1.1. The holonomy Lie algebra hA is the quotient of the free Lie algebra on the dual of A1 , modulo the ideal generated by the image of the transpose of the multiplication map ? : A1 ∧ A1 → A2 : (1) hA = Lie(A? ) ideal (im(?? : A? → A? ∧ A? )). 1 2 1 1

Note that hA depends only on the quadratic closure of A: if we put A = T (V )/(I2 ), then hA = hA . De?nition 1.2. The homotopy Lie algebra gA is the graded Lie algebra of primitive elements in the Yoneda algebra of A: (2) gA = Prim(ExtA (k, k)).

2000 Mathematics Subject Classi?cation. Primary 16E05, 52C35; Secondary 16S37, 55P62. Key words and phrases. Holonomy and homotopy Lie algebras, hyperplane arrangement, supersolvable, hypersolvable, Yoneda algebra, Koszul algebra, Hopf algebra, spectral sequence, homotopy groups. 1 Partially supported by a grant from NSERC of Canada. 2 Partially supported by NSF grant DMS-0311142.
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G. DENHAM AND A. I. SUCIU

In other words, the universal enveloping algebra of the homotopy Lie algebra is the Yoneda algebra: (3) U (gA ) = ExtA (k, k).

The algebra U = ExtA (k, k) is a bigraded algebra; let us write U pq to denote cohomological degree p and polynomial degree q. Then U pq = 0, unless ?q ≥ p. The subalgebra R = p≥0 U p,?p is called the linear strand of U . For convenience, we will p let Uq = U p,?p?q . The lower index q is called the internal degree. Then U is a graded R-algebra, with R = U0 . Note that U+ = q>0 Uq is an ideal in U , with U/U+ ? R. = The relationship between the holonomy and homotopy Lie algebras of A is provided by the following well-known result of L¨fwall. o Lemma 1.3 (L¨fwall [19]). The universal enveloping algebra of the holonomy Lie algeo p bra, U (hA ), equals the linear strand, R = p≥0 U0 , of the Yoneda algebra U = U (gA ). Particularly simple is the case when A is a Koszul algebra. By de?nition, this means the homotopy Lie algebra gA coincides with the holonomy Lie algebra hA , i.e., U = R. ⊥ Alternatively, A is quadratic (i.e., A = A), and its quadratic dual, A! = T (V )/(I2 ), co! = U . For an expository account of Koszul algebras, incides with the Yoneda algebra: A see [13]. As a simple (yet basic) example, take E = V , the exterior algebra on V . Then E is Koszul, and its quadratic dual is E ! = Sym(V ? ), the symmetric algebra on the dual vector space. Moreover, gA = hA is the abelian Lie algebra on V . 1.2. Main result. The computation of the homotopy Lie algebra of a given algebra A is, in general, a very hard problem. Our goal here is to determine gA under certain homological hypothesis. First, we need one more de?nition. Let B = A be the quadratic closure of A. View J = ker(B ? A) as a graded left module over B. De?nition 1.4. The homotopy module of a graded algebra A is (4) MA = ExtB (J, k),

viewed as a bigraded left module over the ring R = U (hA ) = ExtB (k, k) via the Yoneda product. Theorem 1.5. Let A be a graded algebra over a ?eld k, with quadratic closure B = A, and homotopy module M = MA . Assume B is a Koszul algebra, and there exists an integer ? such that Mq = 0 unless ? ≤ q ≤ ? + 1. Then, as graded Hopf algebras, (5) U (gA ) ? T (MA [?2]) ?k U (hA ). =

Here M [q] is the graded R-module with M [q]r = M q+r , while T (M [?2])?k R is the “twisted” tensor product of algebras, with underlying vector space T (M [?2]) ?k R and multiplication (m ? r) · (n ? s) = (?1)|r||n| ((m ? n) ? rs + (m ? nr) ? s). Taking the Lie algebras of primitive elements in the respective Hopf algebras, we obtain the following.

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

3

Corollary 1.6. Under the above hypothesis, the homotopy Lie algebra of A splits as a semi-direct product of the holonomy Lie algebra with the free Lie algebra on the (shifted) homotopy module, (6) gA ? Lie(MA [?2]) ? hA , = where the action of h on Lie(M ) is given by [m, h] = ?hm for h ∈ h and m ∈ M . As pointed out to us by S. Iyengar, Theorem 1.5 implies (under our hypothesis) that the projection map U (gA ) → U (hA ) is a Golod homomorphism. Therefore, the semidirect product structure of gA also follows from results of Avramov [1], [2]. 1.3. Hyperplane arrangements. Let A = {H1 , . . . , Hn } be an arrangement of hyperplanes in C? , with intersection lattice L(A) and complement X(A). The cohomology ring A = H ? (X(A), k) admits a combinatorial description (in terms of L(A)), due to Orlik and Solomon: (7) A = E/I, where E is the exterior algebra over k, on generators e1 , . . . , en in degree 1, and I is the ideal generated by all elements of the form r (?1)q?1 ei1 · · · eiq · · · eir for which q=1 rk(Hi1 ∩ · · · ∩ Hir ) < r; see [22]. The holonomy Lie algebra of the Orlik-Solomon algebra also admits an explicit presentation, this time solely in terms of L≤2 (A). Identify Lie(A? ) with the free Lie algebra 1 over k, on generators xH = e? , H ∈ A. Then: H (8) hA = Lie(A? ) ideal 1 xH ,
H ′ ∈A : H ′ ?F

xH ′

F ∈ L2 (A) and F ? H .

As we shall see in Section 5, the homotopy Lie algebra gA also admits a ?nite presentation, for a certain class of hypersolvable arrangements, to be de?ned below. Question 1.7. Do there exist arrangements for which gA is not ?nitely presented? For which the (bigraded) Hilbert series of U (gA ) is not a rational function? 1.4. Hypersolvable arrangements. An arrangement A is called supersolvable if its intersection lattice admits a maximal modular chain. The OS algebra of a supersolvable arrangement has a quadratic Gr¨bner basis, and thus, it is a Koszul algebra (this result, o implicit in Bj¨rner and Ziegler [4], was proven in Shelton and Yuzvinsky [30]). o An arrangement A is called hypersolvable if it has the same intersection lattice up to rank 2 as that of a supersolvable arrangement. This “supersolvable deformation,” B, is uniquely de?ned, and has the property that the two complements have isomorphic fundamental groups; see Jambu and Papadima [16, 17]. Let A = H ? (X(A), k) and B = H ? (X(B), k) be the respective OS algebras. It is readily seen that B = A; thus, A and B share the same holonomy Lie algebra: h = hA = hB . Furthermore, since B is Koszul, we have gB = h. The hypothesis of Theorem 1.5 holds in two nice situations, which can be checked combinatorially; for precise de?nitions, see §4.2 and §4.3, respectively. Theorem 1.8. Let A be an arrangement, and let A be its Orlik-Solomon algebra. Suppose either

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G. DENHAM AND A. I. SUCIU

(1) A is hypersolvable, and its singular range has length 0 or 1; or (2) A is obtained by ?bred extensions of a generic slice of a supersolvable arrangement. Then gA ? Lie(MA [?2]) ? hA . = An explicit ?nite presentation for gA is given in Theorem 5.1, in the case when A is a generic slice of a supersolvable arrangement. The Eisenbud-Popescu-Yuzvinsky resolution [7] permits us to compute the Hilbert series of MA (and hence, that of gA ) in the case when A is a 2-generic slice of a Boolean arrangement. Theorem 1.8 allows us to distinguish between hyperplane arrangements whose holonomy Lie algebras are isomorphic. In Example 6.2, we exhibit a pair of 2-generic, 4-dimensional sections of the Boolean arrangement in C7 ; the two arrangements have the same fundamental group, the same Poincar? polynomial, and the same holonomy e Lie algebra, yet di?erent homotopy Lie algebras. In Section 7, we provide some topological interpretations. As noted in [5], [24], the holonomy Lie algebra of a supersolvable arrangement equals, up to a rescaling factor, the topological homotopy Lie algebra of the corresponding “redundant” subspace arrangement. We extend this result, and relate the homotopy Lie algebra of an arbitrary hyperplane arrangement to the topological homotopy Lie algebras of the redundant subspace arrangement. As a consequence, we ?nd a pair of codimension-2 subspace arrangements in C8 , whose complements are simply-connected and have the same homology groups, yet distinct higher homotopy groups. 2. Some homological algebra 2.1. The homotopy module. Let A be graded, graded-commutative, connected, locally ?nite algebra. Assume A is generated in degree 1, and its quadratic closure, B = A is a Koszul algebra. Let E be the exterior algebra on A1 = B1 . Let I and J be, respectively, the kernels of the natural surjections E ? B and B ? A, giving the exact sequences (9) (10) 0 0
/I /J /E /B /B /A /0, /0.

In what follows, we will record some homological properties of the ring A, viewed as a B-module. Recall if N is a B-module, the Yoneda product gives ExtB (N, k) the structure of a left module over the ring R = U (hA ) = ExtB (k, k). An object of primary interest for us will be the homotopy module of A, (11) M = MA = ExtB (J, k). This bigraded R-module will play a crucial role in the determination of the homotopy Lie algebra gA . Our grading conventions shall be as follows. Suppose V and W are Z-graded k-vector spaces. Then f ∈ Homk (V, W ) has degree r if f : V q → W q+r for all q. For any Zgraded k-vector space V , we shall let V ? denote the graded k-dual of V . In particular, then, (V ? )q = Homk (V ?q , k). If V has ?nite k-dimension in each graded piece, then (V ? )? ? V . =

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

5

We shall treat all boundary maps in chain complexes as having polynomial degree 0 and homological degree +1. Then, in particular, chain complexes will be regarded as cochain complexes in negative degree. We shall indicate shifts of polynomial grading by de?ning V (q)r = V q+r , and shifts of homological grading by writing V [q] analogously. Following these conventions, M pq = Extp (J, k)q is nonzero only for q ≤ ?p. Then, B p p taking Mq = M p,?p?q (the internal grading), we have Mq = 0 only for q ≥ 0. The p r+p grading is such that, for each ?xed q, the action of R on M satis?es Rr ? Mq → Mq . Lemma 2.1. ExtB (A, k) ? k ⊕ M [?1] as graded R-modules. = Proof. Consider the long exact sequence for ExtB (?, k) applied to (10): (12) ···
/ Extq?1 (J, k) B / Extq (A, k) B / Extq (B, k) B / ···

Since Ext0 (A, k) ? Ext0 (B, k) ? k and Extq (B, k) = 0 for all q > 0, the map = = B B B ExtB (B, k) → ExtB (J, k) is zero. So the long exact sequence breaks into short exact sequences which, using (11), we will write as a single short exact sequence of graded R-modules, (13) 0
/ M [?1] / ExtB (A, k) /k /0.

For each q, one of the two maps is zero and the other is an isomorphism, so the short exact sequence splits. 2.2. Injective resolutions. For any E-module N , let (14) N ? = {a ∈ E : ax = 0 for all x ∈ N }, the annihilator of N in E. Later on, we require explicit, injective resolutions. Lemma 2.2. Suppose the ring B = E/I is an arbitrary quotient of a ?nitely-generated exterior algebra E. If (15) 0o ko B ?k F 0 o B ?k F 1 o
/ B ? ?k (F 1 )?

···

is a minimal, free resolution of k over B, then (16) 0
/k / B ? ?k (F 0 )? / ···

is an injective resolution of k over B. Proof. The resolution (15) is an acyclic complex of E-modules, so its vector space dual (16) is an acyclic complex as well, since each F i has ?nite k-dimension. Now B ? ? I ? (n) as E-modules, via the determinantal pairing in E. On the other = hand, E is injective as a module over itself, so I ? is injective as an E-module; see [29, Prop. 2.27]. Since each F i has ?nite k-dimension, each B ? ?k (F i )? is injective. Lemma 2.3. Let A and B two algebras, with A = B Koszul. Write B = E/I, A = B/J, h = hA = hB , and R = U (h). Then: (1) The complex (17) 0
/k / I ? (n) ?k R0 / I ? (n) ?k R1 / ···

is an injective resolution of k over B, with boundary map described below.

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G. DENHAM AND A. I. SUCIU

(2) Extq (A, k) ? H q (J ? (n) ?k R? ), for all q ≥ 0. = B Proof. The Koszul complex K? = B ?k R? is a free B-module resolution of k, so it is also an acyclic complex of E-modules, with boundary map induced from (18) ? ? : 1 ? x? → ei ? 1. i Then Homk (B ?k R? , k) = B ? ?k R is an injective resolution, by the previous Lemma. To establish (2), it su?ces to note that HomB (A, I ? ) ? J ? . = 3. Proof of the main result Our approach to the proof of Theorem 1.5 is to construct a spectral sequence comparing the minimal resolution and the Koszul complex of A. We show the spectral sequence collapses at E2 under suitable hypotheses in Proposition 3.2, though not in general (Example 3.3). This collapsing is enough to prove the theorem, via Proposition 3.1. 3.1. A spectral sequence. Using the previous notation, A?k U ? → k → 0 is a minimal free resolution of k over A. It is ?ltered by degree, and the linear strand is A ?k R? . That is, there is a short exact sequence of chain complexes (19) 0
/ A ?k R ?
1?ε?

/ A ?k U ?

? / A ?k U+

/0.

Now B ?k R? is a free resolution of k over B, since B is Koszul. Using Lemma 2.1, we ?nd that the homology of the linear strand (Koszul complex) is (20) H? (A ? R? ) ? TorB (A, k) = ? ExtB (A, k)? = ? k ⊕ M [?1]? . = The long exact sequence in homology then reveals that (21) H? (A ?k U ? ) ? M [?2]? =
+

as A-modules. Recall that A acts trivially on M (and hence on M [?2]? ), so ? (22) HomA (H? (A ?k U ? ), k) = M [?2].
+

On the other hand, since our complex is a quotient of a minimal resolution, (23) H? (HomA (A ?k U ? , k)) ? U+ . =
+

Comparing the two gives a universal coe?cients spectral sequence of the form (24)
p+q pq E2 = Extp ((M [?2]? )q , k) ? M [?2]q ?k U p =? U+ . = A

The spectral sequence is used as follows. Proposition 3.1. If E∞ = E2 in the spectral sequence (24), then 0
/ U ?k M [?2]
φ

/U

ε

/R

/0

is exact, and the conclusion of Theorem 1.5 holds.

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

7

Proof. If E∞ = E2 , then U ? M [?2] ? U+ as a (left) U -module. Now U+ = ker ε, = giving the short exact sequence. Since h is a Lie subalgebra of g, R = U (h) is a Hopf subalgebra of U = U (g), so the sequence splits. The isomorphism of Theorem 1.5 can then be obtained by induction. 3.2. Collapsing conditions. In order to show that the higher di?erentials in the spectral sequence (24) vanish, we use a degree argument that begins by considering the E0 term. Since (25) (26) 0
/k / A? ? U 0 k / A? ? U 1 k / ···

is an injective resolution of k over A, (Lemma 2.2) we consider the double complex C pq = HomA (A ?k (U q )? , A? ?k U p ) + ? U q ?k A? ?k U p , =
+

with induced boundary maps ?h and ?v . Then our spectral sequence (24) is obtained by ?ltering C ?? by columns. Checking the grading, we see (27) and (28)
q q ?h : U+ ?k (A? )s ?k U p → U+ ?k (A? )s+1 ?k U p+1 . q q+1 ?v : U+ ?k (A? )s ?k U p → U+ ?k (A? )s+1 ?k U p

By looking at E2 and ?v , we see that we must have E1 = E2 . We ?rst consider the case where the ideal J has a (shifted) linear resolution.
p Proposition 3.2. Suppose A is a hypersolvable arrangement for which Mq = 0 unless q = ?, for some ?xed ?. Then E2 = E∞ .

Proof. In this case, M [?2]q = 0 unless r = ?? 2. Then H q (C p? , ?v )r = 0 unless r = ?? 2. r First we note that (U+ )q = 0 unless t ≥ ? ? 2. This can be seen from the fact that U+ t is a graded subquotient of M [?2] ?k U , from (24): the support of M [?2] is described p above, and Uq = 0 unless q ≥ 0. Regard A? as a chain complex concentrated in homological degree 0. Then observe that the internal degree of a nontrivial cocycle representative in (U+ )q ?k (A? )s is s+t = ??2, t by the ?rst observation above. It follows s ≤ 0 from the inequality above. However, (A? )s = 0 unless 0 ≤ s ≤ ?, so the representative of a nonzero, homogeneous E2 -cocycle in E0 must have s = 0. pq Now suppose x ∈ E2 is such a cocycle, with representative x in C pq . By the above, q ? ) . Then ? (x) = 0 in C p+1,q by (28). This means d (x) = 0, and similarly x ∈ U+ ?k (A 0 2 h for higher di?erentials. Proof of Theorem 1.5. In view of Proposition 3.1, it remains only to show the spectral p sequence collapses when Mq = 0 unless 0 ≤ ??q ≤ 1 for some ?. In this case, let N = M? denote the R-submodule of M of internal degree ?. By the same reasoning as in the proof of Proposition 3.2, N [?2] ? U ? ker dk for k ≥ 2. Now N [?2] ? N [?2] ? U 0 is a submodule of the p = 0 column of E2 . Since it is = (trivially) not in the image of any nonzero di?erentials, N [?2] is an R-submodule of U . Let K denote the Hopf subalgebra of U generated by R and N [?2]. By [20, Theorem 4.4], U is a free K-algebra. It follows that K ? T (N [?2]) ?k R. In the notation of the =

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G. DENHAM AND A. I. SUCIU

previous proposition, any nontrivial di?erential dk with k ≥ 2 would lift in E0 to a map U+ ? (A? )1 ? U → U+ ? (A? )0 ? U . We have shown that the targets of these maps are unchanged between E2 and E∞ , so it follows that the maps themselves must also all be zero. 3.3. A non-collapsing spectral sequence. Calculations with the Macaulay 2 package [15] show that the hypotheses of Theorem 1.5 cannot in general be relaxed: di?erentials in the spectral sequence (24) may not be zero. Example 3.3. Consider the arrangement de?ned by the polynomial Q = xyz(x ? w)(y ? w)(z ? w)(x ? u)(y ? u). Let A be the Orlik-Solomon algebra, and M = MA its homotopy module. It is readily seen that Mq = 0 for q = 3, 4, 5. An Euler characteristic calculation shows that the spectral sequence (24) must have a nonzero di?erential d04 : M [?2]4 ?k U 0 → M [?2]3 ?k U 2 . 5 2 6 It follows that the Hopf algebra U (gA ) will not have the structure we ?nd in Theorem 1.5. 4. Hypersolvable arrangements In this section, we apply our main result to certain classes of hypersolvable arrangements. 4.1. Solvable extensions. We start by reviewing in more detail the notion of a hypersolvable arrangement, introduced by Jambu and Papadima in [16]. Roughly, a hypersolvable arrangement is a linear projection of a supersolvable arrangement that preserves intersections through codimension two. De?nition 4.1 ([16]). An arrangement A is hypersolvable if there exist subarrangements {0} = A1 ? A2 ? · · · ? Am = A, so that each inclusion Ai ? Ai+1 is solvable. In turn, an inclusion of hyperplane arrangements A ? B is called a solvable extension if: (1) There are no hyperplanes H ∈ B \ A and H ′ , H ′′ ∈ A with H ′ = H ′′ and rk(H ∩ H ′ ∩ H ′′ ) = 2; (2) For any H, H ′ ∈ B \ A, there is exactly one H ′′ ∈ A with rk(H ∩ H ′ ∩ H ′′ ) = 2, denoted by f (H, H ′ ); (3) For any H, H ′ , H ′′ ∈ B \ A, one has rk(f (H, H ′ ) ∩ f (H, H ′′ ) ∩ f (H ′ , H ′′ )) ≤ 2. It turns out that if A is hypersolvable with a sequence of solvable extensions as above, then for all i, the rank of Ai and Ai+1 di?er by at most one. If the ranks are equal, the extension is said to be singular; otherwise, the extension is nonsingular (or ?bred, in the sense of Falk and Randell, [11]). If s denotes the number of singular extensions, then, rk A = m ? s. Jambu and Papadima show in [17] that one can replace the singular extensions by nonsingular ones in order to construct a supersolvable arrangement B of rank m that projects onto A, preserving the intersection lattice through rank 2. That is, Theorem 4.2. An arrangement A is hypersolvable i? there exists a supersolvable arrangement B and a linear subspace W for which A = B ∩ W and L(A)≤2 ? L(B)≤2 . =

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

9

Proof. The implication “?” is Theorem 2.4 of [17]. The converse, due to Jambu (private communication), runs as follows. Suppose B is supersolvable and there exists a subspace W as above. By de?nition, B has a maximal modular chain F1 < F2 < · · · < Fm . Putting Bi = BXi gives a sequence of solvable extensions for B, all ?bred. For 1 ≤ i ≤ m, let Ai = Bi ∩ W . Since collinearity relations are preserved, each Ai ? Ai+1 is also a solvable extension, so A is hypersolvable. We remark that, in the above proof, Ai ? Ai+1 is a singular extension if and only if Fi ∩W = Fi+1 ∩W . The arrangement B in called the supersolvable deformation of A. For example, any arrangement A for which no three hyperplanes intersect in codimension three is hypersolvable, and its supersolvable deformation is the Boolean arrangement in Cn , where n = |A|. Lemma 4.3. Suppose A′ ? A is a ?bred extension. The projection p : X(A) → X(A′ ) induces an inclusion A′ ?→ A of the respective Orlik-Solomon algebras which makes A into a free A′ -module of rank k = |A \ A′ |. Proof. The projection p : X → X ′ is a bundle map, with ?ber C\{k points}. As noted by Falk and Randell [11], this bundle admits a section, and thus the Serre spectral sequence collapses at the E2 term. Hence, H ? (X) ? H ? (X ′ ) ? H ? (∨k S 1 ). The result follows. = 4.2. Singular range. We now give some easy to check combinatorial conditions insuring that a hypersolvable arrangement satis?es the hypothesis of Theorem 1.5. We start by attaching a pair of relevant integers to a hypersolvable arrangement. De?nition 4.4. Suppose A is hypersolvable with supersolvable deformation B, and A = B. Let c be the least integer for which L(A)≤c ? L(B)≤c . Since A = B, there is a = largest integer i for which the extension Ai ? Ai+1 is singular. Let d the rank of these two arrangements. We will call the pair (c, d) the singular range of the arrangement A, and |d ? c| the length of this range. Lemma 4.5. If A is hypersolvable with singular range (c, d), then 3 ≤ c ≤ d. Proof. The inequality c ≥ 3 follows from Theorem 4.2. Suppose d < c; then L(A)≤d ? = L(B)≤d . It follows that L(Ad+1 )≤d ? L(Bd+1 )≤d , whence Ad+1 = Bd+1 since the ar= rangements are central. Since d is greater than or equal to the index of the last singular extension, however, Ai = Bi for d + 1 ≤ i ≤ m, so A = B, a contradiction. Let A = H ? (X(A), k) and B = H ? (X(B), k) be the respective Orlik-Solomon algebras. Since L(A)≤2 ? L(B)≤2 , and since the Orlik-Solomon algebra of a supersolvable = arrangement is quadratic, the algebra B = E/I is the quadratic closure of A. Let J = ker(B ? A), and let M = ExtB (J, k), viewed as a module over R = ExtB (k, k). Since B is supersolvable, the algebra B is Koszul (see [30]); thus, R = B ! .
p Lemma 4.6. If A is a hypersolvable arrangement with singular range (c, d), then Mq = 0 unless p ≥ 0 and c ≤ q ≤ d.

Proof. The ideal J has a minimal, (in?nite) free resolution over B of the form (29) 0o Jo B ?k (M 0,? )? o B ?k (M 1,? )? o ···

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G. DENHAM AND A. I. SUCIU

Recall that J is generated by Orlik-Solomon relations. By De?nition 4.4, the least degree p 0 0 of a generator of J is c, so Mc = 0 and Mq = 0 for q < c. Thus Mq = 0 for q < c, establishing the ?rst inequality. p To show Mq = 0 for q > d, too, let i be the largest index of a singular extension ′ Ai ? Ai+1 . let Bi+1 = H ? (X(Bi+1 ), k) and Ai+1 = H ? (X(Ai+1 ), k), and let Bi+1 = ′ H ? (X(Bi+1 ), k) be the cohomology ring of the projectivization (decone) of Bi+1 . Recall ′ from [22] that X(Bi+1 ) = X(Bi+1 ) × C× . From the K¨nneth formula, we obtain the u ′ -modules: following exact sequence of Bi+1 (30) 0
/ B′ i+1 / Bi+1 / B ′ (?1) i+1 / 0.

Let Ji+1 denote the kernel of the canonical projection Bi+1 ? Ai+1 . If we let J ′ = ′ ′ Ji+1 ∩ Bi+1 , then Ji+1 = Bi+1 ?Bi+1 J ′ , as a module over Bi+1 . Since A, B are obtained from Ai+1 , Bi+1 , respectively, by a sequence of ?bred extensions, J = B ?Bi+1 Ji+1 . ′ On the other hand, Bi+1 is a free module over Bi+1 , and by applying Lemma 4.3 ′ inductively, B is free over Bi+1 . Therefore, Bi+1 → B is a ?at change of rings, and it is enough to check that (31) if q > d. By Lemma 2.1, Extp ′ (J ′ , k)q = 0 B
i+1

and (A′ )q = 0 for q > d ? 1, the rank of the arrangement, the groups (31) are zero for i+1 q > d by [18, Lemma 2.2]. The Lemma says, in particular, that the B-module J(?c) has Castelnuovo-Mumford regularity no greater than the length of the singular range, d ? c. Moreover, the Lemma gives a combinatorial condition for the hypotheses of Theorem 1.5 to be satis?ed. Corollary 4.7. If A is hypersolvable and its singular range has length 0 or 1, then gA ? Lie(M [?2]) ? hA . = Example 4.8 (2-generic arrangements of rank 4). Suppose A is a central arrangement in C4 , with the property that no three hyperplanes contain a common plane. Such an arrangement is hypersolvable, by Theorem 4.2, with supersolvable deformation B a Boolean arrangement. From De?nition 4.4 and Lemma 4.5, 3 ≤ c ≤ d ≤ 4, so the singular range has length 0 or 1. On the other hand, the arrangement from Example 3.3 is hypersolvable, with singular range (3, 5), and Corollary 4.7 does not apply (indeed, its conclusion fails). 4.3. Generic slices of supersolvable arrangements. Lemma 4.6 provides bounds on the polynomial degrees of the homotopy module M , which cannot be improved without imposing further restrictions on the arrangement. In general, it is not obvious how to characterize the support of M combinatorially; the problem seems similar to that of characterizing which arrangements have quadratic de?ning ideals, investigated in particular in [10, 6]. To this end, we isolate a class of hypersolvable arrangements for which the situation is more manageable.

Extp ′ (J ′ , k)q Bi+1

′ = Extp+1 (A′ , k)q?1 . Since Bi+1 is Koszul i+1 B′
i+1

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

11

De?nition 4.9. A codimension-k linear space W is said to be generic with respect to an arrangement B if rk(X ∩ W ) = rk X + k for all X ∈ L(B) with rk X ≤ rk B ? k. If B is an essential, supersolvable arrangement of rank m and W is a proper, linear space of dimension ? ≥ 3, then by Theorem 4.2, the arrangement A = B ∩ W is hypersolvable. We call such an arrangement a generic (hypersolvable) slice of rank ?. Not every hypersolvable arrangement is a generic slice, see Example 4.15 from [23]. Lemma 4.10. Let B be a rank m supersolvable arrangement, and let A be a rank ? generic slice. Then the singular range of A is (?, ?). Proof. The assumption of genericity means L(A)≤??1 ? L(B)≤??1 . However, X ∩ W = 0 = for all X ∈ L(B)? , so since W is proper and B is essential, the singular range of A is (?, d) for some d. On the other hand, rk A? = rk Am = ?, so the last m ? ? extensions are all singular, and d = ?. This is to say that, for generic slice arrangements, the module J(??) has a linear resolution. Slightly more generally: Proposition 4.11. Let A be a rank ? hypersolvable arrangement. Suppose there exists a generic slice C and ?bred extensions C = Am?i ? · · · ? Am?1 ? Am = A, for some i ≥ 0. Then the singular range of A is (?, ?). Proof. As in the proof of Lemma 4.6, we may reduce to the case where A = C, a generic slice of rank ?. Let B be the supersolvable deformation of A. Denote by A′ and B ′ the Orlik-Solomon algebras of the respective decones, and let J ′ = ker(B ′ ? A′ ). Let R′ = (B ′ )! , and let K = R′ ?k (B ′ )? be the corresponding Koszul complex. That is, Kq = R′ (?q) ?k (B ′ q )? for q ≥ 0, with di?erential ? : e? ? 1 → 1 ? xi . Since B ′ is a i Koszul algebra, K is a free resolution of k over R′ . Let M ′ be the ?-th syzygy module in the resolution K → k → 0. That is, M ′ is the cokernel of ??+1 , a left R′ -module, which means M ′ has minimal free resolution (32) 0
/ Km
?m

/ ···

/ K?+1

??+1

/ K?

η

/ M′

/ 0.

From this we see that M ′ is concentrated in internal degree ?, and ExtR′ (M ′ , k) ? J ′ , as a = B ′ -module. Since Koszul duality is an involution, ExtB ′ (J ′ , k) ? M ′ as a left R′ -module, = and M ′ is bigraded as claimed. The Proposition gives another criterion for the hypotheses of Theorem 1.5 to be satis?ed. We obtain: Corollary 4.12. If A is obtained by ?bred extensions of a generic slice of a supersolvable arrangement, then gA ? Lie(M [?2]) ? hA . = 4.4. Hilbert series. Expressions for the Hilbert series of the graded module M = ExtB (J, k) are not known in general: compare with [28]. However, a simple formula exists for generic slices, which can be extended to ?bred extensions of generic slices. Let βi denote the ith Betti number of B ′ , so that h(B ′ , t) = m βi ti is its Hilbert i=0 series. The following fact is well-known; see [22].

12

G. DENHAM AND A. I. SUCIU

Lemma 4.13. There exist positive integers 1 = d1 ≤ d2 ≤ · · · ≤ dm for which
m

h(B , t) =



(1 + dj t).
j=2

By taking the Euler characteristic of (32), we note that for a generic slice of dimension ?,
m??

(33)

hR (M, t) = hR′ (M , t) = h(R , t)
i=0





(?1)i βi+? ti .

More generally, a ?bred extension results in the same formula. Under the hypotheses of Theorem 1.5, together with formula (33), we have: Corollary 4.14. If h(U, t, u) = U (gA ), then (34)
p,q p dimk Uq tp uq is the bigraded Hilbert series of U =

h(U, t, u) = h(R, t) 1 ? u?2 hR (M, t, u)

?1

.

In the case of a generic slice of dimension ?,
m?? ?1

(35)

h(U, t, u) = h(R, t) 1 ? t u

2 ?2

h(R, t)
i=0

(?1) βi+? t

i

i

.

5. A presentation for the homotopy Lie algebra For the hypersolvable arrangements satisfying the hypotheses of Theorem 1.8, the problem of writing an explicit presentation for the homotopy Lie algebra gA is equivalent to that of presenting the homotopy module MA = ExtB (J, k). We carry out this computation for generic slices of supersolvable arrangements. Let A = {H1 , . . . , Hn } be a hypersolvable arrangement, with supersolvable deformation B. As usual, let h denote the holonomy Lie algebra, and R = U (h) its enveloping algebra. Recall h has a presentation with n generators x1 , . . . , xn in degree (1, 0), one for each hyperplane Hi ∈ A, and for each ?at F ∈ L2 (A) = L2 (B), relations (36) [xi ,
j∈F

xj ] = 0,

for all i for which i ∈ F (i.e., F ? Hi ). Now assume A is a generic slice of a supersolvable arrangement. Then the resolution (32) gives a presentation of the (deconed) homotopy module M ′ as an R′ -module. In order to use this presentation explicitly, we will choose the basis for B ′? given by identifying it with the ?ag complex of B ′ , for which we refer to [6]. Recall Flp is a free k-module on “?ags” (F1 , . . . , Fp ), where Fi ∈ Li (B ′ ) for 1 ≤ i ≤ p, and Fi < Fi+1 , modulo the following relations: (37)
G: Fi?1 <G<Fi+1

(F1 , . . . , Fi?1 , G, Fi+1 , . . . , Fp ),

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

13

for each i, 1 < i < p. Moreover, the map f : Flp → (B ′p )? given by (38) f : (F1 , . . . , Fp ) →
i∈F1

e? i
i∈F2 ?F1

e? · · · i
i∈Fp ?Fp?1

e? , i

is an isomorphism, cf. [27, dual of (2.3.2)]. Under the identi?cation Fl ? B ′? , the boundary map in the Koszul complex becomes = the following. Given a ?ag F = (F1 , . . . , Fp ) and i ∈ Fp , de?ne an element F ? i ∈ Flp?1 by ?nding the integer j for which i ∈ Fj ? Fj?1 , and letting (39) F ? i = (?1)j?1 (F1 , . . . , Fj?1 , Gj , Gj+1 , . . . , Gp?1 ),

where the sum is taken over all ?ags with the property that i ∈ Gp?1 and Gk < Fk+1 for all k, j ≤ k < p. Then the boundary map is given by extending (40) ? : (F1 , . . . , Fp ) →
i∈Fp

(F ? i) ? xi

R-linearly. For each element F ∈ Fl? , let yF denote the corresponding element of M ′ ; that is, yF = η ? (f ? 1)(F ? 1). In particular, we ?nd a minimal generating set for M ′ by choosing a set of β? ?ags of length ? in L(B) appropriately. In particular, one may construct a basis for Fl? using nbc-sets: see, for example, [6, Lemma 3.2]. Then the relations in M ′ are given by the image of ??+1 in (32). We have, for each ?ag F = (F1 , . . . , F?+1 ), a relation in M ′ of the form (41)
i∈F?+1

yF?i xi .

It follows that in gA , for each ?ag F = (F1 , . . . , F?+1 ), we have a relation (42)
i∈F?+1

[xi , yF?i ].

Now M ′ is the restriction of the module M from R to R′ , so the above gives a presentation for M as well, noting that the central element n xi in R acts trivially. i=1 One can ?nd a minimal set of relations just by taking the ?ags F above to come from a basis of Fl?+1 . We summarize this discussion, as follows. Theorem 5.1. Let A = {H1 , . . . , Hn } be a generic slice of a supersolvable arrangement, and let A be the Orlik-Solomon algebra of A. Then, the homotopy Lie algebra gA has presentation with generators ? xi in degree (1, 0), for each i ∈ [n], ? yF in degree (2, ? ? 2), for each F ∈ Fl? , and relations ? xi , j∈F xj = 0, for each ?at F ∈ L2 (A) and each i ∈ F , ?+1 , ? i∈F?+1 xi , yF?i = 0, for each ?ag F = (F1 , . . . , F?+2 ) ∈ Fl ?
n i=1 xi , yF

= 0, for each F ∈ Fl? .

We illustrate the above with an example.

14

G. DENHAM AND A. I. SUCIU

Example 5.2. Consider the arrangement A de?ned by the polynomial QA = xyz(x ? z)(y ? z)(2x ? y ? 4z)(2x ? y ? 5z)(x + 5y + 2z)(x + 5y + z). This is a generic slice of the supersolvable arrangement B, the cone over the arrangement e de?ned by the polynomial QB′ = vwxy(x ? 1)(y ? 1)(v ? 1)(w ? 1). The Poincar? polynomials of the deconed arrangements are given by π(A′ , t) = 1 + 8t + 24t2 , π(B ′ , t) = (1 + 2t)4 = 1 + 8t + 24t2 + 32t3 + 16t4 . Thus the homotopy module M ′ has 32 generators and 16 relations, which can be described as follows. Label the hyperplanes of B ′ as 00 , 10 , 20 , 30 , 01 , 11 , 21 , 31 , in the order above. A basis of 32 ?ags of length 3 can be constructed by choosing three intersecting hyperplanes ia , jb , kc , with 0 ≤ i < j < k ≤ 3 and a, b, c ∈ {0, 1}, and forming a ?ag by successively intersecting the hyperplanes, from right to left. We will call this ?ag Fia jb kc . A basis of 16 ?ags of length 4 in B ′ is constructed by choosing four intersecting hyperplanes, 0a , 1b , 2c , 3d , for all choices of a, b, c, d ∈ {0, 1}, and forming a ?ag again by successive intersection. Let gA be the holonomy Lie algebra of A. Then gA has one generator xH for each hyperplane H, together with 32 additional generators yia jb kc in degree (2, 1), and relations [x0a , y1b 2c 3d ] ? [x1b , y0a 2c 3d ] + [x2c , y0a 1b 3d ] ? [x3d , y0a 1b 2c ], for each a, b, c, d ∈ {0, 1}, in addition to the holonomy relations (8), and relations xH , yia jb kc
H∈A

for each choice of i, j, k, a, b, c. 6. Two-generic arrangements of rank four We now present a method for computing the Hilbert series of the homotopy Lie algebra of a particularly nice class of arrangements: rank-4 arrangements for which no three hyperplanes contain a common plane. For any rank ? arrangement A with n hyperplanes, let E = k (e1 , . . . , en ) be the exterior algebra, A = E/I the Orlik-Solomon algebra, and S = k[x1 , . . . , xn ] the polynomial algebra. We recall the following. Theorem 6.1 (Eisenbud-Popescu-Yuzvinsky [7]). The complex of S-modules 0o F (A) o A? ? S o ... o A1 ? S o A0 ? S o 0

is exact, where the boundary maps are induced by multiplication by n ei ? xi , and the i=1 S-module F (A) is taken as the cokernel of the map A??1 ? S → A? ? S. It follows from Bernstein-Gelfand-Gelfand duality that, for each p ≥ 0, there is a graded isomorphism of S-modules, (43)
??q Extp (A, k)q = ExtS (F (A), S)p+q . E

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

15

We refer to [28] for the case of the smallest q > 0 for which this is nonzero. Details will appear in further work. Now let A be a 2-generic arrangement. Notice that B = E and U (h) = B ! = S. Then, applying Lemma 2.1 to (43), we obtain (44)
??q+1 p (F (A), S)p+q , Mq = ExtS

for p ≥ 0 and 0 ≤ q ≤ ?. As a result, presentations for the S-modules Mq can be obtained computationally for speci?c examples, using formula (44). We recall from Example 4.8 that, if the rank of the arrangement ? = 4, then A satis?es hypotheses (1) of Theorem 1.8: Mq = 0 unless q = 3 or q = 4, i.e., the singular range of A is (3, 4). Example 6.2. Consider arrangements A1 and A2 de?ned by the polynomials Q1 = xyzw(x + y + z)(y + z + w)(x ? y + z + w), Q2 = xyzw(x + y + z)(y + z + w)(x ? y + z ? w). Both arrangements have 7 hyperplanes and 5 lines that each contain 4 hyperplanes, so the characteristic polynomials are π(A1 , t) = π(A2 , t) = 1 + 7t + 21t2 + 30t3 + 15t4 . Since there are no nontrivial intersections in codimension 2, the fundamental group of both complements is Z7 , and R = U (h) is a polynomial ring. We now use (44) to compute the Hilbert series of the graded modules M3 and M4 (recalling Mq = 0 for q = 3, 4). With the help of Macaulay 2, we ?nd for A1 h(M3 , t) = (5 + 2t)/(1 ? t)3 = 5 + 17t + 36t2 + 62t3 + · · · h(M4 , t) = (2 ? t)(1 + 2t + 2t2 )/(1 ? t)6 = 2 + 15t + 62t2 + 185t3 + · · · while for A2 , h(M3 , t) = (5 + t)/(1 ? t)3 = 5 + 16t + 33t2 + 56t3 + · · · h(M4 , t) = (1 + 6t ? t2 ? t3 )/(1 ? t)6 = 1 + 12t + 56t2 + 175t3 + · · · Using formula (34), this yields expressions for the Hilbert series of U (g1 ) and U (g2 ). Comparing these Hilbert series shows U (g1 ) ? U (g2 ), and hence the two arrangements = must have non-isomorphic homotopy Lie algebras. Example 6.3. In 1946, Nandi [21] showed that there are exactly three inequivalent block designs with parameters (10, 15, 6, 4, 2). We list the blocks of each below. Each block design gives rise to a rank-4 matroid on ten points by taking the dependent sets to be those subsets that either contain one of the blocks or contain at least ?ve elements. D1 {abcd, abef, aceg, adhi, bchi, bdgj, cdfj, afhj, agij, behj, bfgi, ceij, cfgh, defi, degh} D2 {abcd, abef, aceg, adhi, bcij, bdgh, cdfj, afhj, agij, aehj, bfgi, cehi, cfgh, defi, degj} D3 {abcd, abef, acgh, adij, bcij, bdgh, cdef, aegi, afhj, behj, bfgi, cehi, cfgj, degj, dfhi}

16

G. DENHAM AND A. I. SUCIU

By construction, there are no nontrivial, dependent sets of size three, so each arrangement is 2-generic. If we call the corresponding Orlik-Solomon algebras A1 , A2 , and A3 , it is straightforward to calculate that h(Ai , t) = 1 + 10t + 45t2 + 105t3 + 69t4 for i = 1, 2, 3. In each case, the singular range is (3, 4). The ideals J1 , J2 , J3 have di?ering resolutions, however, from which it follows that gA1 , gA2 , and gA3 are pairwise non-isomorphic. 7. Topological interpretations 7.1. Generic slices. A particularly simple situation, analyzed in detail by Dimca and Papadima in [8], is when A is a generic slice of rank ? > 2 of a supersolvable arrangement B. Let A′ and B ′ be the respective decones, with complements X = X(A′ ) and Y = X(B ′ ). The two spaces share the same fundamental group, π, and the same integral holonomy Lie algebra, h. In [8, Theorems 18(ii) and 23], Dimca and Papadima establish the following facts. The universal enveloping algebra U (h) is isomorphic (as a Hopf algebra) to the associated graded algebra grIπ (Zπ), where Zπ is the group ring of π, with ?ltration determined by the powers of the augmentation ideal Iπ. The ?rst non-vanishing higher homotopy group of X is π??1 (X); when viewed as a module over Zπ, it has resolution of the form (45) 0
/ Hm (Y ) ? Zπ / ··· / H? (Y ) ? Zπ / π??1 (X) /0.

Finally, the associated graded module of π??1 (X), with respect to the ?ltration by powers of Iπ, has Hilbert series (46) h(gr? π??1 (X), t) = (?1/t)? Iπ 1?
??1 j j j=0 (?1) βj t m j j j=0 (?1) βj t

,

where βj are the Betti numbers of Y . Consider the integral cohomology rings A = H ? (X, Z) and B = H ? (Y, Z). We have (B i )? = Hi (Y, Z), since the homology of an arrangement complement is torsion-free. Thus, tensoring with k, and passing to the associated graded in resolution (45) recovers resolution (32). As a consequence, we obtain the following. Proposition 7.1. Let A be a generic slice of rank ? > 2 of a supersolvable arrangement, and let X = X(A′ ) be the complement of its decone. The homotopy module of the algebra A = H ? (X, k) is isomorphic to the graded module associated to the the ?rst nonvanishing higher homotopy group of X: (47) MA ? gr? π??1 (X) ? k. = I

7.2. Rescaling. Fix an integer q ≥ 1. The q-rescaling of a graded algebra A is the [q] [q] graded algebra A[q] , with Ai(2q+1) = Ai and Aj = 0 if (2q+1) ? j, and with multiplication rescaled accordingly. When taking the Yoneda algebra of A[q] , the internal degree of the Yoneda algebra of A gets rescaled, while the resolution degree stays unchanged: (48) ExtA[q] (k, k) = ExtA (k, k)[q] .

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

17

Similarly, the q-rescaling of a graded Lie algebra L is the graded Lie algebra L[q] , with [q] [q] L2iq = Li and Lj = 0 if 2q ? j, and with Lie bracket rescaled accordingly. Rescaling works well with the holonomy and homotopy Lie algebras: (49) hA[q] = hA ,
[q] [q]

gA[q] = gA .
[q]

[q]

The Hilbert series of the enveloping algebras of gA and gA are related as follows: (50) h(U (gA ), t, u) = h(U (gA ), tu2q , u2q+1 ). Now let X be a connected, ?nite-type CW-complex. A simply-connected, ?nite-type CW-complex Y is called a q-rescaling of X (over a ?eld k) if the cohomology algebra H ? (Y, k) is the q-rescaling of H ? (X, k), i.e., (51) H ? (Y, k) = H ? (X, k)[q] . Rational rescalings always exist: take a Sullivan minimal model for the 1-connected, ?nite-type di?erential graded algebra (H ? (X, Q)[q] , d = 0), and use [31] to realize it by a ?nite-type, 1-connected CW-complex, Y . The space constructed this way is the desired rescaling. Moreover, Y is formal, i.e, its rational homotopy type is a formal consequence of its rational cohomology algebra. Hence, Y is uniquely determined, up to rational homotopy equivalence, among spaces with the same cohomology ring (though there may be other, non-formal rescalings of X, see [24]). Proposition 7.2. Let X be a ?nite-type CW-complex, with cohomology algebra A = H ? (X; Q). Let Y be a ?nite-type, simply-connected CW-complex with H ? (Y ; Q) ? A[q] . = If Y is formal, then (52) π? (?Y ) ? Q ? gA . =
[q]

Proof. Since Y is formal, the Eilenberg-Moore spectral sequence of the path ?bration ?Y → P Y → Y collapses, yielding an isomorphism of Hopf algebras between the Yoneda algebra of H ? (Y ; Q) and the Pontryagin algebra H? (?Y ; Q). From the rescaling assumption, we obtain (53) Ext [q] (Q, Q) ? H? (?Y ; Q), =
A

By Milnor-Moore [20], we ?nd that gA[q] ? π? (?Y ) ? Q, as Lie algebras. Using (49) = ?nishes the proof. As a consequence, we obtain a quick proof of a special case of Theorem A from [24]. Corollary 7.3 ([24]). Suppose X and Y are spaces as above. If both X and Y are formal and A is Koszul, then (54) π? (?Y ) ? Q ? (gr (π1 X) ? Q)[q] . =
?

Proof. Since A is Koszul, gA = hA . Since X is formal, gr? (π1 X) ? Q ? hA , cf. [31]. The = conclusion follows from (52). Remark 7.4. When X is formal (but not necessarily simply connected), a theorem of Papadima and Yuzvinsky [25] states that the cohomology algebra A = H ? (X; Q) is Koszul if and only if the Bous?eld-Kan rationalization XQ is aspherical. Now, by a

18

G. DENHAM AND A. I. SUCIU

classical result of Quillen [26], U (hA ) ? grIπ Qπ1 (XQ ). More generally, it seems likely = that (55) U (gA ) ? U (π? (?XQ )) ? gr Qπ1 (XQ ), =


in view of a result of F?lix and Thomas [12]. (Here again, Qπ1 (XQ ) acts on the left-hand e factor by the action induced from π1 (XQ ) on the universal cover XQ .) However, if X is a hyperplane arrangement complement, then X is not in general a nilpotent space. This means that we can expect to ?nd such spaces X for which πi (XQ ) ? πi (X) ? Q. The ?rst such example was found by Falk [9], who noted that the = complement X of the D4 re?ection arrangement is aspherical, while its Bous?eld-Kan rationalization XQ is not. In general, then, we know of no way to relate gA with the topological homotopy Lie algebra, π? (?X) ? Q. 7.3. Redundant subspace arrangements. Let A = {H1 , . . . , Hn } be an arrangement ×q ×q of hyperplanes in C? . If q is a positive integer, then A(q) = {H1 , . . . , Hn } is an arq? . For example, if A is the braid arrangement rangement of codimension q subspaces in C in C? , with complement equal to the con?guration space of ? distinct points in C, then the complement of A(q) is the con?guration space of ? distinct points in Cq . Proposition 7.5. Let A be a hyperplane arrangement, with Orlik-Solomon algebra A = H ? (X; Q). Fix q ≥ 1, and let Y = X(A(q+1) ) be the complement of the corresponding subspace arrangement. Then: (56)
[q] π? (?Y ) ? Q ? gA . =

Proof. Clearly, Y is simply-connected. As shown in [5], H ? (Y ; Q) is the q-rescaling of H ? (X; Q). Since A(q+1) has geometric intersection lattice, its complement Y is formal, see [32, Prop. 7.2]. The conclusion then follows from Proposition 7.2. Corollary 7.6. Let A be a hypersolvable arrangement, satisfying either of the hypothesis of Theorem 1.8. Then π? (?Y ) ? Q ? (Lie(MA [?2]) ? hA )[q] . = Example 7.7. Let A1 and A2 be the hyperplane arrangements from Example 6.2. Denote by gi = gAi the respective homotopy Lie algebras, i = 1, 2. Consider the redundant (2) (2) subspace arrangements A1 and A2 . Both are arrangements of 7 codimension-2 com8 . Denoting their complements by Y and Y , respectively, we have plex subspaces of C 1 2 π1 (Y1 ) = π1 (Y2 ) = 0 and H? (Y1 ) ? H? (Y2 ) as graded abelian groups. = Let π? (?Yi ) ? Q be the respective (topological) homotopy Lie algebras. By Proposi[1] tion 7.2, we have π? (?Yi ) ? Q ? gi . Making use of the previous calculations for the = [1] arrangements A1 and A2 , together with formula (50), we ?nd that U (gi )p has rank 1, 0, 7, 0, 28, 0, 84, 5, 210 for 0 ≤ p ≤ 8, for both i = 1, 2. It follows that, for p ≤ 9, the group πp (Yi ) ? Q = 0, except for π3 (Yi ) ? Q ? Q7 and π8 (Yi ) ? Q ? Q5 . = = [1] However, for p = 9, the ranks of U (gi )p are 52 and 51, respectively. Hence, π10 (Y1 ) ? Q ? Q17 and π10 (Y2 ) ? Q ? Q16 , = = and so Y1 ? Y2 .

HOMOTOPY LIE ALGEBRA OF AN ARRANGEMENT

19

Acknowledgment. We thank Srikanth Iyengar for helpful conversations. A substantial portion of this work was carried out while the authors were attending the program “Hyperplane Arrangements and Applications” at the Mathematical Sciences Research Institute in Berkeley, California, in Fall, 2004. We thank MSRI for its support and hospitality during this stay.

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