Lower central series and free resolutions of arrangements Alex Suciu - PDF document

Lower central series and free resolutions of arrangements Alex Suciu (Northeastern) www.math.neu.edu/~suciu joint work with Hal Schenck (Texas A&M) www.math.tamu.edu/~schenck available at: math.AG/0109070 Special Session on Algebraic and Topological Combinatorics A.M.S. Fall Eastern Section Meeting Williamstown, MA October 13, 2001 1

Lower central series G finitely-generated group. • LCS: G = G 1 ≥ G 2 ≥ · · · , G k +1 = [ G k , G ] • LCS quotients: gr k G = G k /G k +1 • LCS ranks: φ k ( G ) = rank(gr k G ) Hyperplane arrangements A = { H 1 , . . . , H n } set of hyperplanes in C ℓ . �� � • Intersection lattice: L ( A ) = H ∈B H | B ⊆ A M ( A ) = C ℓ \ � • Complement: H ∈A H Many topological invariants of M = M ( A ) are determined by the combinatorics of L ( A ). E.g.: • Cohomology ring: A := H ∗ ( M, Q ) = E/I (Orlik-Solomon algebra) • Betti numbers and Poincar´ e polynomial: X ∈ L i ( A ) ( − 1) i µ ( X ) b i := dim H i ( M, Q ) = � P ( M, t ) := Hilb( A, t ) = � ℓ i =0 b i t i 2

G = π 1 ( M ) is not always combinatorially determined. Nevertheless, its LCS ranks are determined by L ( A ). Problem. Find an explicit combinatorial formula for the LCS ranks, φ k ( G ), of an arrangement group G (at least for certain classes of arrangements). LCS formulas • Witt formula A = { n points in C } G = F n (free group on n generators) φ k ( F n ) = 1 d | k µ ( d ) n k/d , or: � k ∞ (1 − t k ) φ k = 1 − nt � k =1 • Kohno [1985] B ℓ = { z i − z j = 0 } 1 ≤ i<j ≤ ℓ braid arrangement in C ℓ G = P ℓ (pure braid group on ℓ strings) ∞ ℓ − 1 (1 − t k ) φ k = � � (1 − jt ) k =1 j =1 3

• Falk-Randell LCS formula [1985] If A fiber-type [ ⇐ ⇒ L ( A ) supersolvable (Terao)] with exponents d 1 , . . . , d ℓ G = F d ℓ ⋊ · · · ⋊ F d 2 ⋊ F d 1 φ k ( G ) = � ℓ i =1 φ k ( F d i ) and so: ∞ (1 − t k ) φ k = P ( M, − t ) � k =1 • Shelton-Yuzvinsky [1997], Papadima-Yuz [99] If A Koszul (i.e., A = H ∗ ( M, Q ) is a Koszul algebra) then the LCS formula holds. Remark. There are many arrangements for which the LCS formula fails. In fact, as noted by Peeva, there are arrangements for which (1 − t k ) φ k � = Hilb( N, − t ) , � k ≥ 1 for any graded-commutative algebra N . 4

LCS and free resolutions We want to reduce the problem of computing φ k ( G ) to that of computing the graded Betti numbers of certain free resolutions involving the OS-algebra A = E/I . The starting point is the following (known) formula: ∞ ∞ (1 − t k ) − φ k = � � b ii t i k =1 i =0 i ( Q , Q ) j is the i th Betti number where b ij = dim Q Tor A (in degree j ) of a minimal free resolution of Q over A : ( e 1 ··· e b 1 ) → ⊕ j A b 2 j ( − j ) − → A b 1 ( − 1) · · · − − − − − − − − → A − → Q → 0 Betti diagram: 0 : 1 b 1 b 22 b 33 . . . ← linear strand 1 : . . b 23 b 34 . . . 2 : . . b 24 b 35 . . . . . . . . . . . . . . . . . . Formula follows from: 5

• Sullivan: M formal = ⇒ assoc. graded Lie algebra of G = π 1 ( M ) ∼ = holonomy Lie algebra of H ∗ = H ∗ ( M, Q ): G k /G k +1 ⊗ Q ∼ � gr G := = k ≥ 1 g := L ( H 1 ) / im( ∇ : H 2 → H 1 ∧ H 1 ) • Poincar´ e-Birkhoff-Witt: ∞ (1 − t k ) − φ k = Hilb( U ( g ) , t ) � k =1 • Shelton-Yuzvinsky: ! U ( g ) = A • Priddy, L¨ ofwall: ! ∼ � Ext i A A ( Q , Q ) i = i ≥ 0 ! Here A = E/I [2] is the quadratic closure of A , and A is its Koszul dual. Remark. If A is a Koszul algebra, i.e., Ext i A ( Q , Q ) j = 0 , for i � = j, then A = A and Hilb( A ! , t ) · Hilb( A, − t ) = 1. This yields the LCS formula of Shelton-Yuzvinsky. 6

Change of rings spectral sequence The idea now is to further reduce the computation to that of a (minimal) free resolution of A over E , → ⊕ j E b ′ → ⊕ j E b ′ · · · − 2 j ( − j ) − 1 j ( − j ) − → E − → A → 0 ij = dim Q Tor E and its Betti numbers, b ′ i ( A, Q ) j . This problem (posed by Eisenbud-Popescu-Yuzvinsky [1999]) is interesting in its own right. Let a j = # { minimal generators of I in degree j } � b 1 � Clearly, a 2 + b 2 = . A 5-term exact sequence 2 argument yields: Lemma. a j = b ′ 1 j = b 2 j , for all j > 2 . Betti diagram: 0 : 1 . . . b ′ b ′ 1 : . a 2 . . . ← linear strand 23 34 b ′ b ′ 2 : . a 3 . . . 24 35 b ′ b ′ ℓ − 1 : . a ℓ . . . 2 ,ℓ +1 3 ,ℓ +2 ℓ : . . . . 7

� Key tool : Cartan-Eilenberg change-of-rings spectral sequence associated to the ring maps E ։ A ։ Q : Tor A Tor E ⇒ Tor E � � j ( A, Q ) , Q = i + j ( Q , Q ) i Tor E Tor A 1 (Tor E Tor A 2 (Tor E Tor A 3 (Tor E 2 ( A, Q ) 2 ( A, Q ) , Q ) 2 ( A, Q ) , Q ) 2 ( A, Q ) , Q ) � ����������������� d 2 , 1 2 Tor E Tor A 1 (Tor E Tor A 2 (Tor E Tor A 3 (Tor E 1 ( A, Q ) 1 ( A, Q ) , Q ) 1 ( A, Q ) , Q ) 1 ( A, Q ) , Q ) � ������������������� � ������������������� d 3 , 0 3 d 2 , 0 d 3 , 0 2 2 Tor A Tor A Tor A Q 1 ( Q , Q ) 2 ( Q , Q ) 3 ( Q , Q ) The (Koszul) resolution of Q as a module over E is linear, with � n + i − 1 � dim Tor E i ( Q , Q ) i = i Thus, if we know Tor E i ( A, Q ), we can find Tor A i ( Q , Q ), provided we can compute the differentials d p,q r . We carry out this program, at least in low degrees. As a result, we express φ k , k ≤ 4, solely in terms of the resolution of A over E . 8

Theorem. For an arrangement of n hyperplanes: φ 1 = n φ 2 = a 2 φ 3 = b ′ 23 � a 2 � + b ′ φ 4 = 34 − δ 4 2 where � µ ( X ) � � a 2 = # { generators of I 2 } = 2 X ∈ L 2 ( A ) b ′ 23 = # { linear first syzygies on I 2 } b ′ 34 = # { linear second syzygies on I 2 } δ 4 = # { minimal, quadratic, Koszul syzygies on I 2 } φ 1 , φ 2 : elementary φ 3 : recovers a formula of Falk [1988] φ 4 : new 9

Decomposable arrangements Let A be an arrangement of n hyperplanes. Recall that φ 1 = n , φ 2 = � X ∈ L 2 ( A ) φ 2 ( F µ ( X ) ) Falk [1989]: � φ k ≥ φ k ( F µ ( X ) ) for all k ≥ 3 (*) X ∈ L 2 ( A ) If the lower bound is attained for k = 3, we say that A is decomposable (or local , or minimal linear strand ). Conjecture (MLS LCS). If A decomposable, equality holds in (*), and so ∞ 1 − µ ( X ) t (1 − t k ) φ k = (1 − t ) n � � 1 − t k =1 X ∈ L 2 ( A ) Proposition. The conjecture is true for k = 4 : µ ( X ) 2 ( µ ( X ) 2 − 1) � φ 4 = 1 4 X ∈ L 2 ( A ) 10

If A decomposable, we compute the entire linear strand of the resolution of A over E . If, moreover, rank A = 3, we compute all b ′ ij from M¨ obius function. Example. A = { H 0 , H 1 , H 2 } pencil of 3 lines in C 2 . OS-ideal generated by ∂e 012 = ( e 1 − e 2 ) ∧ ( e 0 − e 2 ). Minimal free resolution of A over E : ( ∂e 012 ) ( e 1 − e 2 e 0 − e 2 ) − E 2 ( − 2) 0 ← A ← − E ( − 1) − E ← − − − − ← − − − − − − − − −    e 1 − e 2 e 0 − e 2 0     e 1 − e 2 e 0 − e 2  0 − E 3 ( − 3) ← ← − − − − − − − − − − − − − − − − − − − − − − − · · · Thus, b ′ i,i +1 = i , for i ≥ 1, and b ′ i,i + r = 0, for r > 1. Lemma. For any arrangement A : � µ ( X ) + i − 1 � � b ′ i,i +1 ≥ i i + 1 X ∈ L 2 ( A ) � µ ( X ) �� µ ( Y ) � � δ 4 ≤ . 2 2 L 2( A ) ( X,Y ) ∈ ( ) 2 If A is decomposable, then equalities hold. Lemma + Theorem = ⇒ Proposition. 11

Example. X 3 arrangement (smallest non-LCS) ✬ ✩ res of residue field over OS alg ❏ total: 1 6 25 92 325 1138 ❏ 0: 1 6 24 80 240 672 ❏ 1: . . 1 12 84 448 ❏ ✫ ✪ 2: . . . . 1 18 ❏ ❏ res of OS alg over exterior algebra total: 1 4 15 42 97 195 354 595 942 1422 2065 0: 1 . . . . . . . . . . 1: . 3 6 9 12 15 18 21 24 27 30 2: . 1 9 33 85 180 336 574 918 1395 2035 8 i ( i + 1)( i 2 + 5 i − 2). i,i +2 = 1 We find: b ′ i,i +1 = 3 i , b ′ Thus: φ 1 = n = 6 φ 2 = a 2 = 3 φ 3 = b ′ 23 = 6 � a 2 + b ′ � φ 4 = 34 − δ 4 = 3 + 9 − 3 = 9 2 � ∞ k =1 (1 − t k ) φ k = (1 − 2 t ) 3 Conjecture says: 2 ), though definitely G �∼ i.e.: φ k ( G ) = φ k ( F 3 = F 3 2 . 12

Graphic arrangements G = ( V , E ) subgraph of the complete graph K ℓ . (Assume no isolated vertices, so that E determines G .) The graphic arrangement (in C ℓ ) associated to G : A G = { ker( z i − z j ) | { i, j } ∈ E} • G = K ℓ = ⇒ A G braid arrangement • G = A ℓ +1 diagram = ⇒ A G Boolean arrangement • G = ℓ -gon = ⇒ A G generic arrangement Theorem. (Stanley, Fulkerson-Gross) A G is super- solvable ⇐ ⇒ ∀ cycle in G of length > 3 has a chord. Lemma. (Cordovil-Forge [2001], S-S) a j = # { chordless j + 1 cycles } Together with a previous lemma ( a j = b 2 j ), this gives: Corollary. A G supersolvable ⇐ ⇒ A G Koszul. For arbitrary A : = ⇒ true (Shelton-Yuzvinsky) ⇐ = open problem 13