Lower central series, free resolutions, and homotopy Lie algebras of arrangements Alex Suciu

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1 Lower central series, free resolutions, and homotopy Lie algebras of arrangements Alex Suciu Northeastern University NSF-CBMS Regional Research Conference on Arrangements and Mathematical Physics LSU, Baton Rouge January 11,

2 References H. Schenck, A. Suciu, Lower central series and free resolutions of hyperplane arrangements, arxiv:math.ag/ S. Papadima, A. Suciu, Homotopy Lie algebras, lower central series, and the Koszul property, arxiv:math.at/

3 Hyperplane arrangements Let A = {H 1,...,H n } be aset of hyperplanes in C l (passing through 0). Intersection lattice: { L(A) = H } B A H B ranked poset: Complement:, rank codim. M(A) =C l \ H A H a formal space, to an l-dim CW-complex. Main topological invariants associated to M(A): Cohomology ring: A := H (M,Q) =E/I E = exterior algebra, I = Orlik-Solomon ideal determined by L(A) Fundamental group: G = π 1 (M) not always combinatorially determined, though its LCS ranks are (since M formal). 3

4 Lower central series Let G be a finitely-generated group. Set: LCS series: G = G 1 G 2 G k where 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) Chen ranks: θ k (G) =rank(gr k G/G ) Define associated graded Lie algebra: L (G) :=gr G Q = k 1 G k /G k+1 Q with Lie bracket [, ]: L i L j L i+j induced by the group commutator. Problem. Find an explicit combinatorial formula for the LCS ranks φ k (G) ofanarrangement group G (at least for certain classes of arrangements). 4

5 Known LCS formulas Witt formula A = {n points in C}, M(A) n S 1 G = F n (free group of rank n) gr G = L n (free Lie algebra of rank n) Hence: φ 1 = n, (1 t k ) φ k =1 nt k=1 φ k (F n )= 1 k d k µ(d)n k d φ 2 = ( ) n 2, φ3 =2 ( ) n+1 3, φ4 = n2 (n 2 1) 4. Kohno formula [1985] A = {z i z j =0} 1 i<j l braid arrangement in C l G = P l (pure braid group on l strings) (1 t k ) φ k = k=1 l 1 (1 jt) j=1 5

6 Falk-Randell LCS formula [1985] ( ) Terao A fiber-type L(A) supersolvable, with exponents d 1,...,d l. Then: G = F dl F d2 F d1 φ k (G) = l i=1 φ k(f di ) (1 t k ) φ k = P M ( t) k=1 where, for all A: P M (t) := l i=0 b it i, b i = X L i (A) ( 1) i µ(x) Shelton-Yuzvinsky [1997], Papadima-Yuz [99] A Koszul algebra = LCS formula holds Remark. (Peeva) There are arrangements for which (1 t k ) φk Hilb(N, t), k 1 for any graded commutative algebra N. 6

7 LCS and free resolutions (joint work with Hal Schenck) 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. Starting point: (1 t k ) φ k = k=1 b ii t i i=0 where b ij = dim Q Tor A i (Q, Q) j i th Betti number (in degree j) ofaminimal free resolution of Q over A: j A b 2j ( j) j A b 1j ( j) A Q 0 with Betti diagram: 0: 1 b 1 b 22 b linear strand 1:.. b 23 b :.. b 24 b

8 This (known) formula follows from: Sullivan Since M formal: gr G = g:= L(H 1 )/ im( : H 2 H 1 H 1 ) Holonomy Lie algebra of H = H (M,Q) Poincaré-Birkhoff-Witt: (1 t k ) dim g k = Hilb(U(g),t) k=1 Shelton-Yuzvinsky: U(g) =A! Priddy, Löfwall: A! = Ext i A(Q, Q) i i 0 Linear strand in Yoneda Ext-algebra Here A = E/ I 2 is the quadratic closure of A, and A! is its Koszul dual. 8

9 Koszul algebras A connected, graded, graded-commutative Q-algebra. Definition. A is a Koszul algebra if Ext i A(Q, Q) j =0, for i j. i.e., Betti diagram = linear strand. Necessary: A = T/I, T gen in deg 1, I in deg 2. Sufficient: I has quadratic Gröbner basis. Koszul duality: Hilb(A!,t) Hilb(A, t) =1 ( LCS formula of Shelton-Yuzvinsky.) Arrangement interpretation (Shelton-Yuz.) A fiber-type?? H (M(A), Q) Koszul Topological interpretation (Pap-Yuz.) X finite-type CW, X Q Bousfield-Kan rationalization. H (X, Q) Koszul if X formal X Q aspherical 9

10 Resolution of OS-algebra Problem. (Eisenbud-Popescu-Yuzvinsky [1999]) Compute the (minimal) free resolution of A over E, j E b 2j ( j) j E b 1j ( j) E A 0 and its Betti numbers, b ij = dim Q Tor E i (A, Q) j. Let a j = {minimal generators of I in degree j}. Clearly, a 2 = ( b 12 ) b2. Lemma. a j = b 1j = b 2j. Thus, Betti diagram looks like: 0: :. a 2 b 23 b linear strand 2:. a 3 b 24 b l 1:. a l b 2,l+1 b 3,l

11 Change of rings spectral sequence Key tool (Cartan-Eilenberg): Tor A i ( Tor E j (A, Q), Q ) = Tor E i+j(q, Q) Tor E 2 (A, Q) TorA 1 (TorE 2 (A, Q), Q) TorA 2 (TorE 2 (A, Q), Q) TorA 3 (TorE 2 d 2,1 2 Tor E 1 (A, Q) TorA 1 (TorE 1 (A, Q), Q) TorA 2 (TorE 1 d 2,0 2 d 3,0 2 (A, Q), Q) d 3,0 3 Tor A 3 (TorE 1 (A, Q), Q) (A, Q), Q) Q Tor A 1 (Q, Q) TorA 2 (Q, Q) Tor A 3 (Q, Q) The (Koszul) resolution of Q as a module over E is linear, with dim Tor E i (Q, Q) i = ( ) n+i 1 i. Thus, if we know Tor E i (A, Q), wecan find Tor A i (Q, Q), provided we can compute d p,q r. We carry out part of this program mostly in low degrees. As a result, we express φ k, k 4, solely in terms of the resolution of A over E. 11

12 Theorem. For an arrangement of n hyperplanes: φ 1 = n φ 2 = a 2 φ 3 = b 23 φ 4 = ( a2 2 ) + b 34 δ 4 where a 2 = {generators of I 2 } = X L 2 (A) ( ) µ(x) 2 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 12

13 Decomposable arrangements Let A be an arrangement of n hyperplanes. Recall: φ 1 = n, φ 2 = X L 2 (A) φ 2 (F µ(x) ) Falk [1989]: φ k X L 2 (A) φ k (F µ(x) ) for all k 3 (*) Definition. If the lower bound is attained for k =3, A is decomposable (or, minimal linear strand). Conjecture (MLS LCS). If A is decomposable, equality holds in (*), and so (1 t k ) φ k =(1 t) n k=1 X L 2 (A) 1 µ(x)t 1 t 13

14 Example. A = {H 0,H 1,H 2 } pencil of 3 lines in C E = exterior algebra on e 0,e 1,e 2 I = ideal generated by e 012 =(e 1 e 2 ) (e 0 e 2 ) Minimal free resolution of A over E: 0 A E ( e 012) E( 1) (e 1 e 2 e 0 e 2 ) E 2 ( 2) e 1 e 2 e 0 e e 1 e 2 e 0 e 2 E 3 ( 3) Thus, b i,i+1 = i for i 1 b i,i+r =0 for r>1 14

15 More generally, if A decomposable, we compute the entire linear strand of the resolution of A over E. If moreover rank A =3,wecompute all b ij Möbius function. from 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 (X,Y ) ( L 2 (A) 2 ) If A is decomposable, then equalities hold. ). Corollary. The MLS LCS conjecture is true for k =4: φ 4 = 1 4 µ(x) 2 (µ(x) 2 1) X L 2 (A) 15

16 Example. X 3 arrangement res of residue field over OS alg total: : : : res of OS alg over exterior algebra total: : : : We find: b i,i+1 =3i, b i,i+2 = 1 8 i(i + 1)(i2 +5i 2). Thus: φ 1 = n =6 φ 2 = a 2 =3 φ 3 = b 23 =6 φ 4 = ( a 2 2 ) + b 34 δ 4 = 3+9 3=9 Conjecture says: k=1 (1 tk ) φ k =(1 2t) 3 i.e.: φ k (G) =φ k (F 3 2 ), though definitely G = F

17 Graphic arrangements G =(V, E) subgraph of complete graph K l. Set: κ s (G) = {complete subgraphs on s +1vertices} Graphic arrangement (in C κ 0 ) associated to G: A G = {ker(z i z j ) {i, j} E} Theorem. (Stanley, Fulkerson-Gross) A G is supersolvable G is chordal Lemma. (Cordovil-Forge [2001], S-S) a j = {chordless j +1cycles} Together with a previous lemma (a j = b 2j ), we get: Corollary. A G supersolvable A G Koszul. 17

18 Example. Braid arrangement B 4 = A K4 κ 0 =4,κ 1 =6,κ 2 =4,κ 3 =1 e 2 e 0 e 3 e e 4 5 e Free resolution of A over E: 0 A E 1 E 4 ( 2) 2 E 10 ( 3) where 1 = ( e 145 e 235 e 034 ) e 012 and 2 = e 1 e 4 e 1 e e 3 e 0 e 2 e e 2 e 3 e 2 e e 0 e 1 e 0 e e 0 e 3 e 0 e e 1 e 5 e 2 e e 0 e 1 e 0 e 2 e 3 e 5 e 4 e 5 Non-local component in R 1 (B 4 ) 2 extra syzygies Get: b i,i+1 =5i, δ 4 =0. 18

19 Proposition. For a graphic arrangement: b i,i+1 = i(κ 2 + κ 3 ) δ 4 ( κ 2 2 ) 6(κ3 + κ 4 ) Corollary. φ 1 = κ 1 φ 2 = κ 2 φ 3 =2(κ 2 + κ 3 ) φ 4 3(κ 2 +3κ 3 +2κ 4 ) Moreover, if κ 3 =0,equality holds for φ 4. Conjecture (Graphic LCS). or: k=1 φ k = 1 k d k k j=1 k ( ) s ( 1) s j κ s µ(d) j k d j s=j ( 1 t k ) κ φ k 0 1 = j=1 (1 jt) κ 0 1 s=j ( 1) s j( s j) κs 19

20 Resonance Conjectures A = {H 1,...,H n }, A = H (M(A), C), G = π 1 (M) R 1 (A) :={λ C n dim C H 1 (A, λ i e i ) > 0} Then: R 1 (A) = v i=1 L i, where L i linear subspaces Set h r = {L i dim L i = r}. Chen Ranks & Linear Strand of OS-Resolution θ k (G) =b k 1,k for k 2. Resonance Formula for Chen Ranks θ k (G) = r 2 h r θ k (F r ) for k 4. Resonance LCS Formula If φ 4 = θ 4 δ 4 = ( a 2 2 ), then: φ k (G) = r 2 h r φ k (F r ) for k 4. 20

21 Rational homotopy groups and Koszul algebras (joint work with Stefan Papadima) Let k 1beaninteger, and X connected space Y simply-connected space (all spaces to a finite-type CW-complex) Definition. Y is a k-rescaling of X (over R) if: H (Y,R) = H (X, R)[k] as graded rings i.e.: H i (Y,R) = H j (X, R) if i =(2k +1)j, 0 otherwise, and isos are compatible with cup products. (Abstracts a definition of D.Cohen, F.Cohen, M.Xicoténcatl [2000]) Question. When does a homological rescaling pass to a homotopical rescaling? 21

22 Examples of rescalings (over R = Z) X = S 1 Y = S 2k+1 X = n 1 S1 Y = n X = n 1 S1 Y = n 1 S2k+1 1 S2k+1 X =# g 1 S1 S 1 Y =# g 1 S2k+1 S 2k+1 X = C l \ n i=1 H i Y = C (k+1)l \ n i=1 H (k+1) i A = {H 1,...,H n } hyperplane arrangement in C l A k := {H 1 k,...,h n k } redundant subspace arr. (Cohen-Cohen-Xico.) X = S 3 \ n i=1 K i Y = S 4k+3 \ n i=1 K k i K =(K 1,...,K n ) link of oriented circles in S 3 link of S 2k+1 s in S 4k+3 : join of K with k copies of Hopf n (in sense of Koschorke-Rolfsen [1989]) K k 22

23 Rational rescalings Given X and k, there is a k-rescaling Y over Q. Take Y to be the realization of the Sullivan minimal model for dga ( H (X, Q)[k],d =0 ). By construction, this rescaling is formal. In general, k-rescaling is not unique. Eg: X = S 1 S 1 S 4 Y = S 3 S 3 S 12 Z = Sx 3 Sy 3 e 12 [x,[x,[x,[x,y]]]] Nevertheless, k-rescaling is unique /Q, for k large: Shiga-Yagita [1982]: If H >d (X, Q) =0, then X has a unique k-rescaling (up to Q- equivalence), for all k >(d 1)/2. If rescaling Y is unique, it must be formal. 23

24 Simple examples. X = S 1 Y = S 2k+1 X = n 1 S1 Y = n 1 S2k+1 X = n 1 S1 Y = n 1 S2k+1 X =# g 1 S1 S 1 Y =# g 1 S2k+1 S 2k+1 X and Y formal, rescaling unique /Q, for all k Arrangements. X = M(A), Y = M(A k+1 ) Arnol d, Brieskorn: X formal Yuzvinsky [2001]: Y formal Y is unique k-rescaling /Q, for k>(l 1)/2 Links. X = M(K), Y = M(K k ) X may not be formal, but Y is formal (since dim X = 2), so rescaling unique /Q, for all k. 24

25 Rescaling Lie algebras Lie algebras with grading Graded Q-vector space L, with [, ]: L p L q L p+q Lie identities satisfied exactly. E.g.: Graded Lie algebras gr (G) Q L, with Lie identities satisfied up to sign. E.g.: Homotopy Lie algebra π (ΩY ) Q := r 1 π r (ΩY ) Q Y simply-connected space, ΩY loop space [, ] = Samelson product Rescaling L Lie algebra with grading L[k] graded Lie algebra L[k] 2kq = L q and L[k] p =0 otherwise, and Lie bracket rescaled accordingly. 25

26 The Rescaling Formula Theorem A. Let Y be a k-rescaling of X. If H (X, Q) is a Koszul algebra, then: π (ΩY ) Q = gr (π 1 X) Q[k] as graded Lie algebras. Theorem B. Let Y be a formal k-rescaling of X. If Hilb(π (ΩY ) Q,t)=Hilb(gr (π 1 X) Q,t 2k ), then: 1. H (X, Q) is a Koszul algebra. 2. Y is a coformal space (i.e., its Q-homotopy type is determined by its homotopy Lie algebra). 26

27 The Homotopy LCS Formula Set Φ r := rank π r (ΩY )=rank π r+1 (Y ). Theorem C. Let Y be a k-rescaling of X. If H (X, Q) is Koszul, then: (1 t (2k+1)i ) Φ 2ki = P X ( t k ) i 1 and Φ r =0,if2k r. Hence: ΩY Q i 1 K(Q, 2ki)Φ 2ki and so P ΩY (t) =P X ( t 2k ) 1 In fact, by Milnor-Moore: (as Hopf algebras) H (ΩY,Q) = U ( gr (π 1 X) Q [k] ) 27

28 Rescaling arrangements Let A arrangement, X = M(A), Y = M(A k+1 ). Recall: both X, Y formal; Y rescaling of X. Shelton-Yuzvinsky A fiber-type H? (X) Koszul Cohen-Cohen-Xico Thm. A Thm. B.1 Fadell-Neuwirth π (ΩY ) Q = gr (π 1 X) Q [k] Fadell-Husseini Cohen-Gitler? Thm B.2 A braid arr. Y coformal 28

29 Fiber-type arrangements A supersolvable, exp(a) ={d 1,...,d l } X = M(A), Y = M(A k+1 ) φ r = rank gr r (π 1 X), Φ r = rank π r (ΩY ). Classical LCS formula: (1 t r ) φ r = r=1 l (1 d i t) i=1 Homotopy LCS formula: (1 t (2k+1)r ) Φ 2kr = r=1 l (1 d i t 2k+1 ) i=1 Poincaré series of loop space: P ΩY (t) = l (1 d i t 2ki ) 1 i=1 29

30 Generic arrangements If A generic (and non-boolean), then: (a) H (X, Q) not quadratic = not Koszul (b) Rescaling formula fails (c) Y not coformal Failure of (b) and (c) is detected by higher-order Whitehead products. Example. A = {3 generic lines in C 2 } X 2-skeleton of S 1 S 1 S 1 Y 2(2k + 1)-skeleton of S 2k+1 S 2k+1 S 2k+1 Then: gr (π 1 X)=L ab (x 1,x 2,x 3 ) deg x i =1 π (ΩY ) Q = L ab (y 1,y 2,y 3 ) L(w) deg y i =2k, deg w =6k

31 The Mal cev Formula Theorem D. Let Y be a k-rescaling of X. Assume H (X, Q) is a Koszul algebra. Then: X is formal there are filtered group isomorphisms Hom coalg (H (ΩS 2k+1, Q),H (ΩY,Q)) = [ΩS 2k+1, ΩY ] = π 1 X Q ( ) Key ingredient in proof: A result of H. Baues [1981]. Passing to associated graded Lie algebras: Mal cev Formula ( ) Rescaling Formula ( ). ( ) holds for Koszul arrangements. For X = n S 1, Y = n S 2k+1 : recovers a result from T. Sato s thesis [2000]. For A = braid arr., X = M(A), Y = M(A k+1 ): answers a question of Cohen-Gitler [2001]. 31

32 Rescaling links Let K link, X = M(K), Y = M(K k ) G K linking graph of K vertices K i, edges (K i,k j )iflk(k i,k j ) 0 G K connected Markl-Papadima H (X) Koszul Thm. A Thm. B.1 π (ΩY ) Q = gr (π 1 X) Q [k] G K connected X formal Yes No Thm. D [ΩS 2k+1, ΩY ] = π 1 X Q 32

33 K 0 K K 0 = 2 n = Hopf n, n 4 K = 2 nβ, with β (P n 1 ) r 1 \ (P n 1 ) r, r 3 K 0 and K have: G = K n,lk ij =1. Thus, H (X 0, Z) = H (X, Z). Same Milnor µ-invariants. Same Vassiliev invariants, up to order r 2. gr (π 1 X 0 ) = gr (π 1 X). But X 0 formal, X not formal. So: Rescaling Formula holds for both K 0 and K. Mal cev Formula holds for K 0, fails for K. 33