Syzygies, Exterior Algebra, and Hyperplane Arrangements, I P2

P6

P 4

P7 P 1

Resolution 1 · · · · 4 10 15 · · · 6

P3

P5

of OS over E · · · ··· 20 25 30 · · · 25 66 140 · · ·

Resolution of k over OS 1 6 25 90 301 966 3025 · · ·

Hal Schenck Mathematics Department University of Illinois

1

§Syzygies and Betti notation Example 1 Twisted cubic IC ⊆ S = K[x, y, z, w] −z w  2  y −z 2 y −xz yz −xw z −yw −x y 0 −→ S(−3)2 −−−−−−−−−−→ S(−2)3 −−−−−−−−−−−−−−−−−−−−−−−−→ S −→ S/I





Display as a betti table: bij = dimK TorS i (M, K)i+j . total 0 1

1 1 –

3 – 3

2 – 2

For example, b21 = dimK TorS 2 (S/IC , K)3 = 2. The reason for the notation is that the CastelnuovoMumford regularity is indexed by the bottom row of the table: for the twisted cubic, reg(S/IC ) = 1 = deg(IC ) − codim(IC ). 2

§Koszul algebras Definition 2 Quadratic algebra: quotient of T (V ) by I ⊆ V ⊗ V , V a K-vector space. Quadratic algebra has a quadratic dual T (V ∗)/I ⊥: hα ⊗ β | α(a) · β(b) = 0i = I ⊥ ⊆ V ∗ ⊗ V ∗

Definition 3 Quadratic algebra A is Koszul if T oriA(K, K)j = 0, j 6= i A Koszul ↔ minimal free resolution of K over A has matrices with only linear entries. This happens exactly when the betti diagram has nonzero entries only in the top row.

3

Example 4 For S = T (V )/hxi ⊗ xj − xj ⊗ xii, K = S/hx1, . . . , xmi. Clear that I ⊥ = hxi ⊗ xj + xj ⊗ xii, so E = S !. Compute resolution and Hilbert Series. Theorem 5 If A is Koszul, so is A!, and HS(A, t) · HS(A!, −t) = 1 Example 6 If A = Sym(K3), A! = Λ(K3), and 1 3 ⇐ HS(A! , −t) HS(A, t) ⇒ (1−t) 3 · (1 − t) ⇓

1 Example 7 Via upper semicontinuity, can show quadratic GB → Koszul.

Pinched Veronese

(Caviglia): Koszul but no QGB. (see also Eisenbud, Reeves, Totaro) 4

§Bernstein-Gelfand-Gelfand correspondence. Let S = Sym(V ∗) and E = (V ). BGG is an isomorphism between derived categories of V

• bounded cpxs of coherent sheaves on P(V ∗). • bounded cpxs of f.gen’d, graded E–modules. From this, can extract functors R: f.gen’d, graded S-modules −→ linear free E-complexes.

L: f.gen’d, graded E-modules −→ linear free S-complexes. Point: can translate problems to possibly simpler setting. For example, we’ll see this gives a fast way to compute sheaf cohomology, using Tate resolutions.

5

P a f’gend, graded E-module, then L(P ) is the complex ··· /

S ⊗ Pi+1 ·a / S ⊗ Pi ·a / S ⊗ Pi−1 ·a / · · · ,

where a =

n P i=1

xi ⊗ei, so that 1⊗p 7→

P

xi ⊗ei ∧p

Note: elts of V ∗ deg = 1, elts of V deg = −1. Example 8 P = E = 0

/

/

S ⊗ E0

Clearly 1 7→

V

S ⊗ E1

K3. Then we have /

S ⊗ E2

/

S ⊗ E3

/

0.

P3 1 xi ⊗ ei . For d1

e1 7→ −x2e12 − x3e13 e2 7→ x1e12 − x3e23 e3 7→ x1e13 + x2e23 d2 : e12 7→ x3e123, e13 7→ −x2e123 e23 7→ x1e123 Thus, L(E) is 







x1 −x2 x1 0  x2   −x3 0 x1    x3 0 −x3 x2 x3 −x2 x1 −−−−−−−−−−−− → S1 S 1 −−−−−→ S 3 −−−−−−−−−−−−−−−→ S 3 −

The Koszul complex! 6

M a f’gend, graded S-module, then R(M ) is the complex ··· /

ˆ ⊗ Mi−1 ·a / E ˆ ⊗ Mi ·a / E ˆ ⊗ Mi+1 ·a / · · · , E

where a =

n P i=1

ei ⊗ xi, so 1 ⊗ m 7→

P

ei ⊗ xi · m,

ˆ is the K-dual of E: and E ˆ ' E(−n) = HomK(E, K). E Just as L(P ) = S ⊗K P , R(M ) = HomK(E, M ). Example 9 M = K[x0, x1]/hx0x1, x2 0 i. Then 0

/

/

E ⊗ M0

E ⊗ M1

/

E ⊗ M2

/

E ⊗ M3

/

···

1 7→ e0 ⊗ x0 + e1 ⊗ x1 x0 7→ e0 ⊗ x2 0 + e1 ⊗ x0 x1 x1 7→ e0 ⊗ x0x1 + e1 ⊗ x2 1 n + e ⊗ xn+1 xn → 7 e ⊗ x x 0 0 1 1 1 1

Thus, R(M ) is 

 e0       e1 0 e e e 1 1 1 2 1 1 ˆ1 −−−−→ E(−1) ˆ ˆ ˆ E −−−−−−−→ E(−2) −−−−→ E(−3) −−−−→ · · · 7

This complex is exact, except at the second step. Obviously the kernel of h

0 e1

i

is generated by α = [1, 0] and β = [0, e1], with relations im(d1) = β + e0α = 0, e1β = 0, so that H 1(R(M )) ' E(−3)/e0 ∧ e1 Compute this, and compute the free resolution of M . This illustrates Theorem 10 [Eisenbud-Fløystad-Schreyer] H j (R(M ))i+j = T oriS (M, K)i+j . Corollary 11 The regularity of M is ≤ d iff H i(R(M )) = 0 for all i > d.

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For a coherent sheaf F on Pd, there is a f’gend, graded S-module M whose sheafification is F . If F has Castelnuovo-Mumford regularity r, then the Tate resolution of F is obtained by splicing the complex R(M≥r ): /

0

r ˆ ⊗ Mr d / E ˆ ⊗ Mr+1 E /

ˆ ⊗ Mr+2 E /

···,

with a free resolution P• for the kernel of dr : ··· /

P1

/

/

P0

ˆ6 ⊗ Mr E /

ˆ ⊗ Mr+1 E /

···

%

r) ker(d 8

0

)

0

By Corollary 11, R(M≥r ) is exact except at the first step, so this yields an exact complex of free E-modules. Example 12 Since M = S has regularity zero, we obtain Cartan resolutions in both direcb = E(d+1) multitions, with splice map E → E 

plication by e0∧e1∧· · ·∧ed = ker e0, · · · , ed 9

t

.

Theorem 13 [Eisenbud-Fløystad-Schreyer] The ith free module T i in a Tate resolution for F L b i satisfies T = E ⊗ H j (F (i − j)). j

The twisted cubic has regularity one; apply Theorem 13: i h1(F (i)) h0(F (i))

−3 8 0

−2 5 0

−1 2 0

0 0 1

1 0 4

2 0 7

Does this make sense? F = OX = OP1 (3) so h1(F (i)) = h1(OP1 (3i)) = h0(OP1 (−3i − 2)) and h0(F (i)) = h0(OP1 (3i)) = 3i + 1, i ≥ 0

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§Hyperplane Arrangements A hyperplane arrangement in V (usually ' Cn or Rn) is A=

d [ i=1

Hi =

d [

V (li) ⊆ V,

i=1

has complement M = V \ A. In 1983, Hirzebruch wrote: “The topology of the complement of a configuration of lines in the projective plane is very interesting, the investigation of the fundamental group of the complement very difficult.” Orlik and Solomon (Inventiones, 1980) proved H ∗(M, Z) is determined by intersection lattice L(A). V = ˆ 0, 0 = ˆ 1, L(A)1 are hyperplanes. A is central if li homogeneous, and essential if ∩iHi is the origin in V , so is empty in P(V ).

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Example 14 Reflecting hyperplanes of SL(4). [

V (xi − xj ) | 1 ≤ i < j ≤ 4 ⊆ K4,

Share W = Span(1, 1, 1, 1), project to W ⊥ for essential, central A ⊆ K3, view as A ⊆ P2.

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µ

Definition 15 M¨ obius function L(A) −→ Z is given by µ(ˆ 0) =

1

µ(t) = −

P

µ(s), if ˆ 0 < t.

s