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Hyperdeterminants of Polynomials Hyperdeterminants of Polynomials Luke Oeding University of California, Berkeley June 5, 2012 Support: NSF IRFP (#0853000) while at the University of Florence. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 1 / 23 Quadratic polynomials, matrices, and discriminants Square matrix Quadratic polynomial a a f = ax 2 + bx + c A = 1;1 1;2 a2;1 a2;2 discriminant ∆(f ) = ac − b2=4: determinant a1;1a2;2 − a1;2a2;1: vanishes when A is singular- vanishes when f is singular e.g. columns linearly dependent. ∆(f ) = 0 () f has a repeated root. Notice can associate to f a symmetric matrix a b=2 A = f b=2 c and det(Af ) = ∆(f ). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 2 / 23 Quadratic polynomials, matrices, and discriminants Square matrix Quadratic polynomial f = 0a a a ::: a 1 X 2 X 1;1 1;2 1;3 1;n ai;i xi + ai;j 2xi xj Ba2;1 a2;2 a2;3 ::: a2;nC 1≤i≤n 1≤i<j≤n A = B C B . .. C @ . ::: . A l Symmetric matrix Af = an;1 an;2 an;3 ::: an;n 0 1 a1;1 a1;2 a1;3 ::: a1;n determinant det(A): Ba1;2 a2;2 a2;3 ::: a2;nC vanishes when A is singular- B C B . .. C e.g. columns linearly dependent. @ . ::: . A a1;n a2;n a3;n ::: an;n discriminant ∆(f ): vanishes when f is singular i.e. has a repeated root. If f is quadratic, then det(Af ) = ∆(f ). The symmetrization of the determinant of a matrix = the determinant of a symmetrized matrix. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 3 / 23 Binary cubic polynomial f = a1;1;1 a0;1;1 3 2 2 3 b03 x0 + 3b021x0 x1 + 3b012 x0x1 + b13 x1 a1;0;1 a0;0;1 l Symmetric 2 × 2 × 2 tensor a1;1;0 a0;1;0 b13 b0;12 3 6 a1;0;0 a0;0;0 identify same colors / b02;1 b03 2 2 2 2 3 Det(A) = (a000) (a111) + (a100) (a011) 2 2 2 2 2 2 2 2 ∆(f ) = (b03 ) (b13 ) + 3(b021) (b012 ) + (a010) (a101) + (a001) (a110) 2 2 − 6b 3 b 2 b 2 b 3 − 6(b 2 ) (b 2 ) − 2a000a100a011a111 − 2a100a010a011a101 0 0 1 01 1 0 1 01 3 3 − 2a000a010a101a111 − 2a100a001a011a110 + 4b03 (b012 ) + 4(b021) b13 − 2a000a001a110a111 − 2a010a001a101a110 Discriminant vanishes when f is singular. + 4a000a011a101a110 + 4a001a010a100a111 Again Det(Af ) = ∆(f ). 2 × 2 × 2 hyperdeterminant vanishes when A is singular. ...from wikipedia... Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 4 / 23 Ottaviani: 3 × 3 × 3 Hyperdeterminant and plane cubics 3 3 a222 a122 a022 b23 6 b122 b022 6 6 a212 a112 a012 b122 b012 6 2 a202 a102 a002 3 b0 2 6 a221 a121 a021 3 3 b13 b012 a211 a111 a011 identify / 6 a201 a101 a001 same colors b021 a220 a120 a020 3 a210 a110 a010 a200 a100 a000 b03 Specialize the variables, and use Schl¨afli’smethod to express the symmetrized n−1 (degree 36) hyperdeterminant. Boole's formula deg(∆(d);n) = (n)(d − 1) says degree(∆(f )) = 3 · 22 = 12. The determinant of f factors(!) 6 6 Det(Af ) = ∆(f ) · (AR) (deg 36) = (deg 12) · (deg 4) whereLukeAR Oedingis (UC Aronhold's Berkeley) invariant forHyperdets the hypersurface of Polys of Fermat cubics.June 5, 2012 5 / 23 Geometric Version of Ottaviani's Example Hyperdeterminant vanishes = discriminant · Aronhold6 Dual of Segre = Dual of Veronese · (dual of something?)6 Classical geometry and Aronhold's invariant: 2 _ V(AR) = Fermat cubics σ3(v3(P )) V(AR) = completely reducible cubics 3 3 3 2 2 _ = fx + y + z j [x]; [y]; [z] 2 P g. Chow1;1;1 = fξγζ j [ξ]; [γ]; [ζ] 2 (P ) g. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 6 / 23 2 × 2 × 2 × 2 hyperdeterminant and quartic curves a1111 b04 b14 2 a0111 a1110 1 1 a1101 a0101 a0110 3 b0212 3 a1100 a0100 a1001 9 9 a0001 1 b 3 b 3 1 a1011 0 1 2 01 a0011 a1000 a0000 The discriminant of binary quartics has degree 6 - and is a1010 associated to the 3 − 9 − 9 − 3 a0010 chain. What is the other stuff? The 2 × 2 × 2 × 2 hyperdeterminant has degree 24. Schl¨afli’smethod yields 6 6 Det(Af ) = ∆(f )(cat) (deg 24) = (deg 6) · (deg 3) Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 7 / 23 One large example The 3 × 3 × 3 × 3 × 3 hyperdeterminant has degree 68688 and when symmetrized, splits as 10 20 30 60 120 Det(Af ) = (deg 48)(deg 312) (deg 108) (deg 384) (deg 480) (deg 192) And we can identify all the components geometrically. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 8 / 23 Geometry: Segre-Veronese and Chow varieties Let V =∼ Cn. Let λ ` d with #λ = s. jO(λ)j ×s λ1 λs ⊗d Segλ : PV −! P S V ⊗ · · · ⊗ S V ⊆ P V λ1 λs ([a1];:::; [as ]) 7! [a1 ⊗ · · · ⊗ as ]: ×s The image is the Segre-Veronese variety, denoted Segλ (PV ). Pieri formula implies for all λ ` d, there is an inclusion S d V ⊂ S λ1 V ⊗ · · · ⊗ S λs V : Since GL(V ) is reductive, 9! G-invariant complement W λ: W λ ⊕ S d V = S λ1 V ⊗ · · · ⊗ S λs V : Project from this complement λ1 λs d πW λ : P S V ⊗ · · · ⊗ S V 99K PS V λ1 λs λ1 λs [a1 ⊗ · · · ⊗ as ] 7! a1 ··· as : ×s The Chow variety, Chowλ(PV ) = πW λ (Segλ(PV )). ×s Notice Chowλ(PV ) = Segσ(λ)(PV ) for any permutation σ 2 Ss . Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 9 / 23 Dual Varieties, Hyperdeterminants and Projections Incidence ∗ X ⊂ PV I = f(x; H) j Tx X ⊂ Hg ⊂ PV × PV Ð XX _ X _ - variety of tangent hyperplanes to X . Usually a hypersurface. Intuitively, X _ is not a hypersurface only if X has too many lines. Theorem (GKZ) n1−1 nt −1 _ If X = Segµ(P × · · · × P ), then X is a hypersurface iff s X 2ni ≤ nj for all i such that λi = 1: j=1 t Seg((Pn−1)×t )_ = V(Det(A)), hyperdet. hypersurface in Pn −1 n+µ1 n+µs n ×t _ ( µ )···( µ ) Segµ((P ) ) = V(Detµ(A)), µ-hyperdet. hypersurface in P 1 t . n+d n _ ( d ) Seg(d)(P ) = V(∆(f )), discriminant hypersurface in P . Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 10 / 23 Projections and Duals of Chow Varieties Lemma (GKZ) Let X ⊂ PV,W ⊂ V,X 6⊂ PW . Let πW : PV 99K P(V =W ) projection. _ _ ? Then πW (X ) ⊆ X \ PW . ∼ If πW (X ) = X then = holds. Lemma Let X ⊂ PV,U ⊕ W = V . Then (X \ PU)_ ⊆ X _ \ P(V =U)?. P(V =U)? =∼ PU∗. ×s ×t λ1 λl ∗ Fact: If λ ≺ µ, then Segλ(PV ) = Segµ(PV ) \ S V ⊗ · · · ⊗ S V Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 11 / 23 Projections and Duals of Chow Varieties Lemma (GKZ) _ _ ? πW (X ) ⊆ X \ PW Lemma (X \ PU)_ ⊆ X _ \ PU∗. ×s ×t λ1 λs ∗ Fact: If λ ≺ µ, then Segλ(PV ) = Segµ(PV ) \ S V ⊗ · · · ⊗ S V Proposition (O.) _ ×t _ If λ ≺ µ, then Chowλ(PV ) ⊆ Segµ(PV ) Question: What about the other inclusion? Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 11 / 23 Main Results The more general result for Segre-Veronese varieties: Theorem (O.) Let µ be a partition of d ≥ 2, and V be a complex vector space of dimension n ≥ 2. Then ×t _ d ∗ [ _ Segµ PV \ P S V = Chowλ (PV ) ; λ≺µ where λ ≺ µ is the refinement partial order. In particular, Y Mλ,µ V(Sym(∆µ,n)) = Ξλ,n λ≺µ _ where Ξλ,n is the equation of Chowλ (PV ) when it is a hypersurface in d ∗ P(S V ), and the multiplicity Mλ,µ is the number of partitions µ that refine λ. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 12 / 23 Theorem (O.) The n×d -hyperdeterminant of a polynomial (degree d ≥ 2, n ≥ 2 variables) splits as the product Y Mλ V(Sym(Det(A))) = Ξλ,n ; λ n−1 where Ξλ,n is the equation of the dual variety of the Chow variety Chowλ P (n−1+d)−1 when it is a hypersurface in P d , λ = (λ1; : : : ; λs ) is a partition of d, and d the multiplicity Mλ = M d = is the multinomial coefficient. λ,1 λ1,...,λs Which dual varieties of Chow varieties are hypersurfaces? Theorem (O.) m1 mp Suppose d ≥ 2, dim V = n ≥ 2 and λ = (λ1; : : : ; λs ) = (1 ;:::; p ). Then _ Chowλ (PV ) a hypersurface with the only exceptions n = 2 and m1 6= 0 n > 2, s = 2 and m1 = 1 (so λ = (d − 1; 1)). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 2012 13 / 23 If n = 2, get closed formula for degrees of the duals of Chow varieties. Theorem (O.) 1 _ m1 m2 mp P The degree of Chowλ(P ) with λ = (1 ; 2 ;:::; p ), m1 = 0 and m = i mi is m (m + 1) 1m2 2m3 ··· (p − 1)mp m2;:::; mp If n ≥ 2 get a recursion to compute all degrees of the duals of Chow varieties. Theorem (O.) _ Suppose dim V ≥ 2. Let dλ denote deg(Chowλ (PV ) ) when it is a hypersurface and 0 otherwise. Then the vector (dλ)λ is the unique solution to X deg(Detµ) = dλMλ,µ: λ≺µ The degree of Detµ is given by a generating function (see [page 454,GKZ]), the multiplicities Mλ,µ are computable counting number of refinements.
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