Hyperdeterminants of Polynomials

Transcription

1 Hyperdeterminants of Polynomials Luke Oeding University of California, Berkeley June 5, 0 Support: NSF IRFP (#085000) while at the University of Florence. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

2 Quadratic polynomials, matrices, and discriminants Square matrix A = ( ) a, a, a, a, determinant a, a, a, a, : vanishes when A is singular - e.g. columns linearly dependent. Quadratic polynomial f = ax + bx + c discriminant (f ) = ac b /4: vanishes when f is singular (f ) = 0 f has a repeated root. Notice can associate to f a symmetric matrix ( ) a b/ A f = b/ c and det(a f ) = (f ). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

3 Quadratic polynomials, matrices, and discriminants Square matrix a, a, a,... a,n a, a, a,... a,n A = a n, a n, a n,... a n,n determinant det(a): vanishes when A is singular - e.g. columns linearly dependent. Quadratic polynomial f = a i,i xi + a i,j x i x j i n i<j n Symmetric matrix A f = a, a, a,... a,n a, a, a,... a,n a,n a,n a,n... a n,n discriminant (f ): vanishes when f is singular i.e. has a repeated root. If f is quadratic, then det(a f ) = (f ). The symmetrization of the determinant of a matrix = the determinant of a symmetrized matrix. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

4 a,, a 0,, a,0, a 0,0, a,,0 a 0,,0 Binary cubic polynomial f = b 0 x 0 + b 0 x 0 x + b 0 x 0 x + b x Symmetric tensor a,0,0 a 0,0,0 Det(A) = (a 000) (a ) + (a 00) (a 0) + (a 00) (a 0) + (a 00) (a 0) a 000a 00a 0a a 00a 00a 0a 0 b identify same colors b 0, 6 b 0, b 0 (f ) = (b 0 ) (b ) + (b 0 ) (b 0 ) 6b 0 b 0 b 0 b 6(b 0 ) (b 0 ) a 000a 00a 0a a 00a 00a 0a 0 a 000a 00a 0a a 00a 00a 0a 0 + 4a 000a 0a 0a 0 + 4a 00a 00a 00a hyperdeterminant vanishes when A is singular. + 4b 0 (b 0 ) + 4(b 0 ) b Discriminant vanishes when f is singular. Again Det(A f ) = (f )....from wikipedia... Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 4 /

5 Ottaviani: Hyperdeterminant and plane cubics a a a 0 b b 6 b 0 6 a a a 0 b b 0 6 a 0 a 0 a 00 b 6 0 a a a 0 6 a a a 0 b identify b 0 a 0 a 0 a 00 b 6 same colors 0 a 0 a 0 a 00 a 0 a 0 a 00 a 00 a 00 a 000 b 0 Specialize the variables, and use Schläfli s method to express the symmetrized (degree 6) hyperdeterminant. Boole s formula deg( (d),n ) = (n)(d ) n says degree( (f )) = =. The determinant of f factors(!) Det(A f ) = (f ) (AR) 6 (deg 6) = (deg ) (deg 4) 6 where LukeAR Oeding is(uc Aronhold s Berkeley) invariant forhyperdets the hypersurface of Polys of Fermat cubics. June 5, 0 5 /

6 Geometric Version of Ottaviani s Example Hyperdeterminant vanishes = discriminant Aronhold 6 Dual of Segre = Dual of Veronese (dual of something?) 6 Classical geometry and Aronhold s invariant: V(AR) = Fermat cubics σ (v (P )) = {x + y + z [x], [y], [z] P }. V(AR) = completely reducible cubics Chow,, = {ξγζ [ξ], [γ], [ζ] (P ) }. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 6 /

7 hyperdeterminant and quartic curves a b 0 4 b 4 a 00 a 0 a 0 a 00 a 0 b 0 a 000 a 00 a 0 a 00 a b 0 b 0 a 00 a 000 a 0000 a 000 a 00 The discriminant of binary quartics has degree 6 - and is associated to the 9 9 chain. What is the other stuff? The hyperdeterminant has degree 4. Schläfli s method yields Det(A f ) = (f )(cat) 6 (deg 4) = (deg 6) (deg ) 6 Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 7 /

8 One large example The hyperdeterminant has degree and when symmetrized, splits as Det(A f ) = (deg 48)(deg ) 0 (deg 08) 0 (deg 84) 0 (deg 480) 60 (deg 9) 0 And we can identify all the components geometrically. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 8 /

9 Geometry: Segre-Veronese and Chow varieties Let V = C n. Let λ d with #λ = s. Seg λ : PV s O(λ) P ( S λ V S λs V ) P ( V d) ([a ],..., [a s ]) [a λ aλs s ]. The image is the Segre-Veronese variety, denoted Seg λ (PV s ). Pieri formula implies for all λ d, there is an inclusion S d V S λ V S λs V. Since GL(V ) is reductive,! G-invariant complement W λ : Project from this complement W λ S d V = S λ V S λs V. π W λ : P ( S λ V S λs V ) PS d V [a λ aλs s ] a λ aλs s. The Chow variety, Chow λ (PV ) = π W λ (Seg λ (PV s )). Notice Chow λ (PV ) = Seg σ(λ) (PV s ) for any permutation σ S s. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 9 /

10 Dual Varieties, Hyperdeterminants and Projections Incidence X PV I = {(x, H) T x X H} PV PV X 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) X If X = Seg µ (P n P nt ), then X is a hypersurface iff n i s n j for all i such that λ i =. j= Seg((P n ) t ) = V(Det(A)), hyperdet. hypersurface in P nt Seg µ ((P n ) t ) = V(Det µ (A)), µ-hyperdet. hypersurface in P (n+µ µ ) ( n+µs µt ). Seg (d) (P n ) = V( (f )), discriminant hypersurface in P (n+d d ). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 0 /

11 Projections and Duals of Chow Varieties Lemma (GKZ) Let X PV, W V, X PW. Let π W : PV P(V /W ) projection. Then π W (X ) X PW. If π W (X ) = X then = holds. Lemma Let X PV, U W = V. Then (X PU) X P(V /U). P(V /U) = PU. Fact: If λ µ, then Seg λ (PV s ) = Seg µ (PV t ) S λ V S λ l V Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

12 Projections and Duals of Chow Varieties Lemma (GKZ) Lemma π W (X ) X PW (X PU) X PU. Fact: If λ µ, then Seg λ (PV s ) = Seg µ (PV t ) S λ V S λs V Proposition (O.) If λ µ, then Chow λ (PV ) Seg µ (PV t ) Question: What about the other inclusion? Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

13 Main Results The more general result for Segre-Veronese varieties: Theorem (O.) Let µ be a partition of d, and V be a complex vector space of dimension n. Then Seg µ ( PV t ) P ( S d V ) = λ µ Chow λ (PV ), where λ µ is the refinement partial order. In particular, V(Sym( µ,n )) = λ µ Ξ M λ,µ λ,n where Ξ λ,n is the equation of Chow λ (PV ) when it is a hypersurface in P(S d V ), and the multiplicity M λ,µ is the number of partitions µ that refine λ. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

14 Theorem (O.) The n d -hyperdeterminant of a polynomial (degree d, n variables) splits as the product V(Sym(Det(A))) = Ξ M λ λ,n, λ where Ξ λ,n is the equation of the dual variety of the Chow variety Chow λ P n when it is a hypersurface in P (n +d d ), λ = (λ,..., λ s ) is a partition of d, and the multiplicity M λ = M λ, d = ( ) d λ,...,λ s is the multinomial coefficient. Which dual varieties of Chow varieties are hypersurfaces? Theorem (O.) Suppose d, dim V = n and λ = (λ,..., λ s ) = ( m,..., p mp ). Then Chow λ (PV ) a hypersurface with the only exceptions n = and m 0 n >, s = and m = (so λ = (d, )). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

15 If n =, get closed formula for degrees of the duals of Chow varieties. Theorem (O.) The degree of Chow λ (P ) with λ = ( m, m,..., p mp ), m = 0 and m = i m i is ( ) m (m + ) m m (p ) mp m,..., m p If n get a recursion to compute all degrees of the duals of Chow varieties. Theorem (O.) Suppose dim V. Let d λ denote deg(chow λ (PV ) ) when it is a hypersurface and 0 otherwise. Then the vector (d λ ) λ is the unique solution to deg(det µ ) = λ µ d λ M λ,µ. The degree of Det µ is given by a generating function (see [page 454,GKZ]), the multiplicities M λ,µ are computable counting number of refinements. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 4 /

16 Combinatorics: Partitions and Refinement Proposition (5) (,) (,,) (,,) (, ) ( 5 ) (5) (4,) (,) (,,) (,,) (, ) ( 5 ) M λ,µ := number of partitions µ that refine λ. M (d),µ = for all µ = d. M λ,µ = 0 if s > t or if s = t and λ µ i.e. (M λ,µ ) λ,µ is lower triangular. If λ = ( m, m,..., p mp ), then M λ,λ = m! m p!. M λ, d = ( ) ( ) d λ := d λ,...,λ s = d! λ! λ s!, the multinomial coefficient. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 5 /

17 Combinatorics: Naively Counting Refinements Proposition For example, M (,,),(,,,,) = 6. Only φ contribute non-zero to M λ,µ : Let B(t, s) := all surjective maps φ: {,..., t} {,..., s}, and let χ(a, b) = δ a,b. Then M λ,µ = s χ λ i, φ B(t,s) i= j φ (i) Any better formulas? µ j. Note that this construction accounts for the ambiguity in the location of the s in the partition (,, ). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 6 /

18 Combinatorics: generating functions Ring of symmetric functions: [x] = [x, x,... ]. p λ [x] - power-sum symmetric functions, m µ [x] - monomial symmetric functions, m µ (x) = σ x σ.µ, p λ (x) = i λ = (λ,..., λ s ) d. (x λ i + x λ i... ) Proposition (Thanks to Mark Haiman) sum over distinct permutations σ of µ = (µ, µ,..., µ s, 0,... ) and x µ = x µ... x s µs. Suppose λ, µ, p λ and m µ are as above. Then the number of refinements matrix (M λ,µ ) is the change of basis matrix p λ (x) = µ d M λ,µ m µ (x). () The matrix (M λ,µ ) can be quickly computed in any computer algebra system. Compare the coefficients - M λ,µ is the coefficient on the monomial x µ in (). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 7 /

19 Generating function from GKZ for degree of A-discriminants [GKZ] N(κ; µ)z κ = [ κ i ( + z i) j µ ], jz j i j ( + z i) where N(κ; µ) = deg( κ,µ ) is the degree of Seg µ (P k P kt ) and κ Z t >0. We can now compute the degree of the duals of the Chow varieties via the following generating function. Theorem (O.) Suppose dim V. Let d λ denote deg(chow λ (PV ) ) when it is a hypersurface and 0 otherwise. Let µ,n denote the equation of the hypersurface Seg µ (PV t ). The degrees d λ are computed by deg( µ,n )m µ (x) = d λ p λ (x), µ λ where m µ and p λ are respectively the monomial and power sum symmetric functions. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 8 /

20 Example d = 4, n = Consider the system M λ,µ d λ = D µ d (4) d (,) d (,,) = d 6 4 (4 ) The unique solution is D (4) D (,) D (,) D (,,) D ( 4 ) 7 = 7 9. (d (4), d (,), d (,,), d ( 4 )) = (7, 5, 48, 5) The symmetrized hyperdeterminant of format 4 splits as: Sym(Det(A)) = discrim (Chow,) 6 (Chow,,) (Chow,,,) 4 deg 69 = (deg 7) (deg 5) 6 (deg 48) (deg 5) 4. The other µ-discriminants have the same factors, with different multiplicities encoded by M λ,µ. Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 9 /

21 Example d = 5, n = Consider the system M λ,µ d λ = D µ d (5) d (,) d (,,) d (,,) = d (, ) d ( 5 )) D (5) D (4,) D (,) D (,,) D (,,) D (, ) D ( 5 ) = The unique solution is (d (5), d (,), d (,,), d (,,), d (, ), d (5 )) = (48,, 08, 84, 480, 9). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 0 /

22 The first case we consider is n = and d = 7 and solve complicated poset D (7) D (6,) D (5,) D (5,,) 6 D (4,) 6 D (4,,) 48 d (7) D d (4, 7 ) (5,) d (4,) = D 0 (,,) = 84. D (,,), 44 d (,,) D (,,,) 6 D 6 (, ) D 5 50 (,,,) 70 D 5 0 (,, ) 06 D (, ) D ( 7 ) ) The unique solution to M λ,µ d λ = D µ is (d (7), d (5,), d (4,), d (,,) ) = (, 4, 6, 4). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

23 Next we use only a relevant lower-triangular sub-matrix of M λ,µ, when d = 8 and n = where here we have omitted several rows that are unnecessary for computing the degrees of Ξ λ. (d (8) (D (8) 4 d (6,) D (6,) 44 d (5,) D (5,) 6 4 d (4,4) = D (4,,) = 6. d (4,,) D (,,) d (,,) D (,,,) 848 d (,,,) ) D ( 8 )) 600 The unique solution to M λ,µ d λ = D µ is (d (8), d (6,), d (5,), d (4,4), d (4,,), d (,,), d (,,,) ) = (4, 0, 48, 7, 6, 48, 5). Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /

24 Thanks! Luke Oeding (UC Berkeley) Hyperdets of Polys June 5, 0 /