Gram Spectrahedra. Lynn Chua UC Berkeley. joint work with Daniel Plaumann, Rainer Sinn, and Cynthia Vinzant

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1 Gram Spectrahedra Lynn Chua UC Berkeley joint work with Daniel Plaumann, Rainer Sinn, and Cynthia Vinzant SIAM Conference on Applied Algebraic Geometry 017 1/17

2 Gram Matrices and Sums of Squares Let P R n be polytope with vertices in Z n 0. Let m P be a vector of monomials whose exponent vectors are lattice points in P. Let f R[x 1,..., x n ] with Newton polytope Newt(f ) := convex hull of exponents of monomials in f A Gram matrix of f is a real symmetric N N matrix A Sym N, where N := m P, such that f = m t P Am P. Define G(f ) to be the set of all Gram matrices of f. /17

3 Gram Matrices and Sums of Squares Proposition f R[x 1,..., x n ] with Newt(f ) P is a sum of squares iff there is a positive semidefinite Gram matrix A Sym + N such that f = m t P Am P. The Gram spectrahedron of f is G + (f ) = {A Sym + N : mt P Am P = f }. The length of f is the shortest length of any SOS representation of f, or the minimum rank of any matrix in G + (f ). 3/17

4 Gram Spectrahedron Gram matrices are more unique representations than SOS. x + y = (cos(θ)x sin(θ)y) + (sin(θ)x + cos(θ)y) have the same Gram matrix for any θ [0, π]. Two SOS representations f = p p r = q q r are equivalent if they give the same Gram matrix orthogonal (r r)-matrix U such that (q 1,..., q r ) t = U (p 1,..., p r ) t. 4/17

5 Connections to Toric Geometry Theorem (Blekerman, Smith, Velasco, 013) Let P R n be a normal lattice polytope. Suppose every nonnegative f R[x 1,..., x n ] with Newton polytope P is a sum of squares. Then P is, up to translation and an automorphism of the lattice, contained in one of: 1. m 1 = conv{0, e 1,..., e m } R m. conv {([0, d 1 ] e 1 ) ([0, d ] e ) ([0, d m ] e m )} R R m, d i Z 0 3. conv{(0, 0), (, 0), (0, )} R 4. conv{(q {0}) ({0} n 1 )} R m R n, where Q is (1)-(3) and n 1 = conv{0, e 1,..., e n } 5/17

6 Ranks on spectrahedra Proposition (Pataki interval) Let L Sym N be an affine-linear space of dimension m. The rank r of an extreme point of L Sym + N satisfies ( r + 1 ) + m ( N + 1 Furthermore, if L is generic, ( ) N r + 1 m. Proposition For a general polynomial, the rank of all extreme points of its Gram spectrahedron lies in the Pataki interval. ). 6/17

7 Gram spectrahedra of binary forms Proposition Let f R[s, t] be a positive binary form of degree d with distinct roots. G(f ) contains 1 d ) ( d matrices of rank over C. Proof. Writing f as a sum of two squares partitions the roots into two size d subsets: f = p + q = (p + iq)(p iq). Two equivalent representations yield the same partition: f = (cos(θ)p + sin(θ)q) + (sin(θ)p cos(θ)q) = ( e iθ (p iq) ) (e iθ (p + iq) ) The number of such partitions is ( 1 d ) d. 7/17

8 Gram spectrahedra of binary forms Proposition The Gram spectrahedron G + (f ) contains d 1 real matrices of rank. If d is even, there are an additional 1 ( d d/) real indefinite matrices of rank in G(f ). Proof. For the partition of roots to be real, complex conjugation must either fix both blocks of d roots or swap them. If blocks are swapped: f = gg = (Re(g)) + (Im(g)). Each conjugate pair has one element per block, so there are d 1 partitions If blocks are fixed, d is even and there are 1 ( d d/) partitions. f = gh = 1 ((g + h) (g h) ), where g, h are positive. 8/17

9 Binary forms of degrees and 4 Let f = a s + a 1 st + a 0 t be nonnegative, then G + (f ) consists of ( ) 1 a a 1 1 a 1 a 0 If f has a zero, then f = l for the linear form vanishing at the root of f and its Gram matrix has rank 1. If f is strictly positive, then f is the sum of two squares (completing the square) and its Gram matrix has rank. If f is a positive binary form of degree 4, then G + (f ) contains matrices of rank. G + (f ) is a line segment, which is the convex hull of two sum-of-two-squares representations f = p 1 + q 1 = p + q 9/17

10 Binary sextics and Kummer surfaces f = a 6 s 6 6a 5 s 5 t+15a 4 s 4 t 0a 3 s 3 t 3 +15a s t 4 6a 1 st 5 +a 0 t 6 G + (f ) is the set of matrices in Sym + 4 of the form a 6 3a 5 3a 4 + z a 3 y 3a 5 9a 4 z 9a 3 + y 3a + x 3a 4 + z 9a 3 + y 9a x 3a 1 a 3 y 3a + x 3a 1 a 0 The determinant of this matrix is a Kummer surface.. 10/17

11 Kummer surface The Kummer surface is a quartic in P 3 with 16 nodes in a 16 6 configuration. 6 nodes (u i : u i : 1 : 0), where u 1,..., u 6 are the distinct complex roots of f (s, 1). These give rank-3 matrices. 10 nodes correspond to representations of f as a sum of two squares over C. These give rank- matrices. 11/17

12 Kummer surface The dual projective variety of a Kummer surface is again a Kummer surface: F = W F + WF 1 + F 0 F = Y + 4XZ F 1 = a 0 X 3 + 3a 1 X Y + 3a XY + a 3 Y 3 + 3a X Z + 6a 3 XYZ + 3a 4 Y Z + 3a 4 XZ + 3a 5 YZ + a 6 Z 3 F 0 = ( 9(a 0 a a 1)X (a 0 a 3 a 1 a )X 3 Y + 3(5a 0 a 4 a 1 a 3 3a )X Y + 6(a 0 a 5 a a 3 )XY 3 + (a 0 a 6 a 3)Y 4 + 6( a 0 a a 1 a 3 9a )X 3 Z + 6( a 0 a 5 + 1a 1 a 4 11a a 3 )X YZ + ( a 0 a a 1 a 5 9a a 4 8a 3)XY Z + 6(a 1 a 6 a 3 a 4 )Y 3 Z + 9(a 4 a 6 a 5)Z 4 + (a 0 a 6 18a 1 a a a 4 100a 3)X Z + 6( a 1 a 6 + 1a a 5 11a 3 a 4 )XYZ + 3(5a a 6 a 3 a 5 3a 4)Y Z + 6( a a a 3 a 5 9a 4)XZ (a 3 a 6 a 4 a 5 )YZ 3). 1/17

13 SDP over the Gram spectrahedron Proposition The algebraic degree of semidefinite programming over the Gram spectrahedron of a binary sextic is (10, ) in ranks (, 3). Proof. Given a cost vector c = (c 1, c, c 3 ) Q 3, the rank-3 critical points (x, y, z) satisfy c 1 x + c y + c 3 z = F 1(c) ± F 1 (c) 4 F 0 (c) F (c). F (c) and can be expressed over a quadratic field extension of Q. The optimal point is either one of these rank 3-matrices or one of the 10 rank- matrices. 13/17

14 Example f = s 6 s 5 t + 5s 4 t 4s 3 t 3 + 5s t 4 st 5 + t 6 The linear form x z on G + (f ) gives dual coordinates c = (X, Y, Z) = (1, 0, 1). The equation of the dual Kummer surface gives W = ± 5. The corresponding rank-3 Gram matrices over Q[ 5] are , /17

15 The Kummer surface and its dual Figure: The Kummer surface bounding G + (f ), dual Kummer surface, and dual convex body of G + (f ). Both the primal and dual Kummer surfaces have 4 real nodes, and we can compute a real linear transformation from one to the other. If the roots {u 1, u, u 3 } are conjugate to {u 4, u 5, u 6 }, then we can define a real transformation taking the node [0 : 0 : 0 : 1] on the dual surface to the rank- Gram matrix of f corresponding to the representation of f as gg where = (t u 1 )(t u )(t u 3 ). 15/17

16 Ternary Quartics Experimental evidence suggests that a general rational ternary quartic possesses a rational Gram matrix of rank 5. Question What is a formula for the hypersurface dual to the boundary of the Gram spectrahedron of a general ternary quartic? Can we derive an explicit formula analogous to the equation of the dual Kummer surface? 16/17

17 Thank you! Any Questions? 17/17

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