HYPERBOLIC POLYNOMIALS, INTERLACERS AND SUMS OF SQUARES

HYPERBOLIC POLYNOMIALS, INTERLACERS AND SUMS OF SQUARES DANIEL PLAUMANN Universität Konstanz joint work with Mario Kummer (U. Konstanz) Cynthia Vinzant (U. of Michigan) Out[121]= POLYNOMIAL OPTIMISATION Isaac Newton Institute Cambridge 19 July 2013 In[137]:= p1 = ContourPlot3D@x ã 1.6, 8x, -2, 2<, 8z, -4, Mesh Ø None, ContourStyle Ø 8Magenta<, Boxe p2 = ContourPlot3D@z ã 1.5, 8x, -2, 2<, 8z, -4, Mesh Ø None, ContourStyle Ø 8Green<, Boxed Ø Show@8f, p1<d Show@8f, p2<d Show@8f, p1, p2<d Universität Konstanz

Two flavours of real polynomials Let f R[x] d be homogeneous of degree d in x = (x 1,..., x n ). Positive polynomials. f 0 on R n. Ideal representation: Sum of squares f = f1 2 + + fr 2

Two flavours of real polynomials Let f R[x] d be homogeneous of degree d in x = (x 1,..., x n ). Positive polynomials. f 0 on R n. Ideal representation: Sum of squares f = f1 2 + + fr 2 Hyperbolic polynomials. Ideal representation: Determinantal polynomial. f = det(m), M(x) = x 1 M 1 + + x n M n with M 1,..., M n Sym d (R), M(e) positive definite. f R[x] is hyperbolic with respect to e R n if p(e) > 0 and for all v R n f (v te) R[t] has only real zeros in t.

Two flavours of real polynomials Let f R[x] d be homogeneous of degree d in x = (x 1,..., x n ). Positive polynomials. f 0 on R n. Ideal representation: Sum of squares f = f1 2 + + fr 2 Existence only for n 2 or d 2 or (n, d) = (3, 4) (Hilbert 1888) Hyperbolic polynomials. Ideal representation: Determinantal polynomial. f = det(m), M(x) = x 1 M 1 + + x n M n with M 1,..., M n Sym d (R), M(e) positive definite. Existence only for n 3 (Helton-Vinnikov 2004) f R[x] is hyperbolic with respect to e R n if p(e) > 0 and for all v R n f (v te) R[t] has only real zeros in t.

Hyperbolicity cones f hyperbolic with respect to e: f (e) > 0 and f (v te) R[t] has only real roots for all v R n. Hyperbolic polynomials can be thought of as generalised characteristic polynomials.

Hyperbolicity cones In[121]:= f = ContourPlot3D@2 x^4 + y^4 + z^4-3 y^2 x^2-3 z^2 x^2 + y 8x, -2, 2<, 8z, -4, 4<, 8y, -4, 4<, Mesh Ø None, ContourStyle Ø 8 Blue, Opacity@.5D<, Boxed Ø False, Axes Ø f hyperbolic with respect to e: f (e) > 0 and f (v te) R[t] has only real roots for all v R n. Out[121]= Hyperbolic polynomials can be thought of as generalised characteristic polynomials. n = 3, d = 4 C e ( f ) = {v R n All roots of f (v te) are non-negative}, is a closed convex cone, called the hyperbolicity cone. In[137]:= p1 = ContourPlot3D@x ã 1.6, 8x, -2, 2<, 8z, -4, 4<, 8y, -4, 4< Mesh Ø None, ContourStyle Ø 8Magenta<, Boxed Ø False, Axe p2 = ContourPlot3D@z ã 1.5, 8x, -2, 2<, 8z, -4, 4<, 8y, -4, 4< Mesh Ø None, ContourStyle Ø 8Green<, Boxed Ø False, Axes Ø Show@8f, p1<d Show@8f, p2<d Show@8f, p1, p2<d

Optimisation Semidefinite programming relaxations f 0 on R n? f is a sum of squares in R[x]? SDP

Optimisation Semidefinite programming relaxations f 0 on R n? f is a sum of squares in R[x]? SDP Let f be a hyperbolic polynomial. Hyperbolic programme? SDP Optimise over C e ( f )? SDP formulation/relaxation of C e ( f ) Hyperbolic programming. Studied by Güler, Lewis, Renegar, Tunçel and others. Can use log f as a barrier function in interior point methods.

Interlacers Let f be hyperbolic w.r.t. e, deg( f ) = d, irreducible g be hyperbolic w.r.t. e, deg(g) = d 1

Interlacers Let f be hyperbolic w.r.t. e, deg( f ) = d, irreducible g be hyperbolic w.r.t. e, deg(g) = d 1 Call g an interlacer of f if the roots of g(v te) lie between those of f (v te) for all v R n. 2 1-2 -1 0 1 2 3-2 0-1 Quartic with an interlacing cubic α 1 α d roots of f (v te) β 1 β d 1 roots of g(v te) α 1 β 1 α 2 β 2 β d 1 α d

Interlacers Let f be hyperbolic w.r.t. e, deg( f ) = d, irreducible g be hyperbolic w.r.t. e, deg(g) = d 1 Call g an interlacer of f if the roots of g(v te) lie between those of f (v te) for all v R n. 2 1-2 -1 0 1 2 3-2 0-1 Quartic with an interlacing cubic Directional derivatives g = D v f with v C e ( f ) are interlacers. If f = det(m) with M(e) 0, then any symmetric minor of size d 1 of M is an interlacer.

Interlacers Plane quartic curve with two interlacing cubics The product of two interlacers of f is non-negative on V R ( f ) = {v R n f (v) = 0}.

Interlacers Plane quartic curve with two interlacing cubics The product of two interlacers of f is non-negative on V R ( f ) = {v R n f (v) = 0}. Proposition The set Int e ( f ) of interlacers is a convex cone, namely Int e ( f ) = {g R[x] d 1 D e f g f D e g 0 on R n } Reason. If f, g R[t] have real roots with deg(g) = deg( f ) 1, then g interlaces f if and only if the Wronskian f g f g is everywhere positive or negative (Bézout).

Representing the hyperbolicity cone The convex cone of interlacers is Int e ( f ) = {g R[x] d 1 D e f g f D e g 0 on R n }

Representing the hyperbolicity cone The convex cone of interlacers is Int e ( f ) = {g R[x] d 1 D e f g f D e g 0 on R n } A directional derivative D v f is an interlacer if and only if the point v is in the hyperbolicity cone C e ( f ). Corollary Write C e ( f ) = {v R n D e f D v f f D e D v f 0 on R n } W e,v ( f ) = D e f D v f f D e D v f for the family of Wronkians associated with f. This realises the hyperbolicity cone as a linear slice of the cone of non-negative polynomials.

A sums of squares relaxation We obtain an increasing sequence of inner approximations C e (N) ( f ) ={v R n (x 2 1 + + x2 n) N W e,v ( f ) is a sum of squares C e ( f ) for N = 0, 1, 2,... If f is strictly hyperbolic (V R ( f ) smooth), then C e (N) ( f ) = C e ( f ) N 0 by a result of Reznick on sums of squares with denominators. Theorem (Netzer-Sanyal 2012) If f is strictly hyperbolic, then C e ( f ) is a projected spectrahedral cone. Theorem (Parrilo-Saunderson 2012) Derivative cones for determinantal polynomials are projected spectrahedral cones. }

Results on determinantal representations Effective version of the Helton-Vinnikov theorem for Hermitian determinantal representations. Theorem If f R[x, y, z] is hyperbolic with respect to e, we can construct a representation f = det(m) with M(e) 0 and M Hermitian, starting from any interlacer of f. 2 1 0-1 -2-2 -1 0 1 2 3

Results on determinantal representations Theorem If f k = det(m) is a determinantal representation with M(e) 0, k 1, then all the Wronskians W e,v ( f ) with v C e ( f ) are sums of squares.

Results on determinantal representations Theorem If f k = det(m) is a determinantal representation with M(e) 0, k 1, then all the Wronskians W e,v ( f ) with v C e ( f ) are sums of squares. Example The Vámos polynomial h(x 1,, x 8 ) = x i, I ( [8] 4 )/C i I where C = {{1, 2, 3, 4}, {1, 2, 5, 6}, {1, 2, 7, 8}, {3, 4, 5, 6}, {3, 4, 7, 8}} is the bases-generating polynomial of the matroid with circuits in C. The polynomial h is hyperbolic (Wagner-Wei 2009) but no power of h has a definite determinantal representation (Brändén 2011). We show that W e7,e 8 (h) is not a sum of squares, yielding a new proof of Brändén s result.

Results on determinantal representations Theorem If f k = det(m) is a determinantal representation with M(e) 0, k 1, then all the Wronskians W e,v ( f ) with v C e ( f ) are sums of squares. Theorem If f is multi-affine and hyperbolic with respect to e, then f has a determinantal representation f = det(m) with M(e) 0 if and only if the Wronskians W ei,e j are squares in R[x] for all i, j. Example We can use this to reprove a classical result saying that the d-th elementary symmetric polynomial in n variables, which is multi-affine and hyperbolic with respect to (1,..., 1), has a definite determinantal representation if and only if d {1, n 1, n}.

The interlacer cone Example Let f = det(x) where X is a d d symmetric matrix of variables. Int Id (det(x)) = { tr(m X adj ) M Sym + d (R) } Sym+ d (R).

The interlacer cone Example Let f = det(x) where X is a d d symmetric matrix of variables. Int Id (det(x)) = { tr(m X adj ) M Sym + d (R) } Sym+ d (R). Theorem Let f R[x] d be hyperbolic with respect to e R n and assume that the projective variety V C ( f ) is smooth. The algebraic boundary of the cone Int e ( f ) is the irreducible hypersurface in C[x] d 1 given by g C[x] d 1 p P n 1 such that f (p) = g(p) = 0 and rank ( f (p) g(p) ) 1.

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