The Generalized Lax Conjecture: is every hyperbolicity cone spectrahedral?

Hyperbolic polynomials — homogeneous real polynomials whose restriction to every line through a fixed direction has only real roots — define, by a theorem of Gårding, convex semialgebraic sets called hyperbolicity cones, the feasible sets of hyperbolic programming. Every spectrahedral cone, i.e. every feasible set of a semidefinite program given by a linear matrix inequality, is a hyperbolicity cone, and in three variables the converse holds by the Helton–Vinnikov theorem. Hanselka's completely algebraic proof of the ternary case records the boundary of current knowledge: in four or more variables, it is unknown whether every hyperbolicity cone is spectrahedral. The question must be read at the level of cones rather than of the given polynomial: the literal reading — that every hyperbolic polynomial itself admits a definite symmetric determinantal representation — is false in four or more variables by Brändén's Vámos-matroid counterexamples. The source-faithful cone form, commonly called the Generalized Lax Conjecture, asks: for every polynomial $h$ hyperbolic with respect to $e$, does some hyperbolic multiplier $q$ with $\Lambda_+(h,e)\subseteq\Lambda_+(q,e)$ make $q\cdot h$ a definite symmetric determinant? The conjecture remains open: positive results cover special classes, smooth cones are even second-order-cone representable after projection, but no general construction and no counterexample is known.

Background

A homogeneous real polynomial $h\in\mathbb{R}[x_1,\dots,x_n]$ is called hyperbolic with respect to a direction $e\in\mathbb{R}^n$ if $h(e)>0$ and, for every $v\in\mathbb{R}^n$, the univariate polynomial $t\mapsto h(te+v)$ has only real roots. Hyperbolic polynomials were introduced by Gårding in the study of hyperbolic partial differential equations with constant coefficients. He proved that the hyperbolicity cone $\Lambda_+(h,e)$ — the set of all $v$ such that no root of $h(te+v)$ is strictly positive — is a convex semialgebraic cone, equal to the closure of the connected component of $\{h\neq 0\}$ containing $e$, and that $h$ is hyperbolic with respect to every point in its interior. Basic examples are products of linear forms (whose hyperbolicity cones are polyhedral cones) and the determinant on the space of real symmetric matrices (whose hyperbolicity cone is the positive semidefinite cone).

A spectrahedral cone is a set of the form $\{x\in\mathbb{R}^n: A(x)=x_1A_1+\cdots+x_nA_n\succeq 0\}$, where $A_1,\dots,A_n$ are real symmetric matrices and $\succeq 0$ denotes positive semidefiniteness; such a homogeneous linear matrix inequality (LMI) is exactly the constraint format of semidefinite programming. Every spectrahedral cone is a hyperbolicity cone: if $A(e)\succ 0$, then $h=\det A(x)$ is hyperbolic with respect to $e$, and a direct congruence argument shows $\Lambda_+(\det A,e)=\{x: A(x)\succeq 0\}$ — after the substitution $A(e)^{-1/2}A(x)A(e)^{-1/2}$, the roots of $\det(tA(e)+A(x))$ are the negatives of the eigenvalues of that matrix, so no root is strictly positive exactly when all eigenvalues are nonnegative. Consequently, a definite determinantal representation of $h$ — a representation $h=\det(x_1A_1+\cdots+x_nA_n)$ with $A(e)\succ 0$ — realizes the hyperbolicity cone of $h$ exactly as a spectrahedral cone. Hyperbolic programming, i.e. linear optimization over a hyperbolicity cone (introduced by Renegar), generalizes both linear and semidefinite programming, so the question of whether, at the level of feasible sets, hyperbolic programming coincides with semidefinite programming is precisely the question of whether the inclusion of cone classes is an equality.

For three variables the equality is known: Lax conjectured in 1958 that ternary hyperbolic forms admit definite determinantal representations; Vinnikov settled the corresponding bivariate statement in 1993 through self-adjoint determinantal representations of real plane curves; and Helton and Vinnikov proved in 2007 that every ternary form hyperbolic with respect to $e$ admits a linear determinantal representation definite at $e$, using theta functions on the Jacobian variety of the curve $\{h=0\}$. Lewis, Parrilo and Ramana (2005) established the equivalence of the various formulations and showed that this settles Lax's conjecture. Hanselka (2014) gave a new, completely algebraic proof of the Helton–Vinnikov theorem, and records in his abstract the boundary of current knowledge: 'Whether these [spectrahedral cones] are the only examples [of hyperbolicity cones] is unknown in general.'

Beyond three variables, the naive generalization — that every hyperbolic polynomial itself admits a definite determinantal representation — is false: Brändén (2011), relating linear matrix inequality representations to representability of polymatroids, constructed from the Vámos matroid a real-zero polynomial (equivalently, after homogenization, a hyperbolic form of degree four in four variables) none of whose powers admits a symmetric determinantal representation, and Netzer and Thom (2012) showed by a dimension count that almost no real-zero polynomial admits one. The surviving question is therefore genuinely about cones and not about representing the given polynomial: the determinant of the representing pencil may be a multiple of $h$ by a suitable hyperbolic factor. In the equivalent formulation recorded by Kummer (2013), the hyperbolicity cone of $h$ is spectrahedral if and only if there exists a hyperbolic polynomial $q$ with $\Lambda_+(h,e)\subseteq\Lambda_+(q,e)$ such that $q\cdot h=\det(x_1A_1+\cdots+x_nA_n)$ with $A(e)\succ 0$. The multiplier is genuinely necessary, as the same Vámos example shows — no power of it is a determinant — yet Kummer proved that its hyperbolicity cone is nevertheless spectrahedral. This cone-form question is commonly called the Generalized Lax Conjecture; a survey of the surrounding theory is given by Vinnikov (2012).

Problem Statement

Determine whether every hyperbolicity cone in $\mathbb{R}^n$ for $n\geq 4$ is a spectrahedral cone (the Generalized Lax Conjecture in its cone form; the cases $n\leq 3$ are settled affirmatively).

Formally: for every integer $n\geq 4$ and every homogeneous polynomial $h\in\mathbb{R}[x_1,\dots,x_n]$ hyperbolic with respect to a direction $e\in\mathbb{R}^n$, does there exist an integer $m$ and real symmetric $m\times m$ matrices $A_1,\dots,A_n$ such that, writing $A(x)=x_1A_1+\cdots+x_nA_n$,

$\{x\in\mathbb{R}^n : A(x)\succeq 0\} \;=\; \Lambda_+(h,e)\,?$

Here $\Lambda_+(h,e)$ is the closed hyperbolicity cone of $h$, i.e. the set of all $v\in\mathbb{R}^n$ such that no root of the univariate polynomial $t\mapsto h(te+v)$ is strictly positive.

Equivalently, in the multiplier formulation recorded by Kummer (2013): does there exist a hyperbolic polynomial $q$ with $\Lambda_+(h,e)\subseteq\Lambda_+(q,e)$ such that $q\cdot h=\det(x_1A_1+\cdots+x_nA_n)$ with $A(e)\succ 0$?

The representing pencil is not required to compute $h$ itself: by Brändén's Vámos-type examples, some hyperbolic polynomials in four variables have no power admitting a definite determinantal representation, so a hyperbolic multiplier $q$ is unavoidable in general. A complete answer is either a proof that such a pencil (equivalently, such a multiplier) exists for every hyperbolic polynomial in four or more variables, or an explicit hyperbolic polynomial together with a proof that no symmetric linear pencil of any size has positive-semidefinite cone equal to its hyperbolicity cone.

References

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