Homogeneous symmetric trace-positive sextics with vanishing abelianization are sums of Hermitian squares and commutators

Abstract

Let $\mathbb{R}⟨X,Y⟩$ be the free algebra on two symmetric letters. A polynomial is trace-positive if it has non-negative trace on every pair of real symmetric matrices, and it lies in the cone $\Theta$ if it is cyclically equivalent to a sum of Hermitian squares. Burgdorf and Klep proved that the two notions coincide in degree at most four; from degree six onwards they differ. We give a complete, explicit description of both cones on the six-dimensional space $Z$ of symmetric homogeneous sextics in two symmetric letters whose abelianization vanishes identically, and show that on $Z$ they coincide: for $f\in Z$, trace-positivity is equivalent to membership in ${\Theta }_3$, to non-negativity of the trace on pairs of $2\times 2$ symmetric matrices, and to being cyclically equivalent to a sum of four Hermitian squares of cubics with vanishing abelianization. Membership in $\Theta$ reduces to a $4\times 4$ positive semidefinite Gram matrix whose diagonal blocks are fixed by the content-$(4,2)$ and $(2,4)$ parts of $f$; each of those parts ranges over a two-dimensional cone with two linear facets, and the content-$(3,3)$ part is then confined to a rectangle whose sides are the geometric means $\sqrt{p_{+}{q}_{+}}$ and $\sqrt{p_{-}{q}_{-}}$ of the eigenvalues of those blocks. Each facet is certified by one of two families of pairs of $2\times 2$ symmetric matrices. The abelianization hypothesis is natural: every bivariate degree-six example we are aware of -- a trace-positive polynomial outside $\Theta$ -- arises by lifting a non-negative, non-SOS polynomial in two variables to a cyclically sorted polynomial; and that lifting cannot operate when the abelianization vanishes. Symmetry of $f$ is essential: dropping it adds one antisymmetric direction, on which the trace vanishes identically, yet whose non-zero elements all lie outside $\Theta$.