Noncommutative polynomials describing convex sets

TL;DR

This paper characterizes when free semialgebraic sets defined by noncommutative polynomials are convex, providing an algorithm to identify such sets as free spectrahedra and revealing that convex irreducible cases are quadratic and linked to linear pencils.

Contribution

It offers a complete characterization of convex free semialgebraic sets defined by noncommutative polynomials and introduces an efficient algorithm to recognize and construct their minimal linear matrix inequality representations.

Findings

Convexity of $D_f$ implies $f$ has degree at most two for irreducible polynomials.

Provides an algorithm to determine if $D_f$ is a free spectrahedron and to find the corresponding linear pencil.

Shows that convex irreducible polynomials are Schur complements of linear pencils.

Abstract

The free closed semialgebraic set $D_f$ determined by a hermitian noncommutative polynomial $f$ is the closure of the connected component of $\{(X,X^*)\mid f(X,X^*)>0\}$ containing the origin. When $L$ is a hermitian monic linear pencil, the free closed semialgebraic set $D_L$ is the feasible set of the linear matrix inequality $L(X,X^*)\geq 0$ and is known as a free spectrahedron. Evidently these are convex and it is well-known that a free closed semialgebraic set is convex if and only it is a free spectrahedron. The main result of this paper solves the basic problem of determining those $f$ for which $D_f$ is convex. The solution leads to an efficient algorithm that not only determines if $D_f$ is convex, but if so, produces a minimal hermitian monic pencil $L$ such that $D_f=D_L$. Of independent interest is a subalgorithm based on a Nichtsingul\"arstellensatz presented here: given a linear pencil $L'$ and a hermitian monic pencil $L$, it determines if $L'$ takes invertible values on the interior of $D_L$. Finally, it is shown that if $D_f$ is convex for an irreducible hermitian polynomial $f$, then $f$ has degree at most two, and arises as the Schur complement of an $L$ such that $D_f=D_L$.