Pfister's theorem fails in the Hermitian case

TL;DR

This paper demonstrates that Pfister's theorem does not hold in the Hermitian setting by constructing a specific Hermitian polynomial that requires more squares in its sum of Hermitian squares representation than the classical bound.

Contribution

It provides a counterexample showing the failure of Pfister's theorem in the Hermitian case, highlighting a fundamental difference from the real case.

Findings

Counterexample Hermitian polynomial with high sum of squares complexity

Pfister's theorem does not extend to Hermitian polynomials

Shows the minimal number of squares needed exceeds classical bounds

Abstract

We show that the Hermitian analogue of a famous result of Pfister fails. To do so we provide a Hermitian symmetric polynomial $r$ of total degree 2d such that any non-zero multiple of it cannot be written as a Hermitian sum of squares with fewer than $d+1$ squares.