This paper addresses a question posed in [C. Scheiderer, J. Eur. Math. Soc. (JEMS) 18, No. 7, 1495–1513 (2016; Zbl 1354.14036)]: Does there exists a polynomial over the rational numbers \(\mathbb Q[x_1,\dots,x_n]\) defining a smooth complex hypersurface that is strictly positive and cannot be expressed as the sum of squares (sos) of rational polynomial (i.e. not \(\mathbb Q\)-sos)? The author provides an affirmative answer to this question by constructing such a polynomial which must lie in the boundary of the sos-cone.
The construction consists of two parts. First, a specific polynomial \(f\) over \(\mathbb Q\) that is \(\mathbb Q(\sqrt[3]{2})\)-sos but not \(\mathbb Q\)-sos is constructed. This construction is also explained in [S. Laplagne, Math. Comput. 89, No. 322, 859–877 (2020; Zbl 1504.14103)] and [J. Capco, S. Laplagne and C. Scheiderer, “Exact Polynomial sum of squares decomposition”, Preprint 2023]. To prove that \(f\) is not \(\mathbb Q\)-sos, the author finds a linear form in the dual of the sos-cone that vanishes at \(f\) and such that its associated quadratic form has a kernel containing no non-trivial polynomials defined over \(\mathbb Q\).
In the second step, the author constructs a strictly positive polynomial \(g\) that is in the boundary of the sos-cone with a unique Gram matrix (i.e. \(g\) has a unique sos decomposition up to orthogonal equivalence). Finally, \(f+g\) is added to a square polynomial \(r^2\) to obtain the desired example \(h\). The proof that \(h\) is strictly positive and defines a smooth hypersurface is relatively straightforward, and the proof that \(h\) is not \(\mathbb Q\)-sos uses a similar approach as that used for \(f\). The author provides Maple and Singular scripts to verify the steps of the proofs.