Weak Vishik Equivalence and Linear $2$-Basis Blindness for Totally Singular Quadratic Forms

Abstract

Let $K$ be a field of characteristic two, put $k=K^2$, and let $U\subset K$ be the value space of an anisotropic totally singular quadratic form. For $a\in K\setminus k$, the anisotropic dimension after the one-step extension $K(\sqrt a)/K$ is $\oneprofile_U(a)=\dim_{k(a)}k(a)U=\dim_kU-\frac12\dim_k(U\cap aU).$ Thus weak Vishik equivalence becomes an equality of multiplicative-overlap profiles. We derive direct value-space reconstructions of the norm field and the similarity-factor field, and we show that orthogonal complements for a nondegenerate associative pairing have identical one-step profiles. Over $K=\F_2(s,t,u,v)$, we construct explicit nonsimilar eight-dimensional value spaces $U,V$ whose one-step profiles agree for every $a\in K$. In addition, their complete coordinate profiles agree for every $2$-basis obtained from $(s,t,u,v)$ by a matrix in $\GL_4(\F_2)$. Nevertheless, the degree-four intermediate field $k(t,su)$ separates them, with profile values $4$ and $3$; it is coordinate for the nonlinear $2$-basis $(t,su,s,v)$ but for no linear change of the standard basis. Thus the unrestricted implication ``weak Vishik equivalence implies similarity'' fails for anisotropic totally singular quadratic forms in characteristic two. This does not answer Zemkov\'a's Question~Q, which concerns full Vishik equivalence over all field extensions. To the best of our knowledge, our construction gives the first explicit nonsimilar weakly Vishik-equivalent pair in the totally singular characteristic-two setting, as well as the first example showing that all linearly changed coordinate profiles can miss a nonlinear intermediate-field distinction. The only computer-assisted step is an exact symbolic verification over $\F_2(s^2,t^2,u^2,v^2)$ for the $35$ linear degree-four coordinate fields. The results also translate into minimum-rank profiles of symmetric matrices with prescribed diagonal.