Maximal sets of pairwise orthogonal vectors in finite fields (Q2888796)

Let \(\mathbb{F}_q\) be the finite field of \(q\) elements, where \(q\) is an odd prime power. Let \(B\) be a non-degenerate symmetric bilinear form on \(\mathbb{F}_q^n\). Then NEWLINENEWLINE\[NEWLINE B(x,y) = \sum_{i=1}^n a_i x_i y_i,NEWLINE\;a_i \neq 0, NEWLINE\;1 \leq i \leq n, NEWLINE\;x = (x_1, \dots, x_n), NEWLINE\;y = (y_1,\dots y_n) \in \mathbb{F}_q^n.NEWLINE\]NEWLINE NEWLINELet \(\chi\) be the quadratic character of \(\mathbb{F}_q\), and let \(\chi (B) = \prod_{i=1}^n \chi(a_i)\).NEWLINENEWLINEFor any non-degenerate symmetric bilinear form \(B\) on \(\mathbb{F}_q^n\) define \(I(B,\mathbb{F}_q^n)\) as the largest possible cardinality of pairwise \(B\)-orthogonal subsets \(\mathcal{E} \subseteq \mathbb{F}_q^n\).NEWLINENEWLINEThe main result that the author obtains is the following: NEWLINE{\parindent=8mmNEWLINE\begin{itemize}\item[(i)]If \(n\) is odd, then \(I(B,\mathbb{F}_q^n) = q^{(n -1)/2} + (n + 1)/2\). \item[(ii)]If \(n\) is even and \(\chi (B) = \chi (- 1)^{n/2}\), then \(I(B,\mathbb{F}_q^n) = q^{n/2} + n/2\). NEWLINE\item[(iii)]If \(n\) is even and \(\chi (B) = - \chi (- 1)^{n/2}\), then \(I(B,\mathbb{F}_q^n) = q^{n/2 - 1} + n/2 + 1\). NEWLINENEWLINENEWLINE\end{itemize}}NEWLINELemma 2.1 is a well-known result on maximal isotropic subspaces, see e.g. [\textit{E. Artin}, Geometric algebra. New York: Interscience Publishers, Inc.; London: Interscience Publishers Ltd. (1957; Zbl 0077.02101)], in particular Section III.6, Geometry over finite fields.