Orthogonal Lie algebras for degenerate bilinear forms

Question

Let $\mathbb{K}$ be a field such that $\mathrm{char}(\mathbb{K}) \neq 0$, let $V$ be an $n$-dimensional $\mathbb{K}$-vector space, and let $\sigma:V \times V \to V$ be a bilinear form. If $\sigma$ is non-degenerate then we know that the Lie algebra $$ {\frak o}_{\sigma}(V) := [A \in {\frak gl}(V) : \sigma(Av,w) = - \sigma(v,Aw)] $$ is contained in the special linear algebra ${\frak sl}(V)$. Is this true in the case that $\sigma \neq 0$ is degenerate? Does it still hold true for non-zero characteristic?

Answer 1

If $\sigma$ is represented by a matrix $B$, so $\sigma(v,w)=v^\top Bw$, then the condition on $A$ becomes $A^\top B=-BA$. If $\sigma$ is non-degenerate, then we get $\text{tr}(A)=\text{tr}(A^\top)=\text{tr}(-BAB^{-1})=-\text{tr}(A)$ and thus $\text{tr}(A)=0$ if the characteristic is not $2$.

For degenerate $\sigma$, take $B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $A=\begin{bmatrix}0&0\\0&1\end{bmatrix}$. It is clear that $A^\top B=-BA=0$, but $\text{tr}(A)\neq0$, so $A$ is not in $\mathfrak{sl}(\mathbb K^2)$.

For non-degenerate $\sigma$ in characteristic $2$, take $B=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $A=\begin{bmatrix}0&0\\0&1\end{bmatrix}$.