I have troubles understanding parts of the proof of proposition 6.1 of this article. I try to give a short outline of the situation.
We consider a group $H \subset GL(n,\mathbb{C})$, generated by $$A = \begin{pmatrix} 0 & 0 & \cdots & 0 & -A_n\\ 1 & 0 & \cdots & 0 & -A_{n-1}\\ 0 & 1 & \cdots & 0 & -A_{n-2}\\ \vdots & \vdots & \cdots & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & -A_1\\ \end{pmatrix}, B = \begin{pmatrix} 0 & 0 & \cdots & 0 & -B_n\\ 1 & 0 & \cdots & 0 & -B_{n-1}\\ 0 & 1 & \cdots & 0 & -B_{n-2}\\ \vdots & \vdots & \cdots & \vdots & \vdots\\ 0 & 0 & \cdots & 1 & -B_1\\ \end{pmatrix}$$ acting irreducible on $\mathbb{C}^n$. Consider $C=A^{-1}B$. Then $C$ has $n-1$ times the eigenvalue $1$ with independent eigenvectors $e_i$, $i=1,...,n-1$ and an eigenvalue $c$ with eigenvector $v$. Also one has that $C$ is in fact a reflection with special eigenvalue $c$, i.e. one has that $rk(C-Id)=1$. We assume that $c= \pm 1$. Furthermore we have a non-degenerate bilinear form $f$ on $\mathbb{C}^n$, invariant under $H$ and said to be either symmetric or anti-symmetric.
Now my problem is, that in said proposition there is the statement that the form $f$ is symmetric in case $c=-1$ and anti-symmetric for $c=+1$, what i dont see immediately and want to show now. I have an basic idea, but i´m not sure if it works.
My idea:
case $c=-1$:
We assume $f$ to be anti-symmetric, then in particular $f$ is alternating, thus $v^t fv=0$. Also one sees by using the $H$-invariance, that $v^t f e_i = v^t C^t f C e_i = (Cv)^t f (C e_i) = -v^t f e_i$, therefore $v^t f e_i=0$ for all $i=1,...,n-1$. Since $e_1,...,e_{n-1},v$ is a basis of $\mathbb{R}^n$, the set $\{x \in \mathbb{R}^n \mid x^t f y =0, \forall y \in \mathbb{R}^n\}$ is a non-trivial $H$-invariant subspace, contradicting the irreducibility of $H$. Hence $f$ has to be symmetric.
case $c=+1$:
In this case we note, that $C$ is in fact a transvection, i.e. $rk(C-Id)=1$ and $im(C-Id) \subset ker(C-Id)$. I tried using this, to get the definition of anti-symmetry for $f$ but it didnt work out.