Question regarding the existence of an invariant, non-degenerate complex bilinear form. I have a question regarding the proof of proposition 6.1, p.345 of this article, i'm reading and trying to understand right now. In said proof, the author states that

self-duality of the hypergeometric group $H(a;b)$ (i.e. the parameter sets $a,b$ are invariant under $z \mapsto z^{-1}$) implies the existence of an non-degenerate bilinearform $F$ on $V=\mathbb{C}^n$, which is invariant under $H(a;b)$.

Firstly, i understand that "complex bilinear form" here just means a bilinear form on a complex vector space. That is, it is linear in each argument, rather than linear in one argument and conjugate-linear in the other. Which leads me to the following question:
My question:
Well, i want to clarify the statement above.
In theorem 4.3 of the same article, it is shown that if the parameters are invariant under $z \mapsto \overline{z}^{-1}$ there exists a non-degenerate, invariant hermitian form on $V=\mathbb{C}^n$.
Does thrm. 4.3 also hold true for a complex bilinear form rather than a hermitian one, when only assuming the invariance $z \mapsto z^{-1}$ instead of the invariance with the additional complex conjugation? So in particular i want to know, if thrm. 4.3 was used in the proof of 6.1 and in case it was, how exactly. If this is not the case, then why does this bilinear form exist?
I would be very grateful for any answers or comments on this matter. Thank you.
 A: In the language of representation theory, a (finite dim) group representation is self-dual if and only if it has an invariant bilinear form. This is because an invariant non-degenerate bilinear form on a representation $V$ is the same as an injective map of representations $V\to V^*.$
In the language of the paper, a hypergeometric group is specified by matrices $h_0,h_1,h_\infty.$ Since we're working with matrices, the dual transformations are transposes. Applying the inverse transpose map $g\mapsto (g^{-1})^T$ gives another hypergeometric group by $(h_0^{-1})^T,(h_1^{-1})^T,(h_\infty^{-1})^T.$ In the case you're interested in, these two hypergeometric groups have the same parameter sets. So by Theorem 3.5, they are simultaneously conjugate: there exists a matrix $\alpha$ such that $\alpha h_i\alpha^{-1}=(h_i^{-1})^T$ for $i=0,1,\infty.$ The nondegenerate bilinear form given by $(u,v)=u^T\alpha v$ is invariant under each $h_i$:
$$(h_iu,h_iv)=u^Th_i^T\alpha h_iv=u^Th_i^T(h_i^{-1})^T\alpha v=u^T\alpha v=(u,v).$$
