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I am interested in understanding the unitary representations of the symmetric group over $\mathbb{F}_{q^2}$. In general, some comments here are relevant

Unitary representations of finite groups over finite fields

but Lemma 4.4.1 doesn't apply because the Frobenius-Schur indicator for rep'ns of the symmetric group are +1 (not $\circ$ which presumably means zero).

Formula for the Frobenius-Schur indicator of a finite group?

However, this answer seems to imply that in some cases a $G$-invariant symmetric bilinear form does exist. Theorem 1.5 in James' book The Representation Theory of the Symmetric Group implies that irreducible representations of the symmetric group are self-dual. $\chi$ is always real valued, and we find many examples in the appendix with $d(\chi,\phi)=1$.

Is that correct and are there any references with a construction of this form?

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  • $\begingroup$ Are your representations irreducible? Irreducible representations of the symmetric group can always be written over the ground field, and are self-dual. But non-irreducible ones certainly need not be. $\endgroup$
    – Dave Benson
    Commented Oct 21 at 19:52
  • $\begingroup$ Yes, the representations are irreducible (edited). I believe self-duality holds for simple modules as well, but for now even just the non-modular case would be interesting. There is Young's orthogonal form, but those matrices aren't unitary, at least w.r.t the usual form as $AA^T = I$ but $AA^{qT} \ne I$. $\endgroup$
    – Jackson Walters
    Commented Oct 21 at 20:59

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