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Let $G$ be a finite group and let $k$ be an algebraically closed field of characteristic $p \neq 2$.

Let $V$ be a finite-dimensional irreducible $kG$-module. If $V \cong V^*$, then $V$ admits a nonzero $G$-invariant bilinear form $(-,-)$, unique up to scalar, such that $(-,-)$ is alternating or symmetric. Is there a formula or a general method to determine whether $(-,-)$ is going to be alternating or symmetric?

If $k = \mathbb{C}$, then the answer is yes: for a $\mathbb{C}G$-module $V$ with irreducible character $\chi$ we have the Frobenius-Schur indicator $$\nu_2(\chi) = \frac{1}{|G|} \sum_{g \in G} \chi(g^2)$$

which satisfies $\nu_2(\chi) \in \{-1, 0, 1\}$ and

  • $\nu_2(\chi) \neq 0$ iff $V \cong V^*$.
  • $\nu_2(\chi) = 1$ if there is a nonzero $G$-invariant orthogonal bilinear form on $V$.
  • $\nu_2(\chi) = -1$ if there is a nonzero $G$-invariant alternating bilinear form on $V$.

I think similar things should be true when $p \not \mid |G|$. What about in general? Is this an open problem?

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  • $\begingroup$ @Geoff: Do you have a reference? Maybe I am making some easy mistake but I was thinking there might be some problems there. For our irreducible $kG$-module $V \cong V^*$, a nonzero $G$-invariant alternating (symmetric) bilinear form is the same as a $1$-dim submodule of $\wedge^2(V)$ (of $S^2(V)$). Character of $\wedge^2(V)$ and $S^2(V)$ could tell you the composition factors but not the submodule structure (maybe there is a unique $1$-dim submodule, but several $1$-dimensional composition factors). $\endgroup$ – spin Jun 11 '17 at 12:16
  • $\begingroup$ I think you have the form types the wrong way round. $\nu_2(\chi) = 1$ corresponds to a real representation that preserves an orthogonal form, whereas $\mu_2(\chi)=-1$ corresponds to a quaternionic representation that preserves an alternating form. $\endgroup$ – Derek Holt Jun 11 '17 at 12:35
  • $\begingroup$ Sorry, I think I misread the question. I was thinking you were asking whether it was the same in characteristic coprime to $|G|,$ but I see that you weren't. I think W.Willems and P. Sin have written about this question. $\endgroup$ – Geoff Robinson Jun 11 '17 at 13:24
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    $\begingroup$ In char 2, it still holds that the set of invariant bilinear forms is $\le 1$-dimensional, and nonzero iff $V\simeq V^*$; then it always consists of symmetric forms, but there is still a discussion whether it consists of alternating forms (where alternating means $b(x,x)=0$ for all $x$). $\endgroup$ – YCor Jun 11 '17 at 13:31
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    $\begingroup$ @YCor: In char $2$, if $V \cong V^*$ has dimension $> 1$ you can show that you always have a $G$-invariant alternating bilinear form. (follows using fact that if $b$ is a invariant symmetric form, then $\{ v \in V : b(v,v) = 0 \}$ is a submodule). Instead of invariant symmetric forms, I think here one should ask about invariant quadratic forms, and that seems to be a really difficult open problem (which is why I left it out of my question) $\endgroup$ – spin Jun 11 '17 at 13:37
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For one answer, here is a theorem due to Thompson and Willems (Bilinear forms in characteristic $p$ and the Frobenius-Schur indicator, Lecture Notes in Mathematics 1185, pg. 221-230).

For an irreducible self-dual $kG$-module $V$, set $\varepsilon(V) = 1$ if $G$ preserves a nonzero symmetric bilinear form on $V$, and $\varepsilon(V) = -1$ if $G$ preserves a nonzero alternating form on $V$.

Theorem: Let $V$ be an irreducible self-dual $kG$-module with Brauer character $\phi$. There exists an ordinary irreducible character $\chi$ such that

(1) $\chi$ is real valued.

(2) The decomposition number $d(\chi, \phi)$ is odd.

Furthermore, any ordinary irreducible character $\chi$ satisfying (1) and (2) has the property that $\nu_2(\chi) = \varepsilon(V)$.

So assuming you know the ordinary character table of $G$, and the Frobenius-Schur indicators and decomposition numbers of all irreducible characters; you can give an answer. I don't know if there are any other general results.

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The answer to the question can be found in Section 14, Frobeius-Schur indicator for Brauer characters, starting on page 320, of the book "Group Representations, Volume 4" by Gregory Karpilovsky. I don't have easy access to this book myself (our library only has Volume 1, and that is in two parts), but fortunately the Google Books preview is letting me see all 14 pages of that section.

It says on page 321, "It is natural to ask whether there is a Frobenius-Schur indicator for the Brauer character $\beta$ of $G$ afforded by $V$, which allows us to determine whether $V$ is of symmetric of skew symmetric type. It turns out that the answer is positive and the required indicator may be taken to be the Frobenius-Schur indicator of a certain irreducible ${\mathbb C}$-character of $G$ closely related to $\beta$."

The theory of all this follows and takes about 12 pages, so it is not for the faint-hearted.

A Google search for "Frobenius Schur indicator Brauer character" comes up with several relevant pointers to technical papers on this topic, so it seems to be known to specialists.

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    $\begingroup$ I haven't looked at that book for a while: is it possible that Karpilovsky is referring to the work of Thompson and Willems referred to in the answer above? In my experience, his books tend to write chunks of previously published work almost verbatim with minimal connecting narrative $\endgroup$ – Geoff Robinson Jun 14 '17 at 14:26
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    $\begingroup$ Yes, looking again at the part of the book I can see on google books, I see that Karpilovsky is attributing this result to Willems (1981) for the existence of $\chi$ satisfying (1) and (2) in the theorem above, and to Thompson (1984) for the connection with the form-type. (Ideally he would have made this attribution at the beginning of the section, rather than buried in the proofs.) $\endgroup$ – Derek Holt Jun 14 '17 at 14:51

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