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edited Feb 15 2011 at 0:25

F. Ladisch
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EDIT: Ben Webster's answer below still doesn't rule The statement turned out the possibility that a symplectic representation occurs twice in a product of two reps of the same kind, and I don't see how to do this.
For an irreducible $\mathbb{C}G$-module $V$, the type can also seen by looking at $\operatorname{End}_{\mathbb{R}G} be wrong in general (V)$: It's $M_2(\mathbb{R})$ if the type is realsee below), and so the quaternions if $V$ is quaternion. In any case, there original question is $\sigma=\sigma_V\in \operatorname{End}_{\mathbb{R}G}(V)$ with $vz\sigma = v\sigma \bar{z}$ for $z\in \mathbb{C}$ and $v\in V$. In the real case, we may assume $\sigma^2 = 1$, and in the quaternion case we may assume $\sigma^2=-1$. On the tensor product $U\otimes V$ of two modules, we have the element $\tau=\sigma_U\otimes \sigma_V$ with similar properties, and $\tau^2=\pm 1$some sense obsolete. I tried to use A more appropriate questions would have been why this for an argument, but without successFrobenius-Schur indicator grading is there in some (many?) cases.

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edited Feb 7 2011 at 20:45

F. Ladisch
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EDIT: For some hours I thought I had Ben Webster's answer below still doesn't rule out the possibility that a counterexample to this generalizationsymplectic representation occurs twice in a product of two reps of the same kind, but and I don't see how to do thiswas wrong.It seems to be difficult to find a group where all characters have values in a field
For an irreducible $\mathbb{F}$ \mathbb{C}G$-module $V$, the type can also seen by looking at $\operatorname{End}_{\mathbb{R}G} (V)$: It's $M_2(\mathbb{R})$ if the type is real, and such that the endomorphism ring of some simple quaternions if $\mathbb{F}G$-module V$ is a division ring other than quaternion. In any case, there is $\mathbb{F}$ or \sigma=\sigma_V\in \operatorname{End}_{\mathbb{R}G}(V)$ with $vz\sigma = v\sigma \bar{z}$ for $z\in \mathbb{C}$ and $v\in V$. In the quaternions over real case, we may assume $\mathbb{F}$.\sigma^2 = 1$, and in the quaternion case we may assume $\sigma^2=-1$. On the tensor product $U\otimes V$ of two modules, we have the element $\tau=\sigma_U\otimes \sigma_V$ with similar properties, and $\tau^2=\pm 1$. I tried to use this for an argument, but without success.

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edited Jan 25 2011 at 16:53

F. Ladisch
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Edit: NoFor some hours I thought I had a counterexample to this generalization, at least not directlybut this was wrong. Every character of the group$$ C_3 \ltimes (C_9\ltimes C_7) = \langle It seems to be difficult to find a , b, c \mid a^3= b^9= c^7 = 1, c^b = c^2, c^a= c, b^a=b^4 \rangle $$has group where all characters have values in the a field $\mathbb{Q}(\sqrt[3]{1}, \sqrt{-7})$, and there are 12 characters with Schur index 3 over this field (\mathbb{F}$ and degree 3). These divide into two classes according to their associated element in the Brauer group. Comparing with such that the universal grading endomorphism ring of some simple $\operatorname{Irr}(G)$, that \mathbb{F}G$-module is the grading by $\operatorname{Irr}(\mathbf{Z}(G))$, we see that the associated class in the Brauer group can not define a grading. (Here the center is $\langle b^3 \rangle division ring other than $, and there is also one character of trivial Schur index which restricts to a given nontrivial character of \mathbb{F}$ or the center.)quaternions over $\mathbb{F}$.

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edited Jan 25 2011 at 15:40

F. Ladisch
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Let $G$ be a finite group. Then the irreducible complex representations of $G$ come in three sorts: real, complex and symplectic=quaternionic. The type of an irreducible character $\chi$ can be read of from the Frobenius-Schur indicator $$ s_2(\chi) = \frac{1}{|G|}\sum_{g\in G} \chi( g^2 ) \in \{ 1,0,-1 \}. $$ Now the following seems to be true (and has been used by Noah Snyder in his interesting answer to another question) but I can't see why: Suppose all characters of $G$ have real values. (Equivalently, every element of $G$ is conjugate to its inverse.) Then it seems that the Frobenius-Schur indicator defines a grading of the irreducible representations, and thus of the character ring. This means that if $\chi$ and $\psi$ are irreducible characters with $s_2 (\chi) = s_2(\psi)$ , then all the irreducible constitutents of $\chi\psi$ have indicator $1$ , and if $s_2(\chi) = -s_2(\psi)$ , then all constituents of $\chi\psi$ have indicator $-1$ . Why is this actually true? Of course, for example in the first case, $\chi\psi$ is afforded by a real representation, so the symplectic representations must occur with even multiplicity. But why can they not occur at all?

Moreover, I would like to know if this generalizes to other fields than $\mathbb{R}$ , using elements of a Brauer group instead of the Frobenius-Schur indicator.
Edit: No, at least not directly. Every character of the group $$ C_3 \ltimes (C_9\ltimes C_7) = \langle a, b, c \mid a^3= b^9= c^7 = 1, c^b = c^2, c^a= c, b^a=b^4 \rangle $$ has values in the field $\mathbb{Q}(\sqrt[3]{1}, \sqrt{-7})$, and there are 12 characters with Schur index 3 over this field (and degree 3). These divide into two classes according to their associated element in the Brauer group. Comparing with the universal grading of $\operatorname{Irr}(G)$, that is the grading by $\operatorname{Irr}(\mathbf{Z}(G))$, we see that the associated class in the Brauer group can not define a grading. (Here the center is $\langle b^3 \rangle $, and there is also one character of trivial Schur index which restricts to a given nontrivial character of the center.)

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asked Jan 24 2011 at 21:13

F. Ladisch
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Finite groups in which every character has real values: grading the representations