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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: The statement turned out to be wrong in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases.

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: The statement turned out to be wrong in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases.

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: Ben Webster's answer below still doesn't rule 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 $\operatorname{End}_{\mathbb{R}G} (V)$: It's $M_2(\mathbb{R})$ if the type is real, and the quaternions if $V$ is quaternion. In any case, there 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 $\tau^2=\pm 1$. I tried to use this for an argument, but without success.

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: The statement turned out to be wrong in general (see below), so the original question is in some sense obsolete. A more appropriate questions would have been why this Frobenius-Schur indicator grading is there in some (many?) cases.

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: For some hours I thought I had a counterexample to this generalization, but this was wrong. It seems to be difficult to find a group where all characters have values in a field $\mathbb{F}$ and such that the endomorphism ring of some simple $\mathbb{F}G$-module is a division ring other than $\mathbb{F}$ or the quaternions over $\mathbb{F}$.

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: Ben Webster's answer below still doesn't rule 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 $\operatorname{End}_{\mathbb{R}G} (V)$: It's $M_2(\mathbb{R})$ if the type is real, and the quaternions if $V$ is quaternion. In any case, there 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 $\tau^2=\pm 1$. I tried to use this for an argument, but without success.