I have come across this problem earlier, so let me add what I know. I agree with you and ndkrempel that Ben Webster's argument implies only that a symplectic irrep can occur with an even multiplicity in the product of two real irreps or of two symplectic irreps.

[Here I should add a short historical remark. Wigner basically used a similar argument to prove that for simply reducible finite (or even compact) groups the product of two irreps with the same FS-indicator cannot contain a symplectic representation. The term "simply reducible" means that the group has (i) no complex irreps, (ii) and the tensor product of any two irreps decomposes into a sum of irreps with multiplicities not exceeding 1. See E. P. Wigner, "On the Matrices Which Reduce the Kronecker Products of Representations of Simply Reducible Groups", reprinted in: L. C. Biedenharn and H. van Dam, Quantum Theory of Angular Momentum, Academic Press, New York (1965).]

But I don't think that the original form of the statement, that for finite groups with no complex irreps it is generally true that the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, i.e., if the irreps $D_A$ and $D_B$ have the FS indicators $s_2^{A}$ and $s_2^B$, then $D_C$ can only be contained in $D_A \times D_B$ if $s_2^C =s_2^A \cdot s_2^B$. We only know that if $s_2^C \ne s_2^A \cdot s_2^B$ then the multiplicity of $D_C$ in $D_A \times D_B$ has to be even.

However, and here comes maybe the only new thing that I can add to the discussion, this latter statement can also be extended to groups with complex irreps (i.e., having FS indicator $0$). Namely, for every irrep $D_K$ of a group let us introduce a "new" indicator $\tilde{s}^K$, such that it is equal to $s_2^K$ \tilde{s}^K=s_2^K$if$s_2^K= \pm 1$, and is equal to$1$\tilde{s}^K=1$ if $s_2^K= 0$ (or, $\tilde{s}^K=-(s_2^K)^2+s_2^K+1$ if you want :) ). It turns out, that again $\tilde{s}^C \ne \tilde{s}^A \cdot \tilde{s}^B$ implies that the multiplicity of $D_C$ in $D_A \times D_B$ is even. (A proof of this in the context of braided, semi-simple sovereign categories can be found here.)

And finally, a small personal remark :). The whole discussion started from Noah Snyder's remark on a previous question. He stated that $Rep(G)$ is graded basically by its center (this is true), and then he mentioned the non-trivial grading your question refers to, i.e. when there exist symplectic irreps of the group. As I mentioned, I don't think this grading by the FS-indicator is always there for any group. What is true, on the other hand, is that if the the multiplicities of irreps in any product of two irreps are zeros or odd numbers (like in the case of simply reducible groups, where the multiplicity is 0 or 1), then the existence of a symplectic irrep indeed implies the existence of a non-trivial grading of $Rep(G)$ and hence implies a nontrivial center of $G$. I actually noted this down in a very short review of representation rings a few years ago, see here.

All the best, Zoltan

5 added 9 characters in body

I have come across this problem earlier, so let me add what I know. I agree with you and ndkrempel that Ben Webster's argument implies only that a symplectic irrep can occur with an even multiplicity in the product of two real irreps or of two symplectic irreps.

[Here I should add a short historical remark. Wigner basically used a similar argument to prove that for simply reducible finite (or even compact) groups the product of two irreps with the same FS-indicator cannot contain a symplectic representation. The term "simply reducible" means that the group has (i) no complex irreps, (ii) and the tensor product of any two irreps decomposes into a sum of irreps with multiplicities not exceeding 1. See E. P. Wigner, "On the Matrices Which Reduce the Kronecker Products of Representations of Simply Reducible Groups", reprinted in: L. C. Biedenharn and H. van Dam, Quantum Theory of Angular Momentum, Academic Press, New York (1965).]

But I don't think that the original form of the statement, that for finite groups with no complex irreps it is generally true that the Frobenius-Schur indicator gives a $\pm 1$ grading of $Rep(G)$, i.e., if the irreps $D_A$ and $D_B$ have the FS indicators $s_2^{A}$ and $s_2^B$, then $D_C$ can only be contained in $D_A \times D_B$ if $s_2^C =s_2^A \cdot s_2^B$. We only know that if $s_2^C \ne s_2^A \cdot s_2^B$ then the multiplicity of $D_C$ in $D_A \times D_B$ has to be even.

However, and here comes maybe the only new thing that I can add to the discussion, this latter statement can also be extended to groups with complex irreps (i.e., having FS indicator $0$). Namely, if we introduce for every irrep $D_K$ of a group the let us introduce a "new" indicator $\tilde{s}^K$ \tilde{s}^K$, such that it is equal to$s_2^K$if$s_2^K= \pm 1$, and is equal to$1$if$s_2^K= 0$. It turns out, then it is that again true$\tilde{s}^C \ne \tilde{s}^A \cdot \tilde{s}^B$implies that the multiplicity of$D_C$in$D_A \times D_B $is even. (A proof of this in the context of braided, semi-simple sovereign categories can be found here.) And finally, a small personal remark :). The whole discussion started from Noah Snyder's remark on a previous question. He stated that$Rep(G)$is graded basically by its center (this is true), and then he mentioned the non-trivial grading your question refers to, i.e. when there exist symplectic irreps of the group. As I mentioned, I don't think this grading by the FS-indicator is always there for any group. What is true, on the other hand, is that if the the multiplicities of irreps in any product of two irreps are zeros or odd numbers (like in the case of simply reducible groups, where the multiplicity is 0 or 1), then the existence of a symplectic irrep indeed implies the existence of a non-trivial grading of$Rep(G)$and hence implies a nontrivial center of$G$. I actually noted this down in a very short review of representation rings a few years ago, see here. All the best, Zoltan 4 added 241 characters in body Dear F. Ladisch, I have come across this problem earlier, so let me add what I know. I agree with you and ndkrempel that Ben Webster's argument implies only that a symplectic irrep can occur with an even multiplicity in the product of two real irreps or of two symplectic irreps. [Here I should add a short historical remark. Wigner basically used a similar argument to prove that for simply reducible finite (or even compact) groups the product of two irreps with the same FS-indicator cannot contain a symplectic representation. The term "simply reducible" means that the group has (i) no complex irreps, (ii) and the tensor product of any two irreps decomposes into a sum of irreps with multiplicities not exceeding 1. See E. P. Wigner, "On the Matrices Which Reduce the Kronecker Products of Representations of Simply Reducible Groups", reprinted in: L. C. Biedenharn and H. van Dam, Quantum Theory of Angular Momentum, Academic Press, New York (1965).] But I don't think that the original form of the statement, that for finite groups with no complex irreps it is generally true that the Frobenius-Schur indicator gives a$\pm 1$grading of$Rep(G)$, i.e., if the irreps$D_A$and$D_B$have the FS indicators$s_2^{A}$and$s_2^B$, then$D_C$can only be contained in$D_A \times D_B$if$s_2^C =s_2^A \cdot s_2^B $. We only know that if$s_2^C \ne s_2^A \cdot s_2^B $then the multiplicity of$D_C$in$D_A \times D_B$has to be even. However, and here comes maybe the only new thing that I can add to the discussion, this latter statement can also be extended to groups with complex irreps (i.e., having FS indicator$0$). Namely, if we introduce for every irrep$D_K$of a group the indicator$\tilde{s}^K$such that it is equal to$s_2^K$if$s_2^K= \pm 1$, and is equal to$1$if$s_2^K= 0$, then it is again true$\tilde{s}^C \ne \tilde{s}^A \cdot \tilde{s}^B$implies that the multiplicity of$D_C$in$D_A \times D_B $is even. (A proof of this in the context of braided, semi-simple sovereign categories can be found here.) And finally, a small personal remark :). The whole discussion started from Noah Snyder's remark in an other on a previous question(see the beginning of this question), which . He stated that$Rep(G)$is graded basically by its center (this is true), and one then he mentioned the non-trivial grading comes from if your question refers to, i.e. when there are exist symplectic irreps of a the group. As I mentioned, I don't think this grading by the FS-indicator is truealways there for any group. What is true, on the other hand, is that if the the multiplicities of irreps in any product of two irreps are zeros or odd numbers (like in the case of simply reducible groups)groups, where the multiplicity is 0 or 1), then the existence of a symplectic irrep indeed implies the existence of a non-trivial grading of$Rep(G)$and hence implies a nontrivial center of$G\$. I actually noted this down in a very short review of representation rings a few years ago, see here.

All the best, Zoltan