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Dec 28, 2016 at 19:00 comment added Marcel Bischoff Even easier, you can take consider the dual of Noah's finite group example. Then the representation category is equivalent to the symmetric category $\mathrm{Vec}_G$ of $G$-graded vector spaces in which every simple object is self-dual iff $G$ is an elementary abelian 2-group.
Nov 10, 2014 at 14:34 comment added Noah Snyder No problem. For that the easiest example is the trivial corep which is always selfdual.
Nov 10, 2014 at 9:22 comment added JP McCarthy I understand now why I didn't see how this answered my question. I was actually happy that irreducible corepresentations are not necessarily self-dual: what I was really asking was are they ever self-dual. As you said, I failed to think in enough detail about the classical case. After reading your answer, I now understand that there are real irreducible corepresentations --- such as anything on $A=F(S_n)$ --- that are self-dual. Thank you for your help and patience.
Nov 10, 2014 at 9:17 vote accept JP McCarthy
Nov 6, 2014 at 15:56 vote accept JP McCarthy
Nov 6, 2014 at 15:57
Nov 6, 2014 at 15:33 comment added Noah Snyder @JpMcCarthy: Ok, I rewrote it with more detail.
Nov 6, 2014 at 15:32 history edited Noah Snyder CC BY-SA 3.0
More detail
Nov 6, 2014 at 8:29 comment added JP McCarthy i.e. I don't see how this answers my question?
Nov 5, 2014 at 20:18 comment added JP McCarthy I have the matrix elements for $A=F(\mathbb{Z}_3)$ are $\rho_1=1_A$, $\rho_2=(1\,\,\,e^{2\pi i/3}\,\,\,e^{4\pi i/3})^T$ and $\rho_3=(1\,\,\,e^{4\pi i/3}\,\,\,e^{2\pi i/3})^T$. But $\kappa_2$ is not equivalent to $\kappa_3$. Is not $\overline{\kappa_1}=\kappa_2$? Is this (en.wikipedia.org/wiki/Frobenius%E2%80%93Schur_indicator) worth looking at?
Nov 5, 2014 at 18:56 history answered Noah Snyder CC BY-SA 3.0