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Let $V$ be a finite-dimensional irreducible representation (complex or $\ell$-adic) of a group $G$ (compact Lie group or algebraic group etc.). Does there always exist a linear character $\rho$ of $G$, such that $V\otimes\rho$ is a self-dual irrep. of $G?$ Namely $V\otimes\rho\simeq(V\otimes\rho)^*.$ If not, is there any necessary/sufficient conditions on $V$ for it to be "twisted self-dual"?

If this is always the case, then in particular, if $G$ has no non-trivial linear characters (e.g. $G$ is a simply-connected compact Lie group or a perfect finite group), then every irrep. of $G$ is self-dual.

Thanks.

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The question isn't formulated precisely enough at this point to be answered directly. But in any case the complex irreducible representations of a simply connected compact Lie group certainly aren't always self-dual as one sees immediately in rank 2 examples. –  Jim Humphreys Jun 12 '11 at 22:33
    
Isn't the natural representation of $\mathrm{SL}(n)$ a counter example to the first question? –  Bruce Westbury Jun 12 '11 at 22:34
    
@Bruce: Yes, or similarly for compact groups such as $SU(n)$ in the context of the question and my comment. –  Jim Humphreys Jun 12 '11 at 23:35
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This already fails for finite groups. For instance, $PSL(2,7)$ has no nontrivial linear characters, but has two non-isomorphic 3-dimensional irreducible reps which are dual to each other. –  Kevin Ventullo Jun 13 '11 at 6:48

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up vote 9 down vote accepted

If you want a representation $V$ to be self-dual up to a character, then either $S^2V$ or $\Lambda^2V$ (considered as representations of $G$) should have a 1-dimensional summand (corresponding to the isomorphism $V \to V^*\otimes\chi$). But as it was mentioned by Jim there are a lot of representations $V$ for which both $S^2V$ and $\Lambda^2V$ are irreducible. For example, this is the case for $G = \operatorname{SL}(n)$ and $V$ the standard representation — the example suggested by Bruce.

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