Suppose to have a linear irreducible unitary representation $\rho:G\rightarrow U(H)$ on a complex Hilbert space $H$ with $G$ a generic group. Let $A$ be an $\textit{anti}$-linear operator such that $$ A\rho(g)=\rho(g)A\ \ \ \ \forall g\in G $$ What can be said about the operator $A$? Does it hold anything like Schur's lemma?
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The part of Schur's lemma that continues to hold is that any such operator must be invertible or 0, if the representation is irreducible over the reals. I will make no assumption on complex (anti-)linearity from now on, but will assume that all operators are real linear.
The space of real operators commuting with $G$ must be a real division algebra, since it obviously contains the real numbers embedded as scalars. By the Frobenius theorem, there are only 3 possibilities for such algebras:
\begin{align*} &\mathbb{R}, \text{the real numbers}\\ &\mathbb{C}, \text{the complex numbers}\\ &\mathbb{H}, \text{the quaternions}\\ \end{align*} Indeed all of them are realized on real irreducible representations. The examples which admit anti-linear operators must be quaternionic. Amongst Lie groups, the smallest (non-finite) such example is $G=SU(2)$ acting on its 2-dimensional fundamental representation $\mathbb{C}^2$.
PS: Actually the 0-or-invertible property holds even if $H$ is not real-irreducible, since $A$ is a complex linear intertwining map between $H$ and the conjugate representation $\bar H$. It might not come from a division algebra in this case though.
answered Nov 25, 2014 at 14:54
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$\begingroup$ Possibly dumb question: Why can't this real division algebra be something huge, like $\mathrm{Frac} \ \mathbb{R}[x]$? The Gelfand-Mazur theorem rules this out if we restrict to something like compact operators, but the original question doesn't impose anything like this. Trying to figure this out in order to address mathoverflow.net/questions/212950. $\endgroup$ Commented Aug 6, 2015 at 17:55
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$\begingroup$ @DavidSpeyer I think the key idea is that complexifications of real irreps are more than just complex reps, they are complex reps with conjugation, which commutes with the group. Even if the complexification is non-irreducible, this conjugation will help you control the endomorphism ring of the original real irrep. Use this along with the fact that complex Hilbert irreps have only scalar intertwiners, and you should be able to arrive at the 3 finite dimensional possibilities. By the way, I think that was an excellent question, I recommend asking it properly on the site for a better answer. $\endgroup$ Commented Aug 7, 2015 at 0:36
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$\begingroup$ I figured out how to rule out huge division algebras, but I no longer believe that you can show the centralizer algebra is a division algebra. See mathoverflow.net/questions/214412 $\endgroup$ Commented Aug 10, 2015 at 1:17