# About the intertwiners of a real representation and its complex conjugate

i am currently trying to understand a proof in Trautman's "The Spinorial Chessboard", namely theorem 4.2 on page 48. It states the following:

If $\rho:\mathcal{A}\to\operatorname{End}_\mathbb{C} S$ is a complex, faithful and irreducible representation of a finite-dimensional, central simple algebra over $\mathbb{R}$ over a finite-dimensional complex vector space $S$, then there is a $\mathbb{C}$-linear isomorphism $C:S\to\overline{S}$ intertwining $\rho$ and its complex conjugate $\overline{\rho}$. Moreover, $C$ can be chosen so that $\overline{C}C=\pm\operatorname{id}_S$.

The proof is quite straightforward: The complexified algebra $\mathcal{A}_\mathbb{C}$ is also simple and the complexifications $\rho_\mathbb{C}$,$\overline{\rho}_\mathbb{C}$ are also irreducible. Therefore $\rho_\mathbb{C}$ and $\overline{\rho}_\mathbb{C}$ are equivalent via a $\mathbb{C}$-linear isomorphism $C:S\to\overline{S}$ and since $\mathcal{A}\subset\mathcal{A}_\mathbb{C}$, it is clear that $C$ also intertwines $\rho$ and $\overline{\rho}$. Moreover, $\overline{C}C$ is in the commutator of $\rho_\mathbb{C}$ (observe the canonical isomorphism $S\cong\bar{\bar{S}}$), which is equal to $\mathbb{C}\cdot\operatorname{id}_S$, by Schur's Lemma.

But now he assumes $\overline{C}C=\lambda\cdot\operatorname{id}_S$ for $\lambda\in\mathbb{R}$ (!), and i have no idea why this should be true. I suspect one should exploit the property of $\mathcal{A}$,$\mathcal{A}_\mathbb{C}$ being central, since this is not used anywhere else in the proof. I would be grateful if anyone would have some hint for me.

kind regards, Robert Rauch

Ok, I have found a solution: writing $\overline{C}C=\lambda\cdot\operatorname{id}_S$, we conclude $C\overline{C}=\overline{\lambda}\cdot\operatorname{id}_{\overline{S}}$. Multiplying this equation with $C$ from the right then gives $C\overline{C}C=\overline{\lambda}C$, but $C(\overline{C}C)=\lambda C$. Since $C\ne 0$ (we should require $S\ne 0$ for this, which is equivalent to $\mathcal{A}\ne 0$ in our setting), it follows $\lambda=\overline{\lambda}$.

Robert

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