Let $\mathfrak{g}$ be a finite dimensional real Lie algebra and $\pi: \mathfrak{g} \to \mathfrak{gl}(V)$ be a homomorphism of real Lie algebras where $V$ is a finite dimensional real vector space. But $V$ sits inside $V_{\mathbb{C}}$ and $\mathfrak{gl}(V)$ sits inside $\mathfrak{gl}(V_{\mathbb{C}})$ giving us a representation of $\mathfrak{g}$ by $\pi_{\mathbb{C}}:\mathfrak{g} \to \mathfrak{gl}(V_{\mathbb{C}}) $. (Obtained by composing $\pi$ with the inclusion map.)

Clearly if $\pi$ is reducible then $\pi_{\mathbb{C}}$ is reducible. But is the converse true ? i.e. Does $\pi_{\mathbb{C}}$ reducible imply that $\pi$ is reducible ?

I do not think that the answer is yes, but I can not find a counterexample.

Note : By the reducibility of $\pi_{\mathbb{C}}$ I mean that there exists a complex invariant subspace of $V_{\mathbb{C}}$.


Let $G$ be the quaternion group of order 8, let $\mathbb{R}[G]$ be the group algebra over $\mathbb{R}$. This is a Lie algebra, with the Lie bracket defined by $[g,h]=gh-hg$. The standard complex representation $\rho$ of $G$ is not defined over $\mathbb{R}$, but $\rho\oplus\rho$ is, see e.g. Isaacs, Chapters 9-10, in particular Exercise 9.18. It follows that $\rho\oplus\rho$ is an irreducible representation of $\mathbb{R}[G]$ that becomes reducible after base change to $\mathbb{C}$.

Such guys are sometimes called quaternionic, sometimes symplectic. As long as you are in the world of finite groups, they can be detected using Frobenius-Schur indicators.


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