irreducible subgroup of SL(n,R)

Question

Suppose a subgroup of SL(n,R) is irreducible; i.e. R^n contains no proper invariant real subspaces except {0}. Then is it irreducible as a subgroup of SL(n,C)? i.e. Does C^n contain no proper invariant complex subspace except {0}?

Answer 1

A representation $\rho$ over a field $K$ is called absolutely irreducible if for any algebraic field extension $L/K$, the representation $\rho\otimes_K L$ obtained by extension of scalars is irreducible (over $L$). It is enough to check this for the algebraic closure. As damiano's examples in the comments show, this is a much stronger property than irreducibility. Serre's "Linear representations of finite groups" contains a criterion for a real representation of a finite group to be absolutely irreducible.

Here is a way in which non absolutely irreducible representations typically arise. Let $L/K$ be a finite separable field extension of degree $d>1$ and $\sigma$ be an irreducible $n$-dimensional representation of $G$ over $L.$ By restriction of scalars, we obtain an $nd$-dimensional representation $\rho$ of $G$ over $K.$ (In the language of linear group actions, the representation space, which is a vector space over $L,$ is viewed as a vector space over $K$). The representation $\rho$ is not absolutely irreducible because $\rho\otimes_K L$ is isomorphic to the direct sum of $d>1$ Galois conjugates of $\sigma.$ Yet $\rho$ is frequently irreducible. For example, under the restriction of scalars from $\mathbb{C}$ to $\mathbb{R}$, the group $U(1,\mathbb{C})$ becomes $SO(2,\mathbb{R}).$ Therefore, any one-dimensional complex unitary representation (i.e. a character) $\sigma$ of a group $G$ gives rise to a two-dimensional real orthogonal representation $\rho$ whose complexification splits into a direct sum of two representations. This is the construction behind damiano's second and third examples.