faithful irreducible representation of linear algebraic group over reals

Question

Is it true that if a linear algebraic group defined over $\mathbb{R}$ has a faithful irreducible representation, then it is reductive?

Answer 1

Being defined over $\mathbf{R}$ is not really an issue (since $\mathbf{R}$ is perfect). In fact, a linear algebraic group $G$ over a perfect field $k$ is reductive if and only if the group $G_K$ obtained by base change is reductive, where $K$ is an algebraic closure of $k$. So you may as well ask the question for linear algebraic groups over $\mathbf{C}$, or any algebraically closed field $k$.

Now suppose that $V$ is a faithful irreducible representation over the algebraically closed field $k$. Our goal is to argue that the unipotent radical $R$ of $G$ is trivial.

Since $R$ is unipotent, the only irreducible linear representation of $R$ is the trivial representation $k$. It follows that the space of $R$-fixed points $V^R$ is non-zero (this space is just the socle of $V$ as an $R$-module, and the socle is always non-zero). Since $R$ is normal in $G$, the subspace $V^R$ is $G$-invariant. Since $V$ is an irreducible $G$-module, conclude that $V = V^R$, so that $R$ acts trivially on $V$. Since $V$ is a faithful representation for $G$, it now follows that $R$ is trivial, as required.