A two-dimensional irreducible real representation of an abelian Lie algebra

The map $\rho$ is real-linear. Since $\mathfrak{g}$ is abelian, $[at,bt]=0$, and since scalar multiples of $J$ commute, $[aJ,bJ]=0$. Hence $\rho ([at,bt])=[\rho (at),\rho (bt)]$ for all $a,b\in \mathbb{R}$, so [L1] makes $V$ a representation.

Suppose $W$ were a nonzero proper $\mathfrak{g}$-stable subspace of $V$. As $V$ has dimension two, $W=\mathbb{R}v$ for some $v\mathrm{\ne }0$. Stability gives $Jv=\lambda v$ for a real scalar $\lambda$. But $J^2=-I$, so $-v=J^{2}v={\lambda }^{2}v$, forcing ${\lambda }^2=-1$, impossible over $\mathbb{R}$. Thus no such $W$ exists, and [L2] shows that $V$ is irreducible.

The acting Lie algebra is abelian and the verified irreducible module has real dimension two, so it is the required counterexample to any one-dimensional conclusion over $\mathbb{R}$. Over $\mathbb{C}$, the same matrix acquires eigenlines for the eigenvalues $i$ and $-i$, pinpointing the missing field hypothesis. The finite calculation uses no choice principle.