Schur's Lemma in Infinite Dimensional Lie Algebras

Let $$\mathfrak{g}$$ be a $$\mathsf{k}$$-Lie algebra, with $$(\rho, V)$$, $$(\sigma, W)$$ irreducible $$\mathfrak{g}$$-representations. Then the easy part of Schur's Lemma states that a $$\mathfrak{g}$$-linear map $$\phi : V \to W$$ is either the zero map or an isomorphism.

There is an extension to Schur's lemma which states that if $$V$$ or $$W$$ are finite dimensional, and $$\mathsf{k}$$ is algebraically closed, then $$\operatorname{Hom}_{\mathfrak{g}}(V,W)$$ is one-dimensional. One proves this by first considering the case $$V = W$$, and then since $$V$$ is finite dimensional, a linear endomorphism $$\phi \in \operatorname{End}_{\mathsf{g}}(V)$$ has a characteristic polynomial, which thus has a root $$\lambda$$ since $$\mathsf{k}$$ is algebraically closed. Thus there exists some $$0 \neq v \in V$$ with $$\phi(v) = \lambda v$$. It follows that $$\phi - \lambda\operatorname{Id}_V$$ is a linear endomorphism of $$V$$ with non-zero kernel, and so $$\phi - \lambda\operatorname{Id}_V = 0$$ by the easy part of Schur's lemma, and thus $$\phi = \lambda\operatorname{Id}_V$$. Now we drop the supposition that $$V=W$$, and suppose that $$\phi_1, \phi_2 \in \operatorname{Hom}_{\mathfrak{g}}(V,W)$$ are both non-zero. Then once again by the easy part of Schur's lemma, $$\phi_i$$ are both isomorphisms, and so both $$V,W$$ are finite dimensional. Thus $$\psi = \phi_1^{-1} \circ \phi_2$$ is a non-zero $$\mathfrak{g}$$-linear automorphism of finite dimensional $$V$$, and so by the above, there exists some $$\lambda \in \mathsf{k}$$ with $$\psi = \lambda \operatorname{Id}_V$$, and so it follows that $$\phi_2 = \lambda \phi_1$$, and so $$\operatorname{Hom}_{\mathfrak{g}}(V,W)$$ is one-dimensional.

My question here concerns the necessity of the assumption that $$V$$ or $$W$$ is finite dimensional. Clearly this style of proof will not work if $$V$$, or $$W$$ is infinite-dimensional, since then when we assume that $$V=W$$ we cannot guarantee that a linear automorphism of $$V$$ has an eigenvector, even when $$\mathsf{k}$$ is algebraically closed. This makes me suspect that the result does not hold when we drop this finite-dimensional assumption, but I'm not too sure. Could someone help me out with an example?

If you allow arbitrary $$\mathsf{k}$$-Lie algebras, then any field extension $$\mathsf{K}$$ of $$\mathsf{k}$$ can be the endomorphism algebra of some simple representation.
Let $$\mathfrak{g}=\mathsf{K}$$ considered as a commutative $$\mathsf{k}$$-Lie algebra (i.e., with zero Lie bracket). Then since $$\mathfrak{g}$$ is commutative as a Lie algebra and $$\mathsf{K}$$ is commutative as a ring, multiplication in $$\mathsf{K}$$ makes $$\mathsf{K}$$ into a $$\mathfrak{g}$$-module, which is easily checked to be irreducible and to have endomorphism algebra $$\mathsf{K}$$.
• Nice short answer! To be entirely clear, what we would want to do is take some algebraically closed field, $\mathsf{k}$, and take some transcendental extension of $\mathsf{k}$, $\mathsf{K} = \mathsf{k}(t)$ say, then $\mathsf{K}$ is an infinite degree extension of $\mathsf{k}$, and so the Abelian $\mathsf{k}$-Lie algebra you describe $\mathfrak{g} = \mathsf{K}$ is infinite dimensional, and we can endow $\mathsf{K}$ with the structure of a $\mathfrak{g}$-rep by multiplication, and then $\operatorname{End}_{\mathfrak{g}}(\mathsf{K}) \cong \mathsf{K}$? – Adam Higgins May 7 at 16:37