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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?

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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 $\mathsf{K}$.

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  • $\begingroup$ 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 $\operatorname{End}_{\mathfrak{g}}(\mathsf{K}) \cong \mathsf{K}$? $\endgroup$ – Adam Higgins May 7 at 16:37
  • $\begingroup$ @AdamHiggins Yes, exactly. $\endgroup$ – Jeremy Rickard May 7 at 18:01

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