# Does every injective morphism of $\mathfrak{g}$-modules have a left inverse?

Let $$\mathfrak{g}$$ be a finite dimensional Lie algebra and let $$M$$ and $$N$$ be (not necessarily finite dimensional) $$\mathfrak{g}$$-modules over a field $$k$$. Let $$f: M\to N$$ be an injective $$k$$-linear map of $$\mathfrak{g}$$-modules.

We know that $$f$$ has a left inverse in the category of $$k$$-vector spaces but does it have a left inverse in the category of $$\mathfrak{g}$$-modules? Thank you for a proof or a simple counterexample.

By the splitting lemma this is true iff the short exact sequence

$$0 \to M \to N \to N/M \to 0$$

splits. If this is true for every $$M, N$$ then the category of $$\mathfrak{g}$$-modules must be semisimple, which I think never happens unless $$\mathfrak{g} = 0$$ (not even if $$\mathfrak{g}$$ is semisimple; that only guarantees semisimplicity of finite-dimensional representations, and maybe that even requires hypotheses on the field?).

In any case it's easy to give a counterexample: let $$\mathfrak{g} = k$$ be the one-dimensional abelian Lie algebra, so that a $$\mathfrak{g}$$-module is equivalently a $$k[x]$$-module ($$k[x]$$ is the universal enveloping algebra). Then, for example, the $$2$$-dimensional representation given by the Jordan block

$$x \mapsto \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$$

is a nontrivial extension of the trivial representation by itself, hence corresponds to a short exact sequence $$0 \to k \to V \to k \to 0$$ which does not split.

Edit: Here's a proof of the general claim.

Claim: If the category of $$\mathfrak{g}$$-modules is semisimple then $$\mathfrak{g} = 0$$.

Proof. Since the category of $$\mathfrak{g}$$-modules is equivalently the category of $$U(\mathfrak{g})$$-modules, we want to show that the universal enveloping algebra $$U(\mathfrak{g})$$ is semisimple iff $$\mathfrak{g} = 0$$. The PBW theorem implies that $$U(\mathfrak{g})$$ is an integral domain; by the classification of semisimple algebras it follows that $$U(\mathfrak{g})$$ is semisimple iff it is a division algebra over $$k$$. But $$U(\mathfrak{g})$$ admits an augmentation homomorphism to $$k$$ given by sending every element of $$\mathfrak{g}$$ to zero; hence if it's a division algebra over $$k$$ it must be $$k$$ itself. $$\Box$$

• Thank you! This is a very nice answer. Thank you for the counterexample! To prove that the short exact sequence $0\to M\to N\to N/M\to 0$, we do not have to assume that $N/M$ can be identified with a subspace of N, right? I am asking because if M and N are infinite dimensional Lie algebras, then N/M might not be a subspace of $N$. In the special case of an infinite dimensional Lie algebra $\mathfrak{g}$ and an infinite dimensional Lie subalgebra $\mathfrak{n}$, does the short exact sequence $0\to\mathfrak{n}\to\mathfrak{g}\to\mathfrak{g}/\mathfrak{n}\to0$ always split? What fails if not? Jul 18, 2022 at 8:43
• @Flavius: no, we don't have to assume that; you can take a look at the statement of the splitting lemma for more clarification. In what sense do you want that short exact sequence to split? As vector spaces? As Lie algebras? Jul 18, 2022 at 9:33
• @ Qiaochu, Thank you for the confirmation. Yes, I looked at the statement of the splitting lemma. I want the short exact sequence of $\mathfrak{n}$-modules $0\to\mathfrak{n}\to\mathfrak{g}\to\mathfrak{g}/\mathfrak{n}\to$ to split in the category of $\mathfrak{n}$-modules or equivalently in the category of $U(\mathfrak{n})$-modules ($U(\mathfrak{n})$ is the universal enveloping algebra). I think that in the category of vector spaces, it always splits. Jul 18, 2022 at 9:48
• @Flavius: I don't expect that sequence to split in $\mathfrak{n}$-modules in general but I don't know a counterexample off the top of my head. Jul 18, 2022 at 10:22
• Ok, thanks. However, if it would split, would be the isomorphism $\mathfrak{g}\cong\mathfrak{n}\oplus\mathfrak{g}/\mathfrak{n}$ be automatically $\mathfrak{n}$-equivariant? Jul 18, 2022 at 10:25