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.


1 Answer 1


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. $\Box$

  • $\begingroup$ 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? $\endgroup$ Jul 18, 2022 at 8:43
  • $\begingroup$ @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? $\endgroup$ Jul 18, 2022 at 9:33
  • $\begingroup$ @ 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. $\endgroup$ Jul 18, 2022 at 9:48
  • $\begingroup$ @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. $\endgroup$ Jul 18, 2022 at 10:22
  • $\begingroup$ 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? $\endgroup$ Jul 18, 2022 at 10:25

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