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In Humphrey's BGG Category $\mathcal{O}$ Exercise 3.3 he asks:

Does the short exact sequence below always split? $$ 0 \rightarrow M(\lambda) \rightarrow E \rightarrow M(\mu)^\check{} \ \rightarrow 0.$$

Having tried to show this it may possible to prove that it does always split if the map $M(\mu)^\check{} / L(\mu) \cong M(\nu)$ for $\nu$ some lower weight than $\mu$. However I feel the statement is not true in general - both the isomorphism and the splitting of the sequence.

Any help would be greatly appreciated, thanks.

Rhory Ashworth
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    Obviously the sequence could split. It should not always do so though (I just need to try to modify the example I have in mind since that is in the analogous setting of an algebraic group in positive characteristic). A general idea might be to pick $\mu$ such that the Verma is simple (and hence self-dual). – Tobias Kildetoft May 09 '17 at 11:04

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Let the Lie algebra be $\mathfrak{g}:=\mathfrak{sl}_2(\mathbb{C})$ spanned by $\{h,x,y\}$ with $[x,y]=h$, $[h,x]=+2x$, and $[h,y]=-2y$. The category $\mathcal{O}$ for $\mathfrak{g}$ is defined with respect to the Borel subalgebra $\mathfrak{b}:=\mathbb{C}h\oplus \mathbb{C}x$ and with the Cartan subalgebra $\mathfrak{h}:=\mathbb{C}h$. Identify $\mathfrak{h}^*$ with $\mathbb{C}$ via $\lambda\mapsto \lambda(h)$ for every $\lambda\in\mathfrak{h}^*$.

Consider the Verma module $\mathfrak{M}(-2)$. It is simple (i.e., $\mathfrak{M}(-2)\cong\mathfrak{L}(-2)\cong \mathfrak{M}^\vee(-2)$). Tensoring $\mathfrak{M}\left(-1\right)$ (which is a projective module as $-1$ is a dominant weight) with the $2$-dimensional simple module $\mathfrak{L}\left(1\right)$ yields the projective cover $\mathfrak{P}(-2)$ of $\mathfrak{L}(-2)\cong\mathfrak{M}(-2)$. That is, we have a short exact sequence $$0\to \mathfrak{M}(0) \to \mathfrak{P}(-2) \to \mathfrak{M}(-2)\to 0\,,$$ which does not split. Taking the duality of this exact sequence and noting that $\mathfrak{M}(-2)\cong \mathfrak{M}^\vee(-2)$, we obtain the non-splitting exact sequence $$0\to \mathfrak{M}(-2) \to \mathfrak{I}(-2) \to \mathfrak{M}^\vee(0)\to 0\,,$$ where $\mathfrak{I}(-2)=\mathfrak{P}^\vee(-2)$ is the injective hull of $\mathfrak{M}(-2)$.

Batominovski
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    You don't even need to dualize in the end, since the first sequence is also of the desired form given that the final term is self-dual. Also (though this may just be my preference), I would find this answer much easier to read if it used the more standard symbols for the modules (i.e. the regular font rather than the Gothic). This would also make it consistent with the notation used by the OP. – Tobias Kildetoft May 09 '17 at 13:30
  • Is $M(-1) \otimes L(1)$ the projective cover $P(-2) (= \mathfrak{P}(-2)$ here) because of the following? - The weights of the tensor are the sum of the weights of the two modules, so ${-1, -3, -5, \ldots } + {\pm 1}$. So the weight $-2$ and all lower weights are included and thus must cover the simple module $L(-2)$. So we must show the cover is essential (by uniqueness of projective covers). $L(1)$ is simple, so the only possible submodules would be $M' \otimes L(1)$, $M' \subset M(-1)$. But such a tensor would fail to cover $L(-2)$, as we would lose some weight spaces doing this. – Rhory Ashworth Aug 18 '17 at 11:37