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Let $\mathcal{g}=\mathfrak{sl}_2(\mathbb C)$ be the simple lie algebra of $\text{SL}_2$, for any $\lambda \in \mathbb C$ one can consider the corresponding Verma module $M_{\lambda}$, which is infinite dimensional.

It's a direct computation that the dimension of $\text{Hom}_{g}(M_{\lambda},M_{\mu})$ is $1$ if $\lambda = \mu$, is $1$ if $\lambda+\mu=-2$ and $\mu \in \mathbb Z_{\geq 0}$, and is $0$ in other cases.

How to compute the first extension group $\text{Ext}^1(M_{\lambda},M_{\mu})$ inside category $\mathcal{O}$?
When is there a non-split extension of $M_{\lambda}$ by $M_{\mu}$?

I think a necessary condition is $\lambda+\mu=-2$ and $\lambda < \mu$, but don't know whether it's sufficient.

Andrews
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1 Answers1

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Indeed the condition is necessary, see Humphrey's book on BGG category $\mathcal O$ page 113.

We can assume $\lambda = -2, \mu = 0$. To compute this $\text{Ext}^1 $ we use the short exact sequence $$ 0 \to M(0) \to P(-2) \to M(-2) \to 0 $$ where $P(-2)$ is the projective cover of $M(-2)$. To see the existence of such a sequence, you just need to know that $\mathcal O$ has enough projectives and BGG reciprocity.

Taking the long exact sequence associated to $\text{Hom}(-,M(0))$ gives $$ 0 \to \text{Hom}(M(-2),M(0)) \to \text{Hom}(P(-2),M(0)) \to \text{Hom}(M(0),M(0)) \to \text{Ext}^1(M(-2),M(0) \to 0 $$

And since $\text{Hom}(M(-2),M(0)) \cong \Bbb C$ and $\text{Hom}(M(0),M(0)) \cong \Bbb C$ this implies that $\text{Hom}(P(-2),M(0)) \cong \Bbb C$ and $\text{Ext}^1(M(-2),M(0)) \cong \Bbb C$.

Remark : The non trivial class is represented by $P(-2)$.

Remark 2 : Since the short exact sequence is also a projective resolution since $P(0) = M(0)$ one could try to compute $\text{Ext}$ from the resolution, however it boils down to prove that $\text{Hom}(P(-2),M(0)) \cong \Bbb C$ and I don't know how to do it without the long exact sequence. I would be interested to see another argument.

Nicolas Hemelsoet
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