# How to prove $H^2(\mathfrak{g}, J(\mathfrak{g}))\neq0$, where $J(\mathfrak{g})$ is the augmentation ideal of $\mathfrak{g}$?

$\mathfrak{g}$ is a finite-dimensional semisimple Lie algebra over a field $k$ with $\mathrm{char}k=0$. $J(\mathfrak{g})$ is the augmentation ideal of $\mathfrak{g}$. That is, the kernel of $U(\mathfrak{g})\rightarrow k$. The solution to $H^2(\mathfrak{g},J(\mathfrak{g}))\neq0$ shows that a similar statement of Whitehead's second lemma doesn't hold for infinite dimensional $\mathfrak{g}-$module.(Whitehead's second lemma: With the assumptions above, $H^2(\mathfrak{g},M)=0$ if $M$ is a finite dimensional $\mathfrak{g}-$module).

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A dimension shifting argument shows that $\operatorname{Ext}^2_{U(\mathfrak{g})}(J, k) = \operatorname{Ext}^3_{U(\mathfrak{g})}(k,k)$ which is nonzero because of the Killing form. But $H^2(\mathfrak{g}, J)=\operatorname{Ext}^2_{U(\mathfrak{g})}(k,J)$ can't be dealt with this way. – m_t_ Dec 16 '13 at 18:30
repost @ MO: mathoverflow.net/q/83169 – Grigory M Dec 19 '13 at 19:10

This is false. I copy here my answer posted at the MO question linked by Grigory above: consider the example of $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$, and let $J$ be the augmentation ideal of $U(\mathfrak{g})$. Firstly $H^2(\mathfrak{g}, J) = \operatorname{Ext}^2_{U(\mathfrak{g})}(k,J) \cong \operatorname{Ext}^1_{U(\mathfrak{g})}(J,k)$ by Corollary 7.2 of Auslander-Gorenstein Rings by Ajitabh, Smith and Zhang. Then by dimension shifting $\operatorname{Ext}_{U(\mathfrak{g})}^1(J,k) = \operatorname{Ext}_{U(\mathfrak{g})}^2(k,k)$ which is zero by the Whitehead lemmas.