# Partial generalisation to Whitehead's second Lemma

Let $k$ be an algebraically closed field of characteristic $0$ and let $\mathfrak{g}$ be a finite dimensional semisimple $k$-Lie algebra. By Whitehead's second Lemma, we know that $H^{2}(\mathfrak{g}, M)=0$ for any finite dimensional $\mathfrak{g}$-module $M$. Taking $k$ to be the trivial module, we have in particular that $H^{2}(\mathfrak{g},k):=H^{2}(\mathfrak{g})=0$.

It is also known (proved in, I think, Weibel's Introduction to Homological Algebra") that, for any finite dimensional $k$-Lie algebra $\mathfrak{h}$, $(\wedge^{2}\mathfrak{h})^{\mathfrak{h}}\cong H^{2}(\mathfrak{h})$, where $(\wedge^{2}\mathfrak{h})^{\mathfrak{h}}:=\{f\in \wedge^{2}\mathfrak{h}: h\cdot f=0 \forall h\in \mathfrak{h}\}$ (the invariants under the canonical action of $\mathfrak{h}$ on $\wedge^{2}\mathfrak{h}$).

Combining the above, we find that for our finite dimensional semisimple Lie algebra $\mathfrak{g}$, $(\wedge^{2}\mathfrak{g})^{\mathfrak{g}}=0$. In particular, $(\wedge^{2}\mathfrak{g})$ can contain no 1-dimensional $\mathfrak{g}$-submodules.

I'm interested in whether or not there is any way to generalise this: given a finite dimensional simple Lie algebra $\mathfrak{g}$ and an (irreducible, say) $\mathfrak{g}$-module $V$, can the $\mathfrak{g}$-module $(\mathfrak{g}\wedge V+V\wedge \mathfrak{g})$ ever contain a 1-dimensional $\mathfrak{g}$-submodule?

• What does $g \wedge V$ mean? – Matthew Towers Nov 18 '14 at 21:17
• Sorry, by the space $(\mathfrak{g}\wedge V+V\wedge \mathfrak{g})$ I mean the subspace of skew-symmetric elements of $\mathfrak{g}\otimes V+V\otimes \mathfrak{g}$. – Paul Gilmartin Nov 18 '14 at 21:21
• Then isn't this just isomorphic to $\mathfrak{g} \otimes V$? The skew symmetric elements of $(\mathfrak{g} \otimes V) \oplus( V \otimes \mathfrak{g})$ are spanned by those of the form $(x\otimes y,-y\otimes x)$ and you may project this onto the first coordinate to get an iso with $\mathfrak{g}\otimes V$. – Matthew Towers Nov 19 '14 at 11:17
• See also this question. – Dietrich Burde Jan 5 '16 at 12:51