Whitehead's second Lemma and invariants of exterior square

Question

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 $\wedge^{2}\mathfrak{h}$).

EDIT: by $(V\wedge \mathfrak{g} +\mathfrak{g}\wedge V)$ I mean the subspace of skew symmetric elements of $V\otimes \mathfrak{g}+\mathfrak{g}\otimes V$.

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?

Answer 1

The way the question is formulated, it is trivial.

If $V\ne\mathfrak{g}$ ($\mathfrak{g}$ here being the adjoint module, in which case you already know everything), the corresponding sum is clearly direct, and moreover that subspace in question is isomorphic, as a $\mathfrak{g}$-module, to $\mathfrak{g}\otimes V$ (via $g\otimes v-v\otimes g\mapsto g\otimes v$). The latter is isomorphic to $\mathop{\mathrm{Hom}}(V^*,\mathfrak{g})$, and the invariant elements correspond to morphisms of $\mathfrak{g}$-modules. This implies, by Schur lemma, follows that for $V\not\simeq\mathfrak{g}^*$ there are no invariants. Finally, non-degeneracy of Killing form of course implies that $\mathfrak{g}^*\simeq\mathfrak{g}$, so $V\not\simeq\mathfrak{g}^*$ by assumption.