A converse to Whitehead's Second Lemma (and more)

Question

Let $k$ be an algebraically closed field of characteristic $0$ and let $\mathfrak{h}$ be a finite dimensional $k$-Lie algebra. I'm interested in knowing which (Lie algebraic) properties $\mathfrak{h}$ must possess so that $$(\wedge^{2}\mathfrak{h})^{\mathfrak{h}}:=\{f\in \wedge^{2}\mathfrak{h}: h\cdot f=0 \forall h\in \mathfrak{h}\}$$ (where $\cdot$ denotes the canonical action of $\mathfrak{h}$ on $\wedge^{2}\mathfrak{h}$) contains a $1$-dimensional (possibly trivial) $\mathfrak{h}$-submodule.

It is known (proved in, I think, Weibel's ``Introduction to Homological Algebra") that, for any such Lie algebra $\mathfrak{h}$, $$(\wedge^{2}\mathfrak{h})^{\mathfrak{h}}\cong H^{2}(\mathfrak{h})$$

So, we can say a couple of things:

If $\mathfrak{h}$ is semisimple, then $H^{2}(\mathfrak{h})=0$ (Whitehead's second Lemma), so $(\wedge^{2}\mathfrak{h})^{\mathfrak{h}}=0$. What about a converse (i.e. if $\mathfrak{g}$ finite dimensional Lie algebra and $H^{2}(\mathfrak{g})=0$, what can we say about $\mathfrak{g}?$)

If $\mathfrak{h}$ is nilpotent, then the trivial module exists as a $1$-dimensional $\mathfrak{h}$-module of $\wedge^{2}\mathfrak{h}$.

Does a classification of Lie algebras $\mathfrak{h}$ with 1-dimensional $\mathfrak{h}$-submodules in $(\wedge^{2}\mathfrak{h})$ exist?

Answer 1

A converse of Whitehead's first and second lemma has been recently studied by P. Zusmanovich, e.g., see here. One of the results is as follows:

Theorem 0.2 (A converse to the Second Whitehead Lemma). A finite-dimensional Lie algebra over a field of characteristic zero such that its second cohomology with coefficients in any finite-dimensional module vanishes, is one of the following:

(i) an one-dimensional algebra;
(ii) a semisimple algebra;
(iii) the direct sum of a semisimple algebra and an one-dimensional algebra.

If you only require that $H^2$ vanishes for the trivial representation, we have much less, of course. By Dixmier's result, such a Lie algebra cannot be nilpotent. It might be solvable, however, e.g., a solvable, non-nilpotentent strongly rigid Lie algebra. More details are in Zusmanovich's paper.