I believe the following is probably well-known, but so far I couldn't find the answer by myself:
Let $\mathfrak g$ be a real (finite-dimensional) Lie algebra, and $\wedge^2\mathfrak g$ its second exterior power. Then we can define a linear map on decomposable bi-vectors, $$L\colon\wedge^2\mathfrak g\to\mathfrak g,\quad L(X\wedge Y)=[X,Y],$$ which extends by linearity to non-decomposable elements of $\wedge^2\mathfrak g$.
Question: Is $\ker L$ generated (i.e., spanned as a vector subspace) by decomposable elements? In other words, is it true that every element of $\ker L$ can be written as $\sum_i \lambda_i (v_i\wedge w_i)$, where $v_i,w_i\in \mathfrak g$ are such that $[v_i,w_i]=0$ and $\lambda_i$ are constants?
Are there any hypotheses on $\mathfrak g$ that guarantee that this property is satisfied (perhaps assuming $\mathfrak g$ is simple is enough)?
This holds for very basic examples like any abelian Lie algebra (trivially) and also $\mathfrak{su}(2)$ and $\mathfrak u(2)$. Apparently there is an argument in this post on why the above is true if $\dim \mathfrak g=3$. I suspect the answer might be "no" in general (since there are lots of different ways to write an element of $\mathfrak g$ as a bracket of two other elements), but perhaps under some extra condition on $\mathfrak g$ (apart from $\dim=3$) this property still holds. For instance, if $\mathfrak g$ is simple we know that $L$ is surjective (because $[\mathfrak g,\mathfrak g]$ is an ideal), so we even know that $\dim \ker L=\dim\wedge^2\mathfrak g-\dim\mathfrak g$. Then, some sort of counting argument could perhaps work to find a basis of $\ker L$ formed by decomposable elements of $\ker L$?