Do we have $(g \wedge g)^g = 0$? Let $g$ be a simple Lie algebra. Let $(g \wedge g)^g = \{a \wedge b \in g \wedge g: x.(a \wedge b) = [x,a] \wedge b + a \wedge [x,b] = 0\}$ be the set of $g$ invariants under the adjoint action.  Do we have $(g \wedge g)^g = 0$? If $g$ is a semisimple Lie algebra (respectively any Lie algebra), do we have $(g \wedge g)^g = 0$? Any help will be greatly appreciated!
 A: Edit: I am sorry, I was not careful with my first answer, which was wrongly suggesting that the question was about the second Lie algebra cohomology of ${\mathfrak g}$.
The Killing form identifies ${\mathfrak g}\wedge{\mathfrak g}$ with skew-symmetric bilinear forms $\mu: {\mathfrak g}\otimes{\mathfrak g}\to {\mathbb k}$; such a form is ${\mathfrak g}$-invariant precisely if it is associative in that $\mu([X,Y],Z)=\mu(X,[Y,Z])$. Now such associative forms correspond to ${\mathfrak g}$-module homomorphisms ${\mathfrak g}\to{\mathfrak g}^{\ast}$, and by the assumption that $\mathfrak g$ - hence also its dual - are simple ${\mathfrak g}$-modules, such are unique up to scalar. Now, the Killing form is one such morphism, and its associated form is symmetric; consequently, any associative form on ${\mathfrak g}$ must be symmetric, and we conclude that $({\mathfrak g}\wedge {\mathfrak g})^{\mathfrak g} = 0$. 
Remark: Is there a way to deduce the symmetry of associative forms without considering the Killing form?
