Let $\mathfrak{g}$ be a dimension 3 Lie algebra and $[\quad,\quad]$ be a rank 1 map from $\bigwedge^{2}\mathfrak{g} \rightarrow \mathfrak{g}$. In this case, the kernel of $[\quad,\quad]$ is $3 - 1 = 2$ dimensional. Why does this mean that for some $X \in \mathfrak{g}$, the kernel consists of all vectors of the form $X \wedge Y$ with $Y$ ranging over all of $\mathfrak{g}$?
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Any $X\in\mathfrak{g}$ gives a linear form on $\wedge^2\mathfrak{g}$ with values in (1-dimensional) $\wedge^3\mathfrak{g}$; the form is given by $\langle X,\alpha\rangle:=X\wedge\alpha$. This linear map $\mathfrak{g}\to\mathfrak{g}^*\otimes\wedge^3\mathfrak{g}$ has $0$ kernel, so it is a bijection. Your kernel $V\subset\wedge^2\mathfrak{g}$ is 2-dimensional, its annihilator in $\mathfrak{g}$ is 1-dimensional, so there is $0\neq X\in\mathfrak{g}$ s.t. $V=\{\alpha\in\wedge^2\mathfrak{g}; X\wedge\alpha=0\}$. It certainly contains $W=\{X\wedge Y;Y\in\mathfrak{g}\}$, and as $\dim V=\dim W$, $V=W$. (or use the fact that the differential $X\wedge\cdot$ has vanishing cohomology) |
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