# Kernel of the Lie bracket

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)
What do you mean by $\mathfrak{g}^{\ast}$? And why does the map $\mathfrak{g} \rightarrow \mathfrak{g}^{\ast} \otimes \wedge^{3}\mathfrak{g}$ have 0 kernel? – 109230 Apr 2 '12 at 15:41
$\mathfrak{g}^*$ = the dual space of $\mathfrak{g}$. The kernel vanishes: if you complete a $X\neq0$ to a base $X,Y,Z$ of $\mathfrak{g}$ then $\langle X,Y\wedge Z\rangle =X\wedge Y\wedge Z\neq0$. – user8268 Apr 2 '12 at 17:25