Representation theory of simple Lie algebras Let $V$ be a simple complex Lie algebra.
Let $W=\Lambda^2V$ be the second exterior power of $V$. 
Is it possible to find a basis for $W$ that consists of elements of the form
$v \wedge w$, where $v$ and $w$ are a pair of nilpotent elements of an $\mathfrak{sl}_2$ triplet?
 A: Yes, and in fact it suffices to use only the the $\mathfrak{sl}_2$-triple associated with the minimal nilpotent orbit of $V$. 
Let $G$ be the adjoint group of $V$, so that $V={\rm Lie}(G)$. Then $\wedge^2 V$ is spanned by all $G$-conjugates of
$e\wedge f$ where $e=e_{\alpha}$ and $f=e_{-\alpha}$ and $\alpha$ is the highest root of $V$ (we can choose any Borel subgroup of $G$).  Note that $V$ is an irreducible $G$-module where we assume that $G$ acts on $V$ via the adjoint representation. This also induces a natural action of $G$ on $\wedge^2 V$. Although $\wedge^2 V$ is no longer $G$-irreducible (if ${\rm rk}\, V\ge 2$) we still know that it is generated (as a $G$-module) by $e\wedge f$. Indeed, let $W$ be $G$-submodule generated by $e\wedge f$. We first apply positive root vectors to $e\wedge f$ to deduce that $W$ contains $e\wedge v$ for $v\in V$. After that we apply negative root vectors to $e\wedge v$ to deduce by induction (on height of roots involved in $u$) that $u\wedge v\in W$ for all $u,v\in V$.
Since any spanning set contains a basis of $\wedge^2V$, we are done. If $V=\mathfrak{sl}_2$ one just need three $G$-conjugates of $e\wedge f$, say $e\wedge f$, $({\rm Ad}\,g_1)(e)\wedge ({\rm Ad}\,g_1)(f)$ and $({\rm Ad}\,g_2)(e)\wedge ({\rm Ad}\,g_2)(f)$ for some $g_1,g_2$ in $G={\rm SL}_2$. There is plenty of choice for that.
