What is the natural action of $\mathfrak{sl}(4,\Bbb{C})$ on $\wedge^2 \Bbb{C}^4$?

Question

What is the natural action of $\mathfrak{sl}(4,\Bbb{C})$ on $\wedge^2 \Bbb{C}^4$? We know that $\wedge^2 \Bbb{C}^4$ is generated by $\{e_1 \wedge e_2, e_1 \wedge e_3, e_1 \wedge e_4, e_2 \wedge e_3, e_2 \wedge e_4, e_3 \wedge e_4\}$, which has six elements...so I'm not really sure what the natural action of $\mathfrak{sl}(4, \Bbb{C})$ would be...

Answer 1

If we regard $\mathfrak{sl}(4, \mathbb{C})$ as usual as the space of tracefree matrices in $M_4(\mathbb{C})$ and $\mathbb{C}^4$ as the space $M_{4 \times 1}(\mathbb{C})$ of $4 \times 1$ complex matrices, then the standard action of $\mathfrak{sl}(4, \mathbb{C})$ on $\mathbb{C}^4$ is just matrix multiplication, i.e., $$A \cdot z := Az.$$

Anytime you have two representations $V, W$ of a Lie algebra $\mathfrak{g}$, we can form a tensor product representation $V \otimes W$ of $\mathfrak{g}$, with action $$A \cdot (v \otimes w) = (A \cdot v) \otimes w + v \otimes (A \cdot w),$$ and so in our case, this defines a standard action of $\mathfrak{sl}(4, \mathbb{C})$ on $\otimes^2 \mathbb{C}^4$. This representation decomposes as $S^2 \mathbb{C}^4 \oplus \Lambda^2 \mathbb{C}^4$, and the standard action on the latter summand is just the restriction of the above action to that space. In particular, the action is defined by $$A \cdot \left(\sum z_a \wedge w_a\right) = \sum[(A \cdot z_a) \wedge w_a + z_a \wedge (A \cdot w_a)].$$

The above applies more or less immediately to any Lie algebra representation. But something very special happens in the case you asked about: It turns out that $\mathfrak{sl}(4, \mathbb{C})$ is isomorphic to $\mathfrak{so}(6, \mathbb{C})$ (this is one of the exceptional isomorphisms of low-dimensional semisimple Lie algebras). So, we can just as well view our action as some action of $\mathfrak{so}(6, \mathbb{C})$ on $\Lambda^2 \mathbb{C}^4 \cong \mathbb{C}^6$, and this turns out to be just the standard (orthogonal) action on $\mathbb{C}^6$.