As in a previous question, letting $V = \mathbb{C}^4$ with basis $\{e_1, \dots, e_4\}$, and define the representation $\rho$ of $G = \operatorname{SL}_4(\mathbb{C})$ on $\bigwedge^4 V$ given by $g \cdot (v_1 \wedge v_2) = gv_1 \wedge gv_2$.
Set $\Omega = e_1 \wedge e_2 \wedge e_3 \wedge e_4$, and let $B$ be the non-degenerate symmetric bilinear form on $\bigwedge^2 V$ defined by $u \wedge v = B(u,v)\Omega$.
I know that $B(\rho(g) a, \rho(g)b) = B(a,b)$ and $B(d\rho(X)a,b) + B(a,d\rho(X)b) = 0$ for all $g \in G$, $X \in \mathfrak{sl}(4,\mathbb{C})$, and $a, b \in \bigwedge^2 V$.
I need to use the differential $d\rho$ to obtain a Lie algebra isomorphism $\mathfrak{sl}(4, \mathbb{C}) \cong \mathfrak{so}(\bigwedge^2 \mathbb{C}^4,B)$, using the fact that $\mathfrak{sl}(4, \mathbb{C})$ is a simple Lie algebra (a Lie algebra with no proper $\text{ad}(\mathfrak{g})$-invariant subspaces), and without using any knowledge of the general classification of simple Lie algebras/their Dynkin diagrams.
I am not sure how this hint is helpful though.
In trying to find some non-trivial $\text{ad}(\mathfrak{g})$-invariant subspace, I am not going anywhere.
The two identities involving $B$ only tell me that $B$ is invariant under $G$ and $\mathfrak{g}$, but I can draw no further conclusions.
Best Answer
The differential of any Lie group homomorphism is a lie algebra homomorphism.
You already have a Lie group homomorphism $\rho:\mathrm{SL}_4\mathbb{C}\to\mathrm{SO}(\Lambda^2\mathbb{C}^4)$, which means the differential must be a lie algebra homomorphism $\mathrm{d}\rho:\mathfrak{sl}_4\mathbb{C}\to\mathfrak{so}(\Lambda^2\mathbb{C}^4)$. You can check it is a nonzero map, so the kernel is proper, but $\mathfrak{sl}_4\mathbb{C}$ is simple so the kernel must be trivial. Then check dimensions to conclude this is a lie algebra isomorphism (after all, $\mathrm{d}\rho$ is a linear map!).