TITLE: Representation of $\operatorname{SL}_4$ on $\bigwedge^2 \mathbb{C}^4$
QUESTION [5 upvotes]: Letting $V = \mathbb{C}^4$ with basis $\{e_1, \dots, e_4\}$, the representation $\rho$ of $G = \operatorname{SL}_4(\mathbb{C})$ on $\bigwedge^4 V$ is 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 a bilinear form on $\bigwedge^2 V$ defined by $u \wedge v = B(u,v)\Omega$.
I can show that this is a non-degenerate symmetric bilinear form.
I now need to show that $$ \rho: \operatorname{SL}_4(\mathbb{C}) \to \operatorname{SO}\!\big(\!\textstyle\bigwedge^2 V, B\big)$$ is a group homomorphism.
However, I am not even sure that the codomain is correct - why does $\rho(g)$ preserve $B$?
Specifically, I should show that $B(\rho(g) a, \rho(g)b) = B(a,b)$ for all $g \in G$ and $a, b \in \bigwedge^2 V$.
If $g \cdot (a \wedge b ) = ga \wedge gb = a \wedge b$, then I will be done, but I don't think this can be true for all such $a, b \in \bigwedge^2 V$.
Otherwise I have tried decomposing the exterior products, but cannot see where to get equality of scalars.

REPLY [2 votes]: Setting $\eta=a\wedge b$ and $\nu=c\wedge d$ then
$$ g(\eta\wedge\nu)=(g\eta)\wedge(g\nu)=g(a\wedge b)\wedge g(c\wedge d)=(ga)\wedge(gb)\wedge(gc)\wedge(gd)=g(a\wedge b\wedge c\wedge d). $$
Exercise. Check that $g\xi=(\det g)\xi$ for all $\xi\in\Lambda^4\mathbb{C}^4$ and $g\in \mathrm{GL}_4\mathbb{C}$.
Note all $\xi$s are scalar multiples of the orientation element $\omega=e_1\wedge e_2\wedge e_3\wedge e_4$, so really this is the same as showing $g_1\wedge g_2\wedge g_3\wedge g_4=\det(g_1\,g_2\,g_3\,g_4)\omega$ for column vectors $g_i$.