Exterior square of $\operatorname{Sp}(4,\mathbb{C})$ is isomorphic to $\operatorname{SO}(5,\mathbb{C})$ I am studying exceptional isomorphisms recently, which arise due to the coincidence in Dynkin diagram.
I saw two forms of expressing the exceptional isomorphisms, one is isomorphisms between the spin group and the corresponding group, established by spin or half-spin representations; the other is isomorphism from symmetric squares, exterior squares or tensor squares of the relevant group to the special orthogonal group. But in this case, I don't know how to construct the induced symmetric bilinear form on the vector space. And in the case ${\bigwedge}^2\operatorname{Sp}(4,\mathbb{C})\cong \operatorname{SO}(5,\mathbb{C})$, I think the dimension of exterior square of a 4-dimensional vector space is 6, and I am not clear how it gets mapped into $\operatorname{SO}(5)$.
 A: $\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Ker{Ker}$Let's start with the double cover $\SL(4) \to \SO(6)$. Let $V$ be the standard $4$-dimensional representation of $\text{SL}(4)$, and let $\bigwedge^2 V$ be its tensor square.  We have a non-degenerate symmetric bilinear form $\bigwedge^2 V \times \bigwedge^2 V \to \bigwedge^4 V \cong \mathbb{C}$ by $\langle \alpha, \beta \rangle = \alpha \wedge \beta$. (This is symmetric because even degree wedges commute with each other.) So the action of $\SL(4)$ preserves this symmetric bilinear form, and we get a map $\SL(4) \to \SO(6)$. This can be checked to be a double cover (the kernel is $\pm \mathrm{Id}$ and a computation with Lie algebras checks that the map is surjective).
Now, let $\omega : V \times V \to \mathbb{C}$ be a symplectic form and let $\Sp(4)$ be the subgroup of $\SL(4)$ preserving this form. Then $\omega$ induces a linear map $L_{\omega} : \bigwedge^2 V \to \mathbb{C}$, and $\Sp(4)$ preserves the $5$-dimensional space $\Ker(L_{\omega})$ inside $\bigwedge^2 V$. So $\Sp(4)$ maps to $\SO(5)$, sitting inside $\SO(6)$, and this is again a double cover.
