In the coffee hour discussion today, Nick gave me an interesting explicit example of an exceptional isomorphism of Lie groups.

$G = \mathbb{SL}(4, \mathbb C) \to \mathbb{SO}(6, \mathbb C)$

is an isomorphism via the $\bigwedge^2$ map. Let me elaborate. The group $G$ acts naturally on $V = \mathbb C^4$. If $\{e_1, e_2, e_3, e_4\}$ is a basis of $V$, then a basis of $\bigwedge^2 V$ could be taken as $\{ e_1 \wedge e_2, e_1 \wedge e_3, \cdots, e_3 \wedge e_4 \}$. Thus, $\mathbb{GL}(\bigwedge^2 V)$ is $\mathbb {GL}(6, \mathbb C)$ and our $G$-action gives a monomorphism of groups:

$\bigwedge^2 : G=\mathbb{SL}(4,\mathbb C) \to \mathbb{GL}(\bigwedge^2 V) = \mathbb {GL}(6,\mathbb C)$.

The image must actually be inside $\mathbb{SL}(6,\mathbb C)$.

$(\; , \; ) : \bigwedge^2 V \times \bigwedge^2 V \to \mathbb C$,
$(e_i \wedge e_j , e_k \wedge e_l ) \mapsto e_i \wedge e_j \wedge e_k \wedge e_l$
is symmetric (two permutations) and preserves the $G$– action, i.e.,
$(g . \left(e_i \wedge e_j\right) , \left( g . e_k \wedge e_l \right) ) = (e_i \wedge e_j, e_k \wedge e_l)$.
Hence the image of $G$ in $\mathbb{GL}(6, \mathbb C) \simeq \mathbb{GL}(\bigwedge^2 V)$ must also preserve this symmetric bilinear form. Thus, $\bigwedge^2(G) \subseteq \mathbb{SO}(\bigwedge^2 V)$. By dimension consideration, they must be equal.