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).

The quadratic form

(\; , \; ) : \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.

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