Let $\mathbb H = \mathbf H \otimes\_{\mathbf R} \mathbf C$ be the tensor product of the quaternions with $\mathbf C$, and let $\mathcal J\_3(\mathbb H)$ denote the set of $\mathbb H$-hermitian $3 \times 3$ matrices (a $15$-dimensional vector space over $\mathbf C$). I have read that $\mathcal J\_3(\mathbb H)$ is isomorphic to $\bigwedge^2(\mathbf C^6)$ (as algebras), but I don't see what the map is.


What is it? (I'm not much of a representation theorist, so the more explicit the description the better.)



Jordan algebra of $3 \times 3$ quaternionic hermitian matrices
Choose an isomorphism $\iota: \mathbf{H}\otimes \mathbf{C} \simeq M\_2(\mathbf{C})$. For example, one such map is given by $a+bi+cj+dk \mapsto \left(\begin{array}{cc} a+ ib & c+id \\-c+id &a-ib\end{array}\right).$ A way to remember this is is $z + wj \mapsto \left(\begin{array}{cc} z & w \\-w^\* &z^\*\end{array}\right).$


Set $\epsilon$ the $2 \times 2$ matrix $\left(\begin{array}{cc}0 & 1 \\-1 &0\end{array}\right).$ Note that the quaterion conjugation on $M\_2(\mathbf{C})$ is given by $m \mapsto \epsilon m^{t} \epsilon^{-1}$.


Now, suppose $h = \left(\begin{array}{ccc}c\_1 & x\_3 & x\_2^\* \\ x\_3^\* & c\_2 & x\_1 \\ x\_2 & x\_1^\* & c\_3 \end{array}\right)$ is a Hermitian $3 \times 3$ matrix. Define $L(h) \in M\_6(\mathbf{C})$ by applying $\iota$ to every entry of $h$, so that


$L(h) = \left(\begin{array}{ccc}\iota(c\_1) & \iota(x\_3) & \iota(x\_2^\*) \\ \iota(x\_3^\*) & \iota(c\_2) & \iota(x\_1) \\ \iota(x\_2) & \iota(x\_1^\*) & \iota(c\_3) \end{array}\right)$


as a $6 \times 6$ matrix is $2 \times 2$ block form. Now define


$W(h) = \left(\begin{array}{ccc} \epsilon & & \\ & \epsilon & \\ &&\epsilon\end{array}\right) L(h)$.


It follows from the relation $\iota(x)^\* = \epsilon \iota(x)^t \epsilon^{-1}$ that $W(h)$ is antisymmetric, i.e., $W(h)^t = - W(h)$. The map $h \mapsto W(h)$ is what you seek.


