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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.)

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1 Answer 1

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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.

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