Spin(4,1) = Sp(1,1) isomorphism

Question

I am interested in the exceptional isomorphism Spin(4,1) = Sp(1,1). The correspondance is already mentioned here: spin group Spin(4,1) but the explicit isomorphism is not given.

I would like to know where this isomorphism comes from.

Answer 1

Given an associative normed division algebra $\mathbb{K}$, we can give $\mathbb{K}^{p,q}$ a sesquilinear form

$$ \langle u,v\rangle=(\overline{u_1}v_1+\cdots+\overline{u_p}v_p)-(\overline{u_{p+1}}v_{p+1}+\cdots+\overline{u_{p+q}}v_{p+q}).$$

We use the physics convention of conjugate-linearity in the first argument, so that we can have scalars from $\mathbb{K}$ acting from the right and matrices over $\mathbb{K}$ applied from the left commuting with scalar multiplication. Let $x^\ast$ denote conjugate transpose. If we define the row vector

$$ \begin{bmatrix} x \\ y \end{bmatrix}^\dagger:=\begin{bmatrix} x^\ast & -y^\ast \end{bmatrix} $$

(where $x\in\mathbb{K}^p$ and $y\in\mathbb{K}^q$) then this becomes $\langle u,v\rangle=u^\dagger v$. Exercise: compute adjoints,

$$ \begin{bmatrix} A & B \\ C & D \end{bmatrix}^\dagger = \begin{bmatrix} A^\ast & -C^\ast \\ -B^\ast & D^\ast \end{bmatrix} $$

There is a corresponding matrix group $\mathrm{SU}(\mathbb{K}^{p,q})$ of all $X$ with $X^\dagger X=I$ which preserves $\langle -,-\rangle$.

For every unit vector $u$ there is an orthogonal projector $uu^\dagger$. It will be hermitian (i.e. self-adjoint with respect to this $X^\dagger$ operation.) We know $\mathrm{SU}(\mathbb{K}^{p,q})$ acts on $u$s, inducing $uu^\dagger\mapsto X(uu^\dagger)X^\dagger$ on the space of projectors. Indeed, since $X^\dagger=X^{-1}$ for $X\in\mathrm{SU}(\mathbb{K}^{p,q})$, this can be extended to a linear representation of $\mathrm{SU}(\mathbb{K}^{p,q})$ on the space of hermitian matrices by conjugation. Moreoever, this will preserve the trace form $\langle X,Y\rangle=\mathrm{tr}(X^\dagger Y)$.

If $p+q=1$ then the orthogonal projectors $uu^\dagger$ associated to unit vectors $u$ form the unit (quasi)sphere in the space of traceless hermitian matrices, albeit offset by $\frac{1}{2}I$. Examining the signature of this space of matrices with respect to the trace form, we have representations

$$ \begin{array}{|c|c|} \hline & (2,0) & (1,1) \\ \hline \mathbb{R} & \mathrm{SO}(2)\to\mathrm{SO}(2) & \mathrm{SO}(1,1)\to\mathrm{SO}(1,1) \\ \mathbb{C} & \mathrm{SU}(2)\to\mathrm{SO}(3) & \mathrm{SU}(1,1)\to\mathrm{SO}(2,1) \\ \mathbb{H} & \mathrm{Sp}(2)\to\mathrm{SO}(5) & \mathrm{Sp}(1,1)\to\mathrm{SO}(4,1) \\ \hline \end{array} $$