There are all sorts of curios in low-dimensional Lie groups and Lie algebras, many of them due to the presence of the quaternions. There is, I have recently learned, an isomorphism $SO(6,2) \simeq SO(4,\mathbb{H})$ (this *isn't* listed on Wikipedia, for instance). I'm curious to know 



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> is there is any corresponding exceptional isomorphism for $Spin(6,2)$
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analogous to the well-known isomorphism $Spin(5,1)\simeq SL(2,\mathbb{H})$? Clearly one cannot make a simple appeal to the Lie algebra isomorphism, but need to pull out some Clifford algebra or spin representation built from quaternions.


EDIT: bonus imaginary internet points if this exceptional isomorphism is compatible in some way the one for $Spin(5,1)$.



Exceptional isomorphism with Spin(6,2)?
To put the problem to a rest, I add my comment as an answer which is, in a nutshell, $Spin(6,2)\cong Spin(4,\mathbb H)$. The existence of this isomorphism follows from the isomorphism of the Satake diagrams and simple connectedness.


The tricky thing is the definition of a spin group over the quaternions which is explained in much greater generality in section 13 of "The book of involutions" by Knus at al.


The idea is roughly as follows: Given a quadratic space $(V,q)$ over a field $k$ (of characteristic $\ne2$, for simplicity) one constructs the Clifford algebra $Cl(V,q)$. The latter has a $\mathbb Z\_2$-grading $Cl=Cl\_0\oplus Cl\_1$. Then the spin group consists loosely speaking of all $c\in Cl\_0$ preserving $V\subseteq Cl\_1$.


One can do this construction in two steps. First, one assigns to $(V,q)$ the pair $(A,\sigma)$ where $A$ is the algebra $\mathrm{End}(V)$ and $\sigma\in\mathrm{End}(A)$ is taking the adjoint with respect to $q$. This way, one looses information, though: any scalar multiple $cq$ of $q$ gives rise to the same pair $(A,q)$. But the even part $Cl\_0$ depends only on $(A,q)$ and that's where the spin group lives in!


This observed, one can construct an algebra $Cl\_0$ from any pair $(A,\sigma)$ where $A$ is a central simple algebra over $k$ and $\sigma$ is an orthogonal (anti-)involution. This means that $\sigma|\_k=\mathrm{id}\_k$ and $\dim A\_+>\dim A\_-$ where $A\_\pm$ are the $\pm1$-eigenspaces of $\sigma$.


Since $\mathrm{End}(V)=V\otimes V^\*=V^{\otimes2}$, the general philosophy is that any construction involving only even tensor powers of $V$ can be done for any pair $(A,\sigma)$, as well. This is the case for $Cl\_0$. But also $D=V\otimes Cl\_1$ can be defined for any $(A,\sigma)$. Since it possible to formulate the condition "$c\in Cl\_0$ preserves $V\subseteq Cl\_1$" purely in terms of $D$ (and its $Cl\_0$-bimodule structure) alone one arrives at a definition of the group $Spin(A,q)$.


In the case at hand, one takes $k=\mathbb R$, $A=M(4,\mathbb H)$, and $\sigma(X)=a\overline X^ta^{-1}$ with $0\ne a\in\mathbb H$ with $\overline a=-a$. Usually, one works with $a=i$. Observe that conjugation $c\mapsto \overline c$ is a symplectic involution of $\mathbb H$. So, the condition $\overline a=-a$ makes $\sigma$ orthogonal.


EDIT: Let $L\_i$ and $\tilde L\_i$ be the Levi part of the $i$-th maximal parabolic $P\_i$ of $Spin(6,2)$ and $Spin(4,\mathbb H)$, respectively. Then looking at Satake diagrams the groups $L\_1,P\_1,\tilde L\_4,\tilde P\_4$ are defined over $\mathbb R$ and the exceptional isomorphism maps $L\_1$ isomorphically onto $\tilde L\_4$. So also the commutator subgroups $(L\_1,L\_1)=Spin(5,1)$ and $(\tilde L\_4,\tilde L\_4)=SL(2,\mathbb H)$ are mapped to each other.


EDIT2: One thing which puzzled me is the way $L\_1$ and $\tilde L\_4$ projected into $SO(6,2)$ and $SO(4,\mathbb H)$, respectively. In the first case, the image is $SO(5,1)\subseteq SO(6,2)$ in the second it is the $SL(2,\mathbb)\subseteq GL(2,\mathbb H)\subseteq SO(4,\mathbb H)$. The second group is simply connected, the first one is not! The only explanation for this phenomenon is that actually
$$
SO(6,2)\not\cong SO(4,\mathbb H).
$$
So what is the correct isomorphism? The center of $Spin(6,2)$ is the Klein 4-group. Let $c\_+,c\_-,c\_+c\_-$ be its trivial elements such that $Spin(6,2)/c\_+c\_-=SO(6,2)$. Then $Spin(6,2)/c\_\pm=Spin^\pm(6,2)$ are the two "halfspin groups". The correct isomorphisms therefore are
$$
Spin^+(6,2)\cong SO(4,\mathbb H)\cong Spin^-(6,2).
$$


