Exceptional isomorphism with Spin(6,2)?

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

is there is any corresponding exceptional isomorphism for $Spin(6,2)$

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 $Spin(5,1)$.

• I don't remember the answer off the top of my head, but I think you'll find what you want in F. Reese Harvey's book Spinors and Calibrations. I imagine that it's some form of triality, i.e., that you'll get something like an embedding of Spin(6,2) into the product of two copies of SO(4,$\mathbb{H}$), just as Spin(8) appears as embedded in the product of two copies of SO(8). – Robert Bryant Aug 2 '16 at 4:56
• @RobertBryant actually, digging a bit harder I get a suggestion that there is an isomorphism with Sp(4,H), for a sensible definition of this group, or USp(4,4;C) – David Roberts Aug 2 '16 at 5:10
• @RobertBryant And thank you for the pointer! – David Roberts Aug 2 '16 at 5:16
• Just out of curiosity: what is wrong with $Spin(6,2)\cong Spin(2,\mathbb H)$? – Friedrich Knop Aug 6 '16 at 8:13
• @FriedrichKnop what is $Spin(2,\mathbb{H})$? Can you point me to a definition? – David Roberts Aug 6 '16 at 8:27

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 $$Spin^+(6,2)\cong SO(4,\mathbb H)\cong Spin^-(6,2).$$

• Note that your final comment (i.e., that Spin(6,2) occurs embedded into a product of two copies of SO(4,$\mathbb{H}$)) is exactly what I suggested was the case in my comment to the original question. Of course, the projection of Spin(6,2) onto either SO(4,$\mathbb{H}$)-factor is a (nontrivial) double cover. – Robert Bryant Aug 6 '16 at 18:17
• That book, while no doubt extremely enlightening to some, leaves me feeling like when Arthur Dent asked for a cup of tea from the Nutri-Matic Drink Synthesizer: almost - but not quite - entirely unlike something I can understand. I'll try to digest it and see how I go. Presumably there's an inclusion SL(2,H) -> Spin(4,H) covering SO(5,1) -> SO(6,2) I can extract somehow... – David Roberts Aug 6 '16 at 22:27
• I agree that the book of involutions is complete overkill for your problem. If it helps, Tignol has a shorter version on his homepage (perso.uclouvain.be/jean-pierre.tignol/ChinaLectures.pdf) which I found quite readable. There must be an account for real groups somewhere. – Friedrich Knop Aug 7 '16 at 5:01
• @FriedrichKnop (belated) thanks for the pointer! – David Roberts Aug 25 '16 at 4:23