Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm sure someone has already worked-out what all the relevant groups really are; my question is about how signature duality interacts with these groups.

So, by an awful calculation, and choosing a convenient convention, $$ Cl_{3,1} \simeq M_{4\times 4} (\mathbb{R}) $$ and $$ Cl_{1,3} \simeq M_{2\times 2} (\mathbb{H}) $$ while the automorphism groups of these things as ordinary $\mathbb{R}$-algebras are (again, awful calculation) $$ G_{3,1} \simeq PGl_4(\mathbb{R})$$ and $$ G_{1,3} \simeq Gl_2(\mathbb{H}) / \mathbb{R}^\times.$$ My (probably very silly) questions are

  • Are these groups different?
    • If so, whatever does it mean?
    • If not, what's the right way to see their isomorphism?
  • Does this work for all $p,q$ between $Cl_{p,q}$ and $Cl_{q,p}$?
share|cite|improve this question
The calculation of automorphisms of those matrix algebras is noticeably simplified by the Skolem-Noether theorem which tells you that all automorphisms are inner. – Mariano Suárez-Alvarez Aug 9 '12 at 18:19
If your computations of automorphism groups are correct,, then it is clear that the groups are not isomorphic: one of the two is simple while the other has $SO_3$ as a non-trivial quotient. – Mariano Suárez-Alvarez Aug 9 '12 at 18:23
I would be interested in knowing whatever text/paper you are looking at that is discussing this. Thanks! – rschwieb Aug 9 '12 at 18:40
@rschwieb, I'm doing private research; the internet is my library, but there's much too much to know it all. – Jesse C. McKeown Aug 9 '12 at 18:49
@JesseC.McKeown and never enough time to learn it all, it seems! – rschwieb Aug 9 '12 at 18:53
up vote 2 down vote accepted

Too many comments to fit into a comment box:

(1) the initial determination of the isomorphism classes of those Clifford algebras need not be so awful, if one does a two-step induction, then just needing to know how the Hamiltonian quaternions arise, versus the split algebra.

(2) Clifford algebras of non-degenerate quadratic forms on even-dimensional vectorspaces over a field $k$ (at least in characteristic not $2$) are central, simple algebras over $k$ (this is by that same induction), so their centers are just the copies of the field, here $\mathbb R$. Then as Mauriano S.A. notes, the Skolem-Noether theorem shows that all automorphisms are inner. Thus, in either case, the automorphisms are units-of-algebra, modulo $\mathbb R^\times$.

Thus, unless I misunderstand your sense of $PGL(2,\mathbb H)$, your computation in the quaternion case is not correct: I would write the thing just as that $PGL(2,\mathbb H)$, but meaning that the "P" means quotient by the central copy of $\mathbb R^\times$, only, not by $\mathbb H^\times$. (Indeed, there is some problem with what the latter would mean.)

(3) The two groups are not isomorphic. Indeed, c. 1964 in the Indian J. Math, A. Weil's "Algebras with involutions and the classical groups" shows how to make essentially all the rational forms (over whatever groundfield) of the classical groups, meaning types A,B,C,D, excluding the "exceptional" ones $E_6,E_7,E_8, F_4,G_2$. I don't necessarily recommend that source, but it does exactly proceed to manufacture groups by essentially what you are doing.

In particular, the Clifford algebra construction over $\mathbb R$ produces simple algebras over $\mathbb R$. There are two such of each size, namely, the one using the quaternions, and the one not. The resulting groups will have the same dimension, and will be isomorphic when scalars are extended to $\mathbb C$, but they are not isomorphic over $\mathbb R$. These are groups of type $A$.

There are more $\mathbb R$-isomorphism classes of groups of type $A$, namely, unitary groups $U(p,q)$ with $p+q=n$. And $GL(n,\mathbb C)$, when complexified, is two copies of $GL(n,\mathbb C)$, so it fits into this family as well.

(4) How to show non-isomorphism? Many possibilities, but perhaps determining the "real split rank" is most intuitive: this means finding the largest copy of products of $\mathbb R^\times$ inside the things. The "split" version $GL(n,\mathbb R)$ is of rank $4$, while $GL(2,\mathbb H)$ is of rank $2$. Projectifying subtracts $1$ from both, so there is still a disparity. Non-isomorphic.

share|cite|improve this answer
Re "PGL(2,H)", yes, I re-traced my earlier calculations and found the problem with "/H*"; do I understand correctly that the two groups are different integrations of the same Lie Algebra, then? In any case, Thanks for the careful composition! – Jesse C. McKeown Aug 10 '12 at 3:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.