Timeline for Representations of Pin vs. Representations of Clifford
Current License: CC BY-SA 4.0
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| Feb 24, 2023 at 1:27 | history | edited | darij grinberg | CC BY-SA 4.0 |
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| May 11, 2010 at 17:12 | comment | added | S. Carnahan♦ | Correction: My previous comment holds when $\Gamma$ is the Clifford group, which is not in general the group of units. Invertible elements need to satisfy a twisted conjugation condition to lie in the Clifford group, and this condition is typically nontrivial. Since the Clifford algebra is basically a matrix algebra, its group of units looks a lot like a general linear group (which has a lot more representations than a matrix algebra). | |
| May 11, 2010 at 15:43 | vote | accept | darij grinberg | ||
| May 11, 2010 at 14:24 | comment | added | darij grinberg | Thanks, David! I assumed that "representation of an algebra" is generally understood as "representation of the algebra" and not "representation of the unit group", because otherwise it would be a bit pointless to talk about representations of path algebras of acyclic quivers, and similar nilpotent stuff... I'm still wondering: does this work if the Clifford algebra is over a real vector space with a real, positive-definite inner product? | |
| May 11, 2010 at 13:23 | comment | added | S. Carnahan♦ | Representations of the unit group $\Gamma$ are representations of Pin with some rubbish attached. In general, you have a central extension $0 \to K^\times \to \Gamma \to O(V) \to 1$, and Pin is the subgroup of $\Gamma$ with spinor norm 1. In the case above, a representation of Pin becomes a representation of $\Gamma$ once you choose a commuting action of the multiplicative group of positive reals. | |
| May 11, 2010 at 12:52 | history | answered | David E Speyer | CC BY-SA 2.5 |