Timeline for Convention on Clifford Product
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 29, 2016 at 15:43 | comment | added | Andreas Blass | The first advantage that comes to my mind for the $Q$ version of the definition is that it makes obvious the fact that the orthogonal group of $(V,Q)$ acts on the Clifford algebra by algebra-automorphisms. (This is a special case of the fairly general phenomenon that, in linear algebra, things tend to be easier to see when you don't make them depend on a particular basis.) | |
| Jun 29, 2016 at 13:34 | comment | added | abx | I was referring to your edit: yes, the quadratic form $Q$ is definitely useful. | |
| Jun 29, 2016 at 13:27 | comment | added | user21230 | OK, but $1$ and $-1$ exist in any field, aren't they ? Therefore simple definition used by Clifford and me :) still works. | |
| Jun 29, 2016 at 12:57 | comment | added | abx | Clifford algebras also exist (and are very useful!) over other fields than $\Bbb{R}$. | |
| Jun 29, 2016 at 12:49 | history | edited | user21230 | CC BY-SA 3.0 |
added 660 characters in body
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| Jun 29, 2016 at 10:19 | history | answered | user21230 | CC BY-SA 3.0 |