Timeline for Clifford algebra as an adjunction?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 3, 2009 at 22:27 | comment | added | Qiaochu Yuan | Whoops. For some reason I thought I could get away with defining a linear functional to take value 1 at the identity 0 "orthogonal" to the identity, but of course this is nonsense... | |
| Dec 3, 2009 at 19:41 | comment | added | José Figueroa-O'Farrill | I am not sure if this is whole answer, but it seems to be in the right direction. First, $q(x) = - e^* x^2$ seems better, the way I have defined Clifford maps. My main concern is that $e^*$ is not canonically defined. Perhaps one has to add more structure to the algebras... | |
| Dec 3, 2009 at 19:39 | comment | added | Theo Johnson-Freyd | "its dual"? What is the dual to the identity? | |
| Dec 3, 2009 at 18:44 | history | edited | sdcvvc | CC BY-SA 2.5 |
added 30 characters in body
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| Dec 3, 2009 at 18:19 | comment | added | Qiaochu Yuan | Also, I'm not sure if it matters or not whether you want algebra homomorphisms to preserve the identity. | |
| Dec 3, 2009 at 18:11 | comment | added | Qiaochu Yuan | q(x) isn't a quadratic form. But I think if e denotes the identity and e* denotes its dual then defining q(x) = e* x^2 works. | |
| Dec 3, 2009 at 18:08 | history | answered | sdcvvc | CC BY-SA 2.5 |