Timeline for Is there a way to embed Clifford algebras into the corresponding tensor algebra?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Aug 20, 2016 at 16:56 | vote | accept | Chill2Macht | ||
| Aug 1, 2016 at 16:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 2, 2016 at 15:40 | answer | added | Chill2Macht | timeline score: 6 | |
| Jul 2, 2016 at 14:34 | comment | added | Chill2Macht | @KConrad I think you are exactly right. Thank you very much for the reference to that thread; MTS's answer looks like exactly what I wanted. I was just thinking that we might be able to piggyback on the embedding of the exterior algebra regardless of the choice of quadratic form, since for geometric algebras over $\mathbb{R}^n$ the vector space basis is the same as that for the exterior algebra. Maybe this other answer by MTS is relevant mathoverflow.net/questions/60596/… | |
| Jul 2, 2016 at 9:55 | comment | added | KConrad | Might the embedding of a Clifford algebra into the endomorphisms of an exterior algebra, as described in the accepted answer for mathoverflow.net/questions/68378/clifford-algebra-non-zero, be helpful for you? | |
| Jul 2, 2016 at 7:23 | comment | added | Oscar Cunningham | @მამუკაჯიბლაძე Good point! On the other hand $\mathcal S(V)$ and $\Lambda(V)$ have canonical inclusions into it (when $\mathrm{char} k=0$), but these inclusions aren't algebra homomorphisms. | |
| Jul 2, 2016 at 5:44 | comment | added | მამუკა ჯიბლაძე | I don't think ${\cal T}(V)$ has any interesting finite-dimensional subalgebras. | |
| Jul 2, 2016 at 5:09 | history | edited | Chill2Macht | CC BY-SA 3.0 |
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| Jul 1, 2016 at 22:00 | history | edited | Chill2Macht | CC BY-SA 3.0 |
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| Jul 1, 2016 at 21:39 | history | edited | Chill2Macht | CC BY-SA 3.0 |
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| Jul 1, 2016 at 21:09 | comment | added | Chill2Macht | @OscarCunningham Oh I see what you are saying a little more, like $\mathcal{T}(V)\approx \mathcal{Cl}_q(V) \oplus \mathcal{I}_q(V)$? I guess my questions are then: how do we extend the inner product on $V$ defined by polarization of $q$ to all of $\mathcal{T}(V)$? And how do we explicitly identify the homorphism/quotient map of which $\mathcal{I}_q(V)$ is the kernel? | |
| Jul 1, 2016 at 20:29 | comment | added | Oscar Cunningham | There's a canonical quotient map $\mathcal T(V)\rightarrow \mathcal{Cl}_q(V)$. Since $\mathcal T(V)$ is an inner product space, I think that $\mathcal{Cl}_q(V)$ has to be isomorphic to the space orthogonal to the kernel of this map. | |
| Jul 1, 2016 at 20:20 | comment | added | Chill2Macht | @OscarCunningham I suppose we get a canonical inner product by taking the polarization of the quadratic form $q$. Could you elaborate on "taking the adjoint of the quotient map"? I don't understand. | |
| Jul 1, 2016 at 20:16 | comment | added | Oscar Cunningham | If $q$ is an inner product, do $\mathcal T(V)$ and $\mathcal{Cl}_q(V)$ get a canonical inner product? If so we could take the adjoint of the quotient map... | |
| Jul 1, 2016 at 19:44 | history | asked | Chill2Macht | CC BY-SA 3.0 |