Timeline for Dual Clifford module
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2022 at 0:29 | comment | added | Nicholas Todoroff | See my answer here which constructs the canonical isomorphism ${\bigwedge}V \cong \mathrm{Cl}(V)$ without resorting to orthogonal bases (though I do use them to prove the map is bijective). I also want to point out that $\pi$ is actually just the normalized trace: define $\mathrm{tr}(x) = \mathrm{tr}(y \mapsto xy)$ for $x \in \mathrm{Cl}(V)$. Then $\pi(x) = \tfrac1{2^n}\mathrm{tr}(x)$ where $n$ is the dimension of $V$. | |
| Oct 28, 2022 at 18:29 | history | edited | LSpice | CC BY-SA 4.0 |
`\DeclareMathOperator`
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| Oct 28, 2022 at 18:09 | history | edited | domenico fiorenza | CC BY-SA 4.0 |
a silly characteristic zero vs. characteristic different from 2 inaccuracy fixed (thaks Vladimir)
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| Oct 28, 2022 at 17:56 | comment | added | domenico fiorenza | Argh, right! Actually it is true that there is a canonical linear isomorphism if char(k) is different from 2: one writes it in the other direction, i.e., $Cl(V)\to \wedge^\bullet V$ using an orthogonal basis and then showing it is actually basis independent; but if char(k)=0 one can more elegantly write it as $\wedge^\bullet V\to Cl(V)$ with no choice at all. Thanks! Correcting now. | |
| Oct 28, 2022 at 17:23 | comment | added | Vladimir Dotsenko | You divide by $k!$ so it is a bit strange that you just say that the characteristic is different from $2$ ;-) | |
| Oct 28, 2022 at 15:37 | history | answered | domenico fiorenza | CC BY-SA 4.0 |