Timeline for Spin equivariance of the Dirac operator-flat case
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2017 at 8:39 | comment | added | InfiniteLooper | Ok, can you write what the symbol is equivariant means ? It will be pretty much the same for the operator. | |
| S Nov 24, 2017 at 1:40 | history | bounty ended | CommunityBot | ||
| S Nov 24, 2017 at 1:40 | history | notice removed | CommunityBot | ||
| Nov 22, 2017 at 23:40 | comment | added | truebaran | Could you please rewrite it as an answer? I would also like to see some more details (how these actions are defined? Why the equivariance follows for Dirac once we have it for symbol? I'm asking for a standard answer not only for me but also in order to be able to award the bounty | |
| Nov 16, 2017 at 0:58 | comment | added | InfiniteLooper | It is right that given your $spin(n)$ actions the symbol is equivariant as a map $\mathbb R ^n \to End( \mathbb C ^{2^r})$. Dirac is then equivariant by considering the symbol as a map $C^{\infty}(\mathbb R ^n ; \mathbb C ^{2^r}) \to C^\infty(\mathbb R ^n; \mathbb C ^{2^r})$ but with the corresponding $spin(n)$ action : via $spin(n) \to SO(n)$ as you decribed for the first occurence of $\mathbb R^n$ but trivial for the second one. $Spin(n)$ also acts via Clifford on $\mathbb C ^{2^r}$ : $D(g.f\circ g^{-1}) = g.(Df)$ | |
| S Nov 16, 2017 at 0:07 | history | bounty started | truebaran | ||
| S Nov 16, 2017 at 0:07 | history | notice added | truebaran | Canonical answer required | |
| S Nov 10, 2017 at 2:05 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Added paragraph breaks. Shortened first paragraph. Add M.SE link.
|
| Nov 10, 2017 at 1:50 | review | Suggested edits | |||
| S Nov 10, 2017 at 2:05 | |||||
| Nov 10, 2017 at 1:10 | history | asked | truebaran | CC BY-SA 3.0 |