Timeline for Show the Cartan 3-form transgresses to the Killing form in the Weil algebra
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2017 at 9:15 | comment | added | jdc | You're right! I'd forgotten the sign! But pretty much the entire construction of the Weil algebra is forced on you. Picking a connection, you have a map $\chi$ from the exterior factor on the $\theta^i$ into the 1-forms $\omega^i$ on your principal bundle, and you have derivations on either which don't match. The curvature is the difference $\Omega^i = (d\chi - \chi d)\theta^i$, so the only way to fix the map failing to be a DGA map is to say $\delta \theta^i = d \theta^i + u^i$ where $u^i$ is some new element. and $\chi(u^i) = \Omega^i$. Expanding out in coordinates gives the "minus" version. | |
| Jun 12, 2017 at 6:29 | comment | added | José Figueroa-O'Farrill | @jdc Alternatively, you could presumably redefine the differential in the Weyl complex by $F = dA - \tfrac12 [A,A]$ and then keeping the same definition of $x$, you get a different $y$ and hence a different $z$ which is now precisely the Cartan $3$-form. | |
| Jun 11, 2017 at 22:54 | comment | added | jdc | I've always heard that pair of names but never knew that was what their paper said! I calculated $dy$ and got the same signs. One reason for the sign mismatch might be either carelessness on the part of my source or that the "correct" choice is to replace $B$ with $-B$ so that the inner product is positive-semidefinite. Thanks! | |
| Jun 11, 2017 at 22:51 | vote | accept | jdc | ||
| Jun 11, 2017 at 12:00 | history | answered | José Figueroa-O'Farrill | CC BY-SA 3.0 |