Timeline for Can we define exterior derivatives using pushforwards and connections?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 22, 2019 at 20:48 | comment | added | peter | @IvanIzmestiev in besse 1.12, why isn't the second sum 0? we can sum over $i<j$ instead of $i\neq j$ and replace the lie bracket by $[X_i,X_j]+[X_j,X_i]$, or am i missing something? | |
| Oct 9, 2016 at 17:24 | comment | added | Ivan Izmestiev | A covariant derivative on $T^*M$ defines a covariant derivative on $TM$, if we require the Leibniz rule for the canonical pairing: $\nabla(\omega(X)) = (\nabla\omega)(X) + \omega(\nabla X)$. So it does not matter whether you start with an affine connection on $T^*M$ or on $TM$. | |
| Oct 9, 2016 at 1:23 | comment | added | dorebell | Ivan - how did you get that $(\nabla_X \alpha)(Y) = \nabla_X(\alpha(Y)) - \alpha \nabla_X(Y)$? I'm considering $\nabla$ just as a vector bundle connection on the vector bundle $T^* M$, not necessarily as an affine connection on $TM$. | |
| Oct 9, 2016 at 1:19 | history | edited | dorebell | CC BY-SA 3.0 |
added 332 characters in body
|
| Oct 7, 2016 at 9:45 | comment | added | user40276 | @IvanIzmestiev Whoops! I misinterpreted the question. The OP actually wants to recover $d$ itself. | |
| S Oct 7, 2016 at 9:07 | history | suggested | Ivan Izmestiev | CC BY-SA 3.0 |
Replaced X at the beginning by M.
|
| Oct 7, 2016 at 9:06 | comment | added | Ivan Izmestiev | @dorebell What you are writing makes perfect sense. You may consult Besse, Einstein manifolds. Section 1.12 defines the exterior derivative of forms with values in a vector bundle by antisymmetrizing the covariant derivative. Section 1.19 provides a connection between $\nabla$ and $d$ in your context. It is a good exercise to check that the $\nabla$-definition of $d$ coincides with a more usual one. | |
| Oct 7, 2016 at 8:57 | review | Suggested edits | |||
| S Oct 7, 2016 at 9:07 | |||||
| Oct 7, 2016 at 8:54 | comment | added | Ivan Izmestiev | No, there is nothing missing. By Leibniz we have $(\nabla_X\alpha)(Y) = \nabla_X(\alpha(Y)) - \alpha(\nabla_XY)$. Hence $$(\nabla_X\alpha)(Y) - (\nabla_Y\alpha)(X) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha(\nabla_XY - \nabla_YX).$$ The last term is $\alpha$ evaluated on the commutator, provided that $\nabla$ is torsion free. This also answers the question what is going wrong if there is a torsion... | |
| Oct 7, 2016 at 6:27 | comment | added | user40276 | By reading better your question, it seems that the term $- \alpha ([X, Y])$ is missing in your $\omega_{\alpha}$. The general expression should be $d_{\nabla} \alpha (X_1, …, X_{k + 1}) = \sum_i (-1)^{i + 1} \alpha (X_1, …, \hat{X_i}, …, X_{k+ 1}) + \sum_{i > j}(-1)^{i + j} \alpha ([X_i, X_j], …, \hat{X_i}, …, \hat{X_j}, …, X_{k + 1})$ . This is actually the differential for the tangent Lie algebroid with values in a vector bundle (i.e, a representation of this Lie algebroid on a vector bundle). | |
| Oct 7, 2016 at 4:59 | answer | added | Sebastian | timeline score: 6 | |
| Oct 7, 2016 at 4:51 | comment | added | user40276 | I couldn't understand exactly your question, but any connection on a principal bundle can be written locally as $d + A$ where $A$ is a Lie algebra valued form. Furthermore you can extend the connection to higher forms by Leibniz rule in order to get a "complex" (not exactly a complex if your connection is not flat). | |
| Oct 7, 2016 at 3:27 | history | asked | dorebell | CC BY-SA 3.0 |