Timeline for What is the exterior derivative intuitively?
Current License: CC BY-SA 2.5
9 events
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| Jun 8, 2020 at 22:25 | comment | added | Dmitri Pavlov | @FallenApart: Actually, differential 1-forms are precisely Kähler differentials of C^∞(M) if you work with C^∞-rings instead of ordinary commutative rings. See, for example, ncatlab.org/nlab/show/Kähler+differential. | |
| May 26, 2015 at 16:14 | comment | added | Mariano Suárez-Álvarez | @FallenApart, ah. No , I do not mean that $\Omega^1(M)$ is the module of Kähler differentials of $C^\infty(M)$ (mostly, because it isn't! :) ) The operator $d:C^\infty(M)\to\Omega^1(M)$ can be characterized in terms of its functorial properties. This is surely done in detail in the book Natural Operations in Differential Geometry by Kolar, Michor and Slovak. | |
| May 26, 2015 at 15:40 | comment | added | Fallen Apart | @Mariano Suárez-Alvarez The sentence: "The map d:C∞(M)→Ω1(M) itself has a nice characterization as a universal derivation of the algebra C∞(M) of functions satisfying certain rather reasonable conditions" | |
| May 26, 2015 at 15:27 | comment | added | Mariano Suárez-Álvarez | @FallenApart, I don't understand exactly what statement you mean. | |
| May 26, 2015 at 14:24 | comment | added | Fallen Apart | @Mariano Suárez-Alvarez Do you mean that for some subalgebra $A$ of $C^{\infty}(M)$ and $\Omega^1(M)$ truncated to be $A-$module $(A,\Omega^1(M))$ is Kähler differential? Could you give some refrences to your statement? | |
| Feb 21, 2013 at 23:27 | comment | added | Mariano Suárez-Álvarez | Well, my point is that you need only visualize the component in degree zero, as the rest is simply formalities. | |
| Feb 21, 2013 at 21:33 | comment | added | Ben McKay | The definition of a graded derivation was originally just a natural generalisation of $d$, so this approach is almost circular, and I can't visualise it geometrically. | |
| Apr 11, 2010 at 19:18 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
added 235 characters in body
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| Apr 11, 2010 at 19:09 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |