Timeline for Why is the exterior differentiation operator sometimes visualized as the "boundary"?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2021 at 8:33 | comment | added | user76284 | The link is broken. | |
| Mar 31, 2011 at 4:30 | answer | added | user13113 | timeline score: 0 | |
| Mar 31, 2011 at 3:54 | answer | added | Feri | timeline score: 1 | |
| Mar 30, 2011 at 13:42 | answer | added | diverietti | timeline score: 4 | |
| Mar 30, 2011 at 13:14 | comment | added | Joseph O'Rourke | You might learn from this earlier MO question: "Is the boundary $\partial S$ analogous to a derivative?" mathoverflow.net/questions/46252 . See especially Terry Tao's enlightening answer. | |
| Mar 30, 2011 at 12:09 | comment | added | Steve Huntsman | $d$ is really a coboundary operator--the formal adjoint of $\partial$. This fact in the context of the de Rham complex is the content of Stokes' theorem. | |
| Mar 30, 2011 at 12:03 | history | edited | Steve Huntsman | CC BY-SA 2.5 |
added 6 characters in body
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| Mar 30, 2011 at 11:35 | history | asked | Jean-Philippe Burelle | CC BY-SA 2.5 |