Timeline for When is a given matrix of two forms a curvature form?
Current License: CC BY-SA 3.0
13 events
when toggle format | what | by | license | comment | |
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S Apr 1, 2019 at 22:51 | history | suggested | Ali Taghavi |
I add two tags.
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Apr 1, 2019 at 21:07 | review | Suggested edits | |||
S Apr 1, 2019 at 22:51 | |||||
May 2, 2016 at 18:56 | history | edited | Ben McKay | CC BY-SA 3.0 |
formatting
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Aug 23, 2011 at 0:37 | vote | accept | Vamsi | ||
Aug 23, 2011 at 0:25 | vote | accept | Vamsi | ||
Aug 23, 2011 at 0:37 | |||||
Aug 23, 2011 at 0:08 | answer | added | Robert Bryant | timeline score: 25 | |
Aug 22, 2011 at 22:51 | history | edited | Willie Wong | CC BY-SA 3.0 |
removed redundant "enter code here"
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Aug 22, 2011 at 22:08 | comment | added | Robert Bryant | @David: You mean $F$, not $A$, and this closure is guaranteed by the quasi-Bianchi equation $dF = B\wedge F - F\wedge B$. | |
Aug 22, 2011 at 22:03 | comment | added | Vamsi | The Bianchi identity guarantees this ($dtr(F^k) = ktr(dF F^{k-1})=ktr([B,F]F^{k-1})=0$) | |
Aug 22, 2011 at 21:49 | history | edited | Gjergji Zaimi |
edited tags
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Aug 22, 2011 at 21:28 | comment | added | David E Speyer | A necessary condition: All the coefficients of the "characteristic polynomial" of $A$ must be closed differential forms. (If we were working over $\mathbb{C}$, these would be the Chern forms. Not sure what they're called over $\mathbb{R}$, but they should be closed either way.) | |
Aug 22, 2011 at 21:25 | history | edited | Vamsi | CC BY-SA 3.0 |
added 17 characters in body
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Aug 22, 2011 at 21:14 | history | asked | Vamsi | CC BY-SA 3.0 |