Timeline for Locally Riemannian Connection
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 18, 2018 at 13:02 | comment | added | Robert Bryant | @AurelianoSkirzewski: That's close. In dimension $n$, if you make the assumption that the curvature $2$-form $\rho=\mathrm{d}\gamma +\gamma\wedge\gamma$ satisfies the condition that there exists exactly one positive definite, determinant 1, symmetric matrix $H$ such that $H\rho$ is antisymmetric (which is a combination of open and closed first-order conditions on the curvature $\rho$), then you just need to check whether $$ \mathrm{d}H +\tfrac2n\,\mathrm{tr}(\gamma)H-H\gamma-{}^t\gamma H=0.$$ | |
| Jan 18, 2018 at 12:18 | comment | added | Aureliano Skirzewski | Thank you for your response, I suppose the higher dimensional set of conditions is equivalent to finding a non degenerated 0-form $H_{ij}$ such that the 2-form $H_{ki}R^i{}_j$ is antisymmetric in $k$ and $j$. That's enough to reduce the number of independent components of the curvature tensor to the components of a Riemann. It just rests to find when does $H$ exist and check which of the solutions for $H$ satisfies $$\mathrm{d}H + \mathrm{tr}(\gamma)\,H - H\gamma - {}^t\gamma H = 0.$$ Am I missing something? | |
| Jan 18, 2018 at 11:51 | vote | accept | Aureliano Skirzewski | ||
| Jan 18, 2018 at 9:57 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Corrected a counting error in the second derivative equations.
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| Jan 18, 2018 at 2:11 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Cleaned up some of the notation and put in some explanations of various things.
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| Jan 17, 2018 at 23:44 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added an explicit example.
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| Jan 17, 2018 at 15:17 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Clarified a few sentences that were a little vague.
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| Jan 17, 2018 at 14:36 | history | answered | Robert Bryant | CC BY-SA 3.0 |