Timeline for Curvature of a Lie group
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2019 at 8:17 | comment | added | DLIN | Does this imply that the sectional curavture vanishes, i.e. choosing a coordinate $\{x_i\}$, such that $[x_i,x_j]=0$. | |
| Jul 10, 2010 at 15:27 | comment | added | rpotrie | Well, you are right (thanks!), of course it depends on the metric for Levi-Civita conections. The thing about the Lie group is that one can work directly in the Lie algebra and the connection "dissappers" there in the opearator $R$ which can be writen only using brackets. | |
| Jul 10, 2010 at 15:19 | comment | added | José Figueroa-O'Farrill | By Riemann above I mean the curvature operators $R(X,Y)$. Of course, the Riemann tensor $g(R(X,Y)Z,W)$ will. | |
| Jul 10, 2010 at 15:18 | comment | added | José Figueroa-O'Farrill | Well, "the curvature tensor does not depend on the metric" is not quite the case, right?! What happens is that the Levi-Civita connection is invariant under homotheties, hence a constant rescaling of the bi-invariant metric will not change Riemann. | |
| Jul 10, 2010 at 9:01 | comment | added | rpotrie | Yes, the metric is involved in the formula for the sectional curvature (notice that is one quarter of the square of the norm of the bracket). The curvature tensor, $R: \mathcal{X}(M) \times \mathcal{X}(M) \times \mathcal{X}(M) \to \mathcal{X}(M)$ does not depend on the metric (it is defined as $R(X,Y)Z=\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{[X,Y]} Z$. However, the sectional curvature (naturally) starts to depend on the metric: If $X$ and $Y$ are orthonormal, it is defined as: $\langle R(X,Y)X, Y \rangle$. | |
| Jul 10, 2010 at 7:16 | comment | added | Victor Protsak | I don't understand: the formula doesn't involve the metric, but if you rescale the metric, surely the curvature will rescale, too? | |
| Jul 9, 2010 at 21:43 | comment | added | José Figueroa-O'Farrill | Victor, the result is correct at is stands, except that $X,Y,Z$ are supposed to be left-invariant vector fields. | |
| Jul 9, 2010 at 21:26 | comment | added | Victor Protsak | I think it should involve the scalar product with $Z$, not the commutator. | |
| Jul 9, 2010 at 16:03 | vote | accept | Matt | ||
| Jul 9, 2010 at 15:21 | history | edited | rpotrie | CC BY-SA 2.5 |
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| Jul 9, 2010 at 15:02 | history | answered | rpotrie | CC BY-SA 2.5 |