Timeline for $|sec_M| \leqslant C_1$, $|\nabla R|\leqslant C_2$, then frame bundle $|sec_{F(M)}|\leqslant C(C_1,C_2)$?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 3, 2018 at 19:46 | answer | added | Robert Bryant | timeline score: 5 | |
| Apr 30, 2018 at 14:36 | comment | added | Deane Yang | I'm pretty sure that the fibers are totally geodesic, so it's just the sectional curvature of the orthogonal group. The mixed ones are the most confusing. One way is to use the structure equations: $$d\omega^i + \omega^i_j\wedge\omega^j = 0,\ \ d\omega^i_j + \omega^i_k\wedge\omega^k_j = 0$$ The $1$-forms $\omega^i, \omega^i_j$ form an orthonormal frame of $1$-forms on the bundle, and the equations above can be used to infer what the corresponding connection $1$-forms are. From there, you can figure out the curvature $2$-forms and therefore the sectional curvatures. | |
| Apr 30, 2018 at 14:00 | comment | added | mathmetricgeometry | @DeaneYang: Let $X,Y$be orthonormal vector fields on $M$, $\tilde{X},\tilde{Y}$ their lifts. By O'Neil's formula, $$sec_M(X,Y)=sec_{FM}(\tilde{X},\tilde{Y})+\frac34|[\tilde{X},\tilde{Y}]^V|$$. However, for vectors $Z_1,Z_2$ tangent to the fiber, I don't know how to compute $sec_M(X,Z_i)$ and $sec_M(Z_1,Z_2)$. | |
| Apr 29, 2018 at 14:48 | review | Close votes | |||
| May 7, 2018 at 3:04 | |||||
| Apr 29, 2018 at 14:33 | comment | added | Deane Yang | If you know the curvature forms, then you know the sectional curvatures. You could also look up Rienannian submersions. | |
| Apr 29, 2018 at 13:28 | history | asked | mathmetricgeometry | CC BY-SA 3.0 |