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 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 20, 2018 at 7:45 | comment | added | mathmetricgeometry | I don't know why my comment has been deleted, I understand your answer. Thank you very much! | |
| May 11, 2018 at 11:37 | comment | added | Robert Bryant | @mathmetricgeometry: You seem to have deleted your comment that said that you understood my answer. Have you developed doubts or have further questions? | |
| May 5, 2018 at 20:45 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added explicit computations for the case n=2 to aid the OP.
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| May 5, 2018 at 17:11 | comment | added | Robert Bryant | @mathmetricgeometry: I do not understand your question. I think that you must be misunderstanding the difference between the $\theta_{\alpha\beta}$ and the $\omega_{\alpha\beta}$, which are not the same thing at all. I'll add an example at the end of my answer to illustrate exactly what I mean. Maybe that will clarify everything. | |
| May 5, 2018 at 13:40 | comment | added | mathmetricgeometry | Thank you for your answer! I still have a quetion. For $\alpha \leqslant n$, let $\alpha=i$, we have $$ -\sum_{\beta>\alpha} w_{\alpha \beta}\wedge w_{\beta}=dw_{\alpha}=dw_i=-\sum_{j=1}^nw_{ij}\wedge w_j. $$ If $\beta>n$, then $w_{\alpha\beta} \wedge w_{\beta}$ are not of the form $w_{ij}\wedge w_j$. So from this, for $\beta \leqslant n$, let $\beta=j$, we should have $w_{\alpha \beta}=w_{ij}.$ Why do you say "with coeffients that are affine linear combinations of $R_{\gamma \delta}$"? | |
| May 4, 2018 at 13:44 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Fixed some typos and added some explanatory sentences.
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| May 3, 2018 at 19:46 | history | answered | Robert Bryant | CC BY-SA 4.0 |