Timeline for Taylor expansion of the square of the distance function on a Riemannian manifold
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2021 at 15:33 | comment | added | Chee | Morning Otis, thank you for guiding me through it. I got it. $\nabla{_{\cdot}}$ preserves the type of the tensor it is differentiating, and since $T|_{0,t}=0$, the covariant derivative i wrote and you pointed to is $0$ at $(0,t)$ | |
| Oct 13, 2021 at 14:50 | comment | added | Otis Chodosh | Yes that's what I meant. Similarly if the covariant derivative hits E then there's a non differentiated T that vanishes so that term goes away | |
| Oct 13, 2021 at 3:54 | comment | added | Chee | Hi Otis, thank you. Did you mean that since $T|_{0,t}=0$, we must have $(\nabla_{D_{\epsilon}}R)(E,T)E|_{0,t}=0$? | |
| Oct 13, 2021 at 3:40 | comment | added | Otis Chodosh | The covariant derivative term you write vanishes since $T$ does. See the explanation before (13) in the paper. | |
| Oct 13, 2021 at 3:33 | comment | added | Chee | Hi Otis, in the display (24) of Myer's paper, the first equality was obtained by the definition of curvature via 3th-order mixed directional derivatives, and to prove this identity, we should not expand the curvature via its definition and then covariant differentiate each term in the expansion, since this will lead us back to the first equality there in (24). i have tried $(\nabla_{D_{\epsilon}} R) (E,T)E$, i.e., the covariant derivative of the curvature tensor, which equivalently should be $0$ at $(0,t)$ if the identity holds. I was not able to prove that it is $0$ at $(0,t)$ | |
| Oct 13, 2021 at 3:24 | comment | added | Chee | hi Otis, thank you for your reply. I checked every thing in the proof but the identity was the only thing I did not understand. I do not see how $T$ vanishes at $\epsilon=0$ leads to the identity. The left-hand side is the covariant derivative of $R(E,T)E$ whereas the right-hand side is $R(E,\cdot)E$ with $\cdot$ as the covariant derivative of $T$. In a sense, for this specific case, $\nabla_{\epsilon}$ "commutes" with curvature endomorphism. Strange ... | |
| Oct 13, 2021 at 3:05 | comment | added | Otis Chodosh | @Chee, can you be more specific about your issue with this? I think you should use that T vanishes at eps =0 so the given term is the only one that can appear. | |
| Oct 13, 2021 at 2:23 | comment | added | Chee | hi Otis, I am studying this expansion provided by Myer's paper you suggested. There is an equality in display (24) on page 8 of the paper that I do not understand. It is $\nabla_{D_{\epsilon}} R(E,T)E |_{0,t} = R(E, \nabla_{D_{t}}E)E|_{0,t}$, where $\nabla_{D_{t}}E = \nabla_{D_{\epsilon}}T$. Do you know how this is derived? Thanks. | |
| May 26, 2021 at 11:45 | vote | accept | Luis Yanka Annalisc | ||
| May 25, 2021 at 16:19 | history | answered | Otis Chodosh | CC BY-SA 4.0 |