Timeline for Behavior of sectional curvature under metric deformations
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2012 at 22:09 | comment | added | Renato G. Bettiol | @Vitali: Oh, sure, of course. Thank you! | |
| Jan 17, 2012 at 20:29 | comment | added | Vitali Kapovitch | @Rentao: the last term is not there because we are computing at $p$ in normal coordinates so that $\nabla_XY(p)=0$. | |
| Jan 17, 2012 at 17:02 | comment | added | Renato G. Bettiol | @Vitali: If I may, there is one final detail that I just noticed that is not so clear to me. When you use the Koszul formula for $g_t$ to obtain an expression for $S(X,Y,Z)=g_0(S(X,Y),Z)$, shouldn't there also be a term $h(\nabla_X Y,Z)$? More precisely, I get $S(X,Y,Z)=\tfrac12(Xh(Y,Z)+Yh(X,Z)-Zh(X,Y))-h(\nabla_X Y,Z)$, and it seems your formula doesn't have this last term. For the computation of $S(Y,Y,X)$ this doesn't matter since $\nabla_Y Y=0$ at $p$, but I don't see why this extra term $-h(\nabla_X Y,X)$ does not appear on $S(X,Y,X)$. Sorry to bother again with something possibly silly.. | |
| Jan 16, 2012 at 14:41 | comment | added | Vitali Kapovitch | @Renato: Glad you have worked this out and there is no mistake in Topping's paper. BTW, you might want to work out the general formula for the derivative of the curvature that he gives. It is easily obtained using the same method I used above. | |
| Jan 16, 2012 at 5:05 | comment | added | Renato G. Bettiol | I realized that I made a mistake when interpreting Topping's formula (consequently when writing the expression 3) in my original post). Topping's original formula (Prop 2.3.5) is perfectly correct, one just needs to use the Ricci identity to get it to look exactly like Vitali's expression for $k'(0)$. I edited my post to mention this fact, but I'm not sure whether formula 3) should also be updated to be consistent with what actually is in Topping's notes. Regarding expression 1), it seems like the authors actually forgot to mention they are considering $X$ and $Y$ so that $h(R_0(X,Y)Y,X)=0$.. | |
| Jan 15, 2012 at 22:25 | comment | added | Renato G. Bettiol | @Vitali: Thank you for your time and patience to explain so many details. This morning, after finally obtaining the first 3 terms that are common to all three answers, I figured out that the last two terms in (2) where coming from the normalization, as you mentioned. But for some reason I was still missing the last term, $h(R^0(X,Y)Y,X)$, in the formula for $k'(0)$, that now is very clear. I think I was using a poor notation for $\langle,\rangle_t$ and $\langle,\rangle_0$ that eventually got both mixed up... Thank you again for confirming and clarifying everything. | |
| Jan 15, 2012 at 22:22 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Jan 15, 2012 at 22:10 | vote | accept | Renato G. Bettiol | ||
| Jan 15, 2012 at 21:52 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Jan 15, 2012 at 21:37 | history | undeleted | Vitali Kapovitch | ||
| Jan 15, 2012 at 21:37 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
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| Jan 15, 2012 at 19:23 | history | deleted | Vitali Kapovitch | ||
| Jan 15, 2012 at 19:23 | history | answered | Vitali Kapovitch | CC BY-SA 3.0 |