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Jun 15, 2020 at 7:27 history edited CommunityBot
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Sep 23, 2018 at 11:30 comment added Ali Taghavi @RobertBryant How the nth derivative of a function would be defined as a n linear map on the tangent space? I would like to apply this to the following question(To increase the order of Hamiltonian $H$ on the tangent bundle) mathoverflow.net/questions/311221/…
May 20, 2014 at 12:26 comment added Robert Bryant @AliTaghavi: Yes, you are missing something. The expression $X(Y(f))-\mathrm{d}f(\nabla_XY)$ is linear in $Y$ over the smooth functions, so its value at $p$ depends only on the value of $Y$ at $p$. You can see this by replacing $Y$ by $hY$ for some smooth function $h$; the term $X(h)Y(f)$ that is generated by the first term in the formula cancels the term $\mathrm{d}f(X(h)Y) = X(h)Y(f)$ generated by the second term.
May 20, 2014 at 8:57 comment added Ali Taghavi @Renato I can not understand some thing in your formula $g(\nabla_X \nabla f,Y)=X(Y(f))-\mathrm df(\nabla_X Y)$, so I think there is a contradiction in this formula: fix a point $p \in M$. the left side depends only on $Y(p)$ but the right side depends on the value of $Y$ in a neighborhood of $p$ not just on $Y(p)$. Am I missing some thing?
Mar 7, 2013 at 4:07 vote accept Renato G. Bettiol
Mar 5, 2013 at 19:58 answer added Ryan Budney timeline score: 2
Mar 5, 2013 at 18:02 answer added Robert Bryant timeline score: 23
Mar 5, 2013 at 18:01 answer added alvarezpaiva timeline score: 4
Mar 5, 2013 at 17:52 comment added Johannes Hahn I know that the Hessian isn't exactly the differential of something, but shouldn't there be an analogue to closedness of forms for it? Then there would be topological obstructions for the implication "closed => exact" of course.
Mar 5, 2013 at 16:44 history asked Renato G. Bettiol CC BY-SA 3.0