- #1

George Keeling

Gold Member

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\mathrm{1.\ Linearity:}\mathrm{\ }\mathrm{\nabla }\left(T+S\right)=\mathrm{\nabla }T+\mathrm{\nabla }S & \phantom {10000}(1) \nonumber \\

\mathrm{2.\ Leibniz\ rule:}\mathrm{\nabla }\left(T\ \otimes \ \ S\right)=\left(\mathrm{\nabla }T\right)\ \ \otimes \ \ S+T\ \otimes \ \ \left(\mathrm{\nabla }S\right) & \phantom {10000}(2) \nonumber \\

{\mathrm{3.\ Commutes\ with\ contractions:}\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right)={\left(\mathrm{\nabla }T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho } & \phantom {10000}(3) \nonumber\\

{\mathrm{4.\ Reduces\ to\ partial\ derivative\ on\ scalars:}\mathrm{\nabla }}_{\mu }\phi ={\partial }_{\mu }\phi & \phantom {10000}(4) \nonumber \\

\end{align}1,2 and 4 seem reasonable but I cannot understand 3 and he does not seem to use it, even though he implies that he does.

The LHS of (3) seems straight forward\begin{align}

{\mathrm{\nabla }}_{\mu }\left(T^{\lambda }_{\ \ \ \lambda \rho }\right) & ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }+{\mathrm{\Gamma }}^{\lambda }_{\mu \kappa }T^{\kappa }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \lambda }T^{\lambda }_{\ \ \ \kappa \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(5) \nonumber \\

& ={\partial }_{\mu }T^{\lambda }_{\ \ \ \lambda \rho }-{\mathrm{\Gamma }}^{\kappa }_{\mu \rho }T^{\lambda }_{\ \ \ \lambda k} & \phantom {10000}(6) \nonumber \\

\end{align}Which is very like the rule for the covariant derivative of a (0,1) tensor.

I understand that the ##\mathrm{\nabla }T## in (1) and (2) means ##{\mathrm{\nabla }}_{\sigma}T## where ##T## is some tensor. So the RHS of (3) appears to be ##{\left({\mathrm{\nabla }}_{\sigma}T\right)}^{\mathrm{\ \ \ }\lambda}_{\mu \ \ \lambda \rho }## which leaves too many indices on the RHS. Otherwise the RHS is some kind of derivative with one contra- and three co-variant indices. What is that?

Help!