Having proved eqn for covariant derivative covariant vector, want to get eqn for Covariant derivative contravariant vector,using metric tensor $${T^\delta } = {g^{\delta \alpha }}{T_\alpha }$$ $$ \implies {T^\delta }_{;\beta } = {({g^{\delta \alpha }}{T_\alpha })_{;\beta }}$$ $$ = ({g^{\delta \alpha }}_{;\beta }){T_\alpha } + {g^{\delta \alpha }}({T_{\alpha ;\beta }})$$ First term drops out as metric covariantly constant $$ = {g^{\delta \alpha }}({T_{\alpha ,\beta }} - {\Gamma ^\mu }_{\alpha \beta }{T_\mu })$$ $$ = {T^\delta }_{,\beta } - {g^{\delta \alpha }}{\Gamma ^\mu }_{\alpha \beta }{T_\mu }$$ $$ = {T^\delta }_{,\beta } - {g^{\delta \alpha }}{\Gamma ^\delta }_{\alpha \beta }{T_\delta }$$ $$ = {T^\delta }_{,\beta } - {\Gamma ^\delta }_{\alpha \beta }{T^\alpha }$$ Obviously I’ve lost a sign in the connection term, but don’t see how
1 Answer
You've actually got two mistakes in there. First, from the 4th line to the 5th $$ {T^{\delta}}_{ ,\beta}=(g^{\delta \alpha}T_{\alpha})_{,\beta}\neq g^{\delta\alpha}T_{\alpha,\beta} $$ And second, in the 6th line you've got three $\delta$ indices in one term, which is not allowed.
Here's an alternative way to do it. Take any $V_{\alpha}$, then $V_{\alpha}T^{\alpha}$ is a scalar, so acting on it with the covariant derivative is just partial differentiation. Expanding both the partial and the covariant derivative with the chain rule gives $$V_{\alpha,\beta}T^{\alpha}+V_{\alpha}{T^{\alpha}}_{,\beta}=V_{\alpha;\beta}T^{\alpha}+V_{\alpha}{T^{\alpha}}_{;\beta}$$ Now use the formula for $V_{\alpha;\beta}$ that you already now, cancel a term, and use the fact that the result must be valid for all $V_{\alpha}$.
