Is Covariant Derivative Notation Misleading in Vector Calculus?

[Orodruin]

Is that correct now ?

Looks good.

 

[vanhees71]

Looks good.

And what was wrong with my derivation? I don't see any difference.

 

[Orodruin]

And what was wrong with my derivation? I don't see any difference.

I never said anything was wrong with it. I complained about #118.

 

[cianfa72]

I never said anything was wrong with it. I complained about #118.

Yep, my fault sorry.

 

[samalkhaiat]

I cannot believe there are 100 posts here about a simple pure ... issue

This, unfortunately, has become a characteristic feature in here.

In mathematics $\nabla_{\mu}V^{\nu}$ is ill defined

No, it is not. In mathematics we define things. So, on a generic tensor (density) T_{A} \equiv T^{\rho_{1}\cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{s}}, I define the operator \nabla_{\mu} by the rule \nabla_{\mu}T_{A} \equiv \partial_{\mu}T_{A} + \Gamma^{\lambda}_{\mu\nu}[T_{A}]^{\nu}{}_{\lambda} , where [T^{\rho_{1} \cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{s}}]^{\nu}{}_{\lambda} \equiv \sum_{p = 1}^{r} \delta^{\rho_{p}}_{\lambda}T^{\rho_{1}\cdots \rho_{p-1}\nu \rho_{p+1}\cdots \rho_{r}}_{{}{}{}{}\tau_{1} \cdots \tau_{s}} - \sum_{q = 1}^{s} \delta^{\nu}_{\tau_{q}}T^{\rho_{1}\cdots \rho_{r}}_{{}\tau_{1}\cdots \tau_{q-1}\lambda \tau_{q+1}\cdots \tau_{s}} - \delta^{\nu}_{\lambda}T_{A} , with last term is absent when T_{A} is not a density.

Remaks: (1) Notice that [T_{A}]^{\nu}{}_{\lambda} \epsilon^{\lambda}{}_{\nu} is nothing but the change of T_{A} under an infinitesimal \mbox{GL}(n) transformation parametrized by \epsilon^{\lambda}{}_{\nu}. So, for any object \Psi, we define [\Psi]^{\nu}{}_{\lambda} by its infinitesimal transformation under the general linear group \mbox{GL}(n).

(2) You can show that the (above defined) operator \nabla_{\mu} satisfies the Leibniz rule.