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Covariant and controvariant bases derivative

Physics Asked by Gigino on February 12, 2021

How to show that

$overrightarrow{textbf{e}}_sigmacdotpartial_mu overrightarrow{textbf{e}}_nu = overrightarrow{textbf{e}}_sigmacdotpartial^mu overrightarrow{textbf{e}^nu}$

where $overrightarrow{textbf{e}}_nu$ and $overrightarrow{textbf{e}^nu}$ are the the basis and dual basis vector of some manifold?

Any suggestion?

Thank you!

One Answer

Where did you find this claim? The first expression ${bf e}_sigmacdot partial_mu{bf e}_nu$ is not covaraint. If one wrote instead ${bf e}_sigmacdot nabla_mu{bf e}_nu$ then it makes sense because the Christoffel symbol is defined by $$ nabla_mu {bf e}_nu = {bf e}_tau {Gamma^tau}_{numu} $$ giving

$$ {bf e}_sigmacdot nabla_mu{bf e}_nu= g_{sigmaalpha} {Gamma^alpha}_{numu} $$ and with $nabla^mu = g^{mualpha}nabla_alpha $ and with the action of the covariant derivative on a covector being $nabla_alpha {bf e}^nu= - {bf e}^{tau}{Gamma^nu}_{taumu}$ we get $$ {bf e}_sigmacdot nabla^mu{bf e}^nu = {bf e}_sigma (- {bf e}^{tau}){Gamma^nu}_{taubeta}g^{betamu}=-{ Gamma^nu}_{sigma beta}g^{betamu}. $$ So they differ by at least minus sign.

(sorry that I keep editing -- I keep making silly errors)

Correct answer by mike stone on February 12, 2021

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