How to show that
$\overrightarrow{\textbf{e}}_\sigma\cdot\partial_\mu \overrightarrow{\textbf{e}}_\nu = \overrightarrow{\textbf{e}}_\sigma\cdot\partial^\mu \overrightarrow{\textbf{e}^\nu}$
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!
Where did you find this claim? The first expression ${\bf e}_\sigma\cdot \partial_\mu{\bf e}_\nu$ is not covaraint. If one wrote instead ${\bf e}_\sigma\cdot \nabla_\mu{\bf e}_\nu$ then it makes sense because the Christoffel symbol is defined by $$ \nabla_\mu {\bf e}_\nu = {\bf e}_\tau {\Gamma^\tau}_{\nu\mu} $$ giving
$$ {\bf e}_\sigma\cdot \nabla_\mu{\bf e}_\nu= g_{\sigma\alpha} {\Gamma^\alpha}_{\nu\mu} $$ and with $\nabla^\mu = g^{\mu\alpha}\nabla_\alpha $ and with the action of the covariant derivative on a covector being $\nabla_\alpha {\bf e}^\nu= - {\bf e}^{\tau}{\Gamma^\nu}_{\tau\mu}$ we get $$ {\bf e}_\sigma\cdot \nabla^\mu{\bf e}^\nu = {\bf e}_\sigma (- {\bf e}^{\tau}){\Gamma^\nu}_{\tau\beta}g^{\beta\mu}=-{ \Gamma^\nu}_{\sigma \beta}g^{\beta\mu}. $$ So they differ by at least minus sign.
(sorry that I keep editing -- I keep making silly errors)
