The Road to Reality - exercise on scalar product

[martinbn]

Btw, even in $\mathbb R^n$ if one picks a non-standard affine connection (i.e. a non Levi-Civita connection), then $\nabla_{\nu} \nabla_{\mu}f$ doesn't commute as well.

As long as the torsion is not zero.

 

[cianfa72]

Just for practice I did the explicit calculation. From here covariant derivative start from a 1-form $\theta = \theta_{\eta}dx^{\eta}$ written in the chart with coordinates $\{ x^{\eta} \}$.

The gradient $\nabla f$ is by definition the 1-form/covector field $\nabla f = \left ( \partial_{\mu} f \right ) dx^{\mu}$. Hence $$\nabla_{\nu} \nabla_{\mu} f = (\nabla \nabla f)_{\nu \mu} = \left ( \nabla ((\partial_{\eta}f) dx^{\eta}) \right )_{\nu \mu}= \left ( \partial_{\nu} \partial_{\mu}f - (\partial_{\eta}f)\Gamma^{\eta}_{ \mu \nu} \right ) dx^{\nu} \otimes dx^{\mu}$$
From the above we can see that whether the connection $\Gamma$ is symmetric in the lower indices (or vanish) in such coordinates (hence in all coordinates) then $\nabla_{\nu} \nabla_{\mu} f = \nabla_{\mu} \nabla_{\nu} f$ i.e. they commute.