The Road to Reality - exercise on scalar product

  • #36
cianfa72 said:
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.
 
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  • #37
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.
 
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