- #1
Shirish
- 244
- 32
Let's say we have any two covariant derivative operators ##\nabla## and ##\nabla'##. Then there exists a tensor ##C^{\alpha}_{\mu\nu}## such that for all covariant vectors ##\omega_{\nu}##,$$\nabla_{\mu}\omega_{\nu}=\nabla'_{\mu}\omega_{\nu}-C^{\alpha}_{\mu\nu}\omega_{\alpha}$$
Now I'm quoting the relevant section on torsion tensor definition:
Now I'm quoting the relevant section on torsion tensor definition:
I don't understand why this is so. I mean the LHS is can also be notationally represented as ##\nabla_{[\mu}\nabla_{\nu]}\phi##, so either there should be a factor of ##1/2## on the RHS, or the torsion tensor should be defined as ##T^{\alpha}_{\mu\nu}\equiv C^{\alpha}_{[\mu\nu]}##, or am I missing something?What if the no-torsion requirement is dropped? Set ##\omega_{\nu}=\nabla_{\nu}\phi=\nabla'_{\nu}\phi##: (which gives) ##\nabla_{\mu}\nabla_{\nu}\phi=\nabla'_{\mu}\nabla'_{\nu}\phi-C^{\alpha}_{\mu\nu}\nabla_{\alpha}\phi##. Antisymmetrize over ##\mu## and ##\nu##, and assume ##\nabla'## is torsion free, but ##\nabla## is not. In that case ##\nabla_{[\mu}\nabla_{\nu]}\phi=-C^{\alpha}_{[\mu\nu]}\nabla_{\alpha}\phi##. The torsion tensor is defined as ##T^{\alpha}_{\mu\nu}\equiv 2C^{\alpha}_{[\mu\nu]}##, implying that $$(\nabla_{\mu}\nabla_{\nu}-\nabla_{\nu}\nabla_{\mu})\phi=-T^{\alpha}_{\mu\nu}\nabla_{\alpha}\phi$$