Torsion tensor definition doubt

Question

[Ref. Core Principles of Special and General Relativity by Luscombe, page 246]

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:

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 $$(\nabla_{\mu}\nabla_{\nu}-\nabla_{\nu}\nabla_{\mu})\phi=-T^{\alpha}_{\mu\nu}\nabla_{\alpha}\phi$$

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?

Answer 1

If your confusion is with the apparently missing factor of $1/2$, note that

$$\nabla_{[a}\nabla_{b]} \equiv \frac{1}{2}(\nabla_a\nabla_b-\nabla_b\nabla_a)$$

Symmetrization and antisymmetrization brackets come defined with a factor of $1/2$, because they are meant to extract the symmetric and antisymmetric parts of the relevant tensors. With this in mind, your equation becomes

$$\nabla_\mu\nabla_\nu \phi - \nabla_\nu\nabla_\mu\phi = 2\nabla_{[\mu}\nabla_{\nu]}\phi = -2C^\alpha_{[\mu\nu]}\nabla_a \phi \equiv -T^\alpha_{\mu\nu}\nabla_\alpha\phi$$

Answer 2

The if $X$ and $Y$ are (contravariant) vector fields, the torsion tensor $T(X,Y)$ is defined as $$ \nabla_X Y-\nabla_Y X-[X,Y]=T(X,Y) $$ where $T(X,Y)^\lambda= {T^\lambda}_{\mu\nu}X^\mu Y^\nu$, and $[X,Y]^\nu= X^\mu \partial_\mu Y^\nu- Y^\mu\partial_\mu X^\nu$ is the Lie bracket of the vector fields.

If $$ (\nabla_X Y)^\lambda = X^\mu (\nabla_\mu Y^\lambda + {\Gamma^\lambda}_{\mu \nu} Y^\nu) $$ then ${T^\lambda}_{\mu\nu}= {\Gamma^\lambda}_{\mu \nu}-{\Gamma^\lambda}_{\nu \mu}$.

The notation $[\nabla_\mu,\nabla_\nu]$ is potentialy unsafe, although common. This is because $"\nabla_\mu"$ is a acting on a different tenor space depending whether is acts before or after $\nabla_\nu$, so the "commutator" is not really a commutator. That why I prefer to use $\nabla_X$. With this notation $$ [\nabla_X,\nabla_Y]Z- \nabla_{[X,Y]}Z = R(X,Y)Z $$ whether there is torsion or not.