Torsion tensor $T$ associated with an affine connection $\nabla$ on a smooth manifold is defined as a (0,2) tensor field as follows: $$T(X,Y) := \nabla_XY-\nabla_YX-[X,Y]$$ I understand that this is set to zero by the choice of the Levi-Civita connection, but I've read that the following condition (now stated in abstract index notation) is also equivalent to the connection being torsion-free: $$\nabla_a\nabla_bf = \nabla_b\nabla_af,$$ where $f$ is a smooth scalar field.

I can't see how $T = \bf{0}$ is equivalent to the condition above. Could anyone help me out with this?


1 Answer 1


You need $a$ and $b$ to be coordinate indices, so I am more comfortable writing them as $\mu$ and $\nu$. Then (using the MTW convection for the ordering of the indices on the Christoffel symbols) we have $$ \nabla_\mu \nabla_\nu f = \partial_{\mu}\partial_\nu f + (\partial_\lambda f) {\Gamma^\lambda}_{\nu\mu} $$
so $$ \nabla_\mu \nabla_\nu f - \nabla_\nu \nabla_\mu f= (\partial_\lambda f) ({\Gamma^\lambda}_{\nu\mu}-{\Gamma^\lambda}_{\mu\nu}). $$ As $(\partial_\lambda f)$ can be anything, we have your condition requires ${\Gamma^\lambda}_{\nu\mu}-{\Gamma^\lambda}_{\mu\nu}=0$ which is "torsion free".

If you use Vielbein or frame indices then you get your first expression.


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