# Equivalence between two conditions for zero torsion

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?

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".