# Commutator of covariant derivative for rank 2 tensor

I am a newbie at tensor notation and I have been told to prove the identity

$$(\nabla_a\nabla_b - \nabla_b\nabla_a)X^a_{\ \ \ b}=- R^e_{\ \ \ bcd}X^a_{\ \ \ e}+ R^a_{\ \ \ ecd}X^e_{\ \ \ b}$$

I am aware of the definition of the Riemman tensor $$(\nabla_a\nabla_b - \nabla_b\nabla_a)X_c= R_{abc}^{\quad d}X_{d}$$ and using the metric connection $$\nabla_a g_{bc}=0$$ I have already shown $$(\nabla_a\nabla_b - \nabla_b\nabla_a)X^c= -R_{abd}^{\quad c}X^{d}$$ but still I cannot figure out how to obtain the first one... :(

I am working torsion-free and I am aware of the formula that relates the Riemman tensor with the Christoffel symbols.

PS: My guess is to go with $$X^a_{\ \ \ b} = v^a w_b$$ but not sure if that is the right way to go.

PS: My guess is to go with $$X^a_{\ \ \ b} = v^a w_b$$ but not sure if that is the right way to go.
It is. Schematically: $$R\sim\left[\nabla,\nabla\right]$$ Any derivative worth that name satisfies: $$\nabla\left(uv\right)=\left(\nabla u\right)v+u\nabla v$$ Covariant derivatives act like, $$\nabla\sim\partial+\Gamma$$ for contravariant vectors. $$\nabla\sim\partial-\Gamma$$ for covariant vectors.