Commutator of covariant derivative for rank 2 tensor

Question

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.

Answer 1

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.

Finally, the tangent space at a point is a vector space. Is that enough of a clue?