Let $\nabla$ be the covariant derivative associated with a torsionless connection. Prove the Ricci identity for covectors: $$\nabla_a \nabla_b \lambda_c – \nabla_b \nabla_a \lambda_c = -R^d_{\,\,cab}\lambda_d$$
Attempt:
Consider the object $\nabla_a \nabla_b (X^c \lambda_c)$. Then by distributing using the Leibniz rule, we obtain $$\nabla_a (\nabla_b (X^c) \lambda_c + X^c \nabla_b (\lambda_c))$$ and again, this time operating with $\nabla_a$ gives $$(\nabla_a \nabla_b X^c)\lambda_c + \nabla_b(X^c) \nabla_a (\lambda_a) + \nabla_a(X^c)\nabla_b (\lambda_c) + X^c \nabla_a \nabla_b \lambda_c$$
I can then use the Ricci identity for vector fields on the first term and simplify the middle terms to give $$(R^c_{\,\,dab}X^d + \nabla_b \nabla_a X^c)\lambda_c + e_b(X^c) e_a(\lambda_c) + e_a(X^c) e_b(\lambda_c) + X^c\nabla_a \nabla_b \lambda_c$$ using the fact that for a function, $\nabla_X f = X(f)$ where $X$ is a vector field. $\lambda_c$ and $X^c$ are the components of the covector and vector respectively.
I am just not quite sure how to progress. I think I am going to need to symmetrise over the $a$ and $b$ indices to extract another term so I can obtain the required terms in the identity but at the moment I am unsure. Thanks for any help!
Best Answer
You started off with a good idea, but in the last display your thinking went too complicated :-)
Recall, that $X^c \lambda_c$ is a scalar, and therefore for a torsionless connection we have $$ \nabla_a \nabla_b (X^c \lambda_c) = \nabla_b \nabla_a (X^c \lambda_c) $$ which is equivalent to saying that $$ 2 \nabla_{[a} \nabla_{b]} (X^c \lambda_c) = 0 $$ Applying this to your third display, we get $$ 2 (\nabla_{[a} \nabla_{b]} X^c)\lambda_c + 2 X^c \nabla_{[a} \nabla_{b]} \lambda_c = 0 $$ where the two middle terms $\nabla_b(X^c) \nabla_a (\lambda_a) + \nabla_a(X^c)\nabla_b (\lambda_c)$ cancel out under antisymmetrization.
Now your claim easily follows.