Given $\nabla$ a torsionless connection, the Ricci identity for co-vectors is $$\nabla_a \nabla_b \lambda_c - \nabla_b \nabla_a \lambda_c = -R^d_{\,\,cab}\lambda_d.$$ Prove $R^a_{[bcd]} = 0$ by considering the co-vector field $\lambda_c = \nabla_c f$.
By definition, $$R^a_{[bcd]} = 0 = \frac{1}{3!} \left(R^a_{\,\,bcd} + R^a_{\,\,cdb} + R^a_{\,\,dbc} - R^a_{\,\,bdc} - R^a_{\,\,cbd} - R^a_{\,\,dcb}\right)\,\,\,\,(1)$$
Attempt:
Input the given form for the covector into the Ricci identity in the question. Then since $\nabla_c f = e_c(f),$ we have $$\nabla_a \nabla_b e_c(f) - \nabla_b \nabla_a e_c(f) = -R^d_{\,\,cab}e_d(f).$$ True for all functions $f$, so $$\nabla_a \nabla_b e_c - \nabla_b \nabla_a e_c = -R^d_{\,\,cab}e_d.$$ Then since $\nabla_a e_b = \Gamma^d_{ba} e_d$ we can simplify the above to give $$\nabla_a \Gamma^d_{cb}e_d - \nabla_b \Gamma^d_{ca}e_d = -R^d_{\,\,cab}e_d$$ which can then be further rewritten like $$\nabla_a \Gamma^d_{cb} + \Gamma^{\alpha}_{cb}\Gamma^d_{\alpha a} - \nabla_b \Gamma^d_{ca} - \Gamma^{\alpha}_{ca}\Gamma^d_{\alpha b} = -R^d_{\,\,cab}.$$ I was then going to relabel all indices to get terms like that in the equation in $(1)$ and sum them all up and I hoped to get zero, but it is not. Have I made an error in the above somewhere? Thanks!
