Let $R_{abcd}$ be the Riemann curvature tensor defined by a torsion free and metric compatible covariant derivative.
Ultimately I am trying to prove that $R_{abcd} = - R_{abdc}$
The prove in my textbook is as follows: (However I really am unsure as to what is happening here)
We know that $\nabla g_{bc} = 0$ , since we have been told that is metric compatible. If we compute $\nabla_a\nabla_bg_{cd} - \nabla_b\nabla_ag_{cd} $ we have the following: $$0 = \nabla_a\nabla_bg_{cd} - \nabla_b\nabla_ag_{cd} = R_{abc}^{\space \space \space \space \space i}g_{id} + R_{abd} ^{\space \space \space \space \space i}g_{ci} = R_{abcd} +R_{abdc} $$
The part I am struggling to understand is where the term $ R_{abc}^{\space \space \space \space \space i}g_{id} + R_{abd} ^{\space \space \space \space \space i}g_{ci}$ came from.
I am aware that $\nabla_a\nabla_dV_b - \nabla_d\nabla_aV_b = R_{adb}^{\space \space \space \space \space i} V_i$ however I am still unsure as to where the part noted above comes from exactly. Any help would be really appreciated.
I'm not very well versed, but I think the answer to your specific question follows from the metrinilic property of the metric as well the fact that the covariant derivative satisfies Leibniz's law; that is, it should be the case that \begin{align*} 0 &= (\nabla_a\nabla_b- \nabla_b\nabla_a)g_{cd}\\ &= (\nabla_a\nabla_b- \nabla_b\nabla_a)(\textbf{e}_c\cdot \textbf{e}_d)\\ &= ((\nabla_a\nabla_b- \nabla_b\nabla_a)\textbf{e}_c)\cdot\textbf{e}_d + \textbf{e}_c\cdot((\nabla_a\nabla_b- \nabla_b\nabla_a)\textbf{e}_d)\\ &= (R_{abc}^{\space\space\space\space\space\space i}\textbf{e}_{i})\cdot\textbf{e}_d + \textbf{e}_c\cdot(R_{abd}^{\space\space\space\space\space\space i}\textbf{e}_{i})\\ &= R_{abc}^{\space\space\space\space\space\space i}g_{id} + R_{abd}^{\space\space\space\space\space\space i}g_{ci}\\ &= R_{abcd}+R_{abdc}\\ \end{align*}