# Riemann curvature tensor identity proof

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*}
• Huh? What is $g_c$? Oct 18, 2020 at 4:07
• $g_c$ is the covariant basis of the tangent space. Oct 18, 2020 at 4:08
• You mean contravariant? $g$ is the metric tensor, so you need a different letter. Oct 18, 2020 at 4:10
• Sorry, yes the contravariant basis. Also I was not aware that a different letter is required, the text I refer to uses $\textbf{Z}_i$ as the contravariant basis and $Z_{ij}$ as the metric. I can change $g_c$ to $e_c$ in my answer if this is the typical notation? Oct 18, 2020 at 4:23