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.


1 Answer 1


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*}

  • $\begingroup$ Huh? What is $g_c$? $\endgroup$ Oct 18, 2020 at 4:07
  • $\begingroup$ $g_c$ is the covariant basis of the tangent space. $\endgroup$
    – RyanK
    Oct 18, 2020 at 4:08
  • $\begingroup$ You mean contravariant? $g$ is the metric tensor, so you need a different letter. $\endgroup$ Oct 18, 2020 at 4:10
  • $\begingroup$ 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? $\endgroup$
    – RyanK
    Oct 18, 2020 at 4:23

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