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In the context of spacetime, reading Schutz, I'm confused about the symmetries of the Riemann curvature tensor, which I understand are: $$R_{\alpha\beta\gamma\mu}=-R_{\beta\alpha\gamma\mu}=-R_{\alpha\beta\mu\gamma}=R_{\gamma\mu\alpha\beta}.$$

But using the metric to contract the Riemann tensor can't I also say

$$R_{\gamma\mu}=g^{\alpha\beta}R_{\alpha\beta\gamma\mu}=g^{\alpha\beta}R_{\alpha\gamma\beta\mu}?$$ Which leads me to think that $$R_{\alpha\beta\gamma\mu}=R_{\alpha\gamma\beta\mu}.$$ But $R_{\alpha\gamma\beta\mu}$ isn't one of the above listed symmetries. Where am I going wrong?

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  • $\begingroup$ Comment to the question (v2): Also note that different authors may order the four indices of the Riemann curvature tensor differently. $\endgroup$
    – Qmechanic ♦
    Commented Dec 18, 2013 at 19:51

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$g^{\alpha\beta}$ is symmetric in $\alpha$ and $\beta$, while $R_{\alpha\beta\gamma\mu}$ is anti-symmetric in $\alpha$ and $\beta$, so the contraction $g^{\alpha\beta}R_{\alpha\beta\gamma\mu}$ is necessarily $0$, and cannot be $R_{\gamma\mu}$.

Moreover, it is not correct to say, that if the contraction of $2$ tensors with another tensor (here the metric tensor) are equals, then the $2$ tensors are equal. For instance, if you take a tensor $T_1$, and a tensor $T_2$, with $T_2-T_1$ anti-symmetric in lower indices $\alpha, \beta$ and contract the tensors $T_1$ and $T_2$ with a tensor symmetric in upper indices $\alpha, \beta$, you will get the same result.

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    $\begingroup$ Indeed; it may be added that all the metric contractions of the Riemann tensor are either $0$ or $\pm\text{Ric}$, so even the nonzero ones are not necessarily the Ricci tensor. $\endgroup$
    – Stan Liou
    Commented Dec 18, 2013 at 19:59
  • $\begingroup$ Oh, showing my huge ignorance, I thought you could just sort of “cancel” any upper and lower index. I didn't know about symmetric/anti-symmetric. Does that mean that $g^{\alpha\beta}R_{\alpha\beta\gamma\mu}=g^{\alpha\beta}R_{\gamma\mu\alpha\beta}=0$ ? Would that also mean (swapping the metric tensor indices) that $g^{\beta\alpha}R_{\alpha\beta\gamma\mu}=g^{\beta\alpha}R_{\gamma\mu\alpha\beta}=0$ ? So, if I avoid the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, can I state the Ricci tensor as (for example) $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\rho\nu}$ ? $\endgroup$
    – Peter4075
    Commented Dec 18, 2013 at 21:02
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    $\begingroup$ @Peter4075: Yes, Ricci tensor is $R_{\mu\nu} = R^\rho{}_{\mu\rho\nu} = g^{\sigma\rho}R_{\sigma\mu\rho\nu}$. You can switch indices on the metric because it's symmetric and $R_{\alpha\beta\gamma\mu} = R_{\gamma\mu\alpha\beta}$ is the pair-interchange symmetry of the Riemann tensor, so your equations here are correct. $\endgroup$
    – Stan Liou
    Commented Dec 18, 2013 at 22:59
  • $\begingroup$ @StanLiou: Thanks. Final question. Can the Ricci tensor therefore be defined using other index permutations that don't involve the Riemann tensor having the same 1 and 2 or 3 and 4 indices as the metric, ie $R_{\mu\nu}=g^{\sigma\rho}R_{\sigma\mu\nu\rho}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\rho\nu}$ , $R_{\mu\nu}=g^{\sigma\rho}R_{\mu\sigma\nu\rho}$ ? $\endgroup$
    – Peter4075
    Commented Dec 19, 2013 at 6:35
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    $\begingroup$ @Peter4075 : Up to a sign, so with your $3$ examples, with signs $(-1,-1,+1)$, which is obvious from the symmetry properties of $R_{\sigma\mu\nu\rho}$ $\endgroup$
    – Trimok
    Commented Dec 19, 2013 at 8:42

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