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Is it possible to find the Riemann tensor using the Ricci tensor and also to switch from Riemann $(0,4)$ to Riemann $(1,3)$?

I am not clear if these operations can only be carried out in one sense, I will explain better with these two questions.

1) $g^{bd}R_{abcd}=R_{ac}$, where $R_{abcd}$ it is Riemann $(0,4)$ and $R_{ac}$ it is the Ricci tensor, is it now possible to get the Riemann tensor $(0,4)$ again this way: $R_{ac} g_{bd}=R_{abcd}$?

2) $g_{ae}R^a_{bcd}=R_{ebcd}$, where $R^a_{bcd}$ it ise the Riemann $(1,3)$ and $R_{ebcd}$ it is the Riemann $(0,4)$, is it now possible to get the Riemann tensor $(1,3)$ again this way: $g^{ae}R_{ebcd}=R^a_{bcd}$ ?

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1 Answer 1

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Since we're not allowed to use any index more than twice, $R_{ac}g_{bd}=R_{aecf}g^{ef}g_{bd}$, which can't be simplified to $R_{abcd}$, so 1) doesn't work. But 2) is correct because$$g^{ae}R_{ebcd}=g^{ae}g_{fe}R^f_{\:bcd}=\delta^a_fR^f_{\:bcd}=R^a_{\:bcd}.$$You may also find this discussion of $2$- or $3$-dimensional manifolds interesting.

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  • $\begingroup$ Thanks a lot for your answer, but I still don't understand why 2) it works. $\endgroup$
    – user333046
    Commented May 6, 2020 at 8:46
  • $\begingroup$ @exxxit8 See edit. $\endgroup$
    – J.G.
    Commented May 6, 2020 at 10:09

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