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For usual Riemann curvature tensors defined: ##R^i_{qkl},## I read in the book of differential geometry that in 3-dimensional space, Ricci curvature tensors, ##R_{ql}=R^i_{qil}## can determine Riemann curvature tensors by the following relation:

$$R_{\alpha\beta\gamma\delta}=R_{\alpha\gamma}g_{\beta\delta}-R_{\alpha\delta}g_{\beta\gamma}+R_{\beta\delta}g_{\alpha\gamma}-R_{\beta\gamma}g_{\alpha\delta}+\frac{R}{2}(g_{\alpha\delta}g_{\beta\gamma}-g_{\alpha\gamma}g_{\beta\delta})$$

where ##R## is the scalar curvature defined by ##g^{ql}R_{ql}.##

I come up with a way by expanding every Ricci tensor to the linear combinations of Riemann curvature to show the relation holds for ##R_{1212}## and ##R_{1213},## and other situations are up to permutations, which is obviously a very direct and without any beauty. My problem is if there exists any other way to see this fact, more intuitively, or more generally.

Thanks in advances~

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# I Ricci curvatures determine Riemann curvatures in 3-dimension

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