# From Riemann tensor to Ricci tensor and vice versa

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: $$g^{ae}R_{ebcd}=R^a_{bcd}$$ ?

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.