The Riemann tensor is skew-symmetric in its first and last pair of indices, i.e.,
\begin{align} R_{abcd} = -R_{abdc} = -R_{bacd} \end{align}
Can I simply use this to say that, for example, the component $R_{0001} = 0$ because $R_{0001} = -R_{0001}$?
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Yes. All the components where the first two indices are the same, or the last two indices are the same, are zero.
Sometimes it is useful to think of this tensor as a $6\times6$ symmetric matrix where the “indices” are $01$, $02$, $03$, $12$, $13$, and $23$.
However, don’t conclude from this that there are $6+5+4+3+2+1=21$ independent components. There are actually only $20$ because of the algebraic Bianchi identity.
Note that without any of these relations between components, there would be $4^4=256$ components! So the Riemann curvature tensor is about 13 times less complicated than it might appear.
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$\begingroup$ Wow, this tip was very nice! $\endgroup$ Commented Jun 22, 2019 at 20:13
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$\begingroup$ Can you explain me how to use the Bianchi identity to rule out one of the components of the tensor? I applied it to each one of the 21 remaining components, but i wasn't able to vanish anyone of them or to get a relation between two of them. Thanks! $\endgroup$ Commented Jun 22, 2019 at 21:50
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$\begingroup$ The algebraic Bianchi identity is $R_{abcd}+R_{acdb}+R_{adbc}=0$. Take $abcd=0123$ and you will get a relationship between 3 of the 21 components of the matrix I talked about. If you take another permutation of $0123$ you should get an equivalent relation. If you don’t take $abcd$ to all be different, the identity will reduce to one of the other symmetry relations. $\endgroup$– G. SmithCommented Jun 22, 2019 at 21:58