Computing the non-zero, independent components of the Riemann Tensor

The Riemann curvature tensor is a rank four tensor and it is used to describe space-time. $R_{abcd}$ has some peculiar properties i.e

$$R_{abcd} = -R_{bacd}$$
$$R_{abcd} = -R_{abdc}$$
$$R_{abcd} = R_{cdab}$$

i.e it is anti-symmetric in the pairs $a,b$ and $c,d$ but it is symmetric upon the interchange of the pairs $ab$ and $cd$.

$$R_{abcd} +R_{acdb} +R_{adbc} = 0$$
which is referred to as the first Bianchi Identity.

Using the above constraint equation, assuming that $a,b,c,d$ can take values between $0$ and $n$, one can compute that the total number of non-zero and independent components of the Riemann tensor are $\frac{n^2(n^2-1)}{12}$. In (3+1) space-time, there are twenty, in (2+1) space-time dimensions, there are 6 and in (1+1) dimensions, there is only one.

Now that I have introduced what the Riemann tensor and it’s properties are, my question is how does one go about computing these independent, non-zero components i.e what are the different combinations of a,b,c,d!

Naively thinking, one can start with

for a = 0,3
    for b = 0,3
        for c = 0,3
            for d = 0,3
                if a != b
                    if c != d
                        print a,b,c,d

to get the combinations of a,b,c,d for which we get a non-zero Riemann tensor. But the problem now is that we have a lot of redundancy i.e

$$R_{0102} = R_{0201} = -R_{1002} = -R_{0120}$$

I am not able to figure out how to weed out the dependent components of the Riemann tensor and get only the independent ones. If you have a solution, hit me because I haven’t been able to find one online. So far…

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