How to take algebraic Bianchi identity into account when calculating the number of independent components of Riemann curvature tensor I understant that the Riemann curvature tensor has the following symmetry properties:
Antisymmetry in last two indices:
$$R_{a b c d}=-R_{a b d c}$$
Symmetry in first pair of indices with second pair of indices:
$$R_{a b c d}= R_{c d a b}$$
Algebraic Bianchi identity:
$$R_{a b c d}+R_{c a b d} +R_{b c a d}=0$$
In $N$ dimensions, I can have ${ }^{N} C_{2}$
choices for $c$ & $d$ (they cannot be equal otherwise $R_{abcd}$ will be zero).
By combining the first & second symmetry property, we can say that $R_{abcd}$ is antisymmetric in $a$ & $b$ too. So there are ${ }^{N} C_{2}$ choices for the first pair of indices.
Based on the second symmetry property:
We can have ${ }^{N} C_{2}$ choices for the first pair of indicies. We can then have ${ }^{N} C_{2}-1$ different than the first one choices for the second pair. Their is pair swap symmetry, so in total we have $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}-1)$ independent components when the first & second pair differs. We have ${ }^{N} C_{2}$ choices where they are the same. So, not taking tha algebraic Bianchi identity into account we have $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$ independent compontent for the Riemann curvature tensor.
In the algebraic Bianchi identity, all indicies have to be different. There are then ${ }^{N} C_{4}$ different choices we have. This answer just subtracts this number from the earlier result, and gets the right answer:
$$\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1) - { }^{N} C_{4} = \frac{n^{2}\left(n^{2}-1\right)}{12}$$
The equality can be established after a few lines of algebra, or plotting the RHS & LHS and concluding that they are the same. Here is such a plot.
Why can we just subtract this number from the earlier result? This is the part I don't get.
 A: Your thoughts are right about the first part, the two first conditions of the curvature tensor gives you $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$ choices for $a,b,c,d$. Denote by $\mathfrak{B}(V^*)$ the linear subspace of all curvature tensor satisfying your first two conditions, and not the Bianchi identity, and by $\mathfrak{T}(V^*)$ the space of all covariant tensors $(0,4)$.
This means that by your argument, $\mathfrak{B}(V^*)$ has dimension $\frac{1}{2}{ }^{N} C_{2}({ }^{N} C_{2}+1)$. Now consider the linear map $\pi: \mathfrak{B}(V^*) \rightarrow \mathfrak{T}(V^*)$ given by
$$
\pi(T)(w,x,y,z)=\frac{1}{3}(T(w,x,y,z)+T(x,y,w,z)+T(y,w,x,z)).
$$
Hence, you have that the space of all curvature tensors (it is, the space of $(0,4)$ satisfying all the properties you write) is the kernel of this map. So by classical linear algebra theorem, you only have to compute the dimension of the image of that map, which will be ${ }^{N} C_{4} $ (because is the set of alternating $4$-covariant tensors), and substract it to the number you compute to get the result. That is the formal reason why you cand substract that number.
This argument I wrote is part of the general proof you can find in Introduction to Riemannian Manifolds by John M. Lee, proposition 7.21, page 212.
For an easier way to remember this, I use this (NON FORMAL) trick:
I have $n^4$ possibilities, but by the antisymmetry of last two indices, I may substract $n^2$ choices, so I have $n^4-n^2$ in total. But by the second restriction, I must divide this number by two due to they are related $2\leftrightarrow2$, so I have
$$
\frac{n^4-n^2}{2}
$$
and then, by the Bianchi identity, I only have to put off the sum of cyclic permutations of $3$ indixes, so I divide by 6. REMEMBER this is only a trick I use remember the formula, nothing formal.
