Combinatorially counting the Riemann tensor's degrees of freedom One famous way to count the DOFs of $R_{abcd}$ on an $n$-dimensional Riemannian manifold is as follows:

*

*$ab$ has $\binom{n}{2}$ DOFs because $R_{abcd}=-R_{bacd}$;

*Since $R_{abcd}=R_{cdab}$, the Riemann tensor's DOFs number at most$$\binom{\binom{n}{2}+1}{2};$$

*We have one last constraint viz. $R_{a\left[bcd\right]}=0$, so the final answer is$$\binom{\binom{n}{2}+1}{2}-\binom{n}{4}\stackrel{\star}{=}\frac{1}{3!}\binom{n^{2}}{2},$$where $\stackrel{\star}{=}$ follows by chugging through algebra (both sides are $\tfrac{n^2(n^2-1)}{12}$).

My question is whether we can prove the RHS immediately with a different combinatorial argument. (If so, the above equation could be proved by double-counting the tensor's DOFs.)
My first thought was to pair $n^2$ values of $ac$ with $n^2-1$ values of $bd$, then lose a factor of $3!$ through other symmetries. But this won't work, because each choice of $ac$ doesn't disqualify just one value of $bd$; if $R_{abcd}\ne0$, $a\ne b$ and $c\ne d$, and at most $(n-1)^2$ values of $bd$ work.
 A: I'm not sure what exactly you want, but here is another way to find the dimension of the space of curvature tensors. On bottom of page 881 of the paper by Berger, Bryant, Griffths titled "The Gauss Equations and Rigidity of Isometric Embeddings", they state that the space of curvature tensors is the kernel of the surjective linear map
\begin{align*}
\Lambda^2V^*\otimes\Lambda^2V^* &\rightarrow V^*\otimes\Lambda^3V^*\\
R_{abcd} &\mapsto R_{abcd}-R_{bacd}
\end{align*}
Therefore, the dimension of this space is
$$
\left(\frac{n(n-1)}{2}\right)^2 - n\frac{n(n-1)(n-2)}{6}.
$$
A straightforward calculation shows that this is equal to
$$
\frac{n^2(n^2-1)}{12}.
$$
The proof in the paper that the linear map above is surjective and the kernel is the space of curvature tensors follows from the decomposition of tensor products of $V^*$ into irreducible representations of $GL(n)$ and that the space of curvature tensors is an irreducible subspace. However, I think it might be possible to prove this using your "bare hands".
