Geometry of Curved Spacetime 3

Today, some numerology. The Riemann curvature tensor is a tensor $R_{abcd}$ satisfying the identities:

1. $R_{abcd} = -R_{bacd}.$

2. $R_{abcd} = R_{cdba}.$

3. $R_{abcd} + R_{acdb} + R_{adbc} = 0.$ (First Bianchi)

4. $R_{abcd|e} + R_{acec|d} + R_{abde|c} = 0.$ (Second Bianchi)

By 1, the number of independent $ab$ indices is $N = n(n-1)/2$, and similarly for $cd$. By 2, the number of independent pairs of indices is $N(N+1)/2$. Now the cyclic constraint 3 can be written as
$$R_{[abcd]} = 0,$$
and thus constitutes ${n \choose 4}$ equations. So the number of independent components is
$$\begin{align}  N(N+1)/2 - {n \choose 4} &= \frac{n(n-1)((n(n-1)/2+1)}{4} - \frac{n(n-1)(n-2)(n-3)}{24} \\ &= \frac{(n^2-n)(n^2-n+2)}{8} - \frac{(n^2-n)(n^2-5n+6}{24} \\ &= \frac{n^4-2n^3+3n^2+2n}{8} - \frac{n^4-6n^3+11n^2-6n}{24} \\ &= \frac{2n^4-2n^2}{24} \\ &= \frac{n^4-n^2}{12} \\ &= \frac{n^2(n^2-1)}{12} \end{align}$$

Now consider the Weyl tensor $C_{abcd}$ which is defined as the completely trace-free part of the Rienmann tensor. The trace is determined by the Ricci tensor $R_{ab}$ which as $n(n+1)/2$ indepdendent components, so the Weyl tensor has
$$\frac{n^2(n^2-1)}{12} - \frac{n^2-n}{2} = \frac{n^4-7n^2+6n}{12}$$
independent components. Now, for $n = 1$ we see that $R_{abcd}$ has no independent components, i.e. it vanishes identically. In $n=2$, it has only 1 independent component, and so the scalar curvature determines everything. In $n=3$, it has 6 independent components. Note that in this case, the Weyl tensor has no independent components, i.e. it is identically 0. So we see that in $n = 2, 3$ every Riemannian manifold is conformally flat. So things only start to get really interesting in $n=4$, where the Riemann tensor has 20 independent components, and the Weyl tensor has 10.