Geometry of curved spacetime, part 3

This post has been migrated from my old blog, the math-physics learning seminar.

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.