I understand how, if the Riemann tensor is 0 in all its components, since we construct the Ricci tensor by contracting the Riemann, Ricci tensor would be 0 in all components as well.
I've read that vanishing of the Ricci tensor in 3 spacetime dimensions implies the vanishing of the Riemann curvature tensor, but that in higher dimensions that does not hold.
Can somebody explain why is that so? Is it because we have more independent components of the Riemann tensor in 4 spacetime dimension, than in 3 (20 vs 6)?
Also if the number of independent components of Riemann tensor in $n$ spacetime dimensions is
$$N(n)=\frac{n^2(n^2-1)}{12}$$
and since we know that the Riemann tensor has 256 components, does that limit the spacetime dimensions of it's usage? Or does that mean, that for example in 10 spacetime dimensions, there won't be any independent components of Riemann tensor?
Answer
So, let's take your formula, and set $n=3$. This gives you $N = \frac{9\cdot 8}{12} = 6$. Well, how many independent components of the Ricci tensor do you have? Well, since it's a 3x3 symmetric tensor, you've got six independent components. Therefore, there is no room in the Riemann tensor to have additional components. Since, for $n > 3$, you will always have fewer components of the ricci tensor than the riemann tensor (figure out what the fomula for independent components of a symmetric $n\times n$ matrix), for higher dimensions, there will always be additional extra components.
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