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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?

Qmechanic
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dingo_d
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2 Answers2

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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.

Zo the Relativist
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  • So the key is that in the higher dimensions there are more independent components in the Riemann tensor and so I cannot 'cover' it with Ricci tensor, so even if Ricci tensor is 0, some components of the Riemann tensor won't be zero because of it. Is that reasoning right? – dingo_d Oct 09 '13 at 14:38
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    yes. Note that the Schwarzschild solution is Ricci flat, since $T_{ab} = 0$. It is most definitely not Riemann flat. – Zo the Relativist Oct 09 '13 at 14:43
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The Riemann tensor can be decomposed into a trace part and a trace-free part:

$$R_{abcd}=C_{abcd}+\frac{2}{n-2}(g_{a[c}R_{d]b}-g_{b[c}R_{d]a})-\frac{2}{(n-1)(n-2)}Rg_{a[c}g_{d]b},$$

where the traceless object $C_{abcd}$ is the Weyl (conformal) tensor and the remainder is given by the Ricci tensor and its contraction, the Ricci scalar, where $n$ is the number of dimensions and $g_{ab}$ is the metric. In three dimensions, the Weyl tensor vanishes: this implies that when the Ricci tensor is zero in all components, the Riemann tensor is zero as well. However, for $n$ larger than 3, the Ricci tensor can be zero while the Weyl tensor is nonvanishing.

trosos
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Frederic Brünner
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