In general, the 4-dimensional space (or spacetime) has 20 independent components for the Riemann curvature tensor. For the Universe to be flat, one requires all the components to vanish. Only then we can claim that the Universe is flat. Contrary to that, the Universe is called flat if the value of a single scalar parameter $k$ (which appears in the FRW metric) is zero. Why is that?

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    $\begingroup$ A flat universe is spatially flat i.e. if you foliate it along the comoving time axis the spatial submanifolds are flat. The Riemann tensor is not zero. $\endgroup$ May 10, 2017 at 6:58

1 Answer 1


If you assume FRW metric, you greatly restrict the shape of the spacetime. Specifically, FRW metric assumes that the space is isotropic and homogeneous. So the space at a given cosmological time is maximally symmetric Riemannian manifold and these are characterized only by one parameter - curvature. FRW spacetime is just a foliation of such spaces.

On the other hand, general shape of the spacetime can be much more complicated, which is why it takes more parameters to characterize it. For example, Schwarzschild metric describes a black hole in asymptotically flat spacetime and it is obviously not contained in the family of FRW metrics.

  • $\begingroup$ +1 I would add the explicit formula $R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc})$ valid for any MSS, of which FRW is an example. We also have $R= d(d-1)\kappa$, so $R_{abcd}=0$ iff $\kappa=0$. $\endgroup$ May 10, 2017 at 13:46
  • $\begingroup$ @AccidentalFourierTransform NB: FRW is not a maximally symmetric spacetime in general. $\endgroup$
    – gj255
    May 10, 2017 at 15:43
  • $\begingroup$ @gj255 :: facepalm :: yes, I don't know what I was thinking about. $\endgroup$ May 10, 2017 at 16:04

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