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The FLRW metric, in the case of positive scalar curvature, is: $ds^2 =- c^2 dt^2+a(t)^2\left(dw^2+\sin^{2}w(d\theta^2+\sin^2\theta d\phi^2)\right)$.

The Birkhoff’s theorem states that "any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat".

The FLRW metric stated above is obviously spherically symmetric, and also a solution of the vacuum field equation, but it is surely not static nor asymptotically flat (indeed it is not the Schwarzschild solution).

So what is the flaw in my reasoning?

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So what is the flaw in my reasoning?

The flaw is that the FLRW spacetime is not a vacuum solution. Indeed, as you correctly point out the one you wrote down has positive scalar curvature. A vacuum solution has a zero Ricci tensor and therefore a zero scalar curvature.

Physically, the FLRW spacetime represents an isotropic and homogenous distribution of matter. It is not vacuum (except for the trivial solution) anywhere.

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  • $\begingroup$ >A vacuum solution has a zero Ricci tensor and therefore a zero scalar curvature. Ok so the Birkhoff theorem is valid only for a spacetime without cosmological constant, right? Or, equivalently, "vacuum solution" implies that the space is not DeSitter or AdS? $\endgroup$ – Andrea Di Pinto Jul 24 '20 at 20:08
  • $\begingroup$ @AndreaDiPinto yes. Birkhoff’s theorem was proven for a zero cosmological constant (AFAIK). I don’t know if there is a generalization to a spacetime that is vacuum other than a cosmological constant. $\endgroup$ – Dale Jul 24 '20 at 23:13
  • $\begingroup$ Ok thanks, one last thing. "Spherically symmetric" means that the topology of the hypersurface with constant time is a 3-Sphere (in 4 dimension), right? Otherwise shouldn't be also "Spherically symmetric" the Euclidian space in 3 dimension? $\endgroup$ – Andrea Di Pinto Jul 24 '20 at 23:18
  • $\begingroup$ @AndreaDiPinto the spherical symmetry refers to the spatial dimensions. It means that spacetime has three rotational killing vector fields. In other words the space can be foliated as a set of nested 2 spheres. Euclidean space does indeed have this symmetry. $\endgroup$ – Dale Jul 25 '20 at 0:53

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