Main Question or Discussion Point
Why is it needed to consider a 4D Euclidean space to introduce the FLRW metric? Is it because with a fourth parameter, we can set the radius of the 4D sphere formed with the four parametres as constant?
We don't. The FLRW metric is locally Lorentzian, not Euclidean. Why do you think a 4D Euclidean space is needed?Why is it needed to consider a 4D Euclidean space to introduce the FLRW metric?
Ah(Btw, vanhees71 was asking you the same question. He was not answering your question in the OP; he was telling you the question was based on a false assumption.)
Because I found two derivations where the author introduces such a space.Why do you think a 4D Euclidean space is needed?
But you need to read it more carefully. The 4D Euclidean space is introduced in order to define a 3D sphere, it is not the space-time in the FRWL model. That 3D sphere is the constant cosmological time surface. Think of the sphere through all time, that forms the space-time manifold, which is not Euclidean.Because I found two derivations where the author introduces such a space.
One of them is here:
Well, he shows that they have constant curvature and that the curvature and the signature of the metric uniquely determine the space (at least locally and up to an isometry). But to find them he says that since they are unique it suffices to construct them any way we want, and he does the same thing as in the lectures in the link above.Maybe I'm wrongly attributing the derivation to Weinberg (I've to check as soon as I'm back home), but hasn't he also given the derivation of finding systematically the spaces with maximal symmetry via the Killing equation?