If the curvature of the universe is zero, then $$Ω = 1$$ and the Pythagorean Theorem is correct. If instead $$Ω> 1$$ there will be a positive curvature, and if $$Ω <1$$ there will be a negative curvature, in either of these cases, the Pythagorean theorem would be wrong (but the discrepancies are only detectable in the triangles whose lengths its sides are of a cosmological scale). but could think of a curvature of the universe such that $$Ω= a+ib$$ is a complex number? that would mean physically?
I don't think it'll make much sense defining an imaginary curvature parameter.
The curvature (and forget $\Omega$ for now) describes how you "rotate" space vectors at each point, i.e. it gives a "rotation" along the "old" coordinates as well as a possible "rescaling" of the lengths relative to the "old" coordinates. So what it effectively tells you is "this direction is transformed into that one by this amount, that amount and that other amount", etc., for each point of space.
You want the result of such transformation to be a real vector space -- no imaginary components -- because that's what's physically meaningful. Remember this is physics, so in the end it's the math that is subordinate to physics, not the other way around. Just because you can invent an new algebra that doesn't mean that it's physically useful (i.e. translates into the "real world") so be mindful of that when you pick new algebras.
Now back to $\Omega$. Remember that it's a density parameter -- so you'll have to explain us what would mean to you an imaginary density, e.g. an imaginary value for a mass density.
protected by Qmechanic♦ Dec 25 '12 at 16:26
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?