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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?

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Pythogorean theorem is a theory (evident from from the word "theorem") of Euclidean space which is flat. It does not say that the real universe is Euclidean. –  Jus12 Aug 31 '11 at 17:39
    
Ever heard of the Banach-Tarksi paradox!? In mathematics, you can create something that is impossible in the real world (physics). –  mathcal Dec 24 '12 at 13:38
    
Froum our FAQ "You should only ask practical, answerable questions based on actual problems that you face. Chatty, open-ended questions diminish the usefulness of our site and push other questions off the front page. [...] avoid asking subjective questions where [...] we are being asked an open-ended, hypothetical question: “What if ______ happened?”" –  Ebenezer Sklivvze Dec 24 '12 at 14:56
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This is specific enough not to be a "what if?" question. –  David Z Dec 24 '12 at 16:04

1 Answer 1

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.

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protected by Qmechanic Dec 25 '12 at 16:26

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