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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    $\begingroup$ 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. $\endgroup$
    – Jus12
    Aug 31, 2011 at 17:39
  • $\begingroup$ Ever heard of the Banach-Tarksi paradox!? In mathematics, you can create something that is impossible in the real world (physics). $\endgroup$
    – mathcal
    Dec 24, 2012 at 13:38
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    $\begingroup$ 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?”" $\endgroup$
    – Sklivvz
    Dec 24, 2012 at 14:56
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    $\begingroup$ This is specific enough not to be a "what if?" question. $\endgroup$
    – David Z
    Dec 24, 2012 at 16:04

3 Answers 3


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 a 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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    $\begingroup$ There is at least one approach to unifying electromagnetism and gravity that uses a complex metric. In that approach, imaginary mass density equates to charge density. $\endgroup$
    – S. McGrew
    Oct 19, 2020 at 23:06

The density parameter $\Omega$ is a positive real number. The curvature $k$ is normalized to the values $-1$, $0$, or $+1.$ Due to Friedmann, $k$ is negative, zero, or positive according to whether $\Omega$ is less than, equal to, or greater than $1$, respectively. Due to the normalization of $k$, the precise value of $\Omega$ is not computable from it. Generally, one works the other way around, using data on density to inform the curvature. One could find a non-normalized expression for the curvature and look for how it relates to $\Omega$, but the formulas are quite nested in the derivation.

The density parameter $\Omega$ is defined as $$\Omega:=\dfrac{8\pi G\rho}{3H^2}$$ where $G$ is the gravitational constant, $\rho$ is the average density, and $H$ is the Hubble constant. One could think of this as a function of the variable $\rho$, which is approximated based on experimental data. From this viewpoint, an imaginary $\Omega$ would require an imaginary average density, and there is no physical sense of what that would mean in terms of mass per volume, nor is there a sense of negative density.

In differential geometry, curvature can take the value of any real number. From a pure math perspective, I would be surprised if someone has not investigated the notion of complex curvature, but this is not conventional in math, let alone having found a physics application. If it did, one would need to reinterpret Friedman's result about the relationship between $k$ and $\Omega$, but I expect that imaginary $k$ would correspond to negative $\Omega$ due to squaring.

There is a notion of hyperbolic space as a sphere of radius $i$, just as the ordinary sphere has radius $1$, since the sphere is parametrized as the solution set to a quadratic form set equal to the radius. Even there, it's just another way of thinking about negative curvature.


As I said before in response to another question, the notion of complex curvature does exist: see Y. Martinez-Maure, Real and complex hedgehogs, their symplectic area,curvature and evolutes. The author says the paper is to appear in the Journal of Symplectic Geometry.

For instance, a complex circle with (complex) radius R has a (complex) radius of curvature equal to R (see page 17).

One could imagine that there exists a complex metric such that the operator annihilation verifies Dirac or Klein Gordon equations associated with this metric while the operator creation would verify these same equations but for the conjugate metric.


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