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I think the essential problem lies in the difference between the mathematical meaning of curvature, and the way in which we actually describe a manifold, or a curved space (or spacetime).
Although we describe the universe as having spacetime curvature (which is mathematically true), curvature refers to the Riemann curvature tensor, which is a rank-4 tensor, ...
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The spacetime metric of a spatially-flat Friedmann universe — like ours seems to be, on the largest scales — is
$$ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2)$$
where the function $a(t)$ is the Friedmann scale factor describing the expansion of space as a function of cosmological time $t$.
You can calculate its 4D Riemann curvature tensor $R_{\mu\nu\lambda\kappa}$ and ...
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The notion of "spatial curvature" only makes sense when the spacetime geometry is symmetric enough that there is a natural/preferred foliation of it into spacelike slices. You can then talk about the intrinsic curvature of those slices.
The easiest way to understand why the curvatures can be different is to look at a toy cosmological model, like ...
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Assuming (as this chart does) that the universe contains only dust and dark energy, the scale factor satisfies $$\dot a^2 = H_0^2 \left( Ω_{Λ,0}\, a^2 + Ω_{k,0} + Ω_{m,0}\, a^{-1} \right)$$
where $Ω_k = 1 - Ω_m - Ω_Λ$. (The exponents are $-1{-}3w$ where $w$ is the equation of state parameter.)
The boundary of the $κ=\pm1$ regions is just the line $Ω_{k,0}=0$....
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There is a close analogy:
If you jump say 3 feet high, you can calculate your speed as a function of you position between 0 to 3 feet high.
Then you wonder what is your speed at 6 feet high during your jump. The good old high school Newtonian mechanics offers an answer: your speed was imaginary at 6 feet high according to the total energy conservation ...
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What does it mean for a singularity to be resolved into n−1 intersecting two spheres?
Just to be sure, the mathematically precise of idea of "resolving a singularity" is given in the realm of algebraic geometry under the name blow up. Now, what is a blow up intuitively speaking? Algebraic geometers have sets of rules to replace singular spaces ...
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RW assumes that matter is uniformly distributed over the spatial slices, and elliptic, flat, and hyperbolic geometries have very different distributions, in terms of the amount of matter within a given distance of any given point. There's no way you could move the matter around to be homogeneous in a different geometry without violating homogeneity in the ...
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Many geometrical objects degenerate into straight lines as they grow arbitrary large. Some of these lines loop back on themselves as you describe, others do not. The lines of Euclidean and affine spaces do not; the lines of projective, elliptic and spherical spaces do. The lines of hyperbolic spaces can get complicated. Each such space has a characteristic ...
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The axion is an angular degree of freedom
$$ \theta \equiv \frac{a}{f_a}\,,$$
where $f_a$ is the axion decay constant. In particular, the axion parent field $\phi = \frac{1}{\sqrt{2}}\rho e^{{\rm i} \theta}$ acquires a radial VEV $\rho = f_a$.
In the misalignment mechanism, the angular degree of freedom is assumed to have a flat distribution on the interval $...
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Above a certain temperature, quarks get annihilated as quickly as they form. the universe had to cool off a bit to get below that threshold temperature, so the quarks would persist. That point occurred at around 10^-12 seconds.
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Yes.
"The curvature of the universe" is an imprecise term, and describing the curvature of a general four-dimensional spacetime takes 20 numbers at every point. But I'll assume that your phrase should mean the Ricci scalar curvature $R$, which is a single number at each point that is a kind of average curvature of spacetime (where the averaging is ...
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No
The curvature parameter $k$ of the universe remains constant throughout its evolution. If the universe is open ($k < 0$) , it will stay open, and if it is closed ($k > 0$), it will stay closed. That's because the amount of matter-energy of the universe is conserved, so if the density is greater than critical now it will forever be greater than ...
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