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I would like to prove that I can (or can't) change curvature of space, k = 0,+1,-1, via general coordinate transformations, which in principle can mix space and time coordinates together.

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Let's be clear about what you mean by the term "space".

I'm assuming you take your spacetime, and pick out a specific three dimensional submanifold $\Sigma$, say by a function $f(x^{\mu})=$ const. You now compute the intrinsic Ricci curvature scalar for that submanifold. Now you make a coordinate transformation $x'^{\mu} = x'^{\mu}(x^{\nu})$. The new spatial coordinates are allowed to depend on the old $x^0$s. In the new coordinates, the new function specifying the same $\Sigma$ will be different to $f$, but we can still calculate the intrinsic curvature of $\Sigma$ in the new coordinates.

This intrinsic curvature is a geometric property of $\Sigma$ independent of the embedding. The Ricci scalar of $\Sigma$, i.e. $^3R$ assigns a number to each point in $\Sigma$. This number is the same regardless of the values of the coordinate system used on $\Sigma$.

If, however, by "space" you had meant the surface $x^0=$const given by the chosen coordinate system, then when we change coordinates, $x'^0=$const represents a completely different submanifold which may have different intrinsic curvature.

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By the term "space" I meant this: If 4D metric is given by

G_μν = diag[1,-a(t)^2/1-kr^2,-a^2(t)r^2,-a(t)^2r^2sin(Θ)^2],

in some coordinates (t,r,Θ,ϕ), then a 3D "subspace" is simply

G_ij = diag[-a(t)^2/1-kr^2,-a^2(t)r^2,-a(t)^2r^2sin(Θ)^2],

which seems like your other case, surface of t=const.

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  • $\begingroup$ Hi Winnie. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. This markup is very useful for understanding math equations and parameters. Please have a look here for an intro or our FAQ for more info. For example, $\mu$ results $\mu$, $\omega$ inserts $\omega$, etc. It's quite interesting. You can revise your post if you can ;-) $\endgroup$ – Waffle's Crazy Peanut Apr 19 '13 at 13:22

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