Does the curvature parameter $k$ change with coordinate choice (de Sitter spacetime)?

Question

The de Sitter spacetime can be derived from the vacuum Friedmann equations given a choice of $k=0$, where $k$ defines the spatial curvature of the spacetime. The resulting metric in $(t,x,y,z)$ is given by $$ds^{2}=-dt^{2}+e^{2\sqrt{\frac{\Lambda}{3}}t}[dx^{2}+dy^{2}+dz^{2}].$$ The hypersurfaces of constant time therefore correspond to flat space (Minkowski in three dimensions scaled by an exponential term), which is consistent with the value of $k=0$ used in the metric derivation.

I have been motivated to switch to global coordinates $(t,\chi,\theta,\phi)$ since the above metric does not cover the entire manifold. The global de Sitter metric is given by $$ds^{2}=-dt^{2}+\frac{3}{\Lambda}\cosh^{2}\left(\sqrt{\frac{\Lambda}{3}}t\right)[d\chi^{2}+\sin^{2}\chi d\theta^{2}+\sin^{2}\chi\sin^{2}\theta d\phi^{2}]$$

and hypersurfaces of constant time now correspond to 3-spheres with some constant scaling.

Has the value of $k$ changed in switching coordinate systems as the constant time hypersurfaces have shifted from flat to spherical? I am inclined to say no, however I am struggling to justify this apparent inconsistency. Is this simply a result of considering the spacetime globally?

Answer 1

Well, the metric you started with is assuming that $k$, the spatial curvature is zero, $k=0$. So, even when you switch to global coordinates, the spatial curvature is bound to be zero. One way to see it is, the factor (scale factor) sitting in front of the spatial terms has changed, this scaling has affected all three spatial terms equally. Meaning $d\chi$ has not acquired any new terms which could possibly affect the spatial curvature.

A different way to approach is just staring at the spatial metric (or) the constant time hypersurfaces of both forms of the metric. In the first case,

$$d\Sigma_t= dx^2+dy^2+dz^2,$$ it is the usual euclidean space which is spatially flat.

In the second case where the spatial metric is,

$$d\Sigma'_t= d\chi^{2}+\sin^{2}\chi d\theta^{2}+\sin^{2}\chi\sin^{2}\theta d\phi^{2},$$

can be transformed back to Euclidean space by basic transformations. Meaning both the metric spaces are spatially flat as you have started with a flat space by forcing $k=0$.