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Based on the Friedmann equation for a universe with only cosmological constant, $$\left(\frac{\dot{a}}{a}\right)^2 \sim \Lambda$$

I would expect the scale factor $a(t) \sim e^{-it}$ if $\Lambda < 0$. Is this an anti de Sitter universe? How do we interpret the scale factor?

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up vote 3 down vote accepted

Because your solution is mathematically valid but unacceptable physically because $a(t)$ is required to be real, you have proved that the equation has no physical solutions. In other words, a negative-cosmological-constant space can't be sliced into flat spatial slices (you have assumed that the spatial curvature term $k$ is absent as well: with a negative spatial curvature, you could get solutions).

In particular, the AdS space can't be sliced into flat slices. This is also clear if you think about the isometry group. $AdS_4$ has the $SO(3,2)$ isometry group and the Euclidean affine symmetry of $R^3$, $SO(3)$ semidirect product with translations (which must commute with each other), is clearly not a subgroup of it. However, the $H^3$ negatively curved hyperbolic space has the $SO(3,1)$ isometry group (includes "translations" that don't commute) but it is a subgroup of $SO(3,2)$ so you can find solutions to the FRW equations with the negative $k$ added.

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