Explanation for "Squashing" and "Stretching" (Lorentzian Analogue of Berger Spheres) In the paper Anti-de Sitter space, squashed and stretched Bengtsson and Sandin introduce the Lorentzian analogue of the squashed 3-sphere. After looking up Berger spheres, it seems what is meant with "squashing" in the case of $S^3$ is the following: 
Take the Lie algebra $L$ spanned by $z_1$,$z_2$ and $z_3$, which fulfill the necessary relations so $L$ can generate $S^3$. Then take the quotient group $B(\beta) = G/H$ of the group $G = S^3 \bigoplus \mathbb{R}$ by the one-parameter subgroup $H$ generated by (e.g.) $\alpha z_1 + \beta z_2$, where $\alpha^2 + \beta^2 = 1$, $z_1 \in L$ and $z_4$ a left invariant vector field tangent to $\mathbb{R}$. $B(\beta)$ is then called a Berger sphere. I interpret this as "squashing" $S^3$ along $\mathbb{R}$. Why is the condition $\alpha z_1 + \beta z_2$ necessary?
Coming back to the mentioned paper, the authors state that Taub-NUT cosmologies can be described by squashing $S^3$ along Hopf fibres, i.e. $S^1$. This confuses me as the topology is just $S^3 \times \mathbb{R}$, is there a covering space involved?
In the Lorentzian analogue of squashed $S^3$, $AdS_3$ has the topology of $\mathbb{R}^2\times S^1$ and $z_4$ can be either spacelike ($z_4^i z_{4,i} = 1$) or timelike ($z_4^i z_{4,i} = -1$). Consequently there are two ways of of squashing, but along which fibre does one squash in this case? Furthermore, what is meant with "Stretching"?
EDIT: 
So I figured out at least the "stretching" part:
For example at one stage the introduce the metric of $AdS_3$, squashed/stretched along a spacelike fibre:
\begin{align}
   \mathrm{d}s^2_\lambda = \frac{1}{4}\left(-\mathrm{cosh}^2(\omega)\mathrm{d}\tau^2 + \mathrm{d}\omega^2 + \lambda^2 \left(\mathrm{d}\sigma + \mathrm{sinh}(\omega)\mathrm{d}\tau\right)\right)\,,
\end{align}
where $\lambda$ is a real stretching parameter. They simply call the case $\lambda^2 < 1$ "squashing" and $\lambda^2 > 1$ "stretching".
Furthermore, the Hopf-fibres, along which one squashes $AdS_3$, are the geodesic congruences of the spacetime, i.e. $\partial_t$ and $\partial_{\phi}$, which both are compactified do to identifications like e.g. $\phi \sim \phi + 2 \pi$.
The two standing questions are:


*

*Along which fibres is one squashing $S^3$ to get Taub-NUT space?

*Is in the case of the Berger sphere the condition $\alpha^2 + \beta^2 = 1$ necessary as a identification condition, if not why is it necessary?

*Where are the Hopf fibres in the calculation, along which we are squashing?

 A: The subgroup $H$ is not a normal subgroup (or invariant subgroup, in physicist's language), so the quotient $B(\beta)=G/H$ is not naturally a quotient group, only a quotient space. The condition $\alpha^2+\beta^2=1$ is not necessary. Indeed any vector with a nonzero component of $z_4$ will do. Better: any one-dimensional linear subspace of the Lie algebra of $G$ generates a Lie group $H$ which gives a Berger sphere quotient space. Cheeger and Ebin choose to normalize that linear subspace by automorphisms of the 3-sphere, so that they just get finitely many Berger spheres in each isometry class. The Hopf fibers are the curves on which $\omega$ and $\tau$ are constant, parameterized by $\sigma$.
The description in terms of a 4-dimensional Lie group $G$ is not very natural, but fits into a bigger picture in the book of Cheeger and Ebin. It is easier to simply write that the Hopf fibration is given by taking any connected 1-dimensional subgroup $K$ of $S^3$, and then the quotient $S^3\to S^3/K$ is a Hopf fibration. Since the adjoint action of $S^3$ acts transitively on 1-dimensional subspaces of its Lie algebra (i.e. rotations of 3-dimensional Euclidean space act transitively on lines through the origin), $S^3$ acts transitively, by conjugation, on all connected 1-dimensional Lie subgroups. Hence the choice of $K$ is arbitrary. In your basis $z_1,z_2,z_3$ for the Lie algebra of $S^3$, you could just take dual basis, say $\omega_1,\omega_2,\omega_3$, and then your Hopf fibration can have fibers given by setting two linearly independent linear combinations of these two zero, say $0=\omega_1=\omega_2$, so the subgroup $K$ is then tangent to $z_3$. We can make the Berger metric then be anything of the form $\omega_1^2+\omega_2^3+\lambda \omega_3^2$, carried around $S^3$ by left (or right, if you prefer) translation.
In the Lorentzian picture, there are 3 different connected 1-dimensional Lie subgroups of $\operatorname{SL}_2\mathbb{R}$, so there are two different ways to carry out a 1-parameter family of Hopf fibrations.
