# 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?

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.
• Thank you for your helpful answer! Can you maybe elaborate on why one says squashing along "Hopf fibers", when any one-dimensional linear subspace will do? I thought that one had to compactify the fibers so that they count as Hopf fibers, as $S^3$ is (roughly) locally a $S^2$ product with $S^1$. In the case of squashed $AdS_3$ this is done by identifying $\phi \sim \phi + 2\pi$, where can I see this in the Berger sphere or the Taub-NUT case? – horropie Sep 13 '19 at 10:15
• Sorry, in the notation of the paper we have $\sigma \sim \sigma + 2\pi$. – horropie Sep 13 '19 at 10:25