Musings

The FRW universes have spatial slices, $M$, which are either $H_3= SL(2,\mathbb{C})/SU(2)$, $\mathbb{R}^3$ or $S^3=SU(2)$. The corresponding isometry groups of $M$ are $SL(2,\mathbb{C})$, $ISO(3)$ and $SO(4)=SU(2)\times SU(2)/\mathbb{Z}_2$. These spaces are said to be homogeneous and isotropic, which is to say that, given a point $p\in M$, the six-dimensional space of infinitesimal isometries can be broken up into three “rotations” about the point $p$ and three “translations,” which move the point $p$.

Our universe looks locally FRW, but it could just as well be $M/\Gamma$ for $\Gamma$ some freely-acting discrete subgroup of the isometry group. If we want $M/\Gamma$ to be homogeneous, but not isotropic, we should demand that $g\Gamma g^{-1}=\Gamma$, for $g\in G$, a 3-dimensional subgroup of the isometry group of $M$, so that the latter acts transitively on $M/\Gamma$. It’s not really clear how physical that assumption is. Locally, the space is the same as before; it’s the global topology that may or may not admit the three global killing vectors corresponding to “translations.” If (for argument’s sake) the “obstruction” lay outside our current horizon, there’s no way we could know that our Universe was not homogeneous.

Artificial as the restriction may be, it eliminates the possibilities based on $H_3$.

Even among quotients of $\mathbb{R}^3$, there are lots of possibilities. I’ve already mentioned $T^3$, where of course, the “shape” of the $T^3$ is up for grabs. But there are other, more exotic possibilities. Here’s my favourite: the dihedral group,

acts freely on $T^3$ via

So we could just as well have a universe whose spatial slices are $T^3/D$.

For quotients of $S^3=SU(2)$, we can take $\Gamma$ to be any finite subgroup of $SU(2)$ acting (say) on the right, while the $SU(2)$ group of “translations” acts on the left. These finite subgroups have a famous ADE classification. In addition to the two infinite series, $\mathbb{Z}_{n+1}$ and $Q_{n-2}=\langle a,b,c\mid a^2=b^2=c^{n-2}=a b c = -1\rangle$, there are the binary tetrahedral, binary octahedral and binary icosahedral groups (the double covers of the corresponding rotational symmetry groups of the respective Platonic solids). The binary icosahedral group, $I$, has 120 elements and — you guessed it — the “Poincaré Dodecahedral Space” is none other than $S^3/I$.

Now, it’s true that out of the infinite number of finite subgroups of $SU(2)$, most are unsuitable, as they will produce deviations at large multipoles as strongly as they will at small multipoles. Small subgroups, like $\mathbb{Z}_2$ and $\mathbb{Z}_3$ may be problematic because, to get the desired deviation in $l=2,3$, one may need the radius of curvature of $S^3$ to be small enough that the $\Omega_0$ must differ appreciably from 1.

It may well be that $I$ is the unique possibility for a group by which to quotient $S^3$, which is both large enought to allows $\Omega_0$ to be very close to 1 ($\Omega_0\sim 1.013$ is their fit to the data), while still affecting predominantly the low ($l=2,3$) multipoles. But you’d never be able to tell that from their analysis.