$\frac{SO(3)\times SO(2)}{SO(2)}$ = $\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_{2}}{SO(2)\times\mathbb{Z}_{2}}$

algebraic-topologycyclic-groupsgeneral-topologygroup-theorylie-groups

On a manifold $M=S^{2}\times S^{1}$ I'm trying to find the homogenous model/Klein space geometry, and define a reduction of it's structure group. Now since $S^{2}=\frac{SO(3)}{SO(2)}$ we have that:

$$\frac{G}{H}=\frac{SO(3)\times SO(2)}{SO(2)}$$

Where $G=SO(3)\times SO(2)$ is transitive on $M$ and $H=SO(2)$ is the point stabilizer. As far as I can tell everythings all right here; however, in reading about Thurston's Geometerization Conjecture, this isn't one of the eight Thurston geometries for a closed, oriented, 3-manifold. For this manifold it should be of the form:

$$\frac{G}{H}=\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_{2}}{SO(2)\times\mathbb{Z}_{2}}$$

Which is pretty close. I'm guessing these spaces have to be isomorphisms of one another. I can equivalently define the first equation in terms of double covers:

$$\frac{G}{H}=\frac{SU(2)\times U(1)}{U(1)}$$

($U(1)=SO(2)$ double covers itself) and use a hand-wavy argument that $SO(n)\times\mathbb{Z}_{2}=Doublecover(SO(n))$. That still doesn't explain $\mathbb{R}$ versus $SO(2)$ in our transitive group. The former is the universal cover of the latter so maybe that comes into play? Or perhaps ${\mathbb{R} \over \mathbb{Z}_{2}} =SO(2)$?

I'm specifically interested in a reduction of the G-bundle to an H-bundle so how this breaks down is important for me.

Best Answer

Recall that a geometric structure on $M$ is a diffeomorphism $\varphi:X/\Gamma\to M$ where $G$ is a Lie group, $H\subseteq G$ is a compact subgroup, $X=G/H$ is a model geometry, and $\Gamma$ is a discrete subgroup of $G$ acting on $X$ in the canonical way.

The model geometry of $M=S^2\times S^1$ is $X=\frac{O(3)\times\mathbb{R}\times\mathbb{Z}_2}{O(2)\times\mathbb{Z}_2}$, but this does not mean that $M\cong X$. Instead, we have $M\cong X/\Gamma$, and $\Gamma$ is nontrivial in this case (notice that $\Gamma$ is always isomorphic to $\pi_1(M)$). Instead, we have $\Gamma=\mathbb{Z}$, regarded as a subgroup of the $\mathbb{R}$ factor in $O(3)\times\mathbb{R}\times\mathbb{Z}_2$.