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

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.

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 $$O(3)\times\mathbb{R}\times\mathbb{Z}_2$$.
• Aha! I was missing the connection with the fundamental group $\pi_{1}$. So we have a reduction to a $U(1)\times \mathbb{Z}_2$ (H)-structure on $M$. Any idea what kind of invariant object this corresponds to? If it was just $U(1)$ I'd say it's an invariant 2-form similar to the electromagnetic field two-form, but with the $\mathbb{Z}_2$ I'm not sure Commented Aug 14, 2021 at 15:22
• @R.Rankin A geometric structure need not correspond to any particular invariant object, and for a general geometry $X=G/H$ it is possible to come up with all kinds of objects whose stabilizer is $G$. For instance, for $S^2\times\mathbb{R}\cong\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_2}{SO(2)\times\mathbb{Z}_2}$, one can regard the geometry as the stabilizer of $(H,V,g,[\omega])$, where $H$ and $V$ are the horizontal and vertical distributions on the product, $g$ is the standard product metric, and $[\omega]$ is an orientation of the $S^2$ fibers. Commented Aug 14, 2021 at 15:59
• You seem to have things backwards: the manifold $M$ is a quotient of the model geometry $X=G/H$ by a discrete group $\Gamma\subseteq G$ acting freely on $X$. The quotient map $X\to M$ is thus a covering map, and since $X$ is by definition simply connected, we have $\Gamma\cong\pi_1(M)$. This has nothing to do with Postnikov towers or spin geometry; it's just basic covering space theory. Commented Aug 21, 2021 at 5:30