Quotient of $\mathbb R^n\rtimes O\left(n\right)$ by $\mathbb Z^n\rtimes D_8$

Question

What is the coset space $\frac{\mathbb R^n\rtimes O\left(n\right)}{\mathbb Z^n\rtimes D_8}$ as a manifold? I saw a claim that it is the $n$-dimensional torus $\mathbb T^n=S^1\times\cdots\times S^1$? Here the semidirect product is formed using the obvious action of $O\left(n\right)$ on $\mathbb R^n$, and $D_8$ is the dihedral group with $8$ elements, that is the symmetry group of a square.

Answer 1

It is a fibre bundle with fibre $\mathbb{T}^n$ and base $SO(n)/C_4$. The pull back of this bundle along the map $SO(n)\to SO(n)/C_4$ is the trivial bundle.

To see this note that an element of the coset space has the form $(v,A)$ for $A\in O(n), v\in \mathbb{R}^n$, where $(v,A)\sim(x+gv,gA)$, for any $x\in \mathbb{Z}^n$ and $g\in D_8$. Thus without loss of generality we may take $A\in SO(n)$ and $v\in \mathbb{T}^n$. However we still must identify $(v,A)\sim(gv,gA)$ for $g\in C_4$. That is the coset space is $$(\mathbb{T}^n\times SO(n))/C_4.$$

Projecting onto the second factor (mod $C_4$) gives the fibre bundle map: $$p\colon (\mathbb{T}^n\times SO(n))/C_4 \to SO(n)/C_4.$$

If we take a path $\phi$ in $SO(n)$ from the identity to an element $g\in C_4$, and map this path into $SO(n)/C_4$, we get a loop $\gamma$. This loop may be lifted to a homotopy from the identity map $$1\colon p^{-1}(e)\to p^{-1}(e),$$ to the homotopy sending $$((v,e),t)\mapsto (v,\phi(t)).$$Thus $$((v,e),1)\mapsto (v,g)=(g^{-1}(v),e).$$ In other words the monodromy action of $\pi_1(SO(n)/C_4)$ is given by the natural action of $C_4$ on $\mathbb{T}^n$.