Is this quotient always a solid torus? Consider the subgroup $G$ of the unit circle $\mathbb S^1 \subset \mathbb C$ generated by $\xi = \exp(2\pi/n)$, where $n\geq 2$.
Also, consider the closed unit disc $\overline{\mathbb D^2}\subset \mathbb C$ centered at $0$ and the solid torus
$\mathbb T := \overline{\mathbb D^2} \times \mathbb S^1$.
If we take the action $A:G \times \mathbb T \to \mathbb T$
by the formula $$
A(\xi,(z,\gamma)) := (\xi z, \xi^k\gamma),
$$
where $k$ is an integer and $0\leq k \leq n-1$, then we have the quotient $\mathbb T/G$.

Is the quotient $\mathbb T/G$ homeomorphic to $\mathbb T$?

I know the answer is affirmative in two cases:

*

*if $k=0$, then the quotient is $\overline{\mathbb D^2}/G \times \mathbb S^1$ and $\overline{\mathbb D^2}/G$ is a cone, which is homeomorphic to a disc.

*if $k$ and $n$ are coprimes, then the action is free and the quotient is a manifold. Let $l$ be an integer such that $lk = 1 \mod n$ and $0\leq l\leq n-1$. The map $f:\mathbb T/G \to \mathbb T$ given by $f([z,\gamma]) = (\gamma^{-l}z,\gamma^n)$ is a diffeomorphism.


So we can restrict ourselves to the case when $k$ and $n$ are not coprime.

The motivation for this question is that this problem appeared naturally while I was investigating some properties of Seifert manifolds.
 A: We can suppose $1<k\leq n-1$. Consider $d=\gcd(n,k)$ and define $F:\mathbb T \to \mathbb T$ given by $$F(z,\gamma) = (\gamma^{-l}z^d,\gamma^{n/d}),$$ where $l$ is an integer such that $kl = d \mod n$ and $1\leq l\leq n-1$.
It's clear that this map is surjective and continuous. If we have $F(z',\gamma') = F(z,\gamma)$, then $\gamma'^{n/d} = \gamma^{n/d}$ and therefore there is $s \in \mathbb Z$ such that
$$
\gamma' = \exp\Bigg(\frac{2\pi i s}{n/d} \Bigg) \gamma =\exp\Bigg(\frac{2\pi i ds}{n} \Bigg) \gamma.
$$
On the other hand we have $\gamma'^{-l} z'^d = \gamma^{-l} z^d$, and therefore,
$$
z'^d=\exp\Bigg(\frac{2\pi i lds}{n} \Bigg)z^d.
$$
Therefore, there is $t \in \mathbb Z$ such that
$$
z'=\exp\Bigg(\frac{2\pi i ls}{n} \Bigg)\exp\Bigg(\frac{2\pi i t}{d} \Bigg)z.
$$
Note
$$\exp\Bigg(\frac{2\pi i t}{d} \Bigg) = \exp\Bigg(\frac{2\pi i(n/d) t}{n} \Bigg).$$
Taking
$$
\omega = \exp\Bigg(\frac{2\pi i( ls +t(n/d))}{n} \Bigg) = \xi^{ls +t(n/d)}
$$
we have
$$
\omega^k = \exp\Bigg(\frac{2\pi i( ds +tn(k/d))}{n} \Bigg) =\exp\Bigg(\frac{2\pi i ds}{n} \Bigg) =\xi^{ds}.
$$
Therefore,
$$z'=z \omega \quad \text{and}\quad \gamma' = \omega^k \gamma,$$
$$
A(\omega,(z,\gamma)) = (z',\gamma').
$$
Since, $F$ is continuous and $G$-invariant, we have the continuous bijective map $f:\mathbb T /G \to \mathbb T$ given by $$f([z,\gamma]) := F(z,\gamma).$$
Since $\mathbb T/G$ is compact, the map $f$ is homeomorphism.
