Quotients of $S^{2n+1}$ Any sphere $S^{2n+1} = SO(2n+2)/SO(2n+1)$ can be thought to be given as the zero-set in $\mathbb{C}^{n+1}$ of the equation, $\sum_{i=1}^{n+1} \vert z_i \vert ^2 = 1$ 
Now say one wants to quotient it such that the identification is, $\phi_i \sim \phi_i + \alpha_i$ where we think of each $z_i$ to be given as $z_i = r_i e^{i\phi_i}$. 


*

*For this quotienting to be well defined is it necessary that there exist integers $n_i$ such that $\sum_i n_i \alpha_i = 2\pi$?
I vaguely feel that if this condition is not satisfied then the resulting space would become an "orbifold" and may not be a manifold. But I can't make it precise..

*Now consider the special case when $\alpha_1 \neq 0$ and all the others $\alpha$ are $0$.    


*

*In this scenario is is there a discrete group $\Gamma$ (of $SO(2n+1)$) such that the resulting quotient space can be thought of as $\Gamma \backslash SO(2n+2)/SO(2n+1)$?

*Is $\Gamma =  \mathbb{Z}_{n_1=\frac{2\pi}{\alpha_1}}$? If yes, why? 

*Is this quotient space somehow naturally related to $S^{2n}\times S^1$?




Although I don't immediately need it but it would be interesting to know if someone can tell the corresponding statements to the above in the case when more than one $\alpha$ are non-zero. 
 A: Note that I use $n$ where the question uses $n+1$ since I didn't want to carry around all the $+1$'s, so here we are looking at $S^{2n-1} = \mathrm{SO}(2n) / \mathrm{SO}(2n-1) \subset \mathbb{C}^n$.
First note that the map $f(r_1 e^{i \phi_1}, \dots, r_n e^{i \phi_n}) = (r_1 e^{i (\phi_1 + \alpha)}, r_2 e^{i \phi_2}, \dots, r_n e^{i \phi_n})$ is an element of $\mathrm{SO}(2n)$, it is just rotation by $\alpha$ in the $z_1$ plane. It is given by the matrix in $GL(n, \mathbb{C})$, where $I$ is the $(n-1) \times (n-1)$ identity matrix:
$$
\begin{pmatrix}
e^{i \alpha} & 0 \\
0 & I
\end{pmatrix}
$$
Let $A_i$ denote rotation by $\alpha_i$ in the $z_i$ plane. Hence your equivalence relation is the same as quotienting by a certain subgroup of $\mathrm{SO}(2n)$.
Note that the subgroup generated by all the rotations in the $z_i$ planes is a torus, namely all the diagonal matrices in $\mathrm{GL}(n, \mathbb{C})$ with diagonal entries of length $1$, call this subgroup $T$.
I am not sure if you are considering quotienting by the subgroup generated by $\{ A_1, \dots, A_n \}$ or by the subgroup generated by $A = A_1 \cdot \dots \cdot A_n$. Where $A$ is the diagonal matrix:
$$
\begin{pmatrix}
e^{i \alpha_1} \\
& e^{i \alpha_2} \\
&& \ddots\\
&&& e^{i \alpha_n}
\end{pmatrix}
$$
In the first case consider $\alpha_i \in \mathbb{R} \setminus \pi \mathbb{Q}$ for some $i$, you get that each equivalence class $\phi_i + \alpha_i \mathbb{Z}$ is dense in $S^1$, but you still have uncountably many equivalence classes in $S^1$. But all these equivalence classes will be indistinguishable in the quotient topology, in the sense that if you have any open set, it either contains none of them or all of them. Hence the quotient space is not Hausdorff. We can restrict to considerations on $S^1$ like this since, quotienting by the subgroup generated by the individual $A_i$ is the same as the equivalence relations on angles given by $(\phi_1, \dots, \phi_n) \sim (\phi_1 + k_1 \alpha_1, \dots, \phi_n + k_n \alpha_n)$, and we can then fix the $r_i$, and the $k_j$ for $j \neq i$, since we don't risk collapsing equivalence classes in the $\phi_i$ parameter by rotating in the $\phi_j$ parameters. Hence the quotient of the sphere will also be non-Hausdorff.
So in the first case you probably want all the $\alpha_i$ to be rational multiples of $\pi$.
In the second case, for fixed $r_i$ you are looking at an n-dimensional torus and quotienting by the subgroup generated by a single element, the condition you mention looks a lot like $2\pi, \alpha_1, \dots, \alpha_n$ being linearly independent over $\mathbb{Q}$, which is exactly the condition equivalent to the subgroup generated by $A$ not being dense in the torus of rotations. This follows from Kronecker's approximation theorem, near the end of that page is the exactly the specialization that is interesting in this case.
But something stronger than linear dependence over $\mathbb{Q}$ is needed. Think of $n=2$, $\alpha_1 = \alpha_2 = \frac{1}{2}$. This is linearly dependent over $\mathbb{Q}$ but visits $0$ infinitely often. In the sense that for any $m$ and $\epsilon > 0$ we can find $M > m$ such that $(\frac{M}{2}, \frac{M}{2}) \in B_\epsilon(0)$, since $e^{i \frac{1}{2} \mathbb{Z}}$ is dense in $S^1$. And considering $n=3$, $\alpha_1 = \alpha_2 = \frac{1}{2}$, $\alpha_3 = \frac{\pi}{2}$ these parameters fulfill the exact requirement suggested but still give a non-Hausdorff quotient.
We can generalize this argument by considering $\alpha_i \in \mathbb{R} \setminus \pi \mathbb{Q}$ for some $i$. Then the subgroup generated by $A$ is non-finite, consider $H = \overline{\langle A \rangle}$ the closure of the subgroup generated by $A$, this is a closed subgroup of a compact Lie group hence a compact Lie group, it is non-finite and abelian hence the identity components $H_0$ is a torus, at least 1 dimensional. We see $H / \langle A \rangle$ is uncountable but has the trivial topology since $\langle A \rangle$ is dense in $H$. Look at the closed subset $H . p \subset S^{2n-1}$ for $p \in S^{2n-1}$, this is closed since $H$ is compact. If we choose $p$, such that all the $r_i > 0$ then the stabilizer in $T$ of $p$ is $e$. Therefore $H . p$ is a submanifold of $S^{2n-1}$ diffeomorphic to $H$, hence when we quotient by $A$ we get the same topological behavior as on the group. So the quotient of the sphere cannot be Hausdorff.
So we are probably still only interested in rational multiples of $\pi$, assuming the above checks out.
If you consider $\alpha_1 = \frac{p}{q}2\pi$, and $\alpha_i = 0$ for $i > 1$, $\frac{p}{q}$ a reduced fraction. It is now quite easy to see that $A_1$ has order $q$ just by taking higher and higher powers, hence it generates a cyclic subgroup of order $q$. So in this case $\Gamma \simeq \mathbb{Z}_q$.
When $\alpha_i = \frac{p_i}{q_i} 2\pi$ with $\frac{p_i}{q_i}$ reduced and $p_i \neq 0$. We see that $\langle A_1, A_2, \dots, A_n \rangle \simeq \mathbb{Z}_{q_1} \times \dots \times \mathbb{Z_{q_n}}$ whereas $\langle A \rangle \simeq \mathbb{Z}_{\mathrm{lcm}(q_1, \dots, q_n)}$. Hopefully it is now an easy exercise to see what happens with $p_i = 0$ for some of the $i$'s.
I know this doesn't answer all of your questions but hopefully it gives some insight, I will continue thinking about the problem.
