Why is the quotient of a manifold by a group action singular at a point with non-trivial stabilizer? My question is essentially the same as the question here, and I'm not sure if I fully understand that answer. I am reading Don Zagier's chapter in The 1-2-3 of Modular Forms, and the author says

The two points $\omega = \frac{1}{2}(-1 + i \sqrt{3}) = e^{2\pi i /3}$ and $i$ are stabilized by the cyclic subgroups of order $3$ and $2$ generated by $ST$ and $S$ respectively. This means that in the quotient manifold $\Gamma_1 \backslash \mathfrak{h}$, $\omega$ and $i$ are singular. (From a metric point of view, they have neighborhoods which are not discs, but quotients of a disc by these cyclic subgroups, with total angle 120 or 180 instead of 360).

The answer in the linked question seems to directly contradict this by saying that these points are not singular in the quotient. So are these points singular? And if so, can someone explain why, in general, the quotient of a manifold will be singular at a point with nontrivial stabilizer?
 A: As an American president once said, "it all depends on what the meaning of the word `singular' is." There are some ways in which the quotient is singular and some in which it is not:
Why it is nonsingular: The quotient space $Q$ is a topological manifold.  This topological manifold admits a compatible smooth structure. This topological manifold even admits a compatible Riemann surface structure (call it $X$) such that the natural projection $p: H^2\to X$ is holomorphic.
Why it is singular: 1. For every compatible smooth structure on $Q$ the natural projection $p: H^2\to Q$ is not a local diffeomorphism. Similarly, for every compatible Riemann surface structure $X$ on $Q$ the projection $p: H^2\to X$ is not locally biholomorphic.


*From the Riemannian geometry viewpoint (which is what Zagier has in mind, I think): Let $B\subset H^2$ be the (discrete) subset consisting of points where $p$ is not a local homeomorphism (this is the "branch-locus" of $p$). Its image $R=p(B)$  is the set of "ramification points" of $p$. Then there is a Riemannian metric $g$ (of constant curvature $-1$) on $X\setminus R$ such that the projection
$$
p: H^2\setminus B\to (X\setminus R, g)
$$
is a local isometry (where $H^2$ is equipped with the standard hyperbolic metric). However, $g$ does not extend to a Riemannian metric on the entire $X$ (moreover, it does not extend to any point of $R$). However, one can define a singular Riemannian metric on  $X$ extending $g$. Locally, near each point of $R$ the singular metric "looks line a cone". I prefer not to elaborate on this notion, Zagier gives you some intuition behind it.


*$Q$ has a natural orbifold structure ${\mathcal O}$ (which I will not attempt to define here), making the projection $p$ an "orbi-covering map", where each point of $R$ is a singular point of the orbifold ${\mathcal O}$.
