How singular is the metric on an orbifold

I am reading some stuff on orbifolds. I am particularly interested in the metrics on orbifolds. The famous example of one orbifold is the "American football", which is $\mathbb{S}^2$ quotient by the group of rotation by $\pi$. This orbifold $\mathbb{S}^2/\mathbb{Z}_2$ inherits the metric from $\mathbb{S}^2$, which we call them $g_{AF}$ and $g_0$ respectively. If we pull them both back to $\mathbb{R}^2$, one can prove that $$g_{AF}=\frac{1+O(r)}{4r}g_0$$ where $r=\sqrt{x^2+y^2}$. To prove this, we must use some complex analysis technique. Clearly one can see the metic has cone singularity at origin.

Now suppose we have an orbifold of higher dimension $n\geq 3$ with isolated singularity. As we all know a neighborhood of the orbifold tip is diffeomorphic to a ball in $\mathbb{R}^n$ quotient by some finite group $\Gamma$. For simplicity, let us assume the orbifold inherits the manifold of $\mathbb{R}^n$. If we pull back the metric on this manifold, how does it look like? Still be conic?

• A combination of a cone and a Weyl chamber (if part of the finite group is a reflection group). – Peter Michor Feb 22 '18 at 16:15
• Let us assume there is no reflection group. How do you see it is a cone? – Slm2004 Feb 22 '18 at 17:39

There are more complicated things like cones. On the other hand, since the action is linear, each orbit spce has an action of $\mathbb R_{>0}$ (is a cone). (Open sets in) Weyl chambers might be present, for example. One thing that I know, is that the orbit space of a finite groups action is stratified into submanifolds which correspond to orbit types corresponding to cojugacy classes of isotropy groups inside the finite group. There is always one big connected and locally connected open dense stratum, the regular stratum. Besides this, codimension 1 strata (walls) can only come from reflections, if I remember correctly. For this I can point you to sections 29 and 30 of