So i have been reading a little bit about riemann surfaces and at some point the author says that if a group $G$ acts on $X$, our riemann surface, by holomorphic automorphisms, such that :

1) Around each point $p \in X$ we can find an open neighborhood $N$ such that if $q_1,q_2 \in N$ and $g.q_1=q_2$ then $g=id$ and $q_1=q_2$.

2) Suppose $p,q \in X$ wich do not lie in the same orbit, then there are neighborhoods $N_1,N_2$ of $q_1,q_2$ such that $GN_1$ is disjoint from $N_2$.

Now i am trying to see that in fact $X/G$ is a Riemann surface. I think condition 2) will give us the fact that the space is hausdorff, and 1) says that for every element we can find a neighborhood where the projection is a bijection and these are going to be the open sets to construct our riemann surface, but i cant seem to find the coordinate charts, i know i can find coordinate charts in $X$ and i can send $N$ to $X$ but this open set can be partitioned into the different open sets that give $X$ the structure of a riemann surface. So any tips would be appreciated , thanks.

  • $\begingroup$ You can take the neighbourhood of $q_1$ that is homeomorphic to $\mathbb{C}$ and intersect it with $N$ (and possibly shrink it), this is an appropiate neighbourhood of $\pi(q_1)$ that is homoeomorphic to $\mathbb{C}$. $\endgroup$
    – Matthias
    Nov 24, 2019 at 19:26

1 Answer 1


Yes, the second statement implies that the quotient (with quotient topology) is Hausdorff.
You can relax the hypotesis requiring that $g(N_1) \cap N_2 \neq \emptyset$ for at most a finite set of $g \in G$ (this is useful when checking the condition for example on complex tori).

The proof of this is a game of intersecting open subsets. Start with the quotient, recall what open subsets are, see what you need on the preimages such that the intersection of the images is empty.

The first statement is properness (in the sense of covering maps) of the action. It implies that the quotient map $$\pi \colon X \twoheadrightarrow X/G$$ is a covering map and in particular a local homeomorphism.
Let $y = [x] \in X/G$, choose a lift $x \in X$, statement (1) says that there is a $x \in U \subseteq X$ such that $g(U) \cap U \neq \emptyset$ implies $g = id$. This in particular gives that if $g_1 \neq g_2$ then $$g_1(U) \cap g_2(U) = \emptyset.$$

Let $[U] = \pi(U)$ be the image, which is open. Then $$\pi^{-1}([U]) = \bigsqcup_{g \in G} g(U)$$ which shows that $\pi$ is a covering map. In particular the restriction $$\pi_{|} \colon U \to [U]$$ is a homeomorphism, and you use those to push charts.

I have been a bit sketchy, but you can find all this in any book on covering maps.


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