0

Below is an excerpt from an example from Forster's Lectures on Riemann Surfaces. In the example, I cannot figure out why the quotient space $\mathbb{C}/\Gamma$ is Hausdorff. How do we show this fact?

enter image description here

nomadicmathematician
  • 5,511
  • 1
    Follows from https://math.stackexchange.com/questions/91639/x-sim-is-hausdorff-if-and-only-if-sim-is-closed-in-x-times-x?rq=1 – Eric Wofsey Jan 09 '20 at 04:47

1 Answers1

1

Take $u_1,u_2\in \mathbb{C}/\Gamma$. Consider the two representant $x_1,x_2 \in \mathbb{C}$ of $u_1,u_2$ respectively.

Now set $$r = \frac{1}{3} dist(\Gamma x_1, \Gamma x_2)$$ and consider $B_i$ the open ball of center $x_i$ and radius $r$. Obviously $B_1\cap B_2=\emptyset$. Note that $\pi(B_1)$ and $\pi(B_2)$ are open.

Let $u\in \pi(B_1)\cap \pi(B_2)$ be a point. Let $x\in\mathbb{C}$ be a representant of $u$. Then $x$ lie in $\gamma_1B_1\cap \gamma_2B_2$ for some $\gamma_1,\gamma_2\in\Gamma$ (two translations of the initial balls). Let $y_1,y_2$ be the centers of $\gamma_1B_1$ and $\gamma_2B_2$ respectively. Then \begin{eqnarray} dist(y_1,y_2) &\leq& dist(y_1,x)+dist(x,y_2) \\ &<&r+r\\ &\leq& \frac{2}{3} dist(\Gamma y_1,\Gamma y_2)\\ &\leq& \frac{2}{3} dist(y_1, y_2) \end{eqnarray} which is absurd. Hence $\pi(B_1), \pi(B_2)$ are disjoint open neighborhoods of $u_1$ and $u_2$.

Dominique Mattei
  • 890