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Let $(U, G, \phi)$ be a $n$-dimensional complex orbifold chart over $x \in X$, i.e., $x \in \phi(U)$

I want to know if there is a a subset $V$ of $U$ such that $(V, G_x, \phi|_V)$ is also an orbifold chart over $x$. Here is how I began: choose $\tilde{x} \in \phi^{-1}(x)$; for all $g \in G-G_x$ there is a neighborhood $U_g \subset U$ of $\tilde{x}$ such that $U_g \cap g(U_g) = \emptyset$. Define

$$V' := \bigcap_{g \in G-G_x} U_g.$$

Now I need to find an open connected subset $V$ of $V'$ such that $g: V \to V$ is well defined for all $G \in G_x$. I cannot find it. My guess is that it would be a connected component of a good set.

Bruno Suzuki
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1 Answers1

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Your guess is good. Take $\tilde{x}\in\phi^{-1}(x)$ and define $G_{\tilde{x}}:=\{g\in G\mid g.\tilde{x}=\tilde{x}\}$.

Let $U_{\tilde{x}}$ be a neighborhood of $\tilde{x}$ in $U$. For $g\in G_{\tilde{x}}$ we have that $g\cdot U_{\tilde{x}}$ is a neighborhood of $\tilde{x}$ whence :

$$V_1:=\bigcap_{g\in G_{\tilde{x}}}g\cdot U_{\tilde{x}}\text{ is a $G_{\tilde{x}}$-stable neighborhood of }\tilde{x}\text{ in }U $$

Define $V$ to be the connected component of $V_1$ containing $\tilde{x}$. This will be the neighborhood you are looking for. The only thing you need to show is that $V$ is stable by the action of $G_{\tilde{x}}$. Let $g\in G_{\tilde{x}}$, $g\cdot V$ is connected since the action is topological. Since $V_1$ is $G_{\tilde{x}}$-stable, $g\cdot V$ is included in $V_1$. Furthermore $g\cdot V$ is connected so it is included in one connected component of $V_1$. Remark that $\tilde{x}\in V$ by definition, so $\tilde{x}=g\cdot \tilde{x}\in g\cdot V$.

Finally $g\cdot V$ is included in the connected component of $V_1$ containing $\tilde{x}$ which is $V$. We have shown that $V$ is also $G_{\tilde{x}}$-stable.

Clément Guérin
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