I am trying to prove the following "For an orbifold chart $ (\tilde{U},G,\phi)$ the set of non fixed point of $ g : \tilde{U} \rightarrow \tilde{U} $ where $ 1 \neq g \ \in G$ is dense in $\tilde {U}$"
This is the second remark in the page 2 of the paper -
http://www.math.colostate.edu/~renzo/teaching/Orbifolds/pronk.pdf
They say "Indeed, a non trivial automorphism of the connected set $\tilde{U}$ of finite order cannot fix a non empty open set"
This is what I have done so far -
Consider the set $ B = \{ x \in \tilde{U} \vert x $ has a nbd fixed by $g \}$
$B$ is clearly open. If I show $B$ is also closed then I know that $B$ must be either $\tilde{U}$ or empty. Since $g\neq 1$ I see that $B\neq \tilde{U}$ and hence must be empty. This will give me the "Indeed" part.
I am stuck at showing that $B$ is closed. I can show that if $y\in\bar{B}$ then $g(y)=y$. I think the finite order of $g$ must be used some how to show that $y\in B$.
Is the way I am proceeding with it correct so far?
Thank you.
On your orbifold chart $V$ your group $G$ acts (faithfully) by diffeomorphism. The thing is just to say you can choose a Riemaniann metric on $V$ such that $G$ acts isometrically.
This can always be done because you can choose a Riemaniann metric $g$ and take the average of it over $G$ that is :
$$g_0(X,Y):=\frac{1}{|G|}\sum_{\sigma\in G}g(T(\sigma).X,T_{\sigma}.Y)$$
Now suppose $\sigma$ has an open set of fixed points on $V$ and take $v$ to be in this open set. Then you have :
$$\sigma.v=v\text{ and for every tangent vector } X\text{ at } v \text{, } T_v\sigma.X=X$$
The last one holds because $\sigma$ fixes a neighborhood of $v$.
Now we use the fact that an isometry which fixes a point and induce the identity on the tangent space of this fixed point must be the identity on the whole space (provided it is a connected space). And you are done.