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Dec 14, 2019 at 18:45 history edited user108998 CC BY-SA 4.0
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Dec 14, 2019 at 18:34 comment added user108998 @abx ofc u are correct! Sorry about that.
Dec 14, 2019 at 18:31 comment added abx $\mathbb{Z}/p$ acts on $ \mathbb{S}^2 \cong\mathbb{P}^1_{\mathbb{C}}$ by $z\mapsto e^{\frac{2\pi i}{p} }z$, the fixed points are $0$ and $\infty$. But $\chi (\mathbb{S}^2)=2$.
Dec 14, 2019 at 12:12 comment added user108998 @guntram I'm not sure I follow, if P is fixed by \sigma then surely \nu +sigma^{*} \nu singular at P implies \sigma is. What am I missing?
Dec 14, 2019 at 10:41 comment added Guntram After averaging, the singularities may not be disjoint from the fixed points anymore.
Dec 14, 2019 at 10:09 history answered user108998 CC BY-SA 4.0