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Consider $S^3$ defined in $\mathbb{C}$ as $$S^3 = \{(z_0, z_1) \in \mathbb{C}^2 \big\vert |z_0|^2+|z_1|^2=1 \}.$$

Now define for any natural number $n$: $$f_n : S^3 \rightarrow S^3 : (z_0, z_1) \mapsto (e^{n \pi i/7} z_0, e^{3n \pi i/7} z_1).$$

Now it's easy to see that each $f_n$ are homeomorphisms and that $G = \{f_n | n \in \mathbb{N}\}$ is a group.

I now want to compute the fundamental group of $S^3/G$. I don't really know where to start. I have some intuition that I might need the Seifert-Van Kampen theorem, but I can't seem to find a way in.

Geigercounter
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1 Answers1

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Since $S^3$ is simply connected, the fundamental group identifies with $G$. See here for a full proof (+ list of hypotheses to check, all of which hold).

Fundamental Group of a Quotient under Group Action

hunter
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  • I don't have a copy of this book but the point is that $S^3$ is the universal cover of this space and $G$ is the group of deck transformations of the cover, both of which you can check. (I'm reasonably sure that the correspondence between deck transofmrations and the fundamental group are covered.) – hunter Dec 13 '23 at 10:22
  • Could you give any hints on how to prove the hypothesis of the action of $G$ being properly discontinuous? – Geigercounter Jan 08 '24 at 17:58
  • @Geigercounter a finite group acting freely on a Hausdorff space always acts properly discontinuously. (Separate the points $gx$ from each other by open sets using the Hausdorff property -- you need the group to be finite here or else you might have to intersect infinitely many sets.) – hunter Jan 08 '24 at 23:46
  • I haven't been able to prove this. I can't seem to define a $U$ such that $g(U) \cap U = 0$. Could you maybe add this to your answer? – Geigercounter Jan 17 '24 at 17:02
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    @Geigercounter see here for example https://math.ucr.edu/~res/math205C-2011/freeactions.pdf – hunter Jan 17 '24 at 21:48
  • I don't immediately see how this relates to my question of properly discontinuous? – Geigercounter Jan 17 '24 at 21:54
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    @Geigercounter read the proof of the Theorem on the second page. – hunter Jan 17 '24 at 21:58