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i Want to do an exercise out of the book three dimensional Geometry and topology by thursten.

Which states Any isometry of $\mathbb{S}^3$ that has no fixed points is orientation-preserving. Inparticular every elliptic three-manifold (a quotient $\mathbb{S^3}/G,\;\;\;G \subset O(4)$) is orientable.

The first statement isn't to important for my application, but I would need the second statement and I have no clue how to even start proving it.

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    $\begingroup$ The first follows from the Lefschetz Fixed Point Theorem. $\endgroup$ May 5 '17 at 17:51
  • $\begingroup$ Do you see how the first statement implies the second? $\endgroup$ May 6 '17 at 1:34
  • $\begingroup$ no that's the real problem for me the first statement is fine with me. I think i can proof it myself, but I don't have a clue how to start the second one because i really have a problem with how orientability and quotients work together. $\endgroup$ May 6 '17 at 7:08
  • $\begingroup$ The first also follows by explicitly constructing a homotopy to the antipodal map. $\endgroup$
    – Steve D
    Mar 18 '20 at 22:57
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Here's a very useful general (and intentionally vague) principle to have in mind when dealing with quotients: if the action of $G$ on $M$ preserves some structure on $M$, then that structure should descend to the quotient $M/G$. That's super vague, so here's a more concrete claim: if $G$ is finite (so that the quotient map is a local diffeomorphism), then any tensor field on $M$ that is invariant under the action of $G$ descends to $M/G$ (i.e. is the lift of some tensor field on $M/G$).

In this case, the fact that the group action preserves orientation implies that the orientation on $M$ induces an orientation on $M/G$. The precise proof of this will depend on exactly how you're defining orientation. If you use volume forms, you can just average the volume form by the action of $G$ (since it's finite, right?) to get a nonvanishing $G$-invariant volume form on $M,$ which then descends to a volume form on $M/G.$

For $S^3$: Assuming that your group action is free, all non-identity $g \in G$ have no fixed points and thus (by the first statement) are orientation-preserving; and the identity is obviously orientation-preserving.

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  • $\begingroup$ 2 minor points: First, the assumption is that the action is by isometries (with the Riemannian metric implied to be the round metric, from the use of $O(4)$). Such an action will automatically preserve the (Riemannian) volume element - no need to average. Second, the OPs group $G$ is not $S^3$, but rather, it is $O(4)$. So a non-identity $g\in G$ can have fixed points, some can be orientation reversing, etc. $\endgroup$ Mar 18 '20 at 23:26
  • $\begingroup$ @JasonDeVito: good point regarding the averaging. As for the group, it was indeed posed as an arbitrary subgroup of $O(4)$ in the question; but the only subgroups of $O(4)$ that act freely on $S^3$ (and thus yield elliptic manifolds) are in fact subgroups of $SO(4)$ - this is the information obtained from the "first statement" in the OP. So there's no problem. $\endgroup$ Mar 19 '20 at 0:18
  • $\begingroup$ I see I didn't read your last paragraph correctly - we are on the same page now. Thanks! $\endgroup$ Mar 19 '20 at 1:12

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