# "elliptic 3-manifolds are orientable"

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.

• The first follows from the Lefschetz Fixed Point Theorem. May 5 '17 at 17:51
• Do you see how the first statement implies the second? May 6 '17 at 1:34
• 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. May 6 '17 at 7:08
• The first also follows by explicitly constructing a homotopy to the antipodal map. Mar 18 '20 at 22:57

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.
• 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. Mar 18 '20 at 23:26
• @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. Mar 19 '20 at 0:18