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Armstrong question 7.8:

Let $G$ be a finite group which acts as a group of homeomorphisms of a closed surface $S$ in such a way that the only element with any fixed points is the identity.

  1. Show that the orbit space of $S/G$ is a closed surface.
  2. Show that $S$ may be orientable, yet $S/G$ nonorientable.
  3. If $S/G$ is orientable, does $S$ have to be so?

A group action for which only the identity element has any fixed points is said to be fixed-point free, and the group in question is said to act freely.

I have come up with part of the solution, I think.

  1. $G$ fixed-point-free and finite makes each equivalence class in the orbit space a finite subset of $S$. So $S/G$'s neighborhoods are a finite union of $S$'s neighborhoods. Then it can be checked $S/G$'s topology inherits each of the necessary properties from $S$: compact, connected, Hausdorff, every point has a neighborhood homeomorphic to the plane.
  2. $S^2$ is orientable yet $\mathbb{Z}^2$ acting freely on it by identifying antipodes is homeomorphic to the real projective plane.
  3. I think yes but cannot show it with rigor, edited due to Mariano's comments:

    Assume $S$ is a simplical complex, $G$ is a group of simplical homeomorphisms and $c$ is a simple closed curve along edges of $S$. (It takes work to show these assumptions do not lose generality.) $c$ can be identified with a closed curve $c'$ in $S/G$ through the equivalence relation. Consider thickening $c'$ into a subcomplex $C'$ of a subdivision of $S/G$. $S/G$ is orientable so thickening any simple closed loop in $C'$ yields a cylinder. One can cover $c'$ with a collection of such cylinders: call such a covering $\mathcal{C}_{S/G}$. Through the homeomorphisms in $G$, the elements of $\mathcal{C}_{S/G}$ are homeomorphic to a set of cylinders in $S$, call them $\mathcal{C}_S$. $\mathcal{C}_S$ must cover $c$ and its complexes must be compatibly orientable since they came from $\mathcal{C}_{S/G}$. Then some thickening of $c$ is orientable, so $S$ is orientable.

These steps are heavily abridged and the argument is involved compared to the other exercises.

Is this line of reasoning correct? I remain suspicious $\mathbb{Z}_2$ can somehow act on the real projective plane to yield $S^2$.

Christian Chapman
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  • Up to this point the only characterizations of orientation in Armstrong are: (1) a surface is non-orientable if it contains a Mobius strip (2) an orientation is an ordering on a simplex's vertices (3) a combinatorial surface's simplexes can be compatibly oriented. – Christian Chapman Dec 16 '23 at 16:15
  • I can’t think of an example but this does break my line of reasoning, back to trying to come up with a counterexample to the claim – Christian Chapman Dec 16 '23 at 21:49
  • Oops thanks, meant $S$ is a simplical complex – Christian Chapman Dec 17 '23 at 21:43

1 Answers1

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If $S$ is nonorientable in the sense of Armstrong so that it contains a Mobius band, then we can use the fact the "middle loop" of the Mobius band has the property that its normal vector reverses direction when one traverses the loop. If $S$ is a cover of an orientable surface $\bar S$, then the projection of the "reversing" loop to $\bar S$ will be another reversing loop, contradicting its orientability. Hence $S$ must also be orientable.

Mikhail Katz
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  • Far easier, I am sure this is what was intended. Thanks – Christian Chapman Dec 19 '23 at 07:00
  • Note however that this relies on classification results for surfaces, namely that homeomorphism classes are the same as diffeomorphism classes. WIthout a smooth structure one can't construct a metric, and without a metric one can't talk about the normal vector. There may be ways of getting around this and still talk of local two-sidedness, but the argument would become more delicate. – Mikhail Katz Dec 19 '23 at 10:26
  • your first "is orientable" you meant as "is non-orientable" right? – Christian Chapman Dec 27 '23 at 23:31