algebraic-topologycovering-spacesnon-orientable-surfacessurfaces
I need to describe how a 4-genus orientable surface double covers a genus 5-non-orientable surface. I know that in general every non-orientable compact surface of genus $n\geq 1$ has a two sheeted covering by an orientable one of genus n-1. I tried to use the polygonal representation of these surfaces and try to get one from the other by cutting along some side as is done for the torus as a double cover of the Klein bottle. But It got really confused.
Thanks!
Consider the fundamental group of the Klein bottle, $K=\langle x,y\mid y^{-1}xy=x^{-1}\rangle$.
You are looking for a finite index subgroup of $K$ (call it $H$), that is abelian and non-normal. Note that any such covering factors through the orientable double cover, corresponding to the subgroup $\langle x, y^2\rangle$.
Consider $H=\langle xy^{-2}, y^6\rangle$. This is abelian, non-normal, and of index $6$ in $K$. It can be described by the diagram below:
To see that this covering is non-normal, here is a picture of the resulting torus (right before the final gluing): the different colored regions are left invariant by the deck transformations (you can also see there are two distinct paths traced by the black arrows, which border the different colors):
Since Euler characteristic is multiplicative with respect to (finite sheeted) coverings, we see the orientable double cover must have twice the Euler characteristic. This already pins down the homeomorphism type of the cover.
To see it more explicitly, consider the usual embedding of an orientable surface into $\mathbb{R}^3$ (as a long chain-like shape). This embedding can be translated/rotated around so that the reflections in all 3 coordinate planes maps the surface to itself.
With this embedding, there is a $\mathbb{Z}/2$ action on $\mathbb{R}^3$ given by the antipodal map $x \mapsto -x$. This action preserves the embedded surface and acts freely on it so the quotient is a manifold. Since the antipodal map is orientation reversing, the quotient is nonorientable.
Best Answer
Another way to proceed, which is a bit different from polygonal presentations, is to think about connect sums. If $S$ and $T$ are surfaces, then we can form $S \# T$, a new surface, by cutting a disk out of each, and gluing together the resulting circle boundaries. For example, if $S$ and $T$ are both copies of $T^2$ (torus) then $S \# T$ is a copy of the genus two surface. Drawing some pictures will help here.
Let's use $S_{g,n,c}$ to denote the connected, compact surface obtained by taking the connect sum of a two-sphere with $g$ copies of $T^2$ (torus), $n$ copies of $D^2$ (disk), and $c$ copies of $P^2$ (projective plane, aka "cross-cap"). One standard notation is $N_c = S_{0,0,c}$ for the non-orientable surface of "genus" $c$.
Another way to obtain $N_c$ is as follows. Take the two-sphere $S_{0,0,0}$. Cut out $c$ disjoint closed disks to get $S_{0,c,0}$. Identify boundary component with the circle $S^1$. Finally quotient each boundary component by the antipodal map $x \mapsto -x$. This gives $N_c$. (More pictures!) Thus the orientable double-cover of $N_c$ is obtained by gluing two copies of $S_{0,c,0}$ along their boundaries, to get $S_{c-1,0,0}$. (Yet more pictures!)
Here are some closely related questions:
Showing the Sum of $n-1$ Tori is a Double Cover of the Sum of $n$ Copies of $\mathbb{RP}^2$
Covering space of a non-orientable surface
Actually, my answer above is a more abstract version of the second half of https://math.stackexchange.com/a/279249/1307