algebraic-topologycovering-spacesfundamental-groups
I am wondering if there is a covering map of the Torus by the Klein Bottle. I know that there is a covering map of the Klein Bottle by Torus given here Two-sheeted covering of the Klein bottle by the torus.
I also do not need to know what the covering is, just if there is such a covering.
Any help would be greatly appreciated.
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):
I'm not sure where you get $3$ circles from. The model only has two $1$-cells after identifying - horizontal and vertical edges. This coincides with the usual $1$-skeleton for the Klein bottle given by the wedge of two circles.
The picture you should have drawn is a rectangle with $7$ edges; four are vertical, all labelled $a$ and pointing in the same direction, the last three are horizontal, all labelled $b$ and pointing right, left, right as we go from top to bottom of the rectangle.
There are two $2$-cells, both labelled $A$, the top one oriented clockwise, the bottom one oriented anti-clockwise.
Each half is a fundamental domain equal to the usual model of the Klein bottle, and if you remove the central horizontal $1$-cell, gluing the two $2$-cells together along the edge, and forget about identifying the vertical edges with their diagonal partners (so only identifying top-left with top-right, and bottom-left with bottom-right), then this model is the usual model for the torus.
Best Answer
There can be no such covering projection. This follows from standard results on how the fundamental groups are related in a covering projection. (If $K$ is the Klein bottle and $T$ is the torus, then the fundamental group, $\pi_1(K)$ of the Klein bottle is isomorphic to the non-abelian group $\langle x, y \mid x^2 = y^2 \rangle$ while $\pi_1(T)$ is isomorphic to the abelian group $\Bbb{Z} \oplus \Bbb{Z}$. The existence of a covering projection of $K$ onto $T$ would imply that $\pi_1(K)$ was isomorphic to a subgroup of $\pi_1(T)$, which is impossible.)