Cdtjkyj

Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.

Is there a non trivial covering of $K$ by $K$?

The universal cover of $K$ is $Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.

Thank you for any hints and help.

One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $Tto K$ is fairly straightforward.

This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).

The Klein bottle is the quotient of $mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(1-x,y)$ and $v(x,y)=(x,y+1)$Consider $f(x,y)=(x,2y)$ $fcirc u(x,y)=f(1-x,y)=(1-x,2y)=ucirc f(x,y)$.

$fcirc v(x,y)=f(x,y+1)=(x,2y+2)=v^2circ f$. This implies that $f$ induces a continuous map of $mathbb{R}^2/G$ this map is a covering of order $2$.

We can compute many examples by understanding $pi_1(K)$ and its action on $K$'s universal cover. Using van Kampen we can compute

$$pi_1(K) cong langle a, b | bab^{-1} = a^{-1} rangle $$

$langle a rangle$ is a normal subgroup and $pi_1(K)$ is the internal semi-direct product of $langle a rangle$ and $langle b rangle$, so any element $g$ can be written uniquely as $a^kb^j$ for some $k, jin mathbb{Z}$. (In fact if we express $K$ as an $S^1$ bundle over $S^1$ then $langle a rangle$ is the image of $pi_1(F)$ for any fibre $F$.)

Now $mathbb{R}^2$ can be seen as the universal cover via the action of $pi_1(K)$ given by
$$acdot(x, y) = varphi_a(x,y)= (x, y+1)text{ and } bcdot(x, y) = varphi_b (x, y) = (x + 1, - y)$$

You can check that the one relation $varphi_bvarphi_avarphi_b^{-1} = varphi_a^{-1}$ is satisfied. (We could alternatively have computed $pi_1(K)$ by computing deck transformations first.)

Now we can try taking quotients by subgroups with finite index.

Try 1: The subgroup $A_n = langle a^n, b rangle$ with index $n$. Every element looks like $a^k b^j$ where $n$ divides $k$. Considering the conjugations

$$ a(a^kb^j)a^{-1} = a^{k+2} b^jtext{ and } b(a^kb^j)b^{-1} = a^{-k}b^j $$

it follows that $A_n$ is normal iff $n=2$. I managed to convince myself that a fundamental domain of the action is a vertical strip of $n$ boxes, where the tops are identified with the same orientation and the sides are identified with opposite, i.e. $mathbb{R}^2/A_ncong K$ for every $n$. However when $n >2$ I think the group of deck transformations is trivial.

Try 2: The subgroups $B_n = langle a, b^n rangle$. Every element looks like $a^k b^j$ where $n$ divides $j$, and this subgroup is normal for all $n$. Now the fundamental domain is a horizontal strip of $n$ boxes when the top and bottoms are identified with the same orientation, but the way the sides are identified depends on if $n$ is even or odd. In fact the quotient is the torus if $n$ is even and $K$ when $n$ is odd. This produces the coverings given by Rolf Hoyer's answer.

I don't know what the conjugacy classes of subgroups of $pi_1(K)$ with finite index are so there might other interesting examples. Something like $langle a^n, b^m rangle$ will probably produce some combination of the two types of covers given above. In any case, this argument produces coverings $Kto K$ with any number of sheets, and regular coverings with odd numbers of sheets. I wouldn't be surprised if regular self-coverings with an even number of sheets could be ruled out somehow.