Nonpathological nonnormal covering space

Question

A topological covering $p : \tilde{X} \to X$ is normal when the group of deck transformations acts transitively on the fibers of $p$. This is equivalent to the fact that $p_* (\pi (\tilde{X}, \tilde{x}))$ is a normal subgroup of $\pi (X, p (\tilde{x}))$. Such coverings are also known as Galois or regular.

The universal covering is known to be always normal. There are many nonnormal coverings, but all the classical examples that I know give me the impression of being cooked up for the purpose and being more created laboratory than observed in the wild.

The question is whether some nonnormal coverings are known to arise naturally in some mathematical problems or theories. This means that the covering should be motivated by some problem or theory and that unfortunately it turns out not to be normal.

Answer 1

They arise naturally all the time. The Galois correspondence tells us that a covering space is normal if and only if the corresponding subgroup of the fundamental group is normal. So whenever you have a non-normal subgroup, you have a non-normal covering.

As an example of a situation where you can't do without non-normal covers, let's use a nice theorem of Scott.

Theorem(Scott): For any compact surface $\Lambda$ immersing into a surface $\Sigma$ there exists a finite-sheeted covering space $\widehat{\Sigma}\to\Sigma$ so that, after a homotopy, $\Lambda$ lifts to an embedded subsurface of $\widehat{\Sigma}$.

As you can imagine, this is quite useful, because embedded subsurfaces are much easier to handle than immersed ones. However, the theorem just isn't true if you insist that your covering space $\widehat{\Sigma}$ must be normal.

Answer 2

Let $A$, $B$, $C$, $D$, $E$, $F$ be six pairwise-disjoint circles.

Let $p:E\to B$ be a homeomorphism. Let $q:F\to A$ be a homeomorphism.

Let $r:C\to A$ be a double covering. Let $s:D\to B$ be a double covering.

Let $x\in A$, $y\in B$.

Let $\widehat{x}\in F$ be the $q$-preimage of $x$. Let $\widehat{y\,}\in E$ be the $p$-preimage of $y$.

Let $x',x''\in C$ be the $r$-preimages of $x$. Let $y',y''\in D$ be the $s$-preimages of $y$.

Let $M$ be obtained from $A\coprod B$ by gluing, as follows:

Glue $x$ to $y$, obtaining a point $z\in M$.

Let $A_0$ be the image in $M$ of $A$. Let $B_0$ be the image in $M$ of $B$.

Let $N$ be obtained from $C\coprod D\coprod E\coprod F$ by gluing, as follows:

Glue $\widehat{y\,}$ to $x'$ obtaining a point $u\in N$, and

glue $x''$ to $y'$, obtaining a point $v\in N$, and

glue $y''$ to $\widehat{x}$ obtaining a point $w\in N$.

Let $C_0$ be the image in $N$ of $C$. Let $D_0$ be the image in $N$ of $D$.

Let $E_0$ be the image in $N$ of $E$. Let $F_0$ be the image in $N$ of $F$.

Let $\pi:N\to M$ be induced by $p\coprod q\coprod r\coprod s$.

Then $\pi$ is a $3$-to-$1$ covering map, and $\pi^{-1}(z)=\{u,v,w\}$.

We claim that the identity on $N$ is the only deck transformation of $\pi$.

Let $f:N\to N$ be a deck transformation of $\pi$.

We wish to show that $f$ is the identity on $N$.

The action of deck transformations on the total space is free. So it suffices to show that $f$ fixes a point of $N$. So it suffices to show that $f(u)=u$.

Let $I:=[0,1]$. Let $\alpha:I\to B_0$ be a continuous map s.t. $\alpha(0)=z=\alpha(1)$ and s.t. $\alpha|(0,1)$ is 1-1.

Let $\beta:I\to N$ be the $\pi$-lift of $\alpha$ s.t. $\beta(0)=u$. Then $\beta(1)=u$, so $\beta(0)=\beta(1)$. Then $(f\circ\beta)(0)=(f\circ\beta)(1)$.

Let $\gamma:I\to N$ be the $\pi$-lift of $\alpha$ s.t. $\gamma(0)=v$. Then $\gamma(1)=w$, so $\gamma(0)\ne\gamma(1)$.

Let $\delta:I\to N$ be the $\pi$-lift of $\alpha$ s.t. $\delta(0)=w$. Then $\delta(1)=v$, so $\delta(0)\ne\delta(1)$.

Recall that $\pi^{-1}(z)=\{u,v,w\}$. Then the set of $\pi$-lifts of $\beta$ is $\{\beta,\gamma,\delta\}$.

So, since $f\circ\beta$ is a $\pi$-lift of $\alpha$, we get: $f\circ\beta\in\{\beta,\gamma,\delta\}$.

Since $(f\circ\beta)(0)=(f\circ\beta)(1)$, while $\gamma(0)\ne\gamma(1)$, we get $f\circ\beta\ne\gamma$.

Since $(f\circ\beta)(0)=(f\circ\beta)(1)$, while $\delta(0)\ne\delta(1)$, we get $f\circ\beta\ne\delta$.

Then $f\circ\beta=\beta$. Then $(f\circ\beta)(0)=\beta(0)$. Then $f(\beta(0))=\beta(0)$. Then $f(u)=u$.