Why Does the Orbit Space of a Covering Map Not Necessarily Equal the Base Space?

[PsychonautQQ]

Suppose $q: E-->X$ is a covering map (not necessarily normal). Let $E' = E/ Aut_{q}(E)$ be the orbit space, and let $\pi: E-->E'$ be the quotient map. Then there is a covering map $q': E' --->X$ such that $q' * \pi = q$ where $*$ is composition of functions.

I am confused why $E'$ doesn't equal $X$. Isn't $E'$ a space formed by the exact same identifications that $E$ makes on $X$ under the map $q$? Why would these spaces be different at all then? A part of me believes that these spaces can only be equal if $q$ where a normal map, because the action of $Aut_{q}(E)$ is transitive and sooo yeah I need some help filling in my lack of understanding on this.. Thanks!

 

[andrewkirk]

I suspect (but have no proof) that, if $q$ is normal, $E'$ will at least have the same cardinality as $X$, and maybe the same topology too. But it will be a different set.

Consider the usual example of $q:E=\mathbb R\to X=S^1$ given by $q(x)=e^{2\pi i x}$. Then an element of $X$ is a single number in the complex plane, whereas an element $x$ of $E'$ is a set of the form
$$\{\alpha_x+n\ :\ n\in\mathbb Z\}$$
where $\alpha_x\in[0,1)$ uniquely characterises $x$.

In that example the covering map is normal, and the cardinality and topology of $E'$ are the same as those of $S^1$.

But where the covering map is not normal, the set $Aut(q)$ of deck transformations is not transitive, so there may be more than one orbit per fibre. That suggests to me that the cardinality of $Aut(q)$ may be greater, and hence the cardinality of $E'$ may be greater than that of $X$. Differences in topology may also follow.

Perhaps the reason the text said 'not necessarily normal' was because it's in those non-normal cases that $E'$ becomes different from $X$ in more than just set composition, and things become more interesting.

This is all a bit hand-wavy, but hopefully it gives you an idea of the potential differences.

 

[math771]

Hi there,

I can see where your confusion is coming from. The key here is to understand the difference between the quotient space and the orbit space. The quotient space $E'$ is formed by identifying all points in $E$ that are mapped to the same point in $X$ under the covering map $q$. So, in a sense, $E'$ is a "collapsed" version of $E$ where all the points that are "equivalent" under the map $q$ are identified as one point.

On the other hand, the orbit space is formed by identifying all points in $E$ that are mapped to the same point in $X$ under the action of the automorphism group $Aut_q(E)$. So, in this case, the points in $E$ are not necessarily mapped to the same point in $X$ under the covering map $q$, but they are mapped to the same point under some automorphism in $Aut_q(E)$.

To see the difference, let's consider an example. Let $E = \mathbb{R}$ be the real line and let $X = S^1$ be the unit circle. Let $q: \mathbb{R} \to S^1$ be the covering map given by $q(x) = e^{2\pi ix}$. Now, let's consider the point $p = 1$ in $X$. Under the map $q$, there are infinitely many points in $\mathbb{R}$ that are mapped to $p$, namely, all the integers. So, in the quotient space $E'$, all these points will be identified as one point, say $[p]$. On the other hand, the orbit space of $p$ will consist of all the points in $\mathbb{R}$ that are mapped to $p$ under some automorphism in $Aut_q(E)$. In this case, since $Aut_q(\mathbb{R}) = \mathbb{Z}$, the orbit space of $p$ will just be the set of integers, which is different from the quotient space $E'$.

So, to answer your question, $E'$ is not necessarily equal to $X$ because the points in $E'$ are identified based on the covering map $q$, while the points in