1
$\begingroup$

In the book Functions of Several Complex Variables and Their Singularities by Wolfgang Ebeling, coverings are defined as maps $p:Y\to X$ between arbitrary topological spaces such that every point has a neighborhood whose preimage consists of disjoint sets all of which are homeomorphically mapped to said neighborhood.

Universal coverings (of $X$) are then defined as coverings $p:Y\to X$ where $Y$ is simply connected.

It is then said that

If $p:Y\to X$ is a universal covering, $q: Z \to X$ is an arbitrary covering, and $X$ is locally path connected, then by Proposition 1.12 there is a commutative diagram

$\;\; Y\stackrel{\Phi}{\longrightarrow}Z$,

$Y\stackrel{p}{\rightarrow} X\stackrel{q}{\leftarrow}Z$

[sorry, couldn't figure out how to post a triangular commutative diagram. Please identify the two copies of $Y$ and $Z$ to get the original diagram.]

In this sense the universal covering is the "largest" covering of X.

My question How is this property a sign of largness of $p\colon Y\to X$?

For example, if $p:Y:=S^{1}\to X:=S^{1}, x\mapsto x$ is the trivial covering of $S^1$ and $q\colon Z:=\mathbb{R}\to X, t\mapsto \exp(it)$ is the universal covering, then one can also find $\Phi\colon Y\to Z$ as above ($x\mapsto -i\log x \in [0,2\pi)$) but obviously this cannot mean that $p$ is "larger" than $q$, since $q$ is already "larger" than $p$ (and since $p$ is arguably the smallest possible covering).

$\endgroup$
  • 1
    $\begingroup$ Your $\Phi$ isn't continuous, so wouldn't be considered in this context. $\endgroup$ – Dap Mar 1 '18 at 12:54
1
$\begingroup$

In what follows I will restrict to the notion of $\textbf{normal covers}$ and use some nontrivial results from covering space theory (but you can read all of this in Hatchers book on page 70):

There is an algebraic way to think of the size of a cover as follows:

If you have a covering $q \colon Z \to X$ then one can define the group of deck transformations $$\text{Deck}(Z,q):=\{A \colon Z \to Z \mid \ \; q \circ A=q, A\in \text{Homeo}(Z)\}.$$

This group is then isomorphic to the quotient $$\pi_1(X) / q_{\#}(\pi_1(Z)),$$ where $q_{\#} \colon \pi_1(Z) \to \pi_1(X), \{\tilde{\gamma}\} \mapsto \{q \circ \tilde{\gamma}\}$ is the induced map coming from the covering map $q$.

By definition of a universal covering $p \colon Y \to X$ we get $\pi_1(Y)=0$ and in turn our deck transformation group is the whole fundamental group $\pi_1(X)$!

However, if consider an arbitrary covering $$q \colon Z \to X$$ with $\pi_1(Z) \neq0$, then the subgroup $q_{\#}(\pi_1(Z))$ is $\textbf{not}$ equal $0$ and thus the $\text{Deck}(Z,q)$ will not be the whole fundamental group of $\pi_1(X)$.

So in these terms you could say that a universal covering $p \colon Y \to X$ is a covering of $X$ with the largest group of deck transformation possible, i.e. $$\text{Deck}(Y,p) \cong \pi_1(X).$$

$\endgroup$
  • $\begingroup$ I like your answer, but do you also have an explanation that is closer connected to the existence of the diagram? It sounds from the book as if there is a clear connection between the diagram and largeness $\endgroup$ – Bananach Mar 1 '18 at 17:09
  • $\begingroup$ @Bananach Maybe, but I need to think about it. I didn't check your source, but I guess that $\Phi$ satisfies some properties? Could you tell me which ones, if this is the case? $\endgroup$ – noctusraid Mar 2 '18 at 10:38
  • $\begingroup$ The $\Phi$ in Proposition 1.12 is a lifting (the propositions states that maps from simply connected spaces are liftable to any coverings) $\endgroup$ – Bananach Mar 2 '18 at 11:37
0
$\begingroup$

This is quite non-mathematical. But intuitively if $Y$ covers $X$ then $Y$ is atleast as "large" as $X$. The universal cover over $X$ covers all of the covers of $X$, so it is the largest in this vague sense.

Although as you point out self-covers of a space indicate that this is a non-strict ordering on the set of (reasonably nice) topological spaces.

Another intuition from Riemannian geometry states that for a cover of a Riemannian manifold, the pull-back of the metric under an $n$-fold cover has $n$ times the volume of the base manifold (or infinite volume for an infinite degree cover). But this is quite far from topology, and only really make sense if we make some specific curvature restriction (for example in the world of hyperbolic manifolds).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.