1
$\begingroup$

Let (G,e) be a topological group and p : G → X be a covering map. When p can transfer the group structure to make X a topological group? It is clear that if p is a group homomorphism it is done. But we want to find minimal conditions. Moreover, the converse statement happen without extra condition, i.e, if p : X → G is a covering map and G is a topological group, then X is also a topological group and p is a group homomorphism.

$\endgroup$
1
$\begingroup$

Your condition does not make sense : "if $p$ is a group morphism": $X$ has no group structure so $p$ can't be a group morphism !

A necessary condition is clearly : "if $p(x)=p(x'), p(y)=p(y')$, then $p(xy)=p(x'y')$".

Is it sufficient ? Assume we have this condition, then there is only one way to put a group structure on $X$ that makes $p$ a group morphism; but a priori this group structure need not be continuous. That's where the covering hypothesis comes into play.

Indeed let $a,b\in X$, and $V$ a neighbourhood of $ab$. Let also $x,y\in G$ with $p(x)=a, p(y)=b$ (thus $p(xy) = ab$).

Let $W$ be a smaller open neighbourhood of $ab$, and $O$ an open neighbourhood of $xy$ such that $p$ is a homeomorphism $O\to W$ (exists by the covering hypothesis). Multiplication is continuous on $G$, so we can find open neighbourhoods $U, U'$ of $x,y$ such that $UU'\subset O$. Now $p$ is an open map so $p(U)$ is an open neighbourhood of $a$, $p(U')$ of $b$ and so $p(U)p(U') \subset W$ proves that multiplication on $X$ is continuous at $(a,b)$ : hence it is continuous.

One can proceed similarly with the inverse map to show that it's also continuous.

Therefore, our condition was sufficient : it suffices for $p$ to induce a set-theoretic group structure on $X$.

Notice that I haven't actually used that $p$ was a covering map : all I needed was that $p$ be a surjective open map.

$\endgroup$
  • $\begingroup$ Dear Max thanks for spending the time. You are right, if p has the condition it is done. My main question now is whether this condition can be obtained from evenly covered property or not? Can we find a counterexample? $\endgroup$ – Araz Binevli Oct 27 '18 at 16:59
  • $\begingroup$ What do you mean "evenly covered"? $\endgroup$ – Max Oct 27 '18 at 18:20
  • $\begingroup$ I mean the covering maps main property. Let p : X → Y be a surjective map. The space Y is evenly covered by p if for every y in Y, there is an open neighborhood U of y such that U is evenly covered by p (i.e, p^{-1}(U) is a union of disjoint open sets in X, each of which is mapped homeomorphically onto U by p). $\endgroup$ – Araz Binevli Oct 28 '18 at 18:47
  • 1
    $\begingroup$ The condition doesn’t follow from that property; for example consider the torus double-covering the Klein bottle (which cannot admit a group structure). $\endgroup$ – Aleksandar Milivojevic Nov 5 '18 at 4:50

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.