1
$\begingroup$

In wikipedia I read:

In mathematics, a covering group of a topological group $H$ is a covering space $G$ of $H$ such that $G$ is a topological group and the covering map $p : G \rightarrow H$ is a continuous group homomorphism. The map $p$ is called the covering homomorphism. A frequently occurring case is a double covering group, a topological double cover in which $H$ has index 2 in $G$; examples include the Spin groups, Pin groups, and metaplectic groups.

In this definition $H$ is not necessary a subgroup of $G$ (so "$H$ has index 2 in $G$" doesn't make sense in the usual way). So my question is: what is "the index of $H$ in $G$" in this context?

$\endgroup$
  • $\begingroup$ It's just a speculation, perhaps by saying covering map they mean that $p^{-1}(H)$ is homeomorphic to a disjoint union of spaces homeomorphic to $H$. If so the index is the number of such spaces. $\endgroup$ – Yanko Dec 29 '18 at 13:16
  • $\begingroup$ Maybe is $[G:p^{-1}(H)]=2$ $\endgroup$ – asv Dec 29 '18 at 14:36
  • $\begingroup$ that's less likely. If you consider $G=H\times \mathbb{Z}/2\mathbb{Z}$ with the map $p(h,a) = h$ then $p^{-1}(H)=G$ so by your definition the index is $1$ which in my opinion makes less sense than saying that $p^{-1}(H) = H\times\{0\}\bigsqcup H\times\{1\}$ and so the index is $2$ (As the usual index). $\endgroup$ – Yanko Dec 29 '18 at 14:38
  • $\begingroup$ Maybe $[G:p^{-1}(H)]$ is always 1 because, if I'm not wrong, $p$ is surjective by definition then $p^{-1}(H)=G$. Here: en.wikipedia.org/wiki/Covering_space says $p$ is surjective. $\endgroup$ – asv Dec 29 '18 at 14:42
  • 2
    $\begingroup$ The question currently has 4 votes to close. I cannot understand why this is, apart from perhaps that the question is not initially clear. I have tried to make the question clearer (although I am unsure if my edit improve readability - anyone should feel free to edit it!). $\endgroup$ – user1729 Jan 7 at 15:26
3
$\begingroup$

This community wiki solution is intended to clear the question from the unanswered queue.

As Moishe Cohen pointed out in his comment, the phrase "A frequently occurring case is a double covering group, a topological double cover in which $H$ has index $2$ in $G$" appeared for the first time in the January 2009 revision of the Wikipedia article and it is wrong.

In general $H$ is not a subgroup of $G$, and it does not even embed as a subgroup into $G$. Therefore it does not make sense to speak about the index of $H$ in $G$.

Only in the trivial case $G = H \times \mathbb{Z}_2$ there would be a reasonable interpretation.

$\endgroup$

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.