An orbifold structure on some topological space $X$ is a covering of $X$ with local quotient charts $V/G$, where $V$ is some connected manifold and $G$ effectively acts on $V$ via a finite group of diffeomorphisms. These charts should be compatible on the overlap in the sense that for every two of them say $V_1/G_1$ and $V_2/G_2$, and every point on their intersection, say $\overline{x}\in X$, there is some orbifold chart $V_{12}/G_{12}$ around $\overline{x}$, group homomorphisms $h_i:G_{12}\to G_i$, and equivariant embeddings $$ \iota_i:V_{12}\to V_i,\quad i=1,2, $$ such that $(G_{12})_x\cong (G_i)_{\iota_i(x)}$; here $x$ is the preimage of $\overline{x}$ in $V_{12}$ and $G_x$ is the isotropy group at $x$.

An orbifold is a global quotient if it is globally of the form $M/G$ where $M$ is some manifold and $G$ effectively acts on $M$ via a discrete group of diffeomorphisms.

Not every orbifold is a global quotient.

Definition: A global branched covering for an orbifold structure on $X$ consists of a (smooth) manifold $M$ and a continuous map $f:M\to X$, such that for every small enough orbifold chart $V/G$ of $X$, there is a manifold chart $U$ of $M$ such that $f$ lifts to a smooth branched (finite) covering map $f: U\to V$.

Question: Does every compact orbifold posses a global branched covering?

Motivating example: Let $\mathbb{P}^1_{m,n}$ be the weighted $1$-dimensional projective space over the topological space $X=S^2$ with two orbifold charts $\mathbb{C}/\mathbb{Z}_m$ and $\mathbb{C}/\mathbb{Z}_n$. Here the first chart corresponds to the map $z\to z^m$ and the second one corresponds to the map $w\to w^n$, $z^m=(w^n)^{-1}$. If $m\neq n$, $\mathbb{P}^1_{m,n}$ is not a global quotient; however, for $M=\mathbb{P}^1$, the map $f: M \to S^2$ sending $z\to z^{mn}$ is global branched covering.

Hint: Every orbifold is a global quotient $M/G$ where $G$ is some compact lie group with finite stabilizers. Therefore, the question seems to be equivalent to: is there a finite index group $H\subset G$ such that $H\cap G_x=1$, $\forall x\in M$.

  • $\begingroup$ I'm a bit confused. You write "Not every orbifold is a global quotient", but then later you write "every orbifold is a global quotient $M/G$. $\endgroup$ – Ariyan Javanpeykar May 5 '15 at 6:08
  • $\begingroup$ In the algebraic setting every smooth orbifold is a global quotient stack; see Thm 2.18 in arxiv.org/pdf/math/9905049v3.pdf $\endgroup$ – Ariyan Javanpeykar May 5 '15 at 6:09
  • $\begingroup$ @ Ariyan: Think this way, if X is a non simply-connected manifold, instead of orbifold, then such M is simply a finite covering of X. $\endgroup$ – Mohammad F. Tehrani May 8 '15 at 18:56
  • 1
    $\begingroup$ @AriyanJavanpeykar, there's a distinction between "locally a quotient by a finite group" and "locally a quotient by a compact group with finite stabilizers". I don't know how to prove the statement in the hint, but I do know an example: You can easily find actions of $S^1$ on $S^3$ so that the quotient is naturally one of the inadmissible orbifolds $\mathbb{P}^1_{mn}$ mentioned in the problem statement. $\endgroup$ – Dylan Thurston May 10 '15 at 1:46
  • $\begingroup$ Can you specify exactly what you mean by a "branched covering map"? $\endgroup$ – Dylan Thurston May 10 '15 at 1:48

Your Answer

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

Browse other questions tagged or ask your own question.