3
$\begingroup$

Can stabilizer groups in an orbifold have global twisting?

For example, consider the two groups $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$ (where $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ is the unique nontrivial map). Both groups act on $\mathbb R$ through their common quotient $\mathbb Z$, so there are two orbifolds here: $$M_1=[\mathbb R/(\mathbb Z/3\times\mathbb Z)]$$ $$M_2=[\mathbb R/(\mathbb Z/3\rtimes\mathbb Z)]$$ The coarse spaces of $M_1$ and $M_2$ are both $S^1=\mathbb R/\mathbb Z$, and in both $M_1$ and $M_2$, all points have stabilizer group isomorphic to $\mathbb Z/3$. Are $M_1$ and $M_2$ isomorphic as orbifolds?

Why they should be different: Over any orbifold $X$, we think of there being space $E$ with a "nice" map $\pi:E\to X$ such that $\pi^{-1}(\{x\})$ is a $K(\Gamma_x,1)$ (see for example this paper by André Henriques). In the case of $M_1$ and $M_2$, this just means we have a $K(\mathbb Z/3,1)$ bundle over $S^1$, and its monodromy is a homomorphism $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ which should recover the difference between $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$.

This twisting is a subtle point that I don't see mentioned explicitly anywhere, and I didn't realize it could happen until now, when I actually have to do something with orbifolds and the precise definition becomes important.

End note: The answers to these related questions have very good answers in regards to defining orbifolds.

Looking for an introduction to orbifolds

What is meant by smooth orbifold?

$\endgroup$
2
  • $\begingroup$ Doesn't the orbifold fundamental group tell them apart? $\endgroup$ Sep 24, 2017 at 21:36
  • $\begingroup$ @ChrisGerig yes, it does. This question was written what seems like a long time ago to me, when I was just starting to understand what orbifolds are. $\endgroup$ Sep 25, 2017 at 12:28

1 Answer 1

2
$\begingroup$

The short answer is "yes" if you think of orbifolds as groupoids or stacks.

(begin edit) In particular, if you think of orbifolds as groupoids, the "purely ineffective" orbifolds, the issue that seems to be at stake here, can be seen as bundles of groups. That is, they are fiber bundles whose fibers are groups. They are not to be confused with principal bundles, whose fibers are homogeneous spaces (end edit).

Both of your examples are $\mathbb{Z}/3$ bundles gerbes over $S^1$, one trivial, one not. So they are not the same as stacks.

Another example I like comes from an ineffective action of $U(1)$ on $S^3$, say given by $\lambda \cdot z = \lambda^3 z$. The corresponding etale Lie groupoid can be thought of as a nontrivial $\mathbb{Z}/3$ gerbe bundle over $S^2$. Of course there is also a trivial one: $\mathbb{Z}/3$ acting trivially on $S^2$.

$\endgroup$
6
  • $\begingroup$ You mean $\mathbb Z/3$ gerbe, or $B(\mathbb Z/3)$ bundle, definitely not $\mathbb Z/3$ bundle. $\endgroup$ Sep 25, 2017 at 12:27
  • $\begingroup$ I did mean "bundle." Any such bundle is a gerbe. Not all gerbes over a manifold are bundles of groups --- some are extensions of cover groupoids. $\endgroup$ Sep 25, 2017 at 15:46
  • $\begingroup$ A purely ineffective orbifold $\mathcal M$ with, say isotropy group isomorphic to $G$ everywhere, with coarse space $M$, definitely does not give rise to a $G$-bundle over $M$, rather it corresponds to a fibration over $M$ with (unpointed!) $K(G,1)$-fibers. Equivalently, it gives a bundle of groupoids over $M$, but (in general) definitely not of groups. $\endgroup$ Oct 2, 2017 at 16:06
  • $\begingroup$ I have a guess as to why we talking past each other. One can think of orbifolds as objects in the 2-category of (geometric) stacks over the site of manifolds. One can also think of orbifolds as objects in the bicategory of Lie groupoids, bibundles and isos of bibundles. The two viewpoints are equivalent --- there are 2-functors going in both directions. A bundle of finite groups is an object in the bicategory of Lie groupoids. $\endgroup$ Oct 4, 2017 at 18:33
  • $\begingroup$ It is also a 1-morphism between two Lie groupoids with typical fiber the group call it G. (OK, so perhaps you should think of the fiber as a groupoid with one object * and the space of morphisms G). A 1-morphism between groupoids give rise to a map between corresponding stacks. The fiber of this map is BG (or, if you prefer the stack quotient [*//G]). $\endgroup$ Oct 4, 2017 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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