Can stabilizer groups in an orbifold have global twisting? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:47:56Z http://mathoverflow.net/feeds/question/109645 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109645/can-stabilizer-groups-in-an-orbifold-have-global-twisting Can stabilizer groups in an orbifold have global twisting? unknown (google) 2012-10-14T19:52:13Z 2012-10-15T13:38:18Z <p>Can stabilizer groups in an orbifold have global twisting?</p> <p>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?</p> <p><strong>Why they should be different:</strong> Over any orbifold $X$, we think of there being space $E$ with a "nice" map $\pi:E\to X$ such that <code>$\pi^{-1}(\{x\})$</code> is a $K(\Gamma_x,1)$ (see for example <a href="http://arxiv.org/abs/math/0112006" rel="nofollow">this paper</a> 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 $\mathbb Z/3\rtimes\mathbb Z$.</p> <p>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.</p> <p>End note: The answers to these related questions have very good answers in regards to defining orbifolds.</p> <p><a href="http://mathoverflow.net/questions/1861/looking-for-an-introduction-to-orbifolds" rel="nofollow">http://mathoverflow.net/questions/1861/looking-for-an-introduction-to-orbifolds</a></p> <p><a href="http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold" rel="nofollow">http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold</a></p> http://mathoverflow.net/questions/109645/can-stabilizer-groups-in-an-orbifold-have-global-twisting/109666#109666 Answer by Eugene Lerman for Can stabilizer groups in an orbifold have global twisting? Eugene Lerman 2012-10-15T01:13:35Z 2012-10-15T13:38:18Z <p>The short answer is "yes" if you think of orbifolds as groupoids or stacks. </p> <p>(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).</p> <p>Both of your examples are $\mathbb{Z}/3$ bundles <del>gerbes</del> over $S^1$, one trivial, one not. So they are not the same as stacks.</p> <p>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 $\mathbb{Z}/3$ <del>gerbe</del> bundle over $S^2$. Of course there is also a trivial one: $\mathbb{Z}/3$ acting trivially on $S^2$.</p>