What is meant by smooth orbifold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:13:05Z http://mathoverflow.net/feeds/question/19530 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold What is meant by smooth orbifold? AndrewLMarshall 2010-03-27T17:53:14Z 2010-12-29T22:08:16Z <p>There seems to be some confusion over what the tangent space to a singular point of an orbifold is.</p> <p>On the one hand there is the obvious notion that smooth structures on orbifolds lift to smooth $G$-invariant structures on $\mathbb R^n$ ($G$ being the finite group so that the orbifold is locally (about some specific point $x$) the quotient of $\mathbb R^n$ by the action of $G$). One might be tempted to consider cone points as differentiable spaces (that is, subsets of some $\mathbb R^k$, inheriting their differential structure by restriction), however, we are told, for example, that $\mathbb R^2/\mathbb Z_3$ and $\mathbb R^2/\mathbb Z_4$ are distinct as orbifolds, so it is not the case that cone points can be modeled merely with cone-like subsets of some $\mathbb R^k$. The definition in which 'smooth' means 'lifts to $G$-invariant smooth' distinguishes these two cones, as the set of functions with 3-fold symmetry and the set of functions with 4-fold symmetry, in $\mathbb R^2$, are distinct. The third item in Satake's seminal paper [On a Generalization of the Notion of Manifold] corroborates this, giving $C^\infty$ forms of degree $p$ at a singularity $x$ as those $C^\infty$ $p$-forms in $\mathbb R^n$ which are invariant under $G_x$. If we require the same property of vectors, that is, that they lift to $G$-invariant vectors in $\mathbb R^n$, then we have that the dimension of the tangent space of an orbifold is the dimension of the invariant subspace upstairs. In particular the dimension tends to drop at the singular points. For example, the dimension of the tangent space at the singularity in $\mathbb R^2/\mathbb Z_3$ is 0. This notion of vector agrees with the notion of vector as derivation on the germ of smooth functions. In this case smooth functions in $\mathbb R^2$ which have 3-fold symmetry necessarily have vanishing derivatives at the origin.</p> <p>On the other hand, one finds descriptions of smooth orbifolds as objects which have tangent bundle-like structures, which are locally $\mathbb R^n/G$. It is not clear what this means as far as smooth structures go, but the explanation above of $\mathbb R^2/\mathbb Z_3$ having a 0 dimensional tangent space at the cone point seems to contradict the notion that the tangent-like space at the singularity in $\mathbb R^2/\mathbb Z_3$ <em>is</em> $\mathbb R^2/\mathbb Z_3$, whatever that means.</p> <p>It is also said that manifolds with boundary can be viewed as orbifolds, which have isotropy group reflection by $\mathbb Z_2$ along their boundaries. It would be nice to include the note that the differentiable structures are different. Specifically, smooth manifolds with boundaries have tangent spaces along their boundaries which are the same dimension as the manifold. In contrast, the same topological space as an orbifold with $\mathbb Z_2$ structure group along the boundary should have a tangent space which is one dimension less than the dimension at a generic point, if the definition of tangent space follows Satake's guideline. Indeed, smooth functions in $\mathbb R^n$ which locally have symmetry by reflection through a codimension 1 hyperplane, have vanishing partial derivatives in the normal direction.</p> <p>I am asking for concurrence or correction and clarification, since I am still not certain I have the correct notion of tangent space to an orbifold, although I'm fairly confident in the first given here.</p> http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold/19542#19542 Answer by Anton Geraschenko for What is meant by smooth orbifold? Anton Geraschenko 2010-03-27T20:29:11Z 2010-03-27T20:29:11Z <p>Disclaimer: I don't talk to people about orbifolds. This answer may not represent the opinions of orbifolders.</p> <p>As I understand it, the orbifold $\mathbb R^n/G$ is characterized by how manifolds map to it,<sup>&dagger;</sup> not by how it maps to manifolds. In particular, the orbifold <em>is not</em> determined by the ring of smooth functions on it (i.e. the ring of $G$-invariant smooth functions on $\mathbb R^n$). In particular, the ring of 3-fold symmetric smooth functions on $\mathbb R^2$ <em>is isomorphic</em> to the ring of 4-fold symmetric functions on $\mathbb R^2$.</p> <p>Given that the smooth functions on an orbifold don't "remember" everything about it, it seems unreasonable to define the tangent space at a point in terms of derivations of smooth functions. However, we have another construction of the tangent space at a point: equivalence classes of smooth curves through that point. Since this definition has to do with maps <em>into</em> the space rather than maps <em>out of</em> it, we expect it to play well with orbifolds. Any smooth curve through the cone point of $\mathbb R^n/G$ lifts to a curve in $\mathbb R^n$, and any curve in $\mathbb R^n$ induces a curve in $\mathbb R^n/G$, so the tangent space to $\mathbb R^n/G$ should be the same as the tangent space to $\mathbb R^n$. Well, not exactly, since a curve in $\mathbb R^n/G$ can lift to a curve in $\mathbb R^n$ in $G$ different ways. So the tangent space to $\mathbb R^n/G$ at the cone point should really be the quotient of the tangent space of $\mathbb R^n$ by the action of $G$.</p> <p>So far, it seems like I'm arguing that the tangent space to $\mathbb R^n/G$ at the cone point should be <em>the orbifold</em> $\mathbb R^n/G$. But I actually want to say that the tangent space should be the tangent space of $\mathbb R^n$, together with the action of $G$. The reasoning is that the vector space together with the action of the residual group is independent of how you express $\mathbb R^n/G$ as a quotient by a finite group. In other words, even though an isomorphism of orbifolds $\mathbb R^n/G\cong M/G'$ does not induce isomorphisms $\mathbb R^n\cong M$ or $G\cong G'$, it does induce isomorphisms of tangent spaces and residual groups in such a way that respects the actions of the residual groups on the tangent spaces. Once your definition of tangent space is canonically related to the orbifold, you should be welcome to think of it however you like.</p> <p><sup>&dagger;</sup> Roughly, a map from a manifold $M$ to $\mathbb R^n/G$ should be the same thing as a map from $M$ to $\mathbb R^n$, except that maps that differ by the action of $G$ should be regarded as the same map. More precisely, I think a map from $M$ to $\mathbb R^n/G$ should consist of a $G$-fold covering space of $M$ with a $G$-equivariant map to $\mathbb R^n$.</p> http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold/19776#19776 Answer by David Carchedi for What is meant by smooth orbifold? David Carchedi 2010-03-29T21:46:07Z 2010-03-29T21:46:07Z <p>To be quick, just like manifolds, orbifolds have a fixed dimension. This does not vary point to point. This is also true of their tangent spaces. This is actually true for any etale differentiable stack. Here is more explanation:</p> <p>As mentioned in many of the comments, orbifolds are actually instance of differentiable stacks. To make this jump, we have to really make sense of what a smooth map between orbifolds should be. The correct notion is NOT the one first introduced by Satake, but is slightly more refined, so called- strong (or good) maps between orbifolds. These are precisely those maps which induce geometric morphisms between the associated categories of sheaves (See "Orbifolds, Sheaves and Groupoids" by Pronk and Moerdijk). From now on, I will only consider strong maps.</p> <p>If O is an orbifold, and f,g:M->O are smooth maps from an manifold, then, because of the group-structure in the charts, it makes sense to consider when two such maps are isomorphic. So, to every manifold M, we can assign the category Hom(M,O)- which is a groupoid (every map between two smooth maps is an isomorphism). This assignment "M \mapsto Hom(M,O)" is a weak presheaf in groupoids over the category of differentiable manifolds (this is just fancy talk for saying that it's nearly a contravariant functor, but (g^<em>)(f^</em>) and (fg)^* need only be naturally isomorphic rather than equal, and then some needed coherence conditions needed to make things consistent after this). The point is, given another orbifold, L, the category of weak natural transformations between Hom( ,O) and Hom( ,L) is naturally equivalent to Hom(O,L). This means, orbifolds embed fully-faithfully into stacks, so instead of studying the orbifolds themselves, we can study the functors which they represent.</p> <p>The fuctor Hom( ,O) is not just a functor, but it's actually a stack (it's a "sheaf of groupoids", so it satisfies some gluing conditions). </p> <p>To do this in practice, if you start with an orbifold given to you as a topological space with a chart, then from this chart, you can construct an etale proper Lie groupoid G (See "Orbifolds, Sheaves and Groupoids"). If instead, you are given a presentation of a Lie group acting on a manifold with finite stabilizers, simply take G to be the action groupoid. Then the functor Bun_G which assigns every manifold M the groupoid of principal G-bundles over M is the same as the functor Hom( ,O). (In general, a differentiable stack is a weak functor from manifolds to groupoids of the form Bun_G for some Lie groupoid G.)</p> <p>Now, every single stack has a "tangent stack". This can be shown by abstract nonsense (I can elaborate, or you can look at "Vector Fields and Flows on Differentiable stacks" by Richard Hepworth). When the the stack is X=Bun_G, the tangent stack turns out to simply be TX=Bun_TG. (By TG I mean if you look at the diagram expressing G as a groupoid object in manifolds, apply the tangent functor to get a groupoid object in vector bundles, and then apply the forgetful functor to get another groupoid object in manifolds, you get TG). There is a canonical map from TX->X. A vector bundle over a stack X is defined to be a map Y->X of stacks such that if M->X is any map from a manifold, the pullback $Y \times_X M \to M$ is a vector bundle. In general, TX->X is not a vector bundle but a 2-vector bundle, BUT, if G is etale (or Morita equivalent to an etale guy), it IS a vector bundle (See "Vector Fields and Flows on Differentiable stacks"). In particular, the dimension does not change.</p> <p>To be concrete, if G acts on M, and x is in M, then the Lie algebra of G_x acts on T_xM. If p is the image of x in M//G (where here I mean the stacky quotient) then T_p (M//G)=T_x(M)//Lie(G_x). But, in the case of orbifolds, the stabilizers are finite, so, they have no Lie algebras.</p> http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold/30897#30897 Answer by Martin Pinsonnault for What is meant by smooth orbifold? Martin Pinsonnault 2010-07-07T14:37:23Z 2010-07-07T14:37:23Z <p>There is also a diffeological definition of orbifolds which is often useful in concrete geometric problems, see the paper of Iglesias, Karshon, and Zadka:</p> <p>arXiv:math/0501093</p> <p>or</p> <p><a href="http://www.ams.org/journals/tran/2010-362-06/S0002-9947-10-05006-3/home.html" rel="nofollow">http://www.ams.org/journals/tran/2010-362-06/S0002-9947-10-05006-3/home.html</a></p> http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold/50644#50644 Answer by Patrick I-Z for What is meant by smooth orbifold? Patrick I-Z 2010-12-29T11:02:00Z 2010-12-29T22:08:16Z <p>I don't want to discourage you, but I don't think that the tangent space is a good concept for quotient-like spaces, especially for orbifolds. More adapted would be differential forms. Of course, it depends on what we are willing to do with that. But if you consider "symplectic geometry" you can definitely avoid tangent spaces and focus on forms. Moment Maps, for example, don't need any kind of tangent spaces to be defined and computed in even more general situations than orbifolds. </p> <p>First focus on what you want to prove or find, then choose, accordingly, the right tool to use and surprisingly some generalizations coming just from habits of manifolds become obsolete, or drive you in the wrong direction. Let me say straight, what tangent spaces are useful for? Because of the Cauchy-Lipschitz theorem on Hausdorff manifolds: they generate (locally and under some conditions) a one parameter group of diffeomorphisms. So if you are interested in diffeomorphisms of the orbifolds, go directly to the group, and analyze it. For example (taking the diffeological approach) you can check immediately that any diffeomorphism may exchange only points with equivalent structure groups, etc. etc. If you start with vector field you have already a long way before reaching this starting point.</p> <p>Well, these are my two cents :-)</p>