There seems to be some confusion over what the tangent space to a singular point of an orbifold is.


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.


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 $\mathbb R^2/\mathbb Z\_3$ *is* $\mathbb R^2/\mathbb Z\_3$, whatever that means.


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.


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.



What is meant by smooth orbifold?
Disclaimer: I don't talk to people about orbifolds. This answer may not represent the opinions of orbifolders.


As I understand it, the orbifold $\mathbb R^n/G$ is characterized by how manifolds map to it,† not by how it maps to manifolds. In particular, the orbifold *is not* 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 $\mathbb R^2$ *is isomorphic* to the ring of 4-fold symmetric functions on $\mathbb R^2$.


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 *into* the space rather than maps *out of* 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 $G$.


So far, it seems like I'm arguing that the tangent space to $\mathbb R^n/G$ at the cone point should be *the orbifold* $\mathbb R^n/G$. 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.


† 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 $\mathbb R^n$.


