# What tools cannot work for orbifolds?

Consider all of your basic constructions/tools/theorems for manifolds: fundamental group, Euler characteristic, triangulations, orientation, smoothness, bundle structure, cobordisms, etc.. Viewing orbifolds as the natural generalization of manifolds by quotients of group actions (or just locally $\mathbb{R}^n/G_i$), it seems that we can translate all of our notions for manifolds onto orbifolds with slight adjustments in the definitions. I am curious as to the extent of that statement's validity (which is important, because orbifolds are arising everywhere for me):
What basic constructions that exist on manifolds, cannot (or currently do not) have analogs for orbifolds?
Why? ($\leftarrow$ in case there is more to say than "orbifolds have singularities")

As pointed out nicely below, there is the question of whether such constructions will be useful to (all) orbifolds. So this thread also seeks known constructions which are either intractable or trivial.

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One basic remark is that there are (non-manifold) orbifolds which have trivial fundamental group, so are not developable (called "bad orbifolds") (in 2 dimensions, these are the teardrop and football orbifolds with distinct orders of the 2 cone points; it is a deep theorem that a compact 3-orbifold is good if and only if there are no bad 2-suborbifolds). So although the fundamental group is defined, it is not quite as useful as for manifolds (although the theory of good orbifolds are a good analogue). – Ian Agol Jul 9 '13 at 17:41
I am not sure if this qualifies as a "construction" but the cobordism theory for orbifolds is far less developed than the one for manifolds. I might be behind the events, but, I do not think orbifold cobordism rings were completely computed (Andres Angel did some calculations). Same for surgery theory, I think. – Misha Jul 10 '13 at 15:22
There is an example of explicit computations for a particular orbifold. A "divisive" weighted projective space $P(q_0,...,q_n)$ is s.t. the weight $q_i$ divides $q_i+1$ for each i. Its equivariant K-theory and cobordism ring have been recently computed by Harada, Holm, Ray and Williams (arxiv.org/abs/1306.1641). – Al-Amrani Jul 12 '13 at 22:13

I'm not an expert and this might be wrong, but I think that Cerf theory should be impossible for orbifolds, and therefore all that comes from it, e.g. Kirby Calculus. Could somebody who knows please confirm this? I'm guessing that this comes not from orbifolds having singularities, but rather from orbifolds having built-in symmetries which mess up the necessary stratifications of the space of smooth functions to $\mathbb{R}$.

For example, Kirby's Theorem comes from the fact that a generic path between Morse functions in the space of smooth functions to $\mathbb{R}$ involves only finitely many which are not Morse-Smale (these correspond to handleslides). But I think that this is just wrong for smooth orbifolds. In the orbifold case, Morse functions satisfying the Morse-Smale condition (transversality between stable and unstable manifolds at a critical point) are not dense among smooth functions to $\mathbb{R}$, so perhaps a generic path between two Morse functions in the space of smooth functions might contain all kinds of craziness, and I doubt that there exists a sensible finite set of local moves between handle decompositions to parallel the Kirby moves.

Upon further thought, the answer above can be generalized. In topology, there are various constructions of manifolds as cell decompositions of various flavours (e.g. triangulations, surgery presentations...), along with which there are sets of moves telling us how to go between any two such constructions (e.g. Kirby moves, Pachner moves, subdivision...). These give us the possibility to use the constructions in order to construct manifold invariants, for example.

For orbifolds, you might have good analogues of the constructions, but I am not aware of good analogues for the "sets of moves" between them for any construction. I think that the issue is that orbifolds don't really form a category, but rather they form a 2-category of some sort. I suspect that the point now is something like that the relevant notion of a 2-functor makes the relevant diagrams commute not `on the nose', but up to some sort of morphism; and that such a setup can't give rise to a nice set of local moves which relate all of our constructions. I don't know the precise statement though.

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The integral cohomology ring of orbifolds does not satisfy Poincaré duality.
Indeed, orbifolds do not even have finite cohomological dimension.

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Can you comment on what makes it different from other orbifold cohomologies where it does work? (a google search shows me some results on orbifold Poincare duality). – Chris Gerig Jul 11 '13 at 3:44
That's right. There is something called Chen-Ruan orbifold cohomology arxiv.org/abs/math/0004129 that satisfies Poincare duality. It is, however, a rather strange beast: the orbifold needs to be almost complex for these groups to be defined, and the grading is not by the integers, but only by the rationals! – André Henriques Jul 11 '13 at 9:52
Do you know what the crux of the issue is for integer coefficients? – Chris Gerig Jul 11 '13 at 12:55
Orbifold cohomology is a mix of manifold cohomolgoy (finite cohomological dimension) and group cohomology (infinite cohomological dimension). If you work with ℚ or ℝ coefficients, then the cohomology of finite groups vanishes, and you're back in a situation where orbifolds have finite dimensional cohomology and Poincare duality holds. – André Henriques Jul 11 '13 at 15:10
Is there then a way to speak about intersection forms? I guess I'm just trying to see what can be salvaged; if you have any further elaborations on your answer then that would be great! – Chris Gerig Jul 12 '13 at 14:34

A note before: the communities of low-dimensional topologists and categorically minded people have different opinions on what an 'orbifold' is. Let $G$ be a discrete group that acts properly and smoothly on the manifold $M$. As far as I understand, low-dimensional topologists consider the quotient space $M/G$, together with the cleverly packaged information on the isotropy subgroups. For a categorist (and I am following them on this issue), an orbifold is a special type of 'stack', which in turn is an object of a suitable category of functors to groupoids (and can be represented by a certain Lie groupoid; in the case under discussion, it is the action groupoid of the $G$-action on $M$). I think of an orbifold as a space whose points can have automorphisms and (as a side remark) my personal opinion is that it is quite unfortunate that there does not seem to be a definition of a stack that does not appeal to the functor-of-points-philosophy.

Now to the real issue I wish to communicate. The Pontrjagin-Thom construction is fundamental in most branches of manifold topology. The basic case is when $f:E \to B$ is a proper smooth map of manifolds, $dim (E)=dim(B)+d$, $d \in \mathbb{Z}$. There is a stable normal bundle $\nu(f):= f^* TB - TE$ and the PT-construction is a map of spectra $\Sigma^{\infty} B_+ \to Th (\nu(f))$, the Thom spectrum of $\nu(f)$. From this and the Thom isomorphism, one deduces homological invariants, such as the Gysin map $f_!: H^i (E) \to H^{i-d} (B)$, which is moreover functorial , $(f \circ g)_! = f_! \circ g_!$.

My initial thought was that there should be an analogue for orbifolds. The tangent bundles make sense, and one can make sense out of stable vector bundles on a stack. Moreover, there is a homotopy type of stacks. If the stack is $M//G$, its homotopy type $Ho(M//G)$ is the Borel construction $EG \times_G M$, and it comes with a map to $M//G$ that is declared to be a weak homotopy equivalence. One can then define homotopy/(co)homology of stacks using this homotopy type (there are finer theories, due to Henriques-Gepner).

One can pull back stable vector bundles on stacks to the homotopy type and form the Thom spectra there. The expectation is that there is a PT map $\Sigma^{\infty} Ho(B)_+ \to Th (\nu(f))$ when $f$ is proper in a suitable sense. The unique map $\ast //G \to \ast$ should be a proper map.

Unfortunately, this does not work. Together with Jeffrey Giansiracusa, I wrote a paper that gives such a PT-construction, under the restriction that the map $f$ is ''representable'', which in the case $B=\ast$ amounts to saying that $E$ is a closed manifold. In his study of orbifold cobordism, Andres Angel used a different approach than the PT-construction.

Why is such a construction impossible? Well, let us suppose that there is such a construction that works at least for $\ast//G \to \ast$ (the quotient map $\ast \to \ast //G$ does not pose a problem). As in the case of the ordinary PT-construction, one defines the Gysin maps $f_!$ and proves that they are functorial. As the composition $\ast \to \ast//G \to \ast$ is the identity, the composition of the corresponding Gysin maps must be the identity. On the other hand, we know very well what the Gysin map of $\ast \to \ast//G$ does on $H^0$; it is multiplication by $|G|$. This leads us to the contradiction that the identity on $\mathbb{Z}$ factors through the multiplication by $|G|$ map.

Strangely enough, when we take real coefficients, the Gysin map is integration over the fibres, and this seems to work for orbifolds. I can also conceive an integration map in complex $K$-theory, but nevertheless the problem seems to be fundamental and severely restricts the application of manifold methods to the topology of orbifolds.

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My understanding of (higher) category theory and stacks is absent. But viewing the Pontrjagin-Thom construction in terms of $[M,S^n]$ and cobordism classes of framed submanifolds, is there no analog at this level either? – Chris Gerig Jul 11 '13 at 23:39
No, I don't think so. – Johannes Ebert Jul 12 '13 at 7:54
Concerning "it is quite unfortunate that there does not seem to be a definition of a stack that does not appeal to the functor-of-points-philosophy", you might want to have a look at arxiv.org/pdf/math/0112006v1.pdf. In this little note of mine, I defined an orbifold as a continuous map $E\to X$ whose fibers are $K(\pi,1)$'s. Here, $E$ is the homotopy type of the orbifold, and $X$ is the coarse moduli space of the orbifold. – André Henriques Mar 11 at 13:39