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There's a Gauss-Bonnet theorem for compact 2-orbifolds(due to Satake, I think), which gives a relation between the curvature of a Riemannian orbifold and the orbifold topology(i.e. taking into account not just the structure of O as a Hausdorff topological space, but also the structure of the singular points.) The only proof that I've seen was very much like the classical one of the Gauss-Bonnet theorem, with geodesic triangulations, angle defects and so on. This approach doesn't generalise to higher dimensions to prove the Chern-Gauss-Bonnet theorem, and I haven't even found a conjectured Chern-Gauss-Bonnet formula. I'm especially interested in the four-dimensional case, where the integrand is amenable. One reason I'm interested is just to get a feel for how much more complicated orbifolds are than manifolds; on the one hand, many basic definitions seem to go through from the manifold world to the orbifold world, but the generalisation leads to significant complications in practice. On the other hand, Kleiner and Lott posted a paper on the arxiv in which they use Ricci flow to geometrise 3-orbifolds, so orbifolds certainly seem like a good arena in which to generalise differential geometry. Chern-Gauss-Bonnet is a bit of a benchmark for higher-dimensional Riemannian geometry, and I'd like to know the state of the art is in this case.

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I'm a little confused, orbifold Euler characteristic is multiplicative under covering spaces and so are integrals. So why shouldn't the appropriate orbifold Chern-Gauss-Bonnet question/conjecture just be exactly the same statement as Chern's theorem in the orbifold context? –  Ryan Budney Jan 26 '11 at 1:09
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@Ryan: That's a good question, and the observation is definitely correct for good orbifolds. But for bad orbifolds(ones which are not global quotients of a Riemannian manifold of the same dimension,) I'm unsure how you'd come up with the right correction term to take into account the isotropy of the singular strata. –  Gordon Craig Jan 26 '11 at 1:50
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The (upcoming) PhD thesis by Dan Berwick Evans reproved Chern-Gauss-Bonnet using ideas from supersymmetric TQFT. It wouldn't surprise me if the ideas work also for orbifolds, given his set-up. I can ask him, but you should also email him directly. –  Theo Johnson-Freyd Jan 26 '11 at 5:50
    
@Theo: Thanks! I'll get in touch with him. –  Gordon Craig Jan 26 '11 at 20:59
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How about going all the way to noncompact polyhedra? Bei Fang Chen, "The Gram-Sommerville and Gauss-Bonnet theorems and combinatorial geometric measures for noncompact polyhedra", Advances in Mathematics Volume 91, Issue 2, February 1992, Pages 269-291 –  Charlie Frohman Mar 30 '11 at 14:54
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2 Answers

Yes, there is a Gauss-Bonnet-Chern theorem for orbifolds, under a certain technical restriction. The proof is by reduction to the classical case, extending all definitions in the only meaningful way. It has been proven by Satake a long time ago:

http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.jmsj/1261153826

Here is an outline, in my own terms. For me, an orbifold $O$ is a smooth Deligne-Mumford stack (see here: http://arxiv.org/pdf/math/0203100). There is the coarse moduli space $|O|$. If $O$ is a quotient $M//G$ of some proper action of a discrete group, then $|O|$ is the space of orbits. In the definition of orbifolds that is used in $3$-dimensional topology, $|O|$ would be the underlying space which has some atlasses. On orbifolds, we have the following structures available: there is the tangent bundle $TO \to O$; $TO$ is an orbifold as well; this is a vector bundle in the category of stacks; the quotient $|TO| \to |O|$ is not quite a vector bundle.

Let me make the following assumption: There exists a compact oriented manifold $N$ and a finite covering $N \to O$ of degree $d$.

Under these circumstances, we can define the Euler number of the orbifold $\chi(O) \in \mathbb{Q}$ as follows:

$$\chi(O) := \frac{1}{d} \chi(N).$$

Using the multiplicativity of the ordinary Euler number, it is easy to see that $\chi(O)$ only depends on $O$ and not on $N$. Now I define the geometric side of the G-B-C theorem. Orbifolds have tangent bundles and we can talk about differential forms on orbifolds, de Rham complex, connnections on vector bundles and the Levi-Civitta connection in the same way as for manifolds. The cleanest way to define connections might be the framework of connections on principal bundles. The last piece of differential geometry needed is Chern-Weil theory. Conclusion: to any Riemann metric on an $2n$-dimensional oriented orbifold (i.e., the tangent bundle is oriented), we get an Euler-Form $e$, a closed $2n$-form.

This form represents an element in the cohomology of the orbifold, but we have to say what this means. If $O=M //G$ were a global quotient, then $H^* (O) := H^* (EG \times_G M)$, the Borel-equivariant cohomology. If $O$ is not a global quotient, you have to replace the Borel space by the homotopy type of the orbifold.

How do you integrate the $2n$-form? Well, given a closed oriented $2n$-manifold $N$ and a map $f:N \to O$ as before, we define

$$\int_O e := \frac{1}{d} \int_N f^* e,$$

which is the only reasonable way to define integration over orbifolds. Since $N$ is a manifold, we can apply the classical Gauss-Bonnet-Chern theorem and find that

$$\chi(O)=\int_O e.$$

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Thanks very much. The this gives me a second, more abstract way of looking at the problem. –  Gordon Craig Feb 18 '11 at 18:48
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One could probably show that the Gauss-Bonnet integral is a topological invariant of the orbifold (i.e. independent of the Riemannian metric), by doing local deformations of the metric within orbifold atlases. If you take an atlas of the orbifold which is the quotient of a ball by a finite group, then for any two Riemannian metrics which agree in a neighborhood of the boundary, the Gauss-Bonnet integral should be the same (since one may work equivariantly in the ball cover). Any two Riemannian metrics may be connected by a sequence of such local deformations by using a partition of unity, and so the total Gauss-Bonnet integral is a topological invariant. The question is whether this invariant equals the Euler characteristic.

I think this may be shown by reduction to the good orbifold case by a cut and paste trick. If you have a codim-1 suborbifold, one may assume that the Riemannian metric is reflection symmetric in a neighborhood of the codim-1 suborbifold. Cut along the suborbifold, then mirror the resulting boundary to get a new orbifold. Clearly the Gauss-Bonnet integral and the Euler characteristic of the resulting orbifold is the same (the orbifold euler characteristic is additive upon gluing along mirrored suborbifolds). I think one should be able to iterate this to reduce to the case of a good orbifold, which would look like a ball quotient with a right-angled boundary pattern (like in Thurston and Davis' reflection trick). Take a covering by ball-quotients, and assume that their boundaries intersect transversally. One may choose a metric on the orbifold so that the metric is locally a product metric near the boundary of each ball quotient. When several of these intersect, the metric will look locally like a product of a metric on a cube times the metric on the intersection (so that the boundaries meet orthogonally). Then cut along these codim-1 suborbifolds, and mirror the boundary. Davis' argument shows that these orbifolds are good, and thus the Gauss-Bonnet theorem applies. Since each region is contained in a ball quotient, it is good and you can take a manifold cover. Then reflect iteratively in the boundary by doubling along each subset of the boundary coming from a single boundary of a ball quotient. You end up with a closed manifold, so the orbifold is good.

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Thanks very much! –  Gordon Craig Feb 18 '11 at 19:04
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