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In a somewhat limited setting, a Seifert Fibre Space is a 3-manifold $M$ with a "nice" decomposition into circles (http://en.wikipedia.org/wiki/Seifert_fiber_space). That is, $M$ is decomposed into circles in a way such that $M$ has neighbourhoods which are "fibred as a solid tori would be, if these tori are given by a solid cylinders with rational rotations identifying opposite disks".

There is a natural map (the Seifert fibration) from $M$ to the quotient space collapsing each of the circle fibres. Of course, this needn't be a fibration at all. In general one has isolated singular fibres - one may like to view these as having fractional lengths relative to their neighbours. However, it is a fibration when we view the base space as an orbifold (let's call it $B$) instead of just a space.

When given a fibration, it is usual to stick it into a spectral sequence and compute (co)homology. I assume that the same can be done in this setting, but I can't find any discussion of this in the literature (which is understandable - presumably calculating orbifold cohomology with twisted coefficients is almost always more difficult than computing the cohomology of $M$ directly). Of course, one would have to replace cohomology with orbifold cohomology. So, I suppose my question is:

  • Does the cohomology of $M$ fit into a spectral sequence with (twisted) coefficients over the orbifold cohomology of the quotient? I'm pretty certain this will be the case:
  • In which case, there are various flavours of orbifold cohomology. However, I presume that I am still correct in assuming that we should use the cohomology of the classifying space for the orbifold cohomology.
  • Orbifold cohomology agrees with singular cohomology over rational coefficients. With general coefficients, though, it seems usual to get non-trivial cohomology in infinitely many degrees. Of course, the cohomology of $M$ is concentrated in degrees 0 to 3. So my question really is of the nature of this spectral sequence. Either the torsion is killed off in the sequence or never appears because of the original twisting of the coefficients. Is it possible to say which? Is there a simple toy example where the explicit calculations can be seen? I was thinking, for example, of orbifolds associated to quotients of wallpaper groups, which arise naturally from Seifert fibrations.
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I haven't seen anyone take this approach in the literature. The homology can be read off from the fundamental group presentations of these manifolds, which are readily-deduced in Seifert's original work on this topic. The cup product structure in cohomology is more work to get at. It's generally degenerate except in a few cases. Peter Zvengrowski works this out -- just google his name, the papers are easy-enough to find. –  Ryan Budney May 28 '13 at 19:19
Thanks Ryan, My interest is more with understanding the spectral sequence itself than the actual computations. As acknowledged in the question, the spectral sequence isn't that helpful for explicit computations. –  Jamie Walton May 28 '13 at 21:21
Dear Jamie, what you're looking for may be Leray's sheaf-theoretic spectral sequence. Specifically, in this case the fibers vary nicely enough that there is a sheaf on the base space $B$ whose stalks are the cohomology of the fibers, and the spectral sequence starts with the cohomology of $B$ with coefficients in this sheaf. I'm not clear enough on orbifold cohomology to tell you whether it is the same thing, but I'd suspect that it is almost certainly so. There's no magic in this to help the computations, though; you still patch the fibers in manually. –  Tyler Lawson May 29 '13 at 3:40
Thanks Tyler, this seems like a good suggestion. Just to make sure I'm clear on your answer: you say we want a sheaf on $B$ whose stalks are the cohomology of the fibers. We first take some sheaf on $M$ (e.g. the constant sheaf). Then I'm supposing that I need to compute the direct image of this sheaf on $B$, along with its higher direct images? I imagine it turns out that I only get something in degrees zero and one, and that they will presumably all have stalks equal to $\mathbb{Z}$ (this may be what you mean by saying the fibers vary nicely enough). I'll give it a try, thanks again. –  Jamie Walton May 29 '13 at 10:01
Ah, on reflection, it is clear that the direct image (and it's first higher version) of the constant sheaf (on $\mathbb{Z}$) have stalks $\mathbb{Z}$. Indeed, the preimage of a small simply connected open disk is just a solid torus. The first higher direct image won't be constant though since "the stalks of generic fibres are multiples of those of nearby singular fibres", which is the sort of thing I was expecting. –  Jamie Walton May 29 '13 at 10:29

1 Answer 1

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Indeed, you always get a fibration $M\to \hat B$, where $\hat B$ is the Haefliger's classifying space of the orbifold (the space whose cohomology is the orbifold cohomology of $B$). The fiber of this fibration is the principal leaf of your Seifert fibration.

One can write the corresponding spectral sequence. And indeed, if the orbifold is not a manifold you will get cohomology in infinitely many degrees, which has to be killed, when going to the third page of the spectral sequence.

The simplest example is the linear circle action on $M=S^3$ with parameters $(1,p)$, with one exceptional leaf, such that the angle at the quotient point is $2\pi /p$.

Then the spectral sequence is essentially the Gysin sequence of the spherical fibration. You obtain from the sequence that the cohomology of $\hat B$ must be generated by one element $e$ in $H^2(\hat B)= \mathbb Z$, the Euler class class of the fibration. The square $e^2$ of the element $e$ is the geneartor of $H^4 (\hat B)$. It has the same order as all other powers of $e$. And this order is exactly $p$. This you can see only locally and it cannot be seen from the Gysin sequence, since the sequence is the same for all $p$.

The sequence you get is very similar to the spectral sequence of the universal fibration $S^{\infty} \to CP ^{\infty}$. The infinitely many non-zero elements are killed in the same way.

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Beautiful, thanks Alexander. I should have thought about using the Gysin sequence, since we are just looking at circle bundles. I was thinking about twisted coefficients, but I guess that the action of the orbifold fundamental group here is trivial (and this happens if and only if the circle bundle is orientable, which we need for the Gysin sequence). I haven't seen many examples of the sort of cohomology rings you get for these orbifolds. I know that rationally it should be the same as for the base manifold, and I guess this example allows you to work backwards to some extent. –  Jamie Walton Jun 14 '13 at 17:38

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