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.

asheaf 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. $\endgroup$ – Jamie Walton May 29 '13 at 10:01