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.



Seifert Fibrations and their associated Spectral Sequence
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.


