Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Bott & Tu's marvelous book there is a derivation of the spectral sequence for a (smooth) fiber bundle for deRham cohomology done entirely in the realm of the smooth category. Unfortunately, as it's written there it is not clear how to generalize to other cohomology theories (they write down a double complex, and they use very much the fact that cohomology is given as the cohomology of this chain complex.)

Now, in Quillen's paper computing the complex cobordism ring, he introduces a geometric description of complex cobordism as a cohomology theory on smooth manifolds. I've been attempting, to no avail, to come up with a derivation of the Atiyah-Hirzebruch spectral sequence for a smooth fibration without using CW complex-esque techniques. My question is:

Does anyone know of a derivation of the Atiyah-Hirzebruch spectral sequence of a smooth fiber bundle for a generalized cohomology theory that does not leave the realm of manifolds?

So far, the most promising bet I have found is Segal's approach to this spectral sequence in the paper "Classifying spaces and spectral sequences." The trouble appears, however, in his use of a complex $BX_U$ for a covering $U$ of $X$. This is most definitely not a manifold, though it is homotopy equivalent to one for numerable covers. The question in this context, however, reduces to:

Does the natural filtration on $BX_U$ induce a filtration on $X$ that gives rise to the spectral sequence for a covering? If so, is there a nice description of this filtration using, say, just the data of $X$ and the numerable cover?

For this question I should be more specific about $BX_U$. This is defined as the geometric realization of the nerve of the topological category $X_U$ whose points are pairs $(x, U_{\sigma})$ where $U_\sigma$ is a finite intersection of elements of $U$ and $x \in U_\sigma$. Morphisms are inclusions $U_\sigma \subset U_\tau$. The filtration on $BX_U$ is given by looking at the images of $(NX_U)_n \times \Delta^n$ in $BX_U$, where $NX_U$ is the nerve.

share|improve this question
    
re BX_U: "though it is homotopy equivalent to one..." Namely X! Should've mentioned that... –  Dylan Wilson Nov 15 '12 at 5:04
2  
You can construct the AHSS for a single space by filtering the cohomology theory into Eilenberg-MacLane spectra, instead of filtering the space by a CW structure (and for a fibration you can do something similar, thinking of the cohomology of the fiber as a parametrized spectrum over the base). I'm not sure how to do this "geometrically" for something like cobordism, but it might be worth thinking about. –  Eric Wofsey Nov 15 '12 at 7:12
    
@Eric: While this won't help me here since I'm trying to avoid using the Thom spectrum, it's still a really neat fact! Thanks! –  Dylan Wilson Nov 15 '12 at 7:23

1 Answer 1

You can use Morse theory. But I do not know if you will be very happy with that, because in some sense you recover a CW-decomposition of your manifold from this data.

Anyway, if you use a Morse-Smale function on the basis of your bundle and if you suppose that it is self indexing and Morse-Smale you will get a nice filtration of your basis. Then to get the Atiyah-Hirzebruch spectral sequence you proceed by pulling-back this filtration from the basis to the total space of your fiber bundle.

share|improve this answer
    
This makes me quite happy! I have to check that it works (and convince myself that self-indexing, Morse-Smale functions exist for non-compact manifolds...) But this looks great. I'll accept it after thinking for a bit :) –  Dylan Wilson Nov 15 '12 at 6:06
    
Another way to say this: every compact manifold has the structure of a handlebody, and the skeleta of this handle-body are compact manifolds with boundary. The proof of this in the smooth category is given by Morse theory, but it is also true in the PL setting as well. –  John Klein Nov 15 '12 at 23:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.