8
$\begingroup$

For a generalized homology theory $h$ and a Serre fibration $F\rightarrow E\rightarrow B$, we can define an Atiyah-Hirzebruch spectral sequence\begin{equation}E^2_{p,q}=H_p(B,h_q(F))\Rightarrow h_{p+q}(E)\end{equation}Suppose we have determined $E^\infty_{p,q}$ for all relevant $(p,q)$ and want to find $h_n(E)$. We have\begin{equation}\bigoplus_pE^\infty_{p,n-p}=\text{gr }h_n(E)\end{equation}where $\text{gr}$ is the associated graded space for some filtration $F_ph_n$ on $h_n(E)$. This determines $h_n(E)$ up to extension. Are there results about how to solve the extension problem for certain theories $h$?

We sometimes have that $\text{gr }h_n(E)=h_n(E)$, in which case $h_n(E)$ is trivially obtained from $E^\infty_{p,q}$. For which $h$ and $E$ is this true? In my limited experience, this often (though not always?) holds when $h$ is ordinary homology $H$ or oriented cobordism $\Omega^{SO}$ and $E$ is the classifying space $BG$ of a finite abelian group $G$. Meanwhile, spin bordism tends to be less trivial; in particular, suppose $E^\infty_{p,n-p}=\mathbb{Z}_{k_p}$, then $\text{gr }\Omega^{spin}_n(BG)=\Omega^{spin}_n(\text{pt})\bigoplus_{p=0}^{n-1}\mathbb{Z}_{k_p}$ often (always?) becomes $\Omega^{spin}_n(BG)=\Omega^{spin}_n(\text{pt})\oplus\mathbb{Z}_N$ where $N=\prod_{p=0}^{n-1} k_p$. Are these examples supported by general results?

$\endgroup$
  • 4
    $\begingroup$ Whenever $gr h_n(E)$ is a free $h_0$-module, it must be the case that $gr h_n(E) \cong h_n(E)$, since there are no non-trivial extensions of free modules by free modules. I don't think that there is much more that can be said generally. $\endgroup$ – Craig Westerland Jun 26 '14 at 2:07
  • 3
    $\begingroup$ An example in which the extensions are highly non-trivial. $F=pt$, $B=E=BG$ elementary abelian, $h=K^*()$. The situation is easier to describe in cohomology than in homology. In cohomological version of the spectral sequence, $E^{\infty}$ is a vector space over $Z/2$, whereas $K^*(E)$ is torsion-free. So in this example all possible extensions are non-trivial. $\endgroup$ – user43326 Jun 26 '14 at 8:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.