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The Atiyah-Hirzebruch spectral sequence

\begin{equation*}E^2_{p,q}=H_p(B,h_q(F))\Rightarrow h_{p+q}(E),\end{equation*}

computes the generalized homology $h$ of a total space $E$ of a Serre fibration from the ordinary homology $H$ of the base space $B$ with coefficients in the generalized homology of the fiber $F$. In particular, using the fibration $\text{pt}\rightarrow X\rightarrow X$ computes $h_n(X)$ from $h_n(\text{pt})$ and knowledge of the differentials.

For simplicity, suppose that $h$ satisfies $h_0(\text{pt})=\mathbb{Z}$. (This may not be necessary in the end.)

As a generalized homology theory , $h$ corresponds to a topological spectrum (also called $h$). Trivially, the Eilenberg-Maclane spectrum $H$ is a Postnikov truncation of $h$; in particular, the homotopy groups agree at degree $0$ and there are no k-invariants to check.

My intuition is that the spectral sequence refines $H$ by incorporating data from the higher degrees of the Postnikov tower of $h$. The new homotopy groups $\pi_n(h)$ in the Postnikov tower appear as the coefficient modules $h_n(\text{pt})$ in the spectral sequence, while the new k-invariants appear as differentials.

Now, my question. Suppose $h$ and $g$ are topological spectra such that $g$ is a Postnikov truncation of $h$. Is there an analog to the Atiyah-Hirzebruch spectral sequence that computes $h$ from $g$?

(Perhaps Postnikov truncation is too strong or too weak an assumption. Please feel free to interpret my question to make it more interesting!)

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    $\begingroup$ To get a spectral sequence you need an exhaustive filtration. E.g. in AHS there is a filtration by n-truncations, and the fibers of this filtration will be (different) EM-spaces. How do you suppose to build a filtration from a single truncation? $\endgroup$
    – Anton Fetisov
    Commented Oct 27, 2014 at 23:52
  • $\begingroup$ Suppose $\pi_n(g)=0$ for $n>k$. I still have such a descending filtration $F^nh$ of $h$ by $n$-truncations, only I stop when I hit $F^kh=g$. $\endgroup$
    – Alex Turzillo
    Commented Oct 28, 2014 at 0:04
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    $\begingroup$ @AlexTurzillo Your description of the AHSS as arising from the Postnikov filtration of $h$ is valid when $F=*$, but it's not quite that straightforward otherwise. In this case, you do have a spectral sequence with $E^2_{p,q}$ equal to $H_p(B,h_q(*))$ if $q > k$, $g_{p+k}(B)$ if $q=k$, and $0$ otherwise from the associated filtration on $h \wedge B$. $\endgroup$
    – Tyler Lawson
    Commented Oct 28, 2014 at 0:47
  • $\begingroup$ @TylerLawson This is exactly what I needed. Thanks. $\endgroup$
    – Alex Turzillo
    Commented Oct 28, 2014 at 5:59
  • $\begingroup$ Followup question: can I loosen the restriction $h_0(\text{pt})=\mathbb{Z}$? If I'm not mistaken, the usual AHSS works for unoriented bordism $\text{MO}$ which has $\pi_0(\text{MO})=\mathbb{Z}_2$. $\endgroup$
    – Alex Turzillo
    Commented Oct 28, 2014 at 6:02

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