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Let $E$ be a homology theory whose coefficients $E_*(pt)$ are concentrated in even dimensions. This could be complex bordism $MU$, but also complex $K$-theory, $BP\langle n \rangle$, $E(n)$, ...

Let now $X$ be a $CW$-complex whose cells are concentrated in even dimensions. Then the Atiyah-Hirzebruch spectral sequence

$H_*(X; E(pt)) \Rightarrow E(X)$

collapses and we get that $E(X)$ is free as a $E_*(pt)$-module. Examples include $X = \mathbb{CP}^\infty$ or $X = BU$.

If $E_*(pt)$ is torsionfree for a homology theory $E$, then there are also other constructions of spaces with free $E$-homology. For example, one can choose a torsion-element $x\in \pi_{n+k}(S^n)$ and take the cone $C(x)$. Since every simply-connected finite $CW$-complex has torsion in its homotopy groups, this construction can be iterated arbitrarily far. The problem is that this procedure is not explicit at all since homotopy groups are hard to calculate.

So my question is the following:

Is there an explicit infinite $CW$-complex $X$ such that the following three conditions are fulfilled?

  1. Its skeleta have free $E_*$-homology (for one of the examples of $E$ above).
  2. Its skeleta are indecomposable in the stable category.
  3. $X$ is not concentrated in even degrees.
Improve this question
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    $\begingroup$ Note that Atiyah-Hirzebruch differentials are always torsion-valued, so if $H_*(X)$ is free abelian then the AHSS for $MU_*(X)$ will collapse showing that $MU_*(X)$ is free over $MU_*$, and it follow that $E_*(X)$ is free over $E_*$ on the same basis for any $MU$-algebra $E$ (even if $E_*$ is torsion). For example, $MU_*(U(n))$ is free and not in even degrees. However, a theorem of Miller gives a stable splitting of $U(n)$ into Thom spectra each of which is either in even degrees or odd degrees, so we don't exactly get an answer to your question. Other Lie groups might do the trick. $\endgroup$
    – Neil Strickland
    Commented Jan 13, 2012 at 12:43
  • $\begingroup$ Thanks for your comment! Can you tell me why the differentials in the AHSS are torsion-valued? $\endgroup$
    – Lennart Meier
    Commented Jan 13, 2012 at 14:33
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    $\begingroup$ Lennart, there is a very general and precise theorem of Arlettaz that gives bounds on the "order" of the differentials in any AHSS. Here order means the smallest number R so that Rd = 0, where d is the differential. The original, less precise version (which suffices for Neil's statement) goes back to Dold, according to the introduction of Arlettaz's paper. Here's the reference: Arlettaz, Dominique, The order of the differentials in the Atiyah-Hirzebruch spectral sequence. K-Theory 6 (1992), no. 4, 347–361. $\endgroup$
    – Dan Ramras
    Commented Jan 13, 2012 at 19:01
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    $\begingroup$ Ah, I see: because every homology theory is rationally ordinary. $\endgroup$
    – Lennart Meier
    Commented Jan 13, 2012 at 22:30
  • $\begingroup$ $E$=mod-2 K-theory and $X = RP^\infty$? $\endgroup$
    – Tilman
    Commented Jan 15, 2012 at 8:53

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