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Let $X$ be a (finite, say — or maybe of finite type) spectrum with even cells (in other words, $H_\ast(X;\mathbb Z)$ is free and concentrated in even degrees). Let $E^\ast$ be a complex-oriented cohomology theory. Consider the Atiyah-Hirzebruch spectral sequence $H^p(X,E_q) \implies E^{p+q}(X)$.

Question: Does this spectral sequence always collapse at the $E_2$ page?

Notes:

  • The answer is yes if $X = \mathbb C \mathbb P^n$, $BU(n)$, $MU$, etc.

  • The answer is yes if $E$ has homotopy concentrated in even degrees (this is how one shows that such an $E$ is complex-orientable).

  • I don’t know many examples of complex-oriented $E$’s that don’t have homotopy concentrated in even degrees, so in practice the above suffices. But if this is true in general, I expect the proof the be illuminating.

  • There’s an analog: if $E$ is a real-oriented cohomology theory, then the AHSS for $E^\ast(X)$ collapses at the $E_2$ page for any $X$. This is because Thom showed that the unoriented bordism spectrum $MO$ is an $H \mathbb F_2$-algebra. I’m not sure if there’s a more direct proof.

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    $\begingroup$ Just in response to your note, an example "in nature" of a complex-oriented ring that is not even is THH(Z). More generally, THH(R) where R is complex-oriented. Another example would be C^*(X;E) where E is even but X is a space without even cells. Yet another would be the free E_{infinity}-MU-algebra on a class in degree 0. I mention these just because in a few contexts these days people have been spinning their wheels approximating complex-oriented rings by actually even rings. $\endgroup$
    – Jeremy Hahn
    Commented Apr 9 at 11:48

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Yes, this is true.

As you commented, $X$ has finitely generated free homology, and so the $E_2$-term of the AHSS can be identified with $$ H^*(X; E^*) \cong E^* \otimes H^*(X). $$ To show collapse, we need to show that all elements are permanent cycles: $d^r(\alpha \otimes \beta) = 0$ for all $\alpha \in E^*$, $\beta \in H^*(X)$, $r \geq 2$.

Because $E$ is a multiplicative cohomology theory, this spectral sequence is compatible with the left $E^*$-module structure. As a result, $$ d^r(\alpha \otimes \beta) = \alpha \cdot d^r(1 \otimes \beta). $$ This reduces us to showing that the elements $1 \otimes \beta$ are permanent cycles.

Because $E$ is complex orientable, it accepts an $MU$-orientation: a natural transformation $f: MU^* \to E^*$ of multiplicative cohomology theories. The AHSS is natural in the cohomology theory, implying that $$ f_* d^r(\alpha \otimes \beta) = d^r (f_* \alpha \otimes \beta). $$ However, multiplicativity implies that $f$ takes units to units: $f_*(1_{MU}) = 1_E$. Therefore, the result follows from knowing collapse of the AHSS for $MU$-cohomology--which, as you stated, follows from evenness.

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    $\begingroup$ Thanks, this is great! I made a comment along these lines and then deleted it because I was seized with doubt about the naturality of the AHSS along multiplicative maps $F \to E$. I worried this might imply that e.g. the $K(h)_\ast$-AHSS can be completely worked out by knowing the $MU_\ast$-AHSS in general (too strong). But no — if $d_k^{MU}$ doesn’t vanish on some class, then the image $d_k^{K(h)}$ can still vanish on that same class, and then for the subsequent pages we have no information about what happens to that class. But when the $d_k^{MU}$ differentials vanish, we do profit. $\endgroup$
    – Tim Campion
    Commented Apr 9 at 15:46
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    $\begingroup$ The upshot is that we have the implication ($E^\ast$-AHSS for $\mathbb C \mathbb P^\infty$ collapses) $\Rightarrow$ ($E^\ast$-AHSS collapses for all finite-type even-cell complexes collapses). This statement doesn’t refer to $MU$ at all, but the proof requires us to use the universal property of $MU$, and also to show that $MU$ has even homotopy (which is a nontrivial computation). I wonder if there’s a proof of this fact which doesn’t require us to compute $MU_\ast$? $\endgroup$
    – Tim Campion
    Commented Apr 9 at 15:48

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