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Apr 9 at 15:48 comment added Tim Campion 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 $MU_\ast$?
Apr 9 at 15:46 comment added Tim Campion 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.
Apr 9 at 15:22 vote accept Tim Campion
Apr 9 at 3:09 history answered Tyler Lawson CC BY-SA 4.0