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Let $E$ be a ring spectrum with an orientation. Now I want to calculate $E^*(\mathbb{C}P^n)$.

The definition of orientation I am using is: There is an element $x \in E^*(\mathbb{C}P^{\infty})$ such that it's restriction to $\mathbb{C}P^1$ is a generator over $\pi_*(E)$.

To calculate this I want to use the Atiyah-Hirzebruch spectral sequence. But I am unable to figure out why all the differentials vanish. I believe one has to some how use the fact that $E$ has an orientation but I am unable to figure out how. Thank you.

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(I'll use reduced cohomology for this answer for mild convenience.) $\require{AMScd}$For any ring spectrum $E$, the Atiyah-Hirzebruch spectral sequence for (the reduced groups) $E^*(\Bbb{CP}^1)$ degenerates at the $E^2$ page for degree reasons; this can be seen as coming from the (definitional in spectra) isomorphism $\pi_*(\Sigma^2 E) = \pi_{*-2}(E)$. Let the unit be $1 \in E^2(\Bbb{CP}^1) = \pi_0(E)$.

There are now maps of spectral sequences associated to the inclusions $\Bbb{CP}^1 \hookrightarrow \Bbb{CP}^n \hookrightarrow \Bbb{CP}^\infty$. Now assume the ring spectrum is complex oriented. Let $x \in E^2(\Bbb{CP}^\infty)$ restrict to $1 \in E^2(\Bbb{CP}^1).$ We have a commutative diagram

\begin{CD}\pi_*(E) @>>> AH_2^{*,2}(\Bbb{CP}^\infty) @>>> AH_2^{*,2}(\Bbb{CP}^1) \\ @VV=V @VVV @VVV \\ \pi_*(E) @>\sigma \mapsto x \sigma>> E^{2+*}(\Bbb{CP}^\infty) @>>> E^{2+*}(\Bbb{CP}^1) \end{CD}

where the top map identifies $\pi_*(E)$ with the bottom nonzero row in the Atiyah-Hirzebruch spectral sequence. Because the composite map given in the bottom row is an isomorphism, that entire row of the Atiyah-Hirzebruch spectral sequence for $\Bbb{CP}^\infty$ must survive to $AH_\infty$. But it is easy to check using the multiplicative structure of the spectral sequence that if any differential is nonzero, the transgression (whose input lies on $AH_n^{0,n+2}$ and output lies on $AH_n^{n+1,2}$) is nonzero; but if this were true, that would contradict the fact that the entirety of the bottom row survives to the $AH^\infty$. So the spectral sequence for $\Bbb{CP}^\infty$ degenerates on the second page. You see the same for $\Bbb{CP}^n$ by investigating the map of spectral sequences associated to the inclusion $\Bbb{CP}^n \hookrightarrow \Bbb{CP}^\infty$.

So the unreduced cohomology $E^*(\Bbb{CP}^\infty) = \pi_*(E)[x]$, where $|x| = 2$ (I guess this is usually written instead as power series). $E^*(\Bbb{CP}^n)$ is, accordingly, $\pi_*(E)[x]/(x^{n+1})$.


You might see Lurie's brief notes on complex oriented cohomology theories.


Stable cohomotopy is an explicit example of a non-complex-oriented cohomology theory for which the spectral sequence does not degenerate. The Segal conjecture for compact Lie groups (here) implies that after $p$-completion, $\pi_s^0(\Bbb{CP}^\infty_p) = \Bbb Z_p$, the $p$-adics (since the Burnside ring of $S^1$ is just $\Bbb Z$). If the spectral sequence degenerated, then the 0th stable cohomotopy of $\Bbb{CP}^\infty_p$ would have a filtration such that the subquotiens are $\pi_s^{2k}(S)$. But this is not possible, because $\pi_s^{2k}(S)$ can have summands like $(\Bbb Z/p)^2$, and the only groups that can arise as subquotients of $\Bbb Z_p$ are $\Bbb Z_p$ itself or $\Bbb Z/p^k$.

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  • $\begingroup$ There are some minor subtleties about this that need to be edited. I'll do that later tonight. $\endgroup$ – user98602 Apr 2 '17 at 21:57
  • $\begingroup$ I am just a bit confused about the notation.. Are you using rows for the fiber direction? If possible can you please add a snapshot of the $E_2$ page labelled with your notation. Thank you very much for all the details since this was skipped in all the references I looked at. $\endgroup$ – happymath Apr 3 '17 at 7:49
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Your condition should be that the orientation is an element of the reduced $E$ cohomology of $CP^\infty$, restricting to a generator of $\pi_0(E)$.

Look at the reduced AHSS. The elements with bidegree $(2,0)$ are perm cycles in the SS for $CP^1$ for degree reasons. The information about the orientation tells you they have to be perm cycles in the SS for $CP^\infty$ too and, in particular, in the SS for $CP^n$.

Since we have a splitting $* \to CP^n \to *$ the reduced AHSS sits inside the unreduced SS, and in the unreduced SS, elements with $s=0$ are perm cycles.

Now multiply.

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  • $\begingroup$ That middle thing about them being perm cycles in the SS for CP^infty uses the fact that a map of SSs CANNOT lower filtration. The element has to come from somewhere, and it has to be in the same filtration. If you think about it, you'll see that that's the importance of the orientation being a reduced element. $\endgroup$ – Michael Andrews Apr 2 '17 at 22:24
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    $\begingroup$ "Mike Miller forever" was an important part of my answer. $\endgroup$ – Michael Andrews Apr 2 '17 at 22:39

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