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Hello everyone,

I'm having problems understanding a basic fact about complex-orientable cohomology theories:

Let $E^{\ast}$ be a multiplicative cohomology theory and $x\in E^2({\mathbb C}\text{P}^{\infty})$ such that the image of $x$ under

$E^2({\mathbb C}\text{P}^{\infty})\to E^2({\mathbb C}\text{P}^1)\cong E^0(\ast)$

equals $1$. Then the claim is that for any $n\geq 1$ the map

$E^{\ast}[x] / (x^{n+1})\longrightarrow E^{\ast}({\mathbb C}\text{P}^n)$

is an isomorphism (this is lemma 1.4 in Mike Hokpin's Lecture Notes on Complex Orientable Cohomology Theories).

The proof goes via the Atiyah-Hirzebruch spectral sequence, the claim being that the AHSS degenerates at the $E_2$-page $E_2^{p,q} \cong E^{\ast}[x] / (x^{n+1})$. I don't understand why the differentials have to vanish. Could somebody explain this to me in detail? Shouldn't be difficult, but I'm not familiar with the AHSS and don't see it.

Thank you in advance, Hanno

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I'll augment Scott's answer, and point out that you don't even need $E$ to be even. The class $x\in E^* CP^n$ is detected by an element $\bar{x}$ in the $E_2$-term. Because you know the class $x$ exists, $\bar{x}$ survives to $E_\infty$; that is, $d_r(\bar{x})=0$ for all $r\geq2$.

The differentials are derivations of $E_*$-algebras, so every element of the subring of $E_2$ generated by $E^*$ and $\bar{x}$ survives to $E_\infty$. But this is the whole $E_2$-term, so there are no non-trivial differentials.

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Dear Charles Rezk, Sorry for late commenting an old answer but maybe you can clarify a passage for me: How do you conclude that the second page $E_2$ is generated by $\pi_*E$ and $\bar{x}$? The only argument I saw was on Kochman's using UCT for cohomology in the tensor product variant but I'm dubious you can apply it, since $\pi_*E$ needs to be finitely generated abelian group – Riccardo Jun 14 at 16:22

If $E$ is an even cohomology theory (i.e., $E^i(*) = 0$ for odd $i$), then the objects in the $E_2$ page are only nonzero in degree $(p,q)$ with $p$ and $q$ even. In particular, the total degree is even. The differentials of a spectral sequence increment total degree by one, so they only hit entries of degree $(p,q)$ where $p+q$ is odd. Those entries are zero, so the differentials vanish.

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adams grading might help elucidate things a bit here – Sean Tilson May 20 '10 at 5:12
    
Only a small bit, though. The parity problem is concentrated in one variable under Adams grading, but it's still fundamentally a parity proof. Charles's argument is strictly more general, and not much longer. – S. Carnahan May 20 '10 at 5:56

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