# Cohomology of complex projective spaces with coefficientes in a complex-orientable cohomology theory

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.

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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.
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.