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

Question

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

Answer 1

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.

Answer 2

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.