Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.