Cohomology of complex projective spaces with coefficientes in a complex-orientable cohomology theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:26:38Z http://mathoverflow.net/feeds/question/24781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24781/cohomology-of-complex-projective-spaces-with-coefficientes-in-a-complex-orientabl Cohomology of complex projective spaces with coefficientes in a complex-orientable cohomology theory Hanno Becker 2010-05-15T17:11:27Z 2010-05-15T18:44:58Z <p>Hello everyone,</p> <p>I'm having problems understanding a basic fact about complex-orientable cohomology theories:</p> <p>Let <code>$E^{\ast}$</code> be a multiplicative cohomology theory and <code>$x\in E^2({\mathbb C}\text{P}^{\infty})$</code> such that the image of <code>$x$</code> under </p> <p><code>$E^2({\mathbb C}\text{P}^{\infty})\to E^2({\mathbb C}\text{P}^1)\cong E^0(\ast)$</code></p> <p>equals <code>$1$</code>. Then the claim is that for any <code>$n\geq 1$</code> the map</p> <p><code>$E^{\ast}[x] / (x^{n+1})\longrightarrow E^{\ast}({\mathbb C}\text{P}^n)$</code></p> <p>is an isomorphism (this is lemma 1.4 in Mike Hokpin's Lecture Notes on Complex Orientable Cohomology Theories).</p> <p>The proof goes via the Atiyah-Hirzebruch spectral sequence, the claim being that the AHSS degenerates at the <code>$E_2$</code>-page <code>$E_2^{p,q} \cong E^{\ast}[x] / (x^{n+1})$</code>. 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.</p> <p>Thank you in advance, Hanno</p> http://mathoverflow.net/questions/24781/cohomology-of-complex-projective-spaces-with-coefficientes-in-a-complex-orientabl/24795#24795 Answer by S. Carnahan for Cohomology of complex projective spaces with coefficientes in a complex-orientable cohomology theory S. Carnahan 2010-05-15T18:40:06Z 2010-05-15T18:40:06Z <p>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.</p> http://mathoverflow.net/questions/24781/cohomology-of-complex-projective-spaces-with-coefficientes-in-a-complex-orientabl/24797#24797 Answer by Charles Rezk for Cohomology of complex projective spaces with coefficientes in a complex-orientable cohomology theory Charles Rezk 2010-05-15T18:44:58Z 2010-05-15T18:44:58Z <p>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 $r\geq2$. </p> <p>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.</p>