Rational Morava E-theory of cyclic groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:01:15Z http://mathoverflow.net/feeds/question/117794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117794/rational-morava-e-theory-of-cyclic-groups Rational Morava E-theory of cyclic groups Craig Westerland 2013-01-01T18:12:05Z 2013-01-01T19:46:43Z <p>I'm confused about a seeming contradiction that is probably just a reflection of ignorance on my part. Let's try to compute the Morava E-theory of $B \mathbb{Z}/p$ in two different ways.</p> <ol> <li><p>First, following <a href="http://www.math.rochester.edu/people/faculty/doug/mypapers/emspace.pdf" rel="nofollow">Ravenel-Wilson</a> (see also <a href="http://www.math.rochester.edu/people/faculty/doug/mypapers/barc.pdf" rel="nofollow">Hopkins-Kuhn-Ravenel</a>, <a href="http://journals.cambridge.org/action/displayAbstract;jsessionid=6121AED0FBA78F8A056023186B2EB949.journals?fromPage=online&amp;aid=2092580" rel="nofollow">Hunton</a>, etc.), one computes (via knowledge of $K(n)^\ast(\mathbb{C} P^\infty)$ and a Gysin sequence) that the Morava K-theory of $B \mathbb{Z}/p$ is $$K(n)^\ast (B \mathbb{Z}/p) = K(n)_* [x] / x^{p^n}$$ where $x$ is of degree $2$. In particular, it is concentrated in even degrees and free of rank $p^n$ over $K(n)_\ast$. By Bockstein arguments, the Morava E-theory is then also free of the same rank.</p></li> <li><p>Second, using the knowledge that $E_n^\ast (B \mathbb{Z}/p)$ is free over $E_n^\ast$, it must embed into its rationalisation, $E_n^\ast (B \mathbb{Z}/p) \otimes \mathbb{Q}$. Let's try to compute the rank of that rationalisation using the group cohomological Atiyah-Hirzebruch spectral sequence: $$H^*(\mathbb{Z} / p, E_n^\ast \otimes \mathbb{Q}) \implies (E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$$ However, the $E_2$ term of the spectral sequence is rank 1 over $E_n^\ast \otimes \mathbb{Q}$ since $\mathbb{Z} / p$ is a finite group, and must collapse there. Therefore it certainly doesn't contain a sublattice of rank $p^n$.</p></li> </ol> <p>I've indicated my misgivings with the second argument by notating the target of the spectral sequence as $(E_n \otimes \mathbb{Q})^\ast (B \mathbb{Z}/p)$, that is, the value on $B \mathbb{Z}/p$ of the cohomology theory which is the rationalisation of $E_n$ (whose homotopy groups are $E_n^\ast \otimes \mathbb{Q}$). My guess is that this is NOT the same as the rationalisation of the E-theory of $B \mathbb{Z}/p$.</p> <p>So my real question is: is it easy to see why this is the case? More importantly, can one compute the correct answer (i.e., coming from argument 1) through methods based upon argument 2? I'm envisioning some sort of spectral sequence that knows how rationalisation and cohomology may fail to commute.</p> http://mathoverflow.net/questions/117794/rational-morava-e-theory-of-cyclic-groups/117803#117803 Answer by Eric Wofsey for Rational Morava E-theory of cyclic groups Eric Wofsey 2013-01-01T19:33:13Z 2013-01-01T19:46:43Z <p>Your guess is correct, and I think this is a confusion pretty much everyone has when they first see these computations. What's going on here is essentially just the simple algebraic fact that $$\mathbb{Z}[p^{-1}][[x]]\neq\mathbb{Z}[[x]][p^{-1}].$$</p> <p>In computation (1), you can work directly with a Gysin sequence for <code>$E_n$</code>, without going through $K(n)$ and Bocksteins. The Gysin sequence tells you that <code>$$E_n^*(B\mathbb{Z}/p)=E_n^*[[x]]/([p](x)),$$</code> where <code>$[p](x)$</code> is the $p$-series of the formal group law. More generally, this holds for any complex-oriented theory $E$ such that <code>$[p](x)\in E^*[[x]]$</code> is a nonzero divisor.</p> <p>Since the formal group law for <code>$E_n$</code> has height $n$, the first coefficient of <code>$[p](x)$</code> which is a unit is the $x^{p^n}$ coefficient. It follows that <code>$[p](x)$</code>differs from a monic polynomial of degree $p^n$ by a unit in <code>$E_n^*[[x]]$</code>, so <code>$E_n^*(B\mathbb{Z}/p)$</code> is free of rank $p^n$ over <code>$E_n^*$</code>.</p> <p>On the other hand, if we do the same computation with <code>$E_n\otimes \mathbb{Q}$</code> instead of <code>$E_n$</code>, <code>$$\frac{[p](x)}{x}=p+\dots\in(E_n\otimes \mathbb{Q})^*[[x]]$$</code> is now a unit. Thus <code>$(E_n\otimes\mathbb{Q})^*(B\mathbb{Z}/p)$</code> is free of rank 1 over <code>$(E_n\otimes\mathbb{Q})^*$</code>. However, <code>$[p](x)/x$</code> is only a unit in <code>$(E_n\otimes \mathbb{Q})^*[[x]]$</code>, not in <code>$E_n^*[[x]]\otimes\mathbb{Q}$</code>. This is because to construct a power series inverse for it, we need coefficients with arbitrarily large powers of $p$ in the denominator. Thus we find that <code>$(E_n\otimes \mathbb{Q})^*[[x]]/([p](x))$</code> and <code>$E_n^*[[x]]\otimes\mathbb{Q}/([p](x))$</code> look quite different from each other, and this is exactly the discrepancy between your two computations.</p> <p>As for a computation along the lines of (2) that gives the answer from (1), I don't know of anything like that. You could make an AHSS computation for $E_n$ rather than $E_n\otimes\mathbb{Q}$ and only tensor with $\mathbb{Q}$ afterwards, which would give the right answer. However, the AHSS $$H^*(\mathbb{Z}/p,E_n^*)\implies E_n^*(B\mathbb{Z}/p)$$ is quite subtle: I don't know how to compute the differentials in it without already knowing the final answer, and the filtration of the spectral sequence distorts almost all of the structure of $E_n^*(B\mathbb{Z}/p)$ (for instance, the associated graded you get from the spectral sequence consists mostly of $p$-torsion, whereas the actual answer is torsion-free). </p>