Relationship of Bousfield Classes of Morava K-theories - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T22:34:15Z http://mathoverflow.net/feeds/question/89330 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89330/relationship-of-bousfield-classes-of-morava-k-theories Relationship of Bousfield Classes of Morava K-theories Jon Beardsley 2012-02-23T20:55:57Z 2012-02-23T21:55:05Z <p>Suppose we have $\langle K(n)\rangle$ and $\langle K(n-1) \rangle$ for some fixed prime $p$. do we know whether or not $\langle K(n) \rangle \geq \langle K(n-1) \rangle$ or $\langle K(n-1) \rangle \geq \langle K(n) \rangle$?</p> <p>It seems to me that the first relation is accurate if we somehow restricted ourselves to finite spectra (from Ravenel's "Localization at Certain Periodic Homology Theories"). However, we obviously aren't doing that (right?). </p> <p>It also seems to me that such a relationship would be incredibly problematic, mainly because we'd have the following. Assume the first relationship held and we restrict ourselves to operating within the distributive sub-lattice of the Bousfield lattice. Then $\langle K(n)\rangle \wedge\langle K(n-1)\rangle=\langle K(n-1)\rangle = \langle 0\rangle$ and this would seem to me to be incredibly problematic. Is this an accurate assessment of the situation, or have I missed something?</p> http://mathoverflow.net/questions/89330/relationship-of-bousfield-classes-of-morava-k-theories/89333#89333 Answer by Neil Strickland for Relationship of Bousfield Classes of Morava K-theories Neil Strickland 2012-02-23T21:17:25Z 2012-02-23T21:17:25Z <p>It is standard that $K(n)\wedge K(m)=0$ for $n\neq m$. One way to think about this is as follows: if $E$ and $F$ are complex oriented ring spectra then the corresponding formal group laws become isomorphic over $\pi_*(E\wedge F)$, but it is easy to see that formal group laws of different heights can only become isomorphic over the zero ring.</p> <p>On the other hand, as $K(n)$ is a ring spectrum we have maps $K(n)\xrightarrow{\eta}K(n)\wedge K(n) \xrightarrow{\mu} K(n)$ whose composite is the identity, so $K(n)\wedge K(n)$ is nonzero.</p> <p>This means that we cannot have $\langle K(n)\rangle\leq\langle K(m)\rangle$ unless $m=n$. Indeed, if $m\neq n$ then we saw that $K(n)$ is $K(m)$-acyclic. If we had $\langle K(n)\rangle\leq\langle K(m)\rangle$ we could conclude that $K(n)$ was also $K(n)$-acyclic, or in other words $K(n)\wedge K(n)=0$, which is false.</p> <p>It is true that when <code>$n\leq m$</code> we have $$\{K(n)-\text{acyclic finite spectra}\}\supseteq\{K(m)-\text{acyclic finite spectra}\}$$ which might suggest that $\langle K(n)\rangle\leq\langle K(m)\rangle$, but that is only a suggestion and it does not work out to be true.</p> http://mathoverflow.net/questions/89330/relationship-of-bousfield-classes-of-morava-k-theories/89335#89335 Answer by John Palmieri for Relationship of Bousfield Classes of Morava K-theories John Palmieri 2012-02-23T21:55:05Z 2012-02-23T21:55:05Z <p>Neil's answer is great. I just wanted to add that in fact the Bousfield classes of the Morava $K$-theories are minimal non-zero classes in the Bousfield lattice. In particular, $\langle K(n) \rangle$ and $\langle K(n-1) \rangle$ are not comparable.</p>