Compact MU or BP Modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:00:27Z http://mathoverflow.net/feeds/question/116407 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116407/compact-mu-or-bp-modules Compact MU or BP Modules Jon Beardsley 2012-12-14T21:41:32Z 2012-12-15T00:10:12Z <p>Is there a classification of the compact MU or BP modules in any category of spectra? Can the periodicity theorem be finagled to give a MU-module structure on finite spectra?</p> http://mathoverflow.net/questions/116407/compact-mu-or-bp-modules/116415#116415 Answer by Eric Wofsey for Compact MU or BP Modules Eric Wofsey 2012-12-14T23:55:55Z 2012-12-15T00:10:12Z <p>No nonzero finite spectrum admits an $MU$-module structure. Indeed, suppose $F$ is a finite spectrum with an $MU$-module structure. Then for all $n$, $F$ has a map $v_n:\Sigma^{2p^n-2}F\to F$, which induces an isomorphism on $K(n)_*F$ (there's a subtlety here in that it's not obvious that the $v_n$ map on $F$ and the $v_n$ map on $K(n)$ give rise to the same map on $K(n)\wedge F$; see eg the end of the proof of Lemma 7 of <a href="http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf" rel="nofollow">http://math.harvard.edu/~lurie/252xnotes/Lecture33.pdf</a>). Thus for each $n$, $F/v_n$ is $K(n)$-acyclic. But $F/v_n$ is finite, so this implies it is also $K(m)$-acyclic for all <code>$m&lt;n$</code>, and so $v_n$ is also an isomorphism on $K(m)_*F$. But by finiteness of $F$, for any $m$ sufficiently large we can find <code>$n&gt;m$</code> for which $v_n$ must be $0$ on $K(m)_*F$ just for reasons of degree (by the AHSS for $K(m)_*F$). Thus $K(m)_*F=0$ for all sufficiently large $m$, which implies $F=0$.</p> <p>I would also add that even if you did have a finite spectrum with an $MU$-module structure, it could not possibly be compact as an $MU$-module. Indeed, if it were, after smashing with $H\mathbb{Z}$ it would be a compact $H\mathbb{Z}\wedge MU$-module. But <code>$\pi_*(H\mathbb{Z}\wedge MU)$</code> is a polynomial ring on infinitely many generators, and so all but finitely many of those generators have to act non-nilpotently on any compact module (basically, any "finite presentation" of a compact module can only involve finitely many of the polynomial generators). Since <code>$\pi_*(H\mathbb{Z}\wedge F)=H_*(F)$</code> vanishes in all but finitely many degrees for $F$ finite, this is impossible.</p>