Complex orientation of the Adams Summand - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:32:52Z http://mathoverflow.net/feeds/question/59490 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59490/complex-orientation-of-the-adams-summand Complex orientation of the Adams Summand Sean Tilson 2011-03-24T21:54:23Z 2011-03-25T17:48:52Z <p>First lets fix a prime $p$ (I really care about $p=2$ but would be happy to know about other primes as well). When localized at a prime the spectrum $ku$ (Complex connective K-theory) splits as a wedge of suspensions of a spectrum $l$. In their paper on the Cooperation algebra of the Adams Summand, Baker-Richter state that the complex orientation $MU \to l$ factors through the complex orientation of $BP$ and that it can be arranged so that the image of <code>$x_n \in \pi_{2n}MU$</code> and <code>$v_i \in \pi_{2p^i-2}BP$</code> go to zero in <code>$\pi_*l$</code>. Let $BP(n)$ denote the "truncation" of $BP$ so that <code>$\pi_*BP(n) \cong \pi_*BP/(v_{n+1},v_{n+2},...)$</code>. (This may not be the notation in the literature, but I am pretty sure I have seen people mention these objects).</p> <p>My question is can we extend the complex orientation of $l$ over some of these other spectra? Does the structure of $l$ and <code>$l_*BP$</code> as a module over $BP(n)$ change a lot when going from $BP(n)$ to $BP(n+1)$?</p> http://mathoverflow.net/questions/59490/complex-orientation-of-the-adams-summand/59585#59585 Answer by Tyler Lawson for Complex orientation of the Adams Summand Tyler Lawson 2011-03-25T17:33:03Z 2011-03-25T17:48:52Z <p>Yes, the complex orientation can be factored through these truncations of BP. Either classical methods (the Baas-Sullivan theory of manifolds with singularity - see Baas' "On bordism theory of manifolds with singularities") or more modern methods (see e.g. Strickland's "Products on MU-modules") produce truncated Brown-Peterson $BP\langle n\rangle$ as a tower of "quotients" <code>$$MU \to \cdots \to BP \to \cdots \to BP\langle 2\rangle \to \ell \to H\mathbb{Z} \to H\mathbb{Z}/p$$</code> and this produces a sequence of compatible complex orientations on these, provided of course that you've produced compatible multiplicative structures on all of the $BP\langle n\rangle$.</p> <p>The problem doesn't really change if you use $ku$. Also, note that $ku$ and $\ell$ have nicer and more natural multiplicative structures and orientations than any version of $BP\langle n\rangle$ is known to in general.</p>