Changing the orientation of a Landweber exact cohomology theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:36:12Z http://mathoverflow.net/feeds/question/19213 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19213/changing-the-orientation-of-a-landweber-exact-cohomology-theory Changing the orientation of a Landweber exact cohomology theory Johan 2010-03-24T17:16:39Z 2010-03-24T17:38:04Z <p>Let the ring R be a MU<sub><code>*</code></sub>-module via a ring homomorphism &phi; and suppose it satisfies the condition of the Landweber exact functor theorem such that we obtain a cohomology theory <code>$R^*(-) := R \otimes_{MU_*} MU^*(-)$</code>. If &omega; denotes the complex orientation class in $\widetilde{MU}^2(\mathbb{C}P^\infty)$, then R<sup><code>*</code></sup> is oriented by the class $\omega_R := 1 \otimes \omega$.</p> <p>Any other complex orientation of R<sup><code>*</code></sup> is obtainable by homogeneous power series &theta; with leading term x over R: &theta;(&omega;). These power series are in 1-1 correspondence with multiplicative natural transformations <code>$t_\theta\colon MU^*(-) \to R^*(-)$</code>.</p> <p><strong>Question</strong>: Which t<sub>&theta;</sub> restrict to ring homomorphisms which satisfy the Landweber criterion on coefficients? For which theories is this true for any &theta;?</p> <p>The place to start seems to be by noting that if the formal group law associated to R<sup><code>*</code></sup> (with the orientation given by &omega;<sub>R</sub>) is F, then t<sub>&theta;</sub> classifies the FGL <code>$F^\theta(x,y) := \theta\big(F(\theta^{-1}(x),\theta^{-1}(y))\big)$</code> over R. Further, the p-series are related by <code>$[p]_{F^\theta}(x) = \theta\big([p]_{F}(\theta^{-1}(x))\big)$</code>, so it would suffice to show that the sequence of coefficients in the right degrees stay regular under this conjugation by &theta;.</p> <p>This seems to be true for any &theta; as long as <code>$[p]_F(x)$</code> is of the form <code>$\sum_{n \geq1} a_n x^{p^n}$</code> modulo p. In general, it is of the form <code>$\sum_{k\geq1} a_kx^{kp^m}$</code>, where m can be taken to be the height of the FGL (Ravenel's Green Book), but I don't see why it should be true in the general case.</p> <p>I am sure this has been treated by someone, but have yet to see it on print. If anyone has seen question discussed somewhere, please let me know.</p> http://mathoverflow.net/questions/19213/changing-the-orientation-of-a-landweber-exact-cohomology-theory/19214#19214 Answer by Tyler Lawson for Changing the orientation of a Landweber exact cohomology theory Tyler Lawson 2010-03-24T17:22:41Z 2010-03-24T17:38:04Z <p>The property of being Landweber exact is independent of the orientation. In terms of Landweber's criterion, this is generally phrased as saying that the element v<sub>n</sub> is invariant modulo the ideal (p,v<sub>1</sub>,...,v<sub>n-1</sub>), and so any change-of-orientation (which induces a strict isomorphism on the formal group law) does not change the property of v<sub>n</sub> being or not being a zero divisor after modding out the previous terms.</p> <p>This follows from Lemma A2.2.6 in Ravenel's green book, which implies that any endomorphism of the formal group law over an F<sub>p</sub>-algebra R is of the form g(x<sup>p<sup>h</sup></sup>) for some h and some power series g. In particular, the p-series [p]<sub>F</sub>(x) over R/(p,v<sub>1</sub>,...,v<sub>n-1</sub>) has this property, and so the leading coefficient v<sub>n</sub> is invariant under strict isomorphisms.</p> <p>It should be noted that v<sub>n</sub> is not invariant <em>before</em> taking this quotient, but that has no effect on whether these elements form a regular sequence in R.</p>