4
$\begingroup$

Let the ring R be a MU*-module via a ring homomorphism φ and suppose it satisfies the condition of the Landweber exact functor theorem such that we obtain a cohomology theory $R^*(-) := R \otimes_{MU_*} MU^*(-)$. If ω denotes the complex orientation class in $\widetilde{MU}^2(\mathbb{C}P^\infty)$, then R* is oriented by the class $\omega_R := 1 \otimes \omega$.

Any other complex orientation of R* is obtainable by homogeneous power series θ with leading term x over R: θ(ω). These power series are in 1-1 correspondence with multiplicative natural transformations $t_\theta\colon MU^*(-) \to R^*(-)$.

Question: Which tθ restrict to ring homomorphisms which satisfy the Landweber criterion on coefficients? For which theories is this true for any θ?

The place to start seems to be by noting that if the formal group law associated to R* (with the orientation given by ωR) is F, then tθ classifies the FGL $F^\theta(x,y) := \theta\big(F(\theta^{-1}(x),\theta^{-1}(y))\big)$ over R. Further, the p-series are related by $[p]_{F^\theta}(x) = \theta\big([p]_{F}(\theta^{-1}(x))\big)$, so it would suffice to show that the sequence of coefficients in the right degrees stay regular under this conjugation by θ.

This seems to be true for any θ as long as $[p]_F(x)$ is of the form $\sum_{n \geq1} a_n x^{p^n}$ modulo p. In general, it is of the form $\sum_{k\geq1} a_kx^{kp^m}$, 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.

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

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 vn is invariant modulo the ideal (p,v1,...,vn-1), and so any change-of-orientation (which induces a strict isomorphism on the formal group law) does not change the property of vn being or not being a zero divisor after modding out the previous terms.

This follows from Lemma A2.2.6 in Ravenel's green book, which implies that any endomorphism of the formal group law over an Fp-algebra R is of the form g(xph) for some h and some power series g. In particular, the p-series [p]F(x) over R/(p,v1,...,vn-1) has this property, and so the leading coefficient vn is invariant under strict isomorphisms.

It should be noted that vn is not invariant before taking this quotient, but that has no effect on whether these elements form a regular sequence in R.

$\endgroup$
1
  • $\begingroup$ Thanks for your quick comment! A2.2.6 was indeed the result I was referring to above. You remarking that it is an endomorphism over R/(p,v<sub>1</sub>,...,v<sub>n-1</sub>) (and not just R/(p)) really did the trick. $\endgroup$
    – Johan
    Commented Mar 24, 2010 at 18:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .