12
$\begingroup$

Background:

$FGL(R)$ will be the category of formal group laws over the ring $R$. If $(E,x_E)$ is an oriented spectrum (i.e. $x_E \in \tilde{E}^2(\mathbb{P}^{\infty})$ is sent to the unit $1 \in \tilde{E}^2(\mathbb{P}^1)$ by the pullback of the inclusion $i: \mathbb{P}^1 \hookrightarrow \mathbb{P}^\infty$), then the pullback of the multiplication $\mu$ as H-space of $\mathbb{P}^\infty$ define an element in $FGL(E^*(*))$ in the following way:

$ \mu^*:E^*(*)[[x_E]] \cong E^*(\mathbb{P}^\infty) \to E^*(\mathbb{P}^\infty \times \mathbb{P}^\infty) \cong E^*(*)[[x_1,x_2]] $

where $x_i : = p_i^*(x_E)$ and the isomorphism are given by the collapsing of the Atiyah Hirzebruch spectral sequence. The fgl associated to $(E,x_E)$ is by definition $\mu^*(x_E) \in FGL(E^*(*))$. The Landweber exact theorem 'reverse' this contruction: let $f \in FGL(R)$ be a formal group law over $R$, then by certein hypothesis on $f$ the association

$ X \to MU^\bullet(X)\otimes_{\pi_*(MU)} R$

is a cohomology theory on CW-complexes ($MU$ is the complex cobordism spectrum and the $\pi_*(MU)$-module structure on $R$ is given by the classification map $\pi_*(MU) \to R$, since by Quillen's theorem the fgl of the complex cobordism is the universal one). This correspondence given by the Landweber's theorem seems to be only objectwise.

Question:

Given a morphism of formal group laws $\varphi$ over $R$, which is a power series in $R[[x]]$, is there a way to produce a natural transformation of the related cohomology theories induced by $\varphi$? Is there any reference in the literature where i could find an answer?

$\endgroup$
1
  • $\begingroup$ Why does the correspondence seems only objectwise? A map of fgls is the same thing as a map of $\pi_*MU$-algebras $\endgroup$ Jun 4, 2018 at 19:41

1 Answer 1

12
$\begingroup$

Yes, if $E$ and $E'$ are Landweber exact, then ring maps $E\to E'$ biject with isomorphisms of the associated formal groups, suitably interpreted. One version of this (for the context where $E$ and $E'$ are $2$-periodic) is explained in Proposition 8.43 of this Memoir. There is also a rather terse account in Proposition 11 of this lecture of Lurie.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.