26
$\begingroup$

To every complex-oriented ring spectrum $E$ there is associated a formal group law, which is a power series $F_E(x,y)\in E_*[[x,y]]$.

Suppose $E$ and $F$ are two complex-oriented ring spectra and suppose I have an isomorphism of coefficient rings $\phi:E_*\rightarrow F_*$ that carries $F_E(x,y)$ to $F_F(x,y)$.

Does this imply that $E$ and $F$ are homotopy equivalent spectra?

Note that if $F_E(x,y)$ and $F_F(x,y)$ are "Landweber exact" formal group laws then the answer is yes.

$\endgroup$

1 Answer 1

23
$\begingroup$

The following is a communal answer from the algebraic topology Discord [1], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me). Kiran suggested it be recorded here to ease future reference.

The idea is to produce two topological realizations $M$, $N$ of a single $MU_*$–module by finding two distinct resolutions whose effect on homotopy is the same. The two associated square-zero extensions then give a counterexample. We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku_*$–module structure to trivialize.

$\DeclareMathOperator{\Sq}{Sq} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\HFtwo}{H\F_2} \newcommand{\Susp}{\Sigma} \newcommand{\co}{\colon\thinspace}$ The nice homotopy groups of complex $K$–theory, $ku_* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa_{ku}$ is $ku$–linear: in the diagram $$ \begin{array}{ccccc} & & \Susp^4 ku \\ & & u \downarrow \\ \Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\ & & u \downarrow \\ & & ku & \to & H\Z, \end{array} $$ the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear. Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$).

Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$ The bottom $k$–invariant $\kappa_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q_1 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q_1 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$ Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F_2 \oplus \Susp^4 \F_2$, which splits as a $ku_*$–module — hence this $ku_*$–module could alternatively be modeled by $N = \Susp H\F_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map). To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.

$\endgroup$
3
  • $\begingroup$ Are $\Sigma^*\mathbb F_2$ instead of $\Sigma^*H\mathbb F_2$ intentional or just a typo? $\endgroup$ Commented Sep 1, 2021 at 19:31
  • $\begingroup$ Intentional: a bare $\mathbb F_2$ is a graded abelian group concentrated in degree $0$, and $\Sigma^n \mathbb F_2$ is a graded abelian group concentrated in degree $n$, whereas $H \mathbb F_2$ is a spectrum. $\endgroup$ Commented Sep 1, 2021 at 20:55
  • 5
    $\begingroup$ Irakli also comments: these kinds of counterexamples are not possible in $R$-modules where $R_*$ is of global dimension 1, as a universal coefficient spectral sequence can be used to decompose modules into sums of quotients by homotopy elements. This rules out the possibility of similar examples using $KU$ or $ku/p$ in place of $ku$. Bearing this in mind, $ku$ is among the simplest ring spectra with larger homotopy global dimension, so it makes sense to begin a search here. $\endgroup$ Commented Sep 2, 2021 at 1:08

You must log in to answer this question.

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