7
$\begingroup$

Let $E$ be a multiplicative cohomology theory. Fix a prime p. Call a ring map $\psi^{p}:E\rightarrow E$ an Adams operation if it lifts the Frobenius map $E/p\rightarrow E/p$.

It is of course well-known that $K$-theory has Adams operations. (If it weren't for $K$-theory, these operations would have a different name.) In fact, it has an Adams operation for every prime $p$.

My question is: are there examples of multiplicative cohomology theories which have an Adams operation $\psi^{p}$ for only one prime $p$?

By "only one prime" I don't mean that I require that operations at other primes necessarily don't exist, just that they don't necessarily exist. In other words, their (non)existence is much less obvious/clear/explicit than that of $\psi^{p}$.

Motivation: This would turn $E^{*}(X)$ into a $\delta$-ring, which is something some people like to study.

$\endgroup$
  • $\begingroup$ I think $p$-completed $K$-theory should fit the bill. For an interesting application of $\delta$-rings in stable homotopy theory, see here. $\endgroup$ – Tim Campion Dec 7 '19 at 0:06
  • $\begingroup$ @TimCampion I don't think so, because Adams operations at invertible primes are easy to construct. Thanks for the reference, I will have a look! $\endgroup$ – John Greenwood Dec 7 '19 at 0:16
  • $\begingroup$ You're probably right, I really am not as familiar with Adams operations as I should be. Btw in the paper I linked to, the $\delta$-ring stuff first gets going at the beginning of Section 4 (at least I'm guessing it's the same meaning of "$\delta$-ring"). As I understand it, the key property they end up using is that in a $\delta$-ring, all torsion is nilpotent. It's referred to in the abstract as "a certain power operation". $\endgroup$ – Tim Campion Dec 7 '19 at 0:19
  • 1
    $\begingroup$ Not quite what you're asking for, but $\delta$-rings appear very naturally via the power operations on $p$-completed $E$-theory, where they're called $\theta$-rings. Arguably this is a better analogy, since you'd expect the Frobenius to be a map of rings, not a map of modules (as cohomology operations are). $\endgroup$ – Denis Nardin Dec 7 '19 at 7:14
  • $\begingroup$ @TimCampion The notion of $\delta$ ring you refer to is only a "semi-$\delta$-ring". Unfortunately, it does not satisfy in general the multiplicative axiom. In fact, what is the right generalization of $\delta$-ring that is satisfied by the ambidextruous $\delta$-operation and its analogues is still unclear at all, at least to us. Moreover, the main feature used is not the one you mensioned but the fact that there's a unique such structure on $\mathbb{Z}_p$, and it reduces the valuation of every number by exactly $1$ if it is not invertible. $\endgroup$ – S. carmeli Dec 8 '19 at 7:48
5
$\begingroup$

Adams operations exist in quite wide generality. For any even periodic ring spectra $E$ and $F$, we have associated formal groups $G_E$ and $G_F$ over base schemes $S_E$ and $S_F$. There is a moduli scheme $\text{Hom}(G_E,G_F)$ parametrising pairs $(f,\widetilde{f})$ consisting of a map $f\colon S_E\to S_F$ and a homomorphism $\widetilde{f}\:G_E\to f^*G_F$. This contains an open subscheme $\text{Iso}(G_E,G_F)$ consisting of pairs where $\widetilde{f}$ is an isomorphism. There are natural comparison maps $\text{spec}(\text{Ind}(E_0\Omega^\infty F))\to\text{Hom}(G_E,G_F)$ and $\text{spec}(E_0F)\to\text{Iso}(G_E,G_F)$, both of which are isomorphisms when $E$ and $F$ are Landweber exact. By considering the case $(f,\widetilde{f})=(\text{id},k.\text{id})$ we see that $\psi^k$ exists as a ring automorphism of $E$ when $k$ is invertible in $\pi_0(E)$, and as a ring endomorphisms of $\Omega^\infty E$ when $k$ is not invertible. In other words, in the first case $\psi^k$ is an additive and multiplicative stable operation, and in the second case it is an additive and multiplicative unstable operation. This remains true in many cases when $E$ is not Landweber exact, by various less systematic arguments. It will always be easier to produce $\psi^k$ in cases where $k$ is invertible.

$\endgroup$
  • $\begingroup$ Wonderful! Where could i read about this (especially in the case that $E$ is a Morava $E$, or a Morava $K$)? $\endgroup$ – John Greenwood Dec 9 '19 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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