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.

  • $\begingroup$ 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, 2019 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! $\endgroup$ Dec 7, 2019 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, 2019 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). $\endgroup$ Dec 7, 2019 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, 2019 at 7:48

1 Answer 1


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.

  • $\begingroup$ Wonderful! Where could i read about this (especially in the case that $E$ is a Morava $E$, or a Morava $K$)? $\endgroup$ Dec 9, 2019 at 17:32

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