# Cohomology theory with only one Adams operation?

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.

• I think $p$-completed $K$-theory should fit the bill. For an interesting application of $\delta$-rings in stable homotopy theory, see here. – Tim Campion Dec 7 '19 at 0:06
• @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! – John Greenwood Dec 7 '19 at 0:16
• 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". – Tim Campion Dec 7 '19 at 0:19
• 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). – Denis Nardin Dec 7 '19 at 7:14
• @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. – S. carmeli Dec 8 '19 at 7:48

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