Lambda-rings for formula-phobics

In 1956, Alexander Grothendieck (middle) introduced $\lambda$-rings in an algebraic-geometric context to be commutative rings A equipped with a bunch of operations ${\lambda }^i$ (for all numbers $i\in {\mathbb{N}}_{+}$) satisfying a list of rather obscure identities. From the easier ones, such as

${\lambda }^0(x)=1,{\lambda }^1(x)=x,{\lambda }^n(x+y)=\sum _i{\lambda }^i(x){\lambda }^{n-i}(y)$

to those expressing ${\lambda }^n(x.y)$ and ${\lambda }^m({\lambda }^n(x))$ via specific universal polynomials. An attempt to capture the essence of $\lambda$-rings without formulas?

Lenstra’s elegant construction of the 1-power series rings $(\mathrm{\Lambda }(A),\oplus ,\otimes )$ requires only one identity to remember

$(1-at{)}^{-1}\otimes (1-bt{)}^{-1}=(1-abt{)}^{-1}$.

Still, one can use it to show the existence of ringmorphisms ${\gamma }_n:\mathrm{\Lambda }(A)\to A$, for all numbers $n\in {\mathbb{N}}_{+}$. Consider the formal ‘logarithmic derivative’

$\gamma =\frac{tu(t{)}^{\prime }}{u(t)}=\sum _{i=1}^{\mathrm{\infty }}{\gamma }_i(u(t))t^i:\mathrm{\Lambda }(A)\to A[[t]]$

where $u(t{)}^{\prime }$ is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have

$\gamma (u(t)\oplus v(t))=\frac{t(u(t)v(t){)}^{\prime }}{u(t)v(t)}=\frac{t(u(t{)}^{\prime }v(t)+u(t)v(t{)}^{\prime }}{u(t)v(t))}=\frac{tu(t{)}^{\prime }}{u(t)}+\frac{tv(t{)}^{\prime }}{v(t)}=\gamma (u(t))+\gamma (v(t))$

and so all the maps ${\gamma }_n:\mathrm{\Lambda }(A)\to A$ are additive. To show that they are also multiplicative, it suffices by functoriality to verify this on the special 1-series $(1-at{)}^{-1}$ for all $a\in A$. But,

$\gamma ((1-at{)}^{-1})=\frac{t\frac{a}{(1-at{)}^2}}{(1-at)}=\frac{at}{(1-at)}=at+a^2{t}^2+a^3{t}^3+\dots$

That is, ${\gamma }_n((1-at{)}^{-1})=a^n$ and Lenstra’s identity implies that ${\gamma }_n$ is indeed multiplicative! A first attempt :

hassle-free definition 1 : a commutative ring $A$ is a $\lambda$-ring if and only if there is a ringmorphism $s_A:A\to \mathrm{\Lambda }(A)$ splitting ${\gamma }_1$, that is, such that ${\gamma }_1\circ s_A=i{d}_A$.

In particular, a $\lambda$-ring comes equipped with a multiplicative set of ring-endomorphisms $s_n={\gamma }_n\circ s_A:A\to A$ satisfying $s_m\circ s_m=s_{mn}$. One can then define a $\lambda$-ringmorphism to be a ringmorphism commuting with these endo-morphisms.

The motivation being that $\lambda$-rings are known to form a subcategory of commutative rings for which the 1-power series functor is the right adjoint to the functor forgetting the $\lambda$-structure. In particular, if $A$ is a $\lambda$-ring, we have a ringmorphism $A\to \mathrm{\Lambda }(A)$ corresponding to the identity morphism.

But then, what is the connection to the usual one involving all the operations ${\lambda }^i$? Well, one ought to recover those from $s_A(a)=(1-{\lambda }^1(a)t+{\lambda }^2(a)t^2-{\lambda }^3(a)t^3+\dots {)}^{-1}$.

For $s_A$ to be a ringmorphism will require identities among the ${\lambda }^i$. I hope an expert will correct me on this one, but I’d guess we won’t yet obtain all identities required. By the very definition of an adjoint we must have that $s_A$ is a morphism of $\lambda$-rings, and, this would require defining a $\lambda$-ring structure on $\mathrm{\Lambda }(A)$, that is a ringmorphism $s_{AH}:\mathrm{\Lambda }(A)\to \mathrm{\Lambda }(\mathrm{\Lambda }(A))$, the so called Artin-Hasse exponential, to which I’d like to return later.

For now, we can define a multiplicative set of ring-endomorphisms $f_n:\mathrm{\Lambda }(A)\to \mathrm{\Lambda }(A)$ from requiring that $f_n((1-at{)}^{-1})=(1-a^{n}t{)}^{-1}$ for all $a\in A$. Another try?

hassle-free definition 2 : $A$ is a $\lambda$-ring if and only if there is splitting $s_A$ to ${\gamma }_1$ satisfying the compatibility relations $f_n\circ s_A=s_A\circ s_n$.

But even then, checking that a map $s_A:A\to \mathrm{\Lambda }(A)$ is a ringmorphism is as hard as verifying the lists of identities among the ${\lambda }^i$. Fortunately, we get such a ringmorphism for free in the important case when A is of ‘characteristic zero’, that is, has no additive torsion. Then, a ringmorphism $A\to \mathrm{\Lambda }(A)$ exists whenever we have a multiplicative set of ring endomorphisms $F_n:A\to A$ for all $n\in {\mathbb{N}}_{+}$ such that for every prime number $p$ the morphism $F_p$ is a lift of the Frobenius, that is, $F_p(a)\in a^p+pA$.

Perhaps this captures the essence of $\lambda$-rings best (without the risk of getting an headache) : in characteristic zero, they are the (commutative) rings having a multiplicative set of endomorphisms, generated by lifts of the Frobenius maps.