on February 5, 2010 by lieven in geometry, numbers, Comments (1)

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 ~(\Lambda(A),\boxplus,\boxtimes) requires only one identity to remember

~(1-at)^{-1} \boxtimes (1-bt)^{-1} = (1-abt)^{-1}.

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

\gamma = \frac{t u(t)'}{u(t)} = \sum_{i=1}^\infty \gamma_i(u(t))t^i~:~\Lambda(A) \rightarrow \Lambda(A)

where u(t)' is the usual formal derivative of a power series. As this derivative satisfies the chain rule, we have

\gamma(u(t) \boxplus v(t)) = \frac{t (u(t)v(t))'}{u(t)v(t)} = \frac{t(u(t)'v(t)+u(t)v(t)'}{u(t)v(t))} = \frac{tu(t)'}{u(t)} + \frac{tv(t)'}{v(t)} = \gamma(u(t)) + \gamma(v(t))

and so all the maps \gamma_n~:~\Lambda(A) \rightarrow 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^2t^2 + a^3t^3+\hdots

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 \rightarrow \Lambda(A) splitting \gamma_1, that is, such that \gamma_1 \circ s_A = id_A.

In particular, a \lambda-ring comes equipped with a multiplicative set of ring-endomorphisms s_n = \gamma_n \circ s_A~:~A \rightarrow 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 \rightarrow \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+...)^{-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 \Lambda(A), that is a ringmorphism s_{AH}~:~\Lambda(A) \rightarrow \Lambda(\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~:~\Lambda(A) \rightarrow \Lambda(A) from requiring that f_n((1-at)^{-1}) = (1-a^nt)^{-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 \rightarrow \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 \rightarrow \Lambda(A) exists whenever we have a multiplicative set of ring endomorphisms F_n~:~A \rightarrow 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.

1 Comment

  1. Qiaochu Yuan

    February 5, 2010 @ 8:52 pm

    Interesting. My understanding was that lambda-rings are a decategorification of the category of representations of a nice group equipped with the exterior power functors. More generally I guess one can consider the exterior power on vector bundles over a space. What’s the relationship between this perspective and the one you give?

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