Posts tagged Lenstra

Lambda-rings for formula-phobics

Feb 5th

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 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 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 : 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 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.

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fun, Grothendieck, lambda rings, Lenstra, numbers

big Witt vectors for everyone (1/2)

Feb 2nd

Posted by lievenlb in geometry

2 comments

Next time you visit your math-library, please have a look whether these books are still on the shelves : Michiel Hazewinkel‘s Formal groups and applications, William Fulton’s and Serge Lange’s Riemann-Roch algebra and Donald Knutson’s lambda-rings and the representation theory of the symmetric group.

I wouldn’t be surprised if one or more of these books are borrowed out, probably all of them to the same person. I’m afraid I’m that person in Antwerp…

Lately, there’s been a renewed interest in $\lambda$ -rings and the endo-functor W assigning to a commutative algebra its ring of big Witt vectors, following Borger’s new proposal for a geometry over the absolute point.

However, as Hendrik Lenstra writes in his 2002 course-notes on the subject Construction of the ring of Witt vectors : “The literature on the functor W is in a somewhat unsatisfactory state: nobody seems to have any interest in Witt vectors beyond applying them for a purpose, and they are often treated in appendices to papers devoting to something else; also, the construction usually depends on a set of implicit or unintelligible formulae. Apparently, anybody who wishes to understand Witt vectors needs to construct them personally. That is what is now happening to myself.”

Before doing a series on Borger’s paper, we’d better run through Lenstra’s elegant construction in a couple of posts. Let A be a commutative ring and consider the multiplicative group of all ‘one-power series’ over it $\Lambda(A)=1+t A[[t]]$ . Our aim is to define a commutative ring structure on $\Lambda(A)$ taking as its ADDITION the MULTIPLICATION of power series.

That is, if $u(t),v(t) \in \Lambda(A)$ , then we define our addition $u(t) \boxplus v(t) = u(t) \times v(t)$ . This may be slightly confusing as the ZERO-element in $\Lambda(A),\boxplus$ will then turn be the constant power series 1…

We are now going to define a multiplication $\boxtimes$ on $\Lambda(A)$ which is distributively with respect to $\boxplus$ and turns $\Lambda(A)$ into a commutative ring with ONE-element the series $~(1-t)^{-1}=1+t+t^2+t^3+\hdots$ .

We will do this inductively, so consider $\Lambda_n(A)$ the (classes of) one-power series truncated at term n, that is, the kernel of the natural augmentation map between the multiplicative group-units $~A[t]/(t^{n+1})^* \rightarrow A^*$ . Again, taking multiplication in $A[t]/(t^{n+1})$ as a new addition rule $\boxplus$ , we see that $~(\Lambda_n(A),\boxplus)$ is an Abelian group, whence a $\Z$ -module.

For all elements $a \in A$ we have a scaling operator $\phi_a$ (sending $t \rightarrow at$ ) which is an A-ring endomorphism of $A[t]/(t^{n+1})$ , in particular multiplicative wrt. $\times$ . But then, $\phi_a$ is an additive endomorphism of $~(\Lambda_n(A),\boxplus)$ , so is an element of the endomorphism-RING $End_{\Z}(\Lambda_n(A))$ . Because composition (being the multiplication in this endomorphism ring) of scaling operators is clearly commutative ( $\phi_a \circ \phi_b = \phi_{ab}$ ) we can define a commutative RING being the subring of $End_{\Z}(\Lambda_n(A))$ generated by the operators $\phi_a$ .

The action turns $~(\Lambda_n(A),\boxplus)$ into an E-module and we define an E-module morphism $E \rightarrow \Lambda_n(A)$ by $\phi_a \mapsto \phi_a((1-t)^{-1}) = (1-at)^{-a}$ .

All of this looks pretty harmless, but the upshot is that we have now equipped the image of this E-module morphism, say L_n(A) (which is the additive subgroup of $~(\Lambda_n(A),\boxplus)$ generated by the elements $~(1-at)^{-1}$ ) with a commutative multiplication $\boxtimes$ induced by the rule $~(1-at)^{-1} \boxtimes (1-bt)^{-1} = (1-abt)^{-1}$ .

Explicitly, L_n(A) is the set of one-truncated polynomials u(t) with coefficients in such that one can find elements $a_1,\hdots,a_k \in A$ such that $u(t) \equiv (1-a_1t)^{-1} \times \hdots \times (1-a_k)^{-1}~mod~t^{n+1}$ . We multiply u(t) with another such truncated one-polynomial v(t) (taking elements $b_1,b_2,\hdots,b_l \in A$ ) via

$u(t) \boxtimes v(t) = ((1-a_1t)^{-1} \boxplus \hdots \boxplus (1-a_k)^{-1}) \boxtimes ((1-b_1t)^{-1} \boxplus \hdots \boxplus (1-b_l)^{-1})$

and using distributivity and the multiplication rule this gives the element $\prod_{i,j} (1-a_ib_jt)^{-1}~mod~t^{n+1} \in L_n(A)$ . Being a ring-qutient of we have that $~(L_n(A),\boxplus,\boxtimes)$ is a commutative ring, and, from the construction it is clear that L_n behaves functorially.

For rings such that $L_n(A)=\Lambda_n(A)$ we are done, but in general L_n(A) may be strictly smaller. The idea is to use functoriality and do the relevant calculations in a larger ring $A \subset B$ where we can multiply the two truncated one-polynomials and observe that the resulting truncated polynomial still has all its coefficients in .

Here’s how we would do this over $\Z$ : take two irreducible one-polynomials u(t) and v(t) of degrees r resp. s smaller or equal to n. Then over the complex numbers we have $u(t)=(1-\alpha_1t) \hdots (1-\alpha_rt)$ and $v(t)=(1-\beta_1) \hdots (1-\beta_st)$ . Then, over the field $K=\mathbb{Q}(\alpha_1,\hdots,\alpha_r,\beta_1,\hdots,\beta_s)$ we have that $u(t),v(t) \in L_n(K)$ and hence we can compute their product $u(t) \boxtimes v(t)$ as before to be $\prod_{i,j}(1-\alpha_i\beta_jt)^{-1}~mod~t^{n+1}$ . But then, all coefficients of this truncated K-polynomial are invariant under all permutations of the roots $\alpha_i$ and the roots $\beta_j$ and so is invariant under all elements of the Galois group. But then, these coefficients are algebraic numbers in $\mathbb{Q}$ whence integers. That is, $u(t) \boxtimes v(t) \in \Lambda_n(\Z)$ . It should already be clear from this that the rings $\Lambda_n(\Z)$ contain a lot of arithmetic information!

For a general commutative ring we will copy this argument by considering a free overring $A^{(\infty)}$ (with 1 as one of the base elements) by formally adjoining roots. At level 1, consider M_0 to be the set of all non-constant one-polynomials over and consider the ring

$A^{(1)} = \bigotimes_{f \in M_0} A[X]/(f) = A[X_f, f \in M_0]/(f(X_f) , f \in M_0)$

The idea being that every one-polynomial $f \in M_0$ now has one root, namely $\alpha_f = \overline{X_f}$ in $A^{(1)}$ . Further, $A^{(1)}$ is a free A-module with basis elements all $\alpha_f^i$ with $0 \leq i < deg(f)$ .

Good! We now have at least one root, but we can continue this process. At level 2, M_1 will be the set of all non-constant one-polynomials over $A^{(1)}$ and we use them to construct the free overring $A^{(2)}$ (which now has the property that every $f \in M_0$ has at least two roots in $A^{(2)}$ ). And, again, we repeat this process and obtain in succession the rings $A^{(3)},A^{(4)},\hdots$ . Finally, we define $A^{(\infty)} = \underset{\rightarrow}{lim}~A^{(i)}$ having the property that every one-polynomial over A splits entirely in linear factors over $A^{(\infty)}$ .

But then, for all $u(t),v(t) \in \Lambda_n(A)$ we can compute $u(t) \boxtimes v(t) \in \Lambda_n(A^{(\infty)})$ . Remains to show that the resulting truncated one-polynomial has all its entries in A. The ring $A^{(\infty)} \otimes_A A^{(\infty)}$ contains two copies of $A^{(\infty)}$ namely $A^{(\infty)} \otimes 1$ and $1 \otimes A^{(\infty)}$ and the intersection of these two rings in exactly (here we use the freeness property and the additional fact that 1 is one of the base elements). But then, by functoriality of L_n , the element $u(t) \boxtimes v(t) \in L_n(A^{(\infty)} \otimes_A A^{(\infty)})$ lies in the intersection $\Lambda_n(A^{(\infty)} \otimes 1) \cap \Lambda_n(1 \otimes A^{(\infty)})=\Lambda_n(A)$ . Done!

Hence, we have endo-functors $\Lambda_n$ in the category of all commutative rings, for every number n. Reviewing the construction of L_n one observes that there are natural transformations $L_{n+1} \rightarrow L_n$ and therefore also natural transformations $\Lambda_{n+1} \rightarrow \Lambda_n$ . Taking the inverse limits $\Lambda(A) = \underset{\leftarrow}{lim} \Lambda_n(A)$ we therefore have the ‘one-power series’ endo-functor $\Lambda~:~\wis{comm} \rightarrow \wis{comm}$ which is ‘almost’ the functor W of big Witt vectors. Next time we’ll take you through the identification using ‘ghost variables’ and how the functor $\Lambda$ can be used to define the category of $\lambda$ -rings.

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