on February 2, 2010 by lieven in geometry, numbers, Comments (1)

big Witt vectors for everyone (1/2)

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 E 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 A 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 E 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 A 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 A.

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 A 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 A 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 A (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.

1 Comment

  1. James Borger

    February 4, 2010 @ 11:54 am

    Thanks for the mention! That note of Lenstra’s was actually one of the two key things that led to my serious interest in Witt/lambda things. The other was a conversation with Spencer Bloch about his paper “Crystals and de Rham-Witt connections”.

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