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 $\mathrm{\Lambda }(A)=1+tA[[t]]$. Our aim is to define a commutative ring structure on $\mathrm{\Lambda }(A)$ taking as its ADDITION the MULTIPLICATION of power series.

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

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

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

For all elements $a\in A$ we have a scaling operator ${\varphi }_a$ (sending $t\to at$) which is an A-ring endomorphism of $A[t]/(t^{n+1})$, in particular multiplicative wrt. $\times$. But then, ${\varphi }_a$ is an additive endomorphism of $({\mathrm{\Lambda }}_n(A),\oplus )$, so is an element of the endomorphism-RING $En{d}_{\mathbb{Z}}({\mathrm{\Lambda }}_n(A))$. Because composition (being the multiplication in this endomorphism ring) of scaling operators is clearly commutative (${\varphi }_a\circ {\varphi }_b={\varphi }_{ab}$) we can define a commutative RING $E$ being the subring of $En{d}_{\mathbb{Z}}({\mathrm{\Lambda }}_n(A))$ generated by the operators ${\varphi }_a$.

The action turns $({\mathrm{\Lambda }}_n(A),\oplus )$ into an E-module and we define an E-module morphism $E\to {\mathrm{\Lambda }}_n(A)$ by ${\varphi }_a\mapsto {\varphi }_a((1-t{)}^{-1})=(1-at{)}^{-a}$.

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 $({\mathrm{\Lambda }}_n(A),\oplus )$ generated by the elements $(1-at{)}^{-1}$) with a commutative multiplication $\otimes$ induced by the rule $(1-at{)}^{-1}\otimes (1-bt{)}^{-1}=(1-abt{)}^{-1}$.

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

$u(t)\otimes v(t)=((1-a_{1}t{)}^{-1}\oplus \dots \oplus (1-a_k{)}^{-1})\otimes ((1-b_{1}t{)}^{-1}\oplus \dots \oplus (1-b_l{)}^{-1})$

and using distributivity and the multiplication rule this gives the element $\prod _{i,j}(1-a_i{b}_{j}t{)}^{-1}mod{t}^{n+1}\in L_n(A)$.
Being a ring-qutient of $E$ we have that $(L_n(A),\oplus ,\otimes )$ is a commutative ring, and, from the construction it is clear that $L_n$ behaves functorially.

For rings $A$ such that $L_n(A)={\mathrm{\Lambda }}_n(A)$ 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 $\mathbb{Z}$ : 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
$u(t)=(1-{\alpha }_{1}t)\dots (1-{\alpha }_{r}t)$ and $v(t)=(1-{\beta }_1)\dots (1-{\beta }_{s}t)$. Then, over the field $K=\mathbb{Q}({\alpha }_1,\dots ,{\alpha }_r,{\beta }_1,\dots ,{\beta }_s)$ we have that $u(t),v(t)\in L_n(K)$ and hence we can compute their product $u(t)\otimes v(t)$ as before to be $\prod _{i,j}(1-{\alpha }_i{\beta }_{j}t{)}^{-1}mod{t}^{n+1}$. 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, $u(t)\otimes v(t)\in {\mathrm{\Lambda }}_n(\mathbb{Z})$. It should already be clear from this that the rings ${\mathrm{\Lambda }}_n(\mathbb{Z})$ contain a lot of arithmetic information!

For a general commutative ring $A$ we will copy this argument by considering a free overring $A^{(\mathrm{\infty })}$ (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

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

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 $0\le i<deg(f)$.

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 $A^{(3)},A^{(4)},\dots$. Finally, we define $A^{(\mathrm{\infty })}=\underrightarrow{lim}{A}^{(i)}$ having the property that every one-polynomial over A splits entirely in linear factors over $A^{(\mathrm{\infty })}$.

But then, for all $u(t),v(t)\in {\mathrm{\Lambda }}_n(A)$ we can compute $u(t)\otimes v(t)\in {\mathrm{\Lambda }}_n(A^{(\mathrm{\infty })})$. Remains to show that the resulting truncated one-polynomial has all its entries in A. The ring $A^{(\mathrm{\infty })}{\otimes }_A{A}^{(\mathrm{\infty })}$ contains two copies of $A^{(\mathrm{\infty })}$ namely $A^{(\mathrm{\infty })}\otimes 1$ and $1\otimes A^{(\mathrm{\infty })}$ 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
$u(t)\otimes v(t)\in L_n(A^{(\mathrm{\infty })}{\otimes }_A{A}^{(\mathrm{\infty })})$ lies in the intersection ${\mathrm{\Lambda }}_n(A^{(\mathrm{\infty })}\otimes 1)\cap {\mathrm{\Lambda }}_n(1\otimes A^{(\mathrm{\infty })})={\mathrm{\Lambda }}_n(A)$. Done!

Hence, we have endo-functors ${\mathrm{\Lambda }}_n$ in the category of all commutative rings, for every number n. Reviewing the construction of $L_n$ one observes that there are natural transformations $L_{n+1}\to L_n$ and therefore also natural transformations ${\mathrm{\Lambda }}_{n+1}\to {\mathrm{\Lambda }}_n$. Taking the inverse limits $\mathrm{\Lambda }(A)=\underleftarrow{lim}{\mathrm{\Lambda }}_n(A)$ we therefore have the ‘one-power series’ endo-functor
$\mathrm{\Lambda }:\mathbf{comm}\to \mathbf{comm}$
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 $\mathrm{\Lambda }$ can be used to define the category of $\lambda$-rings.