# Polynomials for addition in the Witt vectors

The addition of $p$-typical Witt vectors ($p$ a prime number) is given by universal polynomials $S_n=S_n(X_0,\dots,X_n;Y_0,\dots,Y_n)\in\mathbb{Z}[X_0,X_1,\dots;Y_0,Y_1,\dots]$ determined by the equalities

$\Phi_n(S_0,\dots,S_n)=\Phi_n(X_0,\dots,X_n)+\Phi_n(Y_0,\dots,Y_n)$ for all $n\ge 0$,

where

$\Phi_n(T_0,\dots,T_n)=(T_0)^{p^n}+p(T_1)^{p^{n-1}}+\dots+p^nT_n$.

I guess that whoever sees the Witt vectors for the first time writes down explicitly $S_0=X_0+Y_0$, $S_1=X_1+Y_1+\frac{1}{p}((X_0)^p+(Y_0)^p-(X_0+Y_0)^p)$, maybe $S_2$ if she/he is courageous, and then stops since the computation becomes extremely messy. I think that there is no reasonable explicit expression in general, but patterns seem to exist and my question is about making these patterns more precise. Before I ask, let me illustrate with $S_2$. It is easy to see that there exists a unique sequence of polynomials $R_n\in\mathbb{Z}[X,Y]$, $n\ge 0$, such that

$X^{p^n}+Y^{p^n}=R_0(X,Y)^{p^n}+pR_1(X,Y)^{p^{n-1}}+\dots+p^nR_n(X,Y)$.

For example $R_0=X+Y$ and $R_1=\frac{1}{p}(X^p+Y^p-(X+Y)^p)$. Then:

$S_1=R_0(X_1,Y_1)+R_1(X_0,Y_0)$

$S_2=R_0(X_2,Y_2)+R_1(X_1,Y_1)+R_1(R_0(X_1,Y_1),R_1(X_0,Y_0))+R_2(X_0,Y_0)$.

Can someone make the shape of $S_n$ more precise, e.g. in the form $S_n=P_0+\dots+P_n$ presumably with $P_0=R_0(X_n,Y_n)$, $P_n=R_n(X_0,Y_0)$? The intermediary $P_i$'s are more complicated but should be (uniquely) determined by a condition of the type
"$P_i$ is an iterated composition of $R_i$ involving only the variables $X_0,\dots,X_i$".
Maybe the polynomials $P_i$ should be homogeneous w.r.t. some graduation.

Any hint or relevant reference will be appreciated. Thanks!

-
You mean you're asking for $S_i$ in terms of $R_0$, $R_1$, ..., $R_i$? – darij grinberg Mar 31 '12 at 16:41
The point is that I don't know what "in terms of" should mean: the shape and properties you conjecture for an expression of $S_n$ alters your ability to prove that by induction on $n$. For example, it is not clear to me if just assuming that the property P($n$) : $S_n$ is a polynomial in the $R_i(X_j,Y_j)$ holds is enough to prove that P($n+1$) holds. – Matthieu Romagny Mar 31 '12 at 21:13
ps : I know the question is (unfortunately) a little vague; it is part of the question to make it less vague. The complexity of the polynomials $S_n$ is the cause of the problem, and the interest of the question. – Matthieu Romagny Mar 31 '12 at 21:14

I found a formula for $S_n$ in terms of the previous $S_i$'s and the polynomials $R_i$, which I'm quite happy about. In fact, I need the multivariate version of the $R_i$, which I will construct all at the same time. Let us consider the ring of formal power series in countably many variables $X_1,X_2,\dots$ with integer coefficients, that is $A=\mathbb{Z}[[X_1,X_2,\dots]]$. Note that there are several notions of power series in infinitely many variables; mine is that of Bourbaki, where the underlying module of $A$ is just the product of copies of $\mathbb{Z}$ indexed by (finitely supported!) multiindices. (Some people require that the homogeneous components of a power series be polynomials; this is not the case here.) Then one can see that there exists a unique sequence $(R_n)$ of elements of $A$ such that for all $n\ge 0$ we have

$X_1^{p^n}+X_2^{p^n}+\dots=R_0^{p^n}+pR_1^{p^{n-1}}+\dots+p^nR_n$

where on the left is the sum of all $p^n$-th powers of the variables. This is a straightforward application of Bourbaki, Algèbre Commutative, Chapitre IX, $\S~1$, no 2, prop. 2, c) (phew! reference is finished) with the endomorphism $\sigma:A\to A$ defined by $\sigma(X_i)=X_i^p$ for all $i$. Now if we have finitely many (say $s$) variables, then we set $R_n(X_1,\dots,X_s)=R_n(X_1,\dots,X_s,0,0,\dots)$. Examples:

$R_0(X_1,\dots,X_s)=X_1+\dots+X_s$ and $R_1(X_1,\dots,X_s)=\frac{X_1^p+\dots+X_s^p-(X_1+\dots+X_s)^p}{p}$.

Assuming that the $R_n$ are computable, I have an inductive recipe for $S_n$ which is interesting because it shows that all the $p$-adic congruences implying integrality of the $S_n$ are contained in the $R_n$. The recipe goes like this. For each $i$, the polynomial $S_i$ is a sum of $2i$ terms (it will be obvious below what these terms are) and assuming $S_1,\dots,S_{n-1}$ are known then

$S_n=R_0Z_n+R_1Z_{n-1}+\dots+R_nZ_0+R_1S_{n-1}+R_2S_{n-2}+\dots+R_{n-1}S_1$

where: $Z_j$ is short for the pair of variables $(X_j,Y_j)$, $R_iZ_j$ is short for $R_i(X_j,Y_j)$ (the bivariate $R_i$) and $R_iS_j$ is the ($2j$-variate) polynomial $R_i$ evaluated at the $2j$ terms of $S_j$. I hope the following examples make it clear what this means, and how efficient it is:

$S_1 = R_0 Z_1 + R_1 Z_0$

$S_2=R_0 Z_2 + R_1 Z_1 + R_2 Z_0 + R_1 (R_0 Z_1 , R_1 Z_0 )$

$S_3 = R_0Z_3+R_1Z_2+R_2Z_1+R_3Z_0 \\ \quad + R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0))+ R_2(R_0Z_1,R_1Z_0)$

$S_4 = R_0Z_4+R_1Z_3+R_2Z_2+R_3Z_1+R_4Z_0 \\ \quad + R_1(R_0Z_3,R_1Z_2,R_2Z_1,R_3Z_0,R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)),R_2(R_0Z_1,R_1Z_0)) \\ \quad + R_2(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)) \\ \quad + R_3(R_0Z_1,R_1Z_0)$

The proof that the recipe is correct is an exercise.

-