## Polynomials for addition in the Witt vectors

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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!

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You mean you're asking for $S_i$ in terms of $R_0$, $R_1$, ..., $R_i$? – darij grinberg Mar 31 2012 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 2012 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 2012 at 21:14

## 1 Answer

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

`$S_n=R_0Z_n+R_1Z_2+\dots+R_nZ_0+R_1S_{n-1}+R_2S_{n-2}+\dots+R_{n-1}S_1$`

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.

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