MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n\ge 0$,



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


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



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!

share|cite|improve this question
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
up vote 6 down vote accepted

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


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


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.