Polynomials for addition in the Witt vectors - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:48:41Z http://mathoverflow.net/feeds/question/92750 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92750/polynomials-for-addition-in-the-witt-vectors Polynomials for addition in the Witt vectors Matthieu Romagny 2012-03-31T13:28:15Z 2012-04-04T16:35:20Z <p>The addition of <code>$p$</code>-typical Witt vectors (<code>$p$</code> a prime number) is given by universal polynomials <code>$S_n=S_n(X_0,\dots,X_n;Y_0,\dots,Y_n)\in\mathbb{Z}[X_0,X_1,\dots;Y_0,Y_1,\dots]$</code> determined by the equalities </p> <p><code>$\Phi_n(S_0,\dots,S_n)=\Phi_n(X_0,\dots,X_n)+\Phi_n(Y_0,\dots,Y_n)$</code> for all <code>$n\ge 0$</code>, </p> <p>where </p> <p><code>$\Phi_n(T_0,\dots,T_n)=(T_0)^{p^n}+p(T_1)^{p^{n-1}}+\dots+p^nT_n$</code>. </p> <p>I guess that whoever sees the Witt vectors for the first time writes down explicitly <code>$S_0=X_0+Y_0$</code>, <code>$S_1=X_1+Y_1+\frac{1}{p}((X_0)^p+(Y_0)^p-(X_0+Y_0)^p)$</code>, maybe <code>$S_2$</code> 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 <code>$S_2$</code>. It is easy to see that there exists a unique sequence of polynomials <code>$R_n\in\mathbb{Z}[X,Y]$</code>, <code>$n\ge 0$</code>, such that </p> <p><code>$X^{p^n}+Y^{p^n}=R_0(X,Y)^{p^n}+pR_1(X,Y)^{p^{n-1}}+\dots+p^nR_n(X,Y)$</code>. </p> <p>For example <code>$R_0=X+Y$</code> and <code>$R_1=\frac{1}{p}(X^p+Y^p-(X+Y)^p)$</code>. Then: </p> <p><code>$S_1=R_0(X_1,Y_1)+R_1(X_0,Y_0)$</code> </p> <p><code>$S_2=R_0(X_2,Y_2)+R_1(X_1,Y_1)+R_1(R_0(X_1,Y_1),R_1(X_0,Y_0))+R_2(X_0,Y_0)$</code>. </p> <blockquote> <p>Can someone make the shape of <code>$S_n$</code> more precise, e.g. in the form <code>$S_n=P_0+\dots+P_n$</code> presumably with <code>$P_0=R_0(X_n,Y_n)$</code>, <code>$P_n=R_n(X_0,Y_0)$</code>? The intermediary <code>$P_i$</code>'s are more complicated but should be (uniquely) determined by a condition of the type<br> "<code>$P_i$</code> is an iterated composition of <code>$R_i$</code> involving only the variables <code>$X_0,\dots,X_i$</code>".<br> Maybe the polynomials <code>$P_i$</code> should be homogeneous w.r.t. some graduation. </p> </blockquote> <p>Any hint or relevant reference will be appreciated. Thanks!</p> http://mathoverflow.net/questions/92750/polynomials-for-addition-in-the-witt-vectors/93047#93047 Answer by Matthieu Romagny for Polynomials for addition in the Witt vectors Matthieu Romagny 2012-04-03T21:46:26Z 2012-04-03T21:46:26Z <p>I found a formula for <code>$S_n$</code> in terms of the previous <code>$S_i$</code>'s and the polynomials <code>$R_i$</code>, which I'm quite happy about. In fact, I need the multivariate version of the <code>$R_i$</code>, which I will construct all at the same time. Let us consider the ring of formal power series in countably many variables <code>$X_1,X_2,\dots$</code> with integer coefficients, that is <code>$A=\mathbb{Z}[[X_1,X_2,\dots]]$</code>. Note that there are several notions of power series in infinitely many variables; mine is that of Bourbaki, where the underlying module of <code>$A$</code> is just the product of copies of <code>$\mathbb{Z}$</code> 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 <code>$(R_n)$</code> of elements of <code>$A$</code> such that for all $n\ge 0$ we have </p> <p><code>$X_1^{p^n}+X_2^{p^n}+\dots=R_0^{p^n}+pR_1^{p^{n-1}}+\dots+p^nR_n$</code> </p> <p>where on the left is the sum of all <code>$p^n$</code>-th powers of the variables. This is a straightforward application of Bourbaki, Algèbre Commutative, Chapitre IX, <code>$\S~1$</code>, no 2, prop. 2, c) (phew! reference is finished) with the endomorphism <code>$\sigma:A\to A$</code> defined by <code>$\sigma(X_i)=X_i^p$</code> for all <code>$i$</code>. Now if we have finitely many (say <code>$s$</code>) variables, then we set <code>$R_n(X_1,\dots,X_s)=R_n(X_1,\dots,X_s,0,0,\dots)$</code>. Examples: </p> <p><code>$R_0(X_1,\dots,X_s)=X_1+\dots+X_s$</code> and <code>$R_1(X_1,\dots,X_s)=\frac{X_1^p+\dots+X_s^p-(X_1+\dots+X_s)^p}{p}$</code>. </p> <p>Assuming that the <code>$R_n$</code> are computable, I have an inductive recipe for <code>$S_n$</code> which is interesting because it shows that all the <code>$p$</code>-adic congruences implying integrality of the <code>$S_n$</code> are contained in the <code>$R_n$</code>. The recipe goes like this. For each <code>$i$</code>, the polynomial <code>$S_i$</code> is a sum of <code>$2i$</code> terms (it will be obvious below what these terms are) and assuming <code>$S_1,\dots,S_{n-1}$</code> are known then </p> <p><code>$S_n=R_0Z_n+R_1Z_2+\dots+R_nZ_0+R_1S_{n-1}+R_2S_{n-2}+\dots+R_{n-1}S_1$</code> </p> <p>where: <code>$Z_j$</code> is short for the pair of variables <code>$(X_j,Y_j)$</code>, <code>$R_iZ_j$</code> is short for <code>$R_i(X_j,Y_j)$</code> (the bivariate <code>$R_i$</code>) and <code>$R_iS_j$</code> is the (<code>$2j$</code>-variate) polynomial <code>$R_i$</code> evaluated at the <code>$2j$</code> terms of <code>$S_j$</code>. I hope the following examples make it clear what this means, and how efficient it is: </p> <p><code>$S_1 = R_0 Z_1 + R_1 Z_0$</code> </p> <p><code>$S_2=R_0 Z_2 + R_1 Z_1 + R_2 Z_0 + R_1 (R_0 Z_1 , R_1 Z_0 )$</code> </p> <p><code>$S_3 = R_0Z_3+R_1Z_2+R_2Z_1+R_3Z_0 \\ \quad + R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0))+ R_2(R_0Z_1,R_1Z_0)$</code> </p> <p><code>$S_4 = R_0Z_4+R_1Z_3+R_2Z_2+R_3Z_1+R_4Z_0 \\ \quad + R_1(R_0Z_3,R_1Z_2,R_2Z_1,R_3Z_0,R_1(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)),R_2(R_0Z_1,R_1Z_0)) \\ \quad + R_2(R_0Z_2,R_1Z_1,R_2Z_0,R_1(R_0Z_1,R_1Z_0)) \\ \quad + R_3(R_0Z_1,R_1Z_0)$</code> </p> <p>The proof that the recipe is correct is an exercise.</p>