Monday, January 19, 2009

A very simple example

In my previous contributions to this blog , I have mentioned how the calculations of Dinesh Thakur and Javier Diaz-Vargas suggested that the nonclassical trivial zeroes of characteristic $p$ zeta functions associated to ${\bf F}_q[t]$ should have the following two properties (where nonclassical means that the actual order is higher than what one would expect from classical theory):

1. If a nonclassical trivial zero occurs at $-i$ then the sum of the $p$-adic digits of $i$ must be bounded.

2. The orders of the trivial zeroes should be an invariant of the action of the group $S_{(q)}$ of homeomorphisms of $Z_p$ which permute the $q$-adic digits of a $p$-adic integer.

In my last entry, I discussed Dinesh's remarkable result on valuations of certain basic sums in this game; one key point is that the valuations for arbitrary $d$ iteratively reduced to valuations just involving sums of monics of degree one. Here I want to again use monics of degree one to give a very simple example with properties very similar to 1 and 2 above. We will then draw some conclusions for the relevant theory of nonArchimedean measures.

The example presented here was first mentioned by Warren Sinnott, in the $q=p$ case in Warren's paper "Dirichlet Series in function fields" (J. Number Th. 128 (2008) 1893-1899). The $L$-functions that occur in the theory of Drinfeld modules and the like are functions of two
variables $(x,y)$. If one fixes $x$, the functions in $y\in Z_p$ that one obtains are uniform limits of finite sums of exponentials $u^y$ where $u$ is a $1$-unit. In his paper Warren studies such functions and shows that if $f(y)$ is a nonzero such function, its zero set *cannot* contain an open set (unlike arbitrary continuous functions such as step-functions).

In what follows ALL binomial coefficients are considered modulo $p$ so that the basic lemma of Lucas holds for them.

Lemma: 1. Let $\sigma\in S_{(q)}$. Let $y\in Z_p$ and $k$ a nonnegative integer. Then

$${y \choose k}= {\sigma(y) \choose \sigma (k)} \,.$$

2. Let $i,j$ be two nonnegative integers. Then

$${i +j \choose j}= {\sigma (i) +\sigma (j) \choose \sigma (j)}\,.$$

Proof: 1 is simply $q$-Lucas. For 2 note that if there is carry over of digits in the addition for $i+j$ then there is also in the sum for $\sigma (i)+\sigma (j)$, and vice versa; in this case, both sides are $0$. If there is no carry over the result follows from $q$-Lucas again. QED

As before, let $q=p^m$ and let $y\in Z_p$. Let $A=Fq[t]$ and let $\pi=1/t$; so $\pi$ is a positive uniformizer at the place $\infty$ of ${\bf F}_q(t)$. Define

$$ f(y):= \sum_{g\in A^+(1)} (\pi g)^y \,;$$

where $A^+(1)$ is just the set of monic polynomials of degree $1$. The sum can clearly be rewritten as

$$ f(y)=\sum_{\alpha \in \Fq}(1+\alpha \pi)^y .

Upon expanding out via the binomial theorem, and summing over $\alpha$, we find

$$ f(y)= -\sum_{k \in I} {y \choose k} \pi^k$$

where $I$ is the set of positive integers divisible by $q-1$.

Let $X\subset Z_p$ be the zeroes of $f(y)$; it is obviously closed. When $q=p$, Warren (in his paper and in personal communication) showed that $X$ consists pricisely of those non-negative integers $i$ such that the sum of the $p$-adic digits of $i$ is less than $p$.

Now, in order to show that $f(y) \neq 0$, for a given $y$ in $Z_p$, it is necessary and sufficient to simply show that there is ONE $k \in I$ such that ${y \choose k}$ is nonzero in ${\bf F}_p$. When $q=p$, this is readily accomplished.

However, when $q$ is general it gets much more subtle to make sure that the reduced binomial coefficient is non-zero.

Proposition: The set $X$ is stable under $S_{(q)}$. Moreover, there is an explicit constant $C$ (which depends on $q$) such that the elements of $X$ have their sum of $q$-adic coefficients less than $C$.

(As Warren has remarked, the Proposition then reduces the problem of finding the zero set to checking *finitely many* orbits!)


Let $\sigma \in S_{(q)}$. The first part follows immediately from the first part of the Lemma and the fact that $I$ is stable under $S_{(q)}$.

To see the second part, let $C: = (q-2)(1+2+\cdots+ q-1)=(q-2)(q-1)q/2$. Let $y$ be any $p$-adic integer with the property that its sum of $q$-adic digits is greater than $C$. Then there must be at least one $e$ with $e$ between $0$ and $q-1$ such that $e$ occurs at least $q-1$ times in the expansion of $y$. It is then easy to find $k$ such that the reduction of ${y \choose k}$ is nonzero. QED

There are other important results that arise from the first part of the Lemma. Indeed, upon replacing $k$ with $\sigma^{-1}(t)$, we obtain

$${y \choose \sigma^{-1}(t)}= {\sigma(y) \choose t\,.$$ (*)

This immediately gives the action of $S_{(q)}$ on the Mahler expansion of a continuous function from $Z_p$ to characteristic $p$. One also obviously has

$$\sum_k {\sigma y \choose k} x^k=
\sum_k {\sigma(y) \choose \sigma (k)}x^{\sigma(k)\,.$$

But, by the first part of the Lemma, this then equals

$$\sum {y \choose k}x^{\sigma k}\,,$$

which is a sort of change of variable formula.

As the action of $S_{(q)}$ is continuous on $Z_p$ there is a dual action on measures; if the measures are characteristic $p$ valued, then this action is easy to compute from (*) above.

However, there is ALSO a highly mysterious action of $S_{(q)}$ on the *convolution algebra* of characteristic $p$ valued measures on the maximal compact subrings in the completions of $F_q(T)$ at its places of degree $1$ (e.g, the place at $\infty$ or associated to $(t)$, if the place has higher degree one replaces $S_{(q)}$ with the appropriate subgroup). Indeed, given a Banach basis for the space of $Fq$-linear continuous functions from that local ring to itself, the "digit expansion principle"gives a basis for ALL continuous functions of the ring to itself (see, e.g., Keith Conrad, "The Digit Principle", J. Number Theory 84(2000) 230-257). In the 1980's Greg Anderson and I realized that this gives an isomorphism of the associated convolution algebra of measures with the ring of formal *divided power series* over the local ring.

But let $\sigma \in S_{(q)}$ and define

$$\sigma (z^i/i!):= z^{\sigma (i)}/\sigma(i)! \.$$

The content of the second part of the Lemma is precisely that this definition gives rise to an algebra automorphism of the ring of formal divided power series.


S H Banihashemi said...

One bad thing about being an amateur student of math as myself is that you have no access to all the resources. It is still very curious to ask whether there are any algebraic zeros of the Riemann zeta function as I believe their analog for the finite fields are all algebraic. My guess is that these are all transcendental although I do not have the slightest reason why.

Best Wishes

pages of my diary said...

i have a question , hope that you can help me ...
i'm a M2 stdent of pure math in differential geometry id like to continue my studis in order to get a phd ...anyway there is is prof from clermond ferront who gave us a cours of symbol theories (théorie de symbol) ... it was nice but yet very adanced and unseen ... she gave me the opportunity to continue my PHD with her, and gave me a paper in non commutative geom subject to do about dixmier trace it's connexion with the notion of the residue of pseudo-differentiell operators "... she told me that i would like this math division since i'm a fan of non commutative algebra , and fonctionnel analysis and diferentiel geometry ,workiong with manifolds and stuff like that ...
truth be told it's a great opportunity , but yet as much as i'm exited i'm scared since i'm not so familiar with it and since the whol subject is new for me ...the zeta function , indix theory and sumbols theorie and i'm hesitating about starting this paper, should i stuck to something more familiar!! what's this non commutative geom is about anyway ? what do you advice me ?

Anonymous said...


Just two links concerning french research :