calculus and exponentiation in finite characteristic

In this post, I would like to discuss a really beautiful solution of an open issue in the theory of characteristic p L-series. This solution is in the short paper arXiv:0808.4069 by Sangtae Jeong and is a great illustration of the use of calculus in finite characteristic. I have written this post so that, if you choose, you can just copy it and paste it into a latex file and it will compile.

In my previous post of August 4, 2008, I mentioned the domain ${\mathbb S}_\infty={\mathbb C}_\infty^\ast\times {\mathbb Z}_p$ of characteristic $p$ $L$-series. We write $s\in {\mathbb S}_\infty$ as $(x,y)$. Here ${\mathbb Z}_p$ is used in the following fashion: Let $y\in {\mathbb Z}_p$ and let $u$ be a 1-unit in ${\mathbb F}_q((1/T))$; so $u=1+v$ where $v$ has absolute value strictly less than $1$. Then one simply defines $u^y=(1+v)^y$ by using the binomial expansion (which converges since $v$ is small).

The binomial expansion of $u^y$ obviously shows that the function $u \mapsto u^y$ is analytic on the $1$-units. This analyticity is itself crucial for the analytic continuation of the $L$-series of general Drinfeld modules and the like. Indeed, one writes down an integral for these $L$-series of the form
$$\int u^y\, d\mu_x(u)$$
where we integrate over the $1$-units and where $x$ is our parameter. This integral converges absolutely when $x$ is large. In fact, when $x$ is large the integral will converge if $u^y$ is replaced by {\it any} continuous function in $y$. However, the analyticity of $u^y$ gives very powerful a-priori information about the growth of the expansion coefficients of this function (in a suitable polynomial basis for all continuous functions); indeed, the coefficients go to $0$ quite rapidly. On the other hand, for arbitrary $x$ the measures $\mu_x$ blow up rather slowly (in fact, logarithmically). Putting the two facts together allows for the analytic continuation.

The same argument would work for {\it any} locally-analytic endomorphism of the $1$-units and therefore it is quite reasonable to expect that there are no others. This is what Jeong proves in his note.

One can obtain a proof using formal groups. Jeong's proof, however, seems to work only in the case at hand. Its advantage lies in the fact that you can actually watch the $p$-adic integer $y$ arising out of a series of first order, initial value differential equations that naturally arise.

In fact, Jeong's proof is a finite characteristic reflection of that most famous differential equation $y^\prime =y$ that one learns in first year calculus. Please forgive me for recalling how the differential equation is solved: one first learns that $y=e^x$ is a solution of the equation. One then divides any other solution $h$ by $e^x$ to obtain a function whose derivative (by the quotient rule) is identically $0$, and therefore constant, and so all solutions are multiples of $e^x$. In particular, of course, a solution $y$ is nonzero precisely when $y(0)=y^\prime(0)$ is.

Characteristic $p$ calculus presents many challenges; primarily the fact that having an identically vanishing derivative does NOT guarantee that a function is constant. So, first of all, one differentiates a power series in exactly the same fashion as in first year calculus with the same derivation laws. In particular, then, the derivative of $x^n$ with respect to $x$ is $nx^{n-1}$; this vanishes identically if and only if $p$ divides $n$. So a power series $f(x)$ will have identically vanishing derivative if and only if it can be written as $h(x^p)$, where $h(x)$ is another power series, and therefore we are very far from the classical situation. Similarly, the $p$-th derivative of a power series will vanish identically.

The ancients (such as Hasse, Schmidt and Teichmuller) partially compensated for this as follows: Again from first year calculus, one knows that if $f(x)=\sum a_n x^n$ is a convergent power series over the real numbers then $a_n=\frac{D^n}{n!}f(0)$ where $D=\frac{d~}{dx}$; thus Hasse et al shifted the focus from $D^n$ to the operators $f\mapsto a_n$ which are indeed nonzero in finite characteristic. One writes this ({\it formally now}!) as $a_n=\frac{D^n}{n!}f(0)$ and calls these operators "Hasse derivatives", "hyperderivatives," "divided derivatives" etc. They satisfy many formal properties that may be guessed at from classical theory as well as other special properties arising in finite characteristic.

Let me now briefly sketch Jeong's proof and refer you to his paper for the details. You will see echoes in it of the classical theory of $y=y^\prime$ sketched above. So let $f(1+x)$ be our endomorphism of the $1$-units where $f(1+x)=\sum_i a_ix^i$ for $x$ small.

Step 1. The coefficients $a_i$ are in ${\mathbf F}_p$ for all $i$. Indeed, this follows from the fact $f((1+x)^p)=f(1+x^p)=f(1+x)^p$ as $f$ is an endomorphism.

Step 2. $a_i=0$ if and only if $\frac{D^i}{i!}f(1+x)$ vanishes identically. Indeed, Jeong uses the fact that $f(1+x)$ represents an endomorphism, and some algebra, to show that for all $i$
$$a_if(1+x)(1+x)^{-i}=\frac{D^i}{i!}f(1+x)\,.$$
In particular, if the derivative of $f(1+x)$ vanishes at the origin, then $f(1+x)=g(1+x)^p$ for some endomorphism $g(1+x)$ (recall that the coefficients of $f$ are in ${\mathbb F}_p$).

Step 3: Let $j$ be the largest integer such that $f(1+x)=g(1+x)^{p^j}$ for some endomorphism $g(1+x)$; thus by Step 2, $g^\prime (1)\neq 0$. Let $a\in \{0,1,\ldots,p-1\}$ be in the class mod $p$ given by $g^\prime(1)$. Let $h(1+x)=g(1+x)/(1+x)^a$; clearly $h(1+x)$ is again an endomorphism and the quotient rule shows that $h^\prime(1)=0$. In particular, $h(1+x)$ is a $p$-th power and we may repeat the process.

Step 4: Step 3 allows us to inductively create $y\in {\mathbb Z}_p$ so that the power series for $f(1+x)$ IS $(1+x)^y$. As the $1$-units have no torsion (which is easily seen), one concludes that $f(u)=u^y$ forall $1$-units $u$.