Holomorphic Functions and Residue Theorem on P^1

Author: Eiko

Tags: affinoid, p-adic geometry, affinoid algebra, projective line, residue theorem, holomorphic functions

Time: 2024-09-12 15:06:50 - 2024-09-12 15:10:22 (UTC)

Holomorphic functions on affinoid subsets

Definition 1.

Let $F$ be an affinoid subset of $P$ , we define the ring of Holomorphic functions on $F$ , $O (F)$ be the completion of the $K$ -algebra of rational functions regular on $F$ , taken with respect to the supremum norm $∥ f ∥ = sup_{a \in F} | f (a) | .$ Thus $f$ is holomorphic on $F$ iff it is a uniform limit of rational functions regular on $F$ .
One also defines $O (F)^{\circ} = {f \in O (F) : ∥ f ∥ \leq 1},$ (similar to ring of integers) $O (F)^{\circ \circ} = {f \in O (F) : ∥ f ∥ < 1} .$ (similar to the maximal ideal) $O (F)^{\circ}$ is a $K^{\circ}$ -algebra and $O (F)^{\circ \circ}$ is an ideal. The quotient ring $\overset{―}{O (F)} = O (F)^{\circ} / O (F)^{\circ \circ}$ gives us a $\overset{―}{K}$ -algebra.

Example 1. For $F = {a \in P^{1} : | a | \leq 1}$ , we have the norm defined above happens to be the Gauss norm $‖ \sum a_{n} z^{n} ‖ = sup | a_{n} | .$ The ring of holomorphic functions on $F$ is the completion of the rational functions regular on $F$ with respect to the Gauss norm, which is the Tate algebra $T_{1} = K ⟨ z ⟩$ . $O (F) = K ⟨ z ⟩, O (F)^{\circ} = K^{\circ} ⟨ z ⟩, O (F)^{\circ \circ} = K^{\circ \circ} ⟨ z ⟩ .$ And the quotient $\overset{―}{O (F)} = K^{\circ} ⟨ z ⟩ / K^{\circ \circ} ⟨ z ⟩ = \overset{―}{K} ⟨ z ⟩ .$

Proposition 1. Let $F$ be the closed disk ${a \in P^{1} : | a | \leq 1}$ and $g \in O (F)$ be any function. Define $d$ to be the degree of the normalized polynomial $\overset{―}{c g} \in \overset{―}{K} [z]$ where $c \in K^{*}$ and $∥ c g ∥ = 1$ . Then we can write for any $f \in O (F)$ $f = g q + r$ where $q \in O (F)$ and $r \in O (F)$ with $\deg (r) < d$ , and $∥ f ∥ = max (∥ q ∥ \cdot ∥ g ∥, ∥ r ∥)$ .

Lemma 1. When $g$ is a monic polynomial with $∥ g ∥ = 1$ , and $f \in O (F)$ , there exists $f = g q + r, \deg (r) < \deg (g), ∥ f ∥ = max (∥ q ∥, ∥ r ∥) .$

Proof.

First we investigate the case $(f, g) = (c z^{n}, g)$ with $∥ g ∥ = 1$ and $g = \sum_{i = 0}^{d} c_{i} z^{i}$ a monic ( $| c_{d} | = 1$ , wlog we can assume $c_{d} = 1$ , $| c_{i} | \leq 1$ ) degree $d$ polynomial. Then, we note that the first quotient $q_{n} = c z^{n - d}$ $f = g q_{n} + r_{n}, r_{n} = f - g q_{n}$ satisfies $∥ q_{n} ∥ \leq | c |$ and $∥ r_{n} ∥ \leq ∥ g ∥ \cdot ∥ q_{n} ∥ \leq | c |$ . Repeating the division process should give us $f = g q + r$ with $∥ q ∥ \leq | c |$ and $∥ r ∥ \leq | c |$ . In fact, we have a stronger statement $| c | = ∥ f ∥ = max (∥ q ∥, ∥ r ∥) .$
By the linearity of residue division, we know that when $g$ is monic with $∥ g ∥ = 1$ , with $f = \sum_{i = 0}^{\infty} a_{i} z^{i}, | a_{i} | \to 0$ we should have $q = \sum q_{i}, r = \sum r_{i}$ with $(q_{i}, r_{i}) = Resdiv (a_{i} z^{i}, g)$ , so $| a_{i} | = max (∥ q_{i} ∥, ∥ r_{i} ∥)$ . Therefore $\begin{aligned} ∥ f ∥ & = max | a_{i} | \\ = max_{i} max (∥ q_{i} ∥, ∥ r_{i} ∥) \\ = max (max_{i} ∥ q_{i} ∥, max_{i} ∥ r_{i} ∥) \\ = max (∥ q ∥, max_{i} ∥ r_{i} ∥) \\ \geq max (∥ q ∥, ∥ r ∥) . \end{aligned}$ On the other hand clearly $∥ f ∥ = ∥ g q + r ∥ \leq max (∥ q ∥, ∥ r ∥)$ .

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Now we are ready to prove the proposition.

Proof.

Without loss of generality we can assume $∥ g ∥ = 1$ , and write out $g = g_{0} + g_{1}$ where $g_{0} = \sum_{i = 0}^{d} c_{i} z^{i}$ satisfy $| c_{d} | = 1$ and $| c_{i} | \leq 1$ , $∥ g_{1} ∥ < 1$ . Since $g_{0}$ is monic, we can write $f = q_{0} g_{0} + r_{0} = q_{0} g + r_{0} - q_{0} g_{1}$ with $\deg (r_{0}) < d$ and $∥ f ∥ = max (∥ q_{0} ∥, ∥ r_{0} ∥)$ . We also note that $∥ q_{0} g_{1} ∥ = ∥ q_{0} ∥ \cdot ∥ g_{1} ∥$ is strictly decreasing. Continuing division along $- q_{0} g_{1}$ we can get $\begin{aligned} f & = q_{0} g + r_{0} + q_{1} g + r_{1} + \dots \\ = g (q_{0} + q_{1} + \dots) + (r_{0} + r_{1} + \dots) \end{aligned}$ with $∥ q_{i} ∥ \leq ∥ f ∥ \cdot ∥ g_{1} ∥^{i}$ and $∥ r_{i} ∥ \leq ∥ f ∥ \cdot ∥ g_{1} ∥^{i}$ . The series converges since $∥ g_{1} ∥ < 1$ . In fact $max (∥ q_{1} ∥, ∥ r_{1} ∥) = ∥ q_{0} g_{1} ∥ \leq ∥ f ∥ ∥ g_{1} ∥$ , similarly $max (∥ q_{i} ∥, ∥ r_{i} ∥) \leq ∥ f ∥ ∥ g_{1} ∥^{i}$ . This tells us $∥ f ∥ = max (∥ q_{0} ∥, ∥ r_{0} ∥) \geq max (∥ q ∥, ∥ r ∥),$ and the other direction is derived from the equality $f = g q + r$ by applying the ultrametric inequality.
In the general case where $∥ g ∥ = | c_{g} | \neq 1$ , we can reduce to the case by applying the proposition with $g / c_{g}$ and write $f = (g / c_{g}) (c_{g} q) + r$ where we have $∥ f ∥ = max (∥ c_{g} q ∥, ∥ r ∥) = max (∥ g ∥ \cdot ∥ q ∥, ∥ r ∥)$ .

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Corollary 1. Every non-zero $f \in O (F)$ admits a factorization of the form $f = u \cdot (z - a_{1}) \dots (z - a_{d}), a_{i} \in F$ where $u = c_{f} (1 + s)$ is a ’unit’ in $O (F)$ , $c \in K^{*}$ and $s \in O (F)^{\circ \circ}$ .

Proof. Apply the proposition to $(z^{d}, c_{f}^{- 1} f)$ where $c_{f} \in K^{*}$ is the maximal coefficient of $f$ , i.e. making $d = \deg_{\overset{―}{K}} (\overset{―}{f / c_{f}}), ‖ f / c_{f} ‖ = 1$ and $f / c_{f}$ monic. We should get $z^{d} = q \cdot f / c_{f} + r$ where taking reduction $q$ reduces to $1$ , so $q = 1 + s$ . Clearly $z^{d} - r$ have all roots in $F$ (recall that $K$ is algebraically closed). ◻

Proposition 2. Let $\infty \in F \subset P^{1}$ be a connected affinoid. Write $F^{c} = ⨆_{i = 1}^{n} B (a_{i}, | π_{i} |),$ define $O (F)_{+} := {f \in O (F) : f (\infty) = 0}$ and $F_{i} = B_{i}^{c} = {| \frac{π}{z - a_{i}} | \leq 1}$ . Then

We have $O (F_{i}) = {\sum_{n \geq 0} b_{n} {(\frac{π_{i}}{z - a_{i}})}^{n} : b_{n} \to 0}$ and ${‖ \sum b_{n} {(\frac{π_{i}}{z - a_{i}})}^{n} ‖}_{F_{i}} = max | b_{n} |$ .
$O (F)_{+} = ⨁_{i = 1}^{n} O (F_{i})_{+}$ with $∥ \sum f_{i} ∥_{F} = max_{i} ∥ f_{i} ∥_{F}, ∥ f_{i} ∥_{F} = ∥ f_{i} ∥_{F_{i}}$ .

Residue Theorem

Let $F$ be a connected affinoid set, $ω$ a meromorphic differential on $F$ . We would like to define an interior $F^{\circ}$ and boundary $\partial F$ of $F$ , with some kind of ’integral’ of $ω$ along $\partial F$ so that there is some kind of residue theorem.

Definition 2.

A meromorphic function on an affinoid $F$ is an element of the ring of total quotients of $O (F)$ . If $F$ is connected, this is just an element of the fraction field of $O (F)$ , represented as $\frac{f}{g}$ with $f, g \in O (F)$ having no common zeros.
We can define ${ord}_{a} (f)$ for $f \in O (F), a \in F$ to be the maximal $n$ such that $(z - a)^{n} | f$ . For meromorphic functions, we define ${ord}_{a} (\frac{f}{g}) = {ord}_{a} (f) - {ord}_{a} (g)$ .
For a meromorphic differential $ω = f \cdot d z$ , we define ${ord}_{a} (ω) = {ord}_{a} (f), {ord}_{\infty} (ω) = {ord}_{0} (- \frac{f (t^{- 1})}{t^{2}} d t) .$
Let $t$ be a local parameter at $a$ , which $\in O (D (a, r))$ with ${ord}_{a} (t) = 1$ . Then locally $ω$ can be written as $ω = \sum_{n \in Z} a_{n} t^{n} d t .$ Its residue at $a$ is defined as ${Res}_{a} (ω) = a_{- 1} .$
A ring domain is an affinoid set $F \subset P^{1}$ that is projective isomorphic to ${| z | = 1}$ . A thick ring domain is isomorphic to ${r_{1} \leq | z | \leq r_{2}}$ .

For a ring domain $F$ and a holomorphic differential $ω$ and an invertible holomorphic function $f \in O (F)$ , we want to define what the residues ${Res}_{F} (ω)$ and ${ord}_{F} (f)$ are, similar to the contour integral along circle and the winding number (degree) of a function along a circle.

Example 2. Take the classical $F = {| z | = 1}$ , we know $O (F) = K ⟨ z, z^{- 1} ⟩$ $ω = \sum a_{n} z^{n} d z$ where $a_{n} \overset{| n | \to \infty}{\to} 0$ . We would like to define (not to be confused with the residue at a point) ${Res}_{F} (ω) := a_{- 1} .$ This $a_{- 1}$ in fact will ’sum up’ the residues at all points in $F$ due to the peculiar non-archimedean geometry, imagine an expression like $\frac{A}{z - a} + \frac{B}{z - b} = \frac{1}{z} (\frac{A}{1 - \frac{a}{z}} + \frac{B}{1 - \frac{b}{z}}) = \frac{A + B}{z} + \dots$ where $| a |, | b | < 1$ and the rest terms converges. We see that the $z^{- 1}$ coefficient actually collects the residues at $a, b$ .

Similarly one defines ${ord}_{F} f$ as the unique $n$ such that $f = c z^{n} (1 + s)$ with $c \in K^{*}, s \in O (F)^{\circ \circ}$ i.e. $∥ s ∥ < 1$ (it says that in this case $O (F)$ is a DVR).

Theorem 1 (Simple Residue Theorem). For $F = {| z | = 1}$ , and $ω = f d z$ a meromorphic differential holomorphic on $F$ , we have $\sum_{| a | < 1} {Res}_{a} (ω) = {Res}_{F} (ω),$ and for $f \in O (F)^{\times}$ , we have $\sum_{| a | < 1} {ord}_{a} (f) = {ord}_{F} (f) .$

Theorem 2 (Residue Theorem). Let $F = {(⨆_{i} B_{i})}^{c} \subset P$ be a connected affinoid. Let $q \in F$ and suppose the boundaries $\partial B_{i}$ with respect to $q$ are disjoint and given suitable orientation, counting outside $F$ and inside $B_{i}$ .

Let $ω$ be a meromorphic differential on $F$ which is holomorphic on $\partial B_{i}$ . Then $\sum_{i} {Res}_{\partial B_{i}} (ω) + \sum_{a \in F^{\circ}} {Res}_{a} (ω) = 0.$
Let $f$ be a meromorphic function on $F$ such that $f$ is invertible on all boundaries $\partial B_{i}$ . Then $\sum_{i} {ord}_{\partial B_{i}} (f) + \sum_{a \in F^{\circ}} {ord}_{a} (f) = 0.$

Example 3. Let $F$ be a thick ring domain with boundaries $\partial B_{1}, \partial B_{2}$ , $f \in O (F)$ . Then the number of zeros of $f$ in $F$ is equal to the number of poles of $f$ outside $F$ , $N_{F} (f) = - {ord}_{\partial B_{1}} (f) - {ord}_{\partial B_{2}} (f) .$ where you want the $\partial B_{i}$ to be oriented to count the zeros and poles outside $F$ , inside $B_{i}$ . Let’s expand this formula in the simplest example of $F = {| α | \leq | z | \leq | β |}, f = \sum_{n \in Z} c_{n} z^{n}$ with $| α | < | β | \Leftrightarrow v (1 / α) < v (1 / β)$ . Then we should have, by adjusting to the correct choice of orientation, $N_{F} (f) = - {ord}_{\partial B (0, | α |)} (f) + {ord}_{\partial B (0, | β |)} (f) .$ Then the ring of regular functions are $O (F) = {\sum c_{n} z^{n} : | c_{k} | = o (| β |^{- k}), | c_{- k} | = o (| α |^{k}) as k \to + \infty} .$ or $O (F) = {\sum c_{n} z^{n} : v (c_{k}) - k v (1 / β) \to \infty, v (c_{- k}) - (- k) v (1 / α) \to \infty (k \to + \infty)} .$ You can actually see ${ord}_{B (| β |)} (f) - {ord}_{B (| α |)} (f)$ as the length of the part of the Newton polygon of $\sum_{n \in Z} a_{n} z^{n}$ that has slope in $[v (1 / α), v (1 / β)]$ , and $O (F)$ are functions that converge above a V-shape formed by slopes $v (1 / α), v (1 / β)$ .

Excercise 1. Let $F = {| z | = | z - 1 | = 1}$ , develop residue theorem for this case and a formula for number of zeros $N_{F} (f)$ .