2018/12/28 –Naughie's Notes

$\newcommand\le\leqslant \newcommand\ge\geqslant \newcommand\zbar[1][z]{\overline{#1}} \newcommand\zder{\partial_z} \newcommand\bder{\partial_{\zbar}} \newcommand\xder{\partial_x} \newcommand\yder{\partial_y} \newcommand\ider[1][i]{\partial_{#1}} \newcommand\pfrac[2][]{\frac{\partial #1}{\partial #2}} \newcommand\R{\mathbb{R}} \newcommand\C{\mathbb{C}} \newcommand\bcal{\mathcal{B}} \newcommand\dcal{\mathcal{D}} \newcommand\overk[1][k]{/_{\Large #1}} \DeclareMathOperator\iunit{i} \DeclareMathOperator\Der{Der} \DeclareMathOperator\End{End} \DeclareMathOperator\gl{\mathfrak{gl}} \DeclareMathOperator\id{id}$

Introduction

I have read KASAHARA [KAS] with @kyotomathmath. This book brings me into consideration on the definition of differentiations

$\zder, \bder$ . Let’s take a look at what happens in discussing differentiations of complex functions.

The organization of this article follows that of [KAS]; hence the chapter 1 of this corresponds to the chapter 1 of [KAS], etc.

What are holomorphic functions?

Preliminaries

Let’s define (formal) partial differentials

$\pfrac{z}, \pfrac{\zbar}$ for complex functions.

A linear map $D \colon A \to A$ is called a derivation is $D (a b) = D (a) b + a D (b) \quad (\forall a, b \in A).$

Denote the set of all derivations on $A$ by $\Der_k A$ : $\Der_k A := \{ D \colon A \to A \colon \text{derivation} \}.$

When there is no ambiguity, we denote $\Der_k A$ simply by $\Der A$ .

It’s easy to show that

$\Der A \subset \gl (A)$ is a Lie subalgebra of

$\gl (A)$ .

Let $k = \R$ and $A = \sum_{i = 1}^\infty C^i (\dcal)$ , where $\dcal \subset \R^n$ and $C^i (\dcal) := \{ f \colon \dcal \to \R \colon C^i \ \text{class} \}.$

The partial differentials $\begin{alignat*}{0} \ider \colon &A &&\longrightarrow &&A \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &f &&\longmapsto &&\pfrac{x^i} f \end{alignat*}$ are derivations on $A$ .

Let $k$ be a field, $A, B$ algebras $\overk$ .

$A \oplus B$ is also an algebra $\overk$ .
$\forall D \in \Der A$ , $\forall E \in \Der B$ , $D \oplus E \in \Der (A \oplus B).$
The map $\begin{alignat*}{0} &\Der A \oplus \Der B &&\longrightarrow && \Der (A \oplus B) \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &(D, E) &&\longmapsto && D \oplus E \end{alignat*}$ is an injective Lie algebra hom.

Again, consider the case $k = \R$ , $\dcal \subset \R^n$ , $A = \sum C^i (\dcal)$ . Since there exists an isomorphism $\left( \sum C^i (\dcal) \right)^m \cong \sum C^i (\dcal, \R^m)$ as vector spaces $\overk[\R]$ , where $C^i (\dcal, \R^m) := \{ f \colon \dcal \to \R^m \colon C^i \ \text{class} \},$ and since $A^m$ is an algebra $\overk[\R]$ by the above lemma, $\sum C^i (\dcal, \R^m)$ can be regarded as an algebra $\overk[\R]$ .

Then, e.g. $\ider[i_1] \oplus \dotsb \oplus \ider[i_m] \in \Der \left( \sum C^i (\dcal, \R^m) \right)$ ( $1 \le {}^\forall i_j \le n$ ).

Next, suppose we have an extension of fields

$k \subset K$ . Any vector space

$V$

$\overk$ can be extended to the vector space

$V \otimes_k K$

$\overk[K]$ . Explicitly, the scalar multiplication on

$V \otimes_k K$ is given by

$\mu \cdot (v \otimes \lambda) := v \otimes (\mu \lambda) \quad (v \in V, \lambda, \mu \in K).$

Suppose, furthermore,

$A$ is an algebra

$\overk$ and consider the vector space

$A \otimes_k K$

$\overk[K]$ . This is now an algebra

$\overk[K]$ ; indeed, the ring hom

$k \to A$ induces a ring hom

$K (\cong k \otimes_k K) \to A \otimes_k K$ , which makes

$A \otimes_k K$ into an algebra

$\overk[K]$ .

Let $k \subset K$ be extension of fields, $A$ algebra $\overk$ , $A \otimes_k K$ algebra $\overk[K]$ .

$\forall f \in \End_k A$ , $f \otimes 1 \in \End_K (A \otimes_k K)$ .
$\forall D \in \Der_k A$ , $D \otimes 1 \in \Der_K (A \otimes_k K)$ .

i) It’s obvious that $f \otimes 1$ preserves addition.

For any $x \in A$ , $\lambda, \mu \in K$ , $\begin{align*} (f \otimes 1) (\mu \cdot (x \otimes \lambda) ) &= f (x) \otimes \mu \lambda \\ &= \mu (f (x) \otimes \lambda ) \\ &= \mu ( f \otimes 1 ) (x \otimes \lambda). \end{align*}$

ii) We’ve already seen that $D \otimes 1$ is $K$ -linear. To see that is is also a derivation, note that, for $x, y \in A$ , $\lambda, \mu \in K$ , $(x \otimes \lambda) \cdot (y \otimes \mu) = (x y) \otimes (\lambda \mu).$ Thus, for any $x, y \in A$ , $\lambda, \mu \in K$ , we have $\begin{align*} &(D \otimes 1) ( (x \otimes \lambda) (y \otimes \mu) ) = D (x y) \otimes \lambda \mu \\ ={}& D (x) y \otimes \lambda \mu + x D (y) \otimes \lambda \mu \\ ={}& ( ( D \otimes 1 ) (x \otimes \lambda) ) \cdot (y \otimes \mu) + (x \otimes \lambda) \cdot ( (D \otimes 1) (y \otimes \mu) ). \end{align*}$

In the following discussion, let’s focus on the special case

$\R \subset \C$ , the degree of whose extension is

$2$ .

First, let’s describe the complexification

$A \otimes_\R \C$ more explicitly. Since

$\C = \R \oplus \iunit \R$ , we can write, for any vector space

$V$

$\overk[\R]$ ,

$V \otimes_\R \C = (V \otimes_\R \R) \oplus (V \otimes_\R \iunit \R) = V \oplus \iunit V$ as vector spaces

$\overk[\R]$ .

$A$ is an algebra

$\overk[\R]$ , let

$\begin{alignat*}{0} J \colon &A \otimes_\R \C &&\longrightarrow &&A \otimes_\R \C &&\colon \R \text{-linear,} \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &a \otimes z &&\longmapsto &&a \otimes \iunit z, \end{alignat*}$ so that

$J^2 = - \id$ . Therefore we have a ring hom

$\begin{alignat*}{0} &\C &&\longrightarrow &&\End_\R (A \otimes_\R \C) \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &x + \iunit y &&\longmapsto &&x \cdot \id + y \cdot J. \end{alignat*}$ Endowed with this structure map,

$A \otimes_\R \C$ becomes an algebra

$\overk[\C]$ as described above.

For a linear map

$f \colon V \to W$ between vector spaces

$\overk[\R]$ , the

$\C$ -linear map

$f \otimes 1 \colon V \otimes_\R \C \to W \otimes_\R \C$ is now given by the formula

$\begin{alignat*}{0} &V \oplus \iunit V &&\longrightarrow &&W \oplus \iunit W \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &v + \iunit w &&\longmapsto &&f (v) + \iunit f (w); \end{alignat*}$ hence denoted by

$f \oplus f \colon V \otimes_\R \C \to W \otimes_\R \C$ .

Let $\dcal \subset \R^2$ , $A = \sum C^i (\dcal)$ , $A^2 = \sum C^i (\dcal, \R^2)$ .

By the above lemma, $\ider[j] \oplus \ider[j] \in \Der A^2$ ( $1 \le {}^\forall j \le 2$ ), regarded as an algebra $\overk[\C]$ . Since $\Der A^2$ is a vector space $\overk[\C]$ , $\begin{align*} \zder &:= \frac 12 (\xder \oplus \xder - \iunit \yder \oplus \yder ), \\ \bder &:= \frac 12 (\xder \oplus \xder + \iunit \yder \oplus \yder ) \end{align*}$ are also derivations on $A^2$ : $\zder, \bder \in \Der A^2$ . These satisfies $\begin{align*} \xder \oplus \xder &= \zder + \bder, \\ \yder \oplus \yder &= \iunit ( \zder - \bder ), \end{align*}$ which justifies the notation “ $\zder$ , $\bder$ ”.

$A \otimes_\R \C$ is equipped with another inmortant endmorphism

$S \in \End_\R (A \otimes_\R \C)$ represented by either

$\begin{alignat*}{0} S \colon &A \otimes_\R \C &&\longrightarrow &&A \otimes_\R \C \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ &a \otimes z &&\longmapsto &&a \otimes \zbar, \end{alignat*}$

$\begin{alignat*}{0} S \colon &A \oplus \iunit A &&\longrightarrow &&A \oplus \iunit A \\ &\style{transform: rotate(270deg)}{\in} && &&\style{transform: rotate(270deg)}{\in} \\ & a + \iunit b &&\longmapsto && a - \iunit b. \end{alignat*}$ In general, an endmorphism

$S \in \End_\R V$ of a vector space

$V$

$\overk[\R]$ is called an involution if

$S^2 = \id$ but

$S \neq \id$ .

The following lemma shows the compatibility of the involution

$S$ on

$A \otimes_\R \C$ with that on

$\C$ .

Let $A, B$ be algebras $\overk[\R]$ , $S_A, S_B$ involutions on $A, B$ , resp., given above.

$\forall f \colon A \to B$ : linear, $S_B \circ (f \otimes 1) = (f \otimes 1) \circ S_A.$
$\forall F_1, F_2, G_1, G_2 \colon A \otimes_\R \C \to B \otimes_\R \C$ : $\C$ -linear, $\forall \lambda \in \C$ , if $S_B \circ F_j = G_j \circ S_A \quad (j = 1, 2),$ then $\begin{align*} S_B \circ (F_1 + F_2) &= (G_1 + G_2) \circ S_A, \\ S_B \circ (\lambda F_1) &= (\zbar[\lambda] G_1) \circ S_A \end{align*}$

i) $\forall a \in A$ , $\forall z \in \C$ , we have $\begin{align*} (S_B \circ (f \otimes 1)) (a \otimes z) &= S_B ( f(a) \otimes z) \\ &= f (a) \otimes \zbar, \\ ((f \otimes 1) \circ S_A) (a \otimes z) &= (f \otimes 1) (a \otimes \zbar) \\ &= f (a) \otimes \zbar; \end{align*}$ hence $S_B \circ (f \otimes 1) = (f \otimes 1) \circ S_A$ .

ii) THe first assertion is obvious as $\begin{align*} S_B \circ (F_1 + F_2) &= S_B \circ F_1 + S_B \circ F_2 \\ &= G_1 \circ S_A + G_2 \circ S_A \\ &= (G_1 + G_2) \circ S_A. \end{align*}$

The second follows from that $\forall a \in A$ , $\forall z \in \C$ , we have $\begin{align*} S_B (\lambda F_1) (a \otimes z) &= S_B (F_1 (a \otimes \lambda z)) \\ &= G_1 (S_A (a \otimes \lambda z)) \\ &= G_1 (a \otimes \zbar[\lambda] \zbar) \\ &= \zbar[\lambda] \cdot (G_1 \circ S_A (a \otimes z)). \end{align*}$

Let $\dcal \subset \R^2$ , $A = \sum C^i (\dcal)$ , $A^2 = \sum C^i (\dcal, \R^2)$ .

Since $\xder, \yder \in \Der A \subset \End_\R A$ , by the above lemma, $S_A \circ \bder = \zder \circ S_A,$ which is expressed effectively as $\zbar[\pfrac{\zbar} f] = \pfrac{z} \zbar[f] \quad ({}^\forall f \in A),$ where $\zbar[f]$ denotes the map $\zbar[f] (x, y) := \zbar[f (x, y)].$

Now that we define

$\zder, \bder$ by

$\begin{align*} \zder &:= \frac 12 (\xder \oplus \xder - \iunit \yder \oplus \yder ), \\ \bder &:= \frac 12 (\xder \oplus \xder + \iunit \yder \oplus \yder ), \end{align*}$ and from this follow the relations

$\begin{align*} \zder z = 1, &\quad \zder \zbar = 0, \\ \bder z = 0, &\quad \bder \zbar = 1. \end{align*}$

Conversely, if we start with

$\zder z = 1$ , etc., then we can derive the formula for

$\zder, \bder$ , or equivalently,

$\begin{align*} \xder \oplus \xder &= \zder + \bder, \\ \yder \oplus \yder &= \iunit (\zder - \bder). \end{align*}$

Let

$\dcal \subset \R^2$ ,

$A = \sum C^i (\dcal)$ ,

$A^2 = \sum C^i (\dcal, \R^2)$ , as above, and

$\zder, \bder \in \Der A^2$ satisfy

$\begin{align*} \zder z = 1, &\quad \zder \zbar = 0, \\ \bder z = 0, &\quad \bder \zbar = 1. \end{align*}$

For

$x = (z + \zbar) / 2$ ,

$y = (z - \zbar) / 2 \iunit$ , we have

$\begin{align*} \zder x = \frac 12, &\quad \zder y = \frac 1{2\iunit}, \\ \bder x = \frac 12, &\quad \bder y = - \frac 1{2\iunit}. \end{align*}$

By the composition laws, which we haven’t proved yet, we want

$\begin{align*} \xder z = 1, &\quad \xder \zbar = 1, \\ \yder z = \iunit, &\quad \yder \zbar = -\iunit, \end{align*}$ and

$\forall f \in A^2$ ,

$\begin{align*} \xder f &= (\xder z) (\zder f) + (\xder \zbar) (\bder f) \\ &= \zder f + \bder f, \\ \yder f &= (\yder z) (\zder f) + (\yder \zbar) (\bder f) \\ &= \iunit \zder f - \iunit \bder f. \end{align*}$

So, it’s reasonable to define

$\begin{align*} \xder &:= \zder + \bder, \\ \yder &:= \iunit (\zder - \bder). \end{align*}$

Then, are these operators agree with

$\xder \oplus \xder$ and

$\yder \oplus \yder$ , resp.?

For definiteness, we denote the new operators instead by

$\begin{align*} \xder' &:= \zder + \bder, \\ \yder' &:= \iunit (\zder - \bder). \end{align*}$ What we must show is then

$\forall f \in A$ ,

$\begin{align*} \xder' f = \xder f, \\ \yder' f = \yder f. \end{align*}$

First note that

$\begin{align*} \xder' x = 1, &\quad \xder' y = 0, \\ \yder' x = 0, &\quad \yder' y = 1, \end{align*}$ We can see the linear independence of

$\xder', \yder'$ from these relations: for any

$\lambda, \mu \in \C$ , if

$\lambda \xder' + \mu \yder' = 0$ , then applying

$x$ to the both sies gives

$\lambda = (\lambda \xder' + \mu \yder') (x) = 0,$ and similarly we obtain

$\mu = 0$ .

Since

$\xder', \yder' \in \Der A^2$ , we have

$\xder' = \xder$ and

$\yder' = \yder$ on

$B$ (using the defining properties of derivations). So, if

$\dcal$ is bounded, then, by the Stone-Weierstrass’ theorem,

$B$ is dense in

$A$ ; hence

$\xder' = \xder$ and

$\yder' = \yder$ on

$A$ .

Otherwise, since

$\R^2$ is locally compact, there exists a family

$(\bcal_j)_{j \in I}$ of bounded sets indexed by some set

$I$ such that

$\dcal = \bigcup_{j \in I} \bcal_j.$ Then,

$\xder' = \xder$ and

$\yder' = \yder$ on each

$\sum C^i (\bcal_j)$ , and hence on

$A$ because

$\sum C^i (\dcal) \subset \bigcup_j \sum C^i (\bcal_j).$