\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}
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.
Let’s define (formal) partial differentials \pfrac{z}, \pfrac{\zbar} for complex functions.
Fix a field k, and consider an algebra A over k.
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.
The following lemma is useful to construct a new derivation from old ones.
Let k be a field, A, B algebras \overk.
This lemma can easily be verified by straightforward calculations.
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].
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].
If 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.
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)].
The lemma 1.1.1 of [KAS] is now obvious from above consideration.
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.
Let B := \{ f \in A \colon \text{polynomial} \} = \R [x, y].
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).