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