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The Wirtinger differential operators are defined by:

\begin{equation} \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right) \\ \frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right) \end{equation}

These satisfy the following chain rule:

\begin{equation} \frac{\partial}{\partial z}(f \circ g) = \left(\frac{\partial f}{\partial z}\circ g\right)\frac{\partial g}{\partial z} + \left(\frac{\partial f}{\partial \bar{z}}\circ g\right)\frac{\partial \bar{g}}{\partial z}. \end{equation}

Usually I think about partial derivatives as forming the components of the Jacobian (a.k.a differential/total derivative) and the chain rule for them as a matrix representation of the relation: \begin{equation} \mathbf{D}(f \circ g) = (\mathbf{D}f \circ g)\cdot\mathbf{D}g. \end{equation}

How can one interpret the chain rule for the Writinger differential operators in this light? I would particularly enjoy a formalism that allows me to understand why $\frac{\partial{f}}{\partial \bar{z}} = 0$ iff $f$ is analytic, or makes this seem like a really natural definition to begin investigation of the properties of analytic functions. Alternatively I would like formalism that is closely related to the idea of complexifying the tangent bundle of $\mathbb{R}^2$ with its standard complex structure, with an explanation of the connection.

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Think of $f\colon \mathbb{C} \longrightarrow \mathbb{C}$ as a map $f\colon \mathbb{R}^2 \longrightarrow \mathbb{R}^2$. The jacobian matrix $Jf(p)$ is just the matrix of $Df_p \colon T_p\mathbb{R}^2 \longrightarrow T_{f(p)}\mathbb{R}^2$ with respect to the frame $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$. Complexify the tangent bundle and change the frame to $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$. In this new frame we will have $$ { \large [Df_p] = \begin{pmatrix} \frac{\partial f}{\partial z}(p) & \frac{\partial f}{\partial \bar z}(p)\\ \frac{\partial \bar f}{\partial z}(p)& \frac{\partial\bar f}{\partial \bar z}(p) \end{pmatrix}} $$ Then, by chain rule, $ [D(f\circ g)_p] =[Df_{g(p)}][Dg_p]$ which reads $$ \begin{pmatrix} \frac{\partial f\circ g}{\partial z}(p) & \frac{\partial f\circ g}{\partial \bar z}(p)\\ \frac{\partial \overline{f\circ g}}{\partial z}(p) & \frac{\partial \overline{f\circ g} }{\partial \overline z}(p) \end{pmatrix} = \begin{pmatrix} \frac{\partial f}{\partial z}(g(p)) & \frac{\partial f}{\partial \bar z}(g(p))\\ \frac{\partial \bar f}{\partial z}(g(p)) & \frac{\partial\bar f}{\partial \bar z}(g(p)) \end{pmatrix} \begin{pmatrix} \frac{\partial g}{\partial z}(p) & \frac{\partial g}{\partial \bar z}(p)\\ \frac{\partial \bar g}{\partial z}(p) & \frac{\partial\bar g}{\partial \bar z}(p) \end{pmatrix} $$ And the result follows comparing the entries of the matrices.

Added: In the frame $\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$ we have for $f(x,y) = (u(x,y) , v(x,y))$ $$ Jf = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} &\frac{\partial v}{\partial y} \end{pmatrix} $$ And the base change matrix to $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$ is $$ P = \frac{1}{2}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix} $$ From the relations $\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x} - i\frac{\partial}{\partial y}\right)$ and $\frac{\partial}{\partial \bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)$. Hence the matrix for $Df$ in the $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$ frame is $P \cdot Jf\cdot P^{-1}$, $$ \frac{1}{2}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix} \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} &\frac{\partial v}{\partial y} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix} \begin{pmatrix} \frac{\partial u}{\partial x} + i\frac{\partial u}{\partial y} & \frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} + i\frac{\partial v}{\partial y} & \frac{\partial v}{\partial x} - i\frac{\partial v}{\partial y} \end{pmatrix} = \frac{1}{2}\begin{pmatrix} \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + i\left( \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x} \right) & \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} - i\left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right) \\ \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y} + i\left( \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \right) & \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} - i\left( \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x} \right) \end{pmatrix} $$ that agree with the matrix stated if we write $f = u+iv$ and $z=x+iy$.

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  • $\begingroup$ Great answer, could you just explicitly do the calculation that expresses $Df$ in the $\partial_z$, $\partial_{\bar{z}}$ frame? $\endgroup$ – 11Kilobytes Oct 8 '18 at 6:36
  • $\begingroup$ Just added that. $\endgroup$ – Alan Muniz Oct 8 '18 at 12:30
  • $\begingroup$ Note that the definition of $\frac{\partial}{\partial z}, \frac{\partial}{\partial \bar z}$ you wrote is wrong.. the minus sign is misplaced. $\endgroup$ – Alan Muniz Oct 8 '18 at 12:33

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