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In a lecture about complex analysis, I was introduced to the Wirtinger derivatives $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$. They can be expressed as $\partial_z = \frac{1}{2}(\partial_x-i\partial_y)$ $\partial_\bar{z} = \frac{1}{2}(\partial_x+i\partial_y)$ where $x,y$ are cartesian coordinates.

The lecture notes explains further that "$z,\bar{z}$ are treated as independent variables. But of course we can easily get the conjugate by conjugation, meaning that they are not independent. However, the algebraic rules by Wirtinger, make it as if they were independent, ignoring the complex conjugate." (I'm paraphrasing here)

I believe there is a mixup what independent means. I think independent is a linear algebra term and the basis $(x,y)$ and $(z,\bar{z})$ are different basis for the complex plane related by the matrices $$J = \binom{1,\,\, \, i}{1,-i} \iff J^{-1} = \frac{1}{2}\binom{\,\, \,1, 1}{-i,i}$$.

Are the following statments correct?

$\binom{z}{\bar{z}} = J \binom{x}{y}, \binom{dz}{d\bar{z}} = J \binom{dx}{dy}, (\partial_z, \partial_\bar{z}) = (\partial_x, \partial_y) J^{-1} $

The complex conjugate is not a linear algebraic operation. The same happens also in real analysis where we deal with basis changes, however no one would claim that they are not really independent just because we now how we can get one from the other.

Am I correct? Are they linear independent? Is it about linear independence and is their a mix up with the different meaning of independent in the notes?

theta_phi
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  • Not an answer because I'm writing this in a hurry and because it is self advertisement. You can view them as independent variables (variable meaning, coordinate projection map) of $\mathbb C^2$. Here is a little note that I wrote that clarified things for myself https://www.math.univ-paris13.fr/~michels/files/wirtinger_derivatives.pdf The restrictions to the subspace $\mathbb R^2\cong \mathbb C$ are indeed not independent. – Bart Michels May 12 '24 at 14:55
  • Welcome to MSE! <> Briefly, the proposed equations are not quite correct; $J$ is a real matrix, as are the $1$-forms $dx$ and $dy$, while $dz$ and $d\bar{z}$ are not real. Does that help write down a transition matrix? (If so, please feel free to answer your own question!) – Andrew D. Hwang May 12 '24 at 20:45
  • Alternatively, does this help? https://math.stackexchange.com/questions/314863/what-is-the-intuition-behind-the-wirtinger-derivatives – Andrew D. Hwang May 12 '24 at 20:53
  • Please expand on your command on the J matrix. Do you mean that there are imaginary units missing in the second column? – theta_phi May 12 '24 at 21:17
  • The imaginary units in the second column (and omitting the factor of $\frac{1}{2}$) are what I meant, yes. (We have $z = x + iy$ and $\bar{z} = x - iy$; $J$ just needs to be the resulting coefficient matrix. That's one way to see why there are factors of $\frac{1}{2}$ in the formulas for $\partial_{z}$, etc.) – Andrew D. Hwang May 16 '24 at 23:04
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    @AndrewD.Hwang, I mixed $J$ and it's inverse up. That's why I had the additional factor $2$ when I was quickly writing this down. Also the partial derivatives are dual to the differentials and therefore I transposed them now in the equation above which gives the right formulas by matrix multiplication. I apologise for the hasty formulas, they should be revised now. My question of the mending of independent remains still, is this a domain mistake of functional and algebraic independence? – theta_phi May 17 '24 at 07:51
  • Here is the question I was searching for earlier (when checking if this was a duplicate) but couldn't locate. Particularly, does the comment in my answer address your question here? – Andrew D. Hwang May 17 '24 at 12:45

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