Bicomplex Conjugate Derivative

Question

I have decided to first ask my question and second provide a list of steps I have already considered.

Question: After reading Luna-Elizarrarás, Shapiro, Struppa, and Vajiac - Bicomplex numbers and their elementary functions , I am wondering if the derivative $\frac{\partial F}{\partial Z^{\dagger}}$ could be defined for a function $F: \mathbb{BC} \to \mathbb{BC}$ as a means to generalize the notion of quasiconformality. Unfortunately, the fact that $\bar{z}$ is not holomorphic appears to preclude the usage of the multivariate chain rule to calculate the derivative. I have written the four components of a bicomplex number as $x$, $y$, $k$, and $l$ as a 4-tuple in $\mathbb{R}^4$. Could they be expressed in terms of $Z$ and $Z^{\dagger}$ using an intermediate formula relying on some other notion than the complex conjugate? I wish that the dot product was defined for the underlying four tuples as one could trivially assert $\langle1, 0, 0, 0\rangle \cdot Z = x$.

Existing Thoughts: The notion of complex quasiconformality is defined as follows.

$\mu$ is a complex Lebesgue measure s.t. $\sup\{\mu(x)\}<1$. $$ \frac{\partial f}{\partial\overline z}=\mu(z)\frac{\partial f}{\partial z}. $$

$\mu$-quasiconformality may be ensured if $\frac{\partial f}{\partial\bar z}<\frac{\partial f}{\partial z}$ in the strict sense. The existence of a measure $\mu$ is then guaranteed as no statement of continuity is attested. The above is justified using the operator $\frac\partial{\partial\overline z}=\frac{1}{2}\left(\frac\partial{\partial x}+i\frac\partial{\partial y}\right)$, $x=\frac{z+\overline z}2$, and $y=\frac{z-\overline z}{2i}$. By the multivariate chain rule, we define $\frac{\partial f}{\partial\overline z}$ as follows: $$ \frac{\partial f}{\partial\overline z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\overline z}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\overline z}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)=\frac{1}{2}\left(u_x+i v_x+i u_y-v_y\right). $$

Note that this equation will equal $0$ provided that $u_x-v_y=0$ and $v_x+u_y=0$, which guarantee the Cauchy–Riemann equations.

Naturally, we ask if $\mu$-quasiconformality can be generalized to a bicomplex Lebesgue measure $\mu:\mathbb{BC}\to \mathbb R$ with $\sup\{\mu(Z)\}<1$ in the following sense. $F:\mathbb{BC}\to\mathbb{BC}$ is a bicomplex holomorphic function and $\frac\partial{\partial Z^\dagger}$ is defined by the same algebra as $\frac\partial{\partial\overline z}$. $$ \frac{\partial F}{\partial Z^\dagger}=\mu(Z)\frac{\partial F}{\partial Z}. $$

We consider if such bicomplex quasiconformality guarantees complex quasiconformality on each of the complex components. A notion of a conformal mapping does not exist on $\mathbb R$, so no analogue of this property exists on $\mathbb C$.

Before proceeding, certain terminology will prove to be advantageous. $F(z_1, z_2):\mathbb{BC}\to\mathbb{BC}$ can also be written as $F(x, y, k, l)=(U+i V+j K+i j L):\mathbb R^4 \to \mathbb{BC}$. Our notion of the components of a complex number from linear combinations of the conjugate generalizes to $\mathbb{BC}$ as $z_1=\frac{Z^\dagger + Z}2$, $z_2=\frac{-Z^\dagger+Z}{2j}$, $x=\frac{z_1 + \overline{z_1}}2$, $y=\frac{z_1 - \overline{z_1}}{2i}$, $k=\frac{z_2 + \overline{z_2}}2$, and $l=\frac{z_2 - \overline{z_2}}{2i}$.

Where Things Break Down: The fact that the complex conjugate $\overline z$ is nowhere differentiable presents a particular difficulty when attempting to apply the multivariate chain rule. \begin{multline*} \frac{\partial F}{\partial Z^\dagger}=\frac{\partial F}{\partial x}\left[\frac{\partial x}{\partial z_1}\frac{\partial z_1}{\partial Z^\dagger}+\frac{\partial x}{\partial\overline{z_1}}\frac{\partial\overline{z_1}}{\partial Z^\dagger}\right]+\frac{\partial F}{\partial y}\left[\frac{\partial y}{\partial z_1}\frac{\partial z_1}{\partial Z^\dagger}+\frac{\partial y}{\partial \overline{z_1}}\frac{\partial\overline{z_1}}{\partial Z^\dagger}\right]+ \\ \frac{\partial F}{\partial k}\left[\frac{\partial k}{\partial z_2}\frac{\partial z_2}{\partial Z^\dagger}+ \frac{\partial k}{\partial\overline{z_2}}\frac{\partial\overline{z_2}}{\partial Z^\dagger}\right]+\frac{\partial F}{\partial l}\left[\frac{\partial l}{\partial z_2}\frac{\partial z_2}{\partial Z^\dagger}+\frac{\partial l}{\partial\overline{z_2}}\frac{\partial\overline{z_2}}{\partial Z^\dagger}\right]. \end{multline*}

For example, I cannot possibly see how to define the derivative of $\overline{z_1}=\overline{\frac{Z^\dagger}2}$ considering that the complex conjugate is neither analytic nor holomorphic. (On $\mathbb{BC}$, holomorphicity and analyticity are not logically equivalent.) Does anyone have an alternate idea to define quasiconformality for $\mathbb{BC}$ in a meaningful fashion? Could the relationship to $\bar{z}$ be avoided?

Answer 1

Since $F$ is bicomplex-holomorphic, we necessarily have $\frac{\partial F}{\partial Z^\dagger} = 0$, and your condition, let's call it the bicomplex Beltrami equation, $$ \frac{\partial F}{\partial Z^\dagger}=\mu(Z)\frac{\partial F}{\partial Z} $$ implies that either $\mu$ is the zero measure or $F$ is constant.

By the way, any bicomplex-holomorphic function $F(Z)$ becomes - after a linear change of coordinates - of the form $f_1(w_1) \oplus f_2(w_2)$ for some $\mathbb{C}$-holomorphic functions $f_1$ and $f_2$.

A more appropriate generalization of quasi-conformality is to only require $F: U \to \mathbb{BC}$, $U \subseteq \mathbb{C}^2$ open, to be complex-differentiable in the usual sense as a function of several complex variables or even only real-differentiable as a function of 4 real variables. After all, in the Beltrami equation for $\mathbb{C}$, $f$ is only real-differentiable, which is what makes the problem interesting.