# Bicomplex Conjugate Derivative

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?

• I am not sure I understand your question. If $Z = z_1 + j z_2$ is a bicomplex number, then $Z^\dagger := z_1 - j z_2$. This determines the operator $\partial / \partial Z^\dagger$. Do you want to change the definition of the operation $Z^\dagger$ so as to obtain some notion of bicomplex quasiconformality or are you looking for some notion of bicomplex quasiconformality that corresponds to this definition of $Z^\dagger$?
– M.G.
Apr 8 at 21:05
• In fact, $\partial/\partial Z^\dagger = 1/2(\partial/\partial z_1 + j \partial/\partial z_2)$, hence $\partial \bar{z}_1 / \partial Z^\dagger = 0$, so I don't quite understand your point about things breaking down.
– M.G.
Apr 8 at 21:31
• You are conflating $d/dz$ with $\partial/\partial z = 1/2(\partial/\partial x - i \partial / \partial y)$. Also, $\partial/\partial \bar{z} = 1/2(\partial/\partial x + i \partial / \partial y)$ and both are a consequence of the change of variables $z = x + i y$ and $\bar{z} = x - i y$. In particular, you can apply $\partial / \partial z$ to any function on the complex plane with existing partial derivatives. Likewise for $\partial/ \partial \bar{z}$. OTOH, $df(z)/dz$ exists iff $\partial f(z) / \partial \bar{z} = 0$ (in a nbhd), and then also $df(z) / dz = \partial f(z) / \partial z$.
– M.G.
Apr 8 at 22:46
• To answer your first comment, the expression for $\partial / \partial Z^\dagger$ follows directly from the given change of coordinates (of the corresponding vector fields). In this case, the change of coordinates is linear, so computing the corresponding differential operators is like computing a dual basis b/c $\partial x_i / \partial x_j = \delta_{ij}$. So, if $(Z, Z^\dagger)^T = U. (z_1, z_2)^T$ for some 2x2 matrix $U$ with bicomplex entries, then $(\partial / \partial Z, \partial / \partial Z^\dagger)^T = (U^{-1})^T (\partial / \partial z_1, \partial / \partial z_2)$. $T$ is transposition.
– M.G.
Apr 8 at 22:57
• As far as conformality for functions holomorphic over commutative algebras is concerned, I believe I have seen such notions in the literature, but off top of my head I don't know if they are like your proposed notion. I'll have to check. BTW, the above manipulations are basic multi-variable analysis / basic differential geometry, so I recommend you to look into that. I tried to find the wiki link with the change of coordinates formulae, but it seems butchered for some reason.
– M.G.
Apr 8 at 23:04

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.