Holomorphicity for complex, split and dual functions

Question

The set of complex numbers is: $$\mathbb{C}=\left\{a+bi:a,b\in\mathbb{R},i\notin\mathbb{R},i^{2}=-1\right\}$$ The set of split numbers is: $$\mathbb{D}=\left\{a+bj:a,b\in\mathbb{R},j\notin\mathbb{R},j^{2}=1\right\}$$ The set of dual numbers is: $$\mathbb{E}=\left\{a+b\varepsilon:a,b\in\mathbb{R},\varepsilon\notin\mathbb{R},\varepsilon^{2}=0\right\}$$

The first step should be the same for all three, so we'll generalise. Let $\kappa\in\{i,j,\varepsilon\}$ be one of the units, $z=x+y\kappa$ be a variable with $x,y\in\mathbb{R}$ and $f(z)=u(z)+\kappa v(z)$ be a function. We wish for $f$ to be assigned the title holomorphic. Here's the best I understand it:

The Jacobian of $f$ is $$Df=\begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$$ From what I can gather, $f$ will be holomorphic if the Jacobian takes on the form of the matrix representation of a given number. In $\mathbb{C}$ we may use $$a+bi\cong\begin{pmatrix} a & -b\\ b & a \end{pmatrix}$$ Comparing, we see that the equalities to satisfy are $$\frac{\partial u}{\partial x}=a=\frac{\partial v}{\partial y}$$ $$-\frac{\partial u}{\partial y}=b=\frac{\partial v}{\partial x}$$ which is just the Cauchy-Riemann equations. If we transpose, so switch $b$ and $-b$, nothing changes in the conclusion. Now the total derivative of $f$ is $a+bi$, or rather $$\frac{df}{dz}=\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}+i\frac{\partial v}{\partial x}=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}-i\frac{\partial u}{\partial y}$$

Now the split numbers, where we have $$a+bj\cong\begin{pmatrix} a & b\\ b & a \end{pmatrix}$$ We get the following set of equations: $$\frac{\partial u}{\partial x}=a=\frac{\partial v}{\partial y}$$ $$\frac{\partial u}{\partial y}=b=\frac{\partial v}{\partial x}$$ These are Scheffers' conditions, and again the total derivative of $f$ is $a+bj$, so $$\frac{df}{dz}=\frac{\partial u}{\partial x}+j\frac{\partial u}{\partial y}=\frac{\partial v}{\partial y}+j\frac{\partial v}{\partial x}=\frac{\partial u}{\partial x}+j\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}+j\frac{\partial u}{\partial y}$$ If these look right so far, we can get onto the bit that's giving me pause. The dual numbers. Some people seem to use the convention I'll call convention 1. $$a+b\varepsilon\cong\begin{pmatrix} a & b\\ 0 & a \end{pmatrix}$$ Others seem to use what I'll call convention 2. $$a+b\varepsilon\cong\begin{pmatrix} a & 0\\ b & a \end{pmatrix}$$ With the other two, transposing gave the same conditions. Here, they give two different pairs. The main diagonal has been the same for all of these, so we get the same first equation. $$\frac{\partial u}{\partial x}=a=\frac{\partial v}{\partial y}$$ Now, under convention 1, we are told $$\frac{\partial v}{\partial x}=0$$ $$\frac{\partial u}{\partial y}=b\text{ may be any}$$ Under convention 2, we deduce $$\frac{\partial u}{\partial y}=0$$ $$\frac{\partial v}{\partial x}=b\text{ may be any}$$ I don't know where to go. Is one of them actually a more canonical isomorphism, to do with the alignment and set-up of the Jacobian somehow? Or, since they're both valid isomorphisms, do we take the conjunction, setting both of those to $0$, so $b=0$, and finding that $$\frac{df}{dz}=\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$

Answer 1

This is the kind of trouble you can get into when you try to do everything too explicitly and in coordinates. Here is an abstract general definition (it will show us that convention 2 is the correct convention): let $A$ be a finite-dimensional $\mathbb{R}$-algebra. Then we can talk about the derivative of a function $f : A \to A$ at a point $p \in A$, which (if it exists) will be a linear map $df_p : A \to A$.

Definition: $f$ is (left) $A$-differentiable if, for every point $p \in A$, the derivative $df_p$ exists and is given by left multiplication by some $a \in A$.

(We could also consider right multiplication, which would give us right $A$-differentiability. These are the same condition if $A$ is commutative, which is the case here.)

We can translate this into an explicit condition on matrix entries by choosing a basis of $A$, which gives a matrix representation of left multiplication; importantly, these choices are not independent. For example if $A = \mathbb{C}$ and we pick the basis $\{ 1, i \}$ we get the Cauchy-Riemann equations as you've described, and similarly for the split-complex numbers.

So now let $A = \mathbb{R}[\varepsilon]/\varepsilon^2$ be the dual numbers. If we pick the basis $\{ 1, \varepsilon \}$, then multiplication by $a + b \varepsilon$ is given by the matrix

$$a + b \varepsilon \mapsto \begin{bmatrix} a & 0 \\ b & a \end{bmatrix}$$

which is convention 2. Convention 1 comes from picking the basis $\{ \varepsilon, 1 \}$, which is incompatible with how you've chosen to write a function in components, so it's just the wrong convention here.