Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A function $f: \mathbb{C} \rightarrow \mathbb{C}$ is naturally viewed as mapping $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$. Suppose $f = (u,v)$ is continuously differentiable on $D \subseteq \mathbb{C}$, the the following are equivalent:

  1. the Cauchy Riemann equations $u_x = v_y$, $v_x=-u_y$ hold on $D$
  2. $\frac{df}{dz}$ exists on $D$; that is, $f$ is complex-differentiable on $D$
  3. the differential $df_z$ is complex-linear for each $z \in D$

For example, see pages 50-51 of Theory of Complex Functions by Reinhold Remmert.

Suppose $\mathbb{R}^n$ is given some algebraic structure so that we can multiply vectors. For the sake of discussion say $\mathbb{R}^n = \cal{A}$ and for all $v,w \in \cal{A}$ we have $v*w \in \cal{A}$. We should then define $\cal{A}$-linearity of a real-linear mapping $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ by the additional condition $T(v * w) = v*T(w)$ for all $v,w \in \mathbb{R}^n$. When we consider the matrix $B$ inducing $T$ then the condition of $\cal{A}$-linearity will impose certain equations (call these $\star$) on the components of $B$. If there is a natural theory of differentiation on $\cal{A}$ space then we ought to have some intrinsic concept of $\cal{A}$-differentiability. Moreover, it ought to be the case that continuous differentiability paired with the $\star$-equations on the Jacobian matrix provide sufficient conditions to insure $\cal{A}$-differentiability. These equations would naturally be termed generalized CR equations with respect to $\cal{A}$.

I'm curious is the concept I outline above has been studied systematically for a wider class of algebraic examples. Of course $\cal{A}=\mathbb{C}$ is interesting, but is this idea useful beyond complex variables?

I can calculate the so-called CR-equations from the $\cal{A}$-linearity of the differential for algebras other than $\mathbb{C}$, but for what $\cal{A}$ is the concept of $d/dz$ meaningful for $z$ an $\cal{A}$-valued variable?

Thank you in advance for any insights.

share|improve this question
2  
I think the generalization you are looking for is called Clifford analysis -- en.wikipedia.org/wiki/Clifford_analysis. The Cauchy-Riemann operator of complex analysis is replaced with a -- more general-- Dirac operator. –  Kevin Jul 25 '12 at 22:02
    
Thanks Kevin. That is interesting. However, I wonder if the idea I sketched above is a bit more general than the clifford analysis. If $\cal{A}$ is not a clifford algebra then what is known? –  James S. Cook Jul 26 '12 at 5:20
    
You're right. I probably should have said "One generalization that might interest you is called...". I guess I haven't seen too much on this topic when $\mathcal{A}$ is not a Clifford algebra. One exception is arxiv.org/abs/math/0405471, where differentiability of functions of Cayley-Dickson variables is discussed. –  Kevin Jul 26 '12 at 15:34
    
Great Kevin! The example of Cayley-Dickenson variables is much more in the direction I was looking. Fascinating paper, I think it will take me a while to appreciate the way he has dealt with the nonassociativity. Prop. 2.3 on page 12 is the basic type of result I wanted. I wonder if there are other associative algebras which provide similar results. As best as I can cipher with my first read, the definition of differentiability over the cayley-dickenson algebra $\cal{A}_r$ was given by the condition of $\cal{A}_r$-linearity of the differential (tweaked in the non-associative context) –  James S. Cook Jul 30 '12 at 3:17
    
continuing, is it fair to state that the $\cal{A}$-linearity of the differential is the key feature to abstract? The "natural" theory in the case of Cayley-Dickenson seems to take the $\cal{A}$-linearity of the differential as the defn. of differentiability. –  James S. Cook Jul 30 '12 at 3:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.