11
$\begingroup$

When I first saw the Cauchy-Riemann differential equations they remind me on the conditions for the curl of a function to be zero.

Here some notation I will use: $$\frac{\partial f}{\partial x} = f_x$$

As the curl is: $$\nabla \times \begin{pmatrix} u \\ v \\w \end{pmatrix} = \begin{pmatrix} w_y - v_z \\ u_z - w_x \\ v_x- u_y \\ \end{pmatrix}$$ For the curl to be zero we need \begin{align*} w_y &= v_z\\ u_z &= w_x\\ v_x &= u_y\\ \end{align*} The Cauchy Riemann differential equations are \begin{align*} u_x &= v_y\\ u_y &= -v_x\\ \end{align*} Ok they don't look so similar, but I still think that there is connection between functions with curl $0$ and holomorphic functions, as both are having anti derivatives, and hence path integrals only depends on the start and endpoint, not on the path itself.

Is my Intuition wrong and there is no connection? Or is there one?

$\endgroup$
0

2 Answers 2

12
$\begingroup$

The relationship here is one that comes out in geometric calculus. There, we consider the 2d plane as having 3 kinds of objects: scalars, vectors, and bivectors, which serve the role of the imaginary of complex analysis.

Here's how this works: we have a product of vectors such that $e_x e_y = -e_y e_x$, like the cross product, but $e_x e_x = e_y e_y = 1$, like the dot product. This "geometric product" produces a bivector, which represents an oriented plane, just as vectors represent oriented lines or directions.

Call $i = e_x e_y$, and let $f = u + iv$, in analogy to complex analysis. The condition $\nabla f= 0$ reproduces the Cauchy-Riemann condition. Explicitly, this is

$$\begin{align*} \nabla f &= (e_x \partial_x + e_y \partial_y )f \\ &= e_x \partial_x u + e_x e_x e_y \partial_x v + e_y \partial_y u + e_y e_x e_y \partial_y v \\ &= e_x \partial_x u + e_y \partial_x v + e_y \partial_y u - e_x \partial_y v \\ &= e_x (\partial_x u - \partial_y v) + e_y (\partial_y u + \partial_x v)\end{align*}$$

This is a well-known result: that $\nabla$ corresponds more to $\partial/\partial \bar z$ than it does with $\partial/\partial z$. What's interesting is that the integrability condition of $\nabla f = 0$ applies in higher dimensions also, even when the notion of complex differentiation is no longer clear.

To find the relationship between $f$ and a vector field $F$, multiply by $e_x$ on the right so $F = f e_x = u e_x - v e_y = U e_x + V e_y$, with $V = -v$. This generates the integrability condition

$$\nabla F = (\partial_x U + \partial_y V) + i (\partial_x V - \partial_y U)$$

The scalar part is the divergence and the bivector part is the curl. Both of these must be zero to satisfy the same integrability condition as the Cauchy-Riemann condition.

Edit: you might find that the vector field has a negative $y$-component compared to the complex function, but this is also a known result when converting between, say, velocity fields in 2d and complex functions. Geometric calculus only helps prove it mathematically.

$\endgroup$
3
  • 1
    $\begingroup$ Is the "geometric product" you mentioned the wedge product? If this is the case: Why is $e_xe_x=1$ and not zero? $\endgroup$ Jan 19, 2014 at 23:30
  • 1
    $\begingroup$ No, the geometric product unifies the wedge and dot products. Strictly speaking, the geometric product of two vectors produces both a scalar term and a bivector term; these can be thought of as components of a general "multivector". $\endgroup$
    – Muphrid
    Jan 20, 2014 at 5:49
  • 1
    $\begingroup$ Thanks for your clarification. I have also found the definition for the geometric product at en.wikipedia.org/wiki/Geometric_product $\endgroup$ Jan 20, 2014 at 11:26
7
$\begingroup$

An integral $\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,dt$ can be written as the sum of a real line integral and an imaginary line integral. The vector fields showing up in these line integrals are $(u,-v)$ and $(v,u)$ where $f =u+iv$. The Cauchy-Riemann equations then say that these two vector fields have curl zero.

$\endgroup$
2
  • 2
    $\begingroup$ +1, but I think the part up to and including "Here is what I meant to say" should be removed. The comments to which you refer no longer exist, and anyway it's better to have a clean, correct answer here. $\endgroup$ Jun 24, 2013 at 14:09
  • $\begingroup$ @ˈjuː.zɚ79365 ok now? $\endgroup$
    – BCLC
    Aug 9, 2018 at 0:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .