Understanding the connection between vector fields $F$ with curl=div=0 and the Cauchy-Riemann equations

Question

Let $F(x,y) = M(x,y)i + N(x,y)j$ be a vector field on the real plane.

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Suppose that Div F = $\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} = 0$

It follows that there exists a function $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ (oft called a stream function) such that:

$M = \frac{\partial g}{\partial y}$ and $N = \frac{-\partial g}{\partial x}$

Now, suppose also that Curl F = $\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = 0$

It follows that there exists a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ (oft called a potential function) such that:

$M = \frac{\partial f}{\partial x}$ and $N = \frac{\partial f}{\partial y}$

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Thus, if $F(x,y) = M(x,y)i + N(x,y)j$ has vanishing curl and divergence, we can see that:

$M = \frac{\partial f}{\partial x}=\frac{\partial g}{\partial y}$

$N = \frac{\partial f}{\partial y}=\frac{-\partial g}{\partial x}$

That is it to say, the components of $F$ are equal to the Cauchy-Riemann Equations.

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This is amazing to me. I really would like to understand what's going on more. I understand that vector fields with vanishing curl and divergence are very important in mathematics. Why should we expect that such vector fields have such a direct connection with complex functions? I have a hard time knowing the exact question I want to ask because I don't know what I don't know.

I also recommend Ian's answer in this post: Is there any intuition or meaning regarding Cauchy-Riemann equations?

Where he explains the Cauchy-Riemann equations as the condition for the derivative of a complex function to act as multiplication of a complex number (and thus a linearization of the complex function).

Answer 1

First of all, let us note that $\mathbb{R}^2 \cong \mathbb{C}$. Let us set $z=x+iy$ and $\bar{z}=x-iy$ Furthermore, let us use the language of differential forms.
We can identify a vector field $F:\mathbb{C} \to \mathbb{C}$ with a 1-form $A$ $$ F \mapsto A=A_z dz+A_{\bar{z}}d\bar{z} $$ Vanishing curl is equal to $$ dA=(\partial_{\bar{z}}A_z-\partial_zA_{\bar{z}})d\bar{z} \wedge dz=0 $$ This implies that there exists some $\phi:\mathbb{C} \to \mathbb{C}$ such that $$ d\phi=\partial_{z}\phi dz+\partial_{\bar{z}}d\bar{z}=A. $$ This property is non-trivial and relies on the fact that $\mathbb{C}$ is simply connected. It is the equivalent to every holomorphic function having a primitive, which you might know from your complex analysis class.
Vanishing divergence is expressed via the Hodge dual as
$$ \star dA=0. $$ This implies that $$ (\star d)d \phi=0 $$ Now, if you dont understand the Hodge dual yet, that is fine - that was some heavy machinery I have used. You can show that this is equal to $$ \Delta \phi=0. $$ That implies that the Laplacian of the real and imaginary part vanish. Again, since you are working on a simply connected domain, that means that the real and imaginary part of $\phi$ are the real and imaginary of some holomorphic function. And therefore, they satisfy the Cauchy-Riemann equations.

In more fancy terms: On a map beween simply connected domains of $\mathbb{C}$, every harmonic maps is holomorphic.
And in the fanciest terms: Furthermore, every such vector field such as $A$ can be decomposed into a divergence free, curl free and harmonic field. This is what you call Hodge decomposition, and it has been worked out pretty extensively for Riemannian surfaces (for which $\mathbb{C}$ is an example).

Answer 2

Beware of the minus sign here:

If $u$ and $v$ satisfy the Cauchy-Riemann equations (the OP's $M$ and $N$),
then the vector field $\vec F = \begin{bmatrix} u \\ -v \end{bmatrix} $ is harmonic.
$\quad$ -- $\mathbb{W}$ikipedia Cauchy-Riemann equations Harmonic vector field.

For example, the function $z^2 = (x + iy)^2 = (x^2 - y^2) + (2xy)i $:
$\vec F = \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix} x^2 - y^2 \\ 2xy \end{bmatrix} \ \ $ has Jacobian $\begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix} \ \ $ -- div $4x$, curl $4y$: not harmonic.
$\vec G = \begin{bmatrix} u \\ -v \end{bmatrix} = \begin{bmatrix} x^2 - y^2 \\ -2xy \end{bmatrix} \ $ has Jacobian $\begin{bmatrix} 2x & -2y \\ 2y & 2x \end{bmatrix} \ $ -- div $0$, curl $0$: harmonic.

(Exercise: plot the vector fields $\vec F $ and $\vec G $.)