Coordinates Free Cauchy-Riemann equations

Question

Cauchy-Riemann equations can be seen as a system of 2 PDEs for two functions on the plane:

$$ \begin{align*} L_{\frac{\partial}{\partial x}}(u) &= L_{\frac{\partial}{\partial y}}(v) \\ L_{\frac{\partial}{\partial y}}(u) &= -L_{\frac{\partial}{\partial x}}(v) \end{align*} $$

where $L_X$ denotes the Lie derivative along vector field $X$. Since the vector fields $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ commutes, we have natural integrability conditions:

$$ \begin{align*} \Delta u &= 0\\ \Delta v &= 0, \end{align*} $$

and we know that there exists smooth (even analytic) solutions.

I am surprised that I did not found any results about the natural generalisation of those equations, namely:

$$ \begin{align*} L_A(u)&=L_B(v)\\ L_B(u)&=-L_A(v) \end{align*} $$

where $A$ and $B$ are any linearly independant smooth vector fields on $\mathbb{R}^2$. We can also derive integrability conditions but we obtain two complicate elliptic PDEs and it is not trivial that a solution exists.

Question: Is there references treating this equations? and giving conditions for the existence of $C^{\infty}$ smooth solutions ?

score 5 · Accepted Answer · answered Oct 01 '20 at 19:24

These are the Cauchy-Riemann equations, but with respect to a different complex structure on the plane.

In more detail: the vector fields $A$ and $B$ uniquely determine a Riemannian metric on $\mathbb R^2$, by declaring $(A,B)$ to be an orthonormal frame. For any Riemannian metric in $2$d, in a neighborhood of each point there exist isothermal coordinates, that is, smooth coordinates $(x,y)$ in which the metric has the form $f(x,y)^2(dx^2 + dy^2)$ for some smooth positive function $f$. If necessary, we can interchange $x$ and $y$ so that these coordinates determine the same orientation as the frame $(A,B)$.

In these coordinates, both $(A,B)$ and $(f^{-1}\partial/\partial x, f^{-1}\partial/\partial y)$ are oriented orthornormal frames, so they are related (locally) by a rotation depending smoothly on the point: for some smooth function $\theta$, \begin{align*} A &= f(x,y)^{-1}\left(\cos\theta(x,y) \frac{\partial}{\partial x} - \sin\theta(x,y) \frac{\partial}{\partial y}\right),\\ B &= f(x,y)^{-1}\left(\sin\theta(x,y) \frac{\partial}{\partial x} + \cos\theta(x,y) \frac{\partial}{\partial y}\right). \end{align*} Then a little bit of linear algebra shows that the original pair of equations is equivalent to the Cauchy-Riemann equations in these coordinates.

Thanks! If I understand well, there always exists smooth solutions of my equations since they can be transformed onto the classical CR equations by a smooth transformation? — Athena, Oct 02 '20 at 12:34
Thank you very much! Just for curiosity does it change anything if we consider the same equations in 3D (i.e $A$ and $B$ are vector fields in $\mathbb{R}^3$). — Athena, Oct 07 '20 at 06:37
Yes, it changes everything. In that case, the complex vector field $Z = A-iB$ defines a CR structure, and your system of equations is equivalent to the tangential Cauchy-Riemann equations. There's a famous example due to Louis Nirenberg (described in this Wikipedia article) in which there are no non-constant solutions. — Jack Lee, Oct 07 '20 at 14:51

Coordinates Free Cauchy-Riemann equations

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