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I'm interested in continuous conformal maps $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ that are ONLY angle preserving (not necessarily orientation preserving). I would like to write down a generic system of PDEs such that ALL conformal maps must obey this PDE (or maybe some other suitable functional equation).

It's a well known fact that all holomorphic functions are conformal and a holomorphic function can basically be characterized as a map $D: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with two components $u(x,y),v(x,y)$ such that the cauchy riemann equations hold.

$$ \frac{\partial u}{\partial x} = \frac{ \partial v}{\partial y} \\ \frac{\partial u}{\partial y} = - \frac{ \partial v}{\partial x} $$

But not ALL conformal maps are holomorphic... namely if we move that minus sign to the upper equation we would get an "anti holomorphic" cauchy riemann equation. (See here for an explanation).

So there are at least two different systems of two PDEs whose general solutions result in conformal maps.

Now I haven't seen any result to suggest that harmonic functions are conformal but they certainly seem related and moreover both holomorphic and antiholomorphic functions are harmonic.

So that leads us to the question, can we some characterize ALL conformal maps by a single system of PDEs?

An attempt at an approach:

We try to do this the ugly way... Suppose we have two parametrized curves $\gamma_1 = (\gamma_{1,x}(t), \gamma_{1,y}(t))$ and $\gamma_2 = (\gamma_{2,x}(t), \gamma_{2,y}(t))$

Now suppose these intersect at $t = t_0$ we can then discuss the angle between the curves as

$$ \theta_{\text{original}} = \left| \tan^{-1} \left( \frac{\frac{d\gamma_{2,y}}{dt}}{ \frac{d\gamma_{2,x}}{dt} } \right) - \tan^{-1} \left( \frac{\frac{d\gamma_{1,y}}{dt}}{ \frac{d\gamma_{1,x}}{dt} } \right) \right| @ t = t_0 $$

Now we consider some arbitrary map on $\mathbb{R^2} \rightarrow \mathbb{R^2}$ consisting of two components $u(x,y), v(x,y)$ where the $u$ is our x coordinate and $v$ is our y coordinate.

So we talk about the angle between the two curves after applying this map and set the two things equal to each other. So that angle is given as:

$$ \theta_{\text{new}} = \left| \tan^{-1} \left( \frac{\frac{d v\left( \gamma_{2,x}, \gamma_{2,y} \right) }{dt}}{ \frac{d u\left( \gamma_{2,x}, \gamma_{2,y} \right)}{dt} } \right) - \tan^{-1} \left( \frac{\frac{d v\left( \gamma_{1,x}, \gamma_{1,y} \right)}{dt}}{ \frac{d u\left( \gamma_{1,x}, \gamma_{1,y} \right) }{dt} } \right) \right| @ t = t_0 $$

which we can expand via the chain rule to find

$$ \left| \tan^{-1} \left( \frac{\frac{\partial v }{\partial x } \frac{d \gamma_{2,x}}{dt} + \frac{\partial v}{\partial y} \frac{d \gamma_{2,y}} {dt} }{ \frac{\partial u }{\partial x } \frac{d \gamma_{2,x}}{dt} + \frac{\partial u}{\partial y} \frac{ d\gamma_{2,y}}{dt}} \right) - \tan^{-1} \left( \frac{\frac{\partial v }{\partial x } \frac{d \gamma_{1,x}}{dt} + \frac{\partial v}{\partial y} \frac{d \gamma_{1,y}} {dt} }{ \frac{\partial u }{\partial x } \frac{d \gamma_{1,x}}{dt} + \frac{\partial u}{\partial y} \frac{ d\gamma_{1,y}}{dt}} \right) \right|. $$

So a characterization of all conformal curves maps is the following:

$$ \left| \tan^{-1} \left( \frac{\frac{\partial v }{\partial x } \frac{d \gamma_{2,x}}{dt} + \frac{\partial v}{\partial y} \frac{d \gamma_{2,y}} {dt} }{ \frac{\partial u }{\partial x } \frac{d \gamma_{2,x}}{dt} + \frac{\partial u}{\partial y} \frac{ d\gamma_{2,y}}{dt}} \right) - \tan^{-1} \left( \frac{\frac{\partial v }{\partial x } \frac{d \gamma_{1,x}}{dt} + \frac{\partial v}{\partial y} \frac{d \gamma_{1,y}} {dt} }{ \frac{\partial u }{\partial x } \frac{d \gamma_{1,x}}{dt} + \frac{\partial u}{\partial y} \frac{ d\gamma_{1,y}}{dt}} \right) \right| = \left| \tan^{-1} \left( \frac{\frac{d\gamma_{2,y}}{dt}}{ \frac{d\gamma_{2,x}}{dt} } \right) - \tan^{-1} \left( \frac{\frac{d\gamma_{1,y}}{dt}}{ \frac{d\gamma_{1,x}}{dt} } \right) \right| $$

Of course this definition sucks because of the dependence on the arbitrary $\gamma_1$ and $\gamma_2$ curves, which we would like to remove somehow. It's not even obvious if $u,v$ is holomorphic (meaning obyeing cauchy riemann) that this equation holds true (perhaps some trig identity is required there).

Sidharth Ghoshal
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  • Just checking: 1. Are you allowing critical points (isolated points where a mapping is not conformal)? 2. Are mappings defined on regions, or is the interest only in "entire" functions? <> Separately, the complex conjugate of an "antiholomorphic" map is holomorphic, which should give a good characterization regardless of the answers to 1. and 2. – Andrew D. Hwang Sep 08 '22 at 16:27
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    the interest is only in entire functions, I'm not opposed to critical points (1). if that makes characterizing them in general easier – Sidharth Ghoshal Sep 08 '22 at 16:30
  • Thanks; is the question, then, to write down a single (system of) PDE whose solutions are precisely holomorphic and conjugate-holomorphic functions, or is there also the question of whether there are other conformal maps of the plane (there aren't)? – Andrew D. Hwang Sep 08 '22 at 16:36
  • The question is the latter as well, I assumed there would be conformal maps that aren’t Holomorphic or antiholomorphic but perhaps that assumption is mistaken? – Sidharth Ghoshal Sep 08 '22 at 16:37
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    it becomes a system, $$ u_x^2 + u_y^2 = v_x^2 + v_y^2 $$ $$ u_x v_x + u_y v_y = 0$$ – Will Jagy Sep 08 '22 at 18:39
  • How did u derive that system from the definition of conformality? when I tried to take a determinant of the jacobian I seemed to have messed it up somewhere so that only the second equation came out naturally but not the first. I've undeleted my rightfully downvoted answer showing my approach. – Sidharth Ghoshal Sep 08 '22 at 20:40
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    I wrote $J^T J = \lambda I $ Certainly possible that the usual falls out if you assume $C^2$ and restrict to points where not too many quantities are zero. – Will Jagy Sep 08 '22 at 21:03
  • do u wanna write that as an answer? Your equations hold true even if $\lambda$ is not constant – Sidharth Ghoshal Sep 09 '22 at 03:44
  • @WillJagy I typed it up as an answer but I think i found a typo in your comment (I guess it depends on how one defines the jacobian), i defined my jacobian as same function along each row, same derivative along each column. – Sidharth Ghoshal Sep 20 '22 at 00:07
  • Oh wait I see a way to equate our solutions if we let $JJ^T = \lambda I_2$ vs $J^T J$ then your equation plops out, I think our systems must be equivalent then since orthogonal matrices by definition have both $JJ^T$ and $J^TJ$ equal to scalar times identity. Still kind of weird though... – Sidharth Ghoshal Sep 20 '22 at 00:11
  • If $AB = t I$ we know $B$ and $A$ are nonsingular, also $BA$ is nonsingular. Next, $BABA = B(AB)A = tBA.$ There is an $(BA)^{-1},$ so $BABA (BA)^{-1} = t BA (BA)^{-1} = tI.$ And $BA = tI$ – Will Jagy Sep 20 '22 at 14:45

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@Will Jagy's approach was the following we can write down conformality as meaning the jacobian matrix is a scalar times orthogonal matrix i.e.

$$ J^T J = \lambda(x,y) I_2 $$

We now expand this

$$ \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial v}{\partial x} \\ \frac{\partial u}{\partial y} & \frac{\partial v}{\partial y} \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} = \begin{bmatrix} \lambda(x,y) & 0 \\ 0 & \lambda(x,y) \end{bmatrix} $$

From here we conclude that that by equating the two lambda components:

$$ \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial v}{\partial x} \right)^2 = \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial v}{\partial y} \right)^2 $$

And that

$$ \frac{\partial u}{\partial x} \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \frac{\partial v}{\partial y} = 0$$

This gives us our system of two PDES for conformality

Sidharth Ghoshal
  • 18,359