# Can any two annuli be sub-conformally mapped to each other?

It's a well known result that two annuli can only be conformally mapped to each other under the condition of the ratio of their inner to outer radii are the same. An explanation of this can be found here.

Now a map $$f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ can be called conformal if $$f = (u(x,y),v(x,y))$$ and obeys the following system of two partial differential equations, the derivation of which is here

$$\left( \frac{\partial u}{\partial x}\right)^2 + \left( \frac{\partial v}{\partial x} \right)^2 = \left( \frac{\partial u}{\partial y}\right)^2 + \left( \frac{\partial v}{\partial y} \right)^2 \\ \frac{\partial u}{\partial x} \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x} \frac{\partial v}{\partial y} = 0$$

So a natural question that arises is, what happens if we "weaken" conformality by instead only asking for maps that obey ONE of these two PDES as opposed to the entire system. Say just:

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

For each arbitrary pair of annuli $$A_1,A_2$$ Is it always possible to find a pair of $$u,v$$ obeying that equation which sends $$A_1$$ to $$A_2$$

Though (weak) conformality is equivalent to this system of differential equations, the individual equations are not conformally invariant, just the system. One can see this for both equations by appropriately pre-composing the map $$\mathbb{R}^2 \to \mathbb{R}^2$$ given by $$(x,y) \mapsto (x,0)$$ with rotations. So considering only one differential equation fails to be coherent with respect to the base's conformal structure, which one would ask for in order for a weakening of conformality to be "natural" in any sense.
A more natural weakening of conformality is so-called "quasi-conformality." In the smooth setting, a map $$f: \mathbb{\Omega} \to \mathbb{C}$$ with $$\Omega \subset \mathbb{C}$$ a domain is said to be $$K$$-quasiconformal if $$Df$$ distorts ratios of lengths of major axes to minor axes of ellipses by at most $$K$$.
All conformal annuli, except $$\mathbb{C}-\{p\}$$ and $$\mathbb{C} - B_{1}(0)$$, admit smooth quasiconformal diffeomorphisms between each other. It's worth mentioning that the questions of optimal quasiconformality constants and corresponding extremal mappings (they're exactly the maps you'd hope for) in this setting are a good window into some of the ideas that underlie classical Teichmüller theory.
• Apologies ahead of time, if this gets tedious: so what is your first claim exactly? is it that if we compose $A: (x,y) \rightarrow (x,0)$ with a rotation that the resultant map will NOT obey the single PDE I pointed out, yet both the rotation and this projection map $A$ both DO obey that PDE? If we spell that out we can compose $A$ with a rotation to yield $(x,y) \rightarrow x ( \cos(\theta), \sin(\theta))$ for some choice of $\theta$ and see that it it still obeys the PDE i singled out. Sep 22, 2022 at 20:37
• The second equation says that the dot product $\langle Df(\partial_x), Df(\partial_y) \rangle = 0$. This is true for $A$, but not true for $A \circ R_{\pi/4}$, where $R_{\pi/4}$ is counter-clockwise rotation by $\pi/4$. Similarly, the first equation does not hold for $A$, but does hold for $A \circ R_{\pi/4}$. Sep 22, 2022 at 20:51
• I think I see the point of quasi-conformality, is it true that if $K=1$ then we recover conformality and then letting $K$ get bigger than that gives us a family of options. Sep 22, 2022 at 21:43
• Yes, this is true. Conformal maps are exactly the $1$-quasiconformal homeomorphisms. It's also a conformally invariant condition, since rotating and scaling ellipses preserves ratios of major-to-minor axes. Another motivation is that a conformal structure allows one to measure ratios of lengths of tangent vectors. The quasiconformality constant $K$ of a map measures how much the map distorts these measurements. Sep 22, 2022 at 21:50