# Do there exist generalized conformal maps that preserve elliptic measure?

Let $D_1$ and $D_2$ be two bounded simply connected Jordan domains in $\mathbb{R}^2$. By Carathéodory's Theorem there exists a homeomorphism $f:\bar{D}_1 \to \bar{D}_2$ such that the restriction $f:D_1 \to D_2$ is conformal. If $E\subset \partial D_1$ is Borel measurable then the harmonic extension of $\mathbf{1}_E$ from $\partial D_1$ to $\bar{D}_1$ is called the harmonic measure of $E$ and is denoted by $\omega(x,E;D_1)$ for $x\in \bar{D}_1$. It is well known that $f$ preserves harmonic measure in the sense that $\omega(x,E;D_1)=\omega(f(x),f(E);f(D_1))=\omega(f(x),f(E);D_2)$. Since $\omega(\cdot,E;D_1)$ is really just the solution to the BVP

$\Delta u=0$ in $D_1$
$u=\mathbf{1}_E$ on $\partial D_1$

we can generalize harmonic measure to other elliptic operators $\mathcal{L}$ besides the Laplacian. This generalization is called $\textit{elliptic measure}$ or $\mathcal{L}\textit{-harmonic measure}$, see Diffusions and Elliptic Operators by Bass. My question is whether given $D_1,D_2,$ and $\mathcal{L}$ as above does there exist an injective "generalized conformal map" from $D_1$ onto $D_2$ that preserves elliptic measure and obeys a first order Cauchy-Riemann type system? I'm interested in finding out whether such maps exist and what their Cauchy-Riemann systems are for constant coefficient operators of the form $\mathcal{L}u=\Delta u+\alpha\cdot \nabla u$. I believe that quasiconformal maps give an affirmative answer for a certain class of elliptic operators but I'm not sure these include what I'm looking for.

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One way to think about defining a conformal map from a Jordan domain $D$ to the unit disk $\mathbb{D}$ is to let $u$ be a suitable multiple of the Green's function in $D$ (for Brownian motion or Laplacian depending on your perspective), let $v=v(z_0)+\int_{gamma} \frac{d}{dn} u(z)|dz|$ where $\gamma$ is a curve from $z_0$ to $z$. Then $f=e^{-(u+iv)}$ is a conformal map from $D$ to $\mathbb{D}$. For other diffusions, you still have a Green's function, but I don't know how you'd define $v$ (or if this is in any way the right way to approach things). –  ShawnD Mar 30 '12 at 17:35
Is it obvious that conformal maps won't preserve the generalized harmonic measure! –  Brian Rushton Dec 7 '12 at 2:21