5
$\begingroup$

In this question, a “(closed oriented plane) curve$\Gamma$ will mean a continuous map $f \colon \mathbb{U} \to \mathbb{C}$ where $\mathbb{U} := \{z\in\mathbb{C} : |z|=1\}$ is the unit circle, modulo right-composition by orientation-preserving homeomorphisms $\varphi\colon\mathbb{U}\to\mathbb{U}$. Any such $f$ (i.e. any of the $f\circ\varphi$ defining the curve) will be called a “parametrization” of the curve $\Gamma$.

I'm willing to add any reasonable regularity conditions on $\Gamma$ if they help in answering the question, e.g., piecewise $C^1$ or even $C^\infty$.

Say that a curve $\Gamma$ is harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ (see above) such that the Fourier coefficients $(c_k(f))_{k\in\mathbb{Z}}$ of $f$ are zero for all $k<0$: this then allows us to see $f$ as the restriction to $\mathbb{U} = \partial\Delta$ of a continuous function $F$ on the closed unit disk $\overline{\Delta}$ that is holomorphic on the open unit disk $\Delta$ (namely, $F$ is the Poisson integral of $f$, or equivalently $F(z) = \sum_{k=0}^{+\infty} c_k z^k$; see also this question).

More generally, if $k_0\in\mathbb{Z}$ say that $\Gamma$ is $k_0$-harmonically parametrizable iff it admits a parametrization $f \colon \mathbb{U} \to \mathbb{C}$ that $c_k(f)=0$ for all $k<k_0$. (So “harmonically parametrizable” means “$0$-harmonically parametrizable”, and the larger $k_0$ the stronger the condition.)

For example, if $\Gamma$ is a positively-oriented, sufficiently regular (I think “rectifiable” suffices), Jordan curve, then the Riemann conformal mapping theorem guarantees that we can find $F \colon \Delta \to V$ bijective and conformal, where $V$ is the bounded component of the complement of $\Gamma$, extending continuously to $\partial\Delta = \mathbb{U}$ (assuming $\Gamma$ is sufficiently regular), which shows that $\Gamma$ is harmonically parametrizable.

So, two related questions:

  • Can we find a necessary and sufficient condition on a curve $\Gamma$ (again, assumed sufficiently regular if this is useful) for it to be harmonically parametrizable?

  • What about $k_0$-harmonically parametrizable for $k_0<0$? Can we find, at least, a reasonable sufficient condition analogous to the “Jordan curve” condition for being harmonically parametrizable?

(For $k_0>0$, being $k_0$-harmonically parametrizable seems to say something about the moments of $\Gamma$ for the harmonic measure vanishing, and this is probably less interesting.)

PS: This question seems related to a kind of converse to the one I'm asking (I want to know if there exists a parametrization change which becomes harmonic, that other question is about what happens upon such a change).

$\endgroup$
2
$\begingroup$

First of all, this is a purely topological problem. Let $\gamma:U\to C$ be a curve. Your question is when this curve can be reparameterized so that the new parameterization is by boundary values of an analytic function in the unit disk.

It is necessary that $\gamma$ extends to a topologically holomorphic map of the unit disk to $C$ (topologically holomorphic map, a. k. a. polymersion, is a map which is topologically equivalent to $z\mapsto z^n$ near every point.

This condition is also sufficient. Indeed, once we have a topologically holomorphic map, we can pull back the complex analytic structure to the disk, and then use the Uniformization theorem. So for every topologically holomorphic map $f$ there is a homeomorphism $\phi$ of the disk such that $f\circ\phi$ is holomorphic.

Second, this topological problem is solved in the paper (under some smoothness conditions on the curve):

C. Curley and D. Wolitzer, Btranched immersions of surfaces, Michigan Math. J. 33 (1986) 131-144.

Unfortunately the answer is somewhat complicated and I do not reproduce it here.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.