4

Let $\mathbb{D}^2$ be the closed unit disk, and let $f:\mathbb{D}^2 \to \mathbb{R}^2$ be a harmonic map, satisfying $\det df \neq 0$ almost everywhere on $\mathbb{D}^2$. Suppose also that $$U= \{ p \in \mathbb{D}^2 \, | \, \det(df_p)<0 \, \, \text{and } df_p \text{ is conformal}\}$$

has measure zero in $\mathbb{D}^2$. (By "conformal" here I mean that the two singular values of $df$ are equal.)

Does $\dim_{\mathcal{H}}(U) < 1 $ ? ( $\dim_{\mathcal{H}}(U)$ is the Hausdorff dimension of $U$).

An "easier" question: Is $U$ always finite?

I know that $\dim_{\mathcal{H}}(U) \le 1 $. The question is whether or not strict inequality holds.

($\dim_{\mathcal{H}}(U) \le 1 $ since the domain of conformality is the zero-set of a non-zero real analytic function).

Asaf Shachar
  • 25,967
  • Take $f(z)=\bar{z}$. In this case $U={\mathbb D}^2$. – Moishe Kohan Apr 29 '19 at 20:51
  • The conjugation map $f(x,y)=(x,-y)$ does not satisfy my assumptions. I assumed in advance that the domain where $f$ is conformal and orientation-reversing has measure zero. (Note that in my definition of conformal here I am allowing orientation-reversal, i.e. I only mean conformal in the metric sense). – Asaf Shachar Apr 30 '19 at 06:22
  • I see, I misunderstood the question. – Moishe Kohan Apr 30 '19 at 13:00

1 Answers1

1

For some reason I had hard time parsing your question. First, some terminology: A point $p$ where $f$ has negative Jacobian determinant and is conformal in your sense should be called a point where $f$ is anticonformal as the above conditions are equivalent to $$ \partial f(p)= 0, \bar\partial f(p)\ne 0. $$ (In other words, the function $\bar{f}$ is conformal at $p$ in the traditional sense.) Here $$ \partial= \frac{\partial}{\partial z}, \quad \bar\partial= \frac{\partial}{\partial \bar{z}}. $$ The harmonicity condition for $f$ says $$ \bar\partial \partial f=0, \hbox{in} ~D^2. $$ Now, suppose that $p$ is a point at which $f$ has strictly negative Jacobian. This implies that $$ h= \partial f $$ satisfies $\bar\partial h=0$, i.e. $h$ is holomorphic in a (connected) neighborhood $V$ of $p$. Thus, this function $h$ is either constant in $V$ or has only finitely many zeroes in $V$. In other words, either $V\subset U$ or $V\cap U$ is finite. Since you are assuming that $U$ has empty interior, it follows that $V\cap U$ is finite, i.e. $f$ is anticonformal in a finite subset of $V$. Thus, in view of compactness of $D^2$, your subset $U$ is finite.

Moishe Kohan
  • 111,854
  • Thank you. The only problem that I see is with the boundary. How does your argument work at points on the boundary of the disk? (as far as I can see you can take neighbourhoods around each interior point, but I don't see how this can be applied to boundary points). – Asaf Shachar May 01 '19 at 07:11
  • I asked on this a separate question here: https://math.stackexchange.com/questions/3209301/the-identity-theorem-at-the-boundary-complex-analysis. I appreciate all your help. BTW, I guess that the question could be formulated with $\det df>0$ instead of $\det df < 0$ (i.e. conformal/holomorphic instead of anticonformal ). These two versions are equivalent, since one can always compose the original map $f$ with the map $(x,y) \to (x,-y)$. – Asaf Shachar May 01 '19 at 14:48
  • @AsafShachar: You should be more specific regarding the meaning of harmonicity of $f$ at boundary points: I was assuming that you mean that at each boundary point $p$ the mapping $f$ admits a harmonic extension to a small neighborhood of $p$ in the complex plane. – Moishe Kohan May 01 '19 at 14:58
  • Hmmm... you are right, sorry for not being precise. I really prefer not to assume this in advance. But maybe this follows from the assumption that $f$ is smooth on the closed unit disk (I am assuming that) and harmonic in the interior (in the usual sense)? I am not sure if this comment of yours is relevant here as well...https://math.stackexchange.com/questions/3209301/the-identity-theorem-at-the-boundary-complex-analysis?noredirect=1#comment6604214_3209301 – Asaf Shachar May 01 '19 at 15:47
  • I suggest you revise the question to make it more precise. In particular, would linear measure zero for $U$ be sufficient for your purposes? Otherwise, I think, there are counter-examples, take a look here and see what this construction does for you: http://www.ejournals.eu/pliki/art/8354/ – Moishe Kohan May 01 '19 at 16:25
  • Thank you. I shall try to think some more on what exactly do I need. – Asaf Shachar May 02 '19 at 10:52