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Let $g$ be a smooth Riemannian metric on the closed $n$-dimensional unit disk $\mathbb D^n$.

Let $f: \mathbb D^n \to \mathbb{R}^n$ be a smooth orientation-preserving immersion, and let $\omega :\mathbb D^n \to \mathbb{R}^n$ be the harmonic map corresponding to the Dirichlet problem imposed by $f$, i.e. $\omega|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$.

Must $d\omega$ be invertible outside a set of measure zero in $\mathbb D^n$?

Here is a complete solution for the case where $g$ is real-analytic:

First, we note that $d\omega$ is singular on as set of positive measure if and only if it is singular everywhere on $\mathbb D^n$: Since $\omega$ must be real-analytic, so is $\det d\omega$, and the zero-set of a real-analytic function which is not identically zero has measure zero.

However $d\omega$ cannot be singular everywhere, since

$$ \int_{\mathbb D^n} \det d\omega = \int_{\mathbb D^n} \det df>0,$$

which finishes the job.

I am not sure how to deal with the case where $g$ is smooth but not real-analytic:

One idea I had is to approximate the smooth $g$ with analytic metrics $g(t)$, and to look at the corresponding harmonic maps $\omega(t)$: If $g(t)$ converges to $g$ in a sufficiently strong sense, then the $\omega(t)$ should converge to the harmonic map $\omega$ associated with the original metric $g$. However, I am not sure that this gets us anywhere, since the limiting map may have many more singular points compared to the $\omega(t)$. (The rank can fall in the limit).

Other observations:

We know that near $\partial \mathbb{D}^n$ $\text{rank}(d\omega) \ge n-1$ because $f$ is an immersion. Perhaps there is hope that somewhere near the boundary, $d\omega$ will be invertible.

What happens if we assume that $\text{rank}(d\omega) = n-1$ on a neighbourhood of $\partial \mathbb D^n$? Does this lead to a contradiction?

The following example shows that, at least on all $\partial \mathbb D^n$, the rank can be exactly $n-1$, if we only consider the "local problem": Set

$$\omega(r,\theta)=((r+r^{-1})\cos \theta,(r+r^{-1})\sin \theta)).$$ $\omega$ is harmonic on $\mathbb D^2 \setminus \{(0,0)\}$ and agrees with the immersion $(2x_1,2x_2)$ on $\partial \mathbb D^2$, but its rank on $\partial \mathbb D^2$ is $1$ everywhere.

$\omega$ is not really a counter-example to the question even on $\mathbb D^2 \setminus \{(0,0)\}$, since $d\omega$ is invertible in the interior.

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    $\begingroup$ Have you looked at the paper of Schoen and Uhlenbeck? I believe it answers your question in the affirmative. $\endgroup$
    – Deane Yang
    Jun 6, 2016 at 18:03
  • $\begingroup$ If you mean this paper: projecteuclid.org/euclid.jdg/1214436923, as far as I can see, it only considers aspects of continuity and smoothness. (It does not address the issue of invertibility of the differential). $\endgroup$
    – Asaf Shachar
    Jun 6, 2016 at 20:01
  • $\begingroup$ I think it might be worth digging through it. My guess is that along they way, some of what they do can be used to study your question. $\endgroup$
    – Deane Yang
    Jun 7, 2016 at 3:24
  • $\begingroup$ @AsafShachar Yes I realized it only showed that $d\omega$ couldn't vanish on a set of positive measure. $\endgroup$
    – RBega2
    Feb 17, 2019 at 14:10
  • $\begingroup$ @AsafShachar As you point out in your edit, you are missing a step in the analytic case. The map $\omega(r,\theta)=((r+r^{-1})\cos\theta, (r+r^{-1} \sin \theta)$ is harmonic on $\mathbb{D}^2\backslash \{(0,0\}$ and agrees with the (harmonic) immersion $(2x_1,2 x_2)$ on $\partial \mathbb{D}^2$ but is not an immersion on $\partial \mathbb{D}^2$. This means you are going to have to use that $\omega$ is harmonic on the whole disk. $\endgroup$
    – RBega2
    Feb 17, 2019 at 14:37

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