Points where harmonic functions fail to give a coordinates system Let $\Omega$ be a bounded domain in $\mathbb R^n$ with a smooth boundary and let $g$ be a smooth Riemannian metric on $\Omega$. Let $f_1,f_2,\ldots,f_n$ be non-constant smooth functions on $\partial \Omega$ that are linearly independent from each other. Given any $j=1,2,\ldots,n$, we denote by $u_j$, the unique harmonic function in $\Omega$ (i.e $\Delta_gu=0$ on $\Omega$) with Dirichlet data $f_j$.
Let us consider the set of points $p$ where the set $u_1,\ldots,u_n$ fails to give a coordinate system near $p$ (that is to say, $\det J=0$ where $J_{jk}=\partial_j u_k(p)$). Can we say that there is only a finite number of such points in $\Omega$? If not, can we say that the measure of such points is zero?
 A: To rule out the trivial case that one of the $f_j$ is a nonzero constant, I interpret the linear independence condition as saying that no nontrivial linear combination of the $f_j$ is constant.
The set $\{\det J = 0\}$ is not necessarily finite. Consider for example $u_1 = r\sin(\theta)$ and $u_2 = r^2\sin(2\theta)$ in $B_1 \subset \mathbb{R}^2$, which both vanish on the $x$ axis (hence have parallel gradient there), and have linearly independent boundary data.
In dimension $n = 2$, the set $\{\det J = 0\}$ has measure zero. One can argue as follows: assume not, and let $u,\,v$ be the pair of harmonic functions. Since $u_y + iu_x$ and $v_y + iv_x$ are holomorphic, the sets $\{|\nabla u| = 0\}$ and $\{|\nabla v| = 0\}$ are locally finite, and furthermore, the imaginary part of their ratio,
$$\frac{\nabla^{\perp}u \cdot \nabla v}{|\nabla v|^2},$$
is harmonic. By hypothesis, this function vanishes on a set of positive measure, and by its harmonicity it vanishes identically in $\{|\nabla v| > 0\}$. We conclude in this set that $\nabla u = \lambda(x)\nabla v$. Harmonicity of $u$ and $v$ implies that $\nabla \lambda \cdot \nabla v = 0$, and differentiating the relation $u_x = \lambda v_x$ in $y$, the relation $u_y = \lambda v_y$ in $x$, and subtracting gives $\nabla \lambda \cdot \nabla^{\perp}v = 0$. It follows that $\lambda$ is constant in $\{|\nabla v| > 0\}$ and hence that $u - \lambda v$ is constant, contradicting linear independence of the boundary data.
A: In dimension two, the Rado-Kneser-Choquet theorem explains how to choose boundary data to obtain a non vanishing Jacobian in the interior. Lewy's Theorem shows that harmonic one-to-one mappings have non vanishing jacobians.
J. C. Wood  one page paper called "Lewy's Theorem Fails in Higher Dimensions" gives a counter-example in dimensions larger or equal to three (with an hyperplane where the jacobian cancels). Choosing the boundary data to be the trace of the map on the boundary of the domain of your choice gives you a counter example.
