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?



Points where harmonic functions fail to give a coordinates system
In dimension two, the [Rado-Kneser-Choquet theorem](https://en.wikipedia.org/wiki/Rad%C3%B3%E2%80%93Kneser%E2%80%93Choquet_theorem) explains how to choose boundary data to obtain a non vanishing Jacobian in the interior. [Lewy's Theorem](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-42/issue-10/On-the-non-vanishing-of-the-Jacobian-in-certain-one/bams/1183499319.full) shows that harmonic one-to-one mappings have non vanishing jacobians.


[J. C. Wood](https://www.mscand.dk/article/view/12375/10391) 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.


