Recently I came up with the following problem.


Suppose $U$ is an open subset of $\mathbb{R}^n$ and we are given a continuous map $M:U\to GL(n;\mathbb{R})$. Does anybody know if there are conditions for the existence of a $C^1$ map $f:U\to \mathbb{R}^n$ such that $$Jf(x)=M(x)\quad \forall x \in U$$
(here $Jf(x)$ is the Jacobian matrix of $f$ at $x$)?


I know this question is somehow a little bit general; I will really appreciate even a reference for something related to this problem. :)



Map with prescribed Jacobian
Suppose we can find such an $f$. If $f$ is continuously differentiable, then the i'th row and j'th column of the Jacobian (in the standard basis) is the j'th partial derivative of the the i'th component of $f$. Indeed, the components of $f$ are continuously differentiable if and only if $f$ is. So it suffices to consider each of the components of $f$ along with its corresponding row in M(x) seperately - i.e. we want $\nabla f\_{i}=M\_{i}$ where $M\_{i}$ is the i'th row of $M$. So the ith row of $M$ is given by a scalar potential - e.g. if $M$ is (continuously) differentiable $\omega\_{i}=\sum\_{j}m\_{i,j}dx^{j}$ is exact, where $M(x)=(m\_{i,j}(x))$. A neccesary condition for this to be true is $\partial\_{k}m\_{i,j}=\partial\_{j}m\_{i,k}$ for all $i,j,k\in\{1,..n\}$. The sufficiency of this condition depends on the topology of $U$. See <http://en.wikipedia.org/wiki/Closed_and_exact_differential_forms>.


