# Map with prescribed Jacobian

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. :)

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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.
This is a nice answer, though somehow tautological. Also notice that we are not allowed to take derivatives of the matrix $M$ since it is assumed to depend only continuously on $x.$ – A. Lerario Jul 17 '12 at 15:52
Why is this "somehow tautological"? Also, the derivatives of $M$ do exist in the weak or distributional sense and the answer remains valid using that. – Deane Yang Jul 17 '12 at 16:07
Maybe another thing to notice (maybe its obvious) is that $M$ is invertible on all of $U$. So if such an f exists, the inverse function theorem guarantees that it is a local $C^{1}$ diffeomorphism. – Brian Trundy Jul 17 '12 at 16:07
So if such an $f$ exists, there is (at least locally) a $C^{1}$ map $g$ satifying $Dg(x)=M^{-1}(f(x))$. This suggests that there should be a further restriction on the components of $M$ (Using Cramer's Rule to get the components of $M^{-1}$ in terms of the components of $M$) in terms of weak/distributional derivatives. – Brian Trundy Jul 17 '12 at 16:56