Condition on a differential form arising from the theory of elasticity Let $D$ be the unit $n$-ball (for concreteness). Let $\beta\in\Omega^1(D;R^n)$ be an $R^n$-valued one-form, having full rank (viewed as a section of $T^*D\otimes R^n$). Under what conditions on $\beta$, does there exist a section $Q$ of $SO(n,R)$ (over $D$), such that $Q\circ\beta$ is closed (hence exact)?
The question is non-trivial for the following reason: if there exist such $Q$ and an $f:D\to R^n$, such that $df = Q\circ\beta$, then $\beta^T\circ\beta = df^T\circ df$, and the latter is (up to a musical isomorphism) a flat metric on $D$, whose Riemann curvature tensor vanishes. 
So in a sense, I have an answer to my question. What I am looking for is a more explicit condition; in particular, I wonder whether there exists a condition that is linear in $\beta$.
For the curious, this question came up twice in two different contexts in the theory of elasticity.
 A: This is really a question of computing the curvature of the Levi-Civita connection of the Riemannian metric $g = \beta^T\circ\beta$.  Thus, what one needs to do is first solve the equations
$$
\mathrm{d}\beta = -\theta\wedge\beta\qquad\text{and}\qquad \theta^T+\theta=0
$$
for a $1$-form $\theta$ taking values in skew-symmetric $n$-by-$n$ matrices.  The Fundamental Lemma of Riemannian geometry guarantees that there is always a unique solution to this system of linear algebraic equations for $\theta$.  Then one needs to compute the curvature $2$-form
$$
\Theta = \mathrm{d}\theta + \theta\wedge\theta.
$$
Then a necessary and sufficient condition for the stated problem to have a solution $Q$ is that $\Theta$ vanish identically.  
Necessity follows since, if there exists a $Q$ mapping the ball to $\mathrm{SO}(n)$ such that $Q\beta$ is closed, say, equal to $\mathrm{d}x$ for some $\mathbb{R}^n$-valued function on the ball, then one sees that one must have $\theta = Q^{-1}\mathrm{d}Q$, which implies $\Theta \equiv 0$.
Sufficiency follows since, if $\Theta\equiv0$, then the overdetermined equation $\theta = Q^{-1}\mathrm{d}Q$ can be solved for $Q$, uniquely up to left translation by a constant element of $\mathrm{SO}(n)$, and then $Q\beta$ will be closed.
Note however, that, while $\theta$ is found by solving a system of linear algebraic equations (whose coefficients depend on $\beta$ and $\mathrm{d}\beta$), the expression for $\Theta$ is quadratic in the expression for $\theta$, at least when $n>2$.  Thus, asking for a `linear' condition on $\beta$ that detects $\Theta\equiv0$ is asking for too much.  (By the way, the condition $\Theta\equiv0$ is, of course, exactly the condition that the Riemann curvature tensor of the metric $g$ be identically zero.)
