Vtyjkyuk

Let $D$ be the unit $n$-ball (for concreteness). Let $betainOmega^1(D;R^n)$ be an $R^n$-valued one-form, having full rank (viewed as a section of $T^*Dotimes R^n$). Under what conditions on $beta$, does there exist a section $Q$ of $SO(n,R)$ (over $D$), such that $Qcircbeta$ is closed (hence exact)?

The question is non-trivial for the following reason: if there exist such $Q$ and an $f:Dto R^n$, such that $df = Qcircbeta$, then $beta^Tcircbeta = df^Tcirc df$, 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.

This is really a question of computing the curvature of the Levi-Civita connection of the Riemannian metric $g = beta^Tcircbeta$. Thus, what one needs to do is first solve the equations
$$
mathrmdbeta = -thetawedgebetaqquadtextandqquad theta^T+theta=0
$$
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
$$
Theta = mathrmdtheta + thetawedgetheta.
$$
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 $mathrmSO(n)$ such that $Qbeta$ is closed, say, equal to $mathrmdx$ for some $mathbbR^n$-valued function on the ball, then one sees that one must have $theta = Q^-1mathrmdQ$, which implies $Theta equiv 0$.

Sufficiency follows since, if $Thetaequiv0$, then the overdetermined equation $theta = Q^-1mathrmdQ$ can be solved for $Q$, uniquely up to left translation by a constant element of $mathrmSO(n)$, and then $Qbeta$ will be closed.

Note however, that, while $theta$ is found by solving a system of linear algebraic equations (whose coefficients depend on $beta$ and $mathrmdbeta$), 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 $Thetaequiv0$ is asking for too much. (By the way, the condition $Thetaequiv0$ is, of course, exactly the condition that the Riemann curvature tensor of the metric $g$ be identically zero.)