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When $n>2$, the structure equations show that $\mathcal{I}_+$ is involutive, with the Cartan characters of a regular flag being $s_i = 0$ for $i<n{-}2$, $s_{n-2}=1$, $s_{n-1} = n{-}1$, and $s_n = {n\choose2}-n$. Now apply Cartan-Kähler.

When $n>2$, the structure equations show that $\mathcal{I}_+$, which encodes a system of $n{+}1$ first-order equations on the orthonormal coframing $X$, is involutive, with the Cartan characters of a regular flag being $s_i = 0$ for $i<n{-}2$, $s_{n-2}=1$, $s_{n-1} = n{-}1$, and $s_n = {n\choose2}-n$. Now apply Cartan-Kähler.

I expect that, in dimension $n>3$, the problem is so underdetermined that it is always locally solvable, though I have not yet carried out the analysis. However, see below, where I do complete the analysis in the real-analytic case.

Remark: It is interesting to note that, if $(M^3,g)$ is real-analytic and possesses a real-analytic orthonormal frame field $X = (X_1,X_2,X_3)$ where each of the $X_i$ are divergence free, then $(M,g)$ can be isometrically embedded as a hypersurface in a Calabi-Yau surface $S$ in such a way that $X_i = I_i(N)$ where $N$ is the oriented unit normal and $I_1$, $I_2$, and $I_3$ are the orthogonal parallel complex structures that define the Calabi-Yau structure.

I expect that, in dimension $n>3$, the problem is so underdetermined that it is always locally solvable, though I have not yet carried out the analysis. However, see below, where I do complete the analysis in the real-analytic case.

As Raziel wrote, the local question is whether one can find a local basis of orthonormal vector fields that are divergence free.

I expect that, in dimension $n>3$, the problem is so underdetermined that it is always locally solvable, though I have not yet carried out the analysis.

1. When $n=2$, local solutions exist if and only if the metric is flat.

2. When $n=3$, local solutions always exist if the metric is real-analytic. Moreover, the local solutions depend on 2 arbitrary functions of two variables. If the metric is real-analytic and the scalar curvature is positive, then every local solution is also real analytic.

3. When $n>3$, local solutions always exist if the metric is real-analytic, and the local solutions depend on ${n\choose2}{-}n{-}1$ functions of $n$ variables.

A local orthonormal coframing $X = (X_1,\ldots,X_n)$ defined on an open set $U\subset M$ is simply a section of $F$ over $U$, and it satisfies $X^*\omega_i = \xi_i$ where the $\xi_i$ are the $1$-forms on $U$ dual to the $X_i$. The volume form of the metric is, up to a sign, the wedge product of the $\xi_i$, and so the condition that the $X_i$ be divergence free is that $$ \mathrm{d}\left(\xi_1\wedge\cdots\wedge\widehat{\xi_i}\wedge\cdots\wedge\xi_n\right) = 0 $$ for all $i = 1,\ldots,n$. In other words, defining $$ \Omega_i = (-1)^{i-1}\,\omega_1\wedge\cdots\wedge\widehat{\omega_i}\wedge\cdots\wedge\omega_n\,, $$ we are requiring that $X^*\left(\mathrm{d}\Omega_i\right) = 0$, i.e., that the image of the section $X$ in $F$ should be an integral manifold of the differential ideal $\mathcal{I}$ on $F$ generated by the $n$ $n$-forms $\mathrm{d}\Omega_i$.

Unfortunately, $\mathcal{I}$ is not involutive. However, it turns out that $\mathcal{I}$ can be enlarged to an ideal $\mathcal{I}_+$ as follows: For $i<j$, define $$ \Omega_{ij} = (-1)^{i+j-1}\,\omega_1\wedge\cdots\wedge\widehat{\omega_i}\wedge \cdots\wedge\widehat{\omega_j}\wedge\cdots\wedge\omega_n\,, $$ and set $\Omega_{ii}=0$ while $\Omega_{ji}=-\Omega_{ij}$. Now define the $(n{-}1)$-form $$ \Upsilon = \phi_{ij}\wedge\Omega_{ij}\,. $$ It is not hard to show that $\mathrm{d}\Omega_i = \pm\omega_i\wedge\Upsilon$, and, from this, one concludes that $X_i$ is divergence free for all $i$ if and only iff $X^*(\Upsilon)=0$. Thus, we can let $\mathcal{I}_+$ be the differential ideal generated by $\Upsilon$ and look for integral manifolds of this ideal instead.

When $n=2$, $\Upsilon = 2\phi_{12}$, so $\mathrm{d}\Upsilon = 2R_{1212}\,\omega_1\wedge\omega_2 = 2K\,\mathrm{d}A$, so there are no integral manifolds unless $K=0$.

When $n>2$, the structure equations show that $\mathcal{I}_+$ is involutive. Now apply Cartan-Kähler.

As Raziel wrote, the local question is whether one can find a local basis of orthonormal vector fields that are divergence-free.

I expect that, in dimension $n>3$, the problem is so underdetermined that it is always locally solvable, though I have not yet carried out the analysis. However, see below, where I do complete the analysis in the real-analytic case.

1. When $n=2$, local solutions exist if and only if the metric is flat.

2. When $n=3$, local solutions always exist if the metric is real-analytic. Moreover, the local solutions depend on 2 arbitrary functions of two variables. If the metric is real-analytic and the scalar curvature is positive, then every local solution is also real analytic.

3. When $n>3$, local solutions always exist if the metric is real-analytic, and the local solutions depend on ${n\choose2}{-}n$ functions of $n$ variables.

A local orthonormal coframing $X = (X_1,\ldots,X_n)$ defined on an open set $U\subset M$ is simply a section of $F$ over $U$, and it satisfies $X^*\omega_i = \xi_i$ where the $\xi_i$ are the $1$-forms on $U$ dual to the $X_i$. The volume form of the metric is, up to a sign, the wedge product of the $\xi_i$, and so the condition that the $X_i$ be divergence free is that $$ \mathrm{d}\left(\xi_1\wedge\cdots\wedge\widehat{\xi_i}\wedge\cdots\wedge\xi_n\right) = 0 $$ for all $i = 1,\ldots,n$. In other words, defining the $(n{-}1)$-forms $$ \Omega_i = (-1)^{i-1}\,\omega_1\wedge\cdots\wedge\widehat{\omega_i}\wedge\cdots\wedge\omega_n\,, $$ we are requiring that $\mathrm{d}\left(X^*\Omega_i\right)=X^*\left(\mathrm{d}\Omega_i\right) = 0$, i.e., that the image of the section $X$ in $F$ should be an integral manifold of the differential ideal $\mathcal{I}$ on $F$ generated by the $n$ $n$-forms $\mathrm{d}\Omega_i$.

Unfortunately, $\mathcal{I}$ is not involutive. However, it turns out that $\mathcal{I}$ can be enlarged to an ideal $\mathcal{I}_+$ as follows: For $i<j$, define the $(n{-}2)$-forms $$ \Omega_{ij} = (-1)^{i+j-1}\,\omega_1\wedge\cdots\wedge\widehat{\omega_i}\wedge \cdots\wedge\widehat{\omega_j}\wedge\cdots\wedge\omega_n\,, $$ and set $\Omega_{ii}=0$ while $\Omega_{ji}=-\Omega_{ij}$. Now define the $(n{-}1)$-form $$ \Upsilon = \phi_{ij}\wedge\Omega_{ij}\,. $$ It is not hard to show that $\mathrm{d}\Omega_i = \pm\omega_i\wedge\Upsilon$, and, from this, one concludes that $X_i$ is divergence free for all $i$ if and only iff $X^*(\Upsilon)=0$. Thus, we can let $\mathcal{I}_+$ be the differential ideal generated by $\Upsilon$ (i.e., the exterior ideal generated by $\Upsilon$ and $\mathrm{d}\Upsilon$) and look for integral manifolds of this ideal instead.

When $n=2$, $\Upsilon = 2\phi_{12}$, so $\mathrm{d}\Upsilon = 2R_{1212}\,\omega_1\wedge\omega_2 = 2K\,\mathrm{d}A$, so there are no sections $X$ that are integral manifolds unless $K=0$. (When $K=0$, of course, the sections that are integral manifolds of $\phi_{12}$ are exactly the parallel sections.)

When $n>2$, the structure equations show that $\mathcal{I}_+$ is involutive, with the Cartan characters of a regular flag being $s_i = 0$ for $i<n{-}2$, $s_{n-2}=1$, $s_{n-1} = n{-}1$, and $s_n = {n\choose2}-n$. Now apply Cartan-Kähler.