Cfrgtkky

Let $(M,g)$ be a smooth Riemannian manifold, and let $p in M$.

Is there an open neighbourhood $U$ of $p$ that admits an orthonormal frame of divergence-free vector fields?

Edit: At least for dimension $2$, $M$ admits such a local frame if and only if it's flat (See a proof below). I am not sure about higher dimensions.

Edit 2:

It seems this question is addressed here. However, I don't understand the "counting" of the number of equations and variables:

Robert Bryant writes that this is a system of $n$ first-order PDE for $tfrac12n(n{-}1)$ unknowns. I think that the following argument explains this:

The orthonormality conditions $langle X_i,X_j rangle=delta_{ij}$ form $tfrac12n(n{+}1)$ equations, and $text{div}(X_i)=0$ form additional $n$ equations. So, together we have $tfrac12n(n{+}1)+n$ equations in $n^2$ variables, when we express each $X_i$ in terms of some fixed arbitrary frame.

However, we can reduce the system by restricting the discussion in advance to orthonormal frames:

Indeed, given some fixe orthonormal frame $E_i$, we can represent any other orthonormal frame on $U$ as $QE_i$ where $Q:U to text{O}(n)$ is a smooth map. So, we have an "$ text{O}(n)$" freedom to choose orthonormal frames; More explicitly, since $ dim(text{O}(n))=tfrac12n(n{-}1)$, we can take $Q(p)=text{Id}$ and for $q in U$, $Q(q)$ can be expressed in terms of $tfrac12n(n{-}1)$ functions (If $U$ is small enough so $Q(U)$ is contained in a coordinate chart around $text{Id}$).

A proof for the $2$D case:

Suppose that $X,Y$ are orthonormal and divergence-free. Then
$$ langle X,X rangle=1 Rightarrow langle nabla_YX,X rangle=langle nabla_XX,X rangle=0. tag{1}$$

The divergence-free condition means that
$$ 0=text{div}X=text{trace}(nabla X)=langle nabla_XX,X rangle+langle nabla_YX,Y rangle=langle nabla_YX,Y rangle. tag{2}$$

Combining equations $(1),(2)$ we see that
$$ langle nabla_YX,X rangle=langle nabla_YX,Y rangle=0$$

so $nabla_YX=0$. By symmetry, we also have $nabla_XY=0$, so the symmetry of the connection implies $[X,Y]=0$. This in turn implies $X,Y$ can be realized as coordinate vector fields, but since they are orthonormal this means the metric is flat.

Alternatively, we can proceed from $nabla_YX=nabla_XY=0$ as follows:

Differentiating $langle X,Y rangle=0$ via $X$, we get

$$ 0=langle nabla_XX,Y rangle+langle X,nabla_XY rangle=langle nabla_XX,Y rangle. tag{3}$$

Combining this with $langle nabla_XX,X rangle=0$ (see equation $(1)$ again) we deduce $nabla_XX=0$, which together with $nabla_YX=0$ implies $X$ is parallel. By symmetry, $Y$ is also parallel, so we have a parallel frame for $(TM,nabla)$ which implies $g$ is flat.

Comment: We always have a local orthonormal frame;

Furthermore, there are always divergence-free frames: Indeed, every volume form can be locally written as $dx^1 wedge dots wedge dx^n$, for some coordinates $x_i$. The divergence w.r.t this form is the standard one, i.e. if $V=v^ipartial_i$, then $text{div}V=partial_i v^i$, so in particular the coordinate frame $partial_i$ form a divergence-free frame.

We can apply the Gram-Schmidt process on $partial_i$, but I see now reason why the "divergence-free" property should be preserved.

By divergence of a vector field $X$, I refer to the Riemannian notion:

$text{div} X= text{trace}(nabla X)$, where $nabla$ is the Levi-Civita connection of $(M,g)$. Alternatively, $text{div} X=0 iff L_Xtext{Vol}_g=0$ where $text{Vol}_g$ is the Riemannian volume form of $(M,g)$.