# Does every Riemannian manifold has a local orthonormal divergence free frame of vector fields?

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 $$V=v^i\partial_i$$, 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, $$\text{div} X=0 \iff L_X\text{Vol}_g=0$$ where $$\text{Vol}_g$$ is the Riemannian volume form of $$(M,g)$$.