A question about incompressible vector field

Question

Let $ X $ be a unit vector field on $\mathbb{R}^2$,with canonical metric $g$ and connection $\nabla$ .Show that if

$$divX=0$$

then $$\nabla X=0$$

I tried:

Let $X=(f,g)$,$Y=(-g,f,)$,then $X,Y$ is a orthonormal frame on $\mathbb{R}^2$.Because $divX=0$,I got: $$g(\nabla_{X} X,X)+g(\nabla_{Y} X,Y)=0$$

Because $X$ is unit vector field,hence $g(\nabla_{X} X,X)=0,g(\nabla_{Y} X,X)=0$,so

$$g(\nabla_{Y} X,Y)=0$$

That means $\nabla_{Y} X=0$.But now I need to prove $\nabla_{X} X=0$,so I should prove

$$g(\nabla_{X} X,Y)=0$$.

I can't get it.

So if you can give me an answer need not follow the above,I will appreciate your help.

Answer 1

I'll complete jflipp's answer here.

Let me assume that $X$ is a unit divergence-free vector field on some surface $(M^2,g)$ (with Riemannian connection $\nabla$); at the end I will specialize to $(M^2,g) = (\mathbb{R}^2,\mathrm{can})$ The Poincare lemma states that any closed differential form is exact, at least locally (in $\mathbb{R}^2$ also globally). For $1$-forms in $2$ dimension, this translates to divergence-free fields and gradient fields.

Thus, we can assume (at least locally) $X^\perp = \nabla h$ for some function $h$ (I'm abusing the notation a bit here), where $(X,X^\perp)$ forms a positive orthonormal frame.

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The following could be done slightly easier in $\mathbb{R}^2$, but it still holds for any manifold $(M^n,g)$.

Using the fact that $\nabla$ is a metric connection, we get \begin{align*} \langle \nabla_V \nabla h, W \rangle & = V \langle \nabla h, W \rangle - \langle \nabla h, \nabla_V W \rangle \\ & = V(W h) - (\nabla_V W) h \end{align*} for any vector fields $V,W$. Since $\nabla$ is also torsion-free (i.e. $\nabla_V W - \nabla_W V = [V,W]$), we may exchange $V$ and $W$ in the expression above, hence $$ \langle \nabla_V \nabla h, W \rangle = \langle \nabla_W \nabla h, V \rangle \qquad ( = \operatorname{Hess} h(V,W), \text{ the Hessian}). $$ Applying this and recalling that $\langle \nabla h, \nabla h \rangle = |X|^2 = 1$, \begin{align*} \langle \nabla_{\nabla h} \nabla h, V \rangle & = \langle \nabla_V \nabla h, \nabla h \rangle \\ & = \frac 12 V \langle \nabla h, \nabla h \rangle \\ & = 0 \end{align*} for any vector field $V$. This shows that $\nabla_{\nabla h} \nabla h \equiv 0$, i.e., that integral curves of $\nabla h$ are geodesic.

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Now restrict our attention to $(\mathbb{R}^2,\mathrm{can})$, where geodesic curves are just straight lines. To summarize, we have shown that integral curves of $X^\perp$ are straight lines (even though the reasoning was local, the conclusion is global). Since these curves are not allowed to intersect, they have to be parallel lines, i.e., $X^\perp$ has a fixed direction. Recalling that $X$ is a continuous unit vector field, we conclude that $X$ is constant ($\nabla X \equiv 0$).

Answer 2

I have been thinking about your question for a while, but couldn't find an answer in neat, closed form. Anyway, here's an attempt.

We work on $\mathbb R^2$. I denote the coordinates on $\mathbb R^2$ by $(u,v)$, and I'll use row vectors. So we can dispose of the "fancy" notation from Riemannian geometry, and the metric tensor g is the usual dot product, and the covariant derivative is the total derivative. More specifically, if we put $$ DX := \begin{pmatrix}f_u & g_u \\ f_v & g_v \end{pmatrix},$$ then for some vector field $Z = (r, s)$ we have $$\nabla_Z X = Z\ DX.$$

So, the statement $\nabla X \equiv 0$ is equivalent to $f_u \equiv f_v \equiv g_u \equiv g_v \equiv 0$, or in other words, $X$ is constant.

Now we use the condition $div\ X = 0$. By a "well known fact" from vector calculus, this condition implies that the vector field $Y := (-g, f)$ is a gradient field, i.e. there is a smooth function $h:\mathbb R^2 \rightarrow \mathbb R$ such that $Y = grad\ h$.

Now, since $\|X\|_2 \equiv 1$, we have of course $\|Y\|_2 \equiv 1$. With this, the claim $\nabla X \equiv 0$ would follow if the following statement A were true.

"If $q:\mathbb R^w \rightarrow \mathbb R$ is smooth with $\|grad\ q\|_2 \equiv 1$, then $q$ is affine, i.e. $q(u,v) = \alpha + \beta u + \gamma v$ with some constants $\alpha,\beta,\gamma$ and $\beta^2 + \gamma^2 \equiv 1$."

(Note that "q is affine" is equivalent to "$grad\ q$ is constant".) I couldn't come up with a proof of this statement A, but Google found the following. https://mathoverflow.net/questions/110534/harmonic-function-with-gradient-of-constant-norm-in-hyperbolic-3-space I suggest you read through all of this, and then have a look at a comment from Robert Bryant to the first answer. There, he says "There are global results, such as the fact that for flat n -space, the only hypersurfaces for which the 'flow' does not develop singularities are the hyperplanes." So, assuming these global results are true (which I don't doubt if Robert Bryant says that), then our statement A is true, and we have $\nabla X \equiv 0.$

I have the feeling that a proof of statement A requires a serious amount of analysis and differential geometry.