Do continuous conservative vector fields on $\mathbb{R}^3$ form a semigroup under composition?

Question

Let $\mathcal{F}$ denote the set of conservative vector fields on $\mathbb{R}^3$ that are continuous. That is $$ \mathcal{F}=\{F:\mathbb{R}^3 \rightarrow \mathbb{R}^3: F \text{ is continuous and } F=\nabla \phi \text{ for some } \phi \in C^1(\mathbb{R}^3) \}.$$

Is $\mathcal{F} $ closed under composition?

I suspect it is not since no property like this is mentioned when one studies conservative vector fields (and we do love algebraic structures which appear naturally in analysis, so certainly if this were a semigroup someone would have made mention of it).

Does anyone have a nice example of $F\in\mathcal{F}$ and $G\in\mathcal{F}$, such that $F\circ G \notin\mathcal{F}$?

Answer 1

How about $F(x,y,z)=(y,x,0)$ and $G(x,y,z)=(0,y,0)$ (they're the gradients of $f(x,y,z)=xy$ and $g(x,y,z)=\frac{y^2}{2}$ respectively), but $(F\circ G)(x,y,z)=(y,0,0)$ is not conservative (the corresponding 1-form is $y\,dx$ on $\Bbb{R}^3$ whose exterior derivative is $dy\wedge dx\neq 0$... i.e the curl of $F\circ G$ is non-zero, so it can't be the gradient of some scalar function $\phi$).

Answer 2

Just to record some things I was thinking when I posted this question, here are a few other examples:

• $f(x,y,z)= xy$ and $g(x,y,z)= \frac{3}{2}x^2+y^2$ $\Rightarrow$ $F := \nabla f = \left<y,x,0 \right>$ and $G:=\nabla g = \left<3x,2y,0 \right>$ $\Rightarrow$ $F \circ G: (x,y,z)\mapsto \left<2y,3x,0\right>$ which is not the gradient of any scalar function $\phi$ [Reason: $\nabla \times (F \circ G) = (3-2)\bf{k} \neq \bf{0}$.]

• $f(x,y,z)= xy$ and $h(x,y,z) = \frac{1}{2}x^2y^2$ $\Rightarrow$ $F := \nabla f = \left<y,x,0 \right>$ and $H := \nabla h = \left<xy^2,x^2y,0 \right>$ $\Rightarrow$ $F \circ H: (x,y,z)\mapsto \left<x^2y,xy^2,0\right>$ which is not the gradient of any scalar function $\phi$ [Reason: $\nabla \times (F \circ H) = (y^2-x^2)\bf{k} \neq \bf{0}$.]

• Similiarly, $H \circ G: (x,y,z)\mapsto \left<(3x(2y)^2,(3x)^2(2y),0\right>=\left<(12xy^2,18x^2y,0\right> $ which is not the gradient of any scalar function $\phi$ [Reason: $\nabla \times (F \circ G) = (36xy-24xy){\bf{k}}=12xy {\bf{k}} \neq \bf{0}$.]

Clearly @peek-a-boo has a simpler (and therefore better!) solution which I have upvoted and accepted.

In case anyone is interested, this question was inspired by this sister question: Function class consisting of gradients of real-valued convex functions.