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 $F\circ G \notin\mathcal{F}$?

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    $\begingroup$ Note that to form a group you also need inverses. But a conservative vector field, thought of as a map $\mathbb{R}^3\rightarrow \mathbb{R}^3$, isn't generally injective. $\endgroup$ May 22 at 0:55
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    $\begingroup$ In the tltle of the question you are asking if $\mathcal F$ is a group, but in the body you are only asking if it is closed under composition. Notice these are not the same question. $\endgroup$ May 22 at 0:59
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    $\begingroup$ Do you mean a semigroup? $\endgroup$ May 22 at 1:09
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    $\begingroup$ Depending on the interpretation of vector fields it might make no sense in composing them. According to the most sound physical interpretation, the domain $\mathbb R^3$ is not the same thing as the range. If $p$ is a point in the domain of $F$, then $F(p)$ is an element of the tangent space of $\mathbb R^3$ at $p$, which is isomorphic to $\mathbb R^3$ but is not identical to it. $\endgroup$
    – Ruy
    May 22 at 1:41
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    $\begingroup$ You've got a good answer, but separately: A vector field cannot be viewed as a mapping from $\mathbf{R}^3$ to itself in a coordinate-independent way, so composition doesn't make sense. Instead, a vector field is a mapping $\mathbf{F}:\mathbf{R}^3\to\mathbf{R}^3\times\mathbf{R}^3$ whose value at each point $x$ lies in the tangent space $\{x\}\times\mathbf{R}^3$. ("That's the same thing!" you might exclaim. But it's not the same thing: the base space $\mathbf{R}^3$ does not transform the same way under change of coordinates as the tangent spaces.) $\endgroup$ May 22 at 2:43

2 Answers 2


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 $\phi$).


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.


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