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I'm self-studying multi-variable calculus using MIT's publicly available materials. One of the practice questions for the final exam asks that I determine the truth or falsity of the following statement:

Let $\mathbf{F}$ be a vector field on $\mathbb{R}^3 - \lbrace \mathbf{0}\rbrace$ with $\nabla \times \mathbf{F} = 0$. Then there exists a scalar field on $\mathbb{R}^3 - \lbrace \mathbf{0}\rbrace$ such that $\mathbf{F} = \nabla f$.

Now, as I understand it, this assertion should be false, because $\nabla \times \mathbf{F} = 0$ only implies that $\mathbf{F}$ is conservative on a region if the region in question is star convex, which $\mathbf{R}^3 - \lbrace \mathbf{0} \rbrace$ is not (unless I really don't understand the notion of star convexity). However, to do the exercise well, I need a counterexample, and here I'm having problems.

I am of course familiar with the canonical counterexample in $\mathbf{R}^2$ (and, by extension, in $\mathbf{R}^3$ less the $z$-axis), i.e. $\mathbf{F} = \left(-\frac{y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, 0\right)$. I'm having difficulty extending this to $\mathbf{R}^3 - \lbrace \mathbf{0} \rbrace$, however -- although poking around here and on the Internet generally has led me to suspect that any counterexample is going to be analogous. (The phrase 'de Rham cohomology' keeps popping up....) At any rate, I would appreciate a gentle hint. Am I on the right track? Have I missed something fundamental?

Travis Willse
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solitaireartist
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    The issue is not (star) convexity but simple connectedness, and $\mathbb{R}^3\setminus { 0}$ is simply connected. Try to prove that the integral of $f$ along any path depends only on the endpoints. – Jose27 Mar 03 '15 at 00:15
  • That makes sense. Thanks. – solitaireartist Mar 03 '15 at 00:21

1 Answers1

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So as Jose27 points out, star convexity is not the issue. Instead, I can use Stoke's Theorem, as follows. Let $C$ be a piecewise smooth closed curve bounding an orientable surface $S$. Then we have \begin{align} \oint_C \mathbf{F}\cdot d\mathbf{\alpha} & = \iint\limits_S (\nabla \times \mathbf{F}) \cdot \mathbf{n}dS\\ & = \iint\limits_S 0 dS\\ & = 0 \end{align} Thus $\mathbf{F} = \nabla f$ for some scalar field $f$.

solitaireartist
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