I'm trying to translate these theorems, below, into theorems about vector and scalar fields in $\mathbb R^n\setminus\{0\}$, in the case $n = 2$.

First Theorem:

Let $A = \mathbb R^n\setminus \{0\}$, with $n \ge 1$
a) If $k$ is not equal to $n-1$ then every closed $k$-form on $A$ is exact on $A$
b) There is a closed $n-1$-form $\eta_{\text{initial}}$ on $A$ that is not exact. If $\eta$ is any closed $n-1$ form on $A$, then there is a unique scalar $c$ s.t. $\eta - c\eta_{\text{initial}}$ is not exact.

Second Theorem:

Let $A = \mathbb R^n\setminus\{0\}$, with $n > 1$. If $\eta$ is a closed $n-1$ form in $A$, then $\eta$ is exact in $A$ iff the integral $\int_{S^{n-1}}\eta = 0$.

My version:

For the first theorem, we need to work out some computation, I think, that $\eta$ is not closed, with $\eta \ne \sum_{i = 1}^n (-1)^{i-1} f_i\, dx_1 \wedge \cdots \widehat{dx_i} \cdots \wedge dx_n$ with $f_i(x) = x_i/\|x\|^n$.

For the second theorem, we just assume that $\eta$ is exact and Stokes' theorem follows, but how does this help in answering the question, above?

  • $\begingroup$ Hi, I tried to texify your input, please check whether there are mistakes. $\endgroup$ – martini Apr 25 '12 at 9:03
  • $\begingroup$ Thanks for the edit. All is looking good, except we need an etta initial when we have "etta - c*etta initial" $\endgroup$ – mary Apr 25 '12 at 9:12
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    $\begingroup$ In your first theorem you meant to say that there is a unique $c$ such that ... is exact. $\endgroup$ – Olivier Bégassat Apr 25 '12 at 15:15
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    $\begingroup$ You might consider fixing it ^^ $\endgroup$ – Olivier Bégassat Apr 26 '12 at 3:27
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    $\begingroup$ Do you need them proven? Or just translated into statements about vector fields? $\endgroup$ – Jason DeVito Apr 27 '12 at 12:30

so I'll give it a try. We have $n=2$ (as you write in your first line), so $k$-forms for $k \in \{0,1, 2\}$ then.

  • $0$-forms $f$ are functions $f\colon \mathbb R^2 \setminus\{0\}$, i. e. scalar fields.
  • $1$-forms $\omega$ correspond to vector fields $v\colon \mathbb R^2 \setminus \{0\}\to \mathbb R^2$ via $\omega = v_1 dx^1 + v_2 dx^2$.
  • $2$-forms $\eta$ correspont to scalar fields $\phi\colon \mathbb R^2\setminus\{0\} \to \mathbb R$ via $\eta = \phi dx^1 \wedge dx^2$.

Now we want to determine how the exterior derivative is represented by the above isomorphisms. Given a $0$-form $f$ we have \begin{align*} df &= \frac{\partial f}{\partial x^1} dx^1 + \frac{\partial f}{\partial x^2} dx^2, \end{align*} so $df$ corresponds to the vector field $\nabla f$, the usual gradient.

Given a vector field $v$ which, as noted above, correnponds to the $1$-form $\omega_v = v_1 dx^1 + v_2 dx^2$. We have \begin{align*} d\omega_v &= dv_1 \wedge dx^1 + dv^2 \wedge dx^2\\ &=\frac{\partial v_1}{\partial x^1} dx^1 \wedge dx^1+ \frac{\partial v_1}{\partial x^2} dx^2 \wedge dx^1 + \frac{\partial v_2}{\partial x^1} dx^1 \wedge dx^2+ \frac{\partial v_2}{\partial x^2} dx^2 \wedge dx^2\\ &= \left(\frac{\partial v_2}{\partial x^1} - \frac{\partial v_1}{\partial x^2}\right) dx^1 \wedge dx^2 \end{align*} so $d\omega_v$ correnponds to the scalar field $\frac{\partial v_2}{\partial x^1} - \frac{\partial v_1}{\partial x^2}$.

As $\Omega^3(\mathbb R^2\setminus \{0\}) = 0$, we have $d\eta = 0$ for each $2$-form $\eta$.

Your first theorem reads then (the only $k$ possible is $k = 2$, since you forgot to exclude $k = 0$ IMO as there aren't exact $0$-forms besides the zero form and constants are closed): (a) Given a scalar field $f \colon \mathbb R^2\setminus\{0\} \to \mathbb R$ (seen as a 2-form, which is closed as there aren't any $3$-forms), then there is a $v\colon \mathbb R^2\setminus\{0\}\to \mathbb R^2$ with $f = \frac{\partial v_2}{\partial x^1} - \frac{\partial v_1}{\partial x^2}$. (b) There is a $v_{\text{initial}} \colon \mathbb R^2 \setminus \{0\} \to \mathbb R^2$ such that $\frac{\partial v_{\text{initial},2}}{\partial x^1} - \frac{\partial v_{\text{initial},1}}{\partial x^2}=0$ and $v \neq \nabla f$ for all $f \colon \mathbb R^2\setminus\{0\}\to \mathbb R$. Moreover, each $v\colon \mathbb R^2\setminus\{0\} \to \mathbb R^2$ with $\frac{\partial v_2}{\partial x^1} - \frac{\partial v_1}{\partial x^2}=0$ can then uniquely be written as $v = c v_{\text{initial}} + \nabla g$ for $c\in \mathbb R$ and $g \colon \mathbb R^2\setminus\{0\}$ (note that $g$ isn't unique [only up to a constant], but $\nabla g$ is!).

Do you want the second theorem also for $n=2$? Your question isn't clear to me in this point. If so, the theorem reads: Given a $v \colon \mathbb R^2\setminus \{0\} \to \mathbb R^2$ with $\frac{\partial v_2}{\partial x^1} - \frac{\partial v_1}{\partial x^2} = 0$ then $v$ can be written as $\nabla f$ for some $f \colon \mathbb R^2\setminus\{0\} \to \mathbb R$ iff \begin{align*} 0 &= \int_{S^1} \omega_v\\ &= \int_{S^1} v_1dx^1 + v_2dx^2\\ &= \int_0^{2\pi} v_1(\cos \theta, \sin \theta)\sin\theta - v_2(\cos\theta, \sin\theta)\cos\theta\, d\theta \end{align*}


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