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I'm trying to translate this theorem, below, into theorems about scalar and vector fields in $\mathbb R^3$:

Theorem: Let $A$ be a star-convex open set in $\mathbb R^n$. Let $\omega$ be a closed $k$-form on $A$. If $k > 1$ and if $\eta$ and $\eta_0$ are two $k-1$ forms on $A$, with $d\eta = d\eta_0 = \omega$, then $\eta - \eta_0 = d\vartheta$ for some $k-2$ form $\vartheta$ on $A$.

If $k = 1$, and if $f$ and $f_0$ are 2 $0$-forms on $A$ with $df = df_0 = \omega$, then $f = f_0 + c$, for some constant $c$.

my attempt: I tried considering the cases for $k = 1,2,3$ and not really seeing this geometrically

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This is done in Bott's book, for example. – Mariano Suárez-Alvarez Apr 26 '12 at 6:55
Unfortunately, I don't have a copy of Bott's book, and won't be able to get a hold of it anytime soon. Just out of curiousity, what is the title of the book, when I do get a hold of it? – mary Apr 26 '12 at 7:06
«Differential forms in algebraic topology», by Bott and Tu. In any case: your question is rather weird: it is not the kind of thing one thinks out of thin air: why do you want to do that translation? – Mariano Suárez-Alvarez Apr 26 '12 at 7:08
Extra practice, and ability to understand important theorems clearly – mary Apr 26 '12 at 7:10
How did you know that you were not going to be able to get a hold of Bott's book before even knowing what the title was? :) – Mariano Suárez-Alvarez Apr 26 '12 at 7:53
up vote 4 down vote accepted
  • $k=1$: So $\omega$ corresponds to a vector field $v_\omega\colon A \to \mathbb R^3$ via $\omega = v_{\omega,1}dx^1 + v_{\omega, 2}dx^2 + v_{\omega,3}dx^3$ and your theorem reads: If $f, f_0\colon A \to \mathbb R^3$ are such that $\nabla f = \nabla f_0 = \omega$, then $f = f_0 + c$ for some constant $c$.
  • $k=2$: Here $\omega$ corrensponds to a vector field $w_\omega\colon A \to \mathbb R^3$ via $\omega = w_{\omega, 1}dx^1 \wedge dx^2 + w_{\omega,2} dx^2\wedge dx^3 + w_{\omega, 3} dx^3 \wedge dx^1$. Your theorem reads: If $v_{\eta}$, $v_{\eta,0}$ are such that $\nabla \times v_{\eta} = \nabla \times v_{\eta_0} = w_\omega$, then $v_\eta - v_{\eta_0} = \nabla f$ for some $f \colon A \to \mathbb R^3$.
  • $k=3$ Here $\omega$ corresponds to a $f_\omega\colon A \to \mathbb R^3$ via $\omega = f_\omega dx^1 \wedge dx^2 \wedge dx^3$ and your theorem reads: If $w_\eta, w_{\eta_0}\colon A \to \mathbb R^3$ are such that $\nabla \cdot w_\eta = \nabla \cdot w_{\eta_0} = f_\omega$, that $w_\eta - w_{\eta_0} = \nabla \times v$ for some $v\colon A \to \mathbb R^3$.

Note that for $\omega = \sum_i v_idx^i$ we have \begin{align*} d\omega &= \sum_i dv_i\wedge dx^i\\\ &= \sum_{i,j} \partial_jv_i dx^j \wedge dx^i\\\ &= (\partial_1 v^2 - \partial_2v^1)dx^1 \wedge dx^2 + (\partial_3 v^1 - \partial_1v^3)dx^3 \wedge dx^1 + (\partial_2 v^3 - \partial_3v^2)dx^2 \wedge dx^3 \end{align*}

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This looks complete, but we havent really done stuff with gradients. Can you please elaborate on this furhter? – mary Apr 26 '12 at 7:09
On what exactly? You haven't done gradients, but differential forms? – martini Apr 26 '12 at 7:29

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