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Let $h_1 \mathrm{d}x_1 + h_2 \mathrm{d}x_2$ be a non-vanishing $1$-form on a $2$-dimensional manifold. Why do locally exist smooth functions $f,g$ with $f\mathrm{d}g= h_1 \mathrm{d}x_1 + h_2 \mathrm{d}x_2$?

I think this should follow from some statement about differential equations (and existence of some integrating factor?) in usual analysis, but sadly in all of mathematics differential equations are probably the topic I know the least about.

Edit: It would suffice to prove the following: let $h_1, h_2: \mathbb{R}^2 \rightarrow \mathbb{R}$ be smooth functions. Then exists for all $x \in \mathbb{R}^2$ a neighborhood $U$ and $f,g_1, g_2$ smooth functions on $U$, such that $h_i=f g_i$ and $\partial_{x_1}g_2= \partial_{x_2}g_1$.

user110071
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1 Answers1

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Do you know the Frobenius theorem? Let $\omega=h_1dx_1+h_2dx_2$, at each point $p\in M$ the equation $\omega=0$ defines a 1-dimensional subspace (a line) of the tangent space $T_pM$, which defines a distribution. Then if $\omega$ satisfies the Frobenius conditon, that is, if there exists a 1-form $\sigma$ such that $d\omega=\sigma\wedge\omega$, the distribution is integrable. That is, there exists a function $g$ such that the equation $dg=0$ defines the same subspaces. Thus $\omega(X)=0$ if and only if $dg(X)=0$, hence $\omega=fdg$.

Note For 1-forms they always satisfy the condition: assume $d\omega=Fdx_1\wedge dx_2$, just let $\sigma=\frac{F(h_2dx_1-h_1dx_2)}{h_1^2+h_2^2}$ we will have $\sigma\wedge\omega=d\omega$. So we get the conclusion.

Geometrically it is obvious: $\omega(X)=0$ defines a subspace (line) at each point. If these lines are integrable, there exists a flow $\Phi: \mathbb{R}\times M\rightarrow M$ such that the tangent vector $\dot\Phi_p(0)$ of the curve $\Phi_p(t)=\Phi(t,p)$ lies on $X$. Suppose $\omega=fdg$, then $\omega(X)=0$ implies $dg(X)=0$, that is, the gradient of $g$ is perpendicular to $X$ at each point. Choose a curve $c$ such that $c=c(s)$ intersects the flow perpendicularly and $c(0)=p$, then $s, t$ defines a local coordinate chart. The function $g=s$ has the desired property.

Xipan Xiao
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  • Is this not just a reformulation of the problem? I still have to find smooth functions $f,g_1, g_2$ such that $fg_1=h_1$, $fg_2=h_2$ and $\partial_{x_2} g_1 = \partial_{x_1} g_2$ everywhere locally, right? But for that this suffices, I don't need the Poincaré Lemma, as such $g_1 \mathrm{d}x_1+ g_2 \mathrm{d}x_2$ would have an antiderivative. My problem is finding such $g_i$ though. – user110071 Dec 11 '14 at 23:50
  • Modified the answer now. – Xipan Xiao Dec 14 '14 at 18:29
  • I thought integrability gives me functions from $\mathbb{R}$ into $M$. I guess by some kind of duality, it could also give me a function the other way around, how does that go exactly? It's okay if you can provide me a link. – user110071 Dec 14 '14 at 22:12
  • http://math.stackexchange.com/questions/1071490/local-expression-for-a-1-form-on-a-surface is a shorter but essentially the same answer. – Xipan Xiao Dec 17 '14 at 17:09