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Suppose that $\alpha$ is a non-vanishing 1-form on a 2-dimensional manifold. Why can $\alpha$ locally be written as $\alpha = f \ dg$ for some smooth functions $f$ and $g$?

Balerion_the_black
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1 Answers1

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Given a point $p$ in the manifold, choose a smooth nonvanishing vector field $V$ in a neighborhood of $p$ such that $\alpha(V)\equiv 0$. The straightening lemma implies that there are smooth coordinates $(x,y)$ in a neighborhood of $p$ such that $V = \partial/\partial x$ in these coordinates. In these coordinates, $\alpha = f\,dy$ for some smooth function $f$.

Jack Lee
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    how do we show formally that such $V$ exists? – rmdmc89 Jul 16 '18 at 17:02
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    One way to do it is to choose another $1$-form $\beta$ such that $(\alpha,\beta)$ are linearly independent in a neighborhood of $p$. (For example, $\beta$ can be chosen to have constant coefficients in some smooth coordinate chart.) Then let $(W,V)$ be the local frame for the tangent bundle dual to the coframe $(\alpha,\beta)$, which means in particular that $\alpha(V)\equiv 0$. – Jack Lee Jul 16 '18 at 17:39
  • Oh, I see. One last question: couldn't we say that every $1$-form $\alpha$ in a $3$-dimensional manifold (for example) can be written as $\alpha=fdg$? Similarly to what you did: take linearly independent $\alpha,\beta,\gamma$ (locally) and ${X,Y,Z}$ its dual frame. Then $\alpha(Y)=\alpha(Z)=0$. Choosing coordinates $x,y,z$ with $X=\frac{\partial}{\partial x},Y=\frac{\partial}{\partial y}$, we will have $\alpha=u,dx+v,dy+w,dz$ and, therefore $\alpha=w,dz$, right? – rmdmc89 Jul 16 '18 at 20:37
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    No, if you have more than one vector field, you can't always find coordinates such that they're all coordinate vector fields -- that's only possible if their Lie bracket is zero. In dimensions higher than $2$, a $1$-form can be written locally in the form $\alpha = f,dg$ iff the distribution $\operatorname{Ker}\alpha$ is involutive, as a result of the Frobenius theorem. – Jack Lee Jul 16 '18 at 20:45
  • @Jack Lee: is the neighborhood that you mention in your first sentence (where $\alpha(V) = 0$) necessarily the same neighborhood as we pick later on, on which the coordinates $(x,y)$ exist? – Kamil Mar 19 '20 at 16:26
  • @Kamil: No, you might have to shrink the neighborhood further to get those coordinates. A good example to think about is the $1$-form $d\theta$ on &\mathbb R^2 \setminus {0}$. – Jack Lee Mar 20 '20 at 05:50