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This is a problem on an old qualifying exam.

Let $\omega$ be a smooth, closed $m-1$ form on a smooth $m$-dimensional manifold $M$. If $\omega \neq 0$ at a point $p\in M$ then there is a coordinate system $(x^1,...,x^m)$ on a neighborhood of $p$ in which $$\omega = dx^2 \wedge dx^3 \cdots \wedge dx^m.$$

I am thinking they want us to use the Frobenius theorem but I am having trouble making any progress.

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    $\begingroup$ Your local coordinate system should be $(x^1,\dots,x^m)$, of course. $\endgroup$ – Ted Shifrin Oct 28 '13 at 22:06
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Well, Frobenius, offhand, applies to a differential system generated by a family of $1$-forms. But let's be clever. In standard coordinates $(y^1,\dots,y^m)$, we can write $$\omega = \iota_X dy^1\wedge\dots\wedge dy^m$$ for some vector field $X$ with $X(p)\ne 0$. By the flowbox theorem, we can choose local coordinates $(z^1,z^2,\dots,z^m)$ so that $X = \partial/\partial z^1$, so $\omega = f\,dz^2\wedge\dots\wedge dz^m$ for some nonzero function $f$. Since $d\omega = 0$, we infer that $\dfrac{\partial f}{\partial z^1} = 0$, and so $f=f(z^2,\dots,z^m)$. Now set $$x^2=\int f(z^2,\dots,z^m)\,dz^2\,,$$ so that $$dx^2 = f\,dz^2 + \sum_{j=3}^m \left(\int \dfrac{\partial f}{\partial z^j}\,dz^2\right)\,dz^j\,.$$ It follows that if we set $x^1=z^1, x^3=z^3,\dots, x^m=z^m$, then $\omega = dx^2\wedge\dots\wedge dx^m$.

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    $\begingroup$ Very clever indeed! Can we always write a $k-1$ form as the contraction of a vector field and a $k$ form? $\endgroup$ – Bohring Oct 28 '13 at 23:22
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    $\begingroup$ Sure, in a $k$-dimensional manifold, as, once we choose a volume element, $\Lambda^{k-1}V^*\cong V$. In $\Bbb R^k$, the $(k-1)$-form $\sum (-1)^{i-1}a_i \,dx^1\wedge\dots\wedge\widehat{dx^i}\wedge\dots\wedge dx^k=\iota_X dx^1\wedge\dots\wedge dx^k$, where $X=\sum a_i\dfrac{\partial}{\partial x^i}$. $\endgroup$ – Ted Shifrin Oct 28 '13 at 23:34

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