I want to show that for a nowhere-vanishing $1$-form $\alpha$ on an $n$-dimensional manifold $M$ we have that $\alpha \wedge d\alpha = 0$ if and only if around every point $p$ there exists a coordinate chart $(x^1,\ldots,x^n)$ and a nowhere vanishing function $f$ such that $\alpha = fdx^n$.

One direction is clear to me. If $\alpha = fdx^n$ then $d\alpha = df\wedge dx^n$ so $\alpha \wedge d\alpha = fdx^n\wedge df \wedge dx^n = -fdf\wedge dx^n \wedge dx^n = 0$.

I'm having a hard time going finishing the other direction though. I think the Frobenius theorem should help. I can define a smooth distribution $D \subset TM$ given by $D_p := \mbox{ker}(\alpha_p)$. Then since $\alpha$ is nowhere vanishing, I can choose a local vector field $X$ near $p$ such that $\alpha(X)$ is a non-vanishing function. Then if $Y,Z$ are any sections of $D$ near $p$, I have $0 = \alpha\wedge d\alpha(X,Y,Z) = \alpha(X)d\alpha(Y,Z) - \alpha(X)d\alpha(Z,Y)$ (the other terms vanish since $\alpha(Y) = \alpha(Z) = 0$). Since $\alpha(X) \not= 0$ we have $d\alpha(Y,Z) = d\alpha(Z,Y)$ but by skew-symmetry we also have $d\alpha(Y,Z) = -d\alpha(Z,Y)$. So $d\alpha(Y,Z) = 0$ for any such sections, hence $d\alpha$ annihilates $D$ and $D$ is therefore involutive, hence (by Frobenius' Theorem) integrable. This means there is a flat chart $(x^1,\ldots,x^n)$ for $D$ near $p$ such that the integral manifold of $D$ through $p$ is $\{x^i = 0\}$ for some $i$. Hence the tangent bundle is (locally) $\{dx^i = 0\}$. So $\mbox{ker}(\alpha_q) = \mbox{ker}(dx^i_q)$ for nearby $q$. At this point I can at least conclude that if $\alpha = \sum_j f_jdx^j$ then $f_j(p) = 0$ for $i\not= j$, but is this enough for me to conclude that $\alpha = f_idx^i$ for some non-vanishing $f_i$? Am I even going in the right direction here?

  • $\begingroup$ Yes! If one of the $f_j$ were nonzero near $p$ in this chart, then you could prove that $\text{ker}(\alpha) \neq \langle\partial/\partial x^n\rangle$. You should do so. $\endgroup$ – user98602 Aug 18 '15 at 3:53
  • $\begingroup$ Ah right, I guess I can actually conclude $f_j(q) = 0$ for nearby $q$ by the exact same reasoning by which I conclude $f_j(p) = 0$? $\endgroup$ – user262762 Aug 18 '15 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.