So I have the following problem:

Suppose $\omega=\phi \wedge \theta$ is a closed decomposable 2-form on $M$ a manifold (decomposable just means it can be written as a wedge of 1-forms). Suppose $p\in M$ is a point such that $\omega\neq 0$.

Use the Frobenius Theorem to prove that $\omega=dx^{1}\wedge dx^{2}$ in some coordinate system in a neighborhood around $p$.

So the formulation of the Frobenius theorem that I now is the one about completely integrable and involutive distributions being equivalent. Even if I show that $\omega$ somehow defines an involutive distribution, I don't know how to use the flat chart to get $\omega$ into the desired form.

Thanks for any help!


2 Answers 2


So with some guidance from @TedShifrin I think I got it.

First since $\omega$ is decomposable and closed taking the exterior derivative, $0=d\phi \wedge \theta -(1) \phi \wedge d\theta$. Hence, $d\phi \wedge \theta = \phi \wedge d\theta$. Wedging the expression with $\phi$ gives $d\phi \wedge \theta \wedge \phi=0$, since the right side has $\phi \wedge \phi$ in there. Similarly, $d\theta \wedge \theta \wedge \phi =0 $.

Hence, $d\theta$ and $d\phi$ are linearly dependent on $\theta \wedge \phi$, hence $d\theta, d\phi \in \langle \theta, \phi \rangle $.

The above sentence is incorrect (I have not figured out how to cross it out. I would like to leave it there for instructional purposes).

Rather, I need to prove the following general identity: Let $\omega^1, ... \omega^k$ be locally defining independent one forms for a $n-k$ rank distribution $D$. Then $D$ is involutive iff $d\omega^i \wedge \omega^1 \wedge ... \wedge \omega^k =0$ for all $1\leq i \leq k$. One direction is standard by the $1$-form criterion for involutivity, which says that $D$ is involutive iff for any 1-form $\eta$ annihilating $D$, $d\eta$ annihilates $D$.

Conversely, suppose the above holds. Take the independent $\omega^1, ...\omega^k$ and extend to a smooth local coframe $\{ \omega^1, ...\omega^k, \alpha^{k+1}, ... \alpha^n \}$ with a dual frame $\{ E_1, ... E_k, A_{k+1},...A_n \}$. Note the $A_j$'s span $D$ Then $d\omega^{i}=\Sigma b_{ij}\omega^{i}\wedge \omega^{j} + \Sigma c_{kl} \omega^{k} \wedge \alpha^{l} + \Sigma d_{xy}\alpha^{x}\wedge \alpha^{y}$ with sums over appropriate increasing indices.

Now, wedge this with $\omega^1 \wedge ... \wedge \omega^k$ and evaluate at $(A_x, A_y, E_1, ..., E_k)$. This then shows $d_{xy}=0$. Hence in fact $d\omega^{i}=\Sigma b_{ij}\omega^{i}\wedge \omega^{j} + \Sigma c_{kl} \omega^{k} \wedge \alpha^{l}$ which when evaluated at any $(A_x, A_y)$ is equal to zero. Hence by the 1-form criterion $D$ is involutive.

Now the next paragraph can in fact be skipped.

Thus, if $\lambda=\alpha \wedge \theta + \beta \wedge \phi$, we have $d\lambda=d\alpha \wedge \theta - \alpha \wedge d\theta + d\beta \wedge \phi - \beta \wedge d\phi \in \langle \theta, \phi \rangle$, hence the ideal is a differential ideal.

Thus $D$ defines and involutive distribution, which is integrable by the Frobenius theorem.

Thus there is a flat chart where the distribution, call it $D$ is spanned by $\frac{\partial}{\partial x^{3}}, ... , \frac{\partial}{\partial x^{n}}$ and annihilated by the defining $1$-forms $\theta, \phi$. The only way for this to happen is for $\theta = f_{1}dx^1 + f_{2}dx^2$ and $\phi = g_{1}dx^1 + g_{2}dx^2$ Taking the wedge, we get $\omega =(g_{1}f_{2}-g_{2}f_{1})dx^1 \wedge dx^2=F(x^1, ..., x^n)dx^1\wedge dx^2$.

Now let us take the exterior derivative of $\omega.$ $d\omega=\Sigma \frac{\partial F}{\partial x^k}dx^k \wedge dx^1 \wedge dx^2=0$ implying the partial derivatives vanish. Hence $F=F(x^1, x^2)$

Now define a new coordinate system by $y^1=\int_{0}^{x^1}F(t, x^2)dt, y^2=x^2, ... , y^n=x^n$. The Jacobian of this transform is non-zero. Further $dy^1=Fdx^1+ \frac{\partial}{\partial x^2}(y^1=\int_{0}^{x^1}F(t, x^2)dt)dx^2.$ Hence in this coordinate system $\omega=Fdx^1 \wedge dx^2=dy^1 \wedge dy^2$.

  • $\begingroup$ OK, typographical issue: You have a few $w^i$ that should be $\omega^i$. I think you can make the argument clearer by treating any $k$-form, rather than $d\omega^i$ with all the product rule stuff. More serious point: You do not know that $f_i,g_i$ are functions of $(x^1,x^2)$ only. So you need to repair the end ... $\endgroup$ Jan 2, 2015 at 22:01
  • $\begingroup$ Ah yes, I see. Thank you for all the help! $\endgroup$ Jan 2, 2015 at 22:07
  • $\begingroup$ You've messed up your product rule, now :( But the end is still no good. If I have $\omega = x^3 dx^1\wedge dx^2$, I'm never going to change coordinates to make $\omega = dy^1\wedge dy^2$. Sorry to be a pain :P $\endgroup$ Jan 2, 2015 at 23:08
  • $\begingroup$ Actually, not product rule. The point is that $d\omega^i$ shouldn't have any $d$'s on the right-hand-side. $\endgroup$ Jan 2, 2015 at 23:22
  • $\begingroup$ No, it's all right. I will think about this a bit more. $\endgroup$ Jan 2, 2015 at 23:23

First of all, the Frobenius Theorem is about a system of differential equations given by $1$-forms. (The Cartan-Kähler Theorem addresses the general case.)

Note here that near $p$, $\phi$ and $\theta$ are linearly independent $1$-forms, and the fact that $\omega$ is closed tells us that $d\phi\wedge\theta=\phi\wedge d\theta$. It follows (make sure you work out why) that $d\phi\wedge\theta\wedge\phi = d\theta\wedge\theta\wedge\phi = 0$, and so the ideal $\langle \phi,\theta\rangle$ is a differential ideal, hence defines an involutive distribution. This means that there are local coordinates $(x^1,\dots,x^n)$ on $M$ near $p$ so that $\phi = f_1 dx^1 + f_2 dx^2$ and $\theta = g_1 dx^1+ g_2 dx^2$ for some functions $f_i,g_i$. Can you finish the argument?

  • $\begingroup$ So I wrote down what I think is my solution below, but I thin I made a mistake. I'm not quite sure how $d\theta \wedge \theta \wedge \phi$ implies $d\theta$ and similarly $d\phi$ are in the Ideal. I tried to say something about linear dependence, but that is a pointwise statement... any more hints? $\endgroup$ Jan 2, 2015 at 17:54
  • $\begingroup$ I made some edits to rectify my error. $\endgroup$ Jan 2, 2015 at 18:52
  • $\begingroup$ We're only working locally. Since $\phi$, $\theta$ are linearly independent, we can extend them (on a neighborhood of $p$) to a basis for the $1$-forms. From that you can certainly prove that if $\omega\wedge\phi\wedge\theta = 0$, then $\omega\in\langle \phi,\theta\rangle$. You cannot assert that $\omega\in\langle \phi\wedge\theta\rangle$. $\endgroup$ Jan 2, 2015 at 19:17
  • $\begingroup$ Yup, I think I fixed it below in my answer. Mind taking a look? Though I see what you mean, I could have proven it more nicely without the slightly round about way I did it below. $\endgroup$ Jan 2, 2015 at 19:26
  • $\begingroup$ No need to hurry, thanks :) $\endgroup$ Jan 2, 2015 at 21:29

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