The property of an integral manifold of differential form

Question

Let's define the $1$-form $\eta$ on $\mathbb{R}^3$ by formula: $$\eta = A(x,y,z)\;\mathrm{d}x + B(x,y,z)\;\mathrm{d}y + \mathrm{d}z \text{.}$$ And let's assume that $$\mathrm{d}\eta \wedge \eta = \phi(x,y,z)\;\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z \text{.}$$ How to show that if $M\subset\mathbb{R}^3$ is an integral manifold of $\eta$ (e.g. $M$ is a submanifold of $\mathbb{R}^3$ such that $\forall_{p\in M}\;\forall_{V\in T_pM}\;\eta(V)=0$) then $\phi \equiv 0$ on $M$?

I started with computing $\mathrm{d}\eta$ and $\mathrm{d}\eta\wedge\eta$ to express $\phi$ in other terms, but I got nothing. Can I get some help?

Answer 1

Fix a $p \in M$. Choose a coordinate system near $p$ so that $TM = \text{span} \{ \partial_1, \partial_2 \}$ and $\{ \partial_1, \partial_2, \partial_3 \}_p$ is a positively oriented orthonormal basis for $T_p \mathbb R^3$. Then $\eta(\partial_1) = \eta(\partial_2) = 0$ so

$$\phi (p) = d\eta \wedge \eta (\partial_1, \partial_2, \partial_3)|_p = d\eta(\partial_1, \partial_2 )\eta(\partial_3)=0$$

since $d\eta_{12} = \partial_1 \eta(\partial_2) - \partial_2\eta(\partial_1) = \partial_1 0 - \partial_2 0 = 0$.

Answer 2

Consider the smooth map $\iota:M\to\Bbb R^3$ that is the inclusion of $M$ in $\Bbb R^3$. What can you say about the pullback of $\eta$ by $\iota$? Another question that leads to the same conclusion is to ask yourself what the pullback of the $3$-form $d\eta\wedge\eta$ the the two dimensional manifold $M$ must be.