Existence of a submanifold Is it correct that if the bracket of two vectors $A_{1}$ and $A_{2}$ equals zero, then  a submanifold tangent to $A_{1}$ and $A_{2}$ exists?
 A: More formally, this is Frobenius' theorem : http://en.wikipedia.org/wiki/Frobenius_theorem_(differential_topology) if you have a subbundle, then the subbundle is integrable , i.e., there exists a submanifold whose tangent space coincides with the subbundle, if the Lie bracket of two vectors in the subbundle also takes values in the subbundle. So you want closure of vector fields under taking the Lie bracket.
An interesting case to look at is that of contact manifolds https://en.wikipedia.org/wiki/Contact_geometry , EDIT , in the case of 3-manifolds, contact manifolds are precisely those where the integrability condition is not satisfied at any point.  The contact distribution is then given as the kernel of a 1-form $\omega$ satisfying $ \omega \wedge d\omega \neq 0$ so that, if $\omega$ is given globally, then the manifold is orientable (because $\omega \wedge d\omega$ is an orientation form; notice that the contact forms may be defined locally in such a way that they cannot be coherently patched-up into a globally-defined form).
As pointed out by Jack Lee in the comments, the situation becomes more complicated in higher dimensions, a distribution may be nowhere-integrable and still may not be a contact bundle. The conditions for a higher-dimensional distribution $D$ to be a contact distribution (as pointed out by Jack Lee) is then that the map $(X,Y) \rightarrow [X,Y] mod D$ is a non-degenerate bilinear form.
