This question was asked in assignment of Smooth Manifolds and I need help with the proof.
Question : Let X,Y be two smooth vector fields on $\mathbb{R}^3$ given by $X= \frac{ \partial}{\partial x} + y \frac{\partial} {\partial z}$ and $Y= \frac{\partial} {\partial y}$. Show that there doesn't exist integral submanifold of the distribution generated by {X,Y}.
Definition of a distribution: A k-dimensional distribution on a manifold M is a map $p \to \Delta_p$ , where $\Delta_p \subset T_p M$ is k-dimensional subspace of $T_p M$. We call $\Delta$ a $C^{\infty}$ distribution if for any $p\in M$ we can choose a nbd U of p and k smooth vector fields $X_1,...,X_k$ such that {$X_1(q),...X_k(q)$} is a basis of $\Delta_q$ for each $q\in U$. So, distribution generated by {X,Y}means that $\Delta_p$ is the subspace of $T_p \mathbb{R}^3$ whose basis {X(p), Y(p)} for all $p\in \mathbb{R}^3$.
Given a k-dimensional distribution $\Delta$ on an n-dimensional manifold M, we say that a k-dimensional submanifold $\iota:N\hookrightarrow$ M is an “integral submanifold” of $\Delta$ if $\iota_*\mathcal{T}_pN=\Delta_{\iota(p)}$ for every $p\in N$. That is, if the subspace of $\mathcal{T}_{\iota(p)}M$ spanned by the images of vectors from $\mathcal{T}_pN$ is exactly $\Delta_p$.
Attempt: Let there exists such an integral submanifold of the distribution generated by {X,Y}. It is a map from M to $T_p M$ . I am really sorry but I am badly struck on this problem and would very much appreciate any help. I don't have much to show as attempt except the definitions I applied which are given above. But I am not sure how would I be able to prove that no such integral submanifold exists.
Kindly help!
Are you familiar with the Frobenius Integrability Theorem? This is a simple application of the "easy direction" of the theorem: if a distribution $\Delta$ is integrable, then it is involutive, i.e. $[X,Y]\in\Delta$ for every $X$ and $Y$ in $\Delta$.
Compute the commutator of the two vector fields, you will see that $[X,Y]=-\partial_z$, which does not belong to $\Delta:=\operatorname{span}_{\mathbb{R}}\{X,Y\}$. So the distribution generated by $X$ and $Y$ cannot be integrable.
To see why the theorem holds, assume that there is a submanifold $N$ such that $\Delta=TN$ for every point of $N$. The tangent space is generated by the vector fields tangents to $N$, whose Lie bracket is again a vector field tangent to $N$ (important keyword: naturality of the Lie bracket), so $[TN,TN]\subseteq TN$, and this implies $[\Delta,\Delta]\subseteq\Delta$.