I am having trouble with the following exercise:
Exercise: Let $X= \mathbb{R^3}$ and consider the vector fields $V, W:X \to TX = X \times \mathbb{R^3} $ given by $ V(x,y,z)=(1,0,y)$ and $W(x,y,z)=(0,1,0)$. Let $D$ be the distribution spanned by $V$ and $W$. Show that $D$ is not involutive. This implies $D$ is not integrable.
My attempt: We need to show that if we have two vector fields $A, B \in span\{W,V\}$ then $[A,B]_x \notin D_x$. Fix $f \in C^{\infty}(X)$. By definition of the Lie bracket, we have $[A,B]_x (f)=A_{x}(B(f))-B_x(A(f)).$ Since $A, B$ are vector fields we can express them as: $$A(f) = \sum_{i=1}^{3} a_{i} \frac{\partial f}{\partial x_{i}} , \hspace{3pt} B(f) = \sum_{i=1}^{3} b_{i} \frac{\partial f}{\partial x_{i}},$$ Now, $span\{W,V\} = \{(k,l, ky) : k, l \in \mathbb{R}\}$ so this tells us what $a_i, b_i$ look like. From here, we calculate $[A,B]_x (f)=A_{x}(B(f))-B_x(A(f))$ using the above representation and find it is equal to $0$.
My questions:
- Is what I've done correct? If so, why does the Lie bracket being $0$ imply it is involutive?
- If not, how would we do it?
Thanks a lot.
I don't think your calculation is correct as if I understood correctly it shows that the distribution is involutive, when you want to show that it is not.
In terms of the coordinate vector fields $\partial_x,\partial_y,\partial_z$ we can write these vector fields as $V(x,y,z)=\partial_x+y\partial_z$ and $W(x,y,z)=\partial_y$. Now, $$[\partial_x+y\partial_z,\partial_y]=[\partial_x,\partial_y]+[y\partial_z,\partial_y]=0+y\partial_z\partial_y-\partial_y(y\partial_z)=0+y\partial_z\partial_y-\partial_z-y\partial_y\partial_z=-\partial_z.$$ In particular, the bracket of these two vector fields is not in their span (essentially by inspection) so the distribution is not involutive.