# A question on distributions

Let $\Delta$ be a smooth distribution on a smooth manifold $M$ and let $X,Y$ be 2 vector fields on $M$ which are tangent to $\Delta$ (namely $X(q),Y(q)\in \Delta_q\leq T_qM$ for every $q\in M$. I would like to show that if $Y(p)=0$ for some $p\in M$, then $[X,Y](p)\in \Delta_p$.

To show this I am trying to mimic the proof of one of the implications of Frobenius theorem. Since $[X,Y]$ can be computed using Lie derivatives we have $$[X,Y](p)=L_XY(p)=\lim_{t\to 0} \frac{(\theta^X_{-t})_{\ast}Y(\theta^X_t(p))-Y(p)}{t}=\lim_{t\to 0} \frac{(\theta^X_{-t})_{\ast}Y(\theta^X_t(p))}{t}$$

How does one conclude that the latter limit lies in $\Delta_p$?

• Don't you mean lies in $\Delta_p$? – user99680 Jul 12 '14 at 20:32
• Thank you, I just edited it. – user54631 Jul 12 '14 at 21:10
• You write "Let $X$ be a smooth distribution ...", but maybe you mean "Let $\Delta$ be ... "? – user160609 Jul 12 '14 at 21:31

I'm not sure how to make the flow proof work, but here's how to do it using the algebraic properties of the Lie bracket. Let $e_j$ be a local frame spanning $\Delta$ near $p$ and write $Y = Y^j e_j$. Then we have $$[X,Y] = [X, Y^j e_j] = Y^j [X, e_j] + X(Y^j) e_j.$$ Evaluating this at $p$, the first term vanishes because $Y^j(p)=0$, and the second term lies in $\Delta$ because it is a linear combination of the $e_j$.
The vector $X$ is tangent to $\Delta$ by assumption, and so the curve $\theta^X$ is tangent to $\Delta$. The vector $Y(\theta_t^X(p))$ is in $\Delta,$ and so if we flow it along the curve $\theta$, it stays in $\Delta$.
• Your notation is sloppy here. $\theta^X$ is a flow, not a curve. And your argument is far from clear if the distribution is not integrable. In particular, I cannot see how you're using the hypothesis that $Y(p)=0$. – Ted Shifrin Jul 12 '14 at 21:55