I am working through Schutz's Geometrical methods in mathematical physics. Stuck in 3.8, on Forbenius' theorem. My question is about the very last step.
Let there be an $n$-dimensional manifold $M$ and an open set $U \subset M$. Let there be an $m-1$ dimensional sub-manifold $S'$ and a set of vector fields $\mathbf{Y}_{(a)},\,a=1...m-1$ which form coordinate basis on $S'$ ($m\le n$). Here we only consider $S'$ inside $U$. Next there is a vector field $\mathbf{V}$ such that:
$$ \left[\mathbf{V},\,\mathbf{Y}_{(a)}\right]=0\,\forall a $$
The aim of the author is to build a set of vector fields $\mathbf{Z}_{(a)}$ that would commute with each other and with $\mathbf{V}$. $\mathbf{Z}_{(a)}$ are defined to equal to corresponding $\mathbf{Y}_{(a)}$ on $S'$, and to be Lie-transported along $\mathbf{V}$ in all other points outside $S'$. Also, by definition $\left[\mathbf{V},\,\mathbf{Z}_{(a)}\right]=0$.
I can follow author up to and including proving that:
$$ \mathcal{L}_\mathbf{V}\left[\mathbf{Z}_{(a)},\,\mathbf{Z}_{(b)}\right]=0 $$
But then the author concludes that $\left[\mathbf{Z}_{(a)},\,\mathbf{Z}_{(b)}\right]=0$ everywhere. Not sure about this step. It is certainly true on $S'$, where $\left[\mathbf{Z}_{(a)},\,\mathbf{Z}_{(b)}\right]=\left[\mathbf{Y}_{(a)},\,\mathbf{Y}_{(b)}\right]=0$ and one can see that Lie-transport of that vector ($\left[\mathbf{Y}_{(a)},\,\mathbf{Y}_{(b)}\right]$) along $\mathbf{V}$ will keep it at zero. I am not sure that this implies that commutator of two Lie transported vectors will also be zero: $\left[\mathbf{Z}_{(a)},\,\mathbf{Z}_{(b)}\right]\overset{?}{=}0$
