# Lie transported commutator ?= commutator of Lie-transported vectors [Schutz]

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: $$\left[\mathbf{Z}_{(a)},\,\mathbf{Z}_{(b)}\right]\overset{?}{=}0$$