Meaning of vanishing Lie bracket This is somewhat extension of question in why does Lie bracket of two coordinate vector fields always vanish?
Now i want to understand the meaning of vanishing Lie bracket. 
$i.e$,  For vector field $X$, $Y$ If 
\begin{align}
[X,Y]=0 
\end{align}
for all $Y$ on $M$, 
Of course i know if $X, Y$ are coordinate basis, then $[X,Y]=0$, but here $Y$ can be arbitrary. 
Borrow some logic from usual elementary algebra gives 
$ax=0$ for all $x$ means $a=0$
Can i apply same thing here? 
If $X=0$ then it obviously satisfied $[X,Y]=0$ for all $Y$ but i am uncomfortable with its inverse. 
 A: If you consider an Abelian lie group $G$, then the Lie bracket is trivial. For example check: The Lie algebra of every commutative Lie group carries a trivial lie bracket. Thus from this example, we see that  for a fixed vector field $X$, the condition $[X,Y]=0$ for all vector fields $Y$, does not imply $X=0$.
The Lie bracket of vector fields has a geometric interpretation: If we denote by $\phi_X\colon M\times I\to M$ and $\phi_Y\colon M\times I\to M$ the flows of $X$ and $Y$, then $[X,Y]$ is measuring the infinitesimal difference of the flows. Namely if we consider for $p\in M$  the curve $\phi_Y(-t)\circ\phi_X(-t)\circ\phi_Y(t)\circ\phi_X(t)(p)$, it may not be come back to your starting point $p$. The vector $[X,Y]$ is the limit of this picture when $t\to 0$ (see https://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields).
The condition $[X,Y]=0$ is telling you that the flows of $X$ and $Y$ are commuting for small times $t$. With this fact you can find a chart around $p$ with $X$ and $Y$ as coordinate vector fields.
