# What is wrong with this computation of the Lie derivative (using the pullback definition)

I am trying to understand the Lie derivative, looking at the Wikipedia page. Let's take a trivial example. My manifold is $M = \mathbb R$ with coordinate $m$. I have two constant vector fields $X$ and $Y$, each determined by a number $x$ and $y$ respectively. Viewing these as derivations, we have, for a smooth function $f$, $X(f) = x\frac{d}{dm}f$.

On the one hand, $\mathcal L_X (Y)$ should b $0$, since $Y$ is constant. And also it is clear that $X$ and $Y$ commute, so $[X,Y]=0$.

On the other hand, I think that the flow associated to $X$ should be $\phi_t: m \mapsto m + tx$. Then we should have $$\mathcal L_X (Y) (f)= \frac{d}{dt}|_{t=0}\; \phi_t^* Y(f)\\ = \frac{d}{dt}|_{t=0}\; Y(f \circ \phi_t)\\ =\frac{d}{dt}|_{t=0}\; y \frac{df}{dm}(\phi_t)\frac{d\phi_t}{dm}\\ = \frac{d}{dt}|_{t=0}\; y \frac{df}{dm}(\phi_t)\\ = y\frac{d^2f}{dm^2}(\phi_t)\frac{d\phi_t}{dt} \;|_{t=0}\\ = y \frac{d^2f}{dm^2} x$$ Which seems to be non-zero. Somehow I am only getting one half of the commutator $[X,Y]$, where is the other half to cancel it out?

$\frac{d}{dt}|_{t=0}\; \phi_t^* Y(f)$ computes $\mathcal{L}_X(Y(f))$, not $(\mathcal{L}_XY)(f)$, and Leibniz rule tells you that $\mathcal{L}_X(Y(f))=(\mathcal{L}_XY)(f)+Y(\mathcal{L}_Xf)=(\mathcal{L}_XY)(f)+(YX)f=(YX)f$, which is exactly one half of the commutator.