Lie derivative isn't $\lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}$ in $\mathbb{R^n}$? Why?

Question

The Lie derivative o a vector field $W$ along another vector field $V$ at point p (denoted by $(L_V W)_p$) is defined as the limit when $t \rightarrow 0$ of $\frac{(\theta_{-t})_*W_{\theta_t(p)}-W_p}{t}$ where $(-)_*$ denotes the pushforward and $\theta_t(p)$ the flow of $V$. Suppose we are in $\mathbb{R^n}$. For me, it seems natural that in this case, the Lie derivative would be: $$(L_V W)_p=\lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}.$$ But suppose I take as example $V(x,y)=(3x,4y)$. Then the flow, if we denote a point $p=(x,y)$ is given by $\theta_t(p)=(xe^{3t},ye^{4t})$ so the the pushforward (which as I understand is the Jacobian matrix here) is:
$$D (\theta_{-t}) = \begin{bmatrix} e^{-3t} && 0 \\ 0 && e^{-4t} \end{bmatrix}$$

But then we would have $(L_V W)_p=\lim_{t\rightarrow 0} \frac{D (\theta_{-t})W_{\theta_t(p)}-W_p}{t} \neq \lim_{t\rightarrow 0} \frac{W_{\theta_t(p)}-W_p}{t}$ which is not so intuitive, no?

Answer 1

It is not intuitive if you think about all tangent spaces being "fixed". For example, take the curve $\gamma:t\mapsto(\cos t,\sin t)$. Its tangent vector is given by $V=(-\sin t,\cos t)$.

Imagine that you look above the $\mathbb{R}^2$ plane while $t$ grows: what you see is an arrow, based on points of a circle and tangent to this same circle, making turns around the origin, so the arrow is moving and indeed you tell to yourself that this vector field must have a non zero derivative.

Now, same scene, but you turn with the vector, meaning that you always turn your head in order to keep the vector pointing upward to your eyes. Obvisouly, this times there is no movement of the vector, and so the vector field must have a zero derivative.

The first point of view is not possible in an abstract manifold, since there is not a nice way to see all the tangent vectors based on different points (and so living in different tangent spaces) living in a same ambient vector space. So we need to chose something, for example a curve, to keep track of how we are going to connect all these different spaces. This is what leads to your definition of Lie derivative. For example, you can check that the second illustration that I described to you is correct, by showing that $\mathcal{L}_VV=0$ for all vector field $V$.