Confused about intuition behind Lie derivative I'm trying to fix my intuition behind $\mathcal L_X T$, where $T$ is any tensor field. I'd prefer explanations that are not along the lines of $\mathcal L_XY=[X,Y]$ (I'm not sure how this extends to the case where $Y$ is an arbitrary tensor field).
I have two specific questions, but explanations beyond answering these questions are more than welcome.

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*Denote the flow of $X$ by $\varphi$. Certainly $(\mathcal L_XT)_p$ depends on both $X|_U$ and $T|_U$, where $U$ is an arbitrarily small neighborhood of $p$. But does it depend on the values of $X$ and $T$ on all of $U$, or just their values on the flow line, i.e., $\{\varphi_t(p)\}_{t\in\mathbf R}\cap U$? More specifically, is there an explicit example of $X$, $X'$ (with flow $\varphi'$), and $T$, such that for some $p$, we have $\varphi_t(p)=\varphi'_t(p)$ (i.e., $X=X'$ on the flow line through $p$) but $(\mathcal L_XT)_p\neq(\mathcal L_{X'}T)_p$?


*[Lee's Riemannian Manifolds, Exercise 4-3(b).] Show that there is a vector field on $\mathbf R^2$ that vanishes along the $x$-axis, but whose Lie derivative with respect to $\partial_x$ does not vanish on the $x$-axis.
My failed attempt at this problem: Suppose $V=f\partial_x + g\partial_y$ is such a vector field. Then $f(x,0)=g(x,0)=0$ for all $x$ by hypothesis. Using $\mathcal L_XY=(XY^i-YX^i)\partial_i$, we have $\mathcal L_{\partial_x}V=(\partial_xf-0)\partial_x+(\partial_xg-0)\partial_y$, which is supposed to be nonzero somewhere on the $x$-axis, i.e., there is some $x_0$ such that $\partial_xf(x_0,0)\neq0$ or $\partial_xg(x_0,0)\neq0$. Without loss of generality, assume $\partial_xf(x_0,0)\neq0$. So now we have to come up with some $f:\mathbf R^2\to\mathbf R$ such that $f(x,0)=0$ for all $x$ but $\partial_xf(x_0,0)\neq0$ for some $x$, which is a contradiction. What went wrong?
 A: For the intuition behind the Lie derivative I can maybe add a bit to the comment by Qiaochu Yuan: The simplest way to understand the differentiation is to consider the flows $\phi_t$, use them to pull back $T$ and then evaluate in the point $p$. This lies in the fiber of the appropriate tensor bundle at $p$ for any value of $t$. Hence you obtain a smooth curve in a finite dimensional vector space, and you can simply differentiate this in $t=0$ to obtain an element of that vector space. Work would be needed to deduce that this depends smoothly on $p$, but for the purpose of intuition that's not neccesary. 
This description also implies the answer to question 1, but I can give you a more direct argument: It is true that the Lie derivative is a local operator, so the value of $\mathcal L_XT$ in a point $p$ depends only on the restrictions of $X$ and $T$ to an arbitrarily small neighborhood $U$ of $X$. But much more is true, because the Lie derivative is a first order differential operator. This means that it depends only on the one-jets of $X$ and $T$ in $p$. (There are various ways to make this explicit.) If you look at the coordinate formula for the Lie derivative, you see that to compute $\mathcal L_XT(p)$ you only need the components of $T$ and their derivatives in direction $X(p)$. Hence it is indeed true that the value of $\mathcal L_XT$ in $p$ depends only on the restriction of $T$ to the flow line of $X$ through $p$. 
Question 2 is anserwered in Jack Lee's comment.
