Transporting Tangent Vectors when Taking Lie Derivatives

Question

I am currently working on understanding the (intrinsic) differential geometry underpinning General Relativity, and I think I could benefit from a more intuitive grasp of the process of taking the Lie derivative of a vector field with respect to another vector field.

I ask this question for this reason. Pictorially, I understand what happens to tangent vectors when we parallel transport them along curves when taking covariant derivatives. Therefore, I seek to understand what happens to tangent vectors when we "Lie transport" them along integral curves of vector fields when taking Lie derivatives of vector fields.

To illustrate my point further, consider the following example.

Let $V$ and $W$ be smooth vector fields on a (say smooth) manifold $M$. Let $\gamma_w$ denote an integral curve of $W$ and let $q = \gamma_w (s)$ be an arbitrary point in the image of $\gamma_w$.

Let $\phi^{x}$ be an element of the local one-parameter group of $W$, that is, $\phi^{x}$ is the flow of the vector field $W$ by parameter $x$ along $\gamma_w$.

We then compute the Lie derivative of $V$ with respect to $W$ at $q$, $\mathcal{L}_W V(q)$ (so $\mathcal{L}_W V$ is a vector field on $M$), as follows.

We first let the tangent vector $V(\gamma_w(s+\epsilon)$) "flow" back from $\gamma_w(s+\epsilon)$ to $q$. The resulting tangent vector at $q$ is given by $d\phi^{-\epsilon}(V(\gamma_w(s+\epsilon))$ (here $d\phi^{-\epsilon}$ is the differential of $\phi^{-\epsilon}$). We then subtract $V(q)$ from this tangent vector (this subtraction operation is now well-defined), and divide the result by $\epsilon$. We then take the limit as $\epsilon \rightarrow 0$ to get a genuine derivative of the vector field $V$ along an integral curve of $W$.

That is,

\begin{equation} \mathcal{L}_W V(q) = \lim_{\epsilon \to 0} \frac{d\phi^{-\epsilon}(V(\gamma_w(s+\epsilon))- V(q)}{\epsilon} =\frac{d}{dt} ((d\phi^{-t} \circ V \circ \phi^t) (q))\rvert_{t = 0} \end{equation}

Now, my question is the following. Geometrically/pictorially, what happens to $V(\gamma_w(s+\epsilon)$) when it "flows" from $\gamma_w(s+\epsilon)$ to $q$, and why does the differential $d\phi^{-\epsilon}$ output this Lie transported tangent vector?

Thanks in advance.

Notes

This is a modified version of a couple (now deleted) questions I posted to this site and Math StackExchange. Also, as stated above, I am looking for an intuitive answer, not an algebraic or computational one. As such, I am totally okay with an answer that treats these vector fields as little arrows scattered across the manifold.

I should also say that I have looked at many questions regarding the intuition behind the Lie derivative, specifically how it differs from the covariant derivative. However, I could not find a satisfying answer that answers the above question from a geometrical/intuitive point of view. I guess that I am looking for a physicist's perspective here.

Answer 1

I find that using the notation $d\phi^{-\epsilon}$ and so on quite unhelpful. Instead, given vector fields $X$ and $Y$ I imagine the vector $Y$ at $x$ as a small arrow painted on a fluid whose velocity field is $X$. The tail of the arrow is at $x$ and its head at $x+\eta Y$, where $\eta$ is a small number. After the short time $\epsilon$ the tail of the $Y$ arrow has been carried by along by the fluid to $x+\epsilon X$ and its head is wherever it has been carried to. One subtracts the flow-carried $Y$-arrow from the value of the vector field $Y(x+\epsilon X)$. The latter is represented by the small arrow whose tail is at $x+\epsilon X$ and whose head is at $(x+\epsilon X)+\eta Y(x+\epsilon X)$. Then you divide by $\epsilon$ and by $\eta$. The result is ${\mathcal L}_XY$.