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.


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.

  • $\begingroup$ You can check this answer in mathSE for an intuitive understanding: math.stackexchange.com/questions/2145617/… $\endgroup$ – abhijit975 Aug 15 '20 at 4:35
  • $\begingroup$ You've done a lot of trivial edits to this question, like replacing "a second" with "another". Please don't do that - every edit bumps the question in the "active" queue, and this should only be done when something actually changed. Edits correcting mistakes are fine even when minor, but constantly adjusting formulations to something equivalent is not. $\endgroup$ – ACuriousMind Aug 21 '20 at 14:12

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 ${\mathcal L}_XY$.

  • $\begingroup$ Thank you for your answer @mike stone and I apologize for my late response. I guess the crux of my issue is what happens to the "head" of the $Y$ arrow. I am struggling to visualize where it ends up landing when the $Y$ arrow is carried along by the fluid (whose velocity field is $X$) to $x + \epsilon X$. $\endgroup$ – JG123 Aug 13 '20 at 13:08
  • $\begingroup$ I guess it is at $x^\mu +\epsilon X^\mu +\eta \epsilon Y^\nu \partial_\nu X^\mu$ because $({\mathcal L}_XY)^\mu = X^\nu \partial_\nu Y^\mu- Y^\nu \partial_\nu X^\mu$. $\endgroup$ – mike stone Aug 13 '20 at 13:12
  • $\begingroup$ Is there a way to understand that expression pictorially? $\endgroup$ – JG123 Aug 13 '20 at 13:22
  • $\begingroup$ Yes. The $X$ the flow at at $x+\eta Y$ is $X(x)+\eta Y\partial X$, so each term in $X^\nu \partial_\nu Y^\mu -Y^\nu \partial_\nu X^\mu$ has a pictorial interpretation. $\endgroup$ – mike stone Aug 13 '20 at 13:53
  • $\begingroup$ Forgive me for asking, but what is this "pictorial interpretation" precisely? Or would it be more rewarding to go through the computations myself and see what I can gain from that? $\endgroup$ – JG123 Aug 13 '20 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.