Say I have a family of curves $x_s : [0,1] \longrightarrow M$ where $s \in (-\epsilon, \epsilon)$ is my family's parameter, and $M$ is a manifold (which, for all purposes being, we can assume to be $\mathbb{R}^n$).

Additionally, let $\zeta := \dfrac{\mathrm{d}}{\mathrm{d}s}x_s|_{s = 0}$, which ends up being a vector field along the curve $x$. (More precisely, it is a function $[0,1] \longrightarrow x^*TM$ with appropriate regularity that I care about, but I don't believe it is necessary to go into such details).

I am trying to calculate:

$\dfrac{\mathrm{d}}{\mathrm{d}s}|_{s = 0}\displaystyle\int_{0}^1 x_s^{*}\lambda$

Where $\lambda$ is an arbitrary (integrable) $1$-form defined on my manifold $M$ (which we assume to be $\mathbb{R}^n$).

The source I am reading makes the following claim:

$\boxed{\dfrac{\mathrm{d}}{\mathrm{d}s}|_{s = 0}\displaystyle\int_{0}^1 x_s^{*}\lambda = \displaystyle\int_0^1 x^{*}\mathcal{L}_\zeta\lambda}$

where $\mathcal{L}_\zeta$ denotes the Lie derivative with respect to the vector $\zeta$ defined above. But I'm not entirely sure how to prove it. Here is where my head is at so far:

\begin{align} \dfrac{\mathrm{d}}{\mathrm{d}s}|_{s = 0}\displaystyle\int_{0}^1 x_s^{*}\lambda &= \displaystyle\int_{0}^1 \dfrac{\mathrm{d}}{\mathrm{d}s}|_{s = 0} \, x_s^{*}\lambda \\ &= \displaystyle\int_0^1 \lim\limits_{\varepsilon \to 0} \frac{1}{\varepsilon}(x_\varepsilon^{*}\lambda - x^{*}\lambda) \textit{ (cause } x_0 = x) \end{align}

However, I'm not sure how to formally identify $\lim\limits_{\varepsilon \to 0} \dfrac{1}{\varepsilon}(x_\varepsilon^{*}\lambda - x^{*}\lambda)$ with $x^{*}\mathcal{L}_\zeta \lambda$.

I can see "philosophically" why these should be the same ($\zeta$ is a vector field along the curve $x$, and it is generated by the "curve of curves" $x_s$. So intuitively, the expression on the left sort of corresponds to a derivative along the flow of $\zeta$, which is exactly what the Lie derivative is).

But then, I'm not sure how to prove it formally because the vector field $\zeta$ is only defined along $x$, so what does it even mean to take $\mathcal{L}_\zeta \lambda$? Do we somehow extend $\zeta$ beyond $x$ in an arbitrary way? And then, since we don't care about this extension, maybe that's the reason we take the pullback along $x$, and end up with the expression $x^{*}\mathcal{L}_\zeta\lambda$?

Any help or progress on the question would be much appreciated. Thank you :)

  • 2
    $\begingroup$ This is (one of) the very definition of the Lie derivative, the time derivative of the pullback. You can find the proof on Wikipedia or most differential geometry textbooks. $\endgroup$ Aug 31, 2023 at 18:41
  • $\begingroup$ Well, to me, $\mathcal{L}_\zeta \lambda := \lim\limits_{\varepsilon \to 0} \frac{1}{\varepsilon} (\phi_t^{*} \lambda - \lambda)$, where $\phi_t$ is the flow of the vector field $\zeta$. Here, I want to show that $x^{*}\mathcal{L}_\zeta\lambda = \lim\limits_{\varepsilon \to 0} \frac{1}{\varepsilon} (x_\varepsilon^{*} \lambda - x^{*}\lambda)$. $\endgroup$
    – Azur
    Sep 1, 2023 at 10:42
  • $\begingroup$ I assume it might have to do with somehow factoring out $x = x_0$ from the right-hand side of my last equation above, but I'm not sure how to properly do that, and how to identify that formally with the flow of $\zeta$. (Though it should work, since $(\mathrm{d}/\mathrm{d}s) x_s = \zeta$ $\endgroup$
    – Azur
    Sep 1, 2023 at 10:47

2 Answers 2


If you see $x_s$ as a two variable function that it maps into an open neighborhood of the curve $x_0$ and you can still differentiate with respect to s on that neighborhood. This gives you a natural way to extend $\zeta$ to a neighborhood of $x_0$.

  • $\begingroup$ I see what you mean, I could extend the definition $\zeta = (\mathrm{d}/\mathrm{d}s)x_s$ to a whole neighbourhood of $x_0$. However, I'm still not sure how to conclude $\lim\limits_{\varepsilon \to 0} \frac{1}{\varepsilon}(x_\varepsilon^{*}\lambda - x_0^{*}\lambda) = x^{*}\mathcal{L}_\zeta\lambda$ from there. $\endgroup$
    – Azur
    Aug 31, 2023 at 16:30
  • $\begingroup$ And actually, this only works if my curve is originally embedded in $\mathbb{R}^2$. In higher-dimensional space, all this gives me is a 1-family of curves; which can be good, but which isn't an open neighbourhood $\endgroup$
    – Azur
    Sep 1, 2023 at 11:16

Okay, I think I have reached a partial answer, so let me type it up. I'll explain why it is partial at the end (it has to do with quarague's answer - I still cannot find a way to naturally extend my vector field $\zeta$ to an open neighbourhood of my manifold).

So let me recap the situation. We have a family of curves $x_s$ in a manifold $M = \mathbb{R}^n$, where $s \in (-\epsilon, \epsilon)$ is a parameter, and each $x_s$ is parametrized as $x_s = x_s(t)$ (for $t \in [0,1]$).

As quarague pointed out in their answer, we can view $x$ as a two-argument function $x : (s,t) \mapsto x_s(t)$, and this allows us to parametrize a surface $S = \text{im}(x) \subset M$.

Sadly, this is not an open neighbourhood (unless $M = \mathbb{R}^2$, but I'd like to stay general and keep working in $\mathbb{R}^n$), but still, we have a parametrized surface in our manifold, where all the stuff that we care about happens.

  • So let us restrict our attention to the surface $S$. We define the vector field $\zeta\left(x_s(t)\right) := \dfrac{\mathrm{d}}{\mathrm{d}s} x_s(t)$, which is tangent to the surface $S$. Let $\phi_t$ be the flow of $\zeta$ (we can restrict to a smaller neighbourhood if necessary).

Then, by definition, an integral curve of $\zeta$ (or of the flow $\phi_t$) is a curve $\gamma_p(\cdot)$ such that $\dfrac{\mathrm{d}}{\mathrm{d}s} \gamma_p(s) = \zeta\left(\gamma_p(s)\right)$, where $p$ is some initial point.

Therefore, we can identify the integral curves of $\zeta$ with the curves $s \mapsto x_s(t_0)$, where $t_0$ is here to specify an initial point. In particular, this says that $\boxed{\phi_s \circ x_{s'} = x_{s + s'}}$ (ie, the diffeomorphism $\phi_s$ moves the integral curve $x_{s'}$ to $x_{s + s'}$).

In particular, this means that $x_{s + s'}^* = x_{s'}^* \phi_s^*$ which I will use below.

And from here, the proof of the statement I wanted follows pretty straightforwardly:

I wanted to prove that $x_0^{*}\mathcal{L}_\zeta\lambda = \dfrac{\mathrm{d}}{\mathrm{d}s}|_{s = 0}\,x_s^{*}\lambda$.

So let's start from the left-hand-side. $\lambda$ is an arbitrary $1$-form, and $\mathcal{L}$ denotes the Lie derivative.

Then, by definition, $\mathcal{L}_\zeta\lambda = \lim\limits_{\varepsilon \to 0} \dfrac{1}{\varepsilon} (\phi_\varepsilon^{*} \lambda - \lambda)$. So:

\begin{align} x_0^{*}\mathcal{L}_\zeta\lambda &= \lim\limits_{\varepsilon \to 0} \dfrac{1}{\varepsilon} x_0^{*} (\phi_\varepsilon^{*} \lambda - \lambda) \\ &= \lim\limits_{\varepsilon \to 0} \dfrac{1}{\varepsilon} ( x_0^{*}\phi_\varepsilon^{*}\lambda - x_0^{*} \lambda) \\ &= \lim\limits_{\varepsilon \to 0} \dfrac{1}{\varepsilon} (x_\varepsilon^{*} \lambda - x_0^{*}\lambda) \\ &=: \dfrac{\mathrm{d}}{\mathrm{d}s} x_s|_{s = 0} \end{align}

Which is what I wanted to prove.

Now, why is this only a partial proof?

Well, I am only working on a surface $S = \text{im}\{x_{s}(t) \, \mid \, s,t\} \hookrightarrow M = \mathbb{R}^n$.

So the vector field $\zeta$, as well as the flow $\phi_t$, and even more importantly, the Lie derivative $\mathcal{L}$ are only defined on this surface.

In the original source I was reading, they ascertained the equality $x_0^{*}\mathcal{L}_\zeta\lambda = \dfrac{\mathrm{d}}{\mathrm{d}s} x_s$ on the whole manifold $M$, after simply defining the family of curves $x_s$, and $\zeta$ as $\zeta := \dfrac{\mathrm{d}}{\mathrm{d}s}x_s$.

(Though admittedly, I have no idea how they define their Lie derivative. Maybe they do also restrict to the surface $S$, and simply swept all these details under the rug; though it does seem a bit arbitrary. Or, they have implicitly extended $\zeta$ (somehow) beyond the surface $S$, and then defined its flow on the whole of $M$ (well, on a local neighbourhood, but you know what I mean), but this seems a bit fishy; because I can't think of a natural way to do it).

I'll leave this answer up for a few days like this, in case someone has a comment, or a way to clarify this further. If not, I'll validate it so that it's no longer in the Unanswered!


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