# Differentiating a pullback along a family of curves

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 :)

• 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. Aug 31, 2023 at 18:41
• 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 $x^{*}\mathcal{L}_\zeta\lambda = \lim\limits_{\varepsilon \to 0} \frac{1}{\varepsilon} (x_\varepsilon^{*} \lambda - x^{*}\lambda)$.
– Azur
Sep 1, 2023 at 10:42
• 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 $(\mathrm{d}/\mathrm{d}s) x_s = \zeta$
– Azur
Sep 1, 2023 at 10:47

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 $$x_0$$.

• 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.
– Azur
Aug 31, 2023 at 16:30
• 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
– 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!