I have been studying differential operators on varieties on Liviu Nicolaescu's book "Lectures on the Geometry of Manifolds" and I have managed to understand how to go from differential operators on $\mathbb{R}^n$ to operators on smooth manifolds but, when wanting to apply to a simple exercise, for example, the derivative operator on the variety $S^1$. I don't know how to apply the theory to this operator.
I know that, $S^1$ is a manifold. At $(1,0)\in S^1$ thre exists $T_{(1,0)}(S^1)$, the tangent space at $(1,0)$. Consider the differential operator $d/dx:C^{\infty}(E)\to C^{\infty}(E)$ and define $C^{\infty}(E):=\left\{s:S^1\to E:s \text{ smooth section}\right\}$ and let $\pi:E\to S^1$ be vector bundle.
How can I make sense of this operator? Or can one not make sense of it?
Additionally, the bundle I take is $E=\left\{v\in \mathbb{R}^2: v\perp (1,0)\right\}$
I can't quite understand the operators on varieties when applying it to a simple exercise.
You can use the exponential form $\rho e^{i\theta}$ to express elements in $\mathbb R^2\setminus\{(0,0)\}=\mathbb C\setminus\{0\}$.
Now you can think $\mathbb S^1$ as a subset of $\mathbb C\setminus \{0\}$ defined by $\rho=1$.
Take the map $\pi:\mathbb C\setminus\{0\}\to\mathbb S^1$ s.t. $\pi(\rho e^{i\theta})=e^{i\theta}$. Now we have a structure of vector bundle.
As trivialization take the bijection $$\Phi:\pi^{-1}(\mathbb S^1)\to \mathbb S^1\times \mathbb R$$ $$\rho e^{i\theta}\mapsto (e^{i\theta},\log(\rho)).$$ So if $E:=\mathbb C\setminus \{0\}$, the map $\dfrac{d}{dx}:\mathcal C^{\infty}(E)\to \mathcal C^{\infty}(E)$ is a linear function that satisfies Leibniz's rule.