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My first exposure to the concept of vector fields was in highschool physics courses, which had a simple intuitive idea of being a function which aassociates a point in a given region in space (domain where function is defined) with an arrow pointing in some direction. For example, I can give the famous electric field of point charge centered at origin as vector field:

$$ \vec{E}(r) = \hat{r} \frac{kq}{r^2} $$This is fine.

Recently, I came with a more mathematically sophisiticated definition when going through some lectures on the mathematical side of general relativity. The following definition is given:

A smooth vector field $\chi$ is a smooth map that is a section of the map $TM \xrightarrow{\pi} M$ where $M$ is a topological manifold and $TM$ is the tangent bundle of that manifold. Satifying the law that: $\pi \circ \chi = id_M$ Source 44:45, By prof. Frederic D Schuller

After some few seconds, the professor explains the above definition using the following picture:

enter image description here

He draws on the circle what I know traditionally as a vector field, and then according to length of the arrow of the tangent vector he associates points on the tangent bundle. For example, the bottom most point on the circle has a vector of zero length attached to it, so, on the tangent bundle he choses the point with a height zero above that point.

Note: The prof said that it doesn't matter how we depict the tangent bundle i.e: orient the tangent lines because the tangent bundle in itself has only a structure of a set and nothing else (Maybe I am misinterpretting this point he had mentioned here, please correct me if I am wrong)

This leads me to the following questions:

  1. How exactly does this 'smooth' vector field formalism coincide with the idea of vector fields as function?
  2. What exactly is the significance of this formalism?
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  • $\begingroup$ 2: The significance of this formalism is that a function $S^1\to\Bbb R$ might work to denote vector fields on the circle, but functions $S^2\to \Bbb R^2$ do not work to denote vector fields on the sphere. So we need something else. $\endgroup$
    – Arthur
    May 4, 2021 at 22:31
  • $\begingroup$ I didn't get what you meant here , Could you explain a bit more @Arthur $\endgroup$ Apr 25, 2022 at 23:25

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It's simpler than what it seems: given a surjective function $\pi\colon E \to B$ between sets, we say that a section of $\pi$ is a map $\psi\colon B \to E$ (in the opposite direction) such that $\pi \circ \psi = {\rm Id}_B$. More precisely, if we let $E_b = \pi^{-1}(b)$ be the fiber over $b \in B$, the condition $\pi \circ \psi = {\rm Id}_B$ reads $\psi(b) \in E_b$ for all $b \in B$.

Now, a vector field $X$ on a manifold $M$ satisfies $X_p \in T_pM$ for all $p \in M$. Does that look familiar? We can see $X$ as a map $X\colon M \to TM$, and the previous condition says that $X$ is a section of $\pi\colon TM \to M$ (which takes a tangent vector to its base point). Asking about smoothness of $X$ (again, as a function $M \to TM$) only makes sense once you have introduced a smooth manifold structure on $TM$ as well.

This is all there is to it. The picture may seem a bit misleading because the professor is using a diffeomorphism $T\Bbb S^1 \cong \Bbb S^1 \times \Bbb R$ to give a geometric picture of what a vector field on the circle is. Namely, the tangent line to a circle at a given point is completely determined by the point itself, so the $\Bbb R$ factor in $\Bbb S^1 \times\Bbb R$ only controls the length of the tangent vector, and "negative" lengths just mean that the vector will point in the opposite direction of the one specified by the diffeomorphism $T\Bbb S^1 \cong \Bbb S^1\times\Bbb R$.

The keyword you're looking for is "fiber bundle" (and "vector bundle" and "principal bundle" end up being particular cases). Modern Differential Geometry for Physicists (or something like that) by Chris Isham gives a very nice introduction.

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