Confusion on the vector fields and tangent bundles

Question

Definition of vector fields and tangent vectors I use here:

Let $M$ be a smooth manifold. Given $p\in M$, here the tangent vector $X\in T_pM$ is defined to be a linear map from$C^{\infty}(M)\to \mathbb{R}$ which satisfies $X(fg)=(Xf)g(p)+f(p)(Xg)$. The vector field is a linear map $Y: C^{\infty}(M)\to C^{\infty}(M)$ with the derivation property: $Y(fg)=Y(f)g+f(Xg)$.

Given a set of coordinates $\{x^1,\ldots,x^m\}$ in $\mathbb{R}^m$, then there is a basis for a vector field: $\{\frac{\partial}{\partial x^i}\}^m_{i=1}$. For any vector field $Y\in T\mathbb{R}^m$, $Y(x)=\sum^{m}_{i=1}Y^i(x)\frac{\partial}{\partial x^i}$ ($Y^i\in C^{\infty}(\mathbb{R}^m)$). For example, if $n=3$, then one possible vector field is $x_1\frac{\partial}{\partial x^1}+x_2\frac{\partial}{\partial x^2}+x_3\frac{\partial}{\partial x^1}$.

I am a beginner in Manifolds, some elementary questions confused me:

(1) The tangent bundle is the collection of all tangent spaces ($TM=\sqcup_{x\in M} T_xM$). Why does an element of $TM$ form a vector field? More concretely, what's the elements in the tangent bundle? All the tangent vectors?

(2) Given a vector field $Y$ and $p\in M$, I can understand that every point in a vector field is associated with a value. But in this definition, $Y(p)$ seems to be meaningless since $Y$ is a map from $C^{\infty}(M)\to C^{\infty}(M)$, so it should take a map $f:M\to \mathbb{R}$.

Very appreciated to any help. I've thought it for a long time.

Answer 1

The tangent bundle $TM$ of a manifold $M$ is the set of pairs $(p,\xi)$ consisting of a point $p$ in the manifold $M$ and a tangent vector $\xi \in T_pM$ in the tangent space at $p$. If $\psi \colon U \to \mathbb R^n$ is a diffeomorphism from an open set $U$ of $M$ to (say) an open ball in $\mathbb R^n$ then the tangent spaces at points in $\psi(U)$ are all canonically identified with $\mathbb R^n$, and so the tangent bundle "over $U$" is just a product of $U$ with $\mathbb R^n$. Taking an atlas for $M$ one can then given $TM$ a manifold structure (in fact the structure of a vector bundle over $M$): the compatibility of the charts in the atlas for $M$ induces, by taking the derivative, corresponding maps for $TM$ (here it is easier to work with smooth, rather than say $\mathcal C^k$, manifolds) and these maps are linear when restricted to the tangent spaces themselves.

There are (perhaps unhelpfully) different notions of derivations:

1. One can consider derivations "at a point $p\in M$", that is, maps $D\colon \mathcal C^{\infty}(M) \to \mathbb R$ which satisfy the condition $$ D(f.g) = D(f).g(p)+f(p).D(g) \quad \forall f,g \in \mathbb C^{\infty}.(M) $$ The set of all derivations at a point $p$ form a vector space which can either be canonically identified with $T_pM$ the tangent space of $M$ at $p$, or is just equal to it by definition (depending on your conventions).

2. The other common notion of a derivation in this context is a map $D \colon \mathcal C^{\infty}(M)\to \mathcal C^{\infty}(M)$ which is $\mathbb R$-linear and which satisfies $$ D(f.g)=D(f).g+f.D(g) \quad \forall f,g \in \mathbb C^{\infty}(M) $$ Derivations of this type correspond to vector fields. A vector field $\nu$ however is not an element of the tangent bundle $TM$, it is a "section" of the tangent bundle, that is, it is a right-inverse to the projection map $\pi \colon TM \to M$, so that $\nu\colon M \to TM$ is a smooth map from $M$ to $TM$ and $\pi \circ \nu = \text{id}_M$. Thus $\nu(p) = (p,\xi)$ where we must have $\xi \in T_pM$, and hence $\nu$ gives a choice of tangent vector in $T_pM$ at every point $p \in M$ depending smoothly on the point $p$.

Answer 2

(1): The tangent bundle $TM$ consists of linear derivations at some point of $M$ (this is the Leibniz rule identity that you wrote down). These are the elements. To see a linear derivation $v\in T_mM$ geometrically (i.e. to view elements of the tangent bundle as tangent vectors at a point of $M$), simply pick a coordinate system $(U, \phi)$ centered at $m$ for which $v=d\phi^{-1}\left(\frac{\partial}{\partial x_1}\mid_0\right)$ and observe that $v$ is (in these coordinates) the tangent vector (in the usual geometric sense!) at 0 to the curve $t\mapsto \phi^{-1}(t, 0, \dots, 0)$.

(2): I am not certain what the context is, but it is often the case that an element $p\in M$ is identified with the constant map $x\mapsto p\in C^\infty(M)$. Hopefully that clears things up.