# Confusion on the vector fields and tangent bundles

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.

• An element of $TM$ is not a vector field. A section of $TM$ sounds better. Commented Jul 22 at 17:02

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.
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 $$p$$.
(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.