Equivalence of Definitions of Tangent Vector

Theorem
Let $M$ be a smooth manifold.

Let $m \in M$ be a point.

Let $V$ be an open neighborhood of $m$.

Let $C^\infty \left({V, \R}\right)$ be defined as the set of all smooth mappings $f: V \to \R$.

The following definitions of tangent vector are equivalent:

Definition 2 implies Definition 1
Let $\lambda \in \R$ and $f, g \in C^\infty \left({V, \R}\right)$.

Thus $X_m$ is linear.

Hence $X_m$ satisfies the Leibniz law.

Thus $X_m$ satisfies Definition 1.

Lemma 1
Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1.

Let $V$ be an open neighborhood of $M$.

Let $f \in C^\infty \left({V, \R}\right)$ be constant.

Then $X_m \left({f}\right) = 0$.

Proof of Lemma 1
Let $f \left({m}\right) = 0$.

Then, by constancy, $f = 0$ on $V$.

Hence, by linearity, $X_m \left({0}\right) = 0$.

Let $f \left({m}\right) \ne 0$.

$f$ is constant, $\exists \lambda \in \R : f \left({V}\right) = \left\{ {\lambda} \right\}$  $f = \lambda$.

Let $X_m$ be a tangent vector at $m \in M$ according to Definition 1.

Denote $n := \dim M$.

Let $f \in C^\infty \left({V, \R}\right)$.

Let $(U, \kappa)$ be a chart with $\kappa \left( {m} \right) = 0$.

Let $\kappa^i$ be the $i$th coordinate function of the chart $(U, \kappa)$.

By, Taylor's Theorem/n Variables :

Observe that $\displaystyle \left({f \circ \kappa^{-1} }\right) \left({0}\right) = \left({f \circ \kappa^{-1} }\right) \left( {\kappa \left({m}\right)}\right) = f \left({m}\right) $ is a constant mapping on $V$.

Define $X^i := X_m \left({\kappa ^i}\right)$.

Then by linearity:
 * $\displaystyle X_m \left({f}\right) = X_m \left({f \left({m}\right)}\right) + \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({0}\right) \ X^i + X_m \left({\mathcal O_2 \left({\kappa}\right) }\right)$

Lemma 2

 * $X_m \left({\mathcal O_2 \left({\kappa}\right)}\right) = 0$

Proof of Lemma 2
By Taylor's Theorem/n Variables, for each summand of $\mathcal O _2 \left({\kappa}\right)$ there exists $i \in \left\{ {1, \ldots, n} \right\}$ and an $h \in C^\infty \left({V, \R}\right)$ with $h \left({m}\right) = 0$ such that the summand is $\kappa^i h$.

Thus the sum $\mathcal O_2 \left({\kappa}\right)$ vanishes.

By Lemma 1 :
 * $\displaystyle X_m \left({f \left({m}\right)} \right) = 0$

By Lemma 2 :
 * $\displaystyle X_m \left({\mathcal O_2 \left({\kappa }\right)}\right) = 0$

Hence:
 * $\displaystyle X_m \left({f}\right) = \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1}}\right)} {\partial \kappa^i} \left({0}\right) \ X^i$

Let $\left\{{e_i}\right\}$ be a basis of $\R^n$ such that:
 * $\displaystyle \kappa = \sum_{i \mathop = 1}^n \kappa^i e_i$

Choose a smooth curve $\gamma: I \to M$ with $0 \in I \subseteq \R$ such that $\gamma \left({0}\right) = m$ and:


 * $\dfrac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right) := X^i$

Then:

Hence $X_m$ is a tangent vector according to Definition 2.

This proves the assertion.