Equivalence of Definitions of Tangent Vector

Theorem
Definition 1 and Definition 2 in Definition:Tangent Space are equivalent.

Proof
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, $X_m(0)= 0$.

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

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

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 :
 * $\displaystyle f = f \circ \kappa^{-1} \circ \kappa = \left( { f \circ \kappa^{-1} }\right) \left( { \kappa }\right)

= \left( {f \circ \kappa^{-1} }\right) \left( { 0 }\right)  + \sum_{i=1}^{n} \frac{\partial{ \left( {f \circ \kappa^{-1}  }\right) }}{\partial{ \kappa^i}} \left( { 0 }\right) \, \, \kappa^i + \mathcal O _2 \left( { \kappa }\right)$ where $\mathcal O _2 \left( { \kappa }\right)$ is the $2$nd order error term.

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

Therefore, by linearity :
 * $\displaystyle X_m \left( { f } \right) = X_m \left( { f \left( {m} \right) } \right) + \sum_{i=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 $i \in \left\{ {1, \dots, n}  \right\}$ 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=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=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


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

By $\gamma \left( { 0 }\right) = m$ and the chain rule:


 * $\displaystyle \sum_{i=1}^{n} \frac{\partial{ \left( {f \circ \kappa^{-1} }\right) }}{\partial{

\kappa^i}} \left( { \kappa \left( { m }\right) }\right) \, \, \frac{\mathrm{d}{\kappa^i \circ \gamma} }{\mathrm{d}{\tau} }  \left( { 0 }\right) = \sum_{i=1}^{n} \frac{\partial{ \left( {f \circ \kappa^{-1}  }\right) }}{\partial{ \kappa^i}} \left( { \kappa \circ \gamma \left( { 0 }\right) }\right) \, \, \frac{\mathrm{d}{\kappa^i \circ \gamma} }{\mathrm{d}{\tau} } \left( { 0 }\right) = \sum_{i=1}^{n} \left. {\frac{\partial{ \left( {f \circ \kappa^{-1} }\right) }}{\partial{ \kappa^i}} \left( { \kappa \circ \gamma \left( { \tau }\right) }\right) \, \, \frac{\mathrm{d}{\kappa^i \circ \gamma} }{\mathrm{d}{\tau} } \left( { \tau }\right) } \, \, \right\rvert_0$