# 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 the concept of Tangent Vector are equivalent:

### Definition 1

A tangent vector $X_m$ on $M$ at $m$ is a linear transformation:

$X_m: C^\infty \left({V, \R}\right) \to \R$

which satisfies the Leibniz law:

$\displaystyle X_m \left({f g}\right) = X_m \left({f}\right) \, g \left({m}\right) + f \left({m}\right) \, X_m \left({g}\right)$

### Definition 2

Let $I$ be an open real interval with $0 \in I$.

Let $\gamma: I \to M$ be a smooth curve with $\gamma \left({0}\right) = m$.

Then a tangent vector $X_m$ at a point $m \in M$ is a mapping

$X_m: C^\infty \left({V, \R}\right) \to \R$

defined by:

$X_m \left({f} \right) := \dfrac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, f \circ \gamma \left({\tau}\right)$

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

## Proof

### Definition 2 implies Definition 1

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

 $\displaystyle X_m \left({f + \lambda g} \right)$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({f + \lambda g}\right) \circ \gamma \left({\tau}\right)$ $\quad$ Definition 2 $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({f \circ \gamma \left({\tau}\right)}\right) + \lambda \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, \left({g \circ \gamma} \left({\tau}\right) \right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle X_m \left({f}\right) + \lambda \, X_m \left({g}\right)$ $\quad$ $\quad$

Thus $X_m$ is linear.

 $\displaystyle X_m \left({f g} \right)$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \left({f g}\right) \circ \gamma \left({\tau}\right)$ $\quad$ Definition 2 $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} f \circ \gamma \left({\tau}\right) \, g \circ \gamma \left({0}\right) + f \circ \gamma \left({0}\right) \, \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} g \circ \gamma \left({\tau}\right)$ $\quad$ Product Rule $\quad$ $\displaystyle$ $=$ $\displaystyle X_m \left({f}\right) \, g \left({m}\right) + f \left({m}\right) \, X_m \left({g}\right)$ $\quad$ Definition 2, $\gamma \left({0}\right) = m$ $\quad$

Hence $X_m$ satisfies the Leibniz law.

Thus $X_m$ satisfies Definition 1.

$\Box$

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

 $\displaystyle X_m \left({f}\right)$ $=$ $\displaystyle X_m \left({1 \, f}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle X_m \left({1}\right) \, f \left({m}\right) + X_m \left({f}\right)$ $\quad$ Leibniz law $\quad$ $\displaystyle \iff$  $\displaystyle$ $\quad$ $\quad$ $\displaystyle X_m \left({1}\right)$ $=$ $\displaystyle 0$ $\quad$ $f \left({m}\right) \ne 0$ $\quad$

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

 $\displaystyle \lambda X_m \left({1}\right)$ $=$ $\displaystyle X_m \left({\lambda}\right)$ $\quad$ Definition of Linear Mapping $\quad$ $\displaystyle$ $=$ $\displaystyle X_m \left({f}\right)$ $\quad$ as $f = \lambda$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ as $X_m \left({1}\right) = 0$ $\quad$

$\Box$

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 $(U, \kappa)$.

 $\displaystyle f$ $=$ $\displaystyle f \circ \kappa^{-1} \circ \kappa$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({f \circ \kappa^{-1} }\right) \left({\kappa}\right)$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left({f \circ \kappa^{-1} }\right) \left({0}\right) + \sum_{i \mathop = 1}^n \frac {\partial} {\left({f \circ \kappa^{-1} }\right)} {\partial{\kappa^i} } \left({0}\right) \, \kappa^i + \mathcal O_2 \left({\kappa}\right)$ $\quad$ $\quad$

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 $\kappa^i h$.

 $\displaystyle X_m \left({\kappa^i h}\right)$ $=$ $\displaystyle X_m \left({\kappa^i}\right) h \left({m}\right) + \kappa^i \left({m}\right) X_m \left({h}\right)$ $\quad$ Definition 1 (Leibniz law) $\quad$ $\displaystyle$ $=$ $\displaystyle X_m \left({\kappa^i}\right) 0 + 0 X_m \left({h}\right)$ $\quad$ as $h \left({m}\right) = 0$, $\kappa^i \left({m}\right) = 0$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0$ $\quad$ $\quad$

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

$\Box$

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:

 $\displaystyle X_m \left({f}\right)$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac{\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right)$ $\quad$ as $\kappa \left({m}\right) = 0$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({0}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({0}\right)$ $\quad$ as $m = \gamma \left({0}\right)$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \left. {\frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({\tau}\right)}\right) \frac {\mathrm d \kappa^i \circ \gamma} {\mathrm d \tau} \left({\tau}\right)} \right \rvert_{\tau \mathop = 0}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \left. {\frac {\mathrm d \left({f \circ \kappa^{-1} \circ \kappa \circ \gamma}\right)} {\mathrm d \tau} \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}$ $\quad$ Chain Rule $\quad$ $\displaystyle$ $=$ $\displaystyle \left. {\frac {\mathrm d \left({f \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}$ $\quad$ as $f \circ \kappa^{-1} \circ \kappa = f$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac {\mathrm d} {\mathrm d \tau} {\restriction_0} \, f \circ \gamma \left({\tau}\right)$ $\quad$ $\quad$

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

This proves the assertion.

$\blacksquare$