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 $\map {C^\infty} {V, \R}$ 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:

$X_m: \map {C^\infty} {V, \R} \to \R$

which satisfies the Leibniz law:

$\ds \map {X_m} {f g} = \map {X_m} f \map g m + \map f m \map {X_m} g$

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

$X_m: \map {C^\infty} {V, \R} \to \R$

defined by:

$\map {X_m} f := \map {\dfrac \d {\d \tau} {\restriction_0} } {\map {f \circ \gamma} \tau}$

for all $f \in \map {C^\infty} {V, \R}$.

Proof

Definition 2 implies Definition 1

Let $\lambda \in \R$ and $f, g \in \map {C^\infty} {V, \R}$.

 $\ds \map {X_m} {f + \lambda g}$ $=$ $\ds \map {\frac \d {\d \tau} {\restriction_0} } {\map {\paren {f + \lambda g} \circ \gamma} \tau}$ Definition 2 $\ds$ $=$ $\ds \map {\frac \d {\d \tau} {\restriction_0} } {\map {f \circ \gamma} \tau} + \lambda \map {\frac \d {\d \tau} {\restriction_0} } {\map {g \circ \gamma} \tau}$ $\ds$ $=$ $\ds \map {X_m} f + \lambda \map {X_m} g$

Thus $X_m$ is linear.

 $\ds \map {X_m} {f g}$ $=$ $\ds \frac \d {\d \tau} {\restriction_0} \map {\paren {f g} \circ \gamma} \tau$ Definition 2 $\ds$ $=$ $\ds \frac \d {\d \tau} {\restriction_0} \map {f \circ \gamma} \tau \, \map {g \circ \gamma} 0 + \map {f \circ \gamma} 0 \, \frac \d {\d \tau} {\restriction_0} \map {g \circ \gamma} \tau$ Product Rule $\ds$ $=$ $\ds \map {X_m} f \map g m + \map f m \map {X_m} g$ Definition 2, $\map \gamma 0 = m$

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 $f \in \map {C^\infty} {V, \R}$ be constant.

Then $\map {X_m} f = 0$.

Proof of Lemma 1

Let $\map f m = 0$.

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

Hence, by linearity, $\map {X_m} 0 = 0$.

Let $\map f m \ne 0$.

 $\ds \map {X_m} f$ $=$ $\ds \map {X_m} {1 \, f}$ $\ds$ $=$ $\ds \map {X_m} 1 \map f m + \map {X_m} f$ Leibniz law $\ds \iff$  $\ds$ $\ds \map {X_m} 1$ $=$ $\ds 0$ $\map f m \ne 0$

$f$ is constant, if and only if $\exists \lambda \in \R : f \sqbrk V = \set \lambda$ if and only if $f = \lambda$.

 $\ds \lambda \map {X_m} 1$ $=$ $\ds \map {X_m} \lambda$ Definition of Linear Mapping $\ds$ $=$ $\ds \map {X_m} f$ as $f = \lambda$ $\ds$ $=$ $\ds 0$ as $\map {X_m} 1 = 0$

$\Box$

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

Denote $n := \dim M$.

Let $f \in \map {C^\infty} {V, \R}$.

Let $\struct {U, \kappa}$ be a chart such that $\map \kappa m = 0$.

Let $\kappa^i$ be the $i$th coordinate function of the chart $\struct {U, \kappa}$.

 $\ds f$ $=$ $\ds f \circ \kappa^{-1} \circ \kappa$ $\ds$ $=$ $\ds \map {\paren {f \circ \kappa^{-1} } } \kappa$ $\ds$ $=$ $\ds \map {\paren {f \circ \kappa^{-1} } } 0 + \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } 0 \, \kappa^i + \map {\OO_2} \kappa$

Observe that $\ds \map {\paren {f \circ \kappa^{-1} } } 0 = \map {\paren {f \circ \kappa^{-1} } } {\map \kappa m} = \map f m$ is a constant mapping on $V$.

Define $X^i := \map {X_m} {\kappa^i}$.

Then by linearity:

$\ds \map {X_m} f = \map {X_m} {\map f m} + \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } 0 \ X^i + \map {X_m} {\map {\OO_2} \kappa}$

Lemma 2

$\map {X_m} {\map {\OO_2} \kappa} = 0$

Proof of Lemma 2

By Taylor's Theorem/n Variables, for each summand of $\map {\OO_2} \kappa$ there exists $i \in \set {1, \ldots, n}$ and an $h \in \map {C^\infty} {V, \R}$ with $\map h m = 0$ such that the summand is $\kappa^i h$.

 $\ds \map {X_m} {\kappa^i h}$ $=$ $\ds \map {X_m} {\kappa^i} \map h m + \map {\kappa^i} m \map {X_m} h$ Definition 1 (Leibniz law) $\ds$ $=$ $\ds \map {X_m} {\kappa^i} 0 + 0 \map {X_m} h$ as $\map h m = 0$, $\map {\kappa^i} m = 0$ $\ds$ $=$ $\ds 0$

Thus the sum $\map {\OO_2} \kappa$ vanishes.

$\Box$

By Lemma 1:

$\map {X_m} {\map f m} = 0$

By Lemma 2:

$\map {X_m} {\map {\OO_2} \kappa} = 0$

Hence:

$\ds \map {X_m} f = \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } 0 \ X^i$

Let $\set {e_i}$ be a basis of $\R^n$ such that:

$\ds \kappa = \sum_{i \mathop = 1}^n \kappa^i e_i$

Choose a smooth curve $\gamma: I \to M$ with $0 \in I \subseteq \R$ such that $\map \gamma 0 = m$ and:

$\map {\dfrac {\d \kappa^i \circ \gamma} {\d \tau} } 0 := X^i$

Then:

 $\ds \map {X_m} f$ $=$ $\ds \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m} \map {\frac {\d \kappa^i \circ \gamma} {\d \tau} } 0$ as $\map \kappa m = 0$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} 0} \map {\frac {\d \kappa^i \circ \gamma} {\d \tau} } 0$ as $m = \map \gamma 0$ $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n \valueat {\map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} \tau} \map {\frac {\d \kappa^i \circ \gamma} {\d \tau} } \tau} {\tau \mathop = 0}$ $\ds$ $=$ $\ds \intlimits {\map {\frac {\map \d {f \circ \kappa^{-1} \circ \kappa \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0} {}$ Chain Rule for Derivatives $\ds$ $=$ $\ds \intlimits {\map {\frac {\map \d {f \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0} {}$ as $f \circ \kappa^{-1} \circ \kappa = f$ $\ds$ $=$ $\ds \frac \d {\d \tau} {\restriction_0} \, \map {f \circ \gamma} \tau$

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

This proves the assertion.

$\blacksquare$