Equivalence of Definitions of Tangent Vector

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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.

$\Box$


Definition 1 implies Definition 2

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 $f = \lambda$.


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\(\ds \lambda \map {X_m} 1\) \(=\) \(\ds \map {X_m} \lambda\) 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 $\struct {U, \kappa}$.


By, Taylor's Theorem/n Variables :

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


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 $V$.


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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:

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


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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.

$\blacksquare$


Sources

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