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.
$\Box$
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 $f = \lambda$.
\(\ds \lambda \map {X_m} 1\) | \(=\) | \(\ds \map {X_m} \lambda\) | 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 $V$.
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.
$\blacksquare$
Sources
- 2013: Gerd Rudolph and Matthias Schmidt: Differential Geometry and Mathematical Physics: $\S 1.4$: Proposition $1.4.7$