Equivalence of Definitions of Tangent Vector
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Contents
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.
$\Box$
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 $f \in C^\infty \left({V, \R}\right)$ 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 $f = \lambda$.
$\exists$ in the above? Should it not be $\forall$?
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| \(\displaystyle \lambda X_m \left({1}\right)\) | \(=\) | \(\displaystyle X_m \left({\lambda}\right)\) | $\quad$ 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 $(U, \kappa)$.
By, Taylor's Theorem/n Variables :
| \(\displaystyle f\) | \(=\) | \(\displaystyle f \circ \kappa^{-1} \circ \kappa\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({f \circ \kappa^{-1} }\right) \left({\kappa}\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({f \circ \kappa^{-1} }\right) \left({0}\right) + \sum_{i \mathop = 1}^n \frac {\partial} {\left({f \circ \kappa^{-1} }\right)} {\partial{\kappa^i} } \left({0}\right) \, \kappa^i + \mathcal O_2 \left({\kappa}\right)\) | $\quad$ | $\quad$ |
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 $V$.
The above needs more than observation.
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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:
- $\displaystyle \kappa = \sum_{i \mathop = 1}^n \kappa^i e_i$
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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.
$\blacksquare$
Sources
- 2013: Gerd Rudolph and Matthias Schmidt: Differential Geometry and Mathematical Physics: $\S 1.4$: Proposition $1.4.7$
