Tangent Space is Vector Space
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Theorem
Let $M$ be a smooth manifold of dimension $n \in \N$.
Let $m \in M$ be a point.
Let $\struct {U, \kappa}$ be a chart with $m \in U$.
Let $T_m M$ be the tangent space at $m$.
Then $T_m M$ is a real vector space of dimension $n$, spanned by the basis:
- $\set {\valueat {\dfrac \partial {\partial \kappa^i} } m : i \in \set {1, \dotsc, n} }$
that is, the set of partial derivatives with respect to the $i$th coordinate function $\kappa^i$ evaluated at $m$.
Proof
Let $V$ be an open neighborhood of $m$ with $V \subseteq U \subseteq M$.
Let $\map {C^\infty} {V, \R}$ be the set of smooth mappings $f: V \to \R$.
Let $X_m, Y_m \in T_m M$.
Let $\lambda \in \R$.
Then, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
- $X_m, Y_m$ are linear transformations on $\map {C^\infty} {V, \R}$.
Hence $\paren {X_m + \lambda Y_m}$ are also linear transformations.
Therefore, it is enough to show that $X_m + \lambda Y_m$ satisfies the Leibniz law.
Let $f, g \in \map {C^\infty} {V, \R}$.
Then:
\(\ds \map {\paren {X_m + \lambda Y_m} } {f g}\) | \(=\) | \(\ds \map {X_m} {f g} + \lambda \map {Y_m} {f g}\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {X_m} f \map g m + \map f m \map {X_m} g + \lambda \paren {\map {Y_m} f \map g m + \map f m \map {Y_m} g}\) | Leibniz law for $X_m, Y_m$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {X_m + \lambda Y_m} } f \map g m + \map f m \map {\paren {X_m + \lambda Y_m} } g\) | reordering summands |
It follows that:
- $X_m + \lambda Y_m \in T_m M$
Hence $T_m M$ is a real vector space.
Again, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
- for all $X_m \in T_m M$ there exists a smooth curve:
- $\gamma: I \subseteq \R \to M$
- where $\map \gamma 0 = m$ such that:
\(\ds \map {X_m} f\) | \(=\) | \(\ds \valueat {\map {\frac {\map \d {f \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \valueat {\map {\frac {\map \d {f \circ \kappa^{-1} \circ \kappa \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}\) | $f \circ \kappa^{-1} \circ \kappa = f$, as $\kappa$ is a homeomorphism, in particular a bijection. | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \valueat {\map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} \tau} \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } \tau} {\tau \mathop = 0}\) | Chain Rule for Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0 \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map {\kappa \circ \gamma} 0}\) | rearranging | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \map {\frac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0 \map {\frac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m}\) | as $m = \map \gamma 0$ |
We define:
- $X^i_m := \map {\dfrac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0$
and as above:
- $\valueat {\map {\dfrac \partial {\partial \kappa^i} } m} f := \map {\dfrac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m}$
Therefore:
- $\ds \map {X_m} f = \map {\paren {\sum_{i \mathop = 1}^n X^i_m \valueat {\dfrac \partial {\partial \kappa^i} } m} } f$
- $\ds X_m = \sum_{i \mathop = 1}^n X^i_m \valueat {\frac \partial {\partial \kappa^i} } m$
Hence:
- $\set {\valueat {\dfrac \partial {\partial \kappa^i} } m: i \in \set {1, \dotsc, n} }$
forms a basis.
Hence, by definition of dimension of vector space:
- $\dim T_m M = n = \dim M$
This completes the proof.
$\blacksquare$