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:

$\ds \map {X_m} f = \map {\paren {\sum_{i \mathop = 1}^n X^i_m \valueat {\dfrac \partial {\partial \kappa^i} } m} } f$

if and only if:

$\ds X_m = \sum_{i \mathop = 1}^n X^i_m \valueat {\frac \partial {\partial \kappa^i} } m$

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.

$\blacksquare$