Tangent Space is Vector Space

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:

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:

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.