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 $\left({U, \kappa}\right)$ 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:


 * $\left\{ {\left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m : i \in \left\{{1, \dotsc, n}\right\} }\right\}$

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 $C^\infty \left({V, \R}\right)$ 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 mappings on $C^\infty \left({V, \R}\right)$.

Hence $\left({X_m + \lambda Y_m}\right)$ are also linear mappings.

Therefore, it is enough to show that $X_m + \lambda Y_m$ satisfies the Leibniz law.

Let $f, g \in C^\infty \left({V, \R}\right)$.

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:
 * $\gamma : I \subseteq \R \to M$
 * with $\gamma \left({0}\right) = m$ such that:

We define:
 * $X^i_m := \dfrac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau} \left({0}\right)$

and as above:


 * $\left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m \left({f}\right) := \dfrac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right)$

Therefore:


 * $\displaystyle X_m \left({f}\right) = \left({\sum_{i \mathop = 1}^n X^i_m \left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m}\right) \left({f}\right)$




 * $\displaystyle X_m = \sum_{i \mathop = 1}^n X^i_m \left.{\frac \partial {\partial \kappa^i} }\right\vert_m$

Hence:


 * $\left\{ {\left.{\dfrac \partial {\partial \kappa^i} }\right\vert_m: i \in \left\{ {1, \dotsc, n}\right\} }\right\}$

forms a basis.

Hence, by Definition:Dimension of Vector Space:


 * $\dim T_m M = n = \dim M$

This completes the proof.