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:


 * $\displaystyle \left. { \left\{{\frac \partial {\partial \kappa^i } } \right\vert _{m} : i \in \left\{{1, \dots, n}\right\}  }\right\}$

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 $X_m, Y_m \in T_m M$ and $\lambda \in \R$.

Then, by Definition: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 $f, g \in C^\infty \left( { V, \R } \right)$.

It follows that $X_m + \lambda Y_m \in T_m M$.

Hence $T_m M$ is a real vector space.

Again, by Definition:Tangent Vector and Equivalence of Definitions of Tangent Vector,

$\forall \, X_m \in T_m M \, \exists$ smooth curve $\gamma : I \subseteq \R \to M$ with $\gamma \left( { 0 } \right ) = m $ such that :

We define:


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

and as above:


 * $\displaystyle \left. {   \frac {\partial} {\partial \kappa^i }   } \right\vert _{m} \, \left( { f } \right) := \frac{\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. {   \frac {\partial} {\partial \kappa^i }   } \right\vert _{m}  } \right) \left( { f }\right) $

iff :


 * $\displaystyle X_m  = \sum_{i=1}^{n} X^i_m \,  \left. {   \frac {\partial} {\partial \kappa^i }   } \right\vert _{m} $

Hence:


 * $\displaystyle \left. { \left\{{\frac \partial {\partial \kappa^i } } \right\vert _{m} : i \in \left\{{1, \dots, n}\right\}  }\right\}$

forms a basis.

Hence, by Definition:Dimension of Vector Space:


 * $\mathop{dim} T_m M = n = \mathop{dim} M$

This completes the proof.