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

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

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:

\(\displaystyle \left({X_m + \lambda Y_m}\right) \left({f g}\right)\) \(=\) \(\displaystyle X_m \left({f g}\right) + \lambda Y_m \left({f g}\right)\) Definition of Linear Transformation
\(\displaystyle \) \(=\) \(\displaystyle X_m \left({f}\right) g \left({m}\right) + f \left({m}\right) X_m \left({g}\right) + \lambda \left({Y_m \left({f}\right) g \left({m}\right) + f \left({m}\right) Y_m \left({g}\right)}\right)\) Leibniz law for $X_m, Y_m$
\(\displaystyle \) \(=\) \(\displaystyle \left({X_m + \lambda Y_m} \right) \left({f}\right) g \left({m}\right) + f \left({m}\right) \left({X_m + \lambda Y_m}\right) \left({g}\right)\) 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:
$\gamma : I \subseteq \R \to M$
with $\gamma \left({0}\right) = m$ such that:
\(\displaystyle X_m \left({f}\right)\) \(=\) \(\displaystyle \left. {\frac {\mathrm d \left({f \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right) }\right\rvert_{\tau \mathop = 0}\)
\(\displaystyle \) \(=\) \(\displaystyle \left. {\frac {\mathrm d \left({f \circ \kappa^{-1} \circ \kappa \circ \gamma} \right)} {\mathrm d \tau} \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}\) $f \circ \kappa^{-1} \circ \kappa = f$, as $\kappa$ is a homeomorphism, in particular a bijection.
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \left. {\frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({\tau}\right)}\right) \frac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau} \left({\tau}\right)} \right\rvert_{\tau \mathop = 0}\) Chain Rule for Real-Valued Functions
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \frac {\mathrm d \left({\kappa^i \circ \gamma} \right)} {\mathrm d \tau} \left({0}\right) \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \circ \gamma \left({0}\right)}\right)\) rearranging
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \frac {\mathrm d \left({\kappa^i \circ \gamma}\right)} {\mathrm d \tau} \left({0}\right) \frac {\partial \left({f \circ \kappa^{-1} }\right)} {\partial \kappa^i} \left({\kappa \left({m}\right)}\right)\) as $m = \gamma \left({0}\right)$


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:

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

if and only if:

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

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.

$\blacksquare$