📂Geometry

The Tangent Space on an n-Dimensional Differentiable Manifold is an n-Dimensional Vector Space

Overview

Let $M$ be a $n$-dimensional differential manifold, and let $T_{p}M$ be the tangent space at point $p\in M$. The tangent space becomes a vector space, specifically, a $n$-dimensional vector space. The following set becomes the basis of the tangent space, which is very useful in the study of differential manifolds.

$$\mathcal{B} = \left\{ \left. \dfrac{\partial }{\partial x_{i}} \right|_{p} : 1 \le i \le n \right\}$$

Theorem 1¹

$T_{p}M$ is a $\mathbb{R}-$vector space.

Proof

The conditions for being a vector space are as follows, of which we will prove only a few:

For $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $k, l \in \mathbb{F}$,

(A1) If $\mathbf{u}, \mathbf{v}$ is an element of $V$, then $\mathbf{u}+\mathbf{v}$ is also an element of $V$.

(A2) $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$

(A3) $(\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})$

(A4) For all $\mathbf{u}$ in $V$, there exists $\mathbf{0}$ in $V$ such that $\mathbf{u} + \mathbf{0} = \mathbf{0} + \mathbf{u} = \mathbf{u}$.

(A5) For all $\mathbf{u}$ in $V$, there exists $\mathbf{v}$ in $V$ such that $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} = \mathbf{0}$.

(M1) If $\mathbf{u}$ is an element of $V$, then $k \mathbf{u}$ is also an element of $V$.

(M2) $k(\mathbf{u} + \mathbf{v})=k\mathbf{u} + k\mathbf{v}$

(M3) $(k+l)\mathbf{u}=k\mathbf{u}+ l\mathbf{u}$

(M4) $k(l\mathbf{u})=(kl)(\mathbf{u})$

(M5) For $1\in \mathbb{F}$, $1\mathbf{u} = \mathbf{u}$

Let it be $\mathbf{X}, \mathbf{Y} \in T_{p}M$. Let the set of differentiable functions on $p$ be $\mathcal{D}$.

$$\mathcal{D} := \left\{ f : M \to \mathbb{R} | \text{functions on } M \text{that are differentiable at } p \right\}$$

For being a vector space, the addition between elements and scalar multiplication must be defined. Let’s define the addition and scalar multiplication as follows.

$$(\mathbf{X} + \mathbf{Y}) (f) := \mathbf{X}f + \mathbf{Y}f \\ (r \cdot \mathbf{X}) (f) := r \cdot \mathbf{X}f,\quad r \in \mathbb{R}$$

(A1) Since $\mathbf{X}$ and $\mathbf{Y}$ are functions in $\mathcal{D} \to \mathbb{R}$, $\mathbf{X}f, \mathbf{Y}f \in \mathbb{R}$ holds. The sum of two real numbers is a real number, so $\mathbf{X} + \mathbf{Y} : \mathcal{D} \to \mathbb{R}$ holds. Therefore, $\mathbf{X} + \mathbf{Y} \in T_{p}M$ holds.

(M1) Since $\mathbf{X}f \in \mathbb{R}$ and $r \in \mathbb{R}$, $r \cdot \mathbf{X} f \in \mathbb{R}$ holds. Therefore, $r\mathbf{X} : \mathcal{D} \to \mathbb{R}$ holds, and $r\mathbf{X} \in T_{p}M$ is true.

(A4) Let’s define $\mathbf{0} : \mathcal{D} \to \mathbb{R}$ as $\mathbf{0} f = 0 (\forall f \in \mathcal{D})$. Then

$$(\mathbf{X} + \mathbf{0})(f) = \mathbf{X}f + \mathbf{0}f = \mathbf{X}f$$

(A5) If we define $-\mathbf{X}$ as $(-\mathbf{X})(f) := (-1) \cdot \mathbf{X}f$, by (M1), $(-1)\mathbf{X} \in T_{p}M$ holds, and $\mathbf{X} + (-\mathbf{X}) = \mathbf{0}$ is satisfied.

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Theorem 2

$T_{p}M$ is a $n$dimensional vector space. In particular, let $\mathbf{x} : U \to M$ be the coordinate system for point $p \in M$. Then the set

$$\mathcal{B} = \left\{ \left. \dfrac{\partial }{\partial x_{i}} \right|_{p} : 1 \le i \le n \right\}$$

is the basis of the tangent space $T_{p}M$. In this case, $\left. \dfrac{\partial }{\partial x_{i}} \right|_{p} : \mathcal{D} \to \mathbb{R}$ is defined as follows.

$$\left. \dfrac{\partial }{\partial x_{i}} \right|_{p} f := \left. \dfrac{\partial (f\circ \mathbf{x})}{\partial u_{i}} \right|_{p},\quad f \in \mathcal{D}$$

$(u_{1}, \dots, u_{n})$ are the coordinates of $\mathbb{R}$.

Proof

According to the definition of basis, we need to show that $\mathcal{B}$ is linearly independent, and generates $T_{p}M$.

• Part 1. Linear independence

  For the coordinate system $\mathbf{x} : U \subset \mathbb{R}^{n} \to M$, the coordinate functions $x_{i} : M \to \mathbb{R}$ are as follows:

  $$\mathbf{x}(\mathbf{u}) = p \\ \mathbf{x}^{-1}(p) = (x_{1}(p), \dots, x_{n}(p)) = (u_{1}, \dots, u_{n})$$

  If we let $f = x_{j}$,

  $$f \circ \mathbf{x}(\mathbf{u}) = x_{j} \circ \mathbf{x}(\mathbf{u}) = x_{j}(\mathbf{x}(\mathbf{u})) = x_{j}(p) = u_{j}$$

  Therefore,

  $$\left. \dfrac{\partial }{\partial x_{i}} \right|_{p} x_{j} = \left. \dfrac{\partial (x_{j} \circ \mathbf{x})}{\partial u_{i}} \right|_{p} = \dfrac{\partial u_{j}}{\partial u_{i}} = \delta_{ij}$$

  Here, $\delta_{ij}$ is the Kronecker delta.

  Looking at equation $c_{i}\left. \dfrac{\partial }{\partial x_{i}} \right|_{p} = \mathbf{0}$, if for all $i$, $c_{i} = 0$ holds, then it is linearly independent. By substituting any $x_{j}$ for $1 \le j \le n$,

  $$0 = \mathbf{0}(x_{j}) = c_{i}\left. \dfrac{\partial }{\partial x_{i}} \right|_{p} x_{j} = c_{i}\delta_{ij} = c_{j}$$

  Therefore, for all $j$, $c_{j} = 0$ holds, and no other solution exists. Hence, $\mathcal{B}$ is linearly independent.

• Part 2. Generation

  Let’s denote $f \in \mathcal{D}$, $\mathbf{a} = \mathbf{x}^{-1} (p)$, $F = f \circ \mathbf{x}$. Then $F : \mathbb{R}^{n} \to \mathbb{R}$ holds.

  Taylor’s theorem for multivariable functions

  $$F(\mathbf{x}) = F(\mathbf{a}) + \sum_{i} \dfrac{\partial F}{\partial x_{i}}(\mathbf{a})(x_{i} - a_{i}) + \sum_{i,j}h_{ij}(\mathbf{x})(x_{i} - a_{i}) (x_{j} - a_{j})$$

  Applying the above Taylor’s theorem to $F$, the following holds. For $m \in M$,

  $$\begin{align*} f(m) =&\ F \circ \mathbf{x}^{-1}(m) = F ( \mathbf{x}^{-1}(m)) \\ =&\ F (\mathbf{a}) + \sum_{i} \dfrac{\partial F}{\partial u_{i}}(\mathbf{a})(x_{i}(m) - a_{i}) + \sum_{i,j}h_{ij}(\mathbf{x}^{-1}(m))(x_{i}(m) - a_{i}) (x_{j}(m) - a_{j}) \\ =&\ F \circ \mathbf{x}^{-1}(p) + \sum_{i} \frac{\partial F}{\partial u_{i}}(\mathbf{x}^{-1}(p)) \left(x_{i}(m)-a_{i}\right) +\sum h_{i j}(\mathbf{x}^{-1}(m))\left(x_{i}(m)-a_{i}\right)\left(x_{j}(m)-a_{j}\right) \\ =&\ f(p) + \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \left(x_{i}(m)-a_{i}\right) +\sum h_{i j}(\mathbf{x}^{-1}(m))\left(x_{i}(m)-a_{i}\right)\left(x_{j}(m)-a_{j}\right) \end{align*}$$

  Therefore, $f$ is the following mapping:

  $$f = f(p) + \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \left(x_{i}-a_{i}\right) +\sum (h_{i j} \circ \mathbf{x}^{-1})\left(x_{i}-a_{i}\right)\left(x_{j}-a_{j}\right)$$

  When applied to the tangent vector $\mathbf{X}$,

  $$\mathbf{X}(f) = \mathbf{X}(f(p)) + \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \mathbf{X}\left(x_{i}-a_{i}\right) +\sum \mathbf{X}\left[ (h_{i j} \circ \mathbf{x}^{-1})\left(x_{i}-a_{i}\right)\left(x_{j}-a_{j}\right) \right]$$

  Since the tangent vector is defined as a differential operator, for the constant function $c$, $\mathbf{X}(c) = 0$ holds. Therefore, the first term in the equation above is $0$.

  $$\mathbf{X}_{p}(fg) = f(p)\mathbf{X}_{p}(g) + g(p)\mathbf{X}_{p}(f)$$

  Therefore, if $f(p)=g(p)=0$,

  $$\mathbf{X}_{p}(fg) = 0$$

  Also, by setting $f = (h_{i j} \circ \mathbf{x}^{-1})\left(x_{i}-a_{i}\right)$, $g=\left(x_{j}-a_{j}\right)$, and applying the above lemma, $x_{i}(p) = a_{i}$ holds, so the third term is also concluded to be $0$. Thus, the following is obtained.

  $$\begin{aligned} \mathbf{X}(f) =&\ \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \mathbf{X}\left(x_{i}-a_{i}\right) \\ =&\ \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \left(\mathbf{X}(x_{i}) - \mathbf{X}(a_{i})\right) \\ =&\ \sum_{i} \left( \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \right) \mathbf{X}(x_{i}) \\ =&\ \sum_{i} \mathbf{X}(x_{i}) \left.\frac{\partial }{\partial x_{i}}\right|_{p} f \end{aligned}$$

  $$\implies \mathbf{X} = \sum_{i} \mathbf{X}(x_{i}) \left.\frac{\partial }{\partial x_{i}}\right|_{p}$$

  Therefore, since any $\mathbf{X}$ can be expressed as a linear combination of $\mathcal{B}$, $T_{p}M$ is generated by $\mathcal{B}$.

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