Tangent Space is Vector Space

Prove that the operation of linear combination, as in Definition 2.2.7, makes $T_p(U)$ into an $n$ -dimensional vector space over $\mathbb{R}$ . The zero vector is the infinitesimal curve represented by the constant $p$ . If $\langle s\rangle_p\in T_p(U)$ , then $-\langle s\rangle_p=\langle s^-\rangle_p$ where $s^-(t)=s(-t)$ , defined for all sufficiently small values of $t$ .

Proof:

We verify the axioms of a vector space.

Multiplicative axioms:
* $1\langle s_1\rangle_p=\langle 1s_1+0-(1+0-1)p\rangle_p=\langle s_1\rangle_p$
* $(ab)\langle s_1\rangle_p=\langle abs_1-(ab-1)p\rangle_p$
$\begin{aligned} a(b\langle s_1\rangle_p)&=a\langle bs_1-(b-1)p\rangle_p\\ &=\langle abs_1-(ab-a)p-(a-1)p\rangle_p\\ &=\langle abs_1-(ab-1)p\rangle_p\\ &=(ab)\langle s_1\rangle_p \end{aligned}$

Additive Axioms:
* $\langle s_1\rangle_p+\langle s_2\rangle_p=\langle s_2\rangle_p+\langle s_1\rangle_p=\langle s_1+s_2-p\rangle_p$
* $\begin{aligned} (\langle s_1\rangle_p+\langle s_2\rangle_p)+\langle s_3\rangle_p&=\langle s_1+s_2-p\rangle_p+\langle s_3\rangle_p\\ &=\langle s_1+s_2-p+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
$\begin{aligned} \langle s_1\rangle_p+(\langle s_2\rangle_p+\langle s_3\rangle_p)&=\langle s_1\rangle_p+\langle s_2+s_3-p\rangle_p\\ &=\langle s_1+s_2+s_3-2p\rangle_p \end{aligned}$
* $\langle s\rangle_p+\langle s^-\rangle_p=\langle s+s^- -p\rangle_p$

$\frac{d}{dt}f(s(t)+s(-t)-p)|_{t=0}=0=\frac{d}{dt}f(p)|_{t=0}$

Hence $\langle s\rangle_p+\langle s^-\rangle_p=\langle p\rangle_p$ .
* $\langle s_1\rangle_p+\langle p\rangle_p=\langle s_1+p-p\rangle_p=\langle s_1\rangle_p$

Distributive Axioms:
* $\begin{aligned} a(\langle s_1\rangle_p+\langle s_2\rangle_p)&=a\langle s_1+s_2-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p \end{aligned}$
$\begin{aligned} a\langle s_1\rangle_p+a\langle s_2\rangle_p&=\langle as_1-(a-1)p\rangle_p+\langle as_2-(a-1)p\rangle_p\\ &=\langle a(s_1+s_2)-2(a-1)p-p\rangle_p\\ &=\langle a(s_1+s_2-p)-(a-1)p\rangle_p\\ &=a(\langle s_1\rangle_p+\langle s_2\rangle_p) \end{aligned}$
* $(a+b)\langle s_1\rangle_p=\langle (a+b)s_1-(a+b-1)p\rangle_p$