## 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} $\langle s_1\rangle_p+\langle s_2\rangle_p=\langle s_2\rangle_p+\langle s_1\rangle_p=\langle s_1+s_2-p\rangle_p$ \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} $(a+b)\langle s_1\rangle_p=\langle (a+b)s_1-(a+b-1)p\rangle_p$ $a\langle s_1\rangle_p+b\langle s_1\rangle_p=\langle as_1+bs_1-(a+b-1)p\rangle_p=(a+b)\langle s_1\rangle_p$

Hence $T_p(U)$ is a vector space over $\mathbb{R}$. Since $U\subseteq\mathbb{R}^n$, $T_p(U)$ is $n$-dimensional. 