The problem that I am having is that I am having quite a hard time understanding the ideas of Geometric Tangent vectors and why they are even needed - I mean one already has the usual "Tangent" of a curve, why more definitions?
We let X be an n-dimensional differentiable manifold $p \in X$, and $(U,h,V)$ is a chart for X. The set of all differentiable curves $\gamma_{1},\gamma_{2}$ through p is denoted by $K_{p}$ and $\gamma: (-\epsilon,\epsilon) \rightarrow X, \epsilon > 0$, with $\gamma(0) = p.\space$ Now $\gamma_{1} \sim \gamma_{2}$ if $\dot{\gamma_{1}}(0)_{h} = \dot{\gamma_{2}}(0)_{h}$. Where $\space \dot{\gamma}(0)_{h} \equiv d/dt(h \circ \gamma)(0) \in \mathbb{R}^{n}$
Defn: The geometric Tangent space of a space X at p is the quotient $T_{p}^{geom}X \equiv K_{p} /\sim$ where is the equivalence relation of two curves $\gamma_{1},\gamma_{2}$ through point p.
The equivalence class of $\gamma \in K_{p}$ is denoted $[\gamma] \in T_{p}^{geom}X$ and is called the geometric tangent vector of X at p.
Thanks for any help at getting a better insight.
Brian
Yes one have the usual definition of tangent vector of a curve but the tangent of a curve is defined if the curve lies in some $\mathbb{R}^n$ i.e. if $\gamma$ the curve then $\gamma :(0,1)\rightarrow\mathbb{R}^n $. Now when we are considering the tangent vector of a point in manifold then you have to consider the curve to be in that manifold. So my point is a priori you don't know whether the curve lies in $\mathbb{R}^n$ or not. In other words you don't know whether every smooth manifold can be embedded in $\mathbb{R}^n$ or not.
P.S. The above statement is true. It is called called Whitney embedding theorem.