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.



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.

  • $\begingroup$ Ok.. so we know $p \in X$. Your saying that we don't know if any $\gamma \in K_{p}$ is in the manifold $X$. So the most we can say (without any other Thrm's such as Whitney's) is that we can say that two curves are equivalent i.e. $\gamma_{1} \sim \gamma_{2}$ if $\dot{\gamma_{1}}(0)_{h} = \dot{\gamma_{2}}(0)_{h}$. So we just have an equivalence on the tangent space? So there is just a tangent plane at p in X such that each distinct vector on the tangent plane would represent a different equivalence class $[\gamma]$. I am confused as there are several "tangent spaces" e.g $T_{p}^{alg}X$ $\endgroup$
    – Relative0
    Jun 3 '13 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.