# Reconciling definition of tangent vector with intuition

I'm having trouble getting to grips with what a tangent vector really is.

Let $M$ be an $n$-dimensional manifold, and let $\alpha:(-\epsilon, \epsilon) \to M$ be a curve in $M$ through $p$ (i.e. $\alpha(0) = p$). So for $\forall t \in (-\epsilon, \epsilon), \exists \delta \gt 0$ and a chart $(U, \varphi)$ such that $\alpha(t-\delta, t + \delta) \subseteq U$ and $\varphi \circ \alpha$ is a curve in $\varphi(U)$.

If $f$ is smooth function from $U$ to $\mathbb{R}$ then $f \circ \alpha: (-\epsilon, \epsilon) \to \mathbb{R}$. We define the tangent vector to $\alpha$ at $0$ to be $\alpha'(0): f \mapsto (f \circ \alpha)'(0)$. Thus the tangent vector is an operator. I am further aware that tangent vectors can be seen as differential operators evaluated as $0$.

So let $M = \mathbb{R}^3$ and let $\alpha:(-1,1) \to \mathbb{R}^3$ be defined as $\alpha(t) = (\sin (t), \cos (t), t)$. Then $\alpha'(0) = (1,0,1)$. How do we reconcile this with what I intuitvely expect the map to be: $\alpha'(0): f \mapsto \frac{\partial f}{\partial x_1}\Big|_{t=0} + \frac{\partial f}{\partial x_3}\Big|_{t=0}$?

In this case, the correspondence is reasonably clear, but what if we have a general manifold instead?

• Consider also this question and its answers. – pglpm Mar 30 '18 at 16:41

## 1 Answer

Using $\mathbb{R}^{3}$ as an example for a manifold on which to define tangent spaces is not a helpful start. The tangent spaces corresponding to $p\in\mathbb{R}^{3}$ are all isomorphic to $\mathbb{R}^{3}$ itself. It is more useful to think of a two-dimensional surface in $\mathbb{R}^{3}$ such as $S^{2}$, the surface of the unit ball in $\mathbb{R}^{3}$. A curve $\alpha$ from $(-\epsilon,\epsilon)\rightarrow{}S^{2}$ going through $p\in{}S^{2}$ will have a derivative at $p$, and all curves having the same derivative at $p$ form an equivalence class, which you can define to be a tangent vector at $p$, in other words, $v\in{}T_{p}(S^{2})$, where $v$ is this equivalence class. The set of all such equivalence classes is the tangent space $T_{p}(S^{2})$. $\partial/\partial{}\xi^{i}$ with $i=1,2$ is a basis for this tangent space, where $\xi^{i}$ is a local chart for the manifold $S^{2}$ around $p$, and $\partial/\partial{}\xi^{i}$ is the equivalence class containing a curve which only varies in the coordinate $\xi^{i}$ while the coordinates $\xi^{j}$ remain constant for $j\neq{}i$. Chapter 2 in Manifolds and Differential Geometry (Jeffrey Lee) is a good introduction, and I learned it from a pretty good and dense introduction to tangent vectors in Methods of Information Geometry by Shun-ichi Amari, pages 5-7.