1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Consider also this question and its answers. $\endgroup$ – pglpm Mar 30 '18 at 16:41
1
$\begingroup$

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.

$\endgroup$

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.