Tangent space and tangent vectors

Question

As I have heard, tangent vector to a smooth manifold $M$ in $p \in M$ is the operator $D_{\xi}$:$f \to D_{\xi}f$, where $f$ is a smooth function $f: M \to R$, with the following properties:

$D_{\xi}(f+g)=D_{\xi}f+D_{\xi}g $

$D_{\xi}(fh)=(D_{\xi}f)h(p)+f(p)(D_{\xi}h) $

So, the basis of the tangent space is:

${\frac{\partial }{\partial x^{1}}}, \frac{\partial }{\partial x^{1}}, ..., \frac{\partial }{\partial x^{n}}$, where ${x^{1},...,x^{n}}$ are coordinates of the local map $(U_{p}, \phi_{p})$.

and any vector in tangent space has the following form:

$\xi^{1}\frac{\partial }{\partial x^{1}}+...+\xi^{n}\frac{\partial }{\partial x^{n}} \in TM_{p}$, where the set $\xi^{1},...,\xi^{n}$ is called the coordinates of tangent vector.

How does this definition relate to intuitive graphical interpretation?

Answer 1

Tangent vectors to $\mathbb{R^n}$ (and likewise tangent vectors to embedded surfaces in $\mathbb{R^n}$) can be thought of in this way by identifying a vector $v$ with the directional derivative operator (or "derivation") $D_v = v^i \frac\partial {\partial x^i}$. In fact the directional derivatives are the only operators on functions that satisfy the derivation axioms; so talking about vectors is the same thing as talking about derivations - in a sense derivations are vectors.

When we move to abstract manifolds, we no longer have the affine space structure to provide tangent vectors, but the derivation definition still makes perfect sense. Thus we define vectors to be derivations. An arguably more intuitive definition can be made in terms of equivalence classes of curves if you wish, but they all turn out equivalent in the end.

Answer 2

Exactly as you've set it up. $\partial/\partial x^j$ should be thought of as the tangent vector to the $x^j$ parameter curve in $M$, and $\xi=\sum x^j \partial/\partial x^j$ as the corresponding linear combination in $T_pM$.