TITLE: Understanding tangent vectors
QUESTION [7 upvotes]: I just checked out this thread: help in understanding tangent vectors and I still have some problems understanding this.
The tangent vector on a manifold at point $t_0$ is intuivitely
$$\dot \gamma(t_0) = \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} \gamma(t)\in \mathbb R^n \tag 1$$
Apparently this expression doesn't make sense in some manifolds, so $X_p$, defined
$$X_p :f\in C^\infty(\mathbb R^n) \to \left.\frac {\operatorname d} {\operatorname d t}\right\vert_{t_0} f(\gamma(t)) \in \mathbb R
\tag 2$$
is used as the tangent vector instead. But this isn't equivalent to ($1$) (which is a real valued vector, not an operator), and I have no idea what effect the function $f$ has. I realise it's similar to a directional derivative on the function $f$, but don't see why it is relevant.
If anyone can explain with reference to the sphere, that would probably make it easiest to understand. It's easy to see that the tangent space defined by ($1$) gives the plane that best approximates a point on the sphere, but ($2$) doesn't make sense to me. Thanks.

REPLY [6 votes]: There are quite a few definitions of the tangent space to a manifold. Happily, these are all equivalent in some sense.
One definition that you might like, if you like curves, is the following. Let $M$ be a manifold, and let $P \in M$. Then the tangent space $T_PM$ at $P$ consists of smooth curves $\gamma\colon J \to M$, where $J$ is an open interval in $\mathbf R$ containing $0$, such that $\gamma(0) = P$ under the following equivalence relation: $\gamma_1 \sim \gamma_2$ if there exists a chart $\varphi\colon U \to \mathbf R^n$ around $P$ such that $(\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)$. (This will then hold for every chart around $P$, by the chain rule.)
This has the advantage of being nearly obviously the same as your (1) when $M$ is embedded in Euclidean space, yet working in great generality.
Now, how can we connect this to the operators in (2)? Call a linear map $X \colon C^\infty(M) \to \mathbf R$ a derivation at $P \in M$ if it satisfies the Leibniz rule
$$
X(fg) = g(P)Xf + f(P)Xg
$$
for all $f, g$. You can check that in $\mathbf R^n$, for $v \in \mathbf R^n$, the directional derivative gives a derivation at $P$: define $\gamma\colon (-\varepsilon, \varepsilon) \to \mathbf R^n$ by $\gamma(t) = P + tv$. Then
$$
f \mapsto D_v|_P f = (f \circ \gamma)'(0)
$$
is a derivation.
In general, we get a well-defined linear map $T_PM \to D_P$ by sending the class of $\gamma$ to the derivation $f \mapsto (f \circ \gamma)'(0)$. It should be straightforward to show that this association is injective, since we could take $f$ to be a coordinate function. It's less obvious (to me, at least) that it's surjective. But this does follow from the multivariable Taylor's formula. I won't write out a proof here, but you can find one in the case of $M = \mathbf R^n$ (so, proving that every derivation has the form $D_v|_P$) as Proposition 3.2 in Lee's book. From this the case of general $M$ follows.
What I like about the second definition is that it doesn't mention charts or equivalence relations, and it is obviously a vector space. It also generalizes, with only slight modifications, to more algebraic settings.
I apologize for the lack of mental pictures here. To me, the first definition I gave is the most intuitive, and you have this natural (!) map to another space which turns out to be an isomorphism.