Differential Geometry: directional derivative along a vector field in a manifold

Question

I am trying to understand the definition of directional derivative along a vector field in a manifold, but I am having trouble.
Let $M$ be a smooth manifold, $f:M \rightarrow R$ a smooth function and $X:M \rightarrow TM$ a smooth vector field.
The Lie Derivative page on wikipedia says that the directional derivative of $f$ with respect to $X$ at point $p\in M$ is $\lim_{t \rightarrow 0} (f(p+tX(p))-f(p))/t$ but why does this make sens? $tX(p)$ is a vector, why can I sum it with a point on the manifold?

Answer 1

It looks as if the Wikipedia page is a little sloppy. A better way to say it would be "Let $u$ be an integral curve of $X$ at $p$, i.e., a function with $$ u(0) = p \\ u'(0) = X(p)\\ u'(s) = X(u(s)) $$ for $-\epsilon < s < \epsilon$; the the derivative is $$ \lim_{t \rightarrow 0} \frac{f(u(p))-f(p)}{t} $$

Now if you look closely, if your manifold is embedded in Euclidean space, and you have a smooth extension $\bar{f}$ of $f$ from the manifold to a neighborhood of the manifold in Euclidean space, you have that

$$ f(u(t)) \approx \bar{f}(p + tX(p)) $$

for small $t$; in particular, for $t = 0$, these are equal, and for $t$ small, they agree to first order, i.e., the derivatives of these two are identical. [All this requires some proof, of course!] So you can see why someone being a bit sloppy might write what's on the Wikipedia page.

Answer 2

The long and short of it is that you do everything in coordinates $(u^1,\ldots,u^n)$. This means that you choose a coordinate patch containing your point $p$ and do calculations as if you are in Euclidean space. Here your vector $v$ is replaced by its components $(v_1,\ldots,v_n)$ relative to the standard basis $(\frac{\partial}{\partial u^j})$. Then you show that the result is independent of choosing the patch.

On a more sophisticated level, you can define a vector at a point as a derivation on the space of functions defined on an arbitrarily small neighborhood, and then invariance is obvious.