TITLE: Definition of the Tangent Space
QUESTION [6 upvotes]: I'm watching a series of lectures on differential geometry, and I've run into a bit of a problem on the definition of the tangent space. We first defined a tangent space as $\{(p,v) | v \in \mathbb{R}^n\}$, which makes sense to me: it's the set of all vectors attached at point $p$. We then defined the directional derivative as 
$$
(Df)(p,v) = \lim_{t \rightarrow \infty} \frac{f(p + tv) - f(p)}{t}
$$
We expanded that to this:
$$
(Df)(p,v) = \left( \sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}
\right) f$$
This makes sense to me; we have defined the directional derivative as an operator that is applied to the function. 
Here's the part where I lose the plot. I'm then told that, if I think about it, the portion inside the parentheses is really interchangeable with $(p,v)$. I'm afraid that I've thought about it, and I can't see the equivalence. $\sum_{i = 0}^{n}v_i \left.\frac{\partial}{\partial x_i}\right|_{p}$ is an operator (isn't it?) whereas $(p,v)$ is a ordered pair of elements of $\mathbb{R}^n$. Does that mean that the expression $(p,v)(f)$ makes sense? What would that mean? 
I must be thinking about this the wrong way; can someone clarify?

REPLY [8 votes]: I think the phrase "the portion inside the parentheses is really interchangeable with $(p,v)$" is quite misleading. This is not really true -- as you correctly observed, $(p,v)$ is an ordered pair of elements of $\mathbb R^n$, while the expression in parentheses is an operator on functions.
What is true is that there is a linear map from the set $\{(p,v)|v\in\mathbb R^n\}$ (let's call that the geometric tangent space) into the set of linear differential operators on functions, which takes the pair $(p,v)$ to the "directional derivative operator" that you wrote down. The image of this map is the set of derivations at $\boldsymbol p$, which is the set of all linear maps $X\colon C^\infty(\mathbb R^n)\to \mathbb R$ that satisfy this product rule:
$$
X(fg) = f(p)X(g) + g(p)X(f).
$$
The geometric tangent space is thus canonically isomorphic to the set of derivations at $p$. 
Why does this matter? Because on an abstract manifold, the geometric tangent space doesn't have any coordinate-independent meaning, but the space of derivations at $p$ does. So we take the space of derivations at $p$ as our definition of the tangent space to $M$ at $p$. 
Once you get comfortable with the canonical isomorphism between the geometric tangent space and the space of derivations at $p$, then you might start thinking of them as "interchangeable." But when you're first trying to learn this stuff, it's more productive to think of them as canonically isomorphic.