I'm a physics student with a very loose understanding of the mathematics I use. I'm trying to learn a little more about very basic topology, manifolds, and Riemannian geometry. I'm using Nakahara's Geometry, Topology, and Physics for self-study. I'm trying to read up about vectors on manifolds and the concept of a tangent vector has me thoroughly confused. Up until now I had always pictured the tangent space something like a plane tangent to a point on the surface of a manifold, however if I'm understanding my book correctly the elements of the tangent space seem to be something more like differential operators. I will lay out the definition from my book and then see if perhaps you guys might be able to help me understand exactly what a tangent vector, and what the tangent space is.
My book defines a tangent vector at $c(0)$ as the direction derivative of a function, $f:M \rightarrow \mathbb{R},$ along a curve $c:(a,b) \rightarrow M,$ or $c(t)$ where $a<0$, $b>0$ at $t=0$. The rate of change is
$$\frac{df(c(t))}{dt} \Bigr|_{t=0}.$$
Assuming we've defined all coordinates correctly and stuff this becomes
$$\frac{\partial f}{\partial x^\mu} \frac{d x^\mu(c(t))}{dt} \Bigr|_{t=0}.$$
Defining the differential operator $X$ as
$$X=X^\mu \left( \frac{\partial}{\partial x^\mu} \right),$$ $$X^\mu=\frac{dx^\mu(c(t))}{dt} \Bigr|_{t=0},$$
the directional derivative can be found by applying $X$ to $f$. Putting it all together, this means
$$\frac{df(c(t))}{dt} \Bigr|_{t=0} = X^\mu \left( \frac{\partial f}{\partial x^\mu} \right) = X[f].$$
The book then says, "the last equality defines $X[f]$. It is $X=X^\mu \partial / \partial x^\mu$ which we now define as the tangent vector to $M$ at $p=c(0)$ along the direction given by the curve. This makes the tangent vector a differential operator, it seems like. Intuitively I had thought that the tangent vector would be $X^\mu \partial f / \partial x^\mu$ along the direction of the curve.
The book goes on to say that instead of identifying a tangent vector with a curve you instead identify a tangent vector with the following equivalence class of curves
$$[c(t)] = \left\{ \tilde{c}(t) \, \Bigr| \, \tilde{c}(0)=c(0) \text{ and } \frac{dx^\mu (\tilde{c}(t))}{dt} \Bigr|_{t=0} = \frac{dx^\mu (c(t))}{dt} \Bigr|_{t=0} \right\},$$
and that the tangent space is the collection of all such equivalence classes at $c(0)=p.$ So I'm a little confused. The equation above seems to define a tangent vector as an equivalence class of curves. The previous definition seems to define a tangent vector as a differential operator. My intuition tells me something different altogether. Can someone help me sort through all this? Thanks.
The equivalence relation on the (parametrised) curves is exactly that they give the same differential operator at the point $p$; this reconciles those two definitions. See below for some more details.
As for intuition, that can be a little harder. The thing to recognise is that we need a definition of tangent vectors without reference to an ambient space. In other words, we may have a manifold defined abstractly, rather than embedded in some Euclidean space (in general relativity, this is often the case).
Let $\{y^i\}$ be coordinates on $n$-dimensional Euclidean space. Notice that if we take a 'vector' $\vec{v} = (v^1, \ldots, v^n)$, we can think of this as giving the derivatives of the coordinate functions along the straight line parametrised by $y^i(t) = y^i_0 + v^i t$. Indeed, the directional derivative of any function $f$ along this line is given by $$ \frac{df}{dt} = \sum_i v^i \frac{\partial f}{\partial y^i} $$ So we can associate $v$ with the differential operator $\sum_i v^i\partial/\partial y^i$. You can easily check that this gives a linear isomorphism between the vector space $\mathbb{R}^n$ and the space of operators spanned by $\{\partial/\partial y^i\}$.
Let me return briefly to the 'equivalence classes of curves' bit. Choose some other parametrised curve $\gamma : [0,1] \to \mathbb{R}^n$ passing through the same point, i.e. $y^i(\gamma(0)) = y^i_0$. Then the directional derivative of $f$ along the curve at the point $(y^1_0,\ldots, y^n_0)$ is $$ \left.\frac{df(\gamma(t))}{dt}\right\vert_{t=0} = \sum_i \left.\frac{\partial y^i(\gamma(t))}{\partial t}\frac{\partial f}{\partial y^i}\right\vert_{t=0} $$ We define two curves $\gamma, \gamma'$ to be equivalent iff they give the same directional derivative for every function $f$, i.e. the same differential operator. This gives our isomorphism between equivalence classes of parametrised curves passing through some point and first-order differential operators at the point. Note that rescaling the parameter $t$ will rescale the differential operator, which is why we talk about parametrised curves.
Now suppose we have some embedded manifold $X$, with local coordinates $x^\mu$, and embedding functions $y^i(x^\mu)$. A tangent vector (in the familiar sense) to $X$ just gives the infinitesimal change in the coordinates $y^i$ when we change the coordinates $x^\mu$ by an arbitrary infinitesimal amount $\delta x^\mu$. We have $$ \delta y^i = \sum_\mu \delta x^\mu \frac{\partial y^i}{\partial x^\mu} $$ So the corresponding tangent vector, using the notation we introduced before, is $$ \sum_{i, \mu} \delta x^\mu \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i} $$ The $\delta x^\mu$ are arbitrary, and as we vary them, we map out a linear subspace spanned by the 'vectors' $$ \sum_i \frac{\partial y^i}{\partial x^\mu} \frac{\partial}{\partial y^i} = \frac{\partial}{\partial x^\mu} $$
Now we realise that the operators $\partial/\partial x^\mu$ don't actually depend on the embedding at all, and we have successfully defined tangent vectors in an intrinsic way on any differentiable manifold!
This has become a very long answer, but I hope it's helpful.