From Coordinate System to Curvature (Riemann Tensor in General Relativity)

Coordinate system

A set of coordinates specifies a point, so position can be thought of as a function of the coordinates. $$\mathscr{P} = \mathscr{P}(x^1, x^2, x^3, \dotsc)$$ At each point, a set of basis vectors $\vec{e_i}$ can be defined. Each basis vector is obtained by differentiating position by one coordinate (i.e., holding all other coordinate values constant), which means it is a vector pointing along that coordinate line, where a coordinate line describes a path along which only that one coordinate varies. $$\vec{e_i} = \frac{ \partial \mathscr{P} }{ \partial x^i }$$ Any given vector $\vec{A}$ can now be expressed as a linear combination of the basis vectors. $$\vec{A} = A^i \vec{e_i}$$ Where the $A^i$ are the components of the vector in this basis. Here (and throughout) we are using the Einstein summation convention, $A^i \vec{e_i} = \displaystyle\sum_{i=0}^{N} A^i \vec{e_i}$

Partial derivative of a vector field:

A vector field is a set of vectors, with one vector at each point (e.g. wind speed or electric field at each point). It can be thought of as a vector-valued function of the coordinates: $\vec{A}(x^1, x^2, x^3, \dotsc)$. The (partial) derivative of this function with respect to one of the coordinates, gives a vector that is the rate of change of $\vec{A}$ per unit of that coordinate.

We expand it by the product rule:

$$\frac{ \partial \vec{A} }{ \partial x^i } = \frac{\partial}{ \partial x^i }( A^j \vec{e_j} ) = \frac{\partial A^j}{ \partial x^i } \vec{e_j} + A^j \frac{\partial \vec{e_j} }{ \partial x^i }$$

Define "Christoffel Symbols", which are the rate of change of the basis vectors:
$$\Gamma^{k}{}_{i j} \vec{e_k} = \frac{\partial \vec{e_j} }{ \partial x^i }$$

Substitute that into our formula for the partial derivative: $$\frac{ \partial \vec{A} }{ \partial x^i } = \frac{\partial A^j}{ \partial x^i } \vec{e_j} + A^j \Gamma^{k}{}_{i j} \vec{e_k}$$

In that last term, the summation indices $j$ and $k$ are arbitrarily chosen, so we can replace them with whatever we like. So replace $j$ with $\alpha$ and replace $k$ with $j$. Then the partial derivative of a vector field is:
$$\frac{ \partial \vec{A} }{ \partial x^i } = ( \frac{\partial A^j}{ \partial x^i } + A^{\alpha} \Gamma^{j}{}_{i \alpha} ) \vec{e_j}$$
This is called "covariant derivative" (to distinguish from partial derivatives of components), and has the following notation:
$$\nabla_i \vec{A} = \frac{ \partial \vec{A} }{ \partial x^i }$$

Because $\nabla_i \vec{A}$ is a vector field we can take its partial derivative to get the second partial derivative: $$\nabla_j \nabla_i \vec{A}$$ the partial (covariant) derivative of $\vec{A}$ with respect to coordinate $x^i$ and then with respect to coordinate $x^j$.

Curvature

Curvature is measured by the Riemann curvature tensor, describing the degree to which the second covariant derivative does not commute:
$$( v^b R^a{}_{b c d} ) \vec{e_a} = \nabla_c \nabla_d \vec{v} - \nabla_d \nabla_c \vec{v}$$ So its components are $$R^a{}_{b c d} = (\nabla_c \nabla_d \vec{e_b} - \nabla_d \nabla_c \vec{e_b})^a$$ thus describing how the second partial derivatives of the basis vectors fail to commute. The components are all zero in flat space, because the second derivatives always commute in flat space, no matter what coordinate system is chosen. So the amount of non-commutativity tells us how space is curved.

Next we consider the physical/geometric meaning of the Riemann tensor.