Riemannian geometry:

From this reference.

Riemannian geometry deals with intrinsic geometry on surfaces. The position vector at point P can be expressed on the basis of the extrinsic coordinates, or on the intrinsic coordinates on the manifold (the mesh of space-time). At each point of the manifold (referenced either extrinsically through the Cartesian system; or intrinsically via the curvilinear coordinate lines on the manifold), there will be a tangent vector space (and a dual). However, each point will have its own tangent vector space \(T_P(M).\) Over the manifold, there will be a collection of vector spaces - one for each point, \(T_x(M).\) A single vector can be selected at each point as a function of the coordinates (a function of space-time), creating a vector field:

\[\underbrace{X^\alpha(x)}_{\text{components as f(space-time)}}\underbrace{\quad\partial _\alpha\quad}_{basis}\in\quad T_x(M).\]

The basis of the tangent space will be \(\frac{\partial}{\partial x^i}\) or \(\partial x^i.\)

The vectors in the tangent space and in the dual of the tangent space at every point are related through the metric \(g_{\mu\nu}(x)\,dx^\mu\otimes dx^\nu,\) which is also a function of space time. This gives us an inner product at each point. For each vector in the tangent space \(\left<\vec v,\;\right>\to \mathbb R,\) i.e. filling in the space with any other vector will produce a real number. Hence, $<v,;>V^*, $ linking the vector space with its dual.

However, we are still stuck at any particular point in space-time. As we move differentially along the manifold, the tangent plane changes since the vector space changes with every point. So we need a connector to relate vectors across the field.

The change of coordinates between two points \(P\) and \(Q\) separated by a differential value \(dx\) as expressed in the extrinsic coordinate system (position vectors from the origin) could be encapsulated with the Taylor series. The new extrinsic coordinates in the extrinsic plane will be \(x+dx\) and \(y+dy\), and the height will be a function of these two coordinates:

\[z(x+dx,y+dy)=z(x,y) + \frac{\partial z}{\partial x} dx+ \frac{\partial z}{\partial y} dy\]

The distance vector \(ds\) will be the difference between these two position vectors:

\[ds=z(x+dx,y+dy)- z(x,y)=\frac{\partial z}{\partial x} dx+ \frac{\partial z}{\partial y} dy\]

and the square distance (changing now to covariant notation for covectors)

\[dx^2=ds \cdot ds=g_{ij}\;dx^idy^j\]

with \(g_{ij}\) constituting the metric tensor. In Cartesian coordinates the metric tensor is simply the identity matrix.

XylyXylyX labels the vector that the vector field places at \(x + \delta x,\) i.e. the position in space-time shifted by a small increment,

as \(X^\alpha(x+\delta x),\) with a first grade Taylor expansion:

\[\begin{align} X^\alpha\,(x + \delta x) &= \underbrace{X^\alpha (x)}_{\text{value at }x}+\delta x^b\;\underbrace{\partial_bX^\alpha}_{\partial \text{wrt compon's } \delta x}\\[2ex] &=X^\alpha(x) + \delta x^b\;\frac{\partial X^\alpha}{\partial x^b}\\[2ex] &= X^\alpha(x) + \delta X^\alpha \end{align}\]

Covariant derivative:

Differential changes across a vector field will necessitate a way of moving across tangent spaces