note-math

For a vector field $𝑌$ near $𝑝$, in the $𝑋$ direction, try to calculate the derivative in coordinates $\frac{\partial 𝑌}{\partial 𝑋}=\underset{𝑡\to 0}{\text{ lim }}\frac{1}{𝑡}\left(𝑌\left(𝑝+𝑡𝑋\right)-𝑌\left(𝑝\right)\right)$

However, the #link(<difference-polynomial>)[difference] operation is not linearly compatible with coordinate changes of general diffeomorphisms.

But in a metric-manifold, there are special coordinates — #link(<geodesic-coordinate>)[geodesic coordinates]. The transformation method for different geodesic coordinates at $𝑝$ is $\text{SO}$, which is linear.

In geodesic coordinates at point $𝑝$, the derivative at point $𝑝$, ${\nabla }_{𝑋}𝑌≔\frac{\partial 𝑌}{\partial 𝑋}$

It is also possible to take the derivative of a #link(<tensor>)[] field ${\nabla }_{𝑋}𝑇$. According to the scalar multiplication associated with the tensor structure, calculations can be performed using the #link(<Leibniz-law>)[product-rule] Example ${\nabla }_{𝑋}\left(𝑌\otimes 𝑍\right)=\left({\nabla }_{𝑋}𝑌\right)\otimes 𝑍+𝑌\otimes \left({\nabla }_{𝑋}𝑍\right)$

Compute geodesic coordinates $𝑦$ using general coordinates $𝑥$, then in coordinates $𝑦$, the geodesic derivative is

$$\begin{array}{ccc}{\nabla }_{𝑋}𝑌\left(𝑥\right)& =& \frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)\cdot \frac{\partial 𝑌}{\partial 𝑋}\left(𝑦\right)\\ & =& \frac{\partial }{\partial 𝑋}\left(\frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌\right)-\frac{\partial }{\partial 𝑋}\left(\frac{\partial 𝑥}{\partial 𝑦}\right)\cdot 𝑌& & \text{ by product-rule}\end{array}$$

Using #link(<connection-transformation>)[connection transformation]

$$\begin{array}{ccc}0& =& \mathrm{\Gamma }\left(𝑦\right)\\ & =& \frac{\partial 𝑦}{\partial 𝑥}\cdot \mathrm{\Gamma }\left(𝑥\right)\cdot \frac{\partial 𝑥}{\partial 𝑦}+\frac{\partial 𝑦}{\partial 𝑥}\cdot \frac{\partial }{\partial 𝑦}\left(\frac{\partial 𝑥}{\partial 𝑦}\right)\end{array}$$

==> $\frac{\partial }{\partial 𝑦}\left(\frac{\partial 𝑥}{\partial 𝑦}\right)=-\mathrm{\Gamma }\left(𝑥\right)\cdot \frac{𝑑𝑥}{𝑑𝑦}$

Use $\frac{\partial }{\partial 𝑋}\left(\frac{\partial 𝑥}{\partial 𝑦}\right)=\frac{\partial }{\partial 𝑦}\left(𝑋\right)\left(\frac{\partial 𝑥}{\partial 𝑦}\right)$. Substitute into the calculation of ${\nabla }_{𝑋}𝑌\left(𝑦\right)$

$${\nabla }_{𝑋}𝑌\left(𝑦\right)=\frac{\partial }{\partial 𝑋}\left(\frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌\right)+{𝑋}^{⊺}\cdot \mathrm{\Gamma }\left(𝑥\right)\cdot \frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌$$

The $𝑝$ tangent space linearly transforms ${\nabla }_{𝑋}𝑌\left(𝑦\right)$ to ${\nabla }_{𝑋}𝑌\left(𝑥\right)$ by $\frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)$, but keeps in coordinate $𝑥$, but keep $𝑋,𝑌$ in coordinate $𝑦$

$$\begin{array}{cc}& {\nabla }_{𝑋}𝑌\left(𝑥\right)\\ & =& \frac{\partial }{\partial \left(\frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)\cdot 𝑋\left(𝑦\right)\right)}\left(\frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)\cdot 𝑌\left(𝑦\right)\right)+{\left(\frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)\cdot 𝑋\left(𝑦\right)\right)}^{⊺}\cdot \mathrm{\Gamma }\left(𝑥\right)\cdot \frac{\partial 𝑥}{\partial 𝑦}\left(𝑦\right)\cdot 𝑌\left(𝑦\right)\\ & =& \frac{\partial }{\partial \left(𝑋\left(𝑥\right)\right)}𝑌\left(𝑥\right)+{𝑋\left(𝑥\right)}^{⊺}\cdot \mathrm{\Gamma }\left(𝑥\right)\cdot 𝑌\left(𝑥\right)\end{array}$$

Or written as, in general coordinates, geodesic derivative

$$\nabla =\partial +\mathrm{\Gamma }$$

For coordinate-frame

$${\nabla }_{\frac{\partial }{\partial {𝑥}^{𝑖}}}\frac{\partial }{\partial {𝑥}^{{𝑖}^{\prime }}}={\mathrm{\Gamma }}_{𝑖{𝑖}^{\prime }}^{𝑗}\frac{\partial }{\partial {𝑥}^{𝑗}}$$

Is there a more intuitive explanation, rather than directly using the transformation of connection?

If we only consider linear compatibility, then there are many #link(<principal-bundle-connection>)[linear connections], and the one that coincides with the geodesic-derivative is the metric-connection

Question Similar to the case of vector fields. Use the transformation $𝛼\left(𝑥\right)=𝛼\left(𝑦\right)\cdot \frac{\partial 𝑥}{\partial 𝑦}$ and the product-rule $\frac{\partial }{\partial 𝑋}\left(𝛼\cdot \frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌\right)=\frac{\partial 𝛼}{\partial 𝑋}\cdot \left(\frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌\right)+𝛼\cdot \frac{\partial }{\partial 𝑋}\left(\frac{\partial 𝑥}{\partial 𝑦}\cdot 𝑌\right)$

For co-vector coordinate-frame

$${\nabla }_{\frac{\partial }{\partial {𝑥}^{𝑖}}}𝑑{𝑥}^{{𝑖}^{\prime }}=-{\mathrm{\Gamma }}_{𝑖𝑗}^{{𝑖}^{\prime }}𝑑{𝑥}^{𝑗}$$

Parallel transport as "zero rate of change along the curve" ${\nabla }_{\dot{𝑥}}𝑋=0$ or ${\left(\partial +\mathrm{\Gamma }\right)}_{\dot{𝑥}}𝑋=0$ where $𝑋=𝑋\left(𝑥\left(𝑡\right)\right)$

${\nabla }_{\dot{𝑥}}𝑋=0$ is an ODE

According to calculation (?), covariant derivative can be recovered from parallel transport + difference quotient

Parallel transport of metric-connection preserves the metric

Can be used to construct orthonormal frames

It can be proven that a manifold metric uniquely corresponds to an $\text{SO}$ principal-bundle structure on the manifold

But are there more specific and operational calculation results? Regarding calculating orthonormal frames using parallel transport in geodesic coordinates

A canonical orthonormal frame may be used for simplified some of calculations of spinors on curved manifolds, e.g. Pauli-matrix ${𝜎}^{0,1,2,3}$