Updated and added some to differential geometry.

docs/en/mathematics/differential-geometry/curvature.md

@@ -28,10 +28,10 @@ It then follows from the definition that the curvature operator $\Omega$ can be
 
 ## Curvature tensor
 
-> *Definition 2*: the Riemann curvature tensor $R: \Gamma(\mathrm{T}^*\mathrm{M}) \times \Gamma(\mathrm{TM})^3 \to \mathbb{K}$ is defined as
+> *Definition 2*: the Riemann curvature tensor $\mathbf{R}: \Gamma(\mathrm{T}^*\mathrm{M}) \times \Gamma(\mathrm{TM})^3 \to \mathbb{K}$ is defined as
 >
 > $$
->   R(\bm{\omega}, \mathbf{u}, \mathbf{v}, \mathbf{w}) = \mathbf{k}(\bm{\omega}, \Omega(\mathbf{v}, \mathbf{w}) \mathbf{u}), 
+>   \mathbf{R}(\bm{\omega}, \mathbf{u}, \mathbf{v}, \mathbf{w}) = \mathbf{k}(\bm{\omega}, \Omega(\mathbf{v}, \mathbf{w}) \mathbf{u}), 
 > $$
 >
 > for all $\bm{\omega} \in \Gamma(\mathrm{T}^*\mathrm{M})$ and $\mathbf{u}, \mathbf{v}, \mathbf{w} \in \Gamma(\mathrm{TM})$. 

docs/en/mathematics/differential-geometry/linear-connections.md

@@ -16,6 +16,8 @@ Let $\mathrm{M}$ be a differential manifold with $\dim \mathrm{M} = n \in \mathb
 
 From property 3 it becomes clear that $\nabla_\mathbf{v}$ is an analogue of a directional derivative. The linear connection can also be defined in terms of the cotangent bundle and the dual fiber bundle. 
 
+Note that the first (trivial) element in the notion of the section $\Gamma$ is omitted, generally it should be $\Gamma(\mathrm{M}, \mathrm{TM})$ as the elements of this set are maps from $\mathrm{M}$ to $\mathrm{TM}$. 
+
 ## Covariant derivative
 
 > *Definition 2*: let $\mathbf{v} = v^i \mathbf{e}_i\in \Gamma(\mathscr{B})$ then the **covariant derivative** on $\mathbf{v}$ is defined as
@@ -56,21 +58,88 @@ $$
 
 Will be added later.
 
-## Parallel transport 
+## Intrinsic derivative
 
-> *Definition 3*: let $\mathbf{v} \in \Gamma(\mathrm{TM})$, then **parallel transport** of $\mathbf{v}$ occurs along the manifold $\mathrm{M}$ when
+> *Definition 3*: let $\gamma: \mathscr{D}(\gamma) \to M: t \mapsto \gamma(t)$ be a smooth curve on the manifold parameterized by an open interval $\mathscr{D}(\gamma) \subset \mathbb{R}$ and let $\mathbf{v}: \mathscr{D}(\gamma) \to \mathrm{TM}: t \mapsto \mathbf{v}(t) = \mathbf{u} \circ \gamma(t)$ be a vector field defined along the curve with $\mathbf{u} \in \Gamma(\mathrm{TM})$, the **intrinsic derivative** of $\mathbf{v}$ is defined as
 >
 > $$
->   D_k \mathbf{v} = \mathbf{0}.
+>   D_t \mathbf{v}(t) = \nabla_{\dot\gamma} \mathbf{v}(t),
 > $$
+>
+> for all $t \in \mathscr{D}(\gamma)$. 
 
-For example, a parameterised vector field $\mathbf{v}: x(t) \mapsto \mathbf{v}(x(t)) \in \Gamma(\mathrm{TM})$ is transported parallel if 
+By decomposition of $\dot \gamma = \dot \gamma^i \partial_i$ and $\mathbf{v} = v^i \partial_i$ and using the chain rule we obtain
 
 $$
-    D_t \mathbf{v} = (\partial_k v^i) \dot x^k \partial_i + \Gamma^i_{jk} v^j \partial_i = \mathbf{0},
+\begin{align*}
+    \nabla_{\dot\gamma} \mathbf{v}(t) &= \dot \gamma^i \nabla_{\partial_i} (v^j \partial_j), \\
+                                      &= \dot \gamma^i \big((\partial_i v^j) \partial_j + v^j \Gamma_{ji}^k \partial_k  \big), \\
+                                      &= (\dot \gamma^i \partial_i v^j + \dot \gamma^i \Gamma^j_{ki}v^k) \partial_j, \\
+                                      &= (\dot v^j + \Gamma^j_{ki} v^k \dot \gamma^i) \partial_j,
+\end{align*}
 $$
 
-so $(\partial_k v^i) \dot x^k + \Gamma^i_{jk} v^j = 0$
+for all $t \in \mathscr{D}(\gamma)$. This notion of the intrinsic derivative can of course be extended to any tensor. 
+
+### Parallel transport 
+
+> *Definition 4*: let $\gamma: \mathscr{D}(\gamma) \to M: t \mapsto \gamma(t)$ be a smooth curve on the manifold parameterized by an open interval $\mathscr{D}(\gamma) \subset \mathbb{R}$ and let $\mathbf{v}: \mathbb{R} \to \mathrm{TM}: t \mapsto \mathbf{v}(t) = \mathbf{u} \circ \gamma(t)$ be a vector field defined along the curve with $\mathbf{u} \in \mathrm{TM}$, then **parallel transport** of $\mathbf{v}$ along the curve is defined as
+>
+> $$
+>   D_t \mathbf{v}(t) = \mathbf{0},
+> $$
+>
+> for all $t \in \mathscr{D}(\gamma)$. 
+
+Parallel transport implies the transport of a vector that is held constant along the path; constant direction and magnitude. It then follows that for $\dot \gamma = \dot \gamma^i \partial_i$ and $\mathbf{v} = v^i \partial_i$ parallel transport obtains
+
+$$
+    D_t \mathbf{v}(t) = (\dot v^j + \Gamma^j_{ki} v^k \dot \gamma^i) \partial_j = \mathbf{0},
+$$
+
+obtaining the equations
+
+$$
+    \dot v^j + \Gamma^j_{ki} v^k \dot \gamma^i = 0,
+$$
+
+such that
+
+$$
+    \dot v^j = - \Gamma^j_{ki} v^k \dot \gamma^i,
+$$
+
+for all $t \in \mathscr{D}(\gamma)$. These equations can be solved for $\gamma$, obtaining the curve under which $\mathbf{v}$ stays constant. 
+
+If we let $\mathbf{v} = \dot \gamma^i \partial_i$ be the tangent vector along the curve then parallel transport of $\mathbf{v}$ preserves the tangent vector and we obtain the **geodesic equations** given by
+
+$$
+    \dot v^j + \Gamma^j_{ki} v^k \dot \gamma^i = \ddot\gamma^j + \Gamma^j_{ki} \dot\gamma^k \dot\gamma^i = 0,
+$$
+
+for all $t \in \mathscr{D}(\gamma)$. 
+
+One may interpret a geodesic as a generalization of the notion of a straight line or shortest path defined by $\gamma$. As follows from the following proposition.
+
+> *Proposition 2*: let $\gamma: \mathscr{D}(\gamma) \to M: t \mapsto \gamma(t)$ be a smooth curve on the manifold parameterized by an open interval $\mathscr{D}(\gamma) \subset \mathbb{R}$ and let $\mathscr{L}$ be the Lagrangian defined by
+>
+> $$
+>   \mathscr{L} = \|\dot \gamma\|^2 = g_{ij} \dot \gamma^i \dot \gamma^j, 
+> $$
+>
+> for all $t \in \mathscr{D}(\gamma)$. By demanding [Hamilton's principle]() we obtain the geodesic equations 
+>
+> $$
+>   \ddot\gamma^j + \Gamma^j_{ki} \dot\gamma^k \dot\gamma^i = 0, 
+> $$
+>
+> for all $t \in \mathscr{D}(\gamma)$.
+
+??? note "*Proof*:"
+
+    Will be added later.
+
+It may be observed that by demanding the stationary state of the length of the curve we obtain the geodesic equations. 
 
 ## Contravariant derivative