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

# Linear connections

Let $\mathrm{M}$ be a differential manifold with $\dim \mathrm{M} = n \in \mathbb{N}$ used throughout the section. Let $\mathrm{TM}$ and $\mathrm{T^*M}$ denote the tangent and cotangent bundle, $V$ and $V^*$ the fiber and dual fiber bundle and $\mathscr{B}$ the tensor fiber bundle. 

> *Definition 1*: a **linear connection** on the fiber bundle $\mathscr{B}$ is a map
>
> $$
>   \nabla: \Gamma(\mathrm{TM}) \times \Gamma(\mathscr{B}) \to \Gamma(\mathscr{B}): (\mathbf{v}, \mathbf{T}) \mapsto \nabla_\mathbf{v} \mathbf{T}, 
> $$
>
> satisfying the following properties, if $f,g \in C^\infty(\mathrm{M})$, $\mathbf{v} \in \Gamma(\mathrm{TM})$ and $\mathbf{T}, \mathbf{S} \in \Gamma(\mathscr{B})$ then
>
> 1. $\nabla_{f\mathbf{v}} \mathbf{T} = f \nabla_\mathbf{v} \mathbf{T}$
> 2. $\nabla_\mathbf{v} (f \mathbf{T} + g \mathbf{S}) = (\nabla_\mathbf{v} f) \mathbf{T} + f \nabla_\mathbf{v} \mathbf{T} + (\nabla_\mathbf{v} g) \mathbf{S} + g \nabla_{\mathbf{v}} \mathbf{S}$, 
> 3. $\nabla_\mathbf{v} f = \mathbf{v} f = \mathbf{k}(df, \mathbf{v})$. 

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. 

## 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
>
> $$
>   D_k \mathbf{v} \overset{\text{def}}= \nabla_{\partial_k} \mathbf{v} = (\partial_k v^i) \mathbf{e}_i + v^i \Gamma^j_{ik} \mathbf{e}_j = (\partial_k v^i + \Gamma^i_{jk} v^j)\mathbf{e}_i, 
> $$
>
> with formally $\mathbf{k}(\mathbf{\hat e}^j, \nabla_{\partial_k} \mathbf{e}_i) = \Gamma^j_{ik}$ the **linear connection symbols**, in this case $\nabla_{\partial_k} \mathbf{e}_i = \Gamma^j_{ik} \mathbf{e}_j$. 

The covariant derivative can thus be seen as a linear connection for which only the basis is used of the tangent vector. The covariant derivative can also be applied on higher, mixed rank tensors $\mathbf{T} = T^{ij}_k \mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{\hat e}^k \in \Gamma(\mathscr{B})$ which obtains

$$
    D_l \mathbf{T} = (\partial_l T^{ij}_k) \mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{\hat e}^k + T^{ij}_k (\Gamma_{il}^m\mathbf{e}_m) \otimes \mathbf{e}_j \otimes \mathbf{\hat e}^k + T^{ij}_k \mathbf{e}_i \otimes (\Gamma^m_{jl} \mathbf{e}_m) \otimes \mathbf{\hat e}^k + T^{ij}_k \mathbf{e}_i \otimes \mathbf{e}_j \otimes (\hat \Gamma^k_{ml} \mathbf{\hat e}^m),
$$

with the dual linear connection symbols given by $\mathbf{k}(\nabla_{\partial_k} \mathbf{\hat e}^i, \mathbf{e}_j) = \hat \Gamma^j_{ik}$ with $\nabla_{\partial_k} \mathbf{\hat e}^i = \hat \Gamma^j_{ik} \mathbf{\hat e}^j$. We then have the following proposition such that we can simplify the above expression.

> *Proposition 1*: let $\Gamma^j_{ik}$ be the linear connection symbols of a covariant derivative and let $\hat \Gamma^j_{ik}$ be the dual linear connection symbols given by $\mathbf{k}(\nabla_{\partial_k} \mathbf{\hat e}^i, \mathbf{e}_j) = \hat \Gamma^j_{ik}$, then we have that
>
> $$
>   \hat \Gamma^j_{ik} = - \Gamma^j_{ik},
> $$
>
> for all $(i,j,k) \in \mathbb{N}^3$. 

??? note "*Proof*:"

    Will be added later.

With the result of proposition 1 we may write

$$
    D_l \mathbf{T} = (\partial_l T^{ij}_k + \Gamma_{ml}^i T^{mj}_k + \Gamma_{ml}^j T^{im}_k - \Gamma_{kl}^m T^{ij}_m) \mathbf{e}_i \otimes \mathbf{e}_j \otimes \mathbf{\hat e}^k. 
$$

### Transformation of linear connection symbols

Will be added later.

## Parallel transport 

> *Definition 3*: let $\mathbf{v} \in \Gamma(\mathrm{TM})$, then **parallel transport** of $\mathbf{v}$ occurs along the manifold $\mathrm{M}$ when
>
> $$
>   D_k \mathbf{v} = \mathbf{0}.
> $$

For example, a parameterised vector field $\mathbf{v}: x(t) \mapsto \mathbf{v}(x(t)) \in \Gamma(\mathrm{TM})$ is transported parallel if 

$$
    D_t \mathbf{v} = (\partial_k v^i) \dot x^k \partial_i + \Gamma^i_{jk} v^j \partial_i = \mathbf{0},
$$

so $(\partial_k v^i) \dot x^k + \Gamma^i_{jk} v^j = 0$

## Contravariant derivative

Will be added later.