📂Geometry

Covariant Derivative of Vector Fields

Theorem¹

Let $M$ be a differentiable manifold, and let $\nabla$ be an affine connection on $M$. Then, there exists a unique function $\dfrac{D}{dt} : V \mapsto \dfrac{DV}{dt}$ that maps a vector field $V$ following a differentiable curve $c : I \to M(t\in I)$ to another vector field $\dfrac{D V}{dt}$ following $c$. Such a mapping is referred to as the covariant derivative of $V$ along $c$, and it has the following properties. When $W$ is a vector field following $c$, and $f$ is a differentiable function defined on $I$,

(a) $\dfrac{D}{dt}(V+W) = \dfrac{DV}{dt} + \dfrac{DW}{dt}$

(b) $\dfrac{D}{dt}(fV) = \dfrac{df}{dt}V + f \dfrac{DV}{dt}$

(c) If $V$ is a contraction map of $Y \in \mathfrak{X}(M)$, that is, if $V(t) = Y(c(t))$, then $\dfrac{DV}{dt} = \nabla_{dc/dt}Y$

$\dfrac{D V}{d t}$ is explicitly given as follows.

$$ \dfrac{DV}{dt} = \sum_{j} \left( \dfrac{d v^{j}}{dt} + \sum_{j,k} v^{j}\frac{dc_{k}}{dt} \nabla_{ X_{k}} \right) X_{j} $$

where $V = v^{j}X_{j}$, $X_{j} = \dfrac{\partial }{\partial x_{j}}$.

Explanation

From (a) and (b), there is no reason not to denote such a vector field corresponding to $V$ as $\dfrac{D V}{dt}$, because it has the properties that differentiation should have.

Proof

• Part 1. Uniqueness

  Assume that there exists a mapping $V \mapsto \dfrac{DV}{dt}$ that satisfies (a)~(c). Let $\mathbf{x} : U \to M$ be the coordinates overlapping with curve $c$.

  $$ c(I) \cap \mathbf{x}(U) \ne \varnothing $$

  Let’s denote $c(t) = \mathbf{x}(c_{1}(t), \dots, c_{n}(t))$. Consider $V$ as a vector field.

  $$ V = v^{j}\dfrac{\partial }{\partial x_{j}} = v^{j}X_{j} $$

  Here, $X_{j} = \dfrac{\partial }{\partial x_{j}} = \left.\dfrac{\partial }{\partial x_{j}}\right|_{c(t)}$, $v^{j} = v^{j}(t)$. Then, by properties (a) and (b), the following holds.

  $$ \begin{align*} \dfrac{DV}{dt} =&\ \dfrac{D}{dt}\left( \sum_{j} v^{j}X_{j} \right) \\ =&\ \sum_{j} \dfrac{D}{dt}\left( v^{j}X_{j} \right) & \text{by (a)} \\ =&\ \sum_{j} \left( \dfrac{d v^{j}}{dt}X_{j} + v^{j}\dfrac{D X_{j}}{dt} \right) & \text{by (b)} \\ \end{align*} $$

  Based on property (c) and property 1 of the affine connection, the following is true.

  $$ \begin{align*} \dfrac{D X_{j}}{dt} =&\ \nabla_{dc/dt} X_{j} &\text{by (c)} \\ =&\ \nabla_{\sum_{k} \frac{dc_{k}}{dt} X_{k}} X_{j} \\ =&\ \sum_{k} \frac{dc_{k}}{dt} \nabla_{ X_{k}} X_{j} & \text{by 1.} \\ \end{align*} $$

  By substituting this into the original equation, the following is obtained.

  $$ \begin{align*} \dfrac{DV}{dt} =&\ \sum_{j} \left( \dfrac{d v^{j}}{dt}X_{j} + v^{j}\sum_{k} \frac{dc_{k}}{dt} \nabla_{ X_{k}} X_{j} \right) \\ =&\ \sum_{j} \dfrac{d v^{j}}{dt}X_{j} + \sum_{j,k} v^{j}\frac{dc_{k}}{dt} \nabla_{ X_{k}} X_{j} & \cdots \circledast \end{align*} $$

  Auxiliary theorem

  $(\nabla_{X}Y)(p)$ depends solely on $X(p)$ and $Y(\gamma (t))$.

  According to the auxiliary theorem, $\nabla_{ \frac{\partial }{\partial x_{k}}} \dfrac{\partial }{\partial x_{j}}$ is determined by the coordinates. The rest is uniquely determined by the coordinates as well. Therefore, if it exists, it is unique.

• Part 2. Existence

  Define $\dfrac{DV}{dt}$ as follows; then it satisfies properties (a)~(c).

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