Translation of Vector Field Translation:
Theorem
Let’s call a regular curve on the surface $C^{2}$ through $\boldsymbol{\alpha} (t)$. Let $\tilde{\mathbf{X}} = (\tilde{X}^{1}, \tilde{X}^{2})$ be a vector tangent to $M$ at point $\boldsymbol{\alpha}(t_{0})$. Then, there exists a unique vector field $\mathbf{X}(t)$ parallel to $\boldsymbol{\alpha}(t)$ that satisfies $\mathbf{X}(t_{0}) =\tilde{\mathbf{X}}$.
Definition
The unique vector field $X(t)$ is called the parallel translate of $\tilde{X}$ along $\alpha$.
Proof
Let $\mathbf{x}$ be the coordinate chart mapping for $\boldsymbol{\alpha}(t_{0})$. It can be represented as $\boldsymbol{\alpha}(t) = \mathbf{x} \left( \alpha^{1}(t), \alpha^{2}(t) \right)$.
Now consider the following initial value problem.
$$ \begin{align*} \dfrac{d X^{k}}{dt} =&\ - \sum_{i,j} \Gamma_{ij}^{k} X^{i} \dfrac{d \alpha^{j}}{d t},\quad k=1,2 \\ X^{k}(t_{0}) =&\ \tilde{X}^{k} \end{align*} $$
Auxiliary Lemma: Necessary and Sufficient Condition for Parallel Vector Fields
Let $\boldsymbol{\alpha}(t) = \mathbf{x}\left( \alpha^{1}(t), \alpha^{2}(t) \right)$ be a regular curve on the coordinate chart mapping $\mathbf{x}$. Let $\mathbf{X}(t) = X^{1}\mathbf{x}_{1} + X^{2}\mathbf{x}_{2}$ be a differentiable vector field along $\alpha$. Then, the necessary and sufficient condition for $\mathbf{X}(t)$ to be parallel along $\alpha$ is as follows.
$$ 0 = \dfrac{d X^{k}}{d t} + \sum_{i,j} \Gamma_{ij}^{k} X^{i} \dfrac{d \alpha^{j}}{d t},\quad k=1,2 $$
By Picard’s Theorem, there exists a unique solution near $t_{0}$. Therefore, according to the auxiliary lemma, this solution is a vector field that is parallel along $\alpha$.
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