In my differential geometry textbook, the following definition is given:
If $\alpha$ is path from $P$ to $Q$, we refer to $\mathbf X(Q)$ as the parallel translate of $\mathbf X(P) = \mathbf X_0 \in T_pM$ along $\alpha$.
Where $T_PM$ is the tangent plane of surface $M$ at point $P$.
I feel like I'm not understanding this definition correctly. Any two points in $\mathbb R^3$ have a path between them, so it seems like if you have a point in the tangent plane, then any other point is a valid parallel translate of it. But this seems far too general to be useful.
So what am I misunderstanding here? Or am I understanding correctly, and this definition is more useful than I thought? If so, how?
Actually if you have a curve $\gamma:[t_0,t_1]\to M$ connecting two points $p=\gamma(t_0)$ and $q=\gamma(t_1)$, then there is a unique vector field $X:I\to TM$ along $\gamma$ ( that is, $X(t)\in T_{\gamma(t) M}$ for all $t$) that satisfies $D_t X=0$ and $X(t_0)=X_0$, where $D_t$ is the covariant derivative along $\gamma$. We define the parallel translation $P_{t_0,t_1}:T_p M\to T_q M$ with $P_{t_0,t_1}(X_0)=X(t_1)$ where $X$ is the unique vector field that satisfies $D_t X=0$ and $X(t_0)=X_0$.
In $\mathbb{R}^3$ we can identify $T\mathbb{R}^3$ with $\mathbb{R}^3 \times \mathbb{R}^3$ and a field along $\gamma$ to be a function $X:[t_0,t_1]\to \mathbb{R}^3 \times\mathbb{R}^3$ with $X(t)=(\gamma(t),X^1(t),X^2(t),X^3(t))$ and $D_t X=(\gamma(t),\dot{X}^1(t),\dot{X}^2(t),\dot{X}^3(t))$. A parallel vector field satisfies $D_t X=0$, that is, $\dot{X}^1(t),\dot{X}^2(t),\dot{X}^3(t)$ to be $0$. Then $X^i$ have to be constant.