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Let's say that I know the values to $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\in\mathbb{C}$ along some curve parameterized by real parameter $t$. If at all, how can I use this information to recover the general vector field $\mathbf{F}(\mathbf{r})$ that does not depend on the direction of the tangential vector $\mathbf{r}'(t)$ at each point along the curve? Or, what information can this data give me about the topology of the underlying base space?

Can I can use $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$ to define a vector field along it (a complex number may be thought as a vector with components comprising the real and imaginary parts)? If so, can I extend this field to a unique vector field over the entire $\mathbf{r}$ space to recover information about the 2-dimensional manifold it lives in? If I can, under what conditions can I do so? I am only interested in the case where $\mathbf{r}$ are coordinates of some smooth 2-dimensional manifold.

I am coming from a physics background, and am not sure where exactly to look, nor am I sure what mathematical language I should use to tackle this problem. Some initial thoughts are that I can define a vector field along the curve, and then define a covariant derivative field, and then extend it beyond the curve under some conditions (for example by considering a tubular neighborhood along it). At first I thought analyzing the 2nd order tangent space might help, but it was too abstract for me.

I saw these similar posts, but am not sure how they answer my question directly:

Definition of vector field along a curve

Covariant derivative of a vector field along a curve

Recovering a manifold from its Lie algebra of vector fields

Using bump function in coordinate chart to construct a vector field $V$ along a curve $\gamma$ with particular properties.

Recover vector field from auto-parallel curve

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    You cannot recover $\mathbf F$. Even if you know that the curve is parametrized with speed $1$, so that $\mathbf r'(t)$ is a unit vector, you still cannot. If I tell you the dot product of two vectors $\mathbf u$ and $\mathbf v$ and I tell you the vector $\mathbf v$, then you have a hyperplane's worth of choice for $\mathbf u$. If you know only the direction of $\mathbf v$, you have less information. – Ted Shifrin Oct 07 '20 at 19:45
  • @TedShifrin thank you. – TribalChief Oct 08 '20 at 06:48

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