Turning 2D Trajectories Into 2D Geodesics

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First and foremost let me warn you, potential enlightener to my ignorance, that I'm waaaay out of my element here, so forgive any lack of information or rigor in my statements.

So I have a trajectory of a point in a closed bidimensional plane, that is the $x$ and $y$ coordinates of the 2D-point are bound , like angles.

What I want to do is, I want to twist and deform this plane such that the arbitrary trajectory described by the point is a geodesic in the new twisted crazy plane. Is there a way to figure that out? How do I even put that in mathematical statements?

Then again, forgive me for being so ultra-vague.

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The way I read it, your $x$ and $y$ values are not only bounded, but identified at the boundary, so e.g. in the angle example, $x=180°$ is essentially the same as $x=-180°$. You should make the topology of the “plane” reflect that identification. If you start with a rectangular sheet of rubber, and glue its top edge to its bottom edge, and then the resulting left boundary circle to the right boundary circle, you obtain a torus. At least if you keep edges aligned the same way, otherwise things become more difficult. So topologically you'd describe your setup not as a plane but as a torus, in order to get this wrap-around behavior.

If wrap-around is your only concern, then you could simply choose a simple “flat” metric for the torus and the trajectories should become geodesics. If there are other things affecting your trajectories so that they are no straight lines even when they don't cross the boundary, then you'll have to tell us more about how these trajectories will be shaped. After all, you can't expect a space where every trajectory someone might think of will become a geodesic.

By the way, this concept of “choosing a metric” is just the formal equivalent of your “twist and deform” idea. It is often easier to assume that your space stays fixed and your rulers and other measurement instruments bend and twist depending on location, and the results are largely equiavelent. Unless you need to embed your manifold (i.e. plane, torus, …) into a flat three-dimensional space in such a way that measurements stay true. There are cases where you can choose a metric to satisfy your needs but still find no “isometric embedding” into $\mathbb R^3$. For example, you cannot find an isometric embedding of a topological torus with a flat metric unless you not only bend but actually fold the manifold. The result looks like folded paper, the different sheets incident with the same points in space.