This question already has an answer here:
Here I state and try to answer three variations of the twin paradox
1) "Classical" problem, no acceleration, no turn around
Consider the case where there's a stationary planet, and a moving spaceship moving at close to the speed of light, starting at the left going right at constant velocity. Now imagine at some instant, two beings spawn on earth and in the spaceship. They are both aged 0 at that point. Now as the spaceship continues, both frames remain inertial. Which twin is older at any given point?
According to one of the answers here (the 4th answer down to be exact): How is the classical twin paradox resolved?, there is no solution to this paradox. Either A or B will be older depending on who you ask (person A or B or anyone else).
Is this problem reasonable and is the "solution" correct?
2) Normal problem, no acceleration, yes turn around
Everyone knows this one; the paradox is resolved because the person traveling changes direction. http://www.physicsmatt.com/blog/2017/1/18/the-twin-paradox-in-special-and-general-relativity. Been there, done that.
3) Weird problem, no acceleration, no turn around, curved universe?
Now I was thinking, "what if I combine the two?". Imagine the earth is like an asteroids game, or it's a sphere/torus. Going in a certain direction for long enough means that I eventually end up where I started. Consider the case I stated in (1), where two beings aged 0 appear on the earth and on the spaceship. The spaceship proceeds to go at a constant velocity to the right, never changing direction, never under any gravitational field, but due to the nature of the universe they are in, the spaceship twin ends up reaching the earth again. Now at this point, who is older? The solution for (2) won't work because there's no turning point, right? In the link I provided:
You do this by just putting a sheet above the one you started with (or to the side, if moving in that direction). Nothing wrong with that, and no edges to worry about crossing.
The trick is though, when the traveler gets to this new point, you yell "surprise" and identify the new point with the point in the original sheet using the symmetry of the torus.
So the poor sap thinks they're getting away from their twin, only for you to change the nice new spot they picked out for themselves to the place they started from. It's a mathematical sleight-of-hand, but it is the easiest way to see exactly who is younger in the Twin Paradox on the torus.
And my question that I don't feel like the other question sufficiently addresses:
A specification question: I don't quite see how this particular "mathematical sleight of hand" makes it so that the moving twin is aging less. Sorry if I'm being slow, but could anyone explain that part more thoroughly? I still don't see how even in an inertial constant-velocity frames one twin is aging slower.
Feel free to tell me that this problem is nonsensical and stupid--I'm a complete noob so there is a high chance that I have no idea what I'm talking about (please tell me if that's the case)