Translation Surfaces: Geometric Definition & Billiard Systems

[Joppy]

In this Wiki article, a geometric definition of a translation surface is given.

I'm lost in at the first line were it is stated that a given collection of polygons need not be convex. How is this possible? I am trying to understand translation surfaces from the perspective of dynamical systems, specifically, billiard systems. In this setting we can 'unfold' the trajectory of a point particle. But surely this unfolding process only works for trajectories confined to convex regions?

I suspect my confusion comes from the fact that generating a translation surface from unfolding a billiard trajectory, and generating one given the definition from Wiki are different things. I also don't understand what is meant by $s_j = s_i + \vec{v}_i$. Are we saying that for every side in a plane of polygons, there exist some other side which lies in the same direction?
Thanks

 

[I like Serena]

Hey Joppy,

Here's my 2 cents.

Since it's about billiards, I imagine that we define a set of neighboring rectangles, each representing the table. Now we can shoot the ball, and instead of reflecting it against an edge, we let it roll into the next rectangle.
That does mean that various edges are actually the same, so they are identified with each other through a translation.

And instead of convex rectangles, we might also have non-convex polygons.

 

[Joppy]

Hey Joppy,

Here's my 2 cents.

Since it's about billiards, I imagine that we define a set of neighboring rectangles, each representing the table. Now we can shoot the ball, and instead of reflecting it against an edge, we let it roll into the next rectangle.
That does mean that various edges are actually the same, so they are identified with each other through a translation.

Thanks! Yes that's how I understand it to work for convex polygonal billiards. I am curious as to how this works, if at all, for billiard tables which are not convex.

Now turn those rectangles into rectangular prisms and we have a translation space.

Do you mean we fold up this mesh of rectangles into a prism? For example, if we have a horizontal trajectory inside the unit square, and we let it roll into four squares (so that we have four squares side by side), do we fold them back up to obtain a cube with two faces missing?

Supposedly, using this method of unfolding, some trajectories will yield a torus as the translation surface. I would like to know in what sense the translation surface is a torus (for this case).

 

[I like Serena]

For a billiard table we wouldn't be folding squares up into a cube.
Instead we have 4 neighboring rectangles. And when the ball rolls over a rightmost edge, it will magically appear on the corresponding leftmost edge.Same for top and bottom.

Now imagine a rectangle with a halfsize rectangle removed from a corner. We put again 4 such shapes next each to other in a mirrored layout. There we go.

As for a torus, we effectively get that when we have just a single rectangle. That is, it's topologically equivalent to a torus.

 

[Joppy]

For a billiard table we wouldn't be folding squares up into a cube.
Instead we have 4 neighboring rectangles. And when the ball rolls over a rightmost edge, it will magically appear on the corresponding leftmost edge.Same for top and bottom.

Now imagine a rectangle with a halfsize rectangle removed from a corner. We put again 4 such shapes next each other in a mirrored layout. There we go.

As for a torus, we effectively get that when we have just a single rectangle. That is, it's topologically equivalent to a torus.

Thanks! :)

(I should have read this before asking..)

edit: This article is also nice, if anyone is interested.