Turning a rhombus billiard into an equivalent barrier billiard

Question

I am reading a research paper and the authors map the rhombus billiard (angles $60$-$120$ degrees) to an equivalent barrier billiard. They start with a rhombus standing upright and reflect in a counter clockwise manner:

"after three reflections, we can reflect the third rhombus back onto itself, i.e. the fourth rhombus comes to lie under the third rhombus. If we continue reflecting now, the sixth rhombus will come to lie under the first rhombus. The point 1 (the point shared by all of the rhombus upon reflection), will become a saddlepoint. Continuing the process of reflection we get two planes connected by cuts between the saddle points. This construction projected onto two dimensions entails an orbit looking like a zigzag path. We now construct the fundamental region using six replicas of the rhombus and subsequently tessellate the two dimensional plane by stacking the fundamental regions side by side, exploiting the transitional symmetry. On doing so we will generate a barrier billiard."

Would anyone help me conceptualize what is happening here? I am actually confused by the numbering of the rhombus as well...is the first rhombus the rhombus resulting after one reflection or is it the original rhombus?

Edit: I think this is what's happening:

The pictures above are such that rhombus $4$ is under rhombus $3$ and rhombus $6$ is under rhombus one. We can see that the barrier is formed due to the "reflecting back" of rhombus $4$ under $3$. If we flatten this picture out (project onto the plane?) we get:

Which will generate the barrier billiard proposed if we tessellate by translating this region.

Answer 1

This is a pretty clever trick the authors are using, and one I've never seen before.

The normal approach to studying billiards on a convex polygon is a technique called unfolding. When a billiard ball strikes an edge, instead of having the billiard ball reflect off the edge, you have the billiard ball continue in a straight line and reflect the polygon across the edge. This is equivalent; all you've done is kept the velocity of the ball the same as it was before by changing coordinates on your polygon.

If you reflect the rhombus along all possible sequences of edges, you will tile the plane twice, with one set of rhombi oriented one way and another oriented the other. At each vertex $6$ rhombi meet, so at the vertices with angle $2\pi/3$ the total angle becomes $4\pi$. For comparison, the Riemann surface associated with $z^{1/2}$ is:

Note that the two sheets don't actually overlap, this is just an artifact of drawing the surface in $3$ dimensions. You can see that the total angle around the origin here is also $4\pi$. It so happens that there is a natural complex structure on the unfolding, and at the vertices with angle $2\pi/3$ it is locally equivalent to the Riemann surface of $z^{1/2}$.

The clever idea the authors have is to only perform the unfolding half way. When reflecting the rhombus about the vertices with angle $2\pi/3$, they stop after getting $3$ copies of the rhombus instead of $6$ (and so for one of these edges, the billiard ball will still reflect--hence it is a barrier). Thus the rhombi tile the plane only once, as shown in Figure 4 (although there is an extraneous line near the origin of the figure, which is a mistake).

How does this relate to the Riemann surface point of view? Well, it's equivalent to cutting along the line in the image where the two sheets overlap, and regluing them in the opposite manner, i.e. with both ends of the top sheet connected to the corresponding end of the bottom sheet.