{$T$} is loopless iff {$\Delta T$} is loopless:For any {$T\in\mathcal{B}^+$}, there is a loop on {$T$} iff there is a loop on {$\Delta(T)$}. Proof: We look at the box from the top, with the third dimension as height. A part of a geodesic on a {$a\times b\times c$} box might look like this,  where we use blue for parts on the floor of the box, red for parts on the walls, and green for parts on the top. If we think about the box in which the height is increased by {$a+b$}, the geodesic would wind an additional distance around, {$a+b$}, which would take it half-way around the box.  The result is that the part on top (green) is the same but rotated 180 degrees around the center of the box.  If we continue on the {$a\times b\times c$} box,  and if we also continue on the {$a\times b\times (a+b+c)$} box,  you see that the new segment on the floor (blue) is the same in both---because one was rotated 180 degrees twice. It's easy to see that if one of these connects to become a loop, the other does as well. |