How can I find a medium-length word equal to 1 on a 2-torus? I'm trying to give an example of Dehn's algorithm on a 2-torus and to do so I want to find a word about 20 letters or so in length that is equal to 1 so that I can apply the algorithm. I'm having trouble doing this. I can't seem to find a proper hyperbolic tiling of octagons on the internet so that I could trace out a path on the covering space, and I can't seem to be able to do it just looking at a 2-torus (how would I know if it's null homotopic?). Any ideas on how to generate this? Thanks!
 A: The universal covering space is the hyperbolic plane, which is contractible, so every closed path is null homotopic.
So, "all" you need to do is draw a closed path in the 1-skeleton of the tiling on the hyperbolic plane.
One reason this is hard to visualize is that area grows so swiftly in the hyperbolic plane, in fact it grows exponentially as a function of radius. So you quickly run out of room to visualize, because the count of octagons grows exponentially as you move away from a central octagon. If you start with one central octagon in the tiling, then the number of octagons that touch it is $48$, which is a lot to draw. And then the number of additional octagons that touch those is $272$ (I think, it's hard to count), which is impossible to draw. The number of octagons in each additional layer continues to grow with an exponential base between 5 and 6.
This document has a picture of the 1st layer of 48 octagons. And, as I said, the 2nd layer is impossible to visualize.

If you want an example to play with, you might pick something which doesn't eat up volume so quickly. I would recommend the $(2,3,7)$ tiling of the hyperbolic plane, which is invariant under the Coxeter group
$$\langle a,b,c \mid a^2=b^2=c^2=(ab)^2=(bc)^3=(ca)^7=1\rangle
$$
Its fundamental domain is the triangle with angles $\pi/2$, $\pi/3$, $\pi/7$, with group generators $a,b,c$ being the reflections in the three sides. This tiling has only $14$ tiles touching the central one.

I'll add one more thing, though. Drawing, or attempting to draw, any one of these tilings should lead to an excellent intuition for the proof of Dehn's algorithm.
