On the Usual Orientation of Cubic Graphs in Random Construction of Riemann Surfaces

Solution 1:

The picture that you have is completely right, I think, by the fact that you got the "usual orientation" induced from the orientation of the ambient $3$-dimensional space. Now you can walk around each face by turning always to the left - you get the $6$ left turning paths that the paper speaks about. To get the idea of what other orientation mean, please keep in mind that the paper defines orientation locally as some chosen cyclic order of edges around each vertex. Since you have $3$ edges, and $3$ cyclic shift of thereof, you have only $3!/3=2$ possible orientations around each vertex, and thus $2^8$ possible choices for the cube. Basically, this is what you say on your own: I'm sure you get everything correctly, just hesitate by some reason.

Solution 2:

There is a way of doing so by using permutations associated to vertices and to half edges, but this is a bit too long to describe in detail to be a comment. In short: imagine every vertex has 3 "half-edges" around it. You may think these half edges are numbered anyhow. Then an orientation is a choice of independent 3-cycles, and let S be the product of these cycles. Let us now glue together the half-edges: each glueing gives us a transposition. Let T be the product of these independent transpositions. Then the number of cycles in ST is the number of left-turn paths in the graph (provided T is an imprimitive permutation, as far as I remember, which is necessary for the surface to be connected). I'm not quite sure though if this is the way you want to compute things without drawing ...