For some reason this is reminding me of a very pretty polyhedral surface of genus 3 that does not exist as a subset of R^3 but instead as a subset of the torus R^3/Z^3. Truncate the corners of a unit cube so that what remains is a truncated octahedron (the corners become 8 hexagons and the faces become 6 smaller squares at 45º angles). Now just considering the 14 faces, delete the 6 squares and identify the opposite faces of the original cube so that it becomes the torus T^3 = R^3/Z^3. The truncated octahedron is now a polyhedral surface M_3 of genus 3 inside the torus T^3, tiled by 8 hexagons, 4 per vertex. There is an isometry of the whole M_3 to itself taking a given hexagon to any hexagon in any of 12 dihedral ways, for a total of 96 isometries altogether. —Dan Tomas Rokicki wrote: ----- Or take the eight cube solution and skew two opposite cubes so the visible parts are not rectangles. This introduces the necessary separating edges. -----