The torus T_7 below is the dual to what you get by arranging 6 congruent regular hexagons in the plane around a seventh and then identifying its boundary to itself (by identifying 3 pairs of polygonal arcs via translations; there are only two essentially equivalent ways to do this). To see a way to embed T_7 into the 2-skeleton of the 6-simplex S_6 in R^6:* 1) Label the vertices V of S_7 as V = {p_0, ..., p_6} in cyclic order. 2) For each i = 0, ..., 6 consider the set V_i = V - {p_i}. For each pair of consecutive vertices p_j, p_k in V_j in the inherited cyclic order, draw the triangle whose vertices are p_i, p_j, p_k. 3) This surrounds each vertex with 6 triangles any consecutive pair of which shares an edge. 4) The number of triangles thus drawn is 7*6/3 = 14, because each triangle was drawn thrice, one for each vertex. 5) Thus this is T_7. —Dan _____ * You can think of the 7 vertices of the 6-simplex S_6 as (the endpoints of) the standard basis vectors in R^7, all translated by the vector (-1/7,...,-1/7) so they end up in the subspace R^6 of R^7 defined by sum of coordinates = 0. I wrote: ----- ... Call this torus T_7. Its vertices and edges form the complete graph K_7, and its triangles consist of exactly 2/5 of the possible 35 triangles made of 3 vertices. As a combinatorial torus, it is dual to the 7-hexagon (hexagonal) torus, whose hexagons form a maximally symmetrical 7-coloring of the torus. This shows that T_7 with its intrinsic metric (distances measured along its 14 triangles, not across 6-space) is also a hexagonal torus: One obtained by identifying opposite edges of a regular hexagon. Which suggests the question: Can the 7-hexagon hexagonal torus itself be embedded in R^6 so that each hexagon is a bona fide planar regular hexagon? -----