I'm supposed to compute the number of 1- and 2-dimensional faces of a $k$-simplex after one step of barycentric subdivision. I already figured out that a triangle splits into $6=3!$ triangles as can easily be checked . A triangle has 3 sides which gives $6\cdot3=18$ faces but only 1 side per triangle is not shared by any other triangle i.e. 12 sides are being counted twice giving a total of 6+(18-6)\2=12 lines. So a 2-simplex splits into 6 2d-cells and 12 1d-cells.
In order to get intuition I want to use the same strategy for the tetrahedron. A tetrahedron splits into $4!=24$ 3d-cells each having 4 $2$d-faces of which only one is not shared (correct ?). This gives $24\cdot 4 = 96$ 2d-cells of which $96-24 = 72$ are counted twice. This gives a total of $24+72/2 = 50$ $2$d-faces. However, I struggle from here on because I find it hard to imagine how many $3$d-cells share a common $1$d-cell. I think it's 6 maybe, but I'm not sure. I fear my imagination fails here let alone the $n$-dimensional case.
Can somebody point out to me a clue on how to proceed from here ?
Your quest is to be divided into 3 interlinking steps. You have to consider the subdivision of the outer shell, you have to consider the subdivision of the - topologically open - inner region, and then the summation of both to get the final answer in each dimension.
In 1D the simplex is just a single edge. Thus the subdivision does not change anything wrt. the outer shell. There you just have 2 vertices. The inner region contributes 1 further vertex plus 2 edges. So the total here is 2+1=3 vertices and 2 edges.
In 2D the simplex is a triangle. Thus the subdivision of the shell provides 3*1+3*1=6 vertices (3 for the corners, 3 for the side centers) plus 2*3=6 edges (each side is divided into 2). For the inner region we see that for each of the 3 sides the inner subdivision of the 1D simplex provides the according centri-pyramid and for each of the vertices we get an additional edge plus the body center vertex, for sure. So we get here 1 vertex plus 3+3=6 edges plus 2*3=6 faces. - Note that these numbers are just shifted one dimension up, plus a 1 at the vertex count. - So the total is 6+1=7 vertices plus 6+6=12 edges plus 6 faces.
In 3D the simplex is a tetrahedron. Thus the subdivision of the shell provides 4*1+6*1+4*1=14 vertices (1 each at any vertex, at any edge center, at any face center) plus 4*6+6*2=36 edges (for each face the inner edge count as well as for each edge the inner edge count) plus 4*6=24 faces (for each face the inner face count). - Thus the being used multipliers are nothing but the so far already calculated inner count numbers. - For the inner count numbers we can already apply the afore mentioned rule, i.e. we get 1 further vertex (body center) plus 14 further edges (underneath the outer vertices) plus 36 further faces (underneath the outer edges) plus 24 further cells (underneath the outer faces). - So in total we get 14+1=15 vertices, 36+14=50 edges, 24+36=60 faces, 24 cells.
This could all be applied on and on dimensionally. Accordingly in 4D we get for outer shell: 5*1+10*1+10*1+5*1=30 vertices, 5*14+10*6+10*2=150 edges, 5*36+10*6=240 faces, 5*24=120 cells. For the inner region we get the shifted numbers: 1, 30, 150, 240, 120. Thus in total: 30+1=31, 150+30=180, 240+150=390, 120+240=360, 120.
Cf. - in tabular form - for outer shell so far e.g. $$\begin{array}{l|rrrr} 1D & 2\\ 2D & 6 & 6\\ 3D & 14 & 36 & 24\\ 4D & 30 & 150 & 240 & 120\\ \end{array}$$ or for the inner region $$\begin{array}{l|rrrr} 0D & 1\\ 1D & 1 & 2\\ 2D & 1 & 6 & 6\\ 3D & 1 & 14 & 36 & 24\\ 4D & 1 & 30 & 150 & 240 & 120\\ \end{array}$$
Hence the numbers of the outer shell in 5D clearly would become: $$6*1+15*1+20*1+15*1+6*1=62,\\ 6*30+15*14+20*6+15*2=540,\\ 6*150+15*36*20*6=1560,\\ 6*240+15*120=3240,\\ 6*120=720.$$ Etc.
--- rk