User:Tomruen/Disphenoid

The join operator is:

• Identity element: nullitope: A∨∅ = A
• Commutative : A∨B = B∨A
• Associative : (with both join and sums)
  • A∨B∨C = (A∨B)∨C = A∨(B∨C)
  • A∨B+C = (A∨B)+C = A∨(B+C)
• Supports De Morgan's law with duality: *(A∨B) = (*A)∨(*B)
• rank(A∨B)=rank(A)+rank(B)+1
• Vertex figures:
  • verf(A∨A) = verf(A)∨A
  • verf(A∨B) = verf(A)∨B, A∨verf(B)
  • verf(A∨A∨A) = verf(A)∨A∨A
  • verf(A∨B∨C) = verf(A) ∨B∨C, A∨ verf(B) ∨C, A∨B∨ verf(C)

The join A∨B will be:

• Convex, if A and B are convex.
• self-dual, if A and B are self-dual, or if A and B are duals.
• A simplex, if A and B are simplexes.

When looking at vertices and edges alone as a graph, the join A∨B is the union of graphs A and B, and their connecting complete bipartite graph. It has v_A+v_B vertices, and e_A+e_A+v_A×v_B edges.

Multi-wedges have the vertices of all of the element polytopes. Their edges can be seen as the union of the edges of the element polytopes, and all connections of vertices between elements, as defining in a complete multipartite graph. Higher k-faces exist for all element permutations from nullitope to full polytopes joins.

${\displaystyle {\begin{array}{c}1\\1\quad 1\\1\quad 2\quad 1\\1\quad 3\quad 3\quad 1\\1\quad 4\quad 6\quad 4\quad 1\\1\quad 5\quad 10\quad 10\quad 5\quad 1\\1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1\\1\quad 7\quad 21\quad 35\quad 35\quad 21\quad 7\quad 1\\1\quad 8\quad 28\quad 56\quad 70\quad 56\quad 28\quad 8\quad 1\\\end{array}}}$

Pascal's triangle as simplex f-vectors:
∅
( ) (point)
{ } (segment)
{3} (triangle)
{3,3} (tetrahedron)
{3,3,3} (5-cell)
{3,3,3,3} (5-simplex)
{3,3,3,3,3} (6-simplex)
{3,3,3,3,3,3} (7-simplex)

The f-vector counts the number of k-faces in a polytope, 0..n-1. Extended f-vectors can include end elements -1 and n, both 1. f_-1=1, a nullitope, and f_n=1, the body.

f₀ is the number of vertices, f₁ the number of edges, etc. Regular polygons, f({p})=(1,p,p,1).

If you join only points, f-vectors sum in simplexes as Pascal's triangle as binomial coefficients. A nullitope has f-vector (1). A point, ( ), has f( )=(1,1). Segment, f({ })=(1,2,1). A triangle has f({3})=(1,3,3,1). A tetrahedron has f({3,3})=(1,4,6,4,1).

A self-dual polytope will have f-vectors are forward-reverse symmetric.

k-faces of A∨B are generated by joins of all i-faces of A, and all (k-i)-faces of B. With i=-1 to k.

• The number of vertices are the sum of the vertices of each.
• New edges are edges of A, edges of B, and new edges between vertices of A and vertices of B.
• New faces are generated by all faces of A, all faces of B, and new faces from edges of A to every vertex of B, and edges of B to each vertex of A
• Etc

${\displaystyle {\begin{array}{c}1\quad 4\quad 4\quad 1\\1\quad 5\quad 8\quad 5\quad 1\\1\quad 6\quad 13\quad 13\quad 6\quad 1\\1\quad 7\quad 19\quad 26\quad 19\quad 7\quad 1\\1\quad 8\quad 26\quad 45\quad 45\quad 26\quad 8\quad 1\\1\quad 9\quad 34\quad 71\quad 90\quad 71\quad 34\quad 9\quad 1\\\end{array}}}$

f-vector series for joins with simplices:
2D: {4}∨∅ (square)
3D: {4}∨( ) (square pyramid)
4D: {4}∨{ } (square-segment_di-wedge)
5D: {4}∨{3} (square-triangle_di-wedge)
6D: {4}∨{3,3} (square-tetrahedron di-wedge)
7D: {4}∨{3,3,3} (square-5-cell_di-wedge)

${\displaystyle {\begin{array}{c}1\quad 8\quad 12\quad 6\quad 1\\1\quad 9\quad 20\quad 18\quad 7\quad 1\\1\quad 10\quad 29\quad 38\quad 25\quad 8\quad 1\\1\quad 11\quad 39\quad 67\quad 63\quad 33\quad 9\quad 1\\1\quad 12\quad 50\quad 106\quad 130\quad 96\quad 42\quad 10\quad 1\\1\quad 13\quad 62\quad 156\quad 236\quad 226\quad 138\quad 52\quad 11\quad 1\\\end{array}}}$

f-vector series for joins with simplices:
3D: {4,3}∨∅ (cube)
4D: {4,3}∨( ) (cubic pyramid)
5D: {4,3}∨{ } (cube-segment di-wedge)
6D: {4,3}∨{3} (cube-triangle di-wedge)
7D: {4,3}∨{3,3} (cube-tetrahedron di-wedge)
8D: {4,3}∨{3,3,3} (cube-5-cell di-wedge)

Wedges of the form A∨( )∨( )∨...∨( ) = A∨n+1⋅( ) = A∨{3^n-1}, as a join by a n-simplex.

We can represent as f-vectors as f(A∨n+1⋅( ))=f(A)*(1,1)ⁿ⁺¹ .

This family of wedges has a special property like Pascal's triangle, where each new row has f-vector as neighboring sums of previous row f-vector, starting with A. A∨{ } will have f-vectors of sums, but 2 levels down, and A∨{3} is expressed as sums 3 levels down, A∨{3,3} sums 4 levels down, etc.

These polytopes are self-dual if A is self-dual, i.e. f-vectors are forward-reverse symmetric.

Multi-wedges with points have special names by Jonathan Bowers:^[6] The names come from BSA names of simplices: 2D (scal), 3D:tet, 4D:pen, 5D:hix, 6D:hop, 7D:oca, 8D:ene, 9D: day, 10D: ux, with suffix -ene.^[7]

Joining three or more polytopes allows multiple orthogonal altitudes. Explicit parentheses are needed to differentiate (A∨B)∨_hC from A∨_h(B∨C), with highest level join altitude being expressed, ∨_h, with altitude h.

Multi-wedges can be evaluated in any order of evaluation, as long as the sum of the square of the circum-radius of the polytope elements are less than 1.

We can determine the counts by combinations, ${\displaystyle {\binom {n}{k}}={\frac {n!}{(n-k)!k!}}}$. And with multinomial theorem, it is generalized by ${\displaystyle {\binom {n}{n-a-b,a,b}}={\frac {n!}{(n-a-b)!a!b!}}}$ for 3 partitions where n>a+b.

Altitude, h, case count for n-wedge by pairwise partitioning. If the partition sizes are equal, like 2+2 or 3+3, the combinations are cut in half.

More information , ...

n-wedge	Form	Combinations	Counts
di-wedge+ n≥2	A∨hB	n choose 2	${\displaystyle {\binom {n}{2}}}$	${\displaystyle {\binom {n}{n-2,2}}}$	${\displaystyle {\frac {n!}{(n-2)!2!}}}$
tri-wedge+ n≥3	(A∨B)∨hC	n choose 2+1	${\displaystyle {\binom {n}{2}}{\binom {n-2}{1}}}$	${\displaystyle {\binom {n}{n-3,2,1}}}$	${\displaystyle {\frac {n!}{(n-3)!2!1!}}}$
tetra-wedge+ n≥4	(A∨B)∨h(C∨D)	n choose 2+2	${\displaystyle {\frac {1}{2}}{\binom {n}{2}}{\binom {n-2}{2}}}$	${\displaystyle {\frac {1}{2}}{\binom {n}{n-4,2,2}}}$	${\displaystyle {\frac {1}{2}}{\frac {n!}{(n-4)!2!2!}}}$
	(A∨B∨C)∨hD	n choose 3+1	${\displaystyle {\binom {n}{3}}{\binom {n-3}{1}}}$	${\displaystyle {\binom {n}{n-4,3,1}}}$	${\displaystyle {\frac {n!}{(n-4)!3!1!}}}$
penta-wedge+ n≥5	(A∨B∨C)∨h(D∨E)	n choose 3+2	${\displaystyle {\binom {n}{3}}{\binom {n-3}{2}}}$	${\displaystyle {\binom {n}{n-5,3,2}}}$	${\displaystyle {\frac {n!}{(n-5)!3!2!}}}$
	(A∨B∨C∨D)∨hE	n choose 4+1	${\displaystyle {\binom {n}{4}}{\binom {n-4}{1}}}$	${\displaystyle {\binom {n}{n-5,4,1}}}$	${\displaystyle {\frac {n!}{(n-5)!4!1!}}}$
hexa-wedge+ n≥6	(A∨B∨C)∨h(D∨E∨F)	n choose 3+3	${\displaystyle {\frac {1}{2}}{\binom {n}{3}}{\binom {n-3}{3}}}$	${\displaystyle {\frac {1}{2}}{\binom {n}{n-6,3,3}}}$	${\displaystyle {\frac {1}{2}}{\frac {n!}{(n-6)!3!3!}}}$
	(A∨B∨C∨D)∨h(E∨F)	n choose 4+2	${\displaystyle {\binom {n}{4}}{\binom {n-4}{2}}}$	${\displaystyle {\binom {n}{n-6,4,2}}}$	${\displaystyle {\frac {n!}{(n-6)!4!2!}}}$
	(A∨B∨C∨D∨E)∨hF	n choose 5+1	${\displaystyle {\binom {n}{5}}{\binom {n-5}{1}}}$	${\displaystyle {\binom {n}{n-6,5,1}}}$	${\displaystyle {\frac {n!}{(n-6)!5!1!}}}$

Close

A tri-wedge A∨B∨C has 6 altitudes: A∨B, A∨C, B∨C, (A∨B)∨C, (A∨C)∨B, and (B∨C)∨A.

For example, if A, B, and C are points, it makes a triangle. The first three altitudes correspond to the edge lengths of the triangle, and the next 3 correspond to the 3 altitudes of the triangle.

A tetra-wedge has 6 altitudes A∨B, 12 altitude of form (A∨B)∨C, 3 altitude of form (A∨B)∨(C∨D), and 4 altitudes of form (A∨B∨C)∨D.

For example, if all 4 polytopes are points, this corresponds to a tetrahedron, having with 6 edge lengths, 12 altitudes on the 4 triangular faces, 3 digonal disphenoid altitude of opposite edges, and 4 triangular pyramid altitudes.