Twin Simplex

Geometry Level 5

Let $e_1, \ldots , e_{n+1}$ be the standard basis in $\mathbb R^{n+1}$ . We define an $n$ -dimensional simplex as the convex hull of the endpoints of the vectors $e_1, \ldots , e_{n+1}$ . That is, the $n$ -dimensional simplex can be defined as the convex hull of $n + 1$ affinely independent points.

We can thus clearly see that the standard 1-simplex is a line segment, the standard 2-simplex is an equilateral triangle, the standard 3-simplex is a regular tetrahedron, and so on.

How many 4-dimensional faces does the 197-dimensional simplex have?

Image: orthogonal projection of the 4-simplex (5-cell) performing a simple rotation. Credit for the image goes to Jason Hise.

Bonus: Generalize the problem. How many $k$ -dimensional faces does an $n$ -dimensional simplex have?

The answer is 2410141734.

1 solution

Frank Aiello
Aug 16, 2017

The n-simplex is a generalization of a tetrahedron to N dimensions. The boundary of an n-simplex has $n+1$ 0-dimensional faces (polytope vertices), $n(n+2)/2$ 1-faces (polytope edges), and ${n+1}\choose{k+1}$ faces of dimension k. Additionally, the n-cube has $2^{n-k}$ ${n}\choose{k}$ faces of dimension k and the n-orthoplex (i.e. the dual of the n-cube) has $2^{k+1}$ ${n}\choose{k+1}$ faces of dimension k; these are the only regular n-dimensional polytopes for n > 5.

One way you can prove the given proposition about the k-faces of an n-simplex is from Euler's more general formula for any polytope: $\sum_{k=0}^{n} (-1)^{k} F_k = 1-(-1)^n$ . For the simplex, we have: $\sum_{k=0}^{n-1} (-1)^{k}$ ${n+1}\choose{k+1}$ = $1-(-1)^n$ . For this problem, our arbitrarily chosen dimension for the simplex was 197 and we wanted to find all faces of dimension 4. Hence the correct answer is ${197+1}\choose{4+1}$ = 2,410,141,734. 197 also happens to be a twin prime and the n-simplex is dual to itself (i.e. it's its own "twin"). Hence, the title of the problem: twin simplex.

Twin Simplex

The answer is 2410141734.

1 solution

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