Prove that the number of $k$-simplexes in an $n$-orthoplex is $2^{k+1}\binom{n}{k+1}$ (where $n,k \in \mathbb{N}$ with $0 \leqslant k < n$)?

Question

In the book 'Regular Polytopes' (H.S.M. Coxeter, 1973), Coxeter describes the $n$-orthoplex (which I will refer to from now on as $\beta_{n}$, and is the $n$-dimensional analogue of the octahedron) as the dipyramid with base $\beta_{n-1}$. This description leads to the following equation (Where $N_{k}^{n}$ denotes the number of $k$-simplexes which are elements of $\beta_{n}$)

$$N_{k}^{n}=N_{k}^{n-1} + 2N_{k-1}^{n-1}$$

We have further that $N_{0}^{n} = 2n = 2^{1}\binom{n}{1} \quad \forall n \in \mathbb{N}$.

The author then states the result in the title follows easily by induction, however I am not sure exactly how. Could anyone possibly help?

Answer 1

There is a pretty easy direct argument. The $n$ orthoplex has $n$ pairs of vertices $v_i^{\pm}$ for $i=1,\ldots n,$ and a subset of the vertices represents a face iff it does not contain both a vertex $v^+_i$ and its opposite $v^-_i$. This is easy to see by how the orthoplexes are inductively constructed.

Now a $k$ simplex has $k+1$ vertices, and since it can't contain both vertices in a pair, we must first choose $k+1$ indices which yields $\binom{n}{k+1}$ choices, and then choose on vertex from each pair, which is another $2^{k+1}$ choices.

Answer 2

Start with the given identity, apply the induction hypothesis, factor out $2^{k+1}$, then apply Pascal's rule. \begin{align} N^n_k &= N^{n-1}_k+2N^{n-1}_{k-1} \\&=2^{k+1}\binom{n-1}{k+1}+2\cdot 2^{(k-1)+1}\binom{n-1}{(k-1)+1} \\&=2^{k+1}\left[\binom{n-1}{k+1}+\binom{n-1}k\right] \\&=2^{k+1}\binom{n}{k+1}. \end{align}