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I asked a question to a professor:

Let P be a $d$-dimensional convex polytope with $f$-vector ($f_0, .. , f_{d - 1}$), with $0 ≤ j < k < h ≤ d - 1$. I noticed that for cross polytopes , $f_j$ , $f_k$, and $f_h$ can be expressed in the form $xy$, $yz$ and $xz$, with $x, y, z$ all being positive integers. Why is this so?

Here's what professor answered:

"One can understand why cubes and cross polytopes satisfy your condition because cross polytopes are direct sums of line segments. If + denotes direct sum and $h(P,x)=\sum h_i x^i$ (the h-polynomial of the simplicial polytope, or more generally, simplicial complex, P), then $h(P+Q,x)=h(P,x)h(Q,x)$. Thus for the cross polytope P, $h(P,x)=(1+x)^d$. Then from $\sum f_{i-1}({x-1})^{d-i}$ $=$ $\sum h_i x^{d-i}$ you get for the cross-polytope $\sum f_{i-1} x^{d-i} = (2+x)^d$. This explains your formula for cross-polytopes and therefore for cubes, which are duals of cross polytopes."

I am not sure how the formulas above explain $why$ the $f$ vectors can be expressed in $xy, yz$ and $xz$. Can you give easier and more intuitive explanation on this?

Any help will be greatly appreciated!!!!!

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Cubes and cross polytopes have the same $f$-vectors up to reversal. It is easier to imagine $d$-dimensional cubes. Such a cube has $2^d$ vertices $v_\iota$. For any $v_\iota$ the number of $k$-dimensional faces having $v_\iota$ as a vertex is ${d\choose k}$. Thereby each $k$-face is counted $2^k$ times. It follows that $$f_k=2^{d-k}{d\choose k}\qquad(0\leq k\leq d-1)\ .$$ From this it should not be all to difficult to establish relations of the kind you are hinting at, if true.

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