# Can a convex polytope with $f$ facets have more than $f$ facets when projected into $\mathbb{R}^2$?

Let $P$ be a convex polytope in $\mathbb{R}^d$ with $n$ vertices and $f$ facets. Let $\text{Proj}(P)$ denote the projection of $P$ into $\mathbb{R}^2$.

Can $\text{Proj}(P)$ have more than $f$ facets?

In the general case, each successive projection can increase the number of facets from $f$ to $\left\lfloor \frac{f^2}{2} \right\rfloor$, but I'm wondering if $\mathbb{R}^2$ is a special case.

Consider the polytope in $\mathbb{R}^3$ with $8$ vertices at coordinates $(\pm 1, \pm 2, 1), (\pm 2, \pm 1, -1)$. Geometrically this looks like a cube where the top face is stretched in the direction of the $y$-axis and the bottom face is stretched in the direction of the $x$-axis, but it still has the face structure of a cube and in particular has $6$ facets.
• Thanks for the counterexample. Does requiring $P$ to be facet-transitive and vertex-transitive have an effect? – Pedro Ruiz Jan 20 '17 at 19:05
Your question is essentially about extension complexity. In general, the extension complexity of a polytope $P$ is the minimum number of facets over all polytopes $Q$ which project to $P$. You are interested in the extension complexity of polygons. Fiorini, Rothvoß, and Tiwary proved that regular $n$-gons have extension complexity $O(\log n)$. For lower bounds, they give examples of $n$-gons which have extension complexity $\sqrt{2n}$. It is an open question if there exists for infinitely many $n$, an $n$-gon with extension complexity $\Omega(n)$.
Another, combinatorially minimal, counterexample of such a polytope $P$ (with only five facets) is the convex hull of the six vertices $(\pm2, 0, 0)$, $(\pm1, \pm1, 1)$. Its projection to the $xy$-plane is a hexagon. The minimality of the number of vertices and the number of facets is easy to prove.