On the minimum number of facets of a 2-neighborly polytope

Abstract

Let ${\mu }_{\text{2n}}(d,v)$ (respectively, ${\mu }_{\text{2n}}^{\text{s}}(d,v)$) be the minimal number of facets of a (simplicial) 2-neighborly $d$-polytope with $v$ vertices, $v>d\ge 4$. It is known that ${\mu }_{\text{2n}}(4,v)=v(v-3)/2$, ${\mu }_{\text{2n}}(d,d+2)=d+5$, ${\mu }_{\text{2n}}(d,d+3)=d+7$ for $d\ge 5$, and ${\mu }_{\text{2n}}(d,d+4)\in [d+5,d+8]$ for $d\ge 6$. We show that ${\mu }_{\text{2n}}(5,v)=\Omega (v^{4/3})$, ${\mu }_{\text{2n}}(6,v)\ge v$, and the equality ${\mu }_{\text{2n}}(6,v)=v$ holds only for a simplex and for a dual 2-neighborly 6-polytope (if it exists) with $v\ge 27$. By using $g$-theorem, we get ${\mu }_{\text{2n}}^{\text{s}}(d,v)=\Delta (\Delta (d-3)+3d-5)/2+d+1$, where $\Delta =v-d-1$. Also we show that ${\mu }_{\text{2n}}(d,v)\ge d+7$ for $v\ge d+4$.

10 pages, 1 figure, 1 table