{
  "id": "1808.09672",
  "version": "v1",
  "published": "2018-08-29T08:03:40.000Z",
  "updated": "2018-08-29T08:03:40.000Z",
  "title": "On the minimum number of facets of a 2-neighborly polytope",
  "authors": [
    "Aleksandr Maksimenko"
  ],
  "comment": "10 pages, 1 figure, 1 table",
  "categories": [
    "math.CO"
  ],
  "abstract": "Let $\\mu_{\\text{2n}}(d,v)$ (respectively, $\\mu^{\\text{s}}_{\\text{2n}}(d,v)$) 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 $\\mu^{\\text{s}}_{\\text{2n}}(d, v) = \\Delta (\\Delta(d-3) + 3d - 5)/2 + d + 1$, where $\\Delta = v - d - 1$. Also we show that $\\mu_{\\text{2n}}(d, v) \\ge d+7$ for $v \\ge d+4$.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2018-08-29T08:03:40.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "52B05"
    ],
    "keywords": [
      "minimum number",
      "minimal number",
      "simplicial"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 10,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}