On Walkup's class ${\cal K}(d)$ and a minimal triangulation of $(S^3 \times \rotatebox{90}{\ltimes} S^1)^{\#3}$

Bhaskar Bagchi, Basudeb Datta

Published 2008-04-14, updated 2011-02-26Version 3

For $d \geq 2$, Walkup's class ${\cal K}(d)$ consists of the $d$-dimensional simplicial complexes all whose vertex-links are stacked $(d-1)$-spheres. Kalai showed that for $d \geq 4$, all connected members of ${\cal K}(d)$ are obtained from stacked $d$-spheres by finitely many elementary handle additions. According to a result of Walkup, the face vector of any triangulated 4-manifold $X$ with Euler characteristic $\chi$ satisfies $f_1 \geq 5f_0 - {15/2} \chi$, with equality only for $X \in {\cal K}(4)$. K\"{u}hnel observed that this implies $f_0(f_0 - 11) \geq -15\chi$, with equality only for 2-neighborly members of ${\cal K}(4)$. K\"{u}hnel also asked if there is a triangulated 4-manifold with $f_0 = 15$, $\chi = -4$ (attaining equality in his lower bound). In this paper, guided by Kalai's theorem, we show that indeed there is such a triangulation. It triangulates the connected sum of three copies of the twisted sphere product $S^3 \times {-2.8mm}_{-} S^1$. Because of K\"{u}hnel's inequality, the given triangulation of this manifold is a vertex-minimal triangulation. By a recent result of Effenberger, the triangulation constructed here is tight. Apart from the neighborly 2-manifolds and the infinite family of $(2d+ 3)$-vertex sphere products $S^{d-1} \times S^1$ (twisted for $d$ odd), only fourteen tight triangulated manifolds were known so far. The present construction yields a new member of this sporadic family. We also present a self-contained proof of Kalai's result.