Combinatorics with a Geometric Flavor

NEW

Problem: Reconstructed statement: for a $(d-1)$-dimensional triangulation $K$ with $n$ vertices of a Witt space whose middle intersection homology over a characteristic-zero field vanishes, Kalai’s conjecture asserts in particular the upper-bound conclusion

$$f_i(K)\le f_i(\partial C_d(n))$$

for all $i$, where $C_d(n)$ is the cyclic $d$-polytope. The quoted “strong upper bound conjecture” includes this ordinary face-number bound.

Result: The conjecture is false as stated.

Take the standard 6-vertex triangulation $K$ of $\mathbb{RP}^2$ with vertex set $\{1,\dots,6\}$ and facets

$$\begin{aligned} &123,\ 124,\ 135,\ 146,\ 156,\\ &236,\ 245,\ 256,\ 345,\ 346 . \end{aligned}$$

Every edge of the complete graph $K_6$ occurs in exactly two listed triangles, and every vertex link is a 5-cycle; hence this is a closed triangulated surface. Its $f$-vector is

$$(f_0,f_1,f_2)=(6,15,10),$$

so its Euler characteristic is $6-15+10=1$, and therefore the surface is $\mathbb{RP}^2$.

Now $\mathbb{RP}^2$ is a compact PL manifold, hence a Witt space. For a manifold, intersection homology agrees with ordinary homology. Over $\mathbb Q$,

$$IH_1(\mathbb{RP}^2;\mathbb Q)=H_1(\mathbb{RP}^2;\mathbb Q)=0,$$

so it satisfies the stated “vanishing middle intersection homology” hypothesis.

But for $d=3$, the comparison object is the boundary of the cyclic 3-polytope $C_3(6)$, i.e. a simplicial 2-sphere with 6 vertices. Every simplicial 2-sphere with 6 vertices has

$$f_1=3f_0-6=12.$$

Thus

$$f_1(K)=15>12=f_1(\partial C_3(6)).$$

So even the ordinary upper bound theorem fails, hence the asserted strong upper bound conjecture, and consequently the quoted Conjecture 2.1 as a conjunction, is false.

The defect is the missing orientability / coefficient-sensitive homology condition: $\mathbb{RP}^2$ has vanishing rational middle homology but has nonzero $H_1(-;\mathbb F_2)$. Requiring orientability or vanishing middle homology over all relevant coefficient fields would remove this counterexample.

Citation: No citation is needed for the counterexample beyond the classical 6-vertex triangulation of $\mathbb{RP}^2$; the disproof above is self-contained.