Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 12 / 4

The uniform modular Frankl-Wilson theorem states the following. Let $p$ be prime, let $L\subseteq {\mathbb{F}}_p$ have size $s$, and let $\mathcal{F}$ be a rank-$k$ family on $[N]$, with $0\le s\le \min \{k,N-k\}$. Assume $k\mathrm{mod}p\notin L$ and $|A\cap B|\mathrm{mod}p\in L$ for every distinct $A,B\in \mathcal{F}$. ThenThe restrictions concern distinct pairs; the diagonal size residue is explicitly excluded from $L$.

For each member $A$, over ${\mathbb{F}}_p$ form the intersection polynomialAt the characteristic vector $1_B$, its value vanishes when $B\ne A$ and is nonzero when $B=A$. Thus these functions are linearly independent on the whole rank-$k$ layer: evaluate any linear relation at each family member in turn. Replacing every positive power of $z_i$ by $z_i$ preserves all Boolean evaluations, so the functions are restrictions of multilinear polynomials of degree at most $s$.

We need the sharp low-degree evaluation rank on a uniform layer, not the larger whole-cube dimension ${\sum }_{j\le s}\left(\genfrac{}{}{0ex}{}{N}{j}\right)$. Let $M$ be the integer incidence matrix with rows indexed by subsets $S$ of size at most $s$, columns by rank-$k$ sets $B$, and entries $1_{S\subseteq B}$. Over the rationals, if $|S|=j\le s$, thenThis coefficient counts the possible $T$ inside each containing column set and is positive because $s\le k$. Consequently the degree-$s$ rows span all rows over $\mathbb{Q}$, giving rank at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. Every larger square minor is therefore the integer zero. Reducing those minors modulo $p$ shows the rank over ${\mathbb{F}}_p$ is also at most $\left(\genfrac{}{}{0ex}{}{N}{s}\right)$. This transfer avoids dividing by a binomial coefficient that might vanish modulo $p$. The independent functions $f_A$ lie in this row space, so their number is at most its rank. This proves the theorem in full.

For the requested forbidden-intersection application, split $\mathcal{A}$ into sets containing coordinate one and sets avoiding it. In the first part, delete one. Distinct original sets have intersection size between one and $2p-1$, excluding $p$. Their reduced intersections therefore lie between zero and $2p-2$, excluding $p-1$, so their residues belong to $L=\{0,\dots ,p-2\}$. The reduced set size is $2p-1$, whose residue is $p-1\notin L$. Frankl-Wilson on the ground set of size $4p-1$ bounds this part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$.

For the avoiding part, first complement every set. The complements contain one, retain size $2p$, and satisfyDelete one and apply the same argument. The complement splitting for forbidden midpoint intersections yields the stronger boundThe split is essential: a family may contain complementary pairs with intersection zero, preventing direct use of $L=\{1,\dots ,p-1\}$ on the original layer.

The Borsuk conjecture asserted that a bounded subset of ${\mathbb{R}}^d$ of positive diameter can be partitioned into $d+1$ subsets of strictly smaller diameter. Here is the Kahn-Kalai counterexample to the Borsuk conjecture, implemented by balanced cut vectors. For each $X\in [4p{]}^{(2p)}$, define $v_X\in {\mathbb{R}}^{\left(\genfrac{}{}{0ex}{}{4p}{2}\right)}$ whose coordinate indexed by an unordered pair $\{i,j\}$ is one when exactly one endpoint lies in $X$, and zero otherwise. Complementary sets give the same cut. Choose the representative with $1\in X$, obtaining exactlydistinct points: a cut determines its two parts, and fixing the side containing one removes its only ambiguity. Every cut has $(2p{)}^2=4p^2$ occupied coordinates, so all points lie in the affine hyperplane with coordinate sum $4p^2$. Its dimension isWe can identify this hyperplane isometrically with ${\mathbb{R}}^d$; no claim of full affine span is needed.

For two side sets with $t=|X\cap Y|$, their symmetric difference has $4p-2t$ points. The two cuts disagree exactly on pairs with one endpoint in that symmetric difference, giving the balanced cut-vector distance formulaThe diameter is $2p$, attained exactly when $t=p$; such pairs exist even among representatives containing one. Thus a smaller-diameter part contains no pair with side intersection $p$. Since all representative sets already contain one, deleting that common coordinate and applying the preceding Frankl-Wilson argument bounds each part by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$. There is no extra factor two at this step. This is the cut construction described in the original Kahn-Kalai paper; the distance and part-size calculations above give the required justification directly.

An explicit choice is $p=13$. The point set lies in dimensionThe exact counts areTheir ratio requires at least $\lceil M/B\rceil =1562$ smaller-diameter parts, exceeding $d+1=1326$. In particular,Rescaling the point set to diameter one changes none of these partition counts. Therefore the Borsuk counterexample in dimension 1325 is fully explicit. This is a concrete dimension, not a claim that it is the smallest possible counterexample dimension.