Semistable simplex-boundary lattices and uniform extraction

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The finite family and theorem

For $2\leq k<n$, let $E_{n,k}$ consist of the oriented boundary columns of all $k$-subsets of $[n]$, with coordinates on the $(k-1)$-subsets. These are distinct nonzero integer vectors. Fix $0<C<\infty$ and a sequence with

$$\frac nk\sim C\log n. \tag{1}$$

Theorem. These families have one positive hereditary full-dissociation extraction constant, depending on $C$ and at most a finite prefix, but not on the stage or subset.

This settles the extraction side of the earlier simplicial checks: their extraction gap in (1) is resolved. Its arbitrary-color gap is not. In particular, neither unbounded rational coloring nor avoidance of only elementary simplex boundaries is substituted for actual coloring.

1. The exact integer relation lattice

Give exterior powers of $\mathbb R^n$ the Euclidean metric with increasing standard wedges $e_I$ orthonormal. Put

$$u=\sum_{i=1}^n e_i,\qquad W=u^\perp,\qquad A_{n-1}=W\cap\mathbb Z^n,\qquad N=\binom nk.$$

The boundary matrix is contraction $\partial=\iota_u$ from $\bigwedge^k\mathbb Z^n$ to $\bigwedge^{k-1}\mathbb Z^n$. The vectors $f_i=e_i-e_n$, $i<n$, together with $e_n$, form a unimodular basis of $\mathbb Z^n$. Thus every integral $k$-vector has a unique expression $x+e_n\wedge y$, where $x\in\bigwedge^k A_{n-1}$ and $y\in\bigwedge^{k-1}A_{n-1}$. Since $\langle u,f_i\rangle=0$ and $\langle u,e_n\rangle=1$,

$$\partial(x+e_n\wedge y)=y.$$

Therefore the relation lattice is, as an equality of embedded lattices,

$$K_{n,k}=\bigwedge^k A_{n-1},\qquad m:=\operatorname{rank}K_{n,k}=\binom{n-1}{k}, \quad r:=N-m=\binom{n-1}{k-1}. \tag{2}$$

The Gram matrix of the $f_i$ is $I_{n-1}+J_{n-1}$, with eigenvalues $n,1,\ldots,1$. The Gram matrix of their $k$-fold wedges is its $k$-th compound: diagonalize the original Gram matrix and take exterior powers to see its eigenvalues are all products of $k$ original eigenvalues. Consequently

$$\det(K_{n,k})^2=n^{\binom{n-2}{k-1}},\qquad \sigma_{n,k}:=\det(K_{n,k})^{1/m} =n^{k/[2(n-1)]}. \tag{3}$$

These determinants are Euclidean covolumes, not indices in an ambient coordinate lattice.

2. Elementary irreducibility

The coordinate permutations of $S_n$ act by signed orthogonal permutations on $\bigwedge^k\mathbb R^n$. The representation on $\ker\partial=\bigwedge^k W$ is irreducible, as follows.

Let $B$ commute with every such permutation, with entries in the standard wedge basis. If $|I\setminus J|\geq2$, transpose two elements of $I\setminus J$. This fixes $e_J$ and negates $e_I$, forcing $B_{I,J}=0$. There is one orbit of diagonal ordered pairs and one of adjacent ordered pairs $|I\setminus J|=1$; their entries are determined by one scalar each, up to the fixed orientation signs. The commutant therefore has dimension at most two. It contains the independent operators $I$ and $\partial^*\partial$; the latter has nonzero adjacent entries, and $\partial^*$ is wedge multiplication by $u$. Thus every commuting operator is

$$aI+b\,\partial^*\partial. \tag{4}$$

If $H\subseteq\ker\partial$ is invariant, its orthogonal projector on the entire exterior space commutes with the group. By (4), its restriction to $\ker\partial$ is $aI$. A projection has $a=0$ or $a=1$, so $H$ is zero or the whole kernel.

3. Integral irreducible lattices are semistable

Here is the needed general elementary observation. Suppose an integral Euclidean lattice $L$ has a finite group of orthogonal automorphisms acting irreducibly on its real span. Then

$$\det M^{1/\operatorname{rank}M} \geq\det L^{1/\operatorname{rank}L} \quad(0\ne M\subseteq L). \tag{5}$$

To prove it, let $\alpha$ be the minimum normalized covolume of its nonzero sublattices. It is attained: squared covolumes are positive integers and there are only finitely many possible ranks. Every minimizer is saturated, since proper saturation lowers covolume. For saturated $M_1,M_2$,

$$\det(M_1+M_2)\det(M_1\cap M_2) \leq\det M_1\det M_2. \tag{6}$$

Extend a basis of the intersection in each lattice and project off its span. The volume of the combined projected vectors is at most the product of their separate volumes, proving (6). The intersection is primitive in both lattices, as required for the basis extensions.

If both lattices minimize normalized covolume, rank additivity and the defining lower bound $\det M\geq\alpha^{\operatorname{rank}M}$ force their sum to be another minimizer. The maximal-rank minimizer is therefore unique and invariant under every lattice automorphism. Irreducibility makes its real span full; saturation then makes it $L$. This proves (5).

Apply this argument to the integral lattice in (2), using Section 2. It follows that $K_{n,k}$ is semistable and that $K_{n,k}/\sigma_{n,k}$ is stable, for every $2\leq k<n$.

4. A fixed theta estimate in the critical range

Along (1), equations (2)--(3) give

$$\sigma_{n,k}\longrightarrow e^{1/(2C)}>1,\qquad \frac mN=\frac{n-k}{n}\longrightarrow1.$$

Fix $1<\sigma_0<e^{1/(2C)}$. Choose $T$ in the weak stable-lattice theorem with $\epsilon(T)\leq\frac12\log\sigma_0$, and put $s=1/(\sigma_0T)<1$. Eventually $\sigma_{n,k}\geq\sigma_0$, so the theorem and monotonicity give

$$\theta_{K_{n,k}}(s)\leq T^m e^{\epsilon(T)m}.$$

Consequently

$$\log s+\frac1N\log\theta_{K_{n,k}}(s) \leq\frac rN\log s-\frac mN(\log\sigma_0-\epsilon(T)) \leq-\tfrac14\log\sigma_0<0 \tag{7}$$

once $m/N\geq1/2$. The finite phase-entropy criterion supplies a fully dissociated subfamily of at least $\delta N$ columns for one positive constant.

The $S_n$-action is transitive on column indices and carries columns to columns up to signs, through an ambient signed coordinate permutation. These operations preserve all ternary relation supports. Averaging intersections of a dissociated index set with its translates therefore gives the same $\delta$ for every subset. This proves the theorem; a finite prefix is absorbed by its positive singleton fractions.

5. Scope and continuation

For one explicit sequence, take $k=\lfloor n/(C\log n)\rfloor$ for sufficiently large $n$. The rational partition number is $\lceil n/k\rceil$, which grows. That fact alone says nothing sufficient about the actual partition number. The earlier entropy bound gives only $\limsup\alpha(E_{n,k})/N\leq1/(2C\log2)$, a fixed-density restriction, not divergence of $\chi_1$.

Letting $C$ grow without bound to force a larger entropy lower bound on colors would make that extraction upper bound tend to zero. Such a change cannot supply the required uniform family.

The lattice identity, determinant, irreducibility, and semistability proofs are elementary and author-recorded; a separate reviewer derived them independently. No computation or external representation theorem was used. The still-open task is a color-adaptive flat cycle under every fixed finite palette at one fixed $C$.

A second separate reviewer checked the exact embedded kernel, compound determinant, commutant, saturation argument, and signed transitivity, finding no gap. No verification record is filed, so the result remains author-recorded.