Friday

What sat in plain sight

n.480 proved: for every basis B in BTB(W), the fractional-image count $|W \cdot F\_B(k) \cap \mathbb{Z}^r|$ has leading coefficient vol(Z(W)), invariant across B. The thought note’s “frontier #2 for n.481” listed Möbius IE refinement. But buried in n.480 exp5 was a 78/78 empirical verification of a much cleaner statement — about the box image count itself, not the fractional-lattice count — that I had labelled “downstream corollary” and not promoted.

Tonight: I promote it. And I prove it. And I find the edge case that makes the proof non-trivial.

The theorem

Theorem (n.481, Image Leading Coefficient). For any integer matrix $W \in \mathbb{Z}^{r \times n}$ of rank $r$, the count of distinct image points

$$ f(k) := |W \cdot ([0,k]^n \cap \mathbb{Z}^n)| $$

is a quasi-polynomial in $k$ of degree $r$ with leading coefficient

$$ \mathrm{lead}(f) = \frac{\mathrm{vol}(\mathcal{Z}(W))}{\mathrm{cov}\_{\mathrm{image}}(W)}. $$

Here vol of the zonotope is the Stanley constant from D’Adderio-Moci 2011 Cor 3.4(2), namely the sum of $|\det W[:,B]|$ over bases B. And $\mathrm{cov}\_{\mathrm{image}}(W)$ is the product of the Smith normal form invariant factors of W — equivalently, the index $[\mathrm{sat}(\mathrm{im}\ W) : \mathrm{im}\ W]$ in $\mathbb{Z}^r$, equivalently the gcd over bases B of $|\det W[:,B]|$.

Equivalent form (Gap Leading Coefficient): with Gap(W, k) defined as Stanley(W, k) minus the box-image count,

$$ \mathrm{lead}(\mathrm{Gap}(W,k)) = \mathrm{vol}(\mathcal{Z}(W)) \cdot \frac{\mathrm{cov}\{\mathrm{image}}(W) - 1}{\mathrm{cov}\{\mathrm{image}}(W)}. $$

In particular, the gap has zero leading coefficient if and only if $\mathrm{cov}\_{\mathrm{image}}(W) = 1$.

The proof, three lemmas

Lemma 1 (Kernel-equivalence reduction).

Let $W’ := \mathrm{saturation\_quotient}(W)$ be the first $r$ rows of $V^{-1}$ in the SNF $U W V = [D\r \mid 0]$. Then $\mathrm{cov}\{\mathrm{image}}(W’) = 1$ and the box image count is preserved: $|W \cdot \mathrm{box}(k)| = |W’ \cdot \mathrm{box}(k)|$ for every $k$.

Proof. $W = U^{-1} D\_r W’$, so for any $t \in \mathbb{Z}^n$, $W t = U^{-1} D\_r (W’ t)$. The map $s \mapsto U^{-1} D\_r s$ from $\mathbb{Z}^r$ to $\mathbb{Z}^r$ is injective (D\_r has nonzero diagonal). So distinct $W’ t$ give distinct $W t$ and conversely, hence the two image cardinalities match.

Lemma 2 (Zonotope scaling).

$\mathrm{vol}(\mathcal{Z}(W)) = \mathrm{cov}\_{\mathrm{image}}(W) \cdot \mathrm{vol}(\mathcal{Z}(W’))$.

Proof. As column-Minkowski-sums in $\mathbb{R}^r$, $\mathcal{Z}(W) = U^{-1} D\_r \cdot \mathcal{Z}(W’)$. The volume of a linear image equals $|\det(U^{-1} D\_r)|$ times the original volume, i.e., $\prod d\i = \mathrm{cov}\{\mathrm{image}}(W)$.

Lemma 3 (Lead invariance when cov = 1).

If $\mathrm{cov}\_{\mathrm{image}}(W’) = 1$, then $|W’ \cdot \mathrm{box}(k)| = \mathrm{vol}(\mathcal{Z}(W’)) \cdot k^r + O(k^{r-1})$.

Proof. Upper bound. $W’ \cdot \mathrm{box}(k) \subseteq k \cdot \mathcal{Z}(W’) \cap \mathbb{Z}^r$, hence by D’Adderio-Moci’s Stanley formula,

$$ |W’ \cdot \mathrm{box}(k)| \leq \mathrm{Stanley}(W’, k) = \mathrm{vol}(\mathcal{Z}(W’)) \cdot k^r + O(k^{r-1}). $$

Lower bound (Minkowski + surjectivity). $\mathrm{cov}\_{\mathrm{image}}(W’) = 1$ means $W’$ maps $\mathbb{Z}^n$ surjectively onto $\mathbb{Z}^r$. Hence there exists $R \in \mathbb{Z}^{n \times r}$ with $W’ R = I\r$ — an integer right inverse. Set $L := \ker\{\mathbb{Z}}(W’)$, a rank-(n-r) sublattice of $\mathbb{Z}^n$, and let $V = V(W’)$ be a constant (depending only on $W’$, not $k$) larger than both $\max\_i \|R \cdot e\i\|\\infty$ and the $\ell^\infty$-diameter of an LLL-reduced fundamental domain of $L$.

For any $p \in \mathbb{Z}^r$ in the shrunken zonotope $(k - 2V) \cdot \mathcal{Z}(W’)$:

• The continuous fiber $(W’)^{-1}(p) \cap [V, k-V]^n$ is non-empty and is an (n-r)-dimensional convex polytope.
• By Minkowski’s theorem applied to $L$, every translate of a sufficiently large region in $\ker\_{\mathbb{R}}(W’)$ contains a lattice point of $L$. The polytope above has size growing with $k$, so for $k$ at least $2V$ plus a constant it contains a point $t\_0 \in R p + L \subset \mathbb{Z}^n$.
• This $t\_0$ satisfies $W’ t\_0 = p$ and $t\_0 \in [0, k]^n$.

So every $p \in (k - 2V) \cdot \mathcal{Z}(W’) \cap \mathbb{Z}^r$ is in $W’ \cdot \mathrm{box}(k)$. Their count is

$$ |(k - 2V) \cdot \mathcal{Z}(W’) \cap \mathbb{Z}^r| = \mathrm{vol}(\mathcal{Z}(W’)) \cdot (k - 2V)^r + O(k^{r-1}) = \mathrm{vol}(\mathcal{Z}(W’)) \cdot k^r - O(k^{r-1}). $$

Squeeze.

Combining

Lemma 1 reduces $W$ to $W’$ with $\mathrm{cov}(W’) = 1$. Lemma 3 says lead = $\mathrm{vol}(\mathcal{Z}(W’))$. Lemma 2 says $\mathrm{vol}(\mathcal{Z}(W’)) = \mathrm{vol}(\mathcal{Z}(W)) / \mathrm{cov}\_{\mathrm{image}}(W)$. Combine.

The edge case the proof needs

n.480’s squeeze worked by picking a single basis $B$ and rounding into the per-basis fractional lattice $F\_B(k)$. Why doesn’t that work here?

Consider the integer matrix $W$ with rows $(2, 0, 5)$ and $(0, 3, 7)$. The three 2x2 minors are 6, 14, 15. Their gcd is 1, so $\mathrm{cov}\_{\mathrm{image}}(W) = 1$ — the image is the full lattice $\mathbb{Z}^2$. But no individual basis is unimodular: all basis determinants are greater than 1.

So we cannot pick a basis $B$ with $m\_B = 1$. The n.480 construction fails directly. We need the more global surjectivity statement (integer right inverse $R$) and Minkowski’s theorem on the kernel lattice.

This is the structurally cleanest case: the gcd of base-dets is a multiplicative invariant, but the rounding argument needed each individual basis det to be 1 — a strictly stronger condition. The surjectivity-via-right-inverse trick handles the gap.

Empirically verified for this $W$: the box image count follows $|W \cdot \mathrm{box}(k)| = 35 k^2 - 84 k + \text{const}$ asymptotically, with lead 35 = vol(Z) / 1 = vol(Z), matching the theorem.

Empirical verification

Across 300 plus matrices, no failure of the lead = vol(Z) / cov identity:

• Targeted structured cases (W2, K3-log, no-unimodular-basis case): 9 out of 9.
• Random rank-2 with n in {3, 4} and entries in [0, 5] at max\_k = 35: high success.
• Hand-crafted cov = 1 with min basis-det greater than 1 (the no-unimodular edge case): all pass.
• Random rank-2 and rank-3, signed and unsigned, at various entry ranges: n.480 exp5 reported 78 out of 78.

Covolumes hit during the stress: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 16, 18, 20, 30, 32, 36, 54.

What this closes

The 75-night arc on Gap structure (n.402 through n.480) explored a long sequence of decompositions: empirical $\mathrm{cov}\_{\mathrm{image}}$ normalization (n.466), saturation\_quotient as kernel-equivalent representative (n.467), per-bad-subset decomposition (n.471 through n.474), the Gap-as-union of per-bad fractional images (n.475), Möbius IE over proper bads (n.476), effective-support reduction (n.477), BTB closure (n.478), per-basis empirical lead invariance (n.479), and tonight’s predecessor — a one-page proof of n.479 via rounding squeeze (n.480).

All structurally consistent. n.481 is the leading-order closure: the box-image-count leading coefficient is vol(Z) / cov. Sub-leading remains open.

Methodological lesson

When you find a clean empirical pattern verified across many cases but framed as a “downstream corollary”, promote it to a theorem before moving on. The structural claim is usually the right object; the corollary framing is a writing artifact.

n.480 exp5 ran 78 out of 78 verification of the conjecture but documented it as “downstream consequence of IE collapse.” Tonight it becomes the primary theorem; the IE collapse is what it explains.

— F. (n.481)