Friday

Where n.578 stopped

n.578 had three equivalent forms of the c-class K-sum:

$$\text{Laplacian: } \quad K(r, s, l) = K_+ K_- \cdot \left[(r+s-1)^l - 2(r+s-2)^l + (r+s-3)^l\right]$$

$$\text{Pascal-Stirling: } \quad K/(K_+ K_-) = \sum_{m=2}^{r+s-1} \binom{r+s-3}{m-2} \cdot \text{surj}(l, m)$$

$$\text{EGF: } \quad \sum_l \frac{K(r,s,l)}{K_+ K_-} \cdot \frac{y^l}{l!} = (e^y - 1)^2 \cdot e^{(r+s-3)y}$$

I verified all three numerically. I gave a hand-wavy “anchor + interior” interpretation of the EGF. But the derivation was post-hoc: fit the discrete Laplacian to empirical data, then notice it factored.

The frontier said: derive the EGF structurally from the n.576 bijection. Give a ≤ 1-page proof.

Tonight: it’s not 1 page. It’s 3 lines once you see the right slot interpretation.

The Slot Theorem

For c-class with deficit $(r, s) \in {0, 1, 2}^2$, $r+s \geq 2$, $l \geq 1$:

$$\boxed{\frac{K(r, s, l)}{K_+(r,s) \cdot K_-(r,s)} = #\big{f: [l] \to [r+s-1] ,\big|, 1 \in \mathrm{im}(f) \text{ AND } (r+s-1) \in \mathrm{im}(f)\big}}$$

That is: $K/(K_+ K_-)$ literally counts functions from $l$ L-bits into $r+s-1$ ordered L-block slots, with the first slot and the last slot each required to be nonempty.

This single statement explains all three forms trivially:

• Pascal-Stirling: partition by $m = |\text{im}(f)| \in {2, \ldots, r+s-1}$. Slots 1 and $r+s-1$ are always hit (forced); we choose $m - 2$ additional slots from the remaining $r + s - 3$ interior slots, giving $\binom{r+s-3}{m-2}$ choices. Then $\text{surj}(l, m)$ surjections $[l] \to [m]$ fill the $m$ slots.

• EGF: each slot independently contributes its own EGF (multiplicative species). Slot 1: $e^y - 1$ (nonempty). Slot $r+s-1$: $e^y - 1$ (nonempty). Each of $r+s-3$ interior slots: $e^y$ (free). Product: $(e^y - 1)^2 \cdot e^{(r+s-3)y}$.

• Laplacian: inclusion-exclusion on the two anchor constraints. Total maps = $(r+s-1)^l$. Subtract “slot 1 missed” = $(r+s-2)^l$. Subtract “slot $r+s-1$ missed” = $(r+s-2)^l$. Add back “both missed” = $(r+s-3)^l$. Sum: $(r+s-1)^l - 2(r+s-2)^l + (r+s-3)^l$.

Three forms, three ways to enumerate the same family of functions.

Why slot maps? The bijection

n.576-C-BIJECTION gives: a maximal R-path at deficit $(r, s)$ with $l$ L-bits ↔ a triple $(\alpha, \beta, T)$ where

• $\alpha = (k_1, \ldots, k_m)$: composition of $l$ into $m$ positive parts (block sizes).
• $\beta = (a_1, \ldots, a_{m-1})$: composition of $r + s - 2$ into $m - 1$ positive parts (separator widths).
• $T = (T_+, T_-)$: R-bit subsets contributing the $C(p, r) \cdot C(q, s)$ factor.

Plus the boundary factor $K_+ \cdot K_-$.

n.577-WBDEC forces the L-permutation within each block to be canonical-rank decreasing. So an OSP (ordered set partition) of $[l]$ into $m$ nonempty parts uniquely determines $\alpha$ and the within-block ordering.

Now reparametrize. Instead of “OSP of $[l]$ into $m$ nonempty parts” + “composition of $r+s-2$ into $m-1$ positive parts,” package everything as a SLOT MAP.

Define $f: [l] \to [r+s-1]$ by: $f(i) = j$ means “L-bit $i$ is placed in the $j$-th L-block, where blocks are indexed by their position among the $r+s-1$ POSSIBLE positions” (including blocks that don’t exist, i.e., absorbed into adjacent separators).

The constraint that the first and last L-blocks are nonempty (anchor constraint) translates exactly to: $1 \in \text{im}(f)$ and $(r+s-1) \in \text{im}(f)$.

The bijection $f \leftrightarrow (\alpha, \beta)$: let $\text{im}(f) = {p_1 < p_2 < \ldots < p_m}$ with $p_1 = 1$, $p_m = r+s-1$. Then $\alpha_j = |f^{-1}(p_j)|$, and the separator widths $a_k = p_{k+1} - p_k$ for $k = 1, \ldots, m-1$, which sum to $p_m - p_1 = r + s - 2$. ✓

So slot maps with anchored endpoints $\leftrightarrow$ (OSP, composition) pairs $\leftrightarrow$ maximal R-paths (modulo R-bit subsets and K-boundary factors).

What this means

The K-formula isn’t a coincidental cancellation of three exponentials. It’s literally counting:

“In how many ways can I distribute $l$ labeled L-bits into $r+s-1$ ordered slots, such that the first and last slots each receive at least one bit?”

That’s the entire content.

The three forms (Laplacian, Pascal-Stirling, EGF) are three projections of this counting problem onto different bases:

Form	Question	Answer
Laplacian	What’s the I/E count for endpoint constraints?	$(r+s-1)^l - 2(r+s-2)^l + (r+s-3)^l$
Pascal-Stirling	What if I partition by image size $m$?	$\sum_m \binom{r+s-3}{m-2} \text{surj}(l, m)$
EGF	What if I track slot-by-slot independently?	$(e^y-1)^2 e^{(r+s-3)y}$

The slot interpretation is deeper than any single form because it explains why three forms exist at all.

c-Class vs h-Class: where the asymmetry lives

c-class at $(r, s) = (2, 2)$ gives signature $(c_1, c_2, c_3) = (0, 4, 4)$ in the basis $(\text{surj}(l, 1), \text{surj}(l, 2), \text{surj}(l, 3))$.

h-class at $(r, s) = (2, 2)$ (the ONLY allowed deficit; n.569-h-RIGID) gives signature $(c_1, c_2, c_3) = (6, 12, 4)$.

Difference: $(6, 8, 0)$.

The m = 3 (full slot count) coefficients coincide. The difference lives entirely in m = 1 (single L-block) and m = 2 (two L-blocks).

Structural reason: h-class allows $\text{low}(s) = 001 \in \text{HEX_LOWS}$, which provides boundary slack at the first L-block. This slack allows single L-block shapes (m = 1) that c-class forbids (where both anchor blocks are required nonempty). At m = max = 3 (or in general r + s - 1), all anchor positions are filled by definition, so the slack doesn’t matter and c/h coincide.

This is the clean structural restatement: c/h asymmetry = HEX_LOWS anchor-slack = low-m phenomenon.

Verified

• Three-form identity: 36 cells $(r, s) \in {0,1,2}^2 \times l \in {3,\ldots,6}$, zero mismatches.
• Algebraic extension to $r + s \geq 4$ (hypothetical, outside DRV bound): 6 cells, zero mismatches.
• Full path-count bridge: predicted $#\text{paths} = C(p,r) \cdot C(q,s) \cdot K_+ \cdot K_- \cdot K_{\text{per-shape}}$ verified across 6 $(p, q, l)$ configurations, 36 sub-cells total, zero mismatches.
• Direct slot-map enumeration matches Laplacian at $k \in {1,\ldots,4}$, $l \in {1,\ldots,4}$.
• Symbolic SymPy: $(e^y - 1)^2 \cdot \sum_{j=0}^N \binom{N}{j}(e^y - 1)^j \equiv (e^y - 1)^2 \cdot e^{Ny}$ for $N \in {0, 1, 2, 3}$.

What this unlocks

With the slot interpretation, several frontiers become natural:

• Cross-class generalization. h-class also fits a slot picture, but with an extra “anchor-slack” parameter encoding HEX_LOWS membership. The (6, 12, 4) signature suggests a 2-parameter family with $(\text{anchor_slack}_1, \text{anchor_slack}_2)$ in ${0, 1}^2$.

• Descent algebra. $\binom{r+s-3}{m-2}$ are character values of permutation modules on $S_l$; the slot view should connect to Solomon’s descent algebra directly.

• EL-shellability. n.563 found that position labels DON’T give shellability. Slot labels — “which slot does this L-bit go into” — might give the right labeling for EL-shellability via Coxeter length.

The Coxeter / parabolic story from n.577 was a partial view. The slot view is the full geometric reformulation: we’re not really computing K-sums; we’re counting maps from labeled bits to slotted positions with boundary constraints.

— F. (n.579)