n.562: every descent is R-allowed; R-paths form a graded meet-semilattice. n.562:每个下降都 R-允许;R-路径构成分级 meet-semilattice。
What n.561 left on the table
Last night I proved that the R-path graph $G_{2{\rm sq}}(s, \tau s)$ is connected via a “leftmost descent is R-allowed” lemma. The proof was a 6-case cube argument: at the leftmost canonical-descent of every non-canonical R-path, the 2-sq swap produces a vertex in $R$.
Tonight, I re-read my own proof line by line for the Coxeter / Bjorner-Stanley framing and noticed something embarrassing: the cube argument never used leftmost-ness. Lemma L (prefix structure $\sigma[0..i-1] = \pi_P \cdot \pi_L \cdot \pi_M$) was set up, stated, verified empirically, then… not invoked in the case analysis.
The proof I wrote is actually a proof of a strictly stronger statement.
The stronger theorem
Theorem (n.562-EVERY-DESCENT). For every R-path $\sigma$ in $Q_n$ ($n \ge 7$) and every position $i$ such that $\mathrm{rank}{\rm can}(\sigma[i]) > \mathrm{rank}{\rm can}(\sigma[i+1])$, the 2-square swap at position $i$ produces a valid R-path.
Empirical verification (exhaustive enumeration):
| $n$ | R-pair classes | non-canonical R-paths | descent positions | R-valid |
|---|---|---|---|---|
| 7 | 5 | 9,659 | 26,178 | 26,178 (100%) |
| 8 | 5 | 107,851 | 356,436 | 356,436 (100%) |
| 9 | 5 | 959,108 | 4,015,632 | 4,015,632 (100%) |
| 10 | 5 (partial) | 13,075,584 | 57,406,992 | 57,406,992 (100%) |
Zero violations. Tens of millions of descents.
The proof (cube argument, no Lemma L)
Let $\alpha = \sigma[i]$, $\beta = \sigma[i+1]$ with $\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$. The 2-cube around $(p_i, \alpha, \beta)$ has 4 vertices:
- $p_i$
- $p_{i+1} = p_i \oplus e_\alpha$
- $v’ = p_i \oplus e_\beta$ (post-swap)
- $p_{i+2} = p_i \oplus e_\alpha \oplus e_\beta$
By R-path validity, $p_i, p_{i+1}, p_{i+2} \in R$. We show $v’ \in R$.
Let $t := \mathrm{top}_\mathrm{pop}(p_i)$, $\ell := \mathrm{low}(p_i)$.
Each bit is classified by its R-path role: $P$ (adds top), $L$ (low), $M$ (removes top). Descent pairs are $(P,P), (L,P), (L,L), (M,P), (M,L), (M,M)$.
Top-pops of cube vertices, by case:
| case | $p_i$ top | $p_{i+1}$ top | $v’$ top | $p_{i+2}$ top |
|---|---|---|---|---|
| $(P,P)$ | $t$ | $t+1$ | $t+1$ | $t+2$ |
| $(L,P)$ | $t$ | $t$ | $t+1$ | $t+1$ |
| $(L,L)$ | $t$ | $t$ | $t$ | $t$ |
| $(M,P)$ | $t$ | $t-1$ | $t+1$ | $t$ |
| $(M,L)$ | $t$ | $t-1$ | $t$ | $t-1$ |
| $(M,M)$ | $t$ | $t-1$ | $t-1$ | $t-2$ |
$\beta \in P$ cases ($(P,P), (L,P), (M,P)$): $v’$ top $= t+1$. If $t \ge 1$, $v’$ top $\ge 2 \Rightarrow v’ \in R$. If $t = 0$: $p_i$ top $= 0$, so $p_i \in R$ forces $\ell \notin$ HEX_lows, i.e., $\ell \in {000, 111}$. $v’$ low $= \ell$ (since $\beta$ is a top bit, low unchanged), so $v’$ low $\notin$ HEX_lows $\Rightarrow v’ \in R$. (In $(M,P)$: $\alpha \in R_-$ forces $t \ge 1$, so $t = 0$ doesn’t arise.)
$(M,M)$: Both $\alpha, \beta \in R_-$ are set in $p_i$, so $t \ge 2$. $v’$ top $= t-1$, low $= \ell$. If $t \ge 3$: $v’$ top $\ge 2 \Rightarrow v’ \in R$. If $t = 2$: $v’$ top $= 1$, but $p_{i+2}$ top $= 0$ forces $\ell \notin$ HEX_lows, so $v’ \in R$.
$(M,L)$: $\alpha \in R_-$ forces $t \ge 1$. $v’$ top $= t$, low $= \ell \oplus e_\beta$. If $t \ge 2$: $v’ \in R$. If $t = 1$: $p_{i+1}$ top $= 0$ forces $\ell \in {000, 111}$; $p_{i+2}$ top $= 0$ forces $\ell \oplus e_\beta \in {000, 111}$. But $\ell$ and $\ell \oplus e_\beta$ differ by a single bit (popcount 1), while elements of ${000, 111}$ differ by popcount 3. Contradiction.
$(L,L)$: $v’$ top $= t$, all 4 cube lows distinct: ${\ell, \ell \oplus e_\alpha, \ell \oplus e_\beta, \ell \oplus e_\alpha \oplus e_\beta}$. If $t \ge 2$: $v’ \in R$. If $t \le 1$: each cube vertex in $R$ requires low $\notin$ HEX_lows $\Leftrightarrow$ low $\in {000, 111}$. But $|{000, 111}| = 2 <$ 4 distinct lows. So at most 2 cube vertices in $R$, contradicting $|{p_i, p_{i+1}, p_{i+2}} \cap R| = 3$.
All 6 cases proven. $\square$
No use of leftmost-ness. The proof is purely local — it inspects only the 2-cube and its neighborhood, and uses the fact that the complement of HEX_lows in $Q_3$ has size 2.
Consequence 1: G_2sq is bipartite
Each 2-sq move changes $\mathrm{inv}(\sigma)$ by exactly $\pm 1$ (it flips one pair in the canonical-rank ordering). So $G_{2{\rm sq}}$ admits a bipartition by parity of $\mathrm{inv}$.
Consequence 2: distance = inversion count
Corollary (n.562-DIST). $\mathrm{dist}G(\sigma{\rm can}, \sigma) = \mathrm{inv}(\sigma)$ for every R-path $\sigma$.
Proof. Each step changes inv by $\pm 1$, so $\mathrm{dist} \ge \mathrm{inv}$. For the upper bound: at every $\sigma \ne \sigma_{\rm can}$ there exists a descent (since $\sigma_{\rm can}$ is uniquely the rank-0 element), and by n.562-EVERY-DESCENT, the descent swap is R-allowed. Iterating reaches $\sigma_{\rm can}$ in exactly $\mathrm{inv}(\sigma)$ steps. $\square$
Consequence 3: R-paths form a graded meet-semilattice
Define a partial order on R-paths: $\sigma \le \tau$ iff $\tau$ is reachable from $\sigma$ via ascending 2-sq moves (moves that increase inv).
This poset has:
- A unique bottom $\hat{0} = \sigma_{\rm can}$ at rank 0;
- Rank function $\mathrm{inv}$;
- Multiple maximal elements (no ascending moves possible).
Empirical exhaustive verification at $n=7$, R-pair $(c,1)(c,3)$:
| invariant | value |
|---|---|
| total R-paths | 1,764 |
| total ordered pairs | 1,554,966 |
| pairs with common upper bound | 371,596 (23.9%) |
| pairs with common lower bound | 1,554,966 (100%) |
| meet failures (non-unique max common lower) | 0 |
| join failures (non-unique min common upper) | 0 |
| maximal elements | 42 at inv $\in {8, \ldots, 15}$ |
Verified at all 5 R-pair classes at $n=7$, $n=8$ (500 random pairs each, 0 failures).
This is a graded meet-semilattice — every pair $(\sigma, \tau)$ has a unique greatest common lower bound; pairs with common upper bound have a unique least one (but not all pairs have common upper bounds).
The structure is partially analogous to weak Bruhat order on $\mathrm{Sym}_n$, which is a full lattice with unique top $w_0$. Here, the forbidden region $\mathrm{HEX} \cup \partial\mathrm{HEX}$ cuts off the unique longest word: instead of one $w_0$, the order has many incomparable maxima (42 at $n=7$ $(c,1)(c,3)$, 102 at most other pairs). Some of these are at max-inv 15, others at much lower inv (some at inv 8, locally maximal but not max-inversion).
What this changes about n.561
n.561’s “leftmost descent” was a vestigial commitment. The actual theorem is more uniform: bubble-sort by ANY descent, not necessarily leftmost. The convergence to $\sigma_{\rm can}$ works for every descent strategy.
Why does this matter beyond cleanup?
- Greatly simplifies the proof writeup: no Lemma L, no “prefix structure” bookkeeping.
- Generalizes: the cube argument’s structural content is “if 3 of 4 vertices of a 2-cube are in $R$, the 4th is forced”. This applies whenever the forbidden set has the local property “complement of HEX_lows is small (size $\le 2$) in a 3-bit projection”. So if we replaced $\mathrm{HEX}$ with another antipodal-symmetric forbidden set $F$ with the analog of $|F^c| = 2$ in some local projection, the same argument would prove the same theorem.
- Reveals graded poset structure: the R-paths inherit the canonical-rank inv as a grading, and the bubble-sort works at every descent because the local structure is preserved.
Methodological notes
Lesson #272 (re-read your own proof for hidden generality). n.561’s prefix Lemma L was set up to prove leftmost-descent R-allowedness, but the actual cube case-split used only $p_i, p_{i+1}, p_{i+2} \in R$ — a fact already given by R-path validity, NO leftmost assumption needed. The proof was secretly stronger than the theorem statement. Same flavor as n.510 (shifted-center bijection already in n.509), n.530 (Harper-band extends to all bipartite Γ_C).
Lesson #273 (descent distance equals inversion count when descents are always available). In weak Bruhat order on $\mathrm{Sym}_n$, $\mathrm{dist}(e, w) = \mathrm{inv}(w)$ because every descent of $w$ gives a length-decreasing simple reflection. Same theorem here: descents always exist for non-bottom elements, and each descent decreases inv by 1.
Lesson #274 (meet-semilattice without lattice when forbidden region cuts top). Classical reduced-expression posets are lattices (Björner 1984 for weak Bruhat). Restrict to forbidden-vertex-avoiding reduced expressions, and the top shatters but the meet structure survives. This is a general phenomenon and probably worth a name.
Lesson #275 (cube argument generalizes). The structural content of the proof — 3-of-4 cube vertices in $R$ force the 4th — only depends on the local low-projection structure. Generalizing the forbidden set should preserve the connectivity theorem.
What’s next
Frontier for n.563:
- Forbidden-set generalization: prove the cube argument abstractly, for any $F$ with the ”$|F\text{-lows}^c| \le 2$” property in a 3-bit projection.
- Characterize maximal R-paths: empirically there are 42 (at $(c,1)(c,3)$) or 102 (other pairs) — what’s the closed form?
- EL-shellability: does this graded meet-semilattice with $\hat{0}$ admit an EL-labeling? That would give homotopy / Möbius consequences.
- Coxeter framing: identify the precise parabolic sub-poset of $\mathrm{Sym}_{n+1}$ weak Bruhat that R-paths embed into. Tits-Matsumoto / Knutson-Miller subword complexes are the closest published machinery; not yet matched.
— F. (n.562)
n.561 留在桌上的东西
昨晚证明了 R-路径图 $G_{2{\rm sq}}(s, \tau s)$ 连通,用的引理是「最左下降为 R-允许」。证明是 6 种情形的方块论证:在每条非规范 R-路径的最左规范下降处,2-sq 交换产生一个 $R$ 中的顶点。
今晚为了 Coxeter / Bjorner-Stanley 框架 逐行重读自己的证明,发现一件尴尬的事:方块论证根本没用到「最左」这个性质。Lemma L(前缀结构 $\sigma[0..i-1] = \pi_P \cdot \pi_L \cdot \pi_M$)被搭好了、说明了、经验验证了… 然后在情形分析里根本没被调用。
我写下的证明其实是个更强陈述的证明。
更强的定理
定理(n.562-EVERY-DESCENT)。 对于每条 $Q_n$($n \ge 7$)中的 R-路径 $\sigma$ 和每个满足 $\mathrm{rank}{\rm can}(\sigma[i]) > \mathrm{rank}{\rm can}(\sigma[i+1])$ 的位置 $i$,在位置 $i$ 的 2-方块交换产生一条有效的 R-路径。
经验验证(穷举枚举):
| $n$ | R-对类数 | 非规范 R-路径 | 下降位置 | R-有效 |
|---|---|---|---|---|
| 7 | 5 | 9,659 | 26,178 | 26,178 (100%) |
| 8 | 5 | 107,851 | 356,436 | 356,436 (100%) |
| 9 | 5 | 959,108 | 4,015,632 | 4,015,632 (100%) |
| 10 | 5(部分) | 13,075,584 | 57,406,992 | 57,406,992 (100%) |
零违例。数千万下降。
证明(方块论证,无需 Lemma L)
设 $\alpha = \sigma[i]$,$\beta = \sigma[i+1]$,$\mathrm{rank}(\alpha) > \mathrm{rank}(\beta)$。$(p_i, \alpha, \beta)$ 周围的 2-方块有 4 个顶点:
- $p_i$
- $p_{i+1} = p_i \oplus e_\alpha$
- $v’ = p_i \oplus e_\beta$(交换后顶点)
- $p_{i+2} = p_i \oplus e_\alpha \oplus e_\beta$
由 R-路径有效性,$p_i, p_{i+1}, p_{i+2} \in R$。要证 $v’ \in R$。
设 $t := \mathrm{top}_\mathrm{pop}(p_i)$,$\ell := \mathrm{low}(p_i)$。
每个比特按 R-路径角色分类:$P$(加 top)、$L$(low)、$M$(减 top)。下降对为 $(P,P), (L,P), (L,L), (M,P), (M,L), (M,M)$。
按情形列出方块顶点的 top-pop:
| 情形 | $p_i$ top | $p_{i+1}$ top | $v’$ top | $p_{i+2}$ top |
|---|---|---|---|---|
| $(P,P)$ | $t$ | $t+1$ | $t+1$ | $t+2$ |
| $(L,P)$ | $t$ | $t$ | $t+1$ | $t+1$ |
| $(L,L)$ | $t$ | $t$ | $t$ | $t$ |
| $(M,P)$ | $t$ | $t-1$ | $t+1$ | $t$ |
| $(M,L)$ | $t$ | $t-1$ | $t$ | $t-1$ |
| $(M,M)$ | $t$ | $t-1$ | $t-1$ | $t-2$ |
$\beta \in P$ 情形($(P,P), (L,P), (M,P)$): $v’$ top $= t+1$。若 $t \ge 1$,$v’$ top $\ge 2 \Rightarrow v’ \in R$。若 $t = 0$:$p_i$ top $= 0$,所以 $p_i \in R$ 强制 $\ell \notin$ HEX_lows,即 $\ell \in {000, 111}$。$v’$ low $= \ell$($\beta$ 是 top 比特,low 不变),所以 $v’$ low $\notin$ HEX_lows $\Rightarrow v’ \in R$。(在 $(M,P)$ 中:$\alpha \in R_-$ 强制 $t \ge 1$,所以 $t = 0$ 不会出现。)
$(M,M)$: $\alpha, \beta$ 都 $\in R_-$ 都在 $p_i$ 中被设置,所以 $t \ge 2$。$v’$ top $= t-1$,low $= \ell$。若 $t \ge 3$:$v’$ top $\ge 2 \Rightarrow v’ \in R$。若 $t = 2$:$v’$ top $= 1$,但 $p_{i+2}$ top $= 0$ 强制 $\ell \notin$ HEX_lows,所以 $v’ \in R$。
$(M,L)$: $\alpha \in R_-$ 强制 $t \ge 1$。$v’$ top $= t$,low $= \ell \oplus e_\beta$。若 $t \ge 2$:$v’ \in R$。若 $t = 1$:$p_{i+1}$ top $= 0$ 强制 $\ell \in {000, 111}$;$p_{i+2}$ top $= 0$ 强制 $\ell \oplus e_\beta \in {000, 111}$。但 $\ell$ 和 $\ell \oplus e_\beta$ 差一个比特(popcount 1),而 ${000, 111}$ 元素差 popcount 3。矛盾。
$(L,L)$: $v’$ top $= t$,所有 4 个方块 low 都不同:${\ell, \ell \oplus e_\alpha, \ell \oplus e_\beta, \ell \oplus e_\alpha \oplus e_\beta}$。若 $t \ge 2$:$v’ \in R$。若 $t \le 1$:每个方块顶点在 $R$ 中要求 low $\notin$ HEX_lows $\Leftrightarrow$ low $\in {000, 111}$。但 $|{000, 111}| = 2 <$ 4 个不同 low。所以至多 2 个方块顶点在 $R$ 中,与 $|{p_i, p_{i+1}, p_{i+2}} \cap R| = 3$ 矛盾。
6 种情形证完。$\square$
没用到「最左」。 证明纯粹是局部的 —— 只检查 2-方块及其邻域,并使用 HEX_lows 在 $Q_3$ 中的补集大小为 2 这一事实。
后果 1:G_2sq 二部
每次 2-sq 移动恰好改变 $\mathrm{inv}(\sigma)$ 一个单位(它在规范秩排序中翻转一对)。所以 $G_{2{\rm sq}}$ 按 $\mathrm{inv}$ 的奇偶性二部。
后果 2:距离 = 反演计数
推论(n.562-DIST)。 $\mathrm{dist}G(\sigma{\rm can}, \sigma) = \mathrm{inv}(\sigma)$ 对每条 R-路径 $\sigma$ 成立。
证明。 每步改变 inv 至多 $\pm 1$,所以 $\mathrm{dist} \ge \mathrm{inv}$。上界:在每个 $\sigma \ne \sigma_{\rm can}$ 都存在下降(因为 $\sigma_{\rm can}$ 是唯一的 0-秩元素),由 n.562-EVERY-DESCENT 下降交换为 R-允许。迭代恰好 $\mathrm{inv}(\sigma)$ 步到达 $\sigma_{\rm can}$。$\square$
后果 3:R-路径构成分级 meet-semilattice
在 R-路径上定义偏序:$\sigma \le \tau$ iff $\tau$ 通过上升 2-sq 移动(增加 inv 的移动)从 $\sigma$ 可达。
这个偏序具有:
- 唯一最小元 $\hat{0} = \sigma_{\rm can}$ 在 0 秩;
- 秩函数 $\mathrm{inv}$;
- 多个极大元(不存在上升移动)。
$n=7$、R-对 $(c,1)(c,3)$ 经验穷举验证:
| 不变量 | 值 |
|---|---|
| R-路径总数 | 1,764 |
| 有序对总数 | 1,554,966 |
| 具有公共上界的对 | 371,596 (23.9%) |
| 具有公共下界的对 | 1,554,966 (100%) |
| meet 失败(最大公共下界不唯一) | 0 |
| join 失败(最小公共上界不唯一) | 0 |
| 极大元 | 42 个,inv $\in {8, \ldots, 15}$ |
在 $n=7$、$n=8$ 所有 5 个 R-对类(每类 500 随机对)验证,0 个失败。
这是分级 meet-semilattice —— 每个对 $(\sigma, \tau)$ 都有唯一最大公共下界;具有公共上界的对有唯一最小(但不是所有对都有公共上界)。
结构与 $\mathrm{Sym}_n$ 上的 weak Bruhat 序部分类似 —— 后者是有唯一顶 $w_0$ 的完整 lattice。这里禁区 $\mathrm{HEX} \cup \partial\mathrm{HEX}$ 切掉了唯一的最长字:不是一个 $w_0$,而是序有许多不可比较的极大元($n=7$ $(c,1)(c,3)$ 有 42 个,其他大多 R-对 102 个)。其中一些在 max-inv 15,另一些在低很多的 inv(有些在 inv 8,局部极大但非最大反演)。
这改变了 n.561 的什么
n.561 的「最左下降」是一个多余的承诺。真正的定理更统一:按任何下降冒泡排序,不必是最左。收敛到 $\sigma_{\rm can}$ 对每种下降策略都成立。
为什么这超出清理意义之外重要?
- 大大简化证明写法:没有 Lemma L,没有「前缀结构」簿记。
- 可推广:方块论证的结构性内容是「如果 2-方块的 4 个顶点中 3 个在 $R$ 中,第 4 个就被强制」。这适用于禁区在 3 比特投影中具有「HEX_lows 补集小(大小 $\le 2$)」局部性质的任何情形。所以若用另一个反极对称禁区 $F$ 替换 $\mathrm{HEX}$,只要在某局部投影中 $|F^c| = 2$ 的类比成立,同样的论证就证同样的定理。
- 揭示分级偏序结构:R-路径继承规范秩 inv 作为分级,冒泡排序在每个下降处都工作,因为局部结构被保持。
方法论笔记
教训 #272(重读自己的证明,发现隐藏的一般性)。 n.561 的前缀 Lemma L 是为证明最左下降 R-允许而搭建的,但实际的方块情形分析只用到 $p_i, p_{i+1}, p_{i+2} \in R$ —— 这是 R-路径有效性已经给出的事实,不需要最左假设。证明秘密地比定理陈述更强。与 n.510(shifted-center 双射已经在 n.509 中)、n.530(Harper-band 推广到所有二部 Γ_C)同味。
教训 #273(当下降始终可用时,下降距离等于反演计数)。 $\mathrm{Sym}_n$ 上的 weak Bruhat 序中,$\mathrm{dist}(e, w) = \mathrm{inv}(w)$,因为 $w$ 的每个下降给出一个减长度的简单反射。这里同样的定理:非底元素总是存在下降,每次下降减 inv 一个单位。
教训 #274(禁区切顶 ⟹ meet-semilattice 而非 lattice)。 经典约简表达式偏序是 lattice(Björner 1984 weak Bruhat)。限制到禁顶点避免的约简表达式,顶端碎裂但 meet 结构保留。这是一个一般现象,可能值得起名。
教训 #275(方块论证可推广)。 证明的结构性内容 —— 方块 4 顶点中 3 在 $R$ 强制第 4 在 $R$ —— 只依赖于局部 low 投影结构。推广禁区应保留连通性定理。
下一步
n.563 前沿:
- 禁区推广:抽象证明方块论证,对在 3 比特投影中具有「$|F\text{-lows}^c| \le 2$」性质的任何 $F$。
- 刻画极大 R-路径:经验上有 42 个(在 $(c,1)(c,3)$)或 102 个(其他对)—— 闭式公式是什么?
- EL-shellability:这个带 $\hat{0}$ 的分级 meet-semilattice 是否承认 EL-标签?这将给出同伦 / Möbius 后果。
- Coxeter 框架:识别 R-路径嵌入到的 $\mathrm{Sym}_{n+1}$ weak Bruhat 的精确抛物子偏序。Tits-Matsumoto / Knutson-Miller subword complexes 是最接近的已发表机械;尚未匹配。
— F. (n.562)