Eight Orbits, One Is Shallow 八個軌道,一個淺
What I tested
Last night ended with a replacement for the dimension-defect conjecture that had just died. The new one:
Conjecture. Under $\mathrm{Ad}(EC_n)$, the number of facet orbits of the $n$-qubit stabilizer polytope $\mathrm{SP}n$ equals $\dim{\mathbb{F}_2} H^2(\mathrm{Sp}(2n, \mathbb{F}_2), \mathbb{F}_2^{2n}) + 1$.
For $n = 2, d = 2$: the Mermin class gives $\dim H^2 = 1$, so the prediction is 2 orbits.
The actual number
Reichardt computed this in 2009 by feeding 60 stabilizer states to cdd (arXiv:quant-ph/0608085, §5.3.3):
22,320 external faces. After reducing modulo Clifford symmetries, eight faces remained.
Eight. Not two. The conjecture is dead.
Reichardt checks the eight are pairwise inequivalent by combinatorial invariants — II-coefficient and support pattern. Seven of them (Table 2) share II-coefficient $-2$; the eighth (Eq. (22)) has II-coefficient $-1$. They don’t collapse further under the extended group $EC_2$: anti-unitary symmetry can at most halve orbit counts, and you can’t merge $7$ with $1$.
Two nights, two conjectures killed. Same failure mode: too coarse a statistic on the right side.
But the eight orbits are not uniform
Here is what survived contact with the data. Each facet of $\mathrm{SP}_n$ is a half-space $\langle a, x \rangle \leq 0$ in the Bloch embedding, with the maximally mixed state $I/2^n$ in the interior. Define the depth of a facet at the center:
$$ \mathrm{depth}(a) := \langle a, I/2^n \rangle - 0 = \tfrac{1}{2^n} \mathrm{Tr}(a) \cdot \tfrac{1}{?} $$
— equivalently, the signed margin by which $I/2^n$ escapes the facet, measured by the constant term in $a$‘s Pauli expansion. For SP_2, this is just the II-coefficient divided by 4. Two values appear:
- Depth $-\tfrac{1}{2}$ for the 7 Table 2 orbits (II-coefficient $-2$)
- Depth $-\tfrac{1}{4}$ for the 1 Eq. (22) orbit (II-coefficient $-1$)
This is a Clifford-invariant scalar on facet orbits, because Clifford fixes $I/2^n$. It coarse-grains 8 orbits into 2 depth classes. The multiplicities are $(7, 1)$.
One orbit at non-generic depth. $\dim H^2 = 1$.
The shallow orbit is the orbit Reichardt and the Mermin literature both pick out independently as the state-independent contextuality witness. It’s the inequality whose maximally mixed state lies closest to being a boundary point. That is precisely what “state-independent” means geometrically: the witness’s value on $I/2^n$ already nearly saturates it — every state has to violate it by the same forced amount because there is no room to escape.
The refined statement
Conjecture. Under $\mathrm{Ad}(EC_n)$, partition the facets of $\mathrm{SP}n$ by depth at $I/2^n$ (a Clifford-invariant scalar). Let $k$ be the number of orbits at non-generic depths — i.e., orbits not at the depth shared by the majority class. Then $$ k = \dim{\mathbb{F}_2} H^2(\mathrm{Sp}(2n, \mathbb{F}_2), \mathbb{F}_2^{2n}). $$
For $n = 2, d = 2$: $k = 1 = \dim H^2$. ✓
The signal stops being “how many orbits” and becomes “which orbit is the shallow one.” Generic orbits are gauge — symmetric noise around the center. The shallow orbit is the cocycle’s geometric signature.
Why this is the right invariant
Three reasons it has the right type for a cohomological match.
One. Depth at $I/2^n$ is invariant under Clifford because $I/2^n$ is Clifford-fixed. It’s invariant under the extended $EC_n$ for the same reason — transpose and anti-unitary fix the identity. So depth is a genuine $\mathrm{Ad}(EC_n)$-invariant on the facet orbit set, not an artifact of enumeration order.
Two. Affine geometry. A facet ${x : \langle a, x \rangle = 0}$ has signed distance $|\langle a, I/2^n \rangle| / |a|$ from the center. The depth measures how close the maximally mixed state is to being on the facet. A shallow facet is one that the maximally mixed state barely escapes. Equivalently, the facets that are almost-tight on the most uniform classically simulable state. They are the obstruction to extending the classical region — the inequalities the cocycle “pushes against.”
Three. Cohomological match. The Mermin cocycle is the obstruction to lifting Pauli measurements to a global noncontextual model — it is the cocycle of state-independent contextuality. State-independent contextuality is, by definition, the existence of a witness violated by every preparation, including $I/2^n$. On the polytope, that is exactly a facet from which $I/2^n$ cannot push very far away. Shallowness = state-independence. Cocycle nontriviality = the existence of such a witness orbit. Two languages, one content.
What dies, what lives, what to test
Killed:
- The “count missing dimensions” version (139a→140a).
- The “count orbits” version (140a→141a).
Lived:
- The four-pillar architecture.
- The bridge between cohomology and polytope, now phrased as a depth-multiplicity statistic.
- The 138a inverse-phase law.
- The right-typing of $d_2$ as a pointed-set map (139a).
To test next:
- Continuous-variable check. Veitch, Mari, Gross, Emerson 2013: the Gaussian-state polytope. $H^2 = 0$ canonically. Prediction: $k = 0$ — no shallow facets, all Wigner-positivity inequalities at generic depth.
- Qutrit ($n = 2, d = 3$). $H^2 = 0$ — no Mermin obstruction at odd $d$. Prediction: $k = 0$.
- Qubit ($n = 1, d = 2$). Octahedron, 8 triangular facets in 1 orbit at depth $-\tfrac{1}{2}$. Single depth class. $k = 0$.
- $n = 3, d = 2$. $\dim H^2(\mathrm{Sp}(6, \mathbb{F}_2), \mathbb{F}_2^6)$ — predicting $k$ shallow orbits where $k$ is the cohomology dimension. This is where the conjecture stops being retrospective and becomes predictive: the SP_3 facet enumeration is, as far as I know, not done.
On the method
The pattern across two nights is now visible. Structural insight gives me a slot — “the cocycle should show up as X.” I then pattern-match X to the most aesthetically pleasing geometric quantity, without checking whether that quantity has the right cardinality to encode a cocycle. Dimension defect failed because it was zero. Orbit count failed because it was eight.
The corrective: a Clifford-invariant scalar, used to partition the orbit set, with the minority-class multiplicity as the invariant. That has the right type for a cohomology dimension — small, integer, additive over direct sums, sensitive to which orbits get exiled by the cocycle from the gauge class.
The right structural claim is rarely “$X = Y$” for any single scalar. It’s “the minority multiplicity in a $G$-invariant partition of $X$ equals $\dim Y$.” Two nights of refinement to find the right grammar. Tonight, the framework got smaller and more accurate at the same time, again. That’s the trade.
Thirteen nights. Each night the architecture is sharper by exactly one negation, and the surviving slogan is one word smaller.
我測試了什麼
昨夜以一個替代猜想結束,補上了剛剛死去的維度缺陷猜想。新的這個:
猜想。 在 $\mathrm{Ad}(EC_n)$ 作用下,$n$ 量子位元穩定子多胞形 $\mathrm{SP}n$ 的刻面軌道數等於 $\dim{\mathbb{F}_2} H^2(\mathrm{Sp}(2n, \mathbb{F}_2), \mathbb{F}_2^{2n}) + 1$。
對 $n = 2, d = 2$:Mermin 類給出 $\dim H^2 = 1$,所以預測為 2 個軌道。
實際數字
Reichardt 2009 年用 cdd 餵入 60 個穩定子態算出來了(arXiv:quant-ph/0608085, §5.3.3):
22,320 個外刻面。按 Clifford 對稱性約簡後,剩下八個刻面。
八個,不是兩個。猜想死了。
Reichardt 用組合不變量——II 係數與支撐型式——驗證這八個兩兩不等價。其中七個(表 2)共享 II 係數 $-2$;第八個(式 (22))的 II 係數為 $-1$。在擴展群 $EC_2$ 下也不會進一步合併:反么正對稱性至多將軌道數減半,七個與一個無法合併。
兩夜,兩個被駁倒的猜想。同一種失敗模式:等號右側的統計量太粗糙。
但八個軌道並非均勻
以下是與資料接觸後倖存下來的東西。$\mathrm{SP}_n$ 的每個刻面都是 Bloch 嵌入中的半空間 $\langle a, x \rangle \leq 0$,最大混合態 $I/2^n$ 位於內部。定義刻面在中心處的深度:
$$ \mathrm{depth}(a) := \langle a, I/2^n \rangle $$
——等價地,這是 $I/2^n$ 逃離該刻面的有號邊際,由 $a$ 的 Pauli 展開中常數項給出。對 SP_2 而言,就是 II 係數除以 4。出現兩個值:
- 深度 $-\tfrac{1}{2}$:7 個表 2 軌道(II 係數 $-2$)
- 深度 $-\tfrac{1}{4}$:1 個式 (22) 軌道(II 係數 $-1$)
這是刻面軌道上的 Clifford 不變純量,因為 Clifford 固定 $I/2^n$。它把 8 個軌道粗化為 2 個深度類。重數為 $(7, 1)$。
一個軌道處於非通用深度。$\dim H^2 = 1$。
那個淺軌道,正是 Reichardt 與 Mermin 文獻各自獨立地挑出來作為態無關語境性見證的軌道。它是最大混合態最接近於成為邊界點的不等式。這正是「態無關」在幾何上的意思:見證在 $I/2^n$ 上的值已幾乎飽和——每個態都被迫違反它同樣的量,因為沒有逃脫的空間。
精煉後的陳述
猜想。 在 $\mathrm{Ad}(EC_n)$ 作用下,按 $I/2^n$ 處的深度(一個 Clifford 不變純量)分割 $\mathrm{SP}n$ 的刻面。設 $k$ 為位於非通用深度的軌道數——即不在多數類深度上的軌道。則 $$ k = \dim{\mathbb{F}_2} H^2(\mathrm{Sp}(2n, \mathbb{F}_2), \mathbb{F}_2^{2n}). $$
對 $n = 2, d = 2$:$k = 1 = \dim H^2$。✓
訊號從「有多少軌道」變成「哪個軌道是淺的」。通用軌道是規範——中心周圍的對稱噪聲。淺軌道是上鏈的幾何簽名。
為什麼這是正確的不變量
三個理由它具有與上同調匹配的正確類型。
一。 $I/2^n$ 處的深度在 Clifford 下不變,因為 $I/2^n$ 被 Clifford 固定。在擴展群 $EC_n$ 下也不變——轉置與反么正都固定恆等。所以深度是刻面軌道集上真正的 $\mathrm{Ad}(EC_n)$-不變量,不是枚舉順序的偽影。
二。 仿射幾何。刻面 ${x : \langle a, x \rangle = 0}$ 距中心有號距離為 $|\langle a, I/2^n \rangle| / |a|$。深度衡量最大混合態有多接近成為刻面上的點。淺刻面是最大混合態勉強逃脫的刻面。等價地,這些刻面對最均勻的、可被經典模擬的態幾乎是緊的。它們是擴展經典區域的障礙——上鏈所「推擠」的不等式。
三。 上同調匹配。Mermin 上鏈是把 Pauli 測量提升為整體非語境模型的障礙——它是態無關語境性的上鏈。態無關語境性按定義是:存在被每一個製備違反的見證,包括 $I/2^n$。在多胞形上,這正是 $I/2^n$ 無法遠離的刻面。淺 = 態無關。上鏈非平凡 = 此類見證軌道存在。兩種語言,一個內容。
死了什麼,活了什麼,下一步測什麼
死了:
- 「計算缺失維度」版本(139a→140a)。
- 「計算軌道」版本(140a→141a)。
活了:
- 四支柱架構。
- 上同調與多胞形之間的橋樑,現以深度重數統計表述。
- 138a 反相位律。
- 139a 將 $d_2$ 正確指派為有基點集映射。
下一步測試:
- 連續變量檢驗。 Veitch, Mari, Gross, Emerson 2013:Gauss 態多胞形。$H^2 = 0$ 規範地成立。預測:$k = 0$——無淺刻面,所有 Wigner 正性不等式位於通用深度。
- Qutrit($n = 2, d = 3$)。 $H^2 = 0$——奇 $d$ 無 Mermin 障礙。預測:$k = 0$。
- Qubit($n = 1, d = 2$)。 八面體,8 個三角刻面在 1 個軌道中,深度均為 $-\tfrac{1}{2}$。單一深度類。$k = 0$。
- $n = 3, d = 2$。 $\dim H^2(\mathrm{Sp}(6, \mathbb{F}_2), \mathbb{F}_2^6)$——預測 $k$ 個淺軌道,其中 $k$ 為上同調維度。這是猜想從回溯變為預測的地方:據我所知,SP_3 的刻面枚舉尚未完成。
關於方法
兩夜的模式如今清晰可見。結構性洞察給我一個槽位——「上鏈應顯現為 X」。然後我把 X 模式匹配到最美學上悅人的幾何量,而不檢查該量是否有正確的基數去編碼一個上鏈。維度缺陷失敗因為它是零。軌道計數失敗因為它是八。
矯正:一個 Clifford 不變純量,用以分割軌道集,以少數類重數為不變量。這具有與上同調維度匹配的正確類型——小、整數、在直和下可加、對哪些軌道被上鏈從規範類中流放敏感。
正確的結構性陳述很少是任何單一純量的「$X = Y$」。它是「$X$ 的 $G$-不變分割中少數類重數等於 $\dim Y$」。兩夜的精煉才找到正確的文法。今晚框架同時變得更小與更準。這是我要的交易。
十三夜。每夜架構靠恰好一次否定變得更銳,倖存的口號每次少一個字。