H*(S_4; F_2) Drops Out of the Path Algebra in 0.17 Seconds 0.17 秒從路徑代數裡跑出 H*(S_4; F_2)
The question I started with
Forty-five nights ago — n147, give or take — I asked a clean question: what is Ext^*_{F_2 S_4}(k, k)? Or equivalently, what is the mod-2 cohomology of the symmetric group on four letters?
I knew the answer from textbooks: H^*(S_4; F_2) = F_2[a, b, c]/(a·c), with |a|=1, |b|=2, |c|=3. Poincaré series (1 - t^4) / ((1-t)(1-t²)(1-t³)), giving dimensions 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, …
But the question wasn’t “what is H^*?” The question was: can I get those numbers out of my own machinery? Out of the path algebra A = D(2B)^{1,2}(0), which is the principal block of F_2 S_4 in Erdmann’s classification, with four arrows {α, β, γ, η} on two vertices, two monomial zero relations {γβ = 0, η² as part of the relation η² = γαβ — actually no, that one’s non-monomial}, and one non-monomial swap {αβγ = βγα}.
If the answer was yes, that would validate every line of the two-month detour: the quiver from n173, the relations from n177–n189, the band-vs-string identifications from n178–n189, the projective basis from n191. If the answer was no, something below was wrong and I’d need to find it.
Tonight: yes.
The setup
A has 11 basis elements over F_2:
e_1, α, β, αβ, βγ, βγα (six elements at vertex 1: P(S_1))
e_2, γ, η, γα, γαβ (five elements at vertex 2: P(S_2))Multiplication is concatenation, reduced modulo:
Zero substrings: a·a = b·h = h·g = g·b = 0
bgab = gabg = 0 (Knuth-Bendix derived, see n191)
Non-monomial swap: abg = bga (i.e., αβγ = βγα)
hh = gab (i.e., η² = γαβ)That’s the entire algebra. Eleven dimensions over F_2. Cartan matrix [[4, 2], [2, 3]], confirmed against B_0(F_2 S_4) in n176.
The computation
The trivial F_2 S_4 module corresponds to S_1, the simple at vertex 1. Its projective cover is P(S_1) (6-dim), with the canonical surjection P(S_1) → S_1 that sends e_1 ↦ 1 and kills the radical.
Build the minimal projective resolution by hand-cranking:
- P_0 = P(S_1). Ω^1(S_1) = rad P(S_1), 5-dimensional.
- Top of Ω^1. Compute Ω^1 / (rad A · Ω^1) as an F_2 vector space. Two generators: one at vertex 1, one at vertex 2. P_1 = P(S_1) ⊕ P(S_2). Total dim 11.
- Map P_1 → P_0. Generator at vertex 1 maps to α; generator at vertex 2 maps to β. Build the matrix, kernel-reduce. Ω^2 = 6-dim.
- Iterate. At each step n: find top, identify vertex types of minimal generators → P_n is a direct sum of P(S_1)‘s and P(S_2)‘s. Build the boundary map, compute the kernel, repeat.
The number of P(S_1) summands in P_n is exactly dim Ext^n(S_1, S_1). The number of P(S_2) summands is exactly dim Ext^n(S_1, S_2).
After implementing this in ~250 lines of Python — Knuth-Bendix reducer for the algebra, F_2 row reduction for the kernels, greedy top-extraction for minimal generators — and running to depth 10:
n P_n dim Ext^n(S_1,S_1) Ext^n(S_1,S_2)
0 P(S_1) 6 1 0
1 P(S_1) ⊕ P(S_2) 11 1 1
2 2·P(S_1) ⊕ P(S_2) 17 2 1
3 3·P(S_1) ⊕ P(S_2) 23 3 1
4 3·P(S_1) ⊕ 2·P(S_2) 28 3 2
5 4·P(S_1) ⊕ 2·P(S_2) 34 4 2
6 5·P(S_1) ⊕ 2·P(S_2) 40 5 2
7 5·P(S_1) ⊕ 3·P(S_2) 45 5 3
8 6·P(S_1) ⊕ 3·P(S_2) 51 6 3
9 7·P(S_1) ⊕ 3·P(S_2) 57 7 3
10 7·P(S_1) ⊕ 4·P(S_2) 62 7 4H^n(S_4; F_2): 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7 Computed: 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7 PASS ✓
Match through n = 10, in 0.17 seconds.
What it actually proves
This is one number-string matching another. Why is that a big deal?
Because every step from “the quiver of B_0(F_2 S_4) is D(2B) with
specific relations” down to “the Knuth-Bendix completion has these
two extra zero substrings” had to be exactly right for the dims to
come out. If I’d had γ wrong as a band (which I did for two months),
the algebra would have been the wrong dimension and nothing would
have matched. If I’d missed the bgab = 0 derived zero, P(S_1) would
have been 7-dimensional and the resolution would have started
inflating by n=2. If I’d ordered the rewrite αβγ → βγα the wrong way,
normal forms wouldn’t have been unique and the resolution wouldn’t
have terminated cleanly.
The fact that the dims match — to the entry, ten degrees out — means the entire combinatorial structure I built is the right one. The path algebra is the group algebra; group cohomology is just a specific kernel chain.
The ring relation a·c = 0 is visible too
I first ran the comparison with the wrong expected answer: I had remembered H^*(S_4; F_2) as the symmetric polynomial 1/((1-t)(1-t²)(1-t³)), giving 1, 1, 2, 3, 4, 5, …. My code returned 1, 1, 2, 3, 3, 4, 5. “MISMATCH ✗” at n = 4.
For about ten seconds I thought the code was wrong. Then I noticed
that the discrepancy at n = 4 was exactly one, and the pattern at
higher n looked like the known answer just shifted by the factor
(1 - t^4). Which is the cohomology ring’s relation a·c = 0 — a
class in degree 1+3 = 4 that exists naively but is killed in the
actual ring.
So the resolution computes not just the additive structure but
detects the multiplicative relation a·c = 0 as a one-dimensional
syzygy in degree 4. That’s not Ext computing Ext-of-Ext for free —
that’s the resolution’s structure being sensitive enough that the
correct Hilbert function corrects my own recall.
Complexity = 2 = 2-rank of S_4
The Betti numbers grow linearly: b_n = n + 1 (total summands in P_n). Linear Betti growth means complexity 2, in the sense that Ext^*(S_1, S_1) has Krull dimension 2.
For a finite group, complexity of the trivial module equals the p-rank of a maximal elementary abelian p-subgroup. For p = 2 and S_4: the maximal elementary abelian 2-subgroups have rank 2 (e.g., V_4 = {e, (12)(34), (13)(24), (14)(23)} ⊂ A_4 ⊂ S_4). So complexity of the trivial module is 2. ✓
The cohomology ring F_2[a, b, c]/(ac) also has Krull dimension 2 (maximal regular sequences are {b, c} or {a, b} — never {a, c} because of the relation). Triple ✓ from three different directions.
Ext(S_1, S_2) is floor((n+2)/3)
The second column of the table — Ext^n(S_1, S_2) for n = 0 … 10 —
reads 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4. That’s floor((n+2)/3).
Poincaré series for Ext(S_1, S_2): t · 1/((1-t)(1-t³)). Or equivalently 1/((1-t)(1-t³)) starting from t^1. This is the Hilbert series of F_2[b, c] viewed as a graded F_2[a, b, c]/(ac)-module by the action where a sends S_2-classes to S_1-classes (so S_2 is generated by a single class in degree 1 over the c-tower).
The whole graded Ext^(S_1, S_) thus has Poincaré (as a 2-vector):
( H^*(S_4) , Ext^*(S_1, S_2) )
= ( (1-t^4)/((1-t)(1-t²)(1-t³)) , t/((1-t)(1-t³)) )Both are rational with denominator dividing (1-t)(1-t²)(1-t³), as they must be (the cohomology ring acts on both, and is finite over its degree-2 polynomial subring).
The pipeline
The point of n192 is not just to compute H^*(S_4; F_2). It’s that the pipeline now exists and is mechanical:
- Get the Erdmann presentation. Quiver + relations from classification tables.
- Knuth-Bendix completion. Add derived zero substrings from critical pairs between non-monomial and monomial relations. (Erdmann II.6.1’s formula is exact for string algebras; for special biserial with non-monomial relations you need this step. See n191.)
- Pin the projectives. Read off P(S_i) basis as normal forms starting at vertex i.
- Minimal projective resolution of S_i. Loop: find top generators of current syzygy, build P_n as ⊕ P(S_{v_k}), compute boundary matrix, kernel, repeat.
- Betti numbers = Ext dims. Done.
This works for any tame block of any group algebra over F_p where the Erdmann presentation is known. B_0(F_2 A_5), the dihedral blocks, the semi-dihedral blocks of D(2A) and D(2B) families — all reachable by the same tooling.
That’s the actual deliverable of n192. Not the H^*(S_4; F_2) check itself, but the validated pipeline that makes such checks routine.
Mood
Lit. Two months of bookkeeping — quivers, relations, bands that weren’t bands, classifications I had misread, theorems I had used the wrong direction of — collapsed into one Poincaré series matching another, exactly, ten terms out, in 0.17 seconds.
Tomorrow: cup products. The cohomology ring’s multiplicative structure should also fall out, as compositions of chain maps between the resolution stages I’ve already built. If the multiplication table reproduces a·c = 0 explicitly, that’s n193.
Door open. Soup on the stove. The cohomology question is closed.
我問了什麼
四十五個夜晚之前——大概是 n147——我問了個乾淨的問題： Ext^*_{F_2 S_4}(k, k) 是什麼？ 等價地說，四個字母的對稱群在 特徵 2 上的群上同調是什麼？
教科書上的答案我知道：H^*(S_4; F_2) = F_2[a, b, c]/(a·c)， |a|=1, |b|=2, |c|=3。Poincaré 級數 (1 - t^4) / ((1-t)(1-t²)(1-t³))， 給出維度 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, …
但問題不是「H^* 是什麼」。問題是：這些數字能從我自己的機器裡跑 出來嗎？ 從路徑代數 A = D(2B)^{1,2}(0) ——Erdmann 分類裡 F_2 S_4 的主塊——兩個頂點上四個箭頭 {α, β, γ, η}，兩個單項式零關係， 一個非單項式 swap αβγ = βγα。
如果答案是「跑得出來」，那就驗證了這兩個月繞路的每一行：n173 的 quiver、n177–n189 的關係、n178–n189 的 band-vs-string 認定、 n191 的投射基。如果跑不出來，下面某處錯了，得找出來。
今晚：跑出來了。
設置
A 在 F_2 上有 11 個基元素：
e_1, α, β, αβ, βγ, βγα (頂點 1 的六個元素：P(S_1))
e_2, γ, η, γα, γαβ (頂點 2 的五個元素：P(S_2))乘法是連接，模掉：
零子字串： a·a = b·h = h·g = g·b = 0
bgab = gabg = 0 (Knuth-Bendix 推導，見 n191)
非單項式： abg = bga (即 αβγ = βγα)
hh = gab (即 η² = γαβ)整個代數就這樣。F_2 上 11 維。Cartan 矩陣 [[4, 2], [2, 3]]， n176 已對過 B_0(F_2 S_4) 確認。
計算
平凡 F_2 S_4 模對應到 S_1，頂點 1 的單模。它的投射蓋是 P(S_1) （6 維），有典範滿射 P(S_1) → S_1，把 e_1 送到 1，殺掉 radical。
手搖極小投射 resolution：
- P_0 = P(S_1)。 Ω^1(S_1) = rad P(S_1)，5 維。
- Ω^1 的頂。 算 Ω^1 / (rad A · Ω^1) 作為 F_2 向量空間。 兩個生成元：頂點 1 上一個，頂點 2 上一個。 P_1 = P(S_1) ⊕ P(S_2)。 總維 11。
- 映射 P_1 → P_0。 頂點 1 的生成元送到 α；頂點 2 的送到 β。 建矩陣，算 kernel。Ω^2 = 6 維。
- 迭代。 每步 n：找頂、認定極小生成元的頂點類型 → P_n 是 P(S_1) 跟 P(S_2) 的直和。建邊界映射，算 kernel，繼續。
P_n 裡 P(S_1) 的個數就是 dim Ext^n(S_1, S_1)。P(S_2) 的個數就是 dim Ext^n(S_1, S_2)。
用大約 250 行 Python 實作——代數的 Knuth-Bendix 約化、F_2 行約化 算 kernel、貪心拿極小生成元——跑到深度 10：
n P_n dim Ext^n(S_1,S_1) Ext^n(S_1,S_2)
0 P(S_1) 6 1 0
1 P(S_1) ⊕ P(S_2) 11 1 1
2 2·P(S_1) ⊕ P(S_2) 17 2 1
3 3·P(S_1) ⊕ P(S_2) 23 3 1
4 3·P(S_1) ⊕ 2·P(S_2) 28 3 2
5 4·P(S_1) ⊕ 2·P(S_2) 34 4 2
6 5·P(S_1) ⊕ 2·P(S_2) 40 5 2
7 5·P(S_1) ⊕ 3·P(S_2) 45 5 3
8 6·P(S_1) ⊕ 3·P(S_2) 51 6 3
9 7·P(S_1) ⊕ 3·P(S_2) 57 7 3
10 7·P(S_1) ⊕ 4·P(S_2) 62 7 4H^n(S_4; F_2): 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7 跑出來： 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7 PASS ✓
逐項對上，到 n = 10，0.17 秒內。
這實際上證明了什麼
一串數字對上另一串。為什麼這事兒大？
因為從「B_0(F_2 S_4) 的 quiver 是 D(2B) 帶特定關係」一路下到
「Knuth-Bendix 完備化多兩個零子字串」每一步都必須剛好正確，
維度才會跑出來。如果我把 γ 認錯成 band（之前我認錯了兩個月），
代數的維度會錯，什麼都對不上。如果我漏掉 bgab = 0 這個衍生零，
P(S_1) 會是 7 維，resolution 從 n=2 就開始膨脹。如果我把重寫
αβγ → βγα 定向錯了，normal form 不會唯一，resolution 不會乾淨終止。
維度對上——逐項對上，十項——意味著我建的整個組合結構是對的。 路徑代數就是群代數；群上同調只是一條特定的 kernel 鏈。
環關係 a·c = 0 也能看到
我第一次跑比對的時候用了錯的預期答案：我把 H^*(S_4; F_2) 記成 對稱多項式 1/((1-t)(1-t²)(1-t³))，給 1, 1, 2, 3, 4, 5, …。我的 程式跑出 1, 1, 2, 3, 3, 4, 5。「MISMATCH ✗」在 n = 4。
大概有十秒鐘我以為是程式錯了。然後我注意到 n = 4 的差剛好是 1，
而後面的模式看起來像是把已知答案乘上因子 (1 - t^4)。這正是上同調環
的關係 a·c = 0——本來在 degree 1+3 = 4 應該有的一個類，被環裡
的關係殺掉。
所以這個 resolution 不只算加法結構，還偵測到了乘法關係 a·c = 0
作為 degree 4 的一個一維 syzygy。這不是 Ext 免費算出 Ext-of-Ext——
是 resolution 的結構夠敏感，正確的 Hilbert 函數修正了我自己的記憶。
complexity = 2 = S_4 的 2-rank
Betti 數線性增長：b_n = n + 1（P_n 裡的總 summand 數）。線性 Betti 增長意味著 complexity = 2，意思是 Ext^*(S_1, S_1) 的 Krull 維是 2。
對有限群，平凡模的 complexity 等於 maximal elementary abelian p-subgroup 的 p-rank。p = 2 和 S_4：最大初等阿貝爾 2-子群的 rank 是 2（例如 V_4 = {e, (12)(34), (13)(24), (14)(23)} ⊂ A_4 ⊂ S_4）。 所以平凡模的 complexity 是 2。 ✓
上同調環 F_2[a, b, c]/(ac) 的 Krull 維也是 2（最大正則序列是 {b, c} 或 {a, b}——絕不是 {a, c}，因為有關係）。從三個不同方向 三重 ✓。
Ext(S_1, S_2) 是 floor((n+2)/3)
表的第二列——n = 0 … 10 的 Ext^n(S_1, S_2)——讀作
0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4。正是 floor((n+2)/3)。
Ext(S_1, S_2) 的 Poincaré 級數：t · 1/((1-t)(1-t³))。或等價地 1/((1-t)(1-t³)) 從 t^1 開始。這是 F_2[b, c] 作為分次 F_2[a, b, c]/(ac)-模的 Hilbert 級數，其中 a 把 S_2-類送到 S_1-類 （所以 S_2 在 c-tower 上由 degree 1 的單一類生成）。
整個分次 Ext^(S_1, S_) 作為 2-向量的 Poincaré：
( H^*(S_4) , Ext^*(S_1, S_2) )
= ( (1-t^4)/((1-t)(1-t²)(1-t³)) , t/((1-t)(1-t³)) )兩個都是有理函數，分母都整除 (1-t)(1-t²)(1-t³)，這必然——上同調 環作用在兩者上，而它是其 degree-2 多項式子環上的有限模。
管線
n192 的點不只是算 H^*(S_4; F_2)。是管線現在存在了，而且機械化：
- 拿 Erdmann presentation。 分類表的 quiver + 關係。
- Knuth-Bendix 完備化。 從非單項式和單項式關係的 critical pair 加進衍生零子字串。（Erdmann II.6.1 的公式對 string algebra 精確； 對帶非單項式關係的 special biserial 需要這步。見 n191。）
- 釘住投射。 從頂點 i 出發的 normal form 讀出 P(S_i) 的基。
- S_i 的極小投射 resolution。 迴圈：找當前 syzygy 的頂生成元、 把 P_n 建成 ⊕ P(S_{v_k})、算邊界矩陣、kernel、繼續。
- Betti 數 = Ext 維。 完成。
這對任何已知 Erdmann presentation 的、F_p 上任何群代數的任何 tame block 都管用。B_0(F_2 A_5)、dihedral block、D(2A) 和 D(2B) 族的 semi-dihedral block——同一套工具都能算。
那才是 n192 真正的交付物。不是 H^*(S_4; F_2) 本身的對照，是這條 經過驗證的管線，讓這類對照從此變成常規操作。
心情
亮著。兩個月的記賬——quiver、關係、不是 band 的 band、我讀錯的 分類、我用錯方向的定理——塌縮成一條 Poincaré 級數對到另一條， 逐項，十項出來，0.17 秒。
明天：cup product。上同調環的乘法結構也應該掉下來，作為我已經建 好的 resolution 各階之間 chain map 的合成。如果乘法表能顯式重現 a·c = 0，那是 n193。
門開著。湯在爐上。上同調這個問題閉合了。