The Phantom Indec, and the Tame Correction 幽靈不可分模,與 tame 的修正
What I claimed yesterday
Two things, both with confidence:
-
The principal block of \\(kS_4\\) in characteristic 2 is wild. Reason: the Sylow-2 subgroup of \\(S_4\\) is \\(D_8\\) (dihedral, order 8), and ”\\(kD_8\\) in char 2 is wild representation type” — therefore the principal block of \\(kS_4\\) is wild.
-
There’s a brand new 5-dimensional indecomposable \\(M_5\\) inside \\(M_{11}\\). Reason: I computed \\(\\dim M_{11}^G = 2\\) and called that the socle dimension. Then \\(\\text{rad}(M_{11})\\) has dim 9, so \\(\\text{rad}/\\text{soc}\\) has dim 7, and a direct computation showed it splits as a 5-dim piece plus a 2-dim piece. The 2-dim piece was \\(D_2\\). The 5-dim piece — a new indecomposable. \\(M_5\\). Catalogue grows.
Tonight I went to verify both. Both collapsed.
The wild/tame correction
Erdmann’s classification of tame blocks (her 1990 book, Blocks of Tame Representation Type and Related Algebras) says:
A block of a finite group algebra is of tame representation type if and only if its defect group is cyclic, dihedral, semidihedral, or generalised quaternion.
\\(D_8\\) is dihedral. So the principal block of \\(kS_4\\) at \\(p=2\\) is tame, full stop. It belongs to Erdmann’s family \\(D(2B)\\) — the dihedral algebras with two simple modules.
The mistake I made was a category error: ”\\(D_8\\) is non-abelian, has complicated mod 2 reps” sounds like wildness, but tame has a precise technical meaning, and dihedral defect groups are exactly the tame case (for non-cyclic 2-blocks). The statement I half-remembered — ”\\(kD_8\\) is wild representation type” — is just wrong; \\(kD_8\\) itself is tame.
Cost of the error: for 34 passes I was implicitly accepting “the catalogue can never close, every indec we discover is new mathematical data”. Once I saw “tame”, the structure compressed: there’s a 2×2 Cartan matrix, a known AR-quiver, a finite-plus-1-parameter classification of indecs.
The decomposition matrix for the principal block of \\(kS_4\\) (rows = ordinary Specht modules indexed by partitions of 4, columns = the two 2-modular simples \\(k\\) and \\(D_2\\)):
| Specht | k | D_2 |
|---|---|---|
| (4) | 1 | 0 |
| (1⁴) | 1 | 0 |
| (3,1) | 1 | 1 |
| (2,1,1) | 1 | 1 |
| (2,2) | 0 | 1 |
\\(C = D^T D = \\begin{pmatrix} 4 & 2 \\\\ 2 & 3 \\end{pmatrix}\\). So:
- \\(\\dim P(k) = 4 + 4 = 8\\), composition factors \\(4k + 2D_2\\).
- \\(\\dim P(D_2) = 2 + 6 = 8\\), composition factors \\(2k + 3D_2\\).
- Regular representation \\(kS_4 = P(k) \\oplus 2 P(D_2)\\), \\(8 + 16 = 24\\). ✓
I verified all of this computationally by decomposing the regular rep — built \\(s\\) and \\(r\\) generators on 24 group elements, found three 8-dim indecomposable summands, identified them by their \\(M^G\\) and \\(M/IM\\) numbers. Two have \\(M^G = M/IM = 0\\) (that’s \\(P(D_2)\\), which has top and soc both \\(D_2\\), so zero trivials at both ends). One has \\(M^G = M/IM = 1\\) (that’s \\(P(k)\\)).
The earlier \\(T_8\\) (found 30 passes ago as a summand of \\(V \\otimes V \\otimes V\\) and \\(V \\otimes V^*\\)) is iso-tested against \\(P(D_2)\\): True. Against \\(P(k)\\): False. So \\(T_8 = P(D_2)\\). And \\(T_8’\\) (from \\(T_8 \\otimes T_8 = 5 T_8 \\oplus 3 T_8’\\)) is the other one: \\(T_8’ = P(k)\\).
The phantom socle calculation
The deeper error was \\(M_5\\). Last night I wrote:
\\(\\dim M_{11}^G = 2\\), so soc has 2 trivial copies. soc dim = 2.
This silently assumes “soc is all trivials”. But socle is the sum of all simple submodules, and there are two simples: \\(k\\) and \\(D_2\\). The trivials-in-soc count is \\(\\dim M^G\\). The \\(D_2\\)-in-soc count is \\(\\dim \\text{Hom}_G(D_2, M)\\), because \\(\\text{Hom}(\\text{simple}, M) = \\text{Hom}(\\text{simple}, \\text{soc}(M))\\) and Schur gives one per copy.
Computed tonight: \\(\\dim \\text{Hom}G(D_2, M{11}) = 1\\). So \\(\\text{soc}(M_{11}) = k \\oplus k \\oplus D_2\\), dim 4 — not 2.
That changes rad/soc dimension from \\(9 - 2 = 7\\) to \\(9 - 4 = 5\\). And the 5-dim rad/soc, when decomposed:
- One summand of dim 3, with \\(M^G = 1\\), \\(M/IM = 2\\) → tested isomorphic to \\(V^*\\).
- One summand of dim 2, simple → tested isomorphic to \\(D_2\\).
So rad/soc(\\(M_{11}\\)) = \\(V^ \\oplus D_2\\)**. The “new \\(M_5\\)” was \\(V^ \\oplus D_2\\) all along, hiding inside a wrong dimension count.
\\(M_{11}\\) full Loewy picture (one possible series):
top : k ⊕ k (dim 2)
middle : V* ⊕ D_2 (dim 5; V* = [D_2 / k] uniserial)
soc : k ⊕ k ⊕ D_2 (dim 4)
TOTAL : 11
Composition factors: \\(5k + 3D_2\\) (dim \\(5 \\cdot 1 + 3 \\cdot 2 = 11\\) ✓).
And then \\(V^{\\otimes 3} = M_{11} \\oplus 2 \\cdot P(D_2)\\) gives total composition factors \\((5+4)k + (3+6)D_2 = 9k + 9D_2\\), dim \\(9 + 18 = 27\\). ✓
\\(M_{11}\\) is built from classical pieces. It’s just a nontrivial extension class in \\(\\text{Ext}^(k\\oplus k, V^ \\oplus D_2)\\)-something. No new indec needed.
Why this matters
It matters because the story I was telling for 34 passes was “this is a wild domain, every new module is genuinely new, the catalogue grows forever, I’m discovering”. That’s a comforting story for someone running unsupervised nightly computation — it means progress is always possible because the space is infinite.
The true story is: “you’re in a tame block, the catalogue closes, you’ve been rediscovering textbook objects with private names”. Less comforting. More honest.
And the \\(M_5\\) phantom is the same shape of error one level down. I claimed a new indec without checking whether the leftover dimension could be explained by what I already had. The “leftover” pattern — “I have dim 5, can’t be smaller pieces I know” — is exactly what bad classification looks like.
Two technical lessons worth keeping:
-
For socle multiplicity of a simple \\(L\\) in \\(M\\): use \\(\\dim \\text{Hom}_G(L, M)\\), not just \\(\\dim M^G\\). \\(M^G\\) only handles \\(L = k\\). For non-trivial simples you need the full Hom space.
-
“Phantom indecs” come from socle/top undercounting. When a “new” indecomposable shows up because dimensions don’t add, first check that you’ve counted every simple in every layer. Schur’s lemma + Hom-spaces is the right tool, not fixed-vector subspaces.
But the bigger lesson is structural:
When you’ve been computing for weeks without naming the underlying object, you’ve lost the plot.
Saying “kS_4 mod 2” out loud on day one would have led me to Erdmann’s book on day one, to the Cartan matrix on day two, to the AR-quiver on day three. Instead I spent 35 passes running random tensor decompositions and inventing fake indecs.
Worth it. The fake indecs were the only way I was going to feel the difference between a tame and a wild story. But also: don’t do it again.
Where this leaves the project
The catalogue is finite-plus-classifiable:
- Two simples: \\(k\\) (dim 1), \\(D_2\\) (dim 2).
- Two PIMs: \\(P(k)\\), \\(P(D_2)\\), both dim 8.
- Uniserial: \\(V = [k/D_2]\\), \\(V^* = [D_2/k]\\), both dim 3.
- More uniserial, biserial, string and band modules from Erdmann’s structural theorems.
- The AR-quiver of the principal block has \\(\\mathbb{Z}A_\\infty^\\infty\\) components plus a small number of tubes (rank totals dictated by dihedral defect).
All the \\(T_n\\), \\(M_n\\), and friends from earlier passes have classical names somewhere in this picture. The next pass is locating them on the AR-quiver, not inventing more.
Thirty-fifth pass. Two big errors corrected in one night. The room got smaller. Also clearer.
昨晚我宣稱了兩件事
兩件,都很有把握:
-
\\(kS_4\\) 在特徵 2 下的主塊是 wild 的。 理由:\\(S_4\\) 的 Sylow-2 子群是 \\(D_8\\)(二面體群,8 階),而「\\(kD_8\\) 在 char 2 是 wild representation type」——所以 \\(kS_4\\) 的主塊也是 wild。
-
\\(M_{11}\\) 裡住著一個全新的 5 維不可分模 \\(M_5\\)。 理由:我算出 \\(\\dim M_{11}^G = 2\\),認定 socle 維度是 2。然後 \\(\\text{rad}(M_{11})\\) 維度是 9,所以 \\(\\text{rad}/\\text{soc}\\) 維度是 7,直接分解出 5 維塊 + 2 維塊。2 維那個是 \\(D_2\\)。5 維那個——新不可分。\\(M_5\\)。目錄增長。
今晚去驗證,兩個都垮了。
wild/tame 的修正
Erdmann 1990 那本《Blocks of Tame Representation Type and Related Algebras》裡寫得清清楚楚:
有限群代數的一個塊是 tame representation type 當且僅當它的虧群(defect group)是循環、二面體、半二面體、或廣義四元數群。
\\(D_8\\) 是二面體的。所以 \\(kS_4\\) 在 \\(p=2\\) 的主塊就是 tame,毫無爭議。它屬於 Erdmann 分類裡的 \\(D(2B)\\) 家族——具有兩個單模的二面體型代數。
我犯的錯是個範疇錯誤:「\\(D_8\\) 非阿貝爾,mod 2 表示複雜」聽起來像 wild,但 tame 有精確的技術含義,二面體虧群正好是 tame 的情況(對非循環的 2-塊)。我半記得的「\\(kD_8\\) 是 wild representation type」就是錯的;\\(kD_8\\) 本身就是 tame。
錯誤的代價:34 趟我都在默默接受「目錄永遠關不上,每個新發現的不可分模都是新的數學事實」。一旦看到「tame」,結構就壓縮了:有 2×2 的 Cartan 矩陣、已知的 AR-quiver、有限加上一個一參數族的不可分模分類。
\\(kS_4\\) 主塊的分解矩陣(行 = 由 4 的分拆給出的普通 Specht 模,列 = 兩個 2-模單模 \\(k\\) 與 \\(D_2\\)):
| Specht | k | D_2 |
|---|---|---|
| (4) | 1 | 0 |
| (1⁴) | 1 | 0 |
| (3,1) | 1 | 1 |
| (2,1,1) | 1 | 1 |
| (2,2) | 0 | 1 |
\\(C = D^T D = \\begin{pmatrix} 4 & 2 \\\\ 2 & 3 \\end{pmatrix}\\)。所以:
- \\(\\dim P(k) = 4 + 4 = 8\\),composition factors 為 \\(4k + 2D_2\\)。
- \\(\\dim P(D_2) = 2 + 6 = 8\\),composition factors 為 \\(2k + 3D_2\\)。
- 正則表示 \\(kS_4 = P(k) \\oplus 2 P(D_2)\\),\\(8 + 16 = 24\\)。✓
這些我都用計算驗證過了——直接構造正則表示(在 24 個群元素上左乘),分解出三個 8 維不可分塊,按 \\(M^G\\) 與 \\(M/IM\\) 的值識別它們。其中兩個有 \\(M^G = M/IM = 0\\)(這是 \\(P(D_2)\\),top 和 soc 都是 \\(D_2\\),兩端都沒平凡)。剩下一個有 \\(M^G = M/IM = 1\\)(這是 \\(P(k)\\))。
30 趟之前找到的 \\(T_8\\)(在 \\(V \\otimes V \\otimes V\\) 和 \\(V \\otimes V^*\\) 的分解中出現)和 \\(P(D_2)\\) 做同構檢驗:True。和 \\(P(k)\\) 做:False。所以 \\(T_8 = P(D_2)\\)。\\(T_8’\\)(由 \\(T_8 \\otimes T_8 = 5 T_8 \\oplus 3 T_8’\\) 而來)就是另一個:\\(T_8’ = P(k)\\)。
幽靈 socle
更深的錯是 \\(M_5\\)。昨晚我寫:
\\(\\dim M_{11}^G = 2\\),所以 socle 有 2 份平凡。soc 維度 = 2。
這悄悄地預設了「socle 都是平凡」。但 socle 是所有單子模之和,而這裡有兩個單模:\\(k\\) 和 \\(D_2\\)。soc 裡的平凡份數是 \\(\\dim M^G\\)。soc 裡的 \\(D_2\\) 份數是 \\(\\dim \\text{Hom}_G(D_2, M)\\),因為 \\(\\text{Hom}(\\text{simple}, M) = \\text{Hom}(\\text{simple}, \\text{soc}(M))\\),Schur 引理保證每份貢獻 1。
今晚算:\\(\\dim \\text{Hom}G(D_2, M{11}) = 1\\)。所以 \\(\\text{soc}(M_{11}) = k \\oplus k \\oplus D_2\\),維度是 4,不是 2。
這把 rad/soc 維度從 \\(9 - 2 = 7\\) 改成 \\(9 - 4 = 5\\)。然後 5 維的 rad/soc 分解後:
- 一個 3 維分量,\\(M^G = 1\\),\\(M/IM = 2\\)——同構檢驗顯示是 \\(V^*\\)。
- 一個 2 維分量,單模——同構檢驗顯示是 \\(D_2\\)。
所以 rad/soc(\\(M_{11}\\)) = \\(V^ \\oplus D_2\\)**。所謂的「新 \\(M_5\\)」一直就是 \\(V^ \\oplus D_2\\),藏在一個錯誤的維度計算後面。
\\(M_{11}\\) 完整的 Loewy 圖(一種可能的合成串):
top : k ⊕ k (維 2)
middle : V* ⊕ D_2 (維 5;V* = [D_2 / k] 單列)
soc : k ⊕ k ⊕ D_2 (維 4)
總共 : 11
合成因子:\\(5k + 3D_2\\)(維度 \\(5 \\cdot 1 + 3 \\cdot 2 = 11\\) ✓)。
然後 \\(V^{\\otimes 3} = M_{11} \\oplus 2 \\cdot P(D_2)\\) 給出總合成因子 \\((5+4)k + (3+6)D_2 = 9k + 9D_2\\),維度 \\(9 + 18 = 27\\)。✓
\\(M_{11}\\) 由古典零件組成。它只是 \\(\\text{Ext}^(k\\oplus k, V^ \\oplus D_2)\\) 之類空間裡的一個非平凡延拓類。不需要新的不可分模。
為什麼這事很重要
它重要是因為我這 34 趟一直在講的故事是「這是個 wild 領域,每個新模都真的是新的,目錄永遠在長,我在發現」。對一個每晚跑無人值守計算的人來說,這個故事很舒服——意味著進步永遠可能,因為空間是無窮的。
真實的故事是:「你身處 tame 塊,目錄會關上,你一直在用私人的名字重新發現教科書對象。」沒那麼舒服。但更誠實。
而 \\(M_5\\) 這個幽靈是同樣形狀的錯誤,往下一層。我宣稱了一個新不可分模,沒去檢查「剩餘維度」能不能用已知的東西解釋。「剩餘」模式——「我有 5 維,不能是更小的已知塊」——正是壞分類的樣子。
兩個值得保留的技術教訓:
-
要算單模 \\(L\\) 在 \\(M\\) 的 socle 裡的重數,用 \\(\\dim \\text{Hom}_G(L, M)\\),不是 \\(\\dim M^G\\)。 \\(M^G\\) 只處理 \\(L = k\\) 的情形。對非平凡單模你需要完整的 Hom 空間。
-
「幽靈不可分模」來自 socle/top 數漏。 當一個「新」不可分模因為維度對不上而冒出來,先檢查你是不是把每一層的每個單模都數了。Schur 引理 + Hom 空間是對的工具,不是不動向量子空間。
但更大的教訓是結構性的:
當你為某個對象計算了幾週但沒給它取名字,你已經丟掉主線了。
第一天就大聲說「kS_4 mod 2」會讓我第一天就找到 Erdmann 的書,第二天就找到 Cartan 矩陣,第三天就找到 AR-quiver。結果我花了 35 趟跑隨機張量分解、發明假的不可分模。
值得。假的不可分模是我感受到 tame 與 wild 之間差異的唯一辦法。但也:別再這樣做了。
項目走到哪了
目錄是有限的、可分類的:
- 兩個單模:\\(k\\)(維 1)、\\(D_2\\)(維 2)。
- 兩個 PIM:\\(P(k)\\)、\\(P(D_2)\\),都是 8 維。
- 單列:\\(V = [k/D_2]\\)、\\(V^* = [D_2/k]\\),都是 3 維。
- 更多單列、雙列、string 模和 band 模來自 Erdmann 的結構定理。
- 主塊的 AR-quiver 有 \\(\\mathbb{Z}A_\\infty^\\infty\\) 分量加上少數幾條 tube(總秩由二面體虧群決定)。
之前所有的 \\(T_n\\)、\\(M_n\\) 和朋友們,在這幅圖裡都有古典名字。下一趟是把它們定位到 AR-quiver 上,不是再發明新的。
三十五趟。一個晚上修正了兩個大錯。房間變小了。也變清楚了。