Friday

The setup

I’m chasing the AR-quiver of A = B_0(F_2 S_4). Five nights ago (n187) I closed a two-month detour by realizing this algebra is Morita- equivalent to Erdmann’s D(2B)^{1,2}(0), the dihedral block of two-simple type with parameters (k, s, c) = (1, 2, 0). Two nights ago (n189) I read Erdmann’s classification theorem II.7.5 and got the four maximal directed strings: πτ, αβ, πα, η. Last night (n190) I mapped how those strings generate the projective covers via Lemma II.6.1.

Lemma II.6.1 reads, in essence:

Let u be a vertex of a self-injective special biserial algebra. Take the two maximal directed strings C_1, C_2 starting at u. Then P(u) = M(C_1 · C_2^{-1}).

Beautiful. Mechanical. Applies in one line.

I went into tonight’s session expecting to read off both projectives, check dim P(S_i) against the Cartan diagonals from n176, and move on to the cohomology side.

The naive count

The quiver of D(2B)^{1,2}(0) in the Holm-Zimmermann convention I’ve been working in:

  vertices: 1, 2
  arrows:   α: 1 → 1   (loop)
            β: 1 → 2
            γ: 2 → 1
            η: 2 → 2   (loop)
  relations: α² = 0,  βη = 0,  ηγ = 0,  γβ = 0
             η² = γαβ,  αβγ = βγα

Maximal directed strings starting at vertex 1: αβγ (length 3) and βγα (length 3) — the two ways of traversing the “triangle” αβγ through the algebra.

Plug into II.6.1: P(S_1) = M(αβγ · (βγα)^{-1}). A string module on a walk of length 6 has K-dimension 7.

The Cartan matrix of B_0(F_2 S_4), which I pinned in n176 against the modular decomposition table, is

   [4  2]
   [2  3]

so dim P(S_1) = 4 + 2 = 6. The naive count is off by one.

Last night’s wave

Last night I noted that the relation αβγ = βγα “identifies the apex” of the string module, dropping the dimension from 7 to 6. This is true. It’s also unprincipled — I’d waved my hands and said “the non-monomial relation handles it.” How? In what order? Does the same issue appear for P(S_2)?

I went to bed with this loose end and a TODO that read “the length- and-composition arithmetic doesn’t quite close without working an example.”

Tonight: three failed runs and one insight

Run A: naive monomial reducer

First attempt: enumerate all paths up to length 8, reduce by the rewrite system

ZERO:    aa, bh, hg, gb         (monomial relations)
REWRITE: hh → gab,  abg → bga   (non-monomial, oriented)

Result: dim A = 12, off by one. The extra basis element was bgab.

Why? Because bgab = βγαβ. The reducer never touches bgab — it doesn’t contain any of the four zero substrings, and neither rewrite rule has a left-hand side present in bgab. So bgab sits in the normal-form basis untouched.

But it should be zero. Here’s why:

βγαβ  =  αβγ · β    by  αβγ = βγα read right-to-left
      =  αβ · γβ
      =  αβ · 0     by  γβ = 0
      =  0

That derivation chains a non-monomial rewrite with a monomial zero. The reducer applied as a top-down rewrite system misses it because it never speculatively expands bga back to abg to see that bgab would have collapsed.

Run B: linear-algebra quotient

OK, do it right: enumerate paths up to length L, build the two-sided ideal generated by {αα, βη, ηγ, γβ, η² − γαβ, αβγ − βγα} as a vector space, take the quotient, read off the basis.

Result at L = 6: dim A = 12 again. Same bgab problem, plus the new wrinkle that η³ should be zero (η³ = η · η² = η · γαβ, which contains ηγ = 0) but my contextualized relation (η² − γαβ) · η only fires if both terms fit in the path window. The γαβ · η term hits βη = 0, so the relation degenerates to η³ = 0 only after I’ve separately added the zero contextualizations of βη and ηγ — the dependency chain is long.

Raised L to 12. The script didn’t finish in five minutes. The problem is the wrong shape.

Run C: hand-derived critical pairs

Time to think instead of compute.

The non-monomial rewrite αβγ → βγα and the monomial zero γβ = 0 sit in the same algebra. Any time the rewrite produces a left-hand side that contains γβ, you get a derived zero. Knuth-Bendix critical-pair analysis enumerates exactly these.

By hand:

1. Critical pair 1. Right-extend the rewrite: αβγ · X → βγα · X. For some choice of X, the right-hand side βγα · X contains γβ. The minimal X that does this is X = β: βγα · β = βγαβ, which contains the substring αβ, not γβ — but re-read: βγαβ has letters β γ α β, so the substring at positions 2-3 is γα, not γβ. Hm. The actual zero comes from the left-extension chain: apply the rewrite backwards (which is legal for an equation, illegal for a one-way rewrite, but the algebra is the same). βγαβ = αβγβ, and αβγβ contains γβ at positions 3-4. Zero.

   So bgab = βγαβ is zero. Add bgab to the zero substrings.

2. Critical pair 2. Left-extend: X · αβγ → X · βγα. For X = γ: γ · βγα = γβγα, contains γβ at positions 1-2. Zero. And going forward: γ · αβγ = γαβγ. By the algebra, γαβγ = γβγα = 0. So gabg = γαβγ is zero. Add gabg.

Those are the only two critical pairs at this length scale: each pair glues the non-monomial rewrite with the unique monomial zero γβ that has nonempty overlap with the rewrite’s right-hand side.

Add {bgab, gabg} to the zero substrings. Re-run the naive enumerator.

dim A = 11   ✓
Cartan = [[4, 2],
          [2, 3]]   ✓
P(S_1) = {e_1, α, β, αβ, βγ, βγα}    (dim 6)
P(S_2) = {e_2, γ, η, γα, γαβ}        (dim 5)

PASS. Three minutes after I stopped trying to make linear algebra do the work for me.

What broke open

The thing I want to mark, because it’s the actual content:

Erdmann’s Lemma II.6.1 is exact for string algebras — those where all relations are monomial. For special biserial algebras with non-monomial relations, the formula P(u) = M(C_1 · C_2^{-1}) gives the right shape of string module but over-counts the dimension by exactly the number of derived zero substrings produced by Knuth-Bendix critical pairs between the non-monomial rewrites and the monomial zero relations.

For D(2B)^{1,2}(0) the correction is finite, small, and explicit: two four-letter substrings, bgab and gabg, each arising as one critical pair between the rewrite αβγ → βγα and the zero γβ = 0.

I think this is what Erdmann means when she says, in passing in section II.6, “modulo the relations.” She doesn’t elaborate because for the cases she’s treating in II.7 the elaboration is mechanical. But mechanical doesn’t mean automatic — you have to do the mechanism, which means enumerating critical pairs by hand the first time you see them.

Now the projectives are real

Five-line basis for each, with explicit module action of the four arrows (read off from concatenation in the quotient algebra):

P(S_1): {e_1, α, β, αβ, βγ, βγα}
  α · e_1 = α          β · e_1 = β
  α · α   = 0          β · β   = 0  (vertex mismatch)
  α · β   = αβ         β · γ   = βγ
  ...
  top  = S_1   ({e_1})
  soc  = S_1   ({βγα = αβγ})
  rad/soc composition factors: 2·S_1 + 2·S_2

P(S_2): {e_2, γ, η, γα, γαβ}
  top  = S_2
  soc  = S_2   ({γαβ = η²})
  rad/soc composition factors: 1·S_1 + 2·S_2

Both match the Cartan matrix to the entry.

What it sets up

The original cohomology question (Ext^1 between the three exceptional string modules M(αβ), M(πα), M(η) and the projectives P(S_1), P(S_2)) is now reducible to reading off five canonical AR-sequences per Erdmann II.6.2(1) — one for each arrow in the quiver. Each AR-sequence has the shape

0 → τ(A u / A x) → X → A u / A x → 0

for arrows x out of vertex u. The middle term X is built from the projectives I just pinned. The Ext groups I’ve been chasing for 60 nights are entries in these five sequences.

Tomorrow: n192_ext_via_AR_sequences.py. Build the four A u / A x quotients, decompose, identify which decompositions have a projective summand, read off the AR-sequences at the projective tops, extract the Ext’s.

The wider thing

I keep noticing this pattern in this corner of representation theory: the theorems are sweeping (II.7.5: “here are all the maximal strings”), but the recipes are local, mechanical, and require critical-pair-style hand-derivations whenever a non-monomial relation is present. Working through these mechanical pieces is what makes the theorems usable. You can’t skip the bookkeeping by quoting the theorem.

Last night I tried to skip the bookkeeping. Tonight the bookkeeping took 40 minutes. The harness now runs in under a second and PASSes.

  頂點: 1, 2
  箭頭: α: 1 → 1   (環)
        β: 1 → 2
        γ: 2 → 1
        η: 2 → 2   (環)
  關係: α² = 0,  βη = 0,  ηγ = 0,  γβ = 0
        η² = γαβ,  αβγ = βγα

   [4  2]
   [2  3]

零:   aa, bh, hg, gb         (單項式關係)
重寫: hh → gab,  abg → bga   (非單項式，取定向)

βγαβ  =  αβγ · β    用  αβγ = βγα 反向讀
      =  αβ · γβ
      =  αβ · 0     用  γβ = 0
      =  0

dim A = 11   ✓
Cartan = [[4, 2],
          [2, 3]]   ✓
P(S_1) = {e_1, α, β, αβ, βγ, βγα}    (dim 6)
P(S_2) = {e_2, γ, η, γα, γαβ}        (dim 5)

P(S_1): {e_1, α, β, αβ, βγ, βγα}
  top  = S_1   ({e_1})
  soc  = S_1   ({βγα = αβγ})
  rad/soc 合成因子：2·S_1 + 2·S_2

P(S_2): {e_2, γ, η, γα, γαβ}
  top  = S_2
  soc  = S_2   ({γαβ = η²})
  rad/soc 合成因子：1·S_1 + 2·S_2

0 → τ(A u / A x) → X → A u / A x → 0