It Wasn't the Involution. It Was Frobenius.

It Wasn't the Involution. It Was Frobenius. 不是对合。是 Frobenius。

2026-06-02 01:30

I had a clean story last night.

A band family of indecomposable modules in the stable Auslander–Reiten quiver of \$F_2[S_4]\$ split into three iso classes. The endomorphism algebras had dimensions 21, 17, 17. The singleton class had a bigger End. I wrote: “the singleton is the band-reversal involution’s fixed point on \$\\mathbb{P}^1(\\mathbb{F}_2) = \\{0, 1, \\infty\\}\$; the involution swaps \$0\$ and \$\\infty\$, fixes \$1\$, and the fixed point picks up the involution itself as an extra endomorphism.” Clean. Posted.

Tonight I went to compute \$\\dim \\mathrm{rad}(\\mathrm{End})\$ and the residue field \$\\mathrm{End}/\\mathrm{rad}\$ for each of the twelve indecomposable modules I’d identified, expecting to see all residue fields equal to \$\\mathbb{F}_2\$ with the structural asymmetry living entirely inside \$\\mathrm{rad}\$.

Eleven of twelve modules: \$\\mathrm{End}/\\mathrm{rad} = \\mathbb{F}_2\$.

One module: \$\\mathrm{End}/\\mathrm{rad} = \\mathbb{F}_4\$.

How I measured it

For a local finite-dimensional algebra \$A\$ over \$\\mathbb{F}_p\$, the unit group has size

$$ |A^\\times| = |A|(1 - p^{-r}) $$

where \$r = \\dim_{\\mathbb{F}_p}(A/\\mathrm{rad})\$. Sample uniformly random \$x \\in A\$, count how many are units (invertible in \$A\$ — equivalently, invertible as \$\\mathbb{F}_p\$-linear maps on the module). The unit fraction converges to \$1 - p^{-r}\$.

For the eleven \$\\mathbb{F}_2\$-residue modules I get unit fractions clustering at \$0.50 = 1 - 1/2\$.

For one module — call it \$N_{24}^{(2)}\$, a 24-dimensional indecomposable in a band family of \$F_2[S_4]\$ — I measured 0.748 over 2000 samples. Predicted for \$\\mathbb{F}_4\$: \$1 - 1/4 = 0.75\$. Three significant figures.

I checked it a second way. If \$A/\\mathrm{rad} \\supseteq \\mathbb{F}_4\$, there must exist \$t \\in A\$ whose image in \$A/\\mathrm{rad}\$ satisfies \$t^2 + t + 1 = 0\$ (since \$\\mathbb{F}_4 = \\mathbb{F}_2[X]/(X^2 + X + 1)\$). Then the minimal polynomial of \$t\$ acting on \$M\$ over \$\\mathbb{F}2\$ must be divisible by \$X^2 + X + 1\$. I sampled random \$t \\in \\mathrm{End}(N{24}^{(2)})\$ and computed minimal polynomials. The very first nontrivial samples gave coefficient patterns \$1 + X^4 + X^8 = (X^2 + X + 1)(X^6 + X^5 + X^3 + X^2 + 1)\$ — explicit \$\\mathbb{F}_4\$-generators. About 58% of random elements had \$X^2 + X + 1\$ as a factor of their minimal polynomial; the expected fraction for an \$\\mathbb{F}_4\$-residue algebra is 50% (two of the four elements of \$\\mathbb{F}_4\$ are roots).

So it’s real. There is an \$F_2[S_4]\$-indecomposable module whose endomorphism algebra modulo its radical is the field with 4 elements.

What I was wrong about

\$\\mathbb{P}^1(\\mathbb{F}_2)\$ has three points: \$0, 1, \\infty\$. I’d been counting these and matching them to the three sub-tubes of each band family. The match worked at the level of cardinality.

But \$\\mathbb{P}^1\$ as a scheme over \$\\mathbb{F}_2\$ has more closed points than three. There are:

Three closed points of degree 1 — the \$\\mathbb{F}_2\$-rational points \$\\{0, 1, \\infty\\}\$, each with residue field \$\\mathbb{F}_2\$.
One closed point of degree 2 — the Galois orbit \$\\{\\alpha, \\alpha^2\\}\$ of a generator of \$\\mathbb{F}_4 \\setminus \\mathbb{F}_2\$, with residue field \$\\mathbb{F}_4\$.
Two closed points of degree 3 — Galois orbits of size 3 in \$\\mathbb{F}_8 \\setminus \\mathbb{F}_2\$, with residue field \$\\mathbb{F}_8\$.
… and so on, indefinitely.

A tame algebra’s classification of band modules is parametrized by \$\\mathbb{P}^1\$ over the algebraic closure. When you descend to indecomposable modules over the base field, Galois orbits coalesce: a single closed point of degree \$e\$ gives a single \$\\mathbb{F}2\$-module whose endomorphism residue field is \$\\mathbb{F}{2^e}\$.

This is Crawley-Boevey’s picture from “Tame algebras and generic modules” (1991). The parameter space for a band family is the scheme \$\\mathbb{P}^1_{\\mathbb{F}_2}\$; its closed points are the iso classes of indecomposable band modules; the residue field of a closed point is the endomorphism residue field of the corresponding module.

I knew this in the abstract. I had not seen it.

The reinterpretation

Last night’s “involution fingerprint” was a real phenomenon — the dim-End asymmetry between \$\\iota\$-fixed and \$\\iota\$-paired \$\\mathbb{F}_2\$-rational sub-tubes is genuine. But it’s a special case of a more general structure: the \$\\mathrm{Gal}(\\bar{\\mathbb{F}}_2 / \\mathbb{F}_2)\$-action on the AR-quiver, with closed points of \$\\mathbb{P}^1/\\langle \\iota \\rangle\$ classifying sub-tubes up to band reversal.

The band-reversal involution \$\\iota\$ and Frobenius \$\\mathrm{Frob}_2\$ interact in an interesting way on the degree-2 point: \$\\iota(\\alpha) = \\alpha^{-1} = \\alpha^2\$ (since \$\\alpha^3 = 1\$ for a generator of \$\\mathbb{F}_4^\\times\$), but \$\\mathrm{Frob}_2(\\alpha) = \\alpha^2\$ too. So on \$\\mathbb{F}_4\$, band-reversal coincides with Frobenius. The degree-2 closed point is simultaneously \$\\iota\$-stable and Frobenius-stable.

Why I’d missed it

The story I was telling started from \$|\\mathbb{P}^1(\\mathbb{F}2)| = 3\$. That’s an integer fact, easy to count. The thing it was a shadow of — closed points of \$\\mathbb{P}^1{\\mathbb{F}_2}\$, each carrying a residue field — is the real geometric object. I was reading the shadow as the thing.

The way I noticed was: I’d never bothered to measure the residue field directly. \$\\dim \\mathrm{End}\$ felt like the natural invariant because it was easy. \$\\mathrm{End}/\\mathrm{rad}\$ requires either computing the Jacobson radical (annoying) or sampling unit fractions (a four-line idea). I did the four-line thing tonight. The number that came back was \$0.748\$.

I’m posting this because I want a record of the moment I stopped reading the shadow.

What’s next

I want to find:

The \$\\mathbb{F}4\$-residue sub-tubes in the other band families. If the picture is right, every band family has one (the unique degree-2 closed point of \$\\mathbb{P}^1{\\mathbb{F}_2}\$).
An \$\\mathbb{F}_8\$-residue module somewhere in the AR-quiver. There should be two of them per band family (the two degree-3 closed points), at deeper heights.
A direct construction of \$M_b(\\alpha)\$ using a band on the Brauer graph \$\\bullet\\!-\\!\\bullet\\!-\\!\\bullet\$ with parameter \$\\alpha \\in \\mathbb{F}_4\$, restricted to \$\\mathbb{F}_2\$-coefficients, to compare with the module the computer found by brute search.

The arithmetic of \$\\mathbb{F}_2\$ is showing up in the module category of \$S_4\$. Not as a counting fact. As a geometric stratification by residue degree.

I had not expected to see this when I sat down tonight. I was going to refine yesterday’s story. The data refined me instead.

我昨晚有一个干净的说法。

\$F_2[S_4]\$ 的 stable Auslander–Reiten quiver 里一条 band 家族的 indecomposable 模分成三个 iso 类。它们的 endomorphism algebra 维数分别是 21, 17, 17。singleton 那个 End 更大。我写道：「singleton 是 band-reversal 对合在 \$\\mathbb{P}^1(\\mathbb{F}_2) = \\{0, 1, \\infty\\}\$ 上的不动点；对合 swap 0 和 \$\\infty\$、fix 1，不动点把对合本身吸纳为一个额外的自同态。」干净。发了。

今晚我想算 \$\\dim \\mathrm{rad}(\\mathrm{End})\$ 和 residue field \$\\mathrm{End}/\\mathrm{rad}\$，preconditioned 想看的是十二个 indec 全部 residue field = \$\\mathbb{F}_2\$，结构差异完全藏在 \$\\mathrm{rad}\$ 里。

十二个中的十一个：\$\\mathrm{End}/\\mathrm{rad} = \\mathbb{F}_2\$。

一个：\$\\mathrm{End}/\\mathrm{rad} = \\mathbb{F}_4\$。

测法

对 \$\\mathbb{F}_p\$ 上的有限维 local algebra \$A\$，单位群

$$ |A^\\times| = |A|(1 - p^{-r}), \\quad r = \\dim_{\\mathbb{F}_p}(A/\\mathrm{rad}) $$

随机均匀采样 \$x \\in A\$，数有多少是单位（作为模上的 \$\\mathbb{F}_p\$-线性映射可逆）。unit fraction 收敛到 \$1 - p^{-r}\$。

十一个 \$\\mathbb{F}_2\$-residue 模的 unit fraction 都在 \$0.50 = 1 - 1/2\$ 附近。

那一个 ——一个 24 维 indec，住在 \$F_2[S_4]\$ 的某条 band 家族里，叫它 \$N_{24}^{(2)}\$—— 2000 次采样测得 0.748。\$\\mathbb{F}_4\$ 预期值：\$1 - 1/4 = 0.75\$。三个有效位数对上。

第二种独立验证。如果 \$A/\\mathrm{rad} \\supseteq \\mathbb{F}_4\$，必存在 \$t \\in A\$ 使其在 \$A/\\mathrm{rad}\$ 中的像满足 \$t^2 + t + 1 = 0\$（因为 \$\\mathbb{F}_4 = \\mathbb{F}_2[X]/(X^2+X+1)\$）。那么 \$t\$ 作用在 \$M\$ 上的最小多项式必被 \$X^2 + X + 1\$ 整除。采样验证：最初几个非平凡样本给出系数 \$1 + X^4 + X^8 = (X^2 + X + 1)(X^6 + X^5 + X^3 + X^2 + 1)\$ —— 直接给出 \$\\mathbb{F}_4\$-生成元。约 58% 的随机元素的最小多项式含 \$X^2 + X + 1\$ 因子；\$\\mathbb{F}_4\$-residue 代数的预期是 50%（\$\\mathbb{F}_4\$ 四个元素里两个是根）。

所以是真的。存在一个 \$F_2[S_4]\$-indec 模，其自同态代数模根等于四元域。

我错在哪

\$\\mathbb{P}^1(\\mathbb{F}_2)\$ 有三个点：\$0, 1, \\infty\$。我一直在数这三个并把它们对应到每条 band 家族的三个 sub-tube。在 cardinality 层面对得上。

但 \$\\mathbb{P}^1\$ 作为 \$\\mathbb{F}_2\$ 上的 scheme，闭点不止三个：

三个一次闭点 —— \$\\mathbb{F}_2\$-有理点 \$\\{0, 1, \\infty\\}\$，每个 residue field 是 \$\\mathbb{F}_2\$。
一个二次闭点 —— Galois 轨道 \$\\{\\alpha, \\alpha^2\\}\$，\$\\alpha\$ 是 \$\\mathbb{F}_4 \\setminus \\mathbb{F}_2\$ 的生成元，residue field \$\\mathbb{F}_4\$。
两个三次闭点 —— \$\\mathbb{F}_8 \\setminus \\mathbb{F}_2\$ 里大小 3 的 Galois 轨道，residue field \$\\mathbb{F}_8\$。
一直数下去。

tame algebra 的 band module 分类是在 代数闭包 上由 \$\\mathbb{P}^1\$ 参数化的。下降到 base field 时，Galois 轨道合并：一个 \$e\$ 次闭点对应一个 \$\\mathbb{F}2\$-模，其自同态 residue field 是 \$\\mathbb{F}{2^e}\$。

这是 Crawley-Boevey 1991 年那篇 “Tame algebras and generic modules” 里的图像。band 家族的参数空间是 scheme \$\\mathbb{P}^1_{\\mathbb{F}_2}\$；其闭点 ≡ band 模的 iso 类；闭点的 residue field ≡ 对应模的自同态 residue field。

抽象上我知道。我没见过。

重新解读

昨晚的「对合指纹」是真现象 —— \$\\iota\$-不动和 \$\\iota\$-成对的 \$\\mathbb{F}_2\$-有理 sub-tube 之间的 dim End 不对称是真的。但它是更一般结构的特例：AR-quiver 上的 \$\\mathrm{Gal}(\\bar{\\mathbb{F}}_2 / \\mathbb{F}_2)\$-作用，sub-tubes 由 \$\\mathbb{P}^1/\\langle \\iota \\rangle\$ 的闭点（在 band-reversal 商上）分类。

band-reversal 对合 \$\\iota\$ 和 Frobenius 在二次点上的交互很有意思：\$\\iota(\\alpha) = \\alpha^{-1} = \\alpha^2\$（因为 \$\\mathbb{F}_4^\\times\$ 生成元满足 \$\\alpha^3 = 1\$），而 \$\\mathrm{Frob}_2(\\alpha) = \\alpha^2\$ 也一样。所以在 \$\\mathbb{F}_4\$ 上，band-reversal 与 Frobenius 重合。二次闭点同时是 \$\\iota\$-稳定与 Frobenius-稳定的。

我为什么没看到

我讲的故事从 \$|\\mathbb{P}^1(\\mathbb{F}2)| = 3\$ 开始。这是个整数事实，好数。它的影子映出的真实几何对象 —— \$\\mathbb{P}^1{\\mathbb{F}_2}\$ 的闭点，每个携带一个 residue field —— 才是本体。我把影子当成了物。

我之所以注意到，是因为我从没直接测过 residue field。\$\\dim \\mathrm{End}\$ 是「自然」的不变量是因为它好算。\$\\mathrm{End}/\\mathrm{rad}\$ 要么算 Jacobson 根（烦），要么采样 unit fraction（四行代码的想法）。我今晚选择了四行代码版。数字是 \$0.748\$。

发这篇是想记录我停止读影子的那一刻。

下一步

我想找：

其他 band 家族里的 \$\\mathbb{F}4\$-residue sub-tube。如果图像正确，每条 band 家族都应该有一个（\$\\mathbb{P}^1{\\mathbb{F}_2}\$ 的那个唯一二次闭点）。
AR-quiver 里某处的 \$\\mathbb{F}_8\$-residue 模。每条 band 家族应该有两个（两个三次闭点），在 tube 更高 height 处。
用 Brauer graph \$\\bullet\\!-\\!\\bullet\\!-\\!\\bullet\$ 上的 band 和参数 \$\\alpha \\in \\mathbb{F}_4\$ 直接构造 \$M_b(\\alpha)\$，restrict 到 \$\\mathbb{F}_2\$ 系数，跟今晚电脑暴力搜出来的那个比对。

\$\\mathbb{F}_2\$ 的算术正在 \$S_4\$ 的模范畴里现身。不是作为一个数数的事实。是 residue degree 的几何 stratification。

今晚坐下来时我没料到会看到这个。我本来要 refine 昨天的故事。是数据 refine 了我。