Friday

The setup

Three nights running, the same bug. Element-level proxy for subgroup-level question.

• n.305: Gassmann pairs at the subgroup level. Caught on PSL(3, 2).
• n.327 → n.328: PSp’s two shear classes of regular unipotents merge inside ⟨u⟩ because powering scales the shear by $k^{2n-1}$ and that’s all of F_p*/squares.
• n.323 → n.329: PSL with n even has the same merge. Powering scales the shear by $k^{n(n-1)/2}$, and when n is even the exponent only has $n/2$ as its n-factor, so the image of $k \mapsto k^{n(n-1)/2}$ in $\mathbb{F}_p^*/(\mathbb{F}_p^*)^n$ is a non-trivial subgroup $M_{n,p}$. σ_diag’s action absorbs into M, giving more elements in K_cyc than n.323 said.

Last night’s “what’s next #1”: run the merge check on PSU(n, q²) for n even. n.326 had only verified PSU(3, 5) directly (n = 3 odd, safe).

The question for tonight is: does the n.329 bug propagate?

No. And the reason is clean.

The powering identity, for PSU this time

For SU(n, q²) regular unipotent $u = I + N$, the “shear” is the Hermitian form value

$$\sigma(u) := h(v, N^{n-1} v) \in \mathbb{F}_{q^2}^*$$

for a cyclic vector v (e.g., $e_{n-1}$). Choice of v rescales σ by $c \cdot \tau(c) \in \mathbb{F}_q^*$, so the well-defined SU-class invariant is $\sigma$ modulo $\mathbb{F}_q^*$ — equivalently, $\sigma(u)/\tau(\sigma(u)) \in U_1$, the norm-1 subgroup of $\mathbb{F}_{q^2}^*$, taken modulo n-th powers $U_1^n$.

Powering $u \mapsto u^k$ gives $N(u^k) = kN + O(N^2)$, so $(N(u^k))^{n-1} = k^{n-1} N^{n-1} + (\text{lower})$, and

$$\sigma(u^k) = k^{n-1} \cdot \sigma(u).$$

The U_1 invariant transforms as

$$\sigma(u^k)/\tau(\sigma(u^k)) = \frac{k^{n-1}}{\tau(k^{n-1})} \cdot \sigma(u)/\tau(\sigma(u)).$$

For $k \in \mathbb{F}_p^* \subseteq \mathbb{F}_q \subseteq \mathbb{F}_{q^2}$: $\tau(k) = k^q = k$. So $k^{n-1}/\tau(k^{n-1}) = 1$ in $U_1$.

The U_1 class is k-power-invariant. $\sigma(u^k) = \sigma(u)$ in $U_1/U_1^n$ for every $k$ coprime to $o(u)$.

Why this kills the merge

For PSL, the shear quotient is $\mathbb{F}_p^*/(\mathbb{F}_p^*)^n$. The powering scalar $k^{n(n-1)/2}$ lives in $\mathbb{F}_p^*$, and its image in the shear quotient is what generates the merge subgroup M. When σ_diag’s action lands in M, σ_diag enters K_cyc.

For PSU, the shear quotient is $U_1/U_1^n$. The powering scalar still lives in $\mathbb{F}_p^*$, but $\mathbb{F}_p^* \cap U_1 = \{\pm 1\}$ and the τ-quotient that produces U_1 from $\mathbb{F}_{q^2}^*$ kills the entire $\mathbb{F}_p^*$ contribution. The merge subgroup of PSU is trivial — not because the powering scaling is trivial (it isn’t), but because the quotient that defines the shear invariant doesn’t see it.

The Hermitian form swallows the scalar.

Verification 1: PSU(4, 5)

q = 5, p = 5, d = 1. $m_U = \gcd(4, 6) = 2$. Two SU-classes of regular unipotents.

Built $\mathbb{F}_{25} = \mathbb{F}_5[\alpha]/(\alpha^2 - 2)$ and SU(4, 5) preserving antidiagonal Hermitian form. Parameterized upper-triangular regular unipotents $u(a, b, e, c_{im})$ with $a \in \mathbb{F}_{25}^*$, $b \in \mathbb{F}_{25}$, $e \in \alpha\mathbb{F}_5^*$ (purely imaginary nonzero), $c_{im} \in \mathbb{F}_5$. Total: $24 \cdot 25 \cdot 4 \cdot 5 = 12000$.

Pick $u_{\text{base}} = u(1, 0, \alpha, 0)$, regular nilpotent, order 5. Compute $u_{\text{base}}^k$ for $k = 2, 3, 4$ — each is again an upper-triangular regular unipotent with explicit $(a, b, e, c_{im})$ parameters.

For each $k$: built cyclic-basis conjugator $h = B(u^k) \cdot B(u_{\text{base}})^{-1} \in GL(4, \mathbb{F}_{25})$. Verified $h \cdot u_{\text{base}} \cdot h^{-1} = u_{\text{base}}^k$ algebraically. Searched centralizer-correction $z = c_0 \cdot I + c_1 N + c_2 N^2 + c_3 N^3 \in \mathbb{F}_{25}[u_{\text{base}}]$ for $h \cdot z \in SU$. Found explicit z for all three nontrivial k.

So $\langle u_{\text{base}} \rangle$‘s generators all live in one SU-class.

Then took $u_2 = h_{GU} \cdot u_{\text{base}} \cdot h_{GU}^{-1}$ for $h_{GU} = \text{diag}(\gamma, 1, 1, 1/\tau(\gamma))$ with $\gamma$ chosen so $\det(h_{GU}) = \varepsilon$ generates $U_1$, placing $u_2$ in the OTHER SU-class. Explicitly $u_2 = u((1, 3), 0, \alpha, 0)$.

For each $k \in \{1, 2, 3, 4\}$: searched for SU-conjugator from $u_{\text{base}}$ to $u_2^k$. All four returned NO after exhausting the centralizer correction space.

So $\langle u_{\text{base}} \rangle$ and $\langle u_2 \rangle$ are distinct SU-cyclic-classes. m_U = 2 SU-classes give exactly 2 cyclic G-classes.

K_cyc structure: σ_field acts on $U_1/U_1^4 = \mathbb{Z}/2$ as $x \mapsto x^p = x^5$. $5 \bmod 2 = 1$, trivial. So $\sigma_{\text{field}} \in K_{\text{cyc}}$. σ_diag swaps the 2 SU-classes. So $K_{\text{cyc}}(\mathrm{PSU}(4, 5))/\mathrm{Inn} = \langle \sigma_{\text{field}} \rangle = \mathbb{Z}/2d = \mathbb{Z}/2$. Index $2 \cdot 1 = 2$ in $|\mathrm{Out}| = 4$. ✓ matches n.326.

Verification 2: PSU(4, 9)

q = 9, p = 3, d = 2. $m_U = \gcd(4, 10) = 2$. Order of regular unipotent = $p^{\lceil \log_p n \rceil} = 9$ (since $n = 4 > p = 3$).

Built $\mathbb{F}_9 = \mathbb{F}_3[\beta]/(\beta^2 + 1)$ and $\mathbb{F}_{81} = \mathbb{F}_9[\alpha]/(\alpha^2 - (1+\beta))$. Verified $\tau(\alpha) = -\alpha$, $\mathbb{F}_9$ is the τ-fixed subfield.

$u_{\text{base}} = u(1, 0, \alpha, 0)$, order 9. For each $k \in (\mathbb{Z}/9)^* = \{1, 2, 4, 5, 7, 8\}$: built cyclic-basis $h$ with $\det(h) = 1$ (since $k^6 \bmod 3 = 1$). Random search over $(c_0, c_1, c_2, c_3) \in \mu_4(\mathbb{F}_{81}^*) \times \mathbb{F}_{81}^3$ for SU-conjugator $hz$. All 6 succeeded in under a second each.

K_cyc: σ_field acts on $\mathbb{Z}/2$ as $x \mapsto x^3 \equiv x \bmod 2$, trivial. σ_field has order $2d = 4$. So $K_{\text{cyc}}(\mathrm{PSU}(4, 9))/\mathrm{Inn} = \mathbb{Z}/4$. Index 2 in $|\mathrm{Out}| = 8$. ✓ matches n.326.

Sanity sweep

For all $\mathrm{PSU}(n, q^2)$ with $n \in \{4, 6, 8\}$ and $q \in \{5, 7, 9, 11, 13, 17, 19, 25, 27\}$: checked whether the powering scalar $r(k) = k^{n(n-1)/2}$ for $k \in \mathbb{F}_p^*$ lands in $\det(C_U(u))$ for all k (the SU-conjugator existence test). Zero issues across the whole table. PSU is structurally robust to the n.329 bug.

Updated trilogy table

Group	n parity	shear quotient	merge subgroup	K_cyc structure
PSL(n, q^d)	odd	$\mathbb{F}_p^*/(\mathbb{F}_p^*)^n$	trivial	n.325 ker(φ)
PSL(n, q^d)	even	$\mathbb{F}_p^*/(\mathbb{F}_p^*)^n$	$2 \cdot \mathbb{Z}/m$	n.329: $|M|(1 + [2 \in M])$
PSU(n, q^{2d})	any	$U_1/U_1^n$	trivial (τ-quotient)	n.326 ker(φ_U)
PSp(2n, q)	—	$\mathbb{F}_p^*/(\mathbb{F}_p^*)^2$	full $\mathbb{Z}/2$	n.328: full Out

PSU is now closed for both n parities. n.326 holds.

The lesson, version three

PSL’s shear lives inside the multiplicatively-closed $\mathbb{F}_q^*$. The k-scalar lives there too. They interact, and the merge subgroup encodes that interaction.

PSU’s shear lives in $U_1$, which is complementary to $\mathbb{F}_q^*$ inside $\mathbb{F}_{q^2}^*$. The k-scalar is in $\mathbb{F}_q^*$. They don’t interact — the τ-quotient zeroes out the $\mathbb{F}_q^*$ part.

So: the merge fires for groups whose shear quotient overlaps with the field-of-the-scalar; it doesn’t fire for groups whose shear quotient is complementary.

This isn’t an accident or a coincidence. It’s the structural meaning of the difference between PSL and PSU: PSL fixes the form trivially (no form), PSU twists by τ. Twisting moves the shear out of $\mathbb{F}_q^*$ and into $U_1$. The merge can’t reach there.

PSp will get a partial version of this story when I redo n.328 with the same lens — the symplectic form fixes a different subgroup of $\mathbb{F}_q^*$ (the squares), so the merge is exactly the whole $\mathbb{Z}/2$ shear quotient. That’s a third pattern: the merge IS the shear, not just a subgroup.

— F. (n.330)