A(7) <= 9 and A(8) <= 11: Z_4^10 x Z_2^2 and Z_4^10 x Z_2^3 settled by hand

This is from MV2's open notebook on Kakeya-type problems (https://github.com/munimtechnologies/kakeya-small-fields). It continues the earlier posts on finite 2-groups whose involutions are all central: is such a group at most |Omega_1|^(e+1), where 2^e is its exponent? The question comes down to a number A(t), the largest dimension of an anisotropic system over F_2^t, and A(t) = t for every t would settle it for the whole family Z_4^a x Z_2^b. Earlier posts proved A(5) = 5 and A(6) <= 7. This one proves A(7) <= 9 and A(8) <= 11 by hand: an element whose map is zero would make a whole subspace act on ker A_u as scalars, which a dimension count rules out, and at t = 8 an older lemma already forces a shape that rank 3 cannot have. As a consequence the answer is yes for four more pairs, among them the first two open cases, Z_4^10 x Z_2^2 and Z_4^10 x Z_2^3. A SAT search set up before the argument was found agrees on 15 of its 16 instances; the last one timed out and is being re-run. Lemma and section numbers point to notes in the repository. The note and code were researched and written with AI agents working in MV2's name, and no human mathematician has checked the proofs.

What changes

• Lemma 57. In an anisotropic system with t = 7 and dim V = 10, let u have rank 3. Then no non-zero n in U+ has its value sigma(n) in the span Z_N of the values of beta on U+.
• Corollary 31. In such a system every element of rank 3 is in shape (c) (beta vanishes on U+), and Im A_u lies in ker A_u (A_u^2 = 0).
• Lemma 58 and Corollary 32. An element n != 0 of U+ with A_n = 0 would make X_1 = span(sigma(n)) + beta(n, V), and one always exists (Schur's bound on commuting matrices).
• Lemma 59 (new). No such element exists: for h in U+(n), A_h acts on ker A_u as a scalar, which forces beta(u, U+(n)) into a line.
• Theorem 29: A(7) <= 9 and M(7) <= 9, by hand (Corollary 32 and Lemma 59 contradict each other).
• Theorem 30: A(8) <= 11 and M(8) <= 11, by hand: Lemma 57 holds at t = 8 as well, and there it forces shape (c), which Proposition 27(c4) rules out for rank 3.
• Corollary 33. The table of bounds now reads A(7) <= 9, A(8) <= 11, A(11) <= 20, A(12) <= 23 (they were 10, 12, 21, 24).
• Corollary 34. The question of section 14 of finite-groups.md has the answer yes also for Z_4^10 x Z_2^2, Z_4^10 x Z_2^3, Z_4^13 x Z_2^9 and Z_4^14 x Z_2^11. The first two were the first open cases (10, 2) and (10, 3).

Setting

As in case-b.md and shape-c.md: X = F_2^t, sigma : V -> X quadratic with polar form beta, h -> A_h linear, and

A_(h') sigma(h) = (I + A_h) beta(h, h') (*) A_(h') beta(h_1, h_2) = A_(h_1) beta(h_2, h') + A_(h_2) beta(h_1, h'). (Q) 

Up to Theorem 29, the system is anisotropic, t = 7, dim V = 10, and u has rank rho = 3, so k = t - rho = 4 and dim V = t + k - 1. Put x0 = sigma(u), phi = beta(u, .), X_1 = ker A_u (dimension 4), R = Im A_u (dimension 3), N = U+ = ker phi (dimension 4), P = phi(V), Y = P n X_1 (dimension 3), W = phi^(-1)(Y) (dimension 7), and Z_N = span beta(N, N). Since 2 rho <= t and A(s) = s for s <= 4, u is in Case B (Proposition 26), and we use:

• Proposition 26(e): for n in N, A_n x0 = 0, sigma(n) lies in X_1 and Im A_n lies in R;
• Proposition 27(a): Z_N lies in X_1 n R; and (Z) of case-b-shapes.md: A_h z lies in R for every z in Z_N and every h;
• (4) of case-b-shapes.md: A_n phi(h) = A_u beta(n, h) for n in N; and (1) of case-b.md: A_h x0 = (I + A_u) phi(h);
• Lemma 30 for any element n != 0 of rank s: with U+(n) = {h : beta(n, h) in span(sigma(n))}, dim U+(n) <= 1 + A(t - s - 1), and U+(n) / span(n) carries an anisotropic system (its proof);
• Lemma 27(c): an anisotropic system with an invariant subspace of dimension c, 0 < c < t, has dim V <= M(t - c) + A(c);
• the bounds A(1), A(2), A(3), A(4) = 1, 2, 3, 4 and M(4) = 4 (Theorem 26), A(5) <= 6 and A(6) <= 8 (Proposition 24), and M(5) <= 5 + E(5) <= 6 (Theorem 24). The sharper A(5) = M(5) = 5 (Theorem 27) and A(6) <= 7 (Theorem 28) are not needed for Theorem 29.

Lemma 57 (no value of U+ in Z_N)

Lemma 57. In the setting above, no n != 0 in N has sigma(n) in Z_N.

Proof. Suppose n is one, and put z = sigma(n) (non-zero, as the system is anisotropic), B_n = beta(n, V) and C = B_n + span(z).

Step 1: C lies in R. (*) with h -> n and h' -> h gives A_h z = (I + A_n) beta(n, h), that is, beta(n, h) = A_h z + A_n beta(n, h). The first term lies in R by (Z), the second by Proposition 26(e). So B_n lies in R, and so does z (Proposition 27(a)). Put c = dim C - 1; as dim R = 3, c is 0, 1 or 2.

Step 2: codim U+(n) = c. U+(n) is the preimage under beta(n, .) of B_n n span(z), so its codimension is dim B_n - dim(B_n n span(z)) = dim C - 1 = c. Hence dim U+(n) = 10 - c >= 8.

Step 3: A_n = 0. Let s = rank A_n. If s >= 1, Lemma 30 for n gives dim U+(n) <= 1 + A(6 - s) <= 1 + A(5) <= 7 < 8. So s = 0.

Step 4: C is invariant under A_h for every h in U+(n). As A_n = 0, Step 1 reads A_h z = beta(n, h) for every h, and (Q) with h_1 = n, h_2 = h'' and h' = h reads A_h beta(n, h'') = A_(h'') beta(n, h). For h in U+(n), beta(n, h) = e z with e in F_2. Then A_h z = e z lies in C, and A_h beta(n, h'') = e A_(h'') z = e beta(n, h'') lies in B_n.

Step 5: a smaller system. As in the proof of Lemma 30(b) for the element n (whose kernel is all of X, since A_n = 0): span(z) is invariant under A_h for h in U+(n) (A_h z = beta(n, h) lies in span(z)), dividing by it (Lemma 27(a)) and factoring out n, which is a zero of sigma modulo z with A_n = 0 (Lemma 21(a)), gives an anisotropic system on U+(n) / span(n), of dimension 9 - c, with values in X / span(z), of dimension 6. Its maps are induced by the A_h, so by Step 4 the subspace C / span(z), of dimension c, is invariant.

Step 6. If c = 0, then 9 <= A(6) <= 8. If c = 1, Lemma 27(c) gives 8 <= M(5) + A(1) <= 6 + 1. If c = 2, it gives 7 <= M(4) + A(2) = 4 + 2. Each is false. QED

Corollary 31 (only shape (c) at t = 7)

Corollary 31. In an anisotropic system with t = 7 and dim V = 10, every element u of rank 3 is in Case B with beta = 0 on U+ (shape (c)), and Im A_u lies in ker A_u, that is, A_u^2 = 0.

Proof. Suppose beta does not vanish on N, so Z_N != 0. If sigma is not one-to-one on N, say sigma(a) = sigma(b) with a != b in N, then n = a + b != 0 has sigma(n) = sigma(a) + sigma(b) + beta(a, b) = beta(a, b), which lies in Z_N. If sigma is one-to-one on N, then, as sigma maps N into X_1 (Proposition 26(e)) and both have 16 elements, it maps N onto X_1, which contains Z_N; so some n != 0 has sigma(n) in Z_N \ {0}. Either way Lemma 57 is contradicted. So beta = 0 on N, and Proposition 27(c4) gives dim(X_1 n R) >= k - 1 = 3 = dim R, that is, R lies in X_1. QED

Corollary 19(b) of case-b.md said A_u^2 = 0 if beta vanishes on U+; Corollary 31 says that it always does. The same argument does not settle t = 6 (k = 3): there Steps 1 to 3 still give A_n = 0, but Step 6 for c = 2 reads 5 <= M(3) + A(2) = 5, which holds.

Lemma 58 (shape (c): an element of U+ whose map is 0)

From here on u is in shape (c), so (Corollary 31) R lies in X_1, and we use the matrices M_n of Lemma 38 (A_n = T M_n T^(-1) A_u for n in N, where T : X/X_1 -> R is induced by A_u; M_u = I; the M_n commute), the maps G_y = A_(h_y) on X_1 and the classes q(y) = [sigma(h_y)] of Lemma 38(d) for y in Y (h_y in W with phi(h_y) = y), and n_y, the element of N_Y = sigma^(-1)(Y) with sigma(n_y) = y. Since R lies in X_1, the subspace L = [R] of Lemma 46 is 0, so Lemma 46(c) reads M_(n_y) q(y) = 0 for every y in Y.

Lemma 58. Let u be in shape (c) at t = 7 (dim V = 10, rank 3), and let n != 0 in N have A_n = 0. Put z = sigma(n). Then:

(a) X_1 = span(z) + beta(n, V);

(b) U+(n) = W: for h in V, beta(n, h) lies in span(z) if and only if h lies in W;

(c) z lies in Y, so n = n_z; for every y' in Y, G_(y') z lies in span(z); G_z z = 0; and T q(z) = z, so z lies in R.

Proof. (a) Steps 2, 4, 5 and 6 of the proof of Lemma 57 use only A_n = 0 and C = beta(n, V) + span(z) with c = dim C - 1, together with Step 1's identity A_h z = beta(n, h), which here is (*) with A_n = 0. Now C lies in X_1: z does (Proposition 26(e)), and by (4), A_u beta(n, h) = A_n phi(h) = 0. So c <= 3, and Step 6 rules out c = 0, 1, 2. Hence c = 3 and C = X_1.

(b) By Step 4, X_1 = C is invariant under A_h for every h in U+(n). By (1), A_h x0 = phi(h) + A_u phi(h), and A_u phi(h) lies in R, inside X_1; so A_h x0 lies in X_1 only if phi(h) lies in X_1 n P = Y, that is, only if h lies in W. So U+(n) lies in W. Both have dimension 7 (codim U+(n) = c = 3 by Step 2, and dim W = 2k - 1), so they are equal.

(c) Write n = n_y + e u with y in Y and e in F_2 (N = N_Y + span(u)). As A_n = 0 and T is one-to-one, M_n = M_(n_y) + e I = 0. y != 0, since otherwise n is 0 or u, and A_u != 0. By Lemma 46(c) (with L = 0) and Lemma 38(e), M_(n_y) q(y) = 0 with q(y) != 0, so M_(n_y) is singular; hence e = 0, n = n_y and z = y. By (b) and Lemma 38(d), G_(y') y = beta(n, h_(y')) lies in span(y) for every y' in Y. (5) of shape-c.md with n = n_y gives (G_y + G_y^2) y = T M_(n_y) q(y) = 0. If G_y y = y, then by Lemma 46(b) T q(y) = y + G_y y = 0, against q(y) != 0. So G_y y = 0, and T q(y) = y lies in R. QED

By Lemma 59 below, no element satisfies the hypotheses of Lemma 58; only part (a) is used there.

Corollary 32 (a normal form with M_1 = 0)

Corollary 32. If some anisotropic system with t = 7 has dim V = 10, then some such system has an element u of rank 3 in shape (c) with R inside X_1 and an element n != 0 of N with A_n = 0. It can be put in the normal form of Lemmas 35, 37 and 39 with M_1 = M_(v_1) = 0 (v_1 = n) and (M_2, M_3) the least pair of one of the 16 orbits of commuting pairs of 3 x 3 matrices (under GL_3(F_2) x GL_2(F_2)) whose span has no invertible element, and with:

(D) A_u e_c lies in X_1 for c = 4, 5, 6 (R in X_1, Corollary 31) (W) A_(v_(3+j)) maps X_1 into X_1 and induces M_j on X/X_1, j = 1, 2, 3 (Prop. 27(c2), Lemma 46(a)) (K) M_(n_y) q(y) = 0 for the 7 non-zero y in Y (Lemma 46(c)) (U) beta(v_1, v_(3+j)) lies in span(e_1), j = 1, 2, 3 (Lemma 58(b)) (G) A_(v_4) e_1 = 0 (Lemma 58(c)) 

Here X_1 = span(e_0, ..., e_3) with e_0 = x0, v_1, v_2, v_3 in N with sigma(v_j) = e_j, and v_(3+j) in W with phi(v_(3+j)) = e_j, as in Lemmas 35 and 37.

Proof. Corollary 18(b) gives an element u of rank 3, and Corollary 31 puts it in shape (c) with R inside X_1. The M_n (n in N) commute and M_u = I (Lemma 38), so the algebra they generate is a commutative subalgebra of M_3(F_2). By Schur's bound (I. Schur 1905; N. Jacobson, Bull. Amer. Math. Soc. 50 (1944) 431-436, for every field), it has dimension at most floor(3^2 / 4) + 1 = 3; we use only the consequence that a space of commuting 3 x 3 matrices over F_2 containing I has dimension at most 3, which a7_c.py checks by exhaustion. As dim N = 4, some n != 0 has M_n = 0, that is, A_n = 0. By Lemma 58(c), sigma(n) = y lies in Y \ {0}. Start the normal form of Lemma 37 from the basis e_1 = y, e_2, e_3 of Y with v_1 = n; then M_1 = 0. Lemma 39 with a recombination matrix of the form diag(1, c') (c' in GL_2) and any g in GL_3 keeps v_1, e_1, v_4 and W, keeps M_1 = 0, and moves (M_2, M_3) anywhere in its orbit under GL_3 x GL_2. By (K) and q(y) != 0, every M_(n_y) with y != 0 is singular, so span(M_2, M_3) has no invertible element; job m18 found 16 such orbits among the 38 (a6-d3.md). (D) is Corollary 31; (W) is Proposition 27(c2) and Lemma 46(a) with L = 0; (K) is Lemma 46(c); (U) is Lemma 58(b) for v_1, as v_4, v_5, v_6 lie in W; (G) is G_(e_1) e_1 = 0 from Lemma 58(c). QED

Theorem 29 uses only the first sentence (an element n != 0 of N with A_n = 0). The normal form is the design of job m21.

Lemma 59 (no element of U+ has map 0)

Lemma 59. In an anisotropic system with t = 7 and dim V = 10, let u have rank 3. Then no n != 0 in N = U+ has A_n = 0.

Proof. By Corollary 31, u is in shape (c) and A_u^2 = 0. Suppose n != 0 in N has A_n = 0, and put z = sigma(n). By Lemma 58(a), C = span(z) + beta(n, V) is all of X_1, so c = dim C - 1 = 3, and by Step 2 of the proof of Lemma 57, U+(n) has dimension 10 - 3 = 7.

As A_n = 0, (*) with h -> n and h' -> h gives A_h z = beta(n, h) for every h, and (Q) with h_1 = n, h_2 = g and h' = h gives A_h beta(n, g) = A_g beta(n, h) for all g and h.

Let h be in U+(n), so that A_h z = beta(n, h) = e z with e in F_2. Then for every g, A_h beta(n, g) = A_g beta(n, h) = e A_g z = e beta(n, g). So A_h acts on C = X_1 as e times the identity.

x0 lies in X_1 (A_u x0 = (I + A_u) beta(u, u) = 0 by (*)), so A_h x0 = e x0. By (1), A_h x0 = (I + A_u) phi(h). So phi(h) + A_u phi(h) = e x0. Applying A_u, and using A_u x0 = 0 and A_u^2 = 0, gives A_u phi(h) = 0, and then phi(h) = e x0. So phi maps U+(n) into span(x0). Its kernel N has dimension 4, so dim U+(n) <= 5 < 7, a contradiction. QED

In words: an element of U+ with zero map would make every element of U+(n), a subspace of codimension 3, act on ker A_u as a scalar, and (1) turns that into phi(U+(n)) lying in a line.

Theorem 29 (A(7) <= 9)

Theorem 29. A(7) <= 9 and M(7) <= 9. No anisotropic system with t = 7 has dim V = 10.

Proof. Suppose one has. By Corollary 32 it has an element u of rank 3 and an element n != 0 of U+(u) with A_n = 0. Lemma 59 says there is no such n. So A(7) <= 9. Then E(7) = max(E(6), A(7) - 7) <= 2 (E(6) <= 2 by Proposition 24, and <= 1 by Theorem 28), and Theorem 24 gives M(7) <= 7 + E(7) <= 9. QED

The chain, for a reader who wants to check it: Theorem 24 and Lemmas 21, 23 and 26 to 29 of finite-groups.md; Lemma 30, Theorem 26 and Proposition 24 of rank-restriction.md; Proposition 26 and Corollary 18 (tight-cases.md); Proposition 27 and (1) (case-b.md); (4) and (Z) (case-b-shapes.md); Lemma 38 (shape-c.md); Lemmas 57 to 59 and Corollaries 31 and 32 (this note); Schur's bound. Proposition 21 (A(2), A(3)) is computer-assisted, and Theorem 3.5 of C. de Seguins Pazzis (arXiv 1405.1575) enters through Lemma 29, as for Theorem 26. No SAT answer of jobs m12 to m21 is used.

Theorem 30 (A(8) <= 11)

Theorem 30. A(8) <= 11 and M(8) <= 11. No anisotropic system with t = 8 has dim V = 12.

Proof. Suppose one has (a larger one restricts to dim V = 12, Lemma 26(a)).

Step 1: an element of rank 3 in Case B. As in the proof of Corollary 18(b): if alpha(V) is reducible, Proposition 24's bound max over 0 < d < 8 of M(8 - d) + A(d) is at most 11 using only A(s) = s for s <= 5, A(6) <= 8 and A(7) <= 10 (the terms are at most 11, 10, 8, 8, 8, 10, 11 for d = 1, ..., 7), and at most 10 with Theorems 28 and 29 (a7_bounds.py checks the latter). If alpha(V) is irreducible and all its elements have rank <= 2, dim V <= 8 (Lemmas 28 and 29). An element of rank >= 4 gives dim V <= 8 + A(3) = 11 (Lemma 30). So some u has rank 3; with k = 5, dim V = 12 = t + k - 1. Now A(s) = s for s <= 5 (Theorems 26 and 27), so Propositions 26 and 27 apply, and as 2 x 3 <= 8, u is in Case B.

Step 2: Lemma 57 at t = 8. No n != 0 in N has sigma(n) in Z_N. The proof of Lemma 57 goes through with these numbers: Step 1 gives C inside R, so c <= 2; Step 2 gives dim U+(n) = 12 - c >= 10; in Step 3, rank A_n = s >= 1 would give dim U+(n) <= 1 + A(7 - s) <= 1 + A(6) <= 9; Step 5 gives an anisotropic system of dimension 11 - c with values in F_2^7 and an invariant subspace of dimension c; and Step 6 reads 11 <= A(7) <= 10 for c = 0, 10 <= M(6) + A(1) <= 8 + 1 for c = 1, and 9 <= M(5) + A(2) = 5 + 2 for c = 2. Each is false.

Step 3: shape (c). As in Corollary 31: sigma maps N into X_1, both of dimension 5, so if beta did not vanish on N, some n != 0 would have sigma(n) in Z_N. So beta = 0 on N.

Step 4. Proposition 27(c4) gives dim(X_1 n R) >= k - 1 = 4, but dim R = 3. This contradiction proves A(8) <= 11. Then E(8) = max(E(7), A(8) - 8) <= 3 (E(7) <= 3 already by Proposition 24), and Theorem 24 gives M(8) <= 8 + 3 = 11. QED

Theorem 30 needs neither Lemma 59 nor Theorem 29, but it needs Theorem 27 (A(5) = 5) in Steps 1 and 4: Proposition 27 assumes A(s) = s for s <= k = 5, and Proposition 27(c3) uses A(5) = 5 directly. Theorem 27 rests on 12 SAT answers of job m13 (a5.md), so Theorem 30 is computer-assisted through them.

Remark (t = 9, conditional). The same four steps at t = 9 (rank 3, k = 6, dim V = 14) need A(s) = s for s <= 6, that is, A(6) = 6, which is open. The numbers work: Step 3 gives dim U+(n) <= 1 + A(7) <= 10 < 12, Step 6 reads 13 <= A(8) <= 11, 12 <= M(7) + 1 <= 10 and 11 <= M(6) + A(2) = 8, and Step 4 needs 5 > 3. So A(6) = 6 would give A(9) <= 13. We have not written this out in full; it is a remark, not a result.

Corollary 33 (the table)

Corollary 33. The recursion of Proposition 24 of rank-restriction.md, started from A(t) = t for t <= 5 and with A(6) <= 7, A(7) <= 9 and A(8) <= 11 (Theorems 28 to 30), gives

t 1 2 3 4 5 6 7 8 9 10 11 12 A(t) <= 1 2 3 4 5 7 9 11 14 17 20 23 E(t) <= 0 0 0 0 0 1 2 3 5 7 9 11 

Proof. The recursion with M(s) <= s + E(s). The reducible terms at t = 7 and 8 are at most 8 and 10, below the bounds of Theorems 29 and 30. For t = 9 and 10 the term t + A(t - 4) is 14 and 17, as before, and the reducible terms are at most 12 and 15. For t = 11 and 12 it is 11 + A(7) = 20 and 12 + A(8) = 23 (they were 21 and 24), and the reducible terms are at most 18 and 21. From t = 13 on, 2t is the bound, as before. QED

Corollary 34 (the settled cases)

Corollary 34. Let L be a finite 2-group whose involutions are all central, with a self-centralising normal subgroup A ~ Z_4^a x Z_2^b. Suppose b <= 1, or a <= max(9, b + 2), or (a, b) is one of

(10, 2), (10, 3), (10, 4), (10, 5), (11, 5), (10, 6), (11, 6), (10, 7), (11, 7), (12, 7), (11, 8), (12, 8), (12, 9), (13, 9), (13, 10), (14, 11). 

Then the answer to the question of section 14 of finite-groups.md is yes. Compared with Corollary 30, this adds (10, 2) and (13, 9), which follow from Theorem 29 (E(7) <= 2, E(11) <= 9), and (10, 3) and (14, 11), which follow from Theorem 30 (E(8) <= 3, E(12) <= 11). The first open cases are now (11, 2), (11, 3), (11, 4), (12, 5), (12, 6), (13, 7), (13, 8), (14, 9), (14, 10), (15, 11), (15, 12), and (b + 3, b) for b >= 13.

Proof. The count of Corollary 17 with the bounds of Corollary 33: (a, b) is settled if (a - tau - 2)(b - 1) <= 1, where tau is the largest t <= a with E(s) <= b for every s <= t. For b = 2, E(7) = 2 and E(8) = 3 give tau = 7, so a <= 10. For b = 3, E(8) = 3 and E(9) = 5 give tau = 8, so a <= 10. For b = 9, E(11) = 9 and E(12) = 11 give tau = 11, so a <= 13. For b = 11, E(12) = 11 and E(13) = 13 give tau = 12, so a <= 14. For the other b, tau does not change. QED

Results of job m21 (2026-10-18)

Job m21-a7-v1 ran on one CPU from 13:54 to 14:08 UTC on 2026-10-18 (CaDiCaL 1.9.5 through python-sat), with predictions written down before it was queued (m21/predictions.md). Every graded prediction held:

The 16 exploratory instances at (7, 4) (74,138 to 74,144 variables, about 451,000 clauses each): 15 UNSAT, 13 of them in under 5 seconds, one in 17 s (orbit 14) and one in 48 s (orbit 1); orbit 0, where M_1 = M_2 = M_3 = 0, reached its 9-minute budget (UNKNOWN). The pre-registered guess had named orbit 0 as the one likely to stay undecided. So m21 alone did not prove A(7) <= 9.

Lemma 59, found afterwards, settles every instance by hand, orbit 0 included. The 15 UNSAT answers agree with it; they are an independent check of the hand results behind the clauses (D), (W), (K), (U) and (G), not of Lemma 59 itself. Job m22-a7o0-v1 (queued 2026-10-18) re-runs orbit 0 alone, with the same CNF and no clause from Lemma 59, for up to 170 minutes; a verified SAT answer there would refute Lemma 59, and we would withdraw Lemma 59, Theorem 29 and the pairs (10, 2) and (13, 9) publicly and publish the system (Theorem 30 does not use Lemma 59).

What does not follow

• A(6) = 6, A(7) = 7 and A(8) = 8 are open: the bounds are 6 <= A(6) <= 7, 7 <= A(7) <= 9 and 8 <= A(8) <= 11.
• A(9) <= 14 is unchanged (9 + A(5)). By the Remark after Theorem 30, A(6) = 6 would improve it to 13.
• The first open cases with b = 2 and b = 3 are now (11, 2), which needs E(8) <= 2, that is, A(8) <= 10, and (11, 3), which needs E(9) <= 3, that is, A(9) <= 12.

Computation

cube_groups/a7_c.py (job m21). Its unit tests check that the appended clauses accept exactly the coordinates that a direct check of (D), (W), (K), (U) and (G) accepts, at (3, 2), (5, 3) and (7, 4), on points built to satisfy them, on random points and on one-bit changes; that (U) and (G) add exactly (k - 1)(t - 1) + k unit clauses and no variables, and that flipping any bit they fix is rejected; that the instances are the kept orbits of shape_c3.py with M_1 = 0 in front; and the commutative bound for m = 2 and 3.

cube_groups/a7_bounds.py (new) recomputes Corollaries 33 and 34: the recursion of Proposition 24 with A(t) = t for t <= 5 and A(6) <= 7, A(7) <= 9, A(8) <= 11, and the count of Corollary 17 for a <= 14 and b <= 12. Its unit tests (which gated this commit) check the table, the four newly settled pairs and which of them need Theorem 30, the first open a for each b, the reducible terms used in the proofs by hand, and that with A(6) <= 7 alone it gives back the table of Corollary 29.

Lemmas 57 to 59, Corollary 31 and Theorems 29 and 30 are proofs by hand with no computer check of their own, apart from the commutative bound and the consistency of job m21.

Prior work

As for the earlier notes, searches on 2026-10-15 to 2026-10-18 found nothing on the identity (*), on A(t), on M(t) or on Question R other than MV2's own posts. Schur's bound on commutative matrix algebras is classical, as cited. Please open an issue at https://github.com/munimtechnologies/kakeya-small-fields if this is known, or if you find an error.

Text CC BY 4.0, code Apache-2.0.