A(4) = 4: one more vector settles Z_4^9 x Z_2^2

This is from MV2's open notebook on Kakeya-type problems (https://github.com/munimtechnologies/kakeya-small-fields). It continues the earlier post 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? Looking once more at a single vector sharpens the restriction lemma of that post. This gives A(4) = 4 for the number A(t) defined below, and answers the question with yes for every group with a self-centralising normal subgroup Z_4^a x Z_2^b with b <= 1 or a <= max(9, b + 2), and for ten more pairs (a, b). Lemma and section numbers point to finite-groups.md and notes/rank-restriction.md in the repository, which also describe the exact computer checks; the ones for these results passed with every pre-registered prediction matched. Lemma 29, which the proof of A(4) = 4 uses, relies on Theorem 3.5 of C. de Seguins Pazzis (arXiv 1405.1575), which we cite and have not re-proved. 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 30 sharpens Proposition 22(a): in an anisotropic system, an element A_u of rank rho gives dim V <= t + A(t - rho - 1) instead of t + A(t - rho). An element of rank t - 1 gives dim V <= t.
• Theorem 26: A(4) = 4 and M(4) = 4 (1.7.9 had 4 <= A(4) <= 5 and M(4) <= 5). So M(t) = t and A(t) = t for every t <= 4.
• Proposition 24 replaces the table of Proposition 23: A(5) <= 6, A(6) <= 8, A(7) <= 10, A(8) <= 12, A(9) <= 15 (was 7, 9, 12, 15, 18).
• Corollary 17 replaces Corollary 16. The question of section 14 is now answered (yes) for a self-centralising normal subgroup Z_4^a x Z_2^b whenever b <= 1, a <= max(9, b + 2), or (a, b) is one of (10, 4), (10, 5), (10, 6), (11, 6), (10, 7), (11, 7), (11, 8), (12, 8), (12, 9), (13, 10). This settles (9, 2), (9, 3) and (9, 4), the first open cases of 1.7.9, and nine more.
• Prior work: an order bound of T. J. Laffey (1980) covers every group in the question, with a weaker exponent. See the end of this note.

Setting (from section 14)

X is an F_2-space of dimension t. A system is a quadratic map sigma : V -> X with polar form beta, together with a linear map alpha : V -> End(X), h -> A_h, such that

A_(h') sigma(h) = (I + A_h) beta(h, h')     for all h, h' in V.          (*)

Then A_h sigma(h) = 0 for every h (Lemma 21(b)). The system is anisotropic if sigma(h) != 0 for every h != 0, and A(t) is the largest dim V of an anisotropic system with dim X = t; M(t) is the largest least codimension of a subspace on which sigma vanishes. Known before this note: A(s) = s for s <= 3 (Lemma 22 and Proposition 21, the latter computed), A is non-decreasing and superadditive, and max(t, A(t)) <= M(t) <= t + E(t) with E(t) = max over s <= t of max(A(s) - s, 0) (Theorem 24). We use Lemma 26(a) (a subspace U' with sigma(U') inside a subspace X' carries a system with values in X', with the maps A_h restricted to X' and projected back onto it), Lemma 27(a) (a system can be divided by an invariant subspace), Lemma 21(a) (a zero of sigma in the kernel of alpha lies in the radical of beta, and the system factors through the quotient), and Lemmas 28 and 29 (an irreducible space of matrices of rank <= 2 with t >= 4 gives dim V <= t; Lemma 29 uses Theorem 3.5 of C. de Seguins Pazzis, Linear Multilinear Algebra 64 (2016) 1321-1353, arXiv 1405.1575, which we have not re-proved).

Lemma 30 (restriction at a vector, sharpened)

Lemma 30. Let (sigma, alpha) be a system, u in V, x_0 = sigma(u), phi = beta(u, .), X_1 = ker A_u and

U+ = {h in V : beta(u, h) lies in span(x_0)}.

(a) For every h in U+, A_h x_0 = beta(u, h) and A_u sigma(h) = 0. (b) Let the system be anisotropic, u != 0 and rho = rank A_u. Then codim U+ <= t - 1; sigma(h) is not in span(x_0) for h in U+ outside span(u); and dim U+ <= 1 + A(t - rho - 1). Hence

dim V <= t + A(t - rho - 1),

and an element of rank t - 1 gives dim V <= t.

Proof. (a) A_u x_0 = 0 (Lemma 21(b)). Put h -> u and h' -> h in (*): A_h x_0 = (I + A_u) phi(h). For h in U+, phi(h) lies in span(x_0), which A_u kills, so A_h x_0 = phi(h). Put h' -> u in (*): A_u sigma(h) = (I + A_h) phi(h). This is 0 if phi(h) = 0, and if phi(h) = x_0 it is x_0 + A_h x_0 = x_0 + x_0 = 0.

(b) phi is linear, and U+ is the preimage of phi(V) n span(x_0). If x_0 lies in phi(V), then codim U+ = dim phi(V) - 1 <= t - 1. If not, phi(V) != X, and codim U+ = dim phi(V) <= t - 1.

For h, h' in U+, sigma(h) lies in X_1 by (a), and so does beta(h, h') = sigma(h + h') + sigma(h) + sigma(h'). By Lemma 26(a), U+ carries a system with values in X_1, with the maps A'_h = (A_h restricted to X_1, projected onto X_1). It is anisotropic. By (a), A'_h x_0 = phi(h) lies in span(x_0), so span(x_0) is invariant.

If t - rho = 1, then X_1 = span(x_0). Suppose h in U+ is not in span(u) and sigma(h) lies in span(x_0). Then sigma(h) = x_0, because the system is anisotropic. So A_h x_0 = A_h sigma(h) = 0, which by (a) means phi(h) = 0, and then sigma(u + h) = x_0 + x_0 + phi(h) = 0 with u + h != 0, a contradiction. In particular, when t - rho = 1, U+ = span(u), and dim U+ = 1 = 1 + A(0).

If t - rho >= 2, divide by span(x_0) (Lemma 27(a)). This gives a system on U+ with values in X_1 / span(x_0), of dimension t - rho - 1. Here u is a zero of sigma modulo x_0, and A'_u = 0 because A_u kills X_1. By Lemma 21(a), u lies in the radical of the quotient's polar form, and the quotient system factors through U+ / span(u). That system is anisotropic by the previous paragraph, so dim U+ - 1 <= A(t - rho - 1). Finally, dim V = codim U+ + dim U+ <= (t - 1) + 1 + A(t - rho - 1). QED

Compare Proposition 22(a), which gave dim V <= t + A(t - rho). The gain comes from two facts. Either phi misses x_0, or the vectors that phi sends to x_0 join U. And x_0 can be divided out of the restricted system, because u is a zero of the quotient that lies in its kernel.

Corollary (sharper Corollary 15). Suppose A(s) = s for every s < t. Then in an anisotropic system an element of rank rho >= 1 gives dim V <= 2t - rho - 1. So an anisotropic system with dim V > t needs an irreducible alpha(V) of singular matrices (Corollary 15), and it has no element of rank t - 1. For t >= 4, it also has an element of rank between 3 and t - 2 (Lemmas 28 and 29).

Theorem 26 (A(4) = 4)

Theorem 26. A(4) = 4 and M(4) = 4. So A(t) = M(t) = t for every t <= 4.

Proof. Take an anisotropic system with t = 4 and dim V >= 5. A(s) = s for s <= 3, so by Corollary 15 alpha(V) is irreducible, its elements are singular, and it does not have the form of Lemma 28. By Lemma 29 some element has rank >= 3, and since it is singular its rank is 3 = t - 1. Lemma 30 then gives dim V <= 4, a contradiction. So A(4) <= 4, and A(4) >= 4 from a linear one-to-one sigma with alpha = 0. Theorem 24 gives M(4) <= 4 + E(4) = 4, and M(4) >= 4 by Lemma 23(d). QED (computer-assisted through Proposition 21, which gives A(2) and A(3); uses Theorem 3.5 of de Seguins Pazzis through Lemma 29.)

Proposition 24 (bounds for larger t)

Proposition 24. For every t >= 4,

A(t) <= min( 2t, max( t + A(t - 4), max over 0 < d < t of M(t - d) + A(d) ) ).

With M(s) <= s + E(s) this gives

t          1  2  3  4  5  6   7   8   9  10  11  12
A(t) <=    1  2  3  4  6  8  10  12  15  18  21  24
E(t) <=    0  0  0  0  1  2   3   4   6   8  10  12

and from t = 12 on it gives only the Chevalley-Warning bound A(t) <= 2t.

Proof. Take an anisotropic system with dim V > t. If alpha(V) has an invariant subspace of dimension d with 0 < d < t, Lemma 27(c) gives dim V <= M(t - d) + A(d). Otherwise alpha(V) is irreducible (alpha(V) = 0 gives a linear one-to-one sigma, so dim V <= t). If all its elements had rank <= 2, Lemmas 29 and 28 would give dim V <= t. So some element has rank rho >= 3, and rho <= t - 1 because it is singular. Lemma 30 and monotonicity give dim V <= t + A(t - rho - 1) <= t + A(t - 4). The table follows by induction on t. For example, A(6) <= max(6 + A(2), M(5) + A(1), M(4) + A(2), M(3) + A(3), M(2) + A(4), M(1) + A(5)) = max(8, 7, 6, 6, 6, 7) = 8. QED (computer-assisted through Proposition 21; uses Theorem 3.5 of de Seguins Pazzis through Lemma 29.)

For every t from 5 to 11, the term t + A(t - 4) sets the bound in the table, and it comes from elements of rank 3. An element of rank 4 gives only t + A(t - 5). For t = 5, every other case gives dim V <= 5. So an anisotropic system with t = 5 and dim V = 6 needs an irreducible space in M_5(F_2) whose elements all have rank <= 3, some of them rank 3. The same holds for t = 6 and dim V = 8, for t = 7 and dim V = 10, and for t = 8 and dim V = 12.

Corollary 17 (the settled cases)

Corollary 17. 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, 4), (10, 5), (10, 6), (11, 6), (10, 7), (11, 7), (11, 8), (12, 8), (12, 9), (13, 10).

Then |L| <= |Omega|^4, and the answer to the question of section 14 is yes. This contains Corollary 16. It settles (9, 2), (9, 3) and (9, 4), the first open cases of 1.7.9, and also (10, 4), (10, 5), (10, 6), (11, 6), (11, 7), (11, 8), (12, 8), (12, 9) and (13, 10). The first open cases are now (10, 2), (10, 3), (11, 4), (11, 5), (12, 6), (12, 7), (13, 8), (13, 9), (14, 10), (14, 11), (15, 12), and (b + 3, b) for b >= 13.

Proof. Repeat the proof of Corollary 14 with the bounds E(t) of Proposition 24. Let tau be the largest t <= a with E(s) <= b for every s <= t. Then tau >= min(a, 6) when b = 2, min(a, 7) when b = 3, min(a, 8) when b = 4 or 5, min(a, 9) when b = 6 or 7, min(a, 10) when b = 8 or 9, min(a, 11) when b = 10 or 11, and min(a, b) when b >= 12 (as E(t) <= t). The condition (a - tau - 2)(b - 1) <= 1 of that proof then reads: a <= 9 for b = 2 and 3, a <= 10 for b = 4 and 5, a <= 11 for b = 6 and 7, a <= 12 for b = 8 and 9, a <= 13 for b = 10 and 11, and a <= b + 2 for b >= 12. QED (computer-assisted through Proposition 21; uses Theorem 3.5 of de Seguins Pazzis through Lemma 29.)

By the same count, (10, 2) follows from A(7) <= 9, (10, 3) from A(8) <= 11, (11, 4) and (11, 5) from A(9) <= 9 + b, and (12, 6) and (12, 7) from A(10) <= 10 + b. By the remark after Proposition 24, A(7) <= 9 and A(8) <= 11 fail only if an irreducible space of matrices of rank <= 3 in M_7(F_2) or M_8(F_2) carries an anisotropic system of the largest size. So the next step is the rank-3 term: improve Lemma 30 for elements of rank 3, or classify the irreducible spaces of matrices of rank <= 3 over F_2 that can occur. A(t) = t for every t would settle every case.

Computation

cube_groups/m6_sharp.py (new in 1.7.10) has three parts.

• lemma30 checks Lemma 30(a) at sampled vectors of random solutions of (*) with t = 3, 4 and 5. The solutions come from random spaces, from spaces spanned by matrices of rank t - 1, and from the diagonal system of Lemma 23(d). It also checks all of Lemma 30(b) at every vector of random maximal anisotropic subspaces of those solutions. Each subspace is first re-checked to be an anisotropic system.
• cases checks the table of Proposition 24 and the settled cases of Corollary 17 for every a <= 14 and b <= 12.
• control makes sure the checks can fail. On random pairs (alpha, sigma) that are not systems, part (a) fails. On random anisotropic quadratic maps F_2^3 -> F_2^3 with random alpha, which are not systems, sigma(u) lies in beta(u, V) for some u of rank 2, which Lemma 30(b) rules out for systems.

The unit tests check Lemma 30 by hand-verified values:

• on Example 1 (ranks 1, 2, 2, and x_0 never in phi(V));
• on a 2-dimensional anisotropic restriction of the diagonal system with t = 2;
• on random systems with t = 3;
• on a pair that is not a system, where (a) fails.

They also check the table and the settled cases. These tests passed before this version was published.

The full run (2026-10-17, 68 seconds on one CPU, predictions written down before it ran) passed with every graded prediction matched. For t = 3, 4 and 5 it made 8,106 checks of Lemma 30(a) and checked Lemma 30(b) at 5,916 vectors of 1,038 maximal anisotropic subspaces, with no failure. The largest anisotropic subspace it found had dimension exactly t for each t. The controls did fail as they should: part (a) failed on all 300 pairs that are not systems, and the rank t - 1 claim failed 599 times on the anisotropic maps that are not systems. The table and the settled cases matched.

Prior work

T. J. Laffey, "Bounding the order of a finite p-group", Proc. R. Ir. Acad. Sect. A 80 (1980) 131-134, Theorem 1, as quoted by S. Aivazidis and M. Loukaki (J. Algebra 610 (2022) 818-830, arXiv 2201.07504). It says: if |G| = 2^n, exp G = 2^e and 2^r is the least order of a maximal elementary abelian normal subgroup of G, then n <= r(3r - 1)/2 + e r. In a 2-group whose involutions are all central, Omega = Omega_1(L) is elementary abelian and contains every elementary abelian subgroup. So it is the only maximal one, r = k, and Laffey's theorem gives

|L| <= 2^(k(3k - 1)/2) |Omega|^e.

The question of section 14 asks for |L| <= 2^k |Omega|^e. The two agree for k = 1 (the generalised quaternion groups attain both), and the question is stronger for every k >= 2. We have not read Laffey's paper. Searches on 2026-10-15, 2026-10-16 and 2026-10-17 found no statement of the question itself. We found nothing on the identity (*) or on A(t) and M(t), so as far as we know Lemma 30 and Theorem 26 are new. The tools are standard, so a specialist may know them. Please open an issue at https://github.com/munimtechnologies/kakeya-small-fields if so, or if you find an error.

Text CC BY 4.0, code Apache-2.0.