A(5) = 5: four matrices finish the case t = 5

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. An earlier post proved A(4) = 4. This one proves A(5) = 5: a short linear-algebra argument at four elements rules out the last remaining case at t = 5, and as a consequence the answer is yes for one more pair, Z_4^11 x Z_2^5. Lemma and section numbers point to notes in the repository. The result also rests on 12 answers of a SAT solver saved without proof certificates (job m13), and 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 52 (by hand). No anisotropic system with t = 5 and dim V = 6 has an element in Case A. The proof is short. It uses Lemma 50, one fact of Lemma 51 and the rank bound of Lemma 30 at four elements of V, and no computer search.
• Theorem 27: A(5) = 5 and M(5) = 5. Corollary 24 reduced A(5) = 5 to Case A, so Lemma 52 finishes it. So A(t) = M(t) = t for every t <= 5.
• Corollary 26. The table of Proposition 24 now reads A(5) <= 5 and A(9) <= 14 (it was 6 and 15), with E(5) = 0 and E(9) <= 5.
• Corollary 27. The question of section 14 of finite-groups.md now has the answer yes also for Z_4^11 x Z_2^5. Among the first open cases, (11, 5) is replaced by (12, 5).
• The exact searches m15 and m16 (case-a.md, case-a-radical.md), queued before this proof and not yet run when it was written, are now cross-checks. Lemma 52 says every one of their t = 5 instances is UNSAT. A verified SAT answer there would show that this note contains an error.

Setting

As in case-a-radical.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')                                    (*)

for all h, h' in V. Putting h' = h gives A_h sigma(h) = 0 for every h (Lemma 21(b)). The system is anisotropic if sigma(h) != 0 for every h != 0. A(t) is the largest dim V of an anisotropic system, and M(t) is the number of Theorem 24 of finite-groups.md.

Case A (question-r.md): dim V = t + 1, and u in V has rank A_u = t - 2 and phi = beta(u, .) maps V onto X with kernel span(u). Write x0 = sigma(u), A = A_u, R = Im A, choose a linear section s with phi(s(p)) = p, and put S(p) = sigma(s(p)), B(p, q) = beta(s(p), s(q)), L(p) = A_(s(p)) and psi = B(x0, .), as in Lemma 50. Then z = S(x0) = sigma(s(x0)), and ker A = span(x0, z) with z in R (Proposition 25, Lemma 48).

Lemma 52 (no Case A at t = 5)

Lemma 52. No anisotropic system with t = 5 and dim V = 6 has an element in Case A.

Proof. Take such a system and an element u in Case A, with the notation above. So dim R = 3, ker A = span(x0, z), and z lies in R.

Step 1 (ranks). Every A_h with h != 0 has rank at most 3. It is singular, since A_h sigma(h) = 0 and sigma(h) != 0. An element of rank 4 would give dim V <= 5 + A(0) = 5 by Lemma 30, but dim V = 6.

Step 2 (two coset maps). For p = x0 and p = x0 + z, the map I + L(p) sends X into R. For p = x0 this is (f) of Lemma 50: I + L(x0) = A (I + psi). For p = x0 + z, (d) of Lemma 50 with p = x0 gives L(q) z = (I + L(x0)) psi(q), which lies in R for every q. Then (e) gives L(z) q = L(q) z + A B(z, q), which lies in R too, and I + L(x0 + z) = (I + L(x0)) + L(z). (This is Lemma 51(c) and (d) at the point z of the radical.)

Step 3 (restriction to R). Let h = s(p) + lambda u with p in {x0, x0 + z} and lambda in F_2. Then h != 0, since phi(h) = p != 0, and A_h = L(p) + lambda A = I + N_h, where N_h = (I + L(p)) + lambda A maps X into R by Step 2. So A_h maps R into R, and every x with A_h x = 0 satisfies x = N_h x, so ker A_h lies in R. Let P_h be the restriction of A_h to R, a map R -> R. By Step 1, ker A_h has dimension at least 2, and it equals ker P_h, so P_h has rank at most 1. Also sigma(h) lies in ker A_h = ker P_h. Finally, with a = the restriction of A to R,

P_(s(p)) + P_(s(p) + u) = a       for p = x0 and p = x0 + z,

since A_(s(p) + u) = A_(s(p)) + A.

Step 4 (x0 not in R). Then R meets ker A = span(x0, z) in span(z) only, so a has rank 2 and kernel span(z). Take h_1 = s(x0 + z) and h_2 = h_1 + u, and put P_1 = P_(h_1), P_2 = P_(h_2). Their sum a has rank 2 and each has rank at most 1, so each has rank 1, their images are independent, and ker a = ker P_1 n ker P_2. In particular z lies in ker P_1 and in ker P_2. By Lemma 50,

sigma(h_1) = S(x0 + z) = z',     sigma(h_2) = S(x0 + z) + (x0 + x0 + z) = z' + z,

and these lie in ker P_1 and ker P_2. So z' lies in ker P_1, and z' = (z' + z) + z lies in ker P_2. Then z' lies in ker a = span(z), so sigma(h_1) = z' is 0 or sigma(h_2) = z' + z is 0. This contradicts anisotropy, as h_1 and h_2 are not 0.

Step 5 (x0 in R). Then span(x0, z) lies in R, so a has rank 3 - 2 = 1, and a != 0. By (b) of Lemma 50, L(q) x0 = (I + A) q, and A x0 = 0, so for h = s(p) + lambda u as in Step 3

P_h x0 = (I + A) p + lambda A x0 = p + A p.

For p = x0 this is x0, and for p = x0 + z it is x0 + z (as A z = 0). A map of rank at most 1 that sends x0 to a non-zero vector y has image span(y). So P_(s(x0)) and P_(s(x0) + u) both have image span(x0), and their sum a has image in span(x0). Likewise P_(s(x0+z)) and P_(s(x0+z) + u) both have image span(x0 + z), so a has image in span(x0 + z). Since x0 and z are independent, a = 0, a contradiction. QED

Step 4 is the argument of Lemma 32 of question-r.md, moved to another pair. Proposition 25 applies that lemma to v = s(x0) and w = v + u. There sigma(v) = sigma(w) = z, and the kernels can share the line span(z). Lemma 51(c) puts x0 + z in the coset x0 + Rad, so the same argument also applies to the pair s(x0 + z), s(x0 + z) + u. There sigma takes the two different values z' and z' + z, and each of them must lie in both kernels. In this sense Lemma 51 supplied the missing step, but not through the invariance of Rad that case-a-radical.md was looking for.

Steps 4 and 5 use only the facts about P_h listed in Step 3. Together these facts say something about 3 x 3 matrices over F_2, and a5_bounds.py checks it over all matrices of rank at most 1 (Computation). The proof uses the rank bound only at the four elements s(p) + lambda u. Lemma 51 is used only through z being in its radical, and Corollary 25 is not used.

Theorem 27 (A(5) = 5)

Theorem 27. A(5) = 5 and M(5) = 5. So A(t) = M(t) = t for every t <= 5.

Proof. By Corollary 24(b) of case-a.md, A(5) = 5 holds if and only if no anisotropic system with t = 5 and dim V = 6 has an element of rank 3 in Case A. Lemma 52 says there is none. Theorem 24 gives M(5) <= 5 + E(5) = 5, as A(s) = s for s <= 5, and M(5) >= 5 by Lemma 23(d). QED (computer-assisted through job m13 in Corollary 24(a), with 12 UNSAT answers without proof certificates, and through Proposition 21, which gives A(2) and A(3); uses Theorem 3.5 of C. de Seguins Pazzis, arXiv 1405.1575, through Lemma 29.)

The full 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 of rank-restriction.md; Propositions 25 to 27 (question-r.md, tight-cases.md, case-b.md); Lemmas 35 to 39 (case-b.md, case-b-shapes.md, shape-c.md); Lemma 47 and Corollary 24 of case-a.md, with the m13 answers reported there; Lemma 50 and Lemma 51(c) of case-a-radical.md; and Lemma 52 here.

Corollary 26 (the table)

Corollary 26. The recursion of Proposition 24 of rank-restriction.md, started from A(t) = t for t <= 5, gives

t          1  2  3  4  5  6   7   8   9  10  11  12
A(t) <=    1  2  3  4  5  8  10  12  14  18  21  24
E(t) <=    0  0  0  0  0  2   3   4   5   8  10  12

Proof. As for Proposition 24. For t = 6, 7 and 8 the term t + A(t - 4) is 8, 10 and 12, and the reducible terms max over 0 < d < t of M(t - d) + A(d) are at most 6, 9 and 11, so nothing changes. For t = 9 the term 9 + A(5) is now 14, and the reducible terms are at most max(8 + 4 + 1, 7 + 3 + 2, 6 + 2 + 3, 5 + 4, 4 + 5, 3 + 8, 2 + 10, 1 + 12) = 13 (with M(s) <= s + E(s)). So A(9) <= 14 and E(9) <= 5. For t = 10, 11 and 12 the term t + A(t - 4) is 18, 21 and 24, as before, and it dominates. QED

Two hypotheses elsewhere in the notebook now hold one step further. Corollary 25(d) of case-a-radical.md (Rad is not all of R) now holds for t = 7 as well. Corollary 18(a) of tight-cases.md, which needs A(s) = s for s <= k - 1, now applies whenever k = t - rho <= 6. In particular, Question R at rank 3 for t = 9 now reduces to Case B. tight-cases.md left this open because Proposition 26 needed A(5) = 5.

Corollary 27 (the settled cases)

Corollary 27. 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), (11, 5), (10, 6), (11, 6), (10, 7), (11, 7), (11, 8), (12, 8), (12, 9), (13, 10).

Then the answer to the question of section 14 of finite-groups.md is yes. Compared with Corollary 17, this adds (11, 5). The first open cases are now (10, 2), (10, 3), (11, 4), (12, 5), (12, 6), (12, 7), (13, 8), (13, 9), (14, 10), (14, 11), (15, 12), and (b + 3, b) for b >= 13.

Proof. The count of Corollary 17 with the bounds of Corollary 26. Only E(9) has dropped from a value above 5 to 5, so only b = 5 changes: tau = min(a, 9), and (a - 11)(5 - 1) <= 1 gives a <= 11. This is the remark after Corollary 17 that (11, 5) follows from A(9) <= 14. QED

Computation

cube_groups/a5_bounds.py (new) has two parts, and its unit tests gate the commit that adds it.

• core. Over all 50 matrices of rank at most 1 in M_3(F_2): of the 1,764 pairs (P_1, P_2) whose sum a has rank 2, none has a vector z' outside span(z) (where ker a = span(z)) with P_1 z' = 0 and P_2 (z + z') = 0. This is Step 4. For each of the 49 matrices a of rank 1 and each of the 6 ordered pairs (x0, z) of independent vectors in its kernel, a cannot be written both as P_1 + P_2 with P_1 x0 = P_2 x0 = x0 and as P_3 + P_4 with P_3 x0 = P_4 x0 = x0 + z, with every P_i of rank at most 1. This is Step 5. As a control, both counts are repeated with rank at most 2, where solutions do exist. The pair count 1,764 = 49 x 36 is checked by hand in the tests: x f + y g has rank 2 exactly when x != y and f != g.
• table. The recursion of Proposition 24 started from A(t) = t for t <= 5 (Corollary 26), the count of Corollary 17 with these bounds for a <= 14 and b <= 12 (Corollary 27), the first open a for each b, and the list of newly settled pairs, which is exactly (11, 5). Started from A(t) = t for t <= 4 instead, it gives back the old table of Proposition 24.

Steps 1 to 3 of Lemma 52 have no computer check of their own. If jobs m15 and m16 run, every one of their t = 5 instances should come back UNSAT (or reach its time limit). A SAT answer that passes the brute-force check of (*) and anisotropy would be a counterexample to Lemma 52.

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. The argument is elementary linear algebra, so a specialist may know it. 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.