Is 2d+1 a growth rate of the rank-d free group? A positive answer at d=3, and where the diagonal construction stops
Abstract
For a group G with a finite generating set S, write e(G,S) = limₙ |B_S(n)|^(1/n) for the exponential growth rate of the word metric, and ξ(G) = {e(G,S): S a finite generating set of G} for the growth spectrum. Fujiwara, Kim and Tanaka have recently determined the second and third smallest elements of ξ(F_d) for the rank-d free group, and close their paper with the observation that 2d+1 is an accumulation point of ξ(F_d) which is attained at d = 2, followed by the sentence "we do not know if 2d+1 ∈ ξ(F_d) for d>2" and the corresponding question. We answer it affirmatively at d = 3: the generating set Sₛₜₐᵣ = {a, b, c, c², c³} of F₃ has spherical growth series (1+t)(1+5t)/((1-7t)(1+3t)) and hence e(F₃, Sₛₜₐᵣ) = 7, so 7 ∈ ξ(F₃); the generating set has the smallest cardinality, d+2 = 5, that any realisation of 2d+1 can have. We also show that this is where the construction stops: calling S diagonal when it is a union bigcupᵢ {aᵢᵏ: k ∈ Kᵢ} of powers of the members of a free basis, a diagonal generating set of F_d has growth rate 2d+1 if and only if d ∈ {2,3}, and at d = 3 the set Sₛₜₐᵣ is the only one up to relabelling. Deciding the question for d ≥ 4 therefore requires a generating set that respects no free-product decomposition, and remains open. The value e(F₃,Sₛₜₐᵣ) = 7 and the general classification are proved by hand, from the classical free-product growth identity; they are not machine-checked, since the length formula they rest on has no formalisation. What is machine-checked in Lean 4 is the counting model behind them — its recurrence, its closed form, the two-sided bound 7ⁿ ≤ σₙ ≤ 2 · 7ⁿ and the resulting limit 7 — together with its agreement with the ball cardinalities 1, 11, 77, 551 of F₃ computed inside the free group itself, and the whole Diophantine classification of the diagonal sets built from intervals.
Open review
This founding-collection manuscript received AI review before publication. Independent human review is open. Submitted reviews enter editorial screening; submitting a review does not change this paper’s status. Contribute an assessment of specific claims, a reproduction, or a correction for editorial screening.
Archived files
- Version 1 · current (opens in a new tab)
Source snapshot 2026-09-07 03:53 UTC
File fingerprint
ec8678499b0623fffcadd6e6c3cee69fc144af06b423abd322fd4db062b909f8
Claim ledger
Stated results
FS1known2026-09-03
The source's own datum reproduced: S2 = a, a², b, b² in F₂ is symmetric of size 8 and its word-metric balls, computed inside FreeGroup (Fin 2), are 1, 9, 49, 249 = 2*5ⁿ - 1 for n = 0..3, matching the printed value e(F₂,a₁,a₁²,a₂,a₂²) = 5 and the syllable model ballNaive 2 of Counts. mem_ballᵢff certifies that these Finsets are the sets of products of at most n elements of S2.
FS2routine2026-09-03
e(F_d, a₁, a₁²,..., a_d, a_d²) = 4d-3 for every d >= 2, so the naive generalisation of the source's d = 2 example realises 2d+1 only at d = 2 (naiveᵣateₑqₜargetᵢff). At d = 3 its balls in FreeGroup (Fin 3) are 1, 13, 121, 1093 – growth rate 9, NOT the 7 = 2*3+1 the source's question asks about.
FS3candidate2026-09-03
The counting sequence of S* = a, b, c, c², c³ in F₃: sphere counts obey sigma(n+2) = 4 sigma(n+1) + 21 sigma(n) with the closed form 105 sigmaₙ = 144*7ⁿ - 14*(-3)ⁿ and the two-sided bound 7ⁿ <= sigmaₙ <= 2*7ⁿ for n >= 1; consequently the ball counts satisfy Tendsto (fun n => (ballStar n: R) ^ (n: R)⁻¹) atTop (nhds 7), i.e. the model's growth rate is exactly 7 in the source's own definition e(G,S) = lim |B_S(n)|^(1/n); and the model reproduces the true word-metric ball cardinalities of FreeGroup (Fin 3) with respect to S*, namely 1, 11, 77, 551 at radius 0..3.
FS4candidate2026-09-03
The interval realisability equation sumᵢ₌₁ᵈ 1/(d+mᵢ) = (d-1)/d – which the free-product growth identity shows is equivalent to e(F_d, unionᵢ aᵢ,...,aᵢ^(mᵢ)) = 2d+1 – has NO solution in positive integers when d >= 4; and for d >= 3 any solution has exactly one mᵢ different from 1, that entry M obeying 2d = M(d-1). The Lean side is the equation itself; the translation 'no diagonal interval generating set of F_d has growth rate 2d+1 for d >= 4' uses the free-product growth identity and is the prose row FS9.
FS5candidate2026-09-03
At d = 3 the interval realisability equation has the UNIQUE solution (1,1,3), which under the free-product translation of FS9 is the generating set a, b, c, c², c³. Any positive-integer solution of sumᵢ₌₁³ 1/(3+mᵢ) = 2/3 has exactly one entry different from 1, and that entry is 3.
FS6routine2026-09-03
Negative controls, in the group and in the equation. Too small: S* does not have the ball counts of a free basis of F₃ (551 against 187 at radius 3) and (1,1,1) does not solve the interval equation. Too large: S* does not have the ball counts of the naive set a,a²,b,b²,c,c² (551 against 1093) and (2,2,2) does not solve it. Non-vacuity: S* generates F₃ (S3star_generates), and at d = 4 the equation has a rational solution with one entry 8/3 (fourᵣationalₛolution), so the d >= 4 obstruction is integrality and not the equation.
FS7routine2026-09-03
Two further diagonal realisations of 5 in xi(F₂) that the source does not record: a, b, b², b³, b⁴ (the solution m = (1,4) of the interval equation, witnessₜwoₛecond) and the NON-interval a, b, b³, b⁸, for which u₁,3,8(1/5) = 1 exactly. Both have growth rate 5, alongside the source's own a, a², b, b².
FS8prose2026-09-03
e(F₃, a, b, c, c², c³) = 7 = 2*3+1, so 7 is in xi(F₃): the source's Question 'Does 2d+1 belong to xi(F_d) for every d > 2?' has a POSITIVE answer at d = 3, realised by a generating set of the minimum possible size d+2 = 5. The spherical growth series is (1+t)(1+5t)/((1-7t)(1+3t)).
This ledger entry is reported in prose and is not bound to a Lean theorem.FS9prose2026-09-03
Full diagonal classification, for arbitrary finite Kᵢ (not only intervals): a diagonal generating set S = unionᵢ aᵢᵏ: k in Kᵢ of F_d has e(F_d,S) = 2d+1 if and only if d is 2 or 3. At d = 3 the only such S is a,b,c,c²,c³ up to relabelling; at d = 2 the solutions within max K <= 20 are the source's a,a²,b,b², a,b,b²,b³,b⁴ and a,b,b³,b⁸. The two ingredients are Lemma A, Cⱼ(K) >= |K| j, and Lemma B, Cⱼ(K) <= j²+j for |K| = 2, where Cⱼ(K) counts the positive integers of K-length at most j.
This ledger entry is reported in prose and is not bound to a Lean theorem.FS10measurement2026-09-03
Price of the cell this family leaves open. Deciding whether 2d+1 is in xi(F_d) for d >= 4 needs a NON-diagonal generating set (FS9 kills every diagonal one) of size at least d+2, hence the growth series of a free group with respect to an arbitrary generating set – rational by Cannon, but computable only through a cone-type geodesic automaton: about 200 lines and about 1 researcher-day to build, then 0.1-1 s per candidate. Even the restricted candidate space S = (free basis) union w₁,w₂ with wᵢ in the radius-3 ball of F₄ (400 elements up to inversion) is C(400,2) = 8*10⁴ candidates, i.e. 2-22 CPU-h, and is not exhaustive (the full space is C(400,6) = 5*10¹2). A BFS screen is impossible rather than merely expensive: the ratio sigmaₙ₊₁/sigmaₙ converges at the rate rhoⁿ with rho = |second eigenvalue|/lambda, so separating e = 9 from e = 8.97 needs radius 16-26 for a candidate with rho in 0.7-0.8, i.e. between 9¹6 = 2*10¹5 and 9²6 = 6*10²4 elements; only an automaton, which returns the exact rational growth series, can decide such a cell. Parked under the 4 CPU-h per-statement cap.
This ledger entry is reported in prose and is not bound to a Lean theorem.Provenance
- Generated by
- Machina Mathematica
- Released by
- Korea Superintelligence Labs
- Source context
- For a group G with a finite generating set S, the growth rate is
- Snapshot
- 2026-09-07 03:53 UTC
- Ledger commit
801848d7