ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

UT_4(Z) has ranks three, two, one and growth degree ten

Example

For $G = {UT}_{4} (Z)$ , the subgroup $γ_{i} (G)$ consists of matrices whose superdiagonals of distance less than i vanish. The successive free ranks are 3,2,1, giving h=6 and D=10, and every finite word metric has degree-ten ball growth.

Facts & Assumptions

Given: Write $T_{i j} (a) = I + a E_{i j}$ for $1 \leq i < j \leq 4$ and a integer.

[F1]

Finitely generated nilpotent groups have degree D ball growth (The Bass–Guivarc’h growth degree formula).

[F2]

Lower-central commutators add weights (Lower-central commutators add weights).

[F3]

h and D are the unweighted and weighted rank sums (Bass–Guivarc’h dimension and nilpotent Hirsch length).

Verification

1.1

The matrix-unit rule is $E_{i j} E_{k l} = δ_{j k} E_{i l}$ . Hence $T_{i j} (a)^{- 1} = T_{i j} (- a)$ , $T_{i j} (a) T_{i j} (b) = T_{i j} (a + b)$ , and for $i < j < k$ direct expansion yields $[T_{i j} (a), T_{j k} (b)] = T_{i k} (a b)$ . In that expansion the only surviving cross product is $a b E_{i k}$ , since $E_{i j}^{2} = E_{j k}^{2} = E_{j k} E_{i j} = 0$ .

givenalgebra

2.1

The ordered product $T_{12} (a) T_{23} (b) T_{34} (c) T_{13} (d) T_{24} (e) T_{14} (f)$ has entries (12,23,34,13,24,14) equal to $(a, b, c, a b + d, b c + e, a b c + a e + f)$ . Given entries (A,B,C,D,E,F), the unique exponents are $a = A, b = B, c = C, d = D - A B, e = E - B C, f = F - A E$ . Thus this is a six-integer normal form. The three adjacent transvections with parameter 1 generate: step 1.1 gives T_13(1), T_24(1), and then T_14(1), and all integer powers give every factor.

step 1.1algebra

2.2

Let J_i be the additive group of strictly upper triangular matrices supported at distances at least i, with J_4=0. Matrix-unit multiplication gives $J_{i} J_{j} \subseteq J_{i + j}$ . The groups $F_{i} = 1 + J_{i}$ are closed under multiplication and inversion, since $(1 + u)^{- 1} = 1 - u + u^{2} - u^{3}$ . In the quotient ring by $J_{i + j}$ , the images of u in J_i and v in J_j have both uv=vu=0, so 1+u and 1+v commute. It follows that $[F_{i}, F_{j}] \subseteq F_{i + j}$ . Since G=F_1, induction gives $γ_{i} (G) \subseteq F_{i}$ .

step 1.1algebra

3.1

Conversely F_2 is generated by T_13(1), T_24(1), T_14(1), with arbitrary integer powers: its first superdiagonal is zero and the product has precisely those three independent remaining entries. The first two are adjacent commutators by step 1.1. The third is $[T_{13} (1), T_{34} (1)]$ ; since T_13(1) is in gamma_2, F2 places this commutator in gamma_3, hence also gamma_2. Therefore $F_{2} \leq γ_{2}$ and $F_{3} = ⟨ T_{14} (1) ⟩ \leq γ_{3}$ . Along with step 2.2, this gives $γ_{2} = F_{2}$ , $γ_{3} = F_{3}$ , $γ_{4} = 1$ .

F2step 1.1step 2.2

4.1

Taking the entries on superdiagonal i is an onto homomorphism $F_{i} \to Z^{4 - i}$ with kernel $F_{i + 1}$ : cross products have strictly greater distance. Hence the ranks are 3,2,1; gamma_3 is nontrivial, so the class is three. Thus $h = 3 + 2 + 1 = 6$ and $D = 1 \cdot 3 + 2 \cdot 2 + 3 \cdot 1 = 10$ . Apply F1 using the three explicit finite generators in step 2.1. All six zero exponents give I.

F1F3step 2.1step 3.1∎

Source notes

Druţu–Kapovich, Lectures on Geometric Group Theory (585-page draft), Exercise 10.30, p.282, specialized to n=4 and Z. Draft Exercise 10.30 is specialized to integer 4-by-4 matrices, with both subgroup inclusions and the six-entry normal form calculated.

Depends on

Used by

Hirsch length and growth degree differ Example

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Druţu–Kapovich, Lectures on Geometric Group Theory (585-page draft) (standard reference, not scraped)