Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09
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Both bounds for last-term weighted distortion

Statement

Let G be finitely generated nilpotent of class c1, and H=γc(G). For fixed finite word metrics on G and H, there is C1 with C1hH1/cChGChH1/c+C for every hH. If H is infinite, Δ(n)=max{hH:hH, hGn} lies between positive multiples of nc for all sufficiently large integers n. If H is finite, Δ is bounded.

Facts & Assumptions

Given: H=γc is finitely generated abelian and central. All generating sets are fixed.

[F1]

A short word representing an element of the last term has only last-layer coordinates, bounded by a constant times max(1,n)^c (Weighted collection with finite-order carries).

[F2]

Powers of each fixed last-term element have ambient length at most a constant times the c-th root of the exponent (Power compression in the last lower-central term).

[F3]

Write the finitely generated abelian last term as ZrF with F finite (Integer abelian structure and rank by finite reduction).

Proof

1.1

Choose the decomposition H=z1zrF and finite generators for F, using these as the last-layer coordinate system. Every chosen coordinate generator has a fixed finite H-word. For h1, collect a shortest G-word of length n: all earlier coordinates vanish, each last free exponent is O(nc), and the finitely many residue exponents are bounded. Multiplying fixed H-words for these powers gives hHAnc+B; the same inequality with max(1,n) handles h=1. Increasing A gives hHAmax(1,hG)c. Taking roots gives the required lower ambient bound with an additive constant.

F1F3F4
1.2

For any finite H-generating set V, let L be the maximum absolute free-coordinate entry of a member of VV1, enlarged to at least 1. Projection to each free coordinate is additive, so a shortest H-word gives ajLhH for h=z1a1zrarf. Let M be the maximum G-length of an element of the finite set F. Power compression and subadditivity now give hGjCjaj1/c+M(jCj)L1/chH1/c+M, omitting zero exponents. This proves the other pointwise bound.

F2F3F4
2.1

For each n the defining maximum for Δ(n) exists: the finite alphabet of G has only finitely many words of length at most n, and the identity belongs to the intersection. Step 1.1 gives Δ(n)Amax(1,n)c. If H is infinite then r>=1. The first coordinate estimate in step 1.2 gives z1mHm/L, while F2 gives z1mGKm1/c with K>=1. For m=(n/K)c and sufficiently large n, m is at least (n/K)c/2 and at least 1, so Δ(n)nc/(2LKc).

F2step 1.1step 1.2
3.1

If H is finite, its H-word lengths have a finite maximum, bounding Δ for all n. Its finite ambient and intrinsic diameters are absorbed by the pointwise additive constants. At n=0 the intersection contains only the identity and Δ(0)=0. For c=1 the exponent is one and the same argument applies to H=G. Choose one C larger than all constants in the two pointwise estimates.

step 1.1step 1.2step 2.1

Source notes

Druţu–Kapovich, Geometric Group Theory (837-page edition), Proposition 14.20 and Lemma 14.21, printed pp.504–508 (finite last terms handled separately locally); Druţu–Kapovich, Lectures on Geometric Group Theory (585-page draft), Corollary 12.39, printed pp.322–323. Revised Proposition 14.20 and draft Corollary 12.39 supply the two routes. The infinite-H hypothesis is necessary for positive-power distortion growth; finite H is handled separately.

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