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TheoremStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)
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The Svarc-Milnor lemma

Statement

Let G act geometrically on a geodesic metric space X, fix x0∈X, and choose D≥0 such that every point of X lies within D of G⋅x0. Then G is finitely generated. More precisely, if S:={ g∈G:dX(x0, g⋅x0)≤2D+1 } is the finite generating set obtained from Cobounded proper geodesic actions produce finite generating sets, then the orbit map ϕx0:(G,dS)⟶X,ϕx0(g):=g⋅x0, is a quasi-isometry.

Facts & Assumptions

Given: A geometric action of G on a geodesic metric space X, a point x0∈X, and a real D≥0 such that every point of X lies within distance at most D of the orbit G⋅x0.

[L1]

The set S:={ g∈G:dX(x0, g⋅x0)≤2D+1 } is a finite generating set of G (Cobounded proper geodesic actions produce finite generating sets).

[L2]

For this generating set, the orbit map satisfies dX(g⋅x0, h⋅x0)≤M dS(g,h) for some constant M≥0 (Orbit maps of isometric actions are coarse Lipschitz).

[L3]

A subset is coarsely dense when every point of the space lies within a fixed distance of it, and a quasi-isometry is a coarse Lipschitz map admitting a coarse Lipschitz quasi-inverse (Coarsely dense subsets, quasi-inverses and quasi-isometries).

[L4]

The word metric is dS(g,h)=∣g−1h∣S (The word metric of a group with respect to a generating set).

Proof

technique · direct
1.1L1L2L4

By [L1], the set S is finite and generates G, so dS is a word metric on G. Step [L2] gives the coarse-Lipschitz upper bound for ϕx0.

1.2L3given

The orbit ϕx0(G)=G⋅x0 is D-dense in X by the choice of D, so it is coarsely dense in the sense of [L3].

1.3L1L4algebra

For g∈G, the proof of [L1] writes g as a product of at most m elements of S, where m is the least positive integer with dX(x0, g⋅x0)≤m. Therefore ∣g∣S≤dX(x0, g⋅x0)+1. Applying this to g−1h and using isometricity gives dS(g,h)=∣g−1h∣S≤dX(g⋅x0, h⋅x0)+1.

2.1L1step 1.2construct

The finite generating set S gives a fixed enumeration of G: order words in the finite alphabet S first by length and then lexicographically, and retain the first word representing each group element. The empty word represents the identity. For each x∈X, step 1.2 makes the set of words whose represented element g satisfies dX(x,g⋅x0)≤D nonempty. Let r(x) be the element represented by its first word. This defines one function r:X→G without choosing independently over X, and dX(x,r(x)⋅x0)≤D for every x.

3.1step 1.3step 2.1algebra

For x,y∈X, step 1.3 with g=r(x) and h=r(y) gives dS(r(x),r(y))≤dX(r(x)⋅x0, r(y)⋅x0)+1≤dX(x,y)+2D+1. So r:X→G is coarse Lipschitz.

3.2step 1.3step 2.1algebra

For every x∈X, step 2.1 gives dX(ϕx0(r(x)),x)≤D. For every g∈G, step 1.3 and step 2.1 with x=g⋅x0 give dS(r(g⋅x0),g)≤dX(r(g⋅x0)⋅x0, g⋅x0)+1≤D+1. Thus r∘ϕx0 and id⁡G, and also ϕx0∘r and id⁡X, are at bounded distance.

4.1L3step 1.1step 3.1step 3.2∎

Step 1.1 shows that ϕx0 is coarse Lipschitz, and step 3.2 gives a coarse Lipschitz quasi-inverse. Hence the orbit map is a quasi-isometry by [L3].

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