Baumslag-Solitar-free one-relator groups are hyperbolic

Proof. First observe that, for fixed subgroups K,L of any finitely generated group and R\geq 0, there is a constant D such that

Indeed, if x belongs to the left side, choose k\in K and l\in L within distance R of x. The element d=k^{-1}l lies in the finite ball of radius 2R. For each value d that occurs, fix one pair k_d\in K,l_d\in L with k_d^{-1}l_d=d. Then

Taking a maximum over the finitely many d proves (2.1).

A geodesic with endpoints in K\cap L lies in fixed neighborhoods of both K and L, so (2.1) proves quasiconvexity of their intersection.

Now let \xi\in\Lambda K\cap\Lambda L. Quasiconvexity and limits of geodesic segments show that a ray from the identity to \xi lies in bounded neighborhoods of both subgroups. Applying (2.1) along the ray gives a sequence in K\cap L converging to \xi. Thus \xi\in\Lambda(K\cap L). The reverse inclusion is immediate, and induction gives the finite-family assertions. These facts also appear in [7, Lemmas 2.6 and 2.7].□

Proof. Suppose that c\in K\cap gLg^{-1} has infinite order. Its two fixed boundary points belong to both \Lambda K and \Lambda(gL): use the sequences c^n\in K and c^ng\in gL, for positive and negative n. A geodesic line between those points lies in a uniformly bounded neighborhood of each of K and gL. To see the uniformity, a q-quasiconvex set contains finite geodesics between its points within distance q; passing to limits, and then comparing geodesics with the same ideal endpoints, increases this bound by a constant depending only on the hyperbolicity constant. The quasiconvexity constant of gL equals that of L.

At any point of this line choose nearby vertices k\in K and gl\in gL. Their distance is bounded by a constant depending only on the two quasiconvexity constants and the hyperbolicity constant. Hence k^{-1}gl\in KgL lies in a fixed finite ball. Every double coset under consideration meets that ball, proving finiteness.□

Proof. The subgroup C is quasiconvex. By Lemma 2.2, its commensurator meets only finitely many double cosets CgC. For a representative g in the commensurator, CgC is a union of [C:C\cap g^{-1}Cg] left C-cosets, a finite number. Thus the commensurator is virtually cyclic. A torsion-free virtually cyclic group is cyclic, giving the last assertion.□

Proof. Choose a vertex v and a finite symmetric generating set S of Q. The union of all Q-translates of the finite subtree

is connected: its segments connect the vertices in the orbit Qv according to the Cayley graph of Q. This union is exactly the convex hull of Qv, because its segments join orbit points. Restricting the action to this hull gives a finite quotient, since every edge orbit has a representative in D. The retained vertex and edge stabilizers are unchanged. Bass–Serre theory therefore realizes Q as the fundamental group of a finite connected graph of infinite cyclic groups.

Collapse any non-loop edge whose map to one endpoint group is onto. Such a collapse preserves the fundamental group and leaves a graph of infinite cyclic groups. After finitely many collapses, every remaining non-loop edge has both inclusion indices greater than one.

If a loop remains, the subgraph consisting of that loop and its vertex has fundamental group

Its fundamental group injects into Q by the graph-of-groups normal form theorem.

If a non-loop edge remains, its two-vertex subgraph likewise injects and has fundamental group

The element z=x^m=y^n is central and has infinite order. Modulo \langle z\rangle, this group is C_{|m|}*C_{|n|}, in which the image of xy has infinite order by free-product normal form. Thus the commuting elements z and xy generate an embedded \mathbb Z^2=BS(1,1).

Finally, if no edge remains, the graph has one vertex and Q is cyclic, contrary to the hypothesis. This exhausts the possibilities.□

Proof. Set L=\langle a,g\rangle and choose a vertex v fixed by a. Every element of L commensurates \langle a\rangle. For each h\in L, commensurability supplies nonzero integers r,s with a^r=ha^sh^{-1}; hence a nonzero power of a fixes hv. A common multiple of finitely many such powers fixes any prescribed finite subset of Lv, and therefore fixes its convex hull pointwise. It follows that every finite subtree of \mathcal T_L=\operatorname{Hull}(Lv) is fixed by a nonzero power of a.

Let x be a vertex of \mathcal T_L, and choose N>0 with a^N\in L_x. Every element of L_x commensurates \langle a^N\rangle, now a subgroup of the vertex stabilizer G_x. Thus

Lemma 2.3 makes the last group infinite cyclic, so L_x is infinite cyclic as well. Each edge stabilizer of \mathcal T_L contains a nonzero power of a and is a subgroup of a vertex stabilizer in L; it too is infinite cyclic.

The group L is noncyclic. Otherwise, writing a=c^r and g=c^s in a cyclic group, translation lengths would give \ell(a)=|r|\ell(c)=0 and hence \ell(g)=0, contrary to tree-hyperbolicity of g. Apply Lemma 2.4 to the action of the finitely generated group L on \mathcal T_L.□

Proof. Apply (2.2) with S=E. The identity double coset contributes \bar r(E), so all other nonnegative summands vanish.□

Proof. Every positively oriented edge leaving H is ate for some a\in H; the possible edges are parametrized by aB\in H/B. Its common stabilizer with e is A\cap aBa^{-1}. If this intersection is noncyclic, assumption (i) gives a\in AB. Write a=pb with p\in P and b\in B. Since bt=t\phi^{-1}(b), the edge ate is pte. Moreover,

The common stabilizer is P\cap pBp^{-1}=pIp^{-1}, and conjugating by (pt)^{-1} gives the stated transfer.

A positive predecessor ending at t^{-1}H is t^{-1}ae, with aA\in H/A. After conjugating its pair stabilizer by t, we obtain B\cap aAa^{-1}. Noncyclicity gives a\in BA; writing a=br and p=t^{-1}bt\in P puts the predecessor in the form pt^{-1}e. Two such edges agree exactly when

Their common stabilizer with e is pJp^{-1}, and conjugation by (pt^{-1})^{-1} is the claimed transfer.

At a reversal of traversal sign, the consecutive edge stabilizers, in the common vertex coordinates, are M and aMa^{-1} for the same M\in\{A,B\}. The edges are distinct because the path is reduced, so a\notin M. Assumption (i) makes their intersection at most cyclic. A noncyclic stabilizer of the whole path cannot lie in that intersection. Thus there are no reversals.□

Proof. Build the fixed window by starting with e and adding one edge at an endpoint at a time. The initial stabilizer is P. Suppose W is an existing representative, S=\operatorname{Stab}_G(W)\leq P, and h is a frame at the endpoint edge being extended. In that frame put K=h^{-1}Sh\leq P.

For a forward extension whose stabilizer is noncyclic, Lemma 3.2 restricts the next edge to hpte, with pI\in P/I. The action of S on these choices becomes left multiplication by K. Thus the extension orbits over the orbit of W are indexed by the double cosets KpI, and the extended stabilizer is isomorphic to K\cap pIp^{-1}. Assumption (ii) gives

For backward extension use J and its separately assumed inequality. The stabilizers remain finitely generated: in the current edge coordinates an extension takes the intersection of two finitely generated subgroups of the finite-rank free group P.

The forgetful map from extended windows to the preceding windows is P-equivariant. Its fiber over the orbit of W is precisely the set of extensions modulo S, which is the double-coset set just considered. Summing the displayed inequality over the existing representatives therefore proves the induction step. Windows with cyclic stabilizer may be discarded, since extension only decreases a stabilizer. This proves (3.2). Each noncyclic term contributes at least one, proving the orbit bound. Finally, \operatorname{Stab}_G(e)=P, which identifies the two stated orbit sets.□

Proof. At their first divergence from the common part, the common stabilizer, in the shared endpoint frame, lies in

This intersection is at most cyclic. Indeed, apply (3.1) with S=E. The diagonal double coset already contributes \bar r(E), so every other term has reduced rank zero. Since p^{-1}q\notin E, the displayed intersection is one of these off-diagonal intersections up to conjugacy. If E is cyclic or trivial, the conclusion is immediate as well.□

Proof. Negate Theorem 3.5 with \lambda=2. For each m there is a finite width bound \rho_m and counterannuli of arbitrarily large girth with width at most \rho_m. Choose such a sequence with girth tending to infinity. There are finitely many sign tuples and finitely many bridge tuples of this fixed length and width, so a subsequence fixes both tuples.

The maps f_\epsilon,l_\epsilon are quasi-isometric embeddings. The row equation, with the bridges now fixed, gives constants L\geq1 and D\geq0 such that

at every row of this fixed sequence. Iterating from row 0 bounds all the indicated ratios, since |c_0|_P\to\infty.□

Proof. Continuity follows because \mathcal L_F(1) is compact open. Every compact subset of \partial^2F is covered by finitely many translates of this set. A uniform norm bound therefore bounds the measures of all compact subsets. There are countably many segment cylinders. Choose a diagonal subsequence on their measures. The cylinder identities are preserved: a cylinder is the finite disjoint union of its extensions by one edge at either end. On each compact-open \mathcal L_F(v) these identities define a finite measure, by extension from its clopen cylinders, and the measures agree on overlaps. They consequently define a Radon measure on line space. Invariance and flip invariance pass to the limit. The continuous norm retains its bound. This proves sequential compactness; the countable cylinder coordinates metrize the topology on a bounded set, so compactness follows.□

Proof. Put E=\theta(Q). The subgroup E is finitely generated and hence quasiconvex in F. Thus \theta is a quasi-isometric embedding and extends to a homeomorphism \partial Q\to\Lambda E\subset\partial F. Push a current \nu by this boundary map to an E-invariant measure \widehat\nu on \partial^2\Lambda E, viewed as a measure on \partial^2F, and define

The summand depends only on the coset. It remains to prove that this sum is locally finite and continuous.

Let T_E be the convex hull of \Lambda E in T_F. Its quotient by E is the finite unbased core graph of E. For a fixed vertex v, there are only finitely many cosets pE for which v\in pT_E: indeed such cosets correspond to E-orbits of vertices p^{-1}v in T_E. A coset term in (3.8) that meets a segment cylinder has its hull containing that segment. Hence only finitely many terms contribute to a fixed cylinder. The preimage of that cylinder in \partial^2Q is compact open, because \partial^2\Lambda E is closed in \partial^2F and the boundary map is a homeomorphism onto it. It follows that the sum is Radon and that every cylinder coordinate depends continuously and linearly on \nu. The sum is F-invariant and flip-invariant by construction.

The coset decompositions in (3.8) combine the index sets F/E and E/\theta(\langle w\rangle) into F/\langle\theta(w)\rangle. This proves the assertion on counting currents, including proper powers. The same coset calculation proves functoriality. Indeed, in \psi_*\theta_*\nu the outer representatives range over F'/\psi(F) and the inner representatives over \psi(F)/\psi\theta(Q). Their products form representatives for F'/\psi\theta(Q), and the boundary maps compose. Thus the two nonnegative sums give exactly (\psi\circ\theta)_*\nu.

A nonzero measure has nonzero boundary push, and the summands are nonnegative, so its induction is nonzero. By Lemma 3.7, the norm-one set in \operatorname{Curr}(Q) is compact. The continuous function \nu\mapsto\|\theta_*\nu\|_F on that set is positive, so its minimum and maximum give (3.7). Finally, conjugating E and the boundary map simply reindexes the coset sum; the resulting F-invariant current is unchanged.□

Proof. If Q is cyclic, every nonempty reduced word is already cyclically reduced, so take b empty. Otherwise there are at least four oriented basis letters. Choose a letter b different from the inverse of the last letter of W and from the inverse of its first letter. Then Wb is freely and cyclically reduced. This choice depends only on Q and W, not on \theta.

The image in T_F of a geodesic in T_Q, with the images of successive vertices joined by geodesic segments, is a uniform quasigeodesic. Its constants depend only on \theta and the bases. The axis of Wb in T_Q passes through 1. Its image is therefore a uniform quasigeodesic line passing through 1, with endpoints the fixed points of \theta(Wb). The Morse lemma gives a constant D, independent of W, bounding the distance from 1 to the axis of \theta(Wb) in T_F. Accordingly, if

with z cyclically reduced and the concatenation reduced, then |v|_F\leq D.

Multiplication by \theta(b) changes only a bounded suffix of \operatorname{red}(\theta(W)): at most \max\{|\theta(x)|_F:x\text{ a basis letter of }Q\} letters can be canceled, and at most that many can be added. Removing v and v^{-1} then deletes at most 2D additional letters. For a fixed word u, changing or deleting a bounded number of letters at the ends changes its linear occurrence count by a bounded amount. Passing from a linear word z to its cyclic occurrence count changes the count by at most |u|-1 at each chosen orientation, even when |z|<|u|. These observations prove (3.9), including the inverse occurrences. The same bounded additions and deletions show that |z|_F=|\theta(W)|_F+O_\theta(1); now use (3.6) to obtain (3.10). All constants can be chosen for each member of a finite collection of injections while retaining the same cap.□

Proof. Take the sequence from Lemma 3.6, keeping m, all signs, and all actual bridges fixed. Write its elements as c_s^{(k)} and put L_k=|c_0^{(k)}|_P\to\infty. Lemma 2.1 shows that K_s is finitely generated. It is free because it lies in the free group f_{\epsilon_s}(P), and is nontrivial because it contains f_{\epsilon_s}(c_s^{(k)}). Fix a basis for each K_s.

Represent f_{\epsilon_s}(c_s^{(k)}) by a reduced word W_s^{(k)} in the chosen basis of K_s. Apply Lemma 3.9 to choose a cap b_s^{(k)}, and form

The injection \theta_{s,0} is quasi-isometric between free groups. Consequently |W_s^{(k)}|_{K_s} is at most a fixed multiple of |c_s^{(k)}|_P plus a fixed constant. The ratio bound from Lemma 3.6 shows that \|\nu_s^{(k)}\| is bounded. There are only finitely many s, so after passing to a subsequence all \nu_s^{(k)} converge, to currents \nu_s.

Apply Lemma 3.9 to both injections from K_s. The row equation gives

For every fixed word cylinder, the two induced normalized currents therefore differ from the corresponding normalized symmetrized linear row frequencies by O(1)/L_k. At an interior row, the two neighboring induced limits agree, since both are limits of the same linear frequencies in c_s^{(k)}. Denote the resulting current by \mu_s; at the end rows use their single adjacent transitions. Continuity from Lemma 3.8 gives (3.12).

The length conclusion (3.10) gives

along this subsequence. The middle ratio is 1 and the two end ratios are less than 2, proving (3.11). The errors used here may depend on m, the actual bridges, the bases of K_s, and the test cylinder. They disappear because m and all those data were fixed before L_k\to\infty.□

Proof. Suppress the row index and write f=f_\epsilon, l=l_\delta. Factor an actual bridge as a=f(p)d l(q) with p,q\in P. Conjugation by f(p) identifies K_d=f(P)\cap d l(P)d^{-1} with K_a=f(P)\cap a l(P)a^{-1}. Under this identification, the maps from K_a to the two edge coordinates become

By Lemma 3.8, these inner automorphisms leave the induced currents unchanged. Transporting the transition current along the group isomorphism therefore gives the desired representation. Essentiality is unchanged: for a reversal, membership of the bridge in the common side subgroup depends only on its double coset.

Each of the finitely many maps in (3.13) has upper and positive lower norm bounds from Lemma 3.8. Comparing the two bounds for a type and then taking maxima over the finite type set proves (3.14).

If K_\tau is noncyclic, assumption (i) puts d in the identity double coset. At a reversal that coset was discarded, so the signs must be equal. For equal signs the intersection is I, and direct substitution in (3.13) gives i,j for (+,+,1) and j,i for (-,-,1).□

Proof. For each m take one finite current trajectory from Lemma 3.10, express its transitions using Lemma 3.11, and put

The constants C,D in (3.14) are now independent of m.

Suppose first that D_m is bounded on a subsequence m\to\infty. Keep its original centers. On each fixed finite window the row currents have bounded norm, as do the transition currents by (3.14). The sign and type sets are finite. Successively extract subsequences on the windows [-n,n], fixing the signs and types there and taking limits of the finitely many row and transition currents. A diagonal subsequence gives currents and compatible signs and types for all s\in\mathbb Z. Continuity of the induced maps preserves the two current equalities. Norm continuity preserves \|\mu_0\|=1 and the uniform upper bound.

In the remaining case choose a subsequence with D_m\to\infty, and choose an index j_m where the maximum is attained. Divide all row and transition currents in that finite trajectory by D_m, and reindex so that j_m becomes 0. The new row norms are at most 1, and the central norm is 1. Both original end norms were at most 2. Iterating the adjacent norm comparison therefore gives

Thus

Every fixed bilateral window is eventually present in the recentered trajectory. The same diagonal extraction applies; the transition norms remain uniformly bounded after normalization. It produces a bilateral trajectory with central norm 1 and all row norms at most 1.

In either case, the finite-type equalities and their norm comparisons survive the limit. Applying the lower comparison repeatedly from row 0 gives \|\mu_s\|\geq C^{-|s|}, so none of the row currents vanishes. Compatibility of the two signs of a transition with its neighboring rows survives because these discrete data were fixed on each extracted finite window. The forbidden reversing types were excluded from the type set at the outset. This proves all three conclusions.□

Proof. Translate successive vertices of an atomic line to the identity. All resulting lines have the same positive mass and pass through the identity. That set of lines is compact and has finite measure, so two of the translates coincide. A nontrivial element of the free group therefore preserves the original line. An element reversing a line would fix a vertex or invert an edge, which is impossible for the free action on the Cayley tree. The line stabilizer consequently acts by translations and is infinite cyclic.

For its root r, the number of distinct translates of one orientation through the identity is |r|_{\mathrm{cyc}}. This is the number of vertices in the quotient axis by \langle r\rangle. Flip invariance gives the same mass to the other orientation. The definition of the norm as half the mass of the lines through the identity gives the claimed contribution.□

Proof. For each nontrivial cyclic transition group, include the roots of the images of one generator under both injections. This makes a finite set and proves (i).

For E=I,J, the strong inertia inequality with test subgroup E implies that every off-diagonal intersection E\cap xEx^{-1}, x\notin E, is at most cyclic: the diagonal term already contributes \bar r(E). By Lemma 2.2, only finitely many double cosets ExE have nontrivial intersection. For each such off-diagonal double coset, include the root in P of a generator of its intersection.

Suppose that an axis is carried by p\Lambda E and q\Lambda E, with pE\ne qE. Translate its endpoints by p^{-1} and use Lemma 2.1 to obtain

The intersection subgroup is nontrivial cyclic. If p^{-1}q=e_1xe_2, with e_1,e_2\in E and x a chosen double-coset representative, it is conjugate by e_1 to E\cap xEx^{-1}. Its axis root is therefore one of the included classes. This proves (ii).

Finally, suppose an axis root u has image r^k, where r is a root in P and |k|>1. Write E for the image subgroup, which in a noncyclic transition is I or J. We have r\notin E, since otherwise the inverse image of r would exhibit u as a proper power. The distinct cosets E,rE both carry the axis of r. Assertion (ii) proves (iii).□

Proof. Choose p\in P with p\langle r_i\rangle p^{-1}=\langle r_j\rangle, and put g=h_jph_i^{-1}. Since p fixes e, this element takes the edge h_i e to h_j e. It takes the cyclic subgroup at position i to the subgroup at position j, which is commensurable with the first one.

Let \sigma be the segment from the midpoint of h_i e to the midpoint of h_j e, in the order of the given path. At each join of two consecutive translates g^k\sigma, the incoming and outgoing segments use opposite halves of the same edge. Indeed, g preserves the canonical edge orientation, and the two positions have the same traversal sign. Thus these translates concatenate without backtracking to a bi-infinite geodesic. The element g translates that geodesic a positive distance and is tree-hyperbolic.

The cyclic subgroup at position i is infinite and elliptic. Lemma 2.5 applies to it and to g, and gives the asserted embedded Baumslag–Solitar subgroup.□

Proof. Suppose a row has an atom, and start indexing there. At each forward transition, the coset formula for the induced current has a positive contribution to this atom from an atom of the transition current. Choose one such atom and take its image in the next row. Local finiteness of the coset sum and injectivity of the boundary maps in Lemma 3.8 justify this choice. Lemma 4.1 makes every chosen line a periodic axis. Let r_s be its actual root in the coordinates of the sth edge.

We first develop these matches into an actual reduced tree path, so that the repetition criterion of Lemma 4.3 will apply. For a transition with representative d, use the arriving and departing embeddings f,l:P\to H from the annulus description. A generator k\in K_\tau of the stabilizer of the chosen upstairs axis has images of the form

The bridge

therefore matches r_s^a with r_{s+1}^b. It belongs to the same double coset as d, so it preserves essentiality at a reversal. Develop these bridges using the frame recursion (3.4). They give a reduced path with the prescribed signs, and the framed cyclic subgroups are commensurable at every step. This construction works for arbitrary nonzero exponents, including negative ones. It therefore suffices to find two positions with the same root class and traversal sign.

Call a transition exceptional for this chosen trajectory if its group is cyclic, if either chosen image line is carried by more than one coset of its image subgroup, or if an upstairs root has a proper-power image. By Lemma 4.2, at least one of the adjacent roots at every exceptional transition belongs to \mathcal D. If infinitely many transitions are exceptional, there are infinitely many such root occurrences. Some root class and some traversal sign occur together at two different positions.

If there are only finitely many exceptional transitions, discard an initial segment. On the remaining tail there is exactly one contributing coset on each side of every push formula. Hence both image atoms have exactly the mass of their common upstairs atom. The chosen atoms have one fixed positive mass on this tail. The norm bound and Lemma 4.1 give a uniform bound on |r_s|_{\mathrm{cyc}}. There are finitely many conjugacy classes of roots of bounded cyclic length, so once again a root class repeats at two positions with the same traversal sign.

In either case, Lemma 4.3 gives a Baumslag–Solitar subgroup, a contradiction.□

Proof. A nonzero current on an infinite cyclic group induces an atomic current. Thus no transition in the atomless trajectory is cyclic. Every transition is consequently aligned, all traversal signs are the same, and its maps are i,j or j,i according to that sign. Reverse the indexing if necessary. We then have currents \nu_s on the same group I satisfying

In particular I is nontrivial. Fix a free basis of I for its current norm, and define

The last equality uses positivity. Telescoping gives

First suppose a_N\to\infty. By compactness of the norm-one current space, pass to a subsequence on which S_N/a_N converges to a current \nu of norm one. A uniform norm bound controls the mass of any fixed compact cylinder: such a cylinder is contained in finitely many translates of the set of lines through the identity. The right side of (4.2), divided by a_N, therefore tends to zero on every compact cylinder. Continuity of the induced-current maps yields i_*\nu=j_*\nu. Their positive lower norm bounds show that this common current \mu is nonzero.

Now suppose (a_N) is bounded. The positive bilateral sum

has finite positive norm and defines a Radon measure: every compact line set is covered by finitely many translates of the identity line set. Monotone convergence permits induction through this sum. Reindexing the resulting positive series gives

These currents have finite norm by the upper norm bound for induction, and are nonzero. Equivalently, summability and norm comparison make the endpoint norms in (4.2) tend to zero.

In either case we must still prove atomlessness: a weak limit of atomless currents can have atoms. Use the constant bilateral trajectory with row \mu and transition current \nu. Its transitions are aligned and positive, it is bounded, and Proposition 4.4 applies. Thus \mu is atomless. Every atom of \nu would contribute positive atomic mass through an injective boundary map in the nonnegative coset formula. Thus \nu is atomless as well.□

Proof. Fix a rational endpoint \xi, and orient an axis toward it. Every oriented line with positive endpoint \xi, other than this axis, merges with the axis at a first vertex. Fix a possible merge vertex v, and let D_v be this set of lines. If r is the axis root directed toward \xi, the sets r^{-n}D_v, n\ge0, are pairwise disjoint. They all pass through v, because their merge vertices are r^{-n}v. They have equal measure by invariance, so finiteness of the measure on the compact line set through v gives \mu(D_v)=0. There are countably many merge vertices and rational endpoints. Atomlessness removes the exceptional axes, and flip invariance treats negative endpoints.

An injection preserves rationality by equivariance of its boundary map. To prove reflection, identify its image with a finitely generated subgroup E of a free group. Suppose \xi\in\Lambda E is fixed by a nontrivial element r of the ambient free group. Orient r toward \xi, and choose an axis vertex x. The vertices r^n x on the resulting axis ray are uniformly close to E, by quasiconvexity and the fact that \xi\in\Lambda E. Choose e_n\in E at uniformly bounded distance from r^n x. Among the finitely many possibilities for (r^n x)^{-1}e_n, two agree, say at n>m. It follows that

The inverse image of this nontrivial element fixes the original boundary point, proving reflection.□

Proof. Let X\subset\partial^2P consist of lines with two irrational endpoints. It is a conull Borel set by Lemma 4.6. The coset formulas for i_*\nu and j_*\nu show that almost every line of X is carried by some I coset and some J coset. The preceding uniqueness observation proves (i) wherever the carriers exist.

Let \vartheta:\partial I\to\partial J denote the boundary isomorphism induced by j, with I identified with its included image under i. On the Borel domain

define F_p(L)=\vartheta(p^{-1}L). Rationality preservation and reflection give F_p(D_p)=X\cap\partial^2\Lambda J, a Borel set. For every Borel D\subset D_p, the two push formulas each have just one contributing coset on the indicated sets, and give

The middle term is expressed in the boundary coordinates of I. These maps are homeomorphisms between their domains and images in the corresponding limit-set spaces, so the images in this formula are Borel. The same argument applies to the backward branches. Inverse branches preserve measure as well. Output translations and flip preserve it by the definition of a current. Different representatives of one coset merely insert an output translation.

For completeness, we now choose a single conull set on which all these operations can be iterated. Start in X with the Borel null set

The complement of the support is null because line space has a countable base. Saturate B under the partial forward and backward branches, their inverses, translations, and flip. There are countably many such maps and countably many finite compositions. Equation (4.3) shows that each composition takes null sets to null sets on its domain. Their saturation is therefore Borel and null. Its complement in X is the required \mathcal U. Including X\setminus\Omega among the initial discarded sets ensures that all these iterates remain in the support.□

Proof. In a frame h for the current edge, let pI be the unique forward carrier coset for its line coordinates. Choose the next edge to be hpt e, and use the forward transfer for its line coordinates. For the backward extension, the unique coset pJ gives hpt^{-1}e. Proposition 4.7 permits indefinite iteration in both directions. The frame description in Lemma 3.2 shows that these are aligned positive paths when ordered from left to right. In particular they are reduced.

Changing p to pi_0 in a forward step, with i_0\in I, does not change the next edge, since

It changes its frame on the right by j(i_0)\in P and its line coordinates by the inverse translation. Subsequent carrier cosets change by the same coordinate translation, leaving subsequent actual edges unchanged. The backward calculation uses J and the inverse isomorphism. Arbitrary changes of an output frame by P have the same effect. For example, after a forward step to frame hpt, the new line lies in \partial^2\Lambda J, so its backward choice is the identity J coset. The backward edge is then hptt^{-1}e=hpe=he, with the original line translated into frame hp. This proves (i), including compatibility of the forward and backward constructions.

To prove (ii), build a finite window one end at a time. Initially its stabilizer is P. In the frame at the end being extended, let the current stabilizer be K'. A forward extension intersects it with pIp^{-1}, and a backward extension intersects it with pJp^{-1}. Both subgroups are finitely generated subgroups of a free group. Their intersection is finitely generated, and Lemma 2.1 says that its limit set is the intersection of their limit sets. The current endpoints belong to both, by induction and by the carrier choice. They therefore belong to the new stabilizer limit set. Conjugation by the change of frame transfers this assertion to the next coordinates; its boundary map restricts compatibly to every smaller subgroup. The same intersection computation, without the endpoint assertion, proves finite generation for every finite aligned window. Induction proves the endpoint assertion in the central coordinates. A trivial or cyclic subgroup cannot carry an irrational endpoint, so the stabilizer is noncyclic.

For (iii), Lemma 3.4 gives the cyclic common stabilizer at the first divergence in either direction. The window stabilizers are finitely generated, so their limit sets intersect in the limit set of their intersection. This proves the assertion about irrational endpoints as well. The same proof compares restrictions of longer paths to any common finite shape.

An element fixing the two endpoint edges of a tree segment fixes every edge between them. Thus the stabilizer in (iv) is precisely P\cap hPh^{-1}. Successive coordinate changes along the path compose to conjugation by h^{-1}. Their boundary maps agree with the boundary map on this subgroup, giving the asserted formula. If an irrational line has its endpoints in \Lambda S, every successive carrier needed by this prescribed path exists. Its uniqueness forces the successive branch choices to be exactly those of the path. The measure-preserving branch identities in Proposition 4.7 therefore compose to measure preservation for this partial conjugation. The invariance of \mathcal U gives the last assertion.□

Proof. The window bound, Lemma 3.3, gives finitely many P-orbits of such windows. Choose representatives W_1,\ldots,W_k with stabilizers K_1,\ldots,K_k. For each i, the quotient of \operatorname{Hull}(\Lambda K_i) by K_i is a finite core graph, since K_i is finitely generated. Choose finitely many vertices v_{i,j} representing its vertex orbits.

If the hull associated to pW_i meets the fixed ball, enlarge that ball by one edge if necessary and choose a vertex x in the intersection. Then p^{-1}x=kv_{i,j} for some k\in K_i and some j. Consequently

because k stabilizes W_i. There are finitely many choices of x,i,j, proving the claim.□

Proof. The reduced-rank assertion is Lemma 3.3. In a noncyclic connected core, suppressing degree-two interiors leaves a graph all of whose vertices have degree at least three. If E,V are its edge and vertex counts and r its rank, then

Consequently E\le3(r-1) and V\le2(r-1). Sum these inequalities over the components. This bounds natural edges, not their subdivided lengths.

We describe precisely what a state of a core represents. If the representative window is W and K=K_W, a vertex Kx of C(K) represents the actual window x^{-1}W through e. This description does not depend on the chosen representative of the coset. It is injective: equality x^{-1}W=y^{-1}W says yx^{-1}\in\operatorname{Stab}_P(W)=K, or equivalently Kx=Ky. States in different representative cores cannot represent the same actual window, because that would identify their P-orbits.

A ray starting at Kx lifts in T_P starting at x and remaining in \operatorname{Hull}(\Lambda K). Its label, read from the identity, therefore has endpoint in \Lambda(x^{-1}Kx). If its endpoint is irrational, Proposition 4.8 forbids two distinct represented windows carrying that endpoint. This proves uniqueness of the starting state. Continuation from a state is unique because the core map is an immersion.

For a line in \mathcal U, the endpoints lie in the limit set of its window stabilizer by Proposition 4.8; hence its labels lift to \Gamma_n. Liftability to a fixed finite graph is closed in line space. Indeed, choose a vertex on a limiting line. On each bounded centered segment, sufficiently close lines have the same labels. There are only finitely many possible lift states for the chosen vertex, so a subsequence uses one state for arbitrarily long centered segments. The immersion then gives a lift of the whole line from that state. Since \mathcal U is dense in \Omega, every \Omega-line lifts. This is the only extension from the conull set to the entire support used here; we have not yet asserted that all support lines admit transfers.

The proof applies to any fixed window shape, using the corresponding rank bound and trajectory uniqueness.□

Proof. Suppose a^{-1}Ma\lneq M. No positive power a^k belongs to M: otherwise the nested inclusions

force equality at every stage. Thus the Schreier vertices Ma^{-n}, n\ge0, are distinct. Fix a nontrivial reduced word w\in M, of length L. At each of those vertices the word w labels a closed reduced path, since a^{-n}wa^n\in M.

The Schreier graph consists of its finite unbased core with trees attached. A closed reduced path of length L has its starting vertex at distance at most L/2 from the core: outside it, the path must travel to the core and return along the same tree attachment. The L/2-neighborhood of the finite core has only finitely many vertices. It cannot contain all the distinct Ma^{-n}, proving the first assertion. The reverse direction of conjugate containment follows by replacing a by a^{-1}.

If [Q]\preceq[M]\preceq[Q], suitable representatives give

The inclusion between the first and last groups is equality by the first assertion. Both intermediate inclusions are therefore equalities, and [Q]=[M]. Reflexivity and transitivity are immediate.□

Proof. A nonbridge edge lies on a simple cycle of length at most |E(C)|. If a is a bridge, removing its interior leaves two components, each containing a cycle. Indeed, a component without cycles would be a finite tree with at most one possible vertex of degree one, which is impossible. On each side choose a shortest path from the endpoint of a to a simple cycle, traverse the cycle, and return along that path. The resulting based path is reduced and has length at most twice the number of edges on that side. Combine these two paths with a and a^{-1}. The joins are reduced, including the cyclic join, because the based paths remain on their respective sides of a. The resulting immersed circuit contains a and has length at most 2|E(C)|.□

Proof. Let M be a noncyclic subgroup given by loops at a chosen vertex of a small component. These same loop labels occur at the corresponding state of every selected \Gamma_n. By the state interpretation in Lemma 5.1, M fixes an actual centered window of radius n. The windows obtained for different n have identical restrictions to any common shape, because M is noncyclic. Their compatible restrictions form a bilateral positive path fixed by M. Thus [M]\in\mathcal C.

Now take [Q]\in\mathcal C, and let C(Q) be its fixed unbased core. Put

For every selected n, the group Q is contained in the full stabilizer of its centered radius-n window. After translating to one of our representatives, this inclusion gives a label-preserving immersion

One way to see the map directly is to include the minimal Q-invariant subtree in the minimal subtree of the containing group and take quotients. Conjugating the containing group merely translates its subtree and does not change labels.

Choose n so large that every divergent natural edge of \Gamma_n has length greater than D_Q. Suppose the image of some edge of C(Q) meets the interior of one of those natural edges. By Lemma 5.4, that source edge lies on an immersed circuit of length at most D_Q. Its image remains immersed. At the degree-two interior of a natural edge there is no reduced way to turn around, so the image circuit traverses that entire natural edge. This remains true if its chosen starting point is in the interior: view the traversal cyclically. Its length is then greater than D_Q, a contradiction.

The whole image of C(Q) therefore lies in \Gamma^{\mathrm{sm}}. Since it is connected, it lies in one small component, with subgroup M. The induced inclusion gives [Q]\preceq[M], and M is noncyclic because it contains a noncyclic subgroup. Notice that no compatible choices of conjugating elements were required as n varied: once all long edges exceed D_Q, the argument applies to every possible label-preserving immersion.

There are finitely many small components. A maximal member of their noncyclic subgroup classes is maximal in all of \mathcal C: any larger class lies below another member of the finite family and hence would contradict its maximality there. Conversely a maximal class in \mathcal C equals a small class above it. This proves the final assertion.□

Proof. Choose a frame h for the next edge on the unique path fixed by Q. Transfer gives h^{-1}Qh\le P. Another frame has the form hq, q\in P, and conjugates this output inside P. Conjugating Q inside P translates its path and gives the same output class. Thus F is well defined. Moving backward is its inverse. If Q\le M after choosing representatives, both groups fix the same path by noncyclic uniqueness; the same partial conjugation preserves their inclusion. This proves the order assertions.

An order automorphism permutes maximal elements. There are finitely many of them by Proposition 5.5, so F^p[M]=[M] for some p\ge1. If their number is N, one may choose p\le N. Let h be a frame for the pth forward edge. There exists c\in P with

Set v=hc. Then v has height p and v^{-1}Mv=M, which proves that the return map is an onto automorphism.

Suppose \alpha^k(w)=u^{-1}wu for some nontrivial w\in M, u\in M, and k\ge1. Then

commutes with w. Its height is kp\ne0, whereas w has height zero and infinite order. A relation w^a g^b=1 forces b=0 by height, then a=0. Thus \langle w,g\rangle\cong\mathbb Z^2=BS(1,1), contrary to BS-freeness. This proves atoroidality without using hyperbolicity of G or the flaring statement under investigation.□

Proof. Apply Brinkmann’s theorem to \alpha, and put \gamma=\alpha^{b_0}. If |\gamma(w)|_M\ge\lambda_0|w|_M, applying the inequality to \gamma(w) shows that its expanding direction must again be forward: its backward image has length |w|_M, which is strictly less than \lambda_0|\gamma(w)|_M for w\ne1. Induction gives

The analogous statement holds for a backward initial expanding direction. Choose k\ge1 with \lambda_0^k>4, and set b=b_0k, \beta=\alpha^b, and \lambda=\lambda_0^k. This proves (5.1); it also holds for w=1. Apply it to w^q, divide by q, and let q\to\infty. The identity \lim_{q\to\infty}|w^q|_M/q=\ell_M(w) proves (5.2).

For a counting current \eta_w, the cyclic inequality and \|\eta_w\|_M=\ell_M(w) imply (5.3), with the maximum replaced by the sum. The sum form is linear and therefore holds for all finite nonnegative linear combinations of counting currents. We give the approximation argument extending it to every current.

Fix k\ge2. The directed graph with vertices the reduced words of length k-1 and edges the reduced words of length k has an edge from a word’s prefix to its suffix. The masses \langle u,\sigma\rangle of directed segment cylinders based at the identity and labeled by length-k words u form a nonnegative circulation: both incoming and outgoing sums at a vertex equal its length-(k-1) cylinder mass. Any such circulation is a finite nonnegative sum of directed cycle circulations. Indeed, follow positive-flow edges to find a directed cycle, subtract the minimum edge flow on that cycle, and repeat; at least one positive edge disappears at each step.

Write the circulation as \sum_i a_i C_i. Each directed cycle is the block circulation of a cyclically reduced word w_i, including multiplicities if that word is a proper power. Let C_i^{\mathrm{flip}} be the circulation obtained by reversing words and inverting letters. The counting current \eta_{w_i} has circulation C_i+C_i^{\mathrm{flip}}. Since \sigma is flip-invariant, the current

has exactly the same masses as \sigma on all length-k cylinders. Summing over extensions gives equality on all shorter word cylinders as well. This also covers graph loops and period-one words; proper powers have their prescribed counting multiplicities.

As k\to\infty, the currents \sigma_k converge to \sigma on every cylinder, hence in the current topology. Their norms are equal to that of \sigma, since length-one masses already agree. Continuity of the norm and of the current maps induced by \beta^{\pm1} now proves (5.3).

Finally, the return paths are reduced as observed after Proposition 5.6. Fixing their first and last edges is equivalent to fixing the intervening segment, so their full stabilizers in P are exactly S_+ and S_- above. Since v normalizes M, both contain M. Their partial conjugations are x\mapsto v^{-b}xv^b and x\mapsto v^bxv^{-b}, whose restrictions are \beta and \beta^{-1}. The boundary and measure assertions follow from Proposition 4.8.□

Proof. Use the boundary embedding of M in P to pull back the restriction of \mu to the relatively closed subset \partial^2\Lambda M of \partial^2P. This gives an M-current \sigma: invariance and flip-invariance come from \mu, and the restriction is Radon because the boundary embedding sends compact subsets of \partial^2M to compact subsets of \partial^2P.

Both endpoints are irrational outside a \mu-null set by Proposition 4.7. On those lines the return transfer \Theta_+ is measure-preserving, maps \partial^2\Lambda M onto itself, and agrees with \beta. Thus \beta_*\sigma=\sigma; the inverse has the same property. Equation (5.3) gives

Since \lambda>4, this forces \sigma=0. Hence the restricted measure is zero. Translation invariance of \mu gives the last assertion.□

Proof. Fix Hausdorff bounds between each tightened image line and the images of vertices of L. Choose a vertex u\in L whose a-image is uniformly close to a point of L_a closest to a(q). The lower quasi-isometry bound for a gives a bound on d_Q(q,u) depending only on D and a. The upper bound for b, followed by its line Hausdorff bound, proves the first assertion.

For the order assertion, first consider actual vertices u<v on an oriented source line and one fixed embedding f. The image of v is within a fixed distance of the geodesic from f(u) to f(z) for every source vertex z sufficiently far beyond v. Projecting to the tightened image line and letting z tend to its positive end shows that the projection of f(v) cannot lie more than a fixed distance behind that of f(u). On the other hand, the distance between these projections is at least

for fixed c_f>0,C_f\ge0, by the lower embedding bound and the line Hausdorff bound. Thus sufficiently separated vertices have projections in the correct order, with separation tending to infinity.

Apply this first to a and then to b. Replace q_i by vertices u_i\in L at the uniformly bounded distance obtained in the first paragraph. Their image projections change by uniformly bounded amounts. Sufficient positive separation of the a(q_i) therefore forces u_1<u_2 with large separation, and then forces the asserted order and separation of the b(q_i). All constants came from the fixed embeddings and D.□

Proof. Choose Q bounding the distance of every M-vertex on a geodesic segment or line to its tightened P-image, and bounding the reverse distance from that image to images of vertices. Increase it if necessary. Applying Lemma 6.1 to the two fixed window maps gives a common C_1 such that u=\beta^\epsilon x and z=\beta^\epsilon y are within C_1 of every line in \mathcal A'.

Choose lower embedding constants a>0,b\ge0 with

Let B be a common vertex-to-line bound for the two automorphisms \beta^{\pm1} of M. Choose an integer R such that

Delete R edges at both ends of [u,z] in \mathcal C_M, and call its new endpoints x',y'. The first inequality in (6.3) puts them on the tightened line \beta^\epsilon L. Their P-images lie within Q of [u,z]_P. If v is either new endpoint, its projection to this segment has distance at least aR-b-Q>C_1 from each end. The intervening part of [u,z]_P lies on every line of \mathcal A', by the tree observation. Hence x',y' are still within Q of every such line.

Put

The new length l' satisfies

Choose L_* sufficiently large that, for every l\ge L_*,

Then the opposite direction at the new pair has length strictly less than \lambda l'. Inequality (6.1) forces expansion in the original direction. Measure preservation and membership in \mathcal U follow from the invariant conull transfer construction. Since the maps are fixed boundary homeomorphisms on their domains, the image set is Borel.□

Proof. Choose L_0\ge L_* large enough that the projections of x_0,y_0 to \eta are distinct and occur in that order. This choice is uniform by Lemma 6.1; we will increase L_0 once more below. The cylinder prescribing the directed segment between those projections has positive finite measure, since \eta\in\Omega. Its intersection with \mathcal U is a Borel set \mathcal A_0 of positive finite measure, every line of which is within Q of x_0,y_0.

Attempt to apply \Theta_\epsilon successively to the entire current set. At time n, call the attempt a failure if \mathcal A_n is not contained in \partial^2\Lambda S_\epsilon. Otherwise define \mathcal A_{n+1}=\Theta_\epsilon(\mathcal A_n) and use Lemma 6.2 to obtain the next segment. As long as the process continues, its vertices x_n,y_n lie on \beta^{\epsilon n}L, their distance l_n is at least 2^n l_0, and every line of \mathcal A_n lies within Q of both vertices. Its measure is the fixed number \delta=\mu(\mathcal A_0)>0.

The process cannot continue forever. Choose an M-vertex a_n within one edge of the midpoint of [x_n,y_n]_M and translate \mathcal A_n by a_n^{-1}. The central portion of [x_n,y_n]_P belongs to every line of \mathcal A_n. The image of a_n is uniformly close to this segment, and the lower embedding bound puts its projection arbitrarily far from both endpoints as n increases. That projection therefore lies in the common central portion. Thus the translated sets eventually lie in one compact set \mathcal K of lines meeting a fixed ball. Their measures are still \delta.

Every subsequential limit of these lines belongs to \partial^2\Lambda M. To see this, take a subsequence of the centered M-segments. Both halves tend to infinite length, so a further subsequence converges to a bi-infinite M-geodesic. Its quasigeodesic image in P has two distinct boundary endpoints, and its tightened line is the limit of the growing common central segments. These endpoints lie in \Lambda M.

Consequently every relative neighborhood of \mathcal K\cap\partial^2\Lambda M in \mathcal K contains all sufficiently late translated sets. Otherwise points outside one such neighborhood would have a contrary subsequential limit in the compact set \mathcal K. Outer regularity of the finite measure on \mathcal K now gives

contrary to (6.2).

A failure produces a bounded-period axis. Suppose failure occurs at time n, and choose \ell\in\mathcal A_n\setminus\partial^2\Lambda S_\epsilon. This line has an actual trajectory for the same finite one-sided window. Its stabilizer K is noncyclic, and this actual path differs from the prescribed path: equality would put both endpoints of \ell in \Lambda S_\epsilon.

Both \ell and the tightened image \eta_n of \beta^{\epsilon n}L pass within Q of x_n,y_n. Since x_n,y_n\in M\le S_\epsilon, their distances from \operatorname{Hull}(\Lambda S_\epsilon) equal the fixed number d_P(1,\operatorname{Hull}(\Lambda S_\epsilon)). Set

Deleting H_\epsilon from both ends of [x_n,y_n]_P leaves a segment in \ell, in \eta_n, and in the prescribed stabilizer hull. It also lies in \operatorname{Hull}(\Lambda K) because \ell does. Its length is at least

where a>0,C\ge0 depend only on the fixed maps and embeddings. In particular, no constant from the varying competing subgroup K is needed.

Read this segment in the two finite stabilizer cores. Their fiber product pairs core states, with an edge for each pair of equally labeled edges. At an actual vertex z the corresponding pair of states represents the two paths translated by z^{-1}. They remain distinct. Every loop in the corresponding component of the fiber product therefore fixes both paths, so its fundamental group injects into an at-most-cyclic group, by Proposition 4.8. The component has rank at most one. Since the window is fixed, there are only finitely many core representatives, and the numbers of unoriented edges in all relevant fiber products are bounded by one integer D\ge1. Conjugating a stabilizer preserves its labeled unbased core and changes only the state used for this reading.

Increase L_0 so that aL_0-C>2D+1, taking the maximum of the constants for the two signs. This ensures that every failure reading is long enough to contain a circuit. A reduced path in a finite tree has length at most its number of edges. A connected rank-one finite graph consists of one circuit with trees attached, so a sufficiently long reduced path follows that circuit between initial and terminal pieces of length at most D. Our reading therefore has a central periodic portion of length at least a l_n-C-2D. Extend it in the absolute Cayley tree to an oriented periodic line \zeta. Its primitive P-period is at most D. Unfolding the circuit in each core puts its endpoints in both actual window stabilizer limit sets. Since l_n\ge2^n l_0, this proves (6.5), with constants enlarged uniformly over the two signs. The line \ell belongs to \mathcal U, as required.□

Proof. Translate the given line into \partial^2\Lambda M. Choose successively longer segments on its corresponding M-line and, for each segment, a sign supplied by (6.1). Apply Lemma 6.3, passing to a subsequence with the same sign. Its support line and bounded-period axis share a segment whose length tends to infinity.

Recenter this shared segment by an M-translation chosen near its midpoint on \eta_n. The translated periodic axes have period at most D and meet a fixed ball. There are only finitely many such axes: there are only finitely many cyclic labels of length at most D, and finitely many ways for one of their axes to meet that ball. Along a subsequence the periodic axis is fixed. The translated lines \ell\in\Omega agree with it on segments tending to infinity in both directions, so they converge to it. Closedness of \Omega proves that this periodic line belongs to \Omega.□

Proof. Choose an integer B\ge0 such that the image of any vertex on any line is within B of the tightened image line. Such a uniform constant exists by the quasigeodesic line bound for the fixed map \psi. Put q=2B+1 and C_4=2q.

Write L_i'=\psi L_i for the tightened images and z'=\psi(z). The line L_2' is the axis of z'. If a is the projection of \psi(u) onto this axis, the projection of the other endpoint image is exactly

because projection onto an invariant line commutes with its translation. Both endpoint images lie within B of both output lines. Thus a,z'^ka lie within 2B of L_1'.

The translation length of z' is an integer at least one. The vertices z'^q a and z'^{k-q}a therefore lie farther than 2B from both ends of [a,z'^ka] and occur in the stated order. The tree observation puts their joining segment on L_1', and it already lies on L_2'. It has exactly k-2q=k-C_4 full periods, with both endpoints in the same integer z'-orbit. There is no additional phase loss. Automorphisms preserve being a non-power, proving the final assertion.□

Proof. Number the common positions 0,\ldots,a+b-1. For 0\le i<a, the letter at position i equals the letter at i+b by the second period. By the first period this equals the letter at (i+b)\bmod a. Thus the first period pattern is invariant under addition by b modulo a, and is periodic with period d=\gcd(a,b). The first b common letters then show that the second pattern is the same d-periodic pattern. Both entire sequences therefore agree. A label and an initial vertex determine a unique path in a Cayley tree, so equality of the sequences and the shared segment identifies the actual lines.□

Proof. Let V be the number of vertices in the finite core of M in the P-rose, and take L_D=(V+1)D. The common segment lies in \operatorname{Hull}(\Lambda M). If r is the actual primitive P-translation of \zeta, that segment contains V+1 successive vertices r^i z_0 in one period phase. Two have the same core state:

Cancelling z_0 gives r^{j-i}\in M. Thus a power of the translation of this very axis belongs to M, and the whole line has endpoints in \Lambda M.

If z is its primitive translation in M, it is the smallest positive power of r belonging to M, up to orientation. Hence z=r^q for some 1\le q\le V. Choose embedding constants with

Apply this to powers of z, divide by the exponent, and take the limit to obtain

Thus B_D=AVD suffices. The two primitive translations need not coincide, which is why both groups are specified.□

Proof. Suppose \eta_0\in\Omega is periodic.

Step 1: construct a trajectory and locate a maximal carrier. We first construct one compatible trajectory for \eta_0; uniqueness at its rational endpoints is not assumed.

Approximate \eta_0 by lines of \mathcal U. For each fixed centered window, their stabilizer hulls meet a common finite ball. Lemma 4.9 lets us pass to a subsequence with a constant actual window path. Taking these subsequences successively gives a bilateral aligned path whose every finite window has noncyclic stabilizer. The limiting endpoints belong to each such stabilizer limit set. Choose frames h_s on this path, with h_0=1. Boundary transfer on a fixed window is continuous, so the corresponding conull image lines converge in the successive frames to lines \eta_s\in\Omega. They are periodic: the core-state repetition argument in Lemma 6.7 puts a power of the original axis translation in each fixed window stabilizer, and this power is transferred to a nontrivial element translating the image line.

Choose the primitive P-translation r_s of \eta_s toward its positive endpoint. Write a forward step as h_{s+1}=h_s p_s t q_s, with p_s,q_s\in P. There is a least positive integer m_s with r_s^{m_s}\in p_sIp_s^{-1}. Its image is r_{s+1}^{n_s}, where n_s is the least positive integer with r_{s+1}^{n_s}\in q_s^{-1}Jq_s. Minimality on the second side follows by applying the inverse subgroup isomorphism to its full axis stabilizer. The exponents are positive because we oriented each translation toward the transferred positive endpoint.

Recall the finite ambiguity dictionary of Lemma 4.2: if either minimum exponent exceeds one, or if one of these axes belongs to two distinct domain cosets on its side, one of its primitive root classes lies in a fixed finite set, up to conjugacy and inversion. Infinitely many such indices would repeat a root class along this actual reduced path. Lemma 4.3 would give a tree-hyperbolic element commensurating a nontrivial elliptic cyclic group, contrary to BS-freeness. Therefore only finitely many indices in the entire bilateral trajectory are exceptional in either way.

It follows that one can choose positive integers k_s for all s\in\mathbb Z with

Only finitely many ratios differ from one. Choose k_0 divisible by all the finitely many factors needed when propagating these relations in both directions. Then

is one fixed nontrivial element, independent of s.

Fix s temporarily. The cyclically tightened circuit representing r_s^{k_s} lifts closed to a component of each centered window core \Gamma_n, after recentering the trajectory at h_se: the edges E_j'=h_s^{-1}E_{s+j} have E_0'=e and so belong to the same family of windows defining \Gamma_n. Its length is fixed as the window size tends to infinity. It cannot traverse a natural edge whose length tends to infinity: any immersed closed path using the interior of such an edge traverses the whole edge, cyclically. Thus \eta_s is carried by the fixed small subgraph. If infinitely many positive times were carried only by cyclic small components, their primitive P-roots would again belong to finitely many conjugacy classes and repeat. We may therefore choose a time beyond every exceptional forward index at which the periodic line is carried by a noncyclic small subgroup. Enlarge it to a maximal carrier M and translate the line into \partial^2\Lambda M. If there were no maximal noncyclic carrier, this conclusion itself would be a contradiction.

Step 2: follow expanding return iterates. The subsequent forward path agrees with the path fixed by M. At its first edge, both choices allow the periodic endpoints, and the domain coset is unique at this late time. Transfer and repeat to get agreement at every forward edge. In particular all positive \beta-iterates of the line belong to \Omega and have unique forward choices along the prescribed return windows. This conclusion comes from the constructed trajectory, not from an as yet unproved support-invariance assertion.

Let a_j\ge1 be the primitive cyclic M-length of its jth positive \beta-iterate. At some time one has a_{j+1}\ge\lambda a_j. Otherwise the cyclic version of (6.1), applied at every j\ge1, would give a_{j-1}\ge\lambda a_j, impossible for positive integer lengths. Once a forward expansion occurs, it continues: the backward neighbor is too short to supply the next expanding alternative. Restart at such a time, and call the initial primitive period length d_0.

Let z be the positive primitive M-translation of this initial line. Choose a vertex x_0 on its axis and y_0=z^k x_0, where k is arbitrarily large. Then l_0=kd_0. The cyclic expansion gives \ell_M(\beta z)\ge\lambda d_0, and therefore

This verifies the based-segment hypothesis even if \beta x_0 is off the axis of \beta z. For l_0\ge L_0, Lemma 6.3 now supplies a failure time n, a support line, and a bounded-period axis sharing (6.5) with the nth return image.

Step 3: pull a failure axis back to the initial period. The resulting periodic P-line \zeta has uniformly bounded period, allows both the prescribed and a competing path, and shares with the current M-line a segment of length at least (6.5). Take l_0 large enough that (6.5) exceeds the fixed threshold in Lemma 6.7, already at n=0. That lemma implies that \zeta is itself an M-line, with primitive M-period bounded by (6.7).

The two corresponding M-lines share a segment of length at least c_0 2^n l_0-C_0, for fixed c_0>0,C_0\ge0. Here are the details of this change of metric. Near each end of their common P-segment, choose a vertex on each M-line whose P-image is within the fixed line Hausdorff bound. At each end the two chosen vertices are uniformly close in M, by the lower embedding estimate. The vertices chosen near opposite ends are separated in M by a positive multiple of the P-length minus a constant, by the upper embedding estimate. Choose the pairs in their oriented order, as justified by Lemma 6.1. In a tree, the two joining segments therefore have a common central segment after deleting a fixed amount at both ends. This gives the asserted lower bound.

The uniformly bounded primitive period of \zeta now shows that this common segment contains at least

whole primitive \zeta-periods, for fixed c'>0,C_3\ge0. Apply Lemma 6.5 with \psi=\beta^{-1} successively n times. At time zero at least

whole periods remain. Choose the initial length so large that

Since 2^n\ge n+1, expression (6.10) exceeds d_0+1 for every possible failure time n. In particular all intermediate applications of the pullback lemma have enough periods.

The primitive period of the second line after these pullbacks may be large; call it b\ge1. The number of full periods, rather than its metric length, was controlled. More than d_0+1 periods give a common label segment of length at least (d_0+2)b\ge d_0+b. Lemma 6.6 identifies the original periodic line and the pulled-back \zeta as the same actual M-line. Applying \beta^n identifies their lines at failure, in both the M- and P-trees. This line allows both the prescribed and the competing forward window. That contradicts the unique forward coset choices on the late periodic trajectory. Failure is impossible as well, completing the contradiction.□

Proof. An immersed bi-infinite path in a finite cyclic small component is periodic, and such a line is excluded by Proposition 6.8. A line carried by a noncyclic small component is carried, up to conjugacy, by a maximal noncyclic carrier. Proposition 6.4 would then produce a periodic support line, also impossible. A tree component cannot carry an immersed bi-infinite path. This proves the first assertion about whole lines.

If small-only segments of lifted support lines had unbounded length, center progressively longer such segments at an actual vertex and translate that vertex to the identity in P. The translated lines remain in \Omega and lie in the compact set of lines through the identity. Since the small graph is fixed and finite, a subsequence of the centered lifted segments converges to an immersed bi-infinite path in it. The corresponding limit support line would be carried by the small graph. This contradiction gives the uniform bound B_0.

A rational endpoint is represented by an eventually periodic ray. If such a ray lay on a support line, translating farther and farther along its periodic tail would give a periodic line as a limit of support lines. Closedness of \Omega and Proposition 6.8 exclude this. Both endpoints of every support line are therefore irrational.

The existence of lifts to each finite core was already proved in Lemma 5.1: it follows from conull lifts, density, and closedness of liftability into a fixed finite graph. Each such lift gives an actual window path carrying the line’s endpoints. The irrational-endpoint uniqueness in Proposition 4.8 makes these paths compatible when a window is shortened. They therefore give a unique bilateral trajectory.

It remains to check the support of transferred lines. Approximate a given support line by lines of \mathcal U and fix one window. All approximants meet a common fixed ball, so the hulls of their window stabilizers do too. There are only finitely many possible actual paths. Any path occurring infinitely often carries the limiting endpoints, because its limit set is closed; irrational uniqueness identifies it with the limiting line’s own path. Thus the approximants eventually have that same path. The boundary transfer on its fixed subgroup is continuous. Their images belong to \mathcal U\subset\Omega and converge to the transferred limiting line, which is in \Omega by closedness. This proves the final assertion without discarding any support points.□

Proof. Let \rho be left-special. Both of its predecessor lines lift to each \Gamma_n by Lemma 5.1. Their common endpoint is irrational, so the uniqueness of a starting state for an irrational ray puts both lifts at the same vertex. This vertex has two different incoming edges as well as the outgoing ray edge. It is therefore a natural vertex.

Let B bound the lengths of small-only segments of support lines, as in Corollary 6.9. Starting at this natural vertex, the lift of \rho follows at most B labelled edges in the small subgraph before entering a long natural edge. There are only finitely many possible prefix words of length at most B and finitely many oriented natural edge types. On a subsequence the prefix and the oriented type are fixed. The lengths of these long edges tend to infinity, and their oriented labels have a fixed infinite prefix limit by the construction of \Gamma_n. Thus \rho is the concatenation of one of finitely many prefixes and one of finitely many prescribed infinite words. This proves finiteness.

Suppose now that no occurring ray is left-special. Then there is an integer N such that every occurring word of length N has a unique predecessor letter. Indeed, otherwise arbitrarily long finite words would have two distinct predecessor letters. Passing to a fixed pair of letters and a convergent subsequence of their full-line extensions would give a left-special infinite word. This uses compactness of the space of lines through a vertex and the closedness of \Omega.

Form the finite directed graph whose vertices are occurring words of length N and whose edges are occurring words of length N+1, an edge running from its length-N prefix to its length-N suffix. Every vertex has indegree one and outdegree at least one. The sum of the indegrees is the number of vertices, so every outdegree is also one. The graph is a disjoint union of directed cycles. Every line in \Omega would therefore have a periodic label, contrary to Proposition 6.8.□

Proof. If |f(n)-n| were bounded along an infinite subsequence, then s_n^{-1}\Theta(s_n) takes only finitely many values on that subsequence. The elements s_n escape every finite ball. Hence there are distinct s_n,s_m and d\in P with \Theta(s_n)=s_nd and \Theta(s_m)=s_md. It follows that

The nontrivial element s_ns_m^{-1} has infinite order and commutes with v. Their heights are respectively zero and p\ne0, so they generate a subgroup isomorphic to \mathbb Z^2. This contradicts the Baumslag–Solitar exclusion.

Thus |f(n)-n|\to\infty. The signed sequence f(n)-n has bounded successive increments by (7.2), so it is eventually either positive or negative. The positive case is the assertion. In the negative case choose t_m\in T uniformly close to r_m and integers g(m) such that \Theta^{-1}(t_m) is uniformly close to r_{g(m)}. Because f tends to infinity with bounded increments, a first-crossing choice n(m) satisfies

Then t_m and \Theta(s_{n(m)}) are uniformly close. Applying the inverse quasi-isometry gives

This proves the assertion for the inverse map.□

Proof. For the first assertion, a contrary sequence has a convergent subsequence in the compact space of directed lines through r_s. The limit belongs to \Omega and contains the whole suffix [r_s,\xi). The condition of not containing the fixed finite segment [r_{n_0},r_s] is closed in this compact space. The limit would therefore contradict the unique-past property just proved.

For the domain assertion, each competing line has an actual path for the fixed window. Its stabilizer hull contains that line, hence meets the vertex r_s. There are only finitely many such actual paths by Lemma 4.9. A contrary sequence therefore has a subsequence with one fixed competing path. Its stabilizer limit set contains the limiting positive endpoint \xi. Uniqueness for an irrational endpoint identifies this path with the path defining \Theta, which is a contradiction.□

Proof. Choose a spanning tree T_\Delta of \Delta and extend it to a spanning tree T of \Gamma. The edges outside T give the basis \Sigma, and those in \Delta\setminus T_\Delta give the indicated subset. Collapsing T turns \lambda into a nonempty cyclically reduced word: cancellation between consecutive non-tree edges, including at the cyclic join, would require a closed tree segment between an edge and its inverse. A nonempty such segment cannot be reduced in a tree, while an empty segment would already be backtracking in \lambda.

Suppose that this word used only the letters from \Delta. Between consecutive non-tree edges of \lambda, its tree segment is the unique tree path between vertices of \Delta. That path lies in T_\Delta. Every segment of \lambda would therefore lie in \Delta, contrary to the hypothesis. The omitted-letter condition follows. The injection is the Freiheitssatz; see [18, Theorem 1].□

Proof. Choose a spanning tree downstairs. We may then index the lifted vertices by (v,n) so that tree edges preserve n, while each oriented non-tree edge has an integral increment d_e. These increments generate \mathbb Z. Let D_N be the subgraph induced by all vertex types at levels [-N,N]. For sufficiently large N, every base edge occurs and its occurrences form an interval.

We verify connectivity, including near the boundary of the interval. Choose a word in the signed increments whose sum is 1, and bound the absolute values of all its partial sums by L. Its translates connect consecutive levels whenever those levels are sufficiently far from the ends of [-N,N]; the lifted spanning trees connect the different vertex types at each level. Choose also a nonzero increment, with absolute value a. From a level near the lower boundary, repeated steps of size a in the positive direction reach the interior, and from the upper boundary repeated negative steps do the same. For N>L+a, these steps remain in the interval and reach the connected interior. Thus D_N is connected. The identical argument applies to the interval [-N+1,N], whose induced subgraph is D_N\cap\tau D_N.

Among domains with connected nonempty adjacent overlap, choose D with the fewest edges. If D\cap\tau D contains a circuit, choose a simple one and translate it upward as long as it remains in D. Every position reached remains in D\cap\tau D, because its preceding translate lies in D. At the last position, some edge e of the circuit has \tau e\notin D. Since e\in D\cap\tau D, its preceding translate \tau^{-1}e does lie in D. Delete the open edge e, retaining its endpoints. It was the highest occurrence of its base edge, so all occurrence intervals remain nonempty intervals. Moreover,

because \tau e\notin D. Both graphs remain connected: the deleted edge lies on a circuit in their common overlap. This contradicts minimality. The overlap is therefore a nonempty tree.□

Proof. We first construct U and prove the complexity decrease, using two cases according to the projection of the lifted support. We then verify the HNN splitting and the simultaneous Magnus bases for both constructions. This is the domain method of [12, Propositions 4.9 and 4.11].

Case 1: the lifted support does not project injectively on vertices. Take D from Lemma 8.2 and let

where m\geq0 is the least integer for which this union contains some closed lift \widetilde\lambda. Such an integer exists: translate a fixed finite lifted circuit far enough in the positive direction, and then take a sufficiently long interval of translates of D. The graph U is connected and its cell occurrences are intervals. Indeed, an occurrence interval [a,b] in D becomes [a,b+m] in U. Figure 3 shows how taking adjacent overlaps trims the endpoints of this interval.

If m=0, both adjacent overlaps are trees. If m\geq1, the interval calculation gives

Minimality of m shows that neither contains a closed lift of \lambda. Their intersection is

It is connected and nonempty. In the case m=0, a tree cannot contain the nontrivial immersed circuit, so the overlaps are again Magnus subgraphs. The lifted circuit has the same length as \lambda and strictly more support vertices. Thus the first coordinate of (8.1) has decreased.

Case 2: the lifted support projects injectively on vertices. It then projects isomorphically onto S_\lambda: two lifts of the same oriented edge cannot have the same initial vertex. The character therefore vanishes on \pi_1S_\lambda, and every lift of this support is a copy of it. Choose a spanning tree T_S of S_\lambda and extend it to a spanning tree of \Gamma; in the resulting level coordinates each support copy lies at one level.

Contract the support downstairs and every support copy upstairs. This gives the connected cyclic cover of the quotient graph \Gamma/S_\lambda corresponding to the still-surjective character. Apply Lemma 8.2 there. Expand every occurrence of the contracted vertex in the resulting domain by a copy of T_S. At one chosen occurrence, add all the remaining edges of S_\lambda. Call the resulting graph U. It is a connected domain: the support-tree cells have the same occurrence interval as the contracted vertex, and each added support edge has a singleton occurrence interval. The overlap U\cap\tau U is a tree, obtained by expanding vertices of a tree into trees; none of the added support edges lies in it. The full support at the chosen occurrence contains a closed lift \widetilde\lambda. Both overlaps are trees and hence Magnus subgraphs.

Here the first complexity coordinate is unchanged and the second decreases. To see this directly, for each base open cell its number of occurrences in U exceeds its number of occurrences in U\cap\tau U by one. Counting vertices and edges gives

Since the overlap is a tree, \chi(U)=\chi(\Gamma)+1, and hence

The HNN splitting. Both constructions now have the required overlaps and complexity decrease. We recover X from the translates of Z, as in [12, Proposition 4.3]. Since \phi([\lambda])=0, the graph cover extends over the cells to the cyclic cover Y\to X. The translates \tau^iZ cover Y, and their two-cell interiors are distinct. Form a line of disjoint copies of Z, joining each adjacent pair by the cylinder on its intersection. Denote this space by W; collapsing these cylinders gives an equivariant map W\to Y. For a finite interval of copies, the intersection of a newly adjoined copy with the preceding union is precisely its adjacent overlap. Indeed, if a graph cell lies in the new copy and in an earlier copy, its occurrence interval also puts it in the immediately preceding copy. Distinct copies have no common two-cell interior, and cell boundaries consist of graph cells, so this checks the entire intersection. Van Kampen therefore shows inductively that the cylinder-collapse map induces a fundamental-group isomorphism over every finite interval. Every loop and every homotopy uses only finitely many copies, so W\to Y induces an isomorphism on fundamental groups as well.

Taking the quotient by translation gives a graph of spaces with one vertex space Z and one edge space, and a map W/\langle\tau\rangle\to X. The two covering exact sequences have quotient \mathbb Z; the map is an isomorphism on their kernels and on this quotient. It follows that it induces an isomorphism on fundamental groups. The edge maps are injective by Lemma 8.1, because each overlap omits a traversed relator edge. Thus this graph of spaces gives the HNN extension (8.2), and the normal form theorem gives injection of H into \pi_1X.

Simultaneous Magnus bases and the edge isomorphism. It remains to choose the two Magnus subsets simultaneously and specify their basepoint paths. If the overlap groups are trivial, this is immediate. Otherwise put U_0=U_-\cap U_+, which is connected and nonempty by the construction. Extend a spanning tree of U_0 to spanning trees of U_- and U_+. Their intersection is exactly the chosen tree of U_0, so their union is a tree. Extend that union to a spanning tree of U. With basepoint in U_0, the non-tree edges of the two overlaps are then subsets of a single free basis of \pi_1U. Their common letters freely generate C=\pi_1U_0, which is a free factor in each side group.

To specify the stable letter in this nontrivial-edge case, choose x\in U_0 and a path q\subset U_+ from x to \tau x. The latter vertex belongs to U_+ because x\in U_-. Then t=[p(q)] has height one, and the edge isomorphism sends a loop a in U_- to q\,(\tau a)\,q^{-1} in U_+. Its projection is tat^{-1}, giving the specified isomorphism in (8.2). When the overlap groups are trivial, choose instead any x\in U_- and a path q\subset U from x to \tau x. Again t=[p(q)] has height one, and the edge relation is empty. In either case, q in the copy Z closes in the quotient graph of spaces by crossing the cylinder to the copy \tau Z at \tau x. It crosses the quotient loop edge exactly once, so it is a stable letter; collapsing the cylinder sends it to p(q).

Finally, \pi_1U injects into \pi_1\Gamma, by inclusion into the graph cover and its covering map. If the lifted relator were a proper power in \pi_1U, its image would be a proper power in \pi_1\Gamma. This proves the remaining assertion.□

Proof. We induct on the nonnegative lexicographic complexity (8.1). In graph rank one, the non-power relator generates the infinite cyclic fundamental group, and the quotient is trivial. Rank zero cannot support the stipulated circuit.

In higher rank, apply Lemma 8.3 to obtain

The vertex group H embeds in G, and so is BS-free. It has a non-power graph presentation of smaller complexity, hence is hyperbolic by induction. The one-relator torsion theorem makes it torsion-free [18, Theorem 2]. The side groups A and B are finitely generated free Magnus subgroups of H; by [13, Theorem 4.7], they are quasiconvex in H.

Use the simultaneous bases from Lemma 8.3. Collins’s conjugate-intersection theorem [4, Theorem 2], which permits equal Magnus subgroups, says that for M,N\in\{A,B\} and g\in H,

This is the first overlap hypothesis of Theorem 3.1. Collins’s direct-intersection theorem [4, Theorem 1] gives

where C is generated by the common basis letters and the second expression is an internal free product with an infinite cyclic group. Thus I and its transported copy J=t^{-1}It are finitely generated. If the edge group is nontrivial, the two Magnus subgraphs U_-,U_+ have connected intersection by the construction. Linton’s strong-inertia lemma [12, Lemma 2.10] states that the intersection of the fundamental groups of two such Magnus subgraphs is strongly inert in either side group. It follows that I is strongly inert in P=A and in B. Isomorphisms preserve reduced ranks and the indexing double cosets, so transporting the latter assertion through the edge isomorphism makes J strongly inert in P as well. These are precisely the two strong reduced-rank inertia inequalities in Theorem 3.1. When the edge group is trivial, the same requirements hold directly.

All hypotheses of Theorem 3.1 have now been verified. Its conclusion makes G hyperbolic, completing the induction.□

Proof of Theorem 1.1. Freely and cyclically reduce the defining relator; conjugating or inverting it does not change its normal closure. If it becomes trivial, the group is a finitely generated free group and is hyperbolic.

If the cyclically reduced relator is a proper power, write it as w^k, where w is a nontrivial cyclically reduced non-power and k\geq2. Newman’s spelling theorem [20], in the form [18, Theorem 4], says that every nonempty freely reduced null word contains a segment of a cyclic conjugate of w^{\pm k} of length greater than (k-1)|w|. This is strictly more than half the relator length, including when k=2. Replacing that segment by the inverse complementary segment strictly shortens the word and uses one defining relation. Repeating, with free reduction after each replacement, fills a null word with at most its original length in relator cells. These shortening rules give a finite Dehn presentation (and a linear isoperimetric inequality). The group is therefore word-hyperbolic [2, Theorem III.\Gamma.2.6].

In the remaining case, realize the cyclically reduced non-power relator as an immersed circuit in the basis rose. Proposition 8.4 applies and proves hyperbolicity under the stated exclusion of BS(m,n) for all nonzero integers m,n.□

Proof of Theorem 1.2. Under its non-power graph-presentation hypotheses, Proposition 8.4 makes the ambient group hyperbolic. For any connected subgraph omitting a traversed edge, Lemma 8.1 identifies its fundamental group with an injective Magnus subgroup in a one-relator presentation. Linton’s theorem [13, Theorem 4.7] makes that subgroup quasiconvex. Basepoint paths only conjugate it, so the assertion holds for every choice of such paths.□

Proof. If G is finite, take H=1 and F_d=F_0=1, with ambient group \mathbb Z. Otherwise Theorem 1.1 makes G hyperbolic, and Theorem 1.1 of [21] makes it virtually compact special. The free group F(X) is the fundamental group of a compact connected planar surface with |X|+1 boundary components. Kielak–Linton’s Corollary 1.3 [10] therefore makes G virtually free-by-cyclic. The surface may be a disk, and their relator is arbitrary; trivial relators and proper powers are not excluded.

For a hyperbolic virtually compact special group, Theorem 1.1 of [10] equates being virtually free-by-cyclic with a finite-index embedding in a free-by-cyclic group with finitely generated free kernel. In the rational cohomological dimension two branch of their proof the ambient quotient is \mathbb Z. In the dimension-at-most-one branch they obtain a finite-index free subgroup, which has finite rank here and embeds in its direct product with \mathbb Z. In either branch the ambient extension splits over \mathbb Z, giving the displayed semidirect product.

Identify H with its image and restrict the projection onto \mathbb Z. Its kernel is K=H\cap F_d, which is free as a subgroup of a free group, and its image is a subgroup Q\leq\mathbb Z. This is the asserted exact sequence. Finite generation of the ambient kernel F_d does not imply finite generation of K.□