A group is a set of symmetries together with a way of composing them.
The simplest infinite groups are the free groups. Fix a set X of
"letters" and give each letter an inverse. The free group F(X) on X
is the set of all finite words in these letters and their inverses,
where a letter next to its own inverse cancels. For example, with
X = {a,b}, the words "a", "ab", and "abb^{-1}a^{-1}" all count, and
the last cancels to the empty word, the identity. There is no
relation among the letters beyond the cancellation of x with x^{-1}:
this is what "free" means. Free groups are the most spread-out
groups there are.
A one-relator group is what you get by imposing a single extra
equation. Choose one more word r in the letters and declare it equal
to the identity. Formally,
G = <X | r> = F(X) / N,
where N is the smallest normal subgroup containing r (the "normal
closure" of r). The notation reads: G is generated by the letters X,
subject to the single relation r = 1. Concretely, two words name the
same element of G exactly when one can be turned into the other by
cancelling inverses and by inserting or deleting copies of r and its
conjugates.
Two examples show the range:
* If r is empty, then G = F(X) is free.
* With X = {a,t} and r = t a t^{-1} a^{-1}, the relation says
t a t^{-1} = a, so t commutes with a. This group is Z^2, the
integer lattice plane, generated by two commuting translations.
One equation, many possible shapes. A central theme of group theory
is that one equation is very little constraint, so a one-relator
group is usually close to being free. The question
is exactly how close.
The relator r has a normal form worth remembering. Do two things:
first cancel inverses inside the word (free reduction); then cut off
any common prefix and suffix so that the remaining word, read
cyclically, has no cancellation across its ends (cyclic reduction).
If after cyclic reduction the word becomes w^k for some k >= 2 and
some non-power w, r is called a proper power. Otherwise r is a
non-power. The three cases -- trivial relator, proper power, non-power
-- behave very differently, and the proof treats them separately.
A hyperbolic group, defined next, must not contain a copy of Z^2 (a
flat plane), because negative curvature forbids flat pieces. But Z^2
is only the beginning of a family of almost-flat subgroups.
Fix two nonzero integers m and n. The Baumslag-Solitar group BS(m,n)
is
BS(m,n) = <a,t | t a^m t^{-1} = a^n>.
Read the relation: conjugating a by t raises it, in a twisted way, to
the n-th power, while the m-th power of a is what gets conjugated.
Familiar groups live here:
* BS(1,1) = Z^2. Here t commutes with a.
* BS(1,n) is the "solenoid" group, with t a t^{-1} = a^n. It acts
on the Euclidean line by an expanding map plus a translation, and
it is solvable but not virtually abelian.
The word flat is not a metaphor. Each BS(m,n) acts on a tree in a way
that produces a line with a nondiscrete stabilizer, and none of these
groups is negatively curved. Every BS(m,n) fails to be word-
hyperbolic. This is why they are the basic obstructions to a one-
relator group being hyperbolic, and why they prompted a conjecture.
To state what the conjecture predicts, negative curvature for a group
must be defined.
Pick a group G and a finite set S of generators. Build the Cayley
graph: the vertices are the group elements, and two vertices g and gs
are joined by an edge whenever s is in S. This graph carries the word
metric: the distance between g and h is the length of the shortest
word in S spelling g^{-1}h. Different finite generating sets change
distances by bounded amounts, so shortest-path geometry is robust.
A geodesic triangle is three shortest paths joining three vertices.
In the Euclidean plane a geodesic triangle can have a fat middle. In
a tree, every geodesic triangle is thin: each side lies inside a
fixed-width fattening of the other two. A group is word-hyperbolic
when, for some (equivalently any) finite generating set, there is a
constant delta such that every geodesic triangle in its Cayley graph
is delta-thin. Trees give delta = 0; free groups are hyperbolic. The
surface groups of genus at least two are hyperbolic. The group Z^2 is
not. The notion is M. Gromov's (1987); its stability under quasi-isometry
is what makes it intrinsic.
Three facts about hyperbolic groups recur throughout the proof.
(a) Hyperbolicity is invariant under quasi-isometry. If two
groups' Cayley graphs are the same shape at large scale, the
groups are hyperbolic together.
(b) A hyperbolic group cannot contain Z^2. The flat lattice
washes out negative curvature.
(c) A hyperbolic group obeys a linear isoperimetric inequality: a
loop that fills with k tiles can be spread on the boundary of
about k tiles. A concrete form is a Dehn algorithm, a finite
set of local shortening moves that reduces any loop
representing the identity to nothing while shortening it at
every step.
Fact (c) is how the proper-power case of the main theorem is handled:
a spelling theorem provides such shortening moves, hence hyperbolicity.
In 1992 Steve Gersten asked whether the obvious obstruction is the
only one: are the Baumslag-Solitar subgroups the sole reason a one-
relator group can fail to be hyperbolic? The answer is yes.
(local pdf: linton-nyberg-brodda-survey.pdf; https://arxiv.org/abs/2501.18306)
MAIN THEOREM. Let X be a finite set and r an element of the free
group F(X). If the one-relator group G = <X | r> contains no
subgroup isomorphic to BS(m,n) for any nonzero integers m and n,
then G is word-hyperbolic.
Call a group Baumslag-Solitar-free (or BS-free) if it contains no
such subgroup. The statement includes the cases r = 1 (a free group)
and r a proper power. It includes BS(1,1) = Z^2 in the forbidden
list. Excluding every BS(m,n) with nonzero m,n is equivalent to the
more customary statement that excludes every BS(1,n); a small lemma
records this, and it lets the proof work with the cleaner all-(m,n)
statement.
The theorem is a positive answer, but a hard one. It is not that a
BS-free one-relator group trivially looks negatively curved. It is
that the only possible failure of curvature would manufacture a
Baumslag-Solitar subgroup, and the hypothesis forbids that.
The proof has two large pieces.
* A combination theorem. One-relator groups split, in a
controllable way, as an amalgam or HNN extension of a smaller
one-relator group glued along free subgroups. The proof shows
that if the gluing has certain "no bad intersection" properties
and avoids Baumslag-Solitar subgroups, the result is hyperbolic.
This is where curvature is won.
* A Magnus induction. The splitting reduces a one-relator group to
a one-relator group of smaller complexity, built from a graph
inside a cyclic cover. Inducting on complexity climbs down to
free groups, where hyperbolicity is known.
Combine the two and the main theorem follows. Everything below is
either a tool for the combination theorem or an ingredient of the
induction.
The Magnus step descends from Wilhelm Magnus's 1930 Freiheitssatz: a
one-relator group embeds the subgroup generated by all but one relator
letter, and the induction here is a geometric reworking of Magnus's proof.
A related body of work (Kielak-Linton, together with a separate theorem
proving virtual compact specialness for hyperbolic one-relator groups)
turns hyperbolicity into an explicit structural statement. A group H is
virtually free-by-cyclic if it has a finite-index subgroup that fits
into a short exact sequence with free kernel and cyclic quotient. The
main theorem, together with virtual compact specialness, proves that
every BS-free one-relator group is virtually free-by-cyclic.
Before the main argument, three tools are assembled. The
first controls how quasiconvex subgroups of a hyperbolic group meet.
The second manufactures a Baumslag-Solitar subgroup from an action on
a tree, which is how the hypothesis "BS-free" is turned into a
geometric statement. The third is a rank inequality inside a free
group that keeps the later bookkeeping finite.
A subgroup K of a hyperbolic group H sits inside H's Cayley graph.
Call K quasiconvex if every geodesic between two points of K stays
within a fixed distance of K, the bound being independent of the two
points. Quasiconvex subgroups are exactly the nicely embedded ones:
* K is quasiconvex if and only if K is finitely generated and the
inclusion K -> H is a quasi-isometric embedding.
Infinite cyclic subgroups, and finite subgroups, are quasiconvex. So
are the vertex groups produced by the relevant splittings.
A subgroup K also has a limit set Lambda K in the boundary of H: the
endpoints of all geodesic rays that eventually stay close to K.
LEMMA (quasiconvex intersections). If K and L are quasiconvex
subgroups of a finitely generated hyperbolic group H, then K
intersected with L is quasiconvex and
Lambda(K intersected with L) = Lambda K intersected with Lambda L.
The same holds for any finite family of quasiconvex subgroups.
Why it is true. The engine is a metric observation. For fixed
subgroups K, L and a radius R there is a bigger radius D with
N_R(K) intersected with N_R(L) contained in N_D(K intersected with L),
where N_R means the R-neighborhood. The point: if x is within R of
both K and L, pick k in K and l in L near x. Then k^{-1}l lies in the
ball of radius 2R, which is finite. For each of the finitely many
elements d that actually occur, fix one witness pair k_d, l_d. If
k^{-1}l equals some d, then z = k k_d^{-1} = l l_d^{-1} lies in both
K and L, and x is close to z by a bound depending only on that finite
list. This proves the observation.
Now a geodesic with endpoints in K intersected with L lies close to
both K and L (they are quasiconvex), so the observation puts it close
to K intersected with L: that is quasiconvexity. For the boundary, a
ray to a point xi of Lambda K intersected with Lambda L stays close to
both K and L; applying the observation along the ray produces a
sequence in K intersected with L converging to xi. So xi lies in
Lambda(K intersected with L). The reverse inclusion is immediate, and
finite families follow by induction.
The next fact counts intersections of conjugates.
LEMMA (finite double-coset lists). Let K, L be quasiconvex subgroups
of a hyperbolic H. Only finitely many double cosets KgL have
K intersected with gLg^{-1} containing an element of infinite order.
If H is torsion-free, only finitely many have nontrivial
intersection at all.
Why it is true. If c has infinite order and lies in
K intersected with gLg^{-1}, then the two fixed boundary points of c
lie in Lambda K and in Lambda(gL). A geodesic line joining them lies
in a uniformly bounded neighborhood of both K and gL, the bound
depending only on the quasiconvexity constant. At any point of the
line pick k in K and gl in gL nearby; their distance is uniformly
bounded, so k^{-1}gl lies in KgL and within a fixed ball. Every such
double coset therefore meets that fixed finite ball, and a ball holds
only finitely many group elements. So only finitely many double
cosets occur.
This finiteness is what later makes a set of carriers finite.
Groups often act on trees, and Bass-Serre theory (due to Bass and Serre)
turns a tree action into a description of the group as a graph of
groups. One conversion is needed: a tree action whose stabilizers are
as simple as possible forces a Baumslag-Solitar subgroup.
A tree action is without inversions if no group element exchanges the
two ends of an edge. An element is elliptic if it fixes a vertex and
tree-hyperbolic if it translates along a bi-infinite geodesic (its
axis). The translation length is how far it moves points of its axis.
LEMMA (graphs of cyclic groups). Let Q be a finitely generated
noncyclic group acting on a simplicial tree, without inversions, so
that every vertex and edge stabilizer is infinite cyclic. Then Q
contains a subgroup isomorphic to BS(m,n) for some nonzero m,n.
Why it is true. Take a vertex v and a finite symmetric generating set
S. The union D of the segments [v, sv] for s in S is a finite
subtree, and the union of its translates is exactly the convex hull of
the orbit Qv. The action on this hull has only finitely many edge
orbits (each has a representative inside D), so Bass-Serre theory
realizes Q as the fundamental group of a finite connected graph of
infinite cyclic groups.
Now simplify the graph. Collapse any non-loop edge whose map into one
endpoint group is onto; this does not change the fundamental group.
After finitely many such collapses, every remaining non-loop edge has
both gluing indices greater than one.
* If a loop remains, the subgraph of that loop and its vertex has
fundamental group
<x,t | t x^m t^{-1} = x^n> = BS(m,n), m,n nonzero.
It injects into Q by the graph-of-groups normal form theorem.
* If a non-loop edge remains, its two-vertex subgraph has
fundamental group <x,y | x^m = y^n> with |m|,|n| > 1. The element
z = x^m = y^n is central of infinite order. Modulo <z> the group
is the free product C_{|m|} * C_{|n|}, in which xy has infinite
order. The commuting elements z and xy therefore generate
Z^2 = BS(1,1), embedded in Q.
* If no edge remains, Q has one vertex and is cyclic, contradicting
the hypothesis.
This exhausts the possibilities.
The second lemma converts commensuration into a Baumslag-Solitar
subgroup. Two subgroups are commensurable if their intersection has
finite index in both.
LEMMA (elliptic commensuration). Suppose G acts on a simplicial tree
without inversions, with torsion-free hyperbolic vertex stabilizers.
If an infinite-order elliptic element a has its cyclic subgroup
commensurated by a tree-hyperbolic element g, then G contains a
subgroup isomorphic to BS(m,n).
Why it is true. Let L = <a,g> and let v be a vertex fixed by a. Every
element of L commensurates <a>, so for each h in L there are nonzero
integers r,s with a^r = h a^s h^{-1}; hence a nonzero power of a fixes
hv. A common multiple of finitely many such powers fixes any finite
set of orbit points Lv, hence fixes the convex hull T_L pointwise. So
every finite subtree of T_L is fixed by some nonzero power of a.
Now take a vertex x of T_L and choose N with a^N in the vertex
stabilizer L_x. Every element of L_x commensurates <a^N>, a subgroup of
the vertex stabilizer G_x. So
<a^N> is contained in L_x is contained in Comm_{G_x}(<a^N>).
The commensurator of an infinite cyclic subgroup in a torsion-free
hyperbolic group is infinite cyclic (the cyclic-commensurator lemma
below), so L_x is infinite cyclic. Edge stabilizers of T_L contain a
power of a and are cyclic too. Finally L is noncyclic: otherwise
a = c^r and g = c^s, and comparing translation lengths would make g
elliptic, contrary to hypothesis. So L is a finitely generated
noncyclic group acting on T_L with all stabilizers infinite cyclic,
and the graphs-of-cyclic-groups lemma applies.
For completeness, here is the cyclic-commensurator lemma used above.
The commensurator Comm_H(C) of C in H is the set of h with C and
hCh^{-1} commensurable; it is a subgroup.
LEMMA. If C is infinite cyclic in a hyperbolic group H, then C has
finite index in Comm_H(C). If H is torsion-free, Comm_H(C) is
infinite cyclic.
Why: C is quasiconvex, so by finite double-coset lists its
commensurator meets only finitely many double cosets CgC. A
representative g in the commensurator makes CgC a finite union of left
C-cosets, so the commensurator is virtually cyclic; a torsion-free
virtually cyclic group is cyclic.
The rest of the proof lives in a free group P (one of the edge
groups). The proof will repeatedly intersect a fixed subgroup E of P with a
family of its translates and add up the resulting ranks. A single
inequality keeps that sum from growing.
For a finitely generated free group K put
rbar(K) = max(rank(K) - 1, 0).
Subtracting 1 makes a cyclic group count as 0, matching the intuition
that cyclic groups are thin.
DEFINITION. A finitely generated subgroup E of a finitely generated
free group P is strongly inert if, for every finitely generated
subgroup S of P,
sum over (P,E)-double-cosets SpE in S of rbar(E intersected with p^{-1}Sp)
<= rbar(S),
where the sum runs over the (P,E)-double-cosets SpE in S and p is any
representative of its double coset.
Intuition: E meets its translates thinly, so total reduced rank does
not exceed the rank of the test subgroup S. Only finitely many of the
intersections are nontrivial, so the sum is finite.
The inequality has an immediate and heavily used corollary.
COROLLARY. Let E be strongly inert in the free group P, and let g be
an element of P not in E. Then E intersected with gEg^{-1} is
trivial or cyclic.
Why: apply the inequality with S = E. The identity double coset
already uses up all of rbar(E), so every other summand must vanish;
that is, every off-diagonal intersection has reduced rank 0, hence is
cyclic or trivial.
This is exactly the inertia hypothesis that the restricted
combination theorem imposes on the two edge subgroups I
and J. For Magnus splittings it is supplied later by a strong-inertia
theorem of Linton.
This is the heart of the argument. A one-relator group is about to be
built as an HNN extension: take a known hyperbolic group H, choose two
free subgroups A and B, identify A with B by an isomorphism phi, and
add a stable letter t that implements the identification. The question
is when the result is hyperbolic.
The strategy is a proof by contradiction carried by currents (measures
on the boundary of a free group). If the glued group is not
hyperbolic, a combination theorem of Bestvina and Feighn says that
some annulus of bounded width and huge girth fails to flare. Such
annuli, stacked by length, produce a bounded two-sided sequence of
currents. The rest of the argument rules out any such sequence when the
group is BS-free.
(local: Bestvina-Feighn, A combination theorem for negatively curved groups, https://doi.org/10.4310/jdg/1214458531)
Here is the object and the theorem. Recall that a free group is
quasiconvex in any hyperbolic group containing it, and torsion-free.
THEOREM (restricted combination theorem). Let H be a torsion-free
word-hyperbolic group, let A and B be finitely generated free
quasiconvex subgroups of H, and let phi: A -> B be an isomorphism.
Set
G = <H,t | t a t^{-1} = phi(a) for a in A>,
P = A, I = A intersected with B, J = t^{-1} I t (a
subgroup of P).
Assume:
(i) for M,N in {A,B} and g in H, the intersection
M intersected with g N g^{-1} is at most cyclic unless g lies
in MN;
(ii) I and J are finitely generated, and each E in {I,J} is
strongly inert in P (the rank inequality).
If G contains no subgroup isomorphic to BS(m,n) for any nonzero m,n,
then G is word-hyperbolic.
The hypotheses are restrictions by design: this is a restricted class
of HNN extensions, and that is all the Magnus application needs.
Throughout, write i: I -> P for the inclusion c -> c and j: I -> P for
j(c) = t^{-1} c t, with image J. Fix a free basis of P and finite
generating sets. If P is trivial then G = H * Z is hyperbolic, so
assume P is nontrivial.
Bass-Serre tree. Let T be the Bass-Serre tree of the HNN extension.
It is a tree on which G acts, with one edge for each coset of H in G,
glued at H-vertices. The height homomorphism G -> Z sends t to 1 and
H to 0. Orient the canonical edge e from t^{-1}H to H and orient all
translates consistently; its stabilizer is P. A frame for the edge he
is the element h; replacing h by hp with p in P changes the
coordinates on the stabilizer but not the edge. A path traverses an
edge positively if it follows the orientation.
The first lemma pins down which edges a path with a fat (noncyclic)
stabilizer can use.
LEMMA (neighboring edges and frames). A reduced path in T with
noncyclic pointwise stabilizer has constant traversal sign. The
positive successors of e that can occur in such a path are pte, with
pI in P/I; the positive predecessors are pt^{-1}e, with pJ in P/J.
Their common stabilizers with e are p I p^{-1} and p J p^{-1}. For a
positive successor the coordinate transfer is c -> t^{-1}p^{-1}cpt;
for a positive predecessor it is c -> t p^{-1} c p t^{-1}.
Why it is true. Every positively oriented edge leaving H is ate for
some a in H, and these are parametrized by aB in H/B. Its common
stabilizer with e is A intersected with aBa^{-1}. If that is noncyclic,
hypothesis (i) forces a into AB. Write a = pb with p in P, b in B.
Since bt = t phi^{-1}(b), the edge ate equals pte. Two of them agree,
pte = qte, exactly when t^{-1}q^{-1}pt lies in A, that is q^{-1}p lies
in I; so the cosets are indexed by P/I, the common stabilizer is
p I p^{-1}, and conjugating by (pt)^{-1} gives the transfer formula.
Predecessors are handled by conjugating by t. At a reversal of sign,
the two consecutive edge stabilizers, in common vertex coordinates,
are M and a M a^{-1} for the same M in {A,B}; the path is reduced so a
is not in M, and hypothesis (i) makes the intersection at most cyclic.
A noncyclic stabilizer of the whole path cannot sit inside a cyclic
group, so there is no reversal.
Call two consecutive positive edges aligned when their bridge lies in
the double coset AB in common vertex coordinates. The lemma says a
path with noncyclic stabilizer is aligned once its traversal sign is
fixed.
Next, a window bound. A window is a finite block of consecutive edges.
LEMMA (rank bound for a fixed window). Fix l,r >= 0. Consider the
positively traversed reduced paths with edges indexed -l,...,r, with
edge 0 equal to e, and with noncyclic pointwise stabilizer. Then
sum over P-orbits [W] of such paths of rbar(Stab_G(W))
<= rbar(P).
Every stabilizer in the sum is finitely generated and free, and
there are at most rbar(P) such P-orbits.
Why it is true. Build the window by starting from e and adding one
edge at an endpoint at a time; the stabilizer only shrinks. At one
extension the new stabilizer is an intersection K intersected with
p I p^{-1} (forward) or K intersected with p J p^{-1} (backward), with
K the previous stabilizer in suitable coordinates. These intersections
are finitely generated because they are intersections of finitely
generated subgroups of a finite-rank free group. The strong inertia of
I and of J says the reduced ranks of all the choices at one step add
up to at most the reduced rank of the previous stabilizer. Summing
this inequality over the representatives present at each step proves
the displayed bound. Cyclic stabilizers may be discarded, since
extension only decreases a stabilizer, and each noncyclic term
contributes at least 1, which gives the orbit count. This window shape
(l,r) is fixed; the lemma makes no claim about the union over all
lengths.
LEMMA (distinct paths). Two distinct aligned paths of the same fixed
window shape, with the same distinguished edge e and the same
traversal sign, have at most cyclic common pointwise stabilizer.
Why: at the first divergence, the common stabilizer, in the shared
endpoint frame, lies in p E p^{-1} intersected with q E q^{-1} for
distinct cosets of E in {I,J}. The strong inertia of E with test
subgroup E kills every off-diagonal intersection (they get reduced
rank 0), so the intersection is cyclic or trivial.
Now the geometric criterion the theorem rests on. An algebraic
annulus packages a proposed tree path together with a single group
element that fixes all of it, row by row.
Definitions first. For a sign epsilon in {+,-} define two maps into H:
f_+(c) = c, f_-(c) = t c t^{-1},
l_+(c) = t c t^{-1}, l_-(c) = c.
An algebraic annulus with 2m+1 rows consists of signs epsilon_s in
{+,-} and elements c_s in P minus {1} for -m <= s <= m, together with
bridges a_s in H for -m <= s < m, satisfying the row equation
f_{epsilon_s}(c_s) = a_s l_{epsilon_{s+1}}(c_{s+1}) a_s^{-1}.
Its width is max_s |a_s|_H and its girth is |c_0|_P. It is essential
if a (+, -) transition has a_s not in A and a (-, +) transition has
a_s not in B; at equal-sign transitions there is no restriction. It is
lambda-hyperbolic if
max(|c_{-m}|_P, |c_m|_P) >= lambda |c_0|_P.
Why this is a tree path. Write f_{epsilon_s}(c) = u^{-1} c u and
l_{epsilon_{s+1}}(c) = v^{-1} c v with u,v in {1, t^{-1}}, and start
from a frame h_s. Set h_{s+1} = h_s u a_s v^{-1}. The two edges meet at
h_s u H = h_{s+1} v H, and the row equation says
h_s c_s h_s^{-1} = h_{s+1} c_{s+1} h_{s+1}^{-1}. So one nontrivial
element fixes the whole developed path. At the two reversing
transitions the consecutive geometric edges agree exactly when a_s is
in A (or B), so essentiality is exactly absence of backtracking.
THEOREM (algebraic annular flaring). Let H be word-hyperbolic, and
let A, B be finitely generated subgroups of H with an isomorphism
phi: A -> B such that both attaching maps A -> H and B -> H are
quasi-isometric embeddings. Set
G = <H,t | t a t^{-1} = phi(a), a in A>.
Then G is word-hyperbolic if there are lambda > 1 and an integer
m >= 1 such that, for every rho > 0, there is h(rho) such that every
essential algebraic annulus with 2m+1 rows, width at most rho, and
girth at least h(rho), is lambda-hyperbolic.
This is the algebraic form of the Bestvina-Feighn combination
theorem, quoted from Kapovich's paper. The condition "girth big
relative to width forces flaring" is a type of negative curvature: a
long thin loop cannot close up too flat.
(local pdf: kapovich-nonquasi.pdf, https://arxiv.org/abs/math/9704203; Bestvina-Feighn addendum, https://doi.org/10.4310/jdg/1214458531)
The theorem's contrapositive gives the sequence of currents that
drives the contradiction.
LEMMA (counterannuli). If G is not hyperbolic, then for every
m >= 1 there is a sequence of essential annuli with 2m+1 rows whose
signs and bridges are fixed throughout, such that
|c_0|_P -> infinity, |c_{-m}|_P, |c_m|_P < 2|c_0|_P,
and for each fixed s the ratios |c_s|_P / |c_0|_P stay bounded.
Why: negate the flaring theorem with lambda = 2. For each m there is a
finite width bound and counterannuli of arbitrarily large girth with
width below it. Choose a sequence with girth tending to infinity. Only
finitely many sign tuples and finitely many bridge tuples of that
fixed length and width exist, so a subsequence fixes both. The maps
f_epsilon and l_epsilon are quasi-isometric embeddings, so the row
equation with fixed bridges bounds |c_{s+1}| in terms of |c_s| and
back, by fixed constants; iterating from row 0 bounds all ratios
because |c_0| tends to infinity.
The technical engine is a current: a way to measure how a typical
long word in a free group looks. Fix a free group F of finite rank
with a basis, and let T_F be its Cayley tree.
The boundary dF is the set of infinite reduced words, the
ends of the tree. The doubled boundary
d^2 F = (dF x dF) minus { (xi,xi) : xi in dF }
parametrizes oriented bi-infinite geodesic lines: the pair (xi, zeta)
is the line directed from xi to zeta. For a finite nondegenerate
oriented segment gamma in T_F, the cylinder Cyl_F(gamma) is the set of
lines containing that segment with that orientation. These cylinders
form a basis for the topology on line space.
A current on F is a nonnegative, F-invariant,
flip-invariant Radon measure on d^2 F. Radon means finite on compact
sets; flip-invariant means reversing every orientation gives the same
measure. The cone of currents is Curr(F), topologized by convergence
on cylinders. Its norm is
||mu||_F = (1/2) mu(L_F(1)),
where L_F(1) is the compact set of lines through the identity vertex.
The factor of one half is bookkeeping: a line through the identity is
seen from the identity twice, once in each orientation direction. The
norm is additive on the cone and vanishes only on the zero current,
because translates of L_F(1) cover all of line space.
(local pdf: kapovich-currents.pdf; https://arxiv.org/abs/math/0412128)
For w in F minus {1}, the counting current eta_w is
the sum of point masses at the two orientations of the translates of
the axis of w, indexed by F over <w>. Indexing by <w> (not by the set
of powers of w) keeps the multiplicity when w is a proper power.
Counting the translated axes that pass through the identity gives
||eta_w||_F = |w|_cyc,
where |w|_cyc is the cyclic word length: length of w, then shortened
by any cancellation across its two ends. For a segment labeled by a
reduced word u, the mass eta_w gives counts the occurrences of u and
u^{-1} in the cyclic word for w, with multiplicity, including
occurrences of u longer than w itself when w is short and repeats.
Currents form a compact cone at fixed norm.
LEMMA (compactness). The norm is continuous on Curr(F), and each
closed norm-bounded subset is compact. In particular every bounded
sequence of currents has a convergent subsequence.
Why: continuity holds because L_F(1) is compact and open. Every
compact subset of d^2 F is covered by finitely many translates of
L_F(1), so a uniform norm bound bounds the mass of every compact set.
There are countably many segment cylinders; a diagonal subsequence
converges on all of them. The cylinder identities (a cylinder is the
finite disjoint union of its one-edge extensions at either end)
survive the limit and define a finite Radon measure, F-invariant and
flip-invariant. This gives sequential compactness; the countably many
cylinder coordinates metrize the topology on a bounded set, so
compactness follows.
The key operation transports a current along an injection,
spreading it over cosets so that the result is invariant.
LEMMA (induction of currents). Let theta: Q -> F be an injection of
nontrivial finite-rank free groups. It defines a continuous linear
map theta_*: Curr(Q) -> Curr(F) which sends eta_w to eta_{theta(w)}
and every nonzero current to a nonzero current. For the fixed bases
there are constants 0 < c_theta <= C_theta < infinity with
c_theta ||nu||_Q <= ||theta_* nu||_F <= C_theta ||nu||_Q.
Postcomposing theta with an inner automorphism of F does not change
theta_*. Induction is functorial: (psi circ theta)_* = psi_* circ
theta_* for another injective psi.
Why: put E = theta(Q). E is finitely generated, hence quasiconvex in
F, so theta extends to a homeomorphism dQ -> Lambda E, a closed subset
of dF. Push nu by this boundary map to an E-invariant measure on
d^2 Lambda E, view it as a measure on d^2 F, and set
theta_* nu = sum over pE in F/E of p_*(that measure).
The summand depends only on the coset. The sum is locally finite: the
convex hull of Lambda E, call it T_E, has finite quotient by E, so
only finitely many cosets pE put a fixed vertex inside pT_E, and only
finitely many summands meet any fixed cylinder. So theta_* nu is a
Radon current, continuous and linear in nu. A nonzero measure pushes to
a nonzero measure, and the norm is continuous and positive on the
compact norm-one set, giving the two norm bounds. Inner automorphisms
just reindex the coset sum. Functoriality is a coset calculation:
representatives for F/E and E/theta(Q) multiply to representatives for
F/(theta(Q)).
The girth of an annulus uses based length, but the norm
of a counting current uses cyclic length; conjugating by a long tail
can make these very different. Capping by one letter fixes the gap.
LEMMA (capping linear words). Fix an injection theta: Q -> F of
nontrivial finite-rank free groups with fixed bases. For each
nonempty reduced word W in Q there is a suffix b of length at most
one such that Wb is cyclically reduced. For every fixed nonempty
reduced word u in F there is a constant D_{theta,u}, independent of
W, such that the difference of eta_{theta(Wb)}(Cyl_F([1,u])) and
N_u^sym(red(theta(W))) has absolute value at most D_{theta,u}.
There is also a constant D_theta, independent of W, such that
| ||eta_{theta(Wb)}||_F - |theta(W)|_F | <= D_theta.
Here N_u^sym counts occurrences of u and of u^{-1} as contiguous
subwords. The same cap b works for any finite collection of fixed
injections out of Q.
Why: if Q is cyclic, every nonempty reduced word is already cyclically
reduced, so take b empty; otherwise pick a letter b different from the
inverse of the last letter and the inverse of the first letter of W,
and then Wb is freely and cyclically reduced. The image of a geodesic
of Q in F is a uniform quasigeodesic, and the axis of Wb passes
through the identity, so its image is a uniform quasigeodesic line
through the identity; the Morse lemma bounds the distance from the
identity to the axis of theta(Wb) in F by a constant D. Writing
red(theta(Wb)) = v z v^{-1} with z cyclically reduced and the
concatenation reduced, this gives |v| <= D. Capping changes only a bounded
suffix of theta(W) (bounded by the largest theta(x) over basis
letters), and stripping v and v^{-1} deletes at most 2D more letters.
For a fixed u, changing or deleting a bounded number of letters at the
ends changes the occurrence count by a bounded amount, and passing
from linear to cyclic counting costs at most |u|-1. That gives the
frequency bound; the same bounded change shows |z| = |theta(W)| plus a
constant, and the norm identity gives the length bound.
The counterannuli become, window by window, a two-sided
sequence of currents. That sequence is the object the rest of the argument
analyzes.
LEMMA (a current limit for each finite window). Suppose G is not
hyperbolic and fix m >= 1. There are currents mu_s in Curr(P) for
-m <= s <= m with
||mu_0|| = 1, ||mu_{-m}||, ||mu_m|| <= 2,
and a fixed essential sign-and-bridge tuple, such that each
consecutive pair is induced by a current on its intersection
subgroup: for
K_s = f_{epsilon_s}(P) intersected with a_s l_{epsilon_{s+1}}(P) a_s^{-1}
there is nu_s in Curr(K_s) with
(theta_{s,0})_* nu_s = mu_s, (theta_{s,1})_* nu_s = mu_{s+1}.
All K_s are nontrivial finitely generated free groups.
Why: take the counterannulus sequence with m, the signs,
and the bridges fixed. Its elements c_s^k have |c_0^k| = L_k tending
to infinity. Each K_s is finitely generated (quasiconvex intersection)
and nontrivial (it contains f_{epsilon_s}(c_s^k)). Represent
f_{epsilon_s}(c_s^k) by a reduced word W_s^k in the chosen basis of K_s,
cap it to W_s^k b_s^k by the capping lemma, and normalize
nu_s^k = (1/L_k) eta_{W_s^k b_s^k}.
The row equation gives theta_{s,0}(W_s^k) = c_s^k and
theta_{s,1}(W_s^k) = c_{s+1}^k. Applying the capping lemma to both
injections shows the two induced normalized currents differ from the
normalized linear row frequencies of c_s^k and c_{s+1}^k by O(1)/L_k on
every fixed cylinder. So at an interior row the two neighboring induced
limits agree, giving mu_s; an end row uses its single neighbor. Norms
converge to |c_s^k|/L_k, which is 1 in the middle and below 2 at the
ends.
A finite list of transitions. The bridges a_s are not yet controlled as
m grows. Replace each by its double coset, of which there are only
finitely many.
For each ordered sign pair (epsilon, delta), choose representatives d
for the double cosets of (f_epsilon(P), l_delta(P)) in H whose intersection
K_{epsilon,delta,d} = f_epsilon(P) intersected with d l_delta(P) d^{-1}
is nontrivial. There are finitely many, by finite double-coset lists
and torsion-freeness of H. Use d = 1 for the identity double coset
when its intersection is nontrivial; for reversing sign pairs, discard
that double coset, since it is exactly the forbidden pinch. The
remaining triples tau = (epsilon, delta, d) are the transition types,
with groups K_tau and maps
theta_{tau,0} = f_epsilon^{-1},
theta_{tau,1} = l_delta^{-1} composed with (c -> d^{-1} c d).
LEMMA (uniformity of transition types). Every consecutive pair of
currents above can be written with a current on one fixed K_tau.
There are constants C >= 2 and D >= 1, independent of m, with
C^{-1}||mu_s|| <= ||mu_{s+1}|| <= C||mu_s||,
||nu_s|| <= D||mu_s||.
If K_tau is noncyclic, the two signs are equal and d = 1; the
positive aligned type has maps i, j and the negative aligned type
has j, i.
Why: factor an actual bridge as a = f(p) d l(q). Conjugating by f(p)
identifies K_d with K_a; under this identification the two maps from
K_a to the edge coordinates are inner twists of theta_{tau,0} and
theta_{tau,1}, which the induction lemma says do not change the
currents. So a transition current on K_a transports to one on K_tau.
Each of the finitely many theta maps has upper and positive lower norm
bounds, and comparing them over the finite type set gives the two
uniform constants. Noncyclicity puts d in the identity double coset by
hypothesis (i) of the combination theorem; a reversal there was
discarded, so the signs coincide.
Now pass from a fixed window to all integer positions.
PROPOSITION (the current trajectory forced by nonflaring). Under the
hypotheses of the combination theorem, if G is not hyperbolic, there
are currents mu_s in Curr(P) for all integers s, signs epsilon_s, and
transition types tau_s with signs (epsilon_s, epsilon_{s+1}), such
that
(i) ||mu_0|| = 1, every mu_s is nonzero, and sup_s ||mu_s|| is
finite;
(ii) for every s there is nu_s in Curr(K_{tau_s}) with
mu_s = (theta_{tau_s,0})_* nu_s and
mu_{s+1} = (theta_{tau_s,1})_* nu_s;
(iii) all types are essential: a reversing type uses a double coset
outside the common side subgroup.
The uniform constants C, D apply, so in particular
||mu_s|| >= C^{-|s|} for all s.
Why: for each m take one finite trajectory and express it with the
transition types. Let D_m be the largest ||mu_s|| on -m <= s <= m.
If D_m is bounded on a subsequence, keep the original centers. On each
fixed window [-n,n] the row and transition currents have bounded norm
and only finitely many sign-and-type patterns; extract a subsequence on
[-n,n] that fixes the pattern there and the limiting currents, then
diagonalize over n. Continuity of the induced maps keeps the
compatibility equalities, and norm continuity keeps ||mu_0|| = 1 and the
uniform bound.
If instead D_m tends to infinity on a subsequence, pick an index j_m
where the maximum is attained, divide every row and transition current
by D_m, and reindex so that j_m becomes 0. Now row norms are at most 1
and the central norm is 1; both original end norms were at most 2.
Iterating the adjacent comparison gives D_m <= 2 C^{j_m+m} and
D_m <= 2 C^{m-j_m}, hence
min(j_m+m, m-j_m) >= log_C(D_m/2) -> infinity.
So every fixed two-sided window is eventually present in the recentered
trajectory, and the same diagonal extraction applies, producing a
two-sided trajectory with central norm 1 and all norms at most 1.
In either case the lower comparison applied repeatedly from row 0 gives
||mu_s|| >= C^{-|s|}, so no row vanishes. This proves all three
conclusions.
There are two distinct limits here, and their order matters. First send
the original annular girth to infinity with the bridges fixed, so the
O(1)/L_k errors vanish. Only afterward replace bridges by finitely many
types and let the window length grow. The recentering estimate makes
sure neither half of the trajectory is lost when the window slides, and
norm continuity preserves nonzero mass. The rest of the argument excludes
the trajectory of this proposition under BS-freeness.
A bounded two-sided sequence of currents was obtained, with
transition currents linking consecutive rows. What follows turns the
sequence into a single current satisfying i_* nu = j_* nu = mu, and
shows that almost every line can be transferred along a unique path
in the Bass-Serre tree. Two jobs: eliminate atoms, then average.
An atom of a current is a single line with positive mass. In a free
group the only positive-mass lines are periodic axes.
A root of an axis in P is a generator of its stabilizer in P. Such a
generator is not a proper power, but it need not belong to a free
basis. Roots are taken up to conjugacy and inversion, equivalently
conjugacy classes of maximal cyclic subgroups.
LEMMA (every atom is a periodic axis). Every atom of a free-group
current is a periodic axis. If one oriented axis has mass b > 0 and
root r, then its translate orbit, together with the reversed
orientations, contributes b |r|_cyc to the norm.
Why: translate successive vertices of an atomic line to the identity.
All the resulting lines have the same positive mass and pass through
the identity. That set of lines is compact with finite measure, so two
of the translates coincide, and a nontrivial element of the free group
preserves the original line. An element reversing a line would fix a
vertex or invert an edge, impossible for the free action; so the line
stabilizer acts by translations and is infinite cyclic. For its root
r, the number of distinct translates of one orientation through the
identity equals |r|_cyc, the number of vertices in the quotient of the
axis by <r>. Flip invariance gives the other orientation the same
mass, and the norm is half the mass through the identity.
Recall the transition types tau = (epsilon, delta, d),
each with a group K_tau and two injections theta_{tau,0},
theta_{tau,1}. A noncyclic type is aligned: its signs agree, its
representative is 1, and the maps are i, j in one order or the other.
LEMMA (finite exceptional roots). There is a finite set D of root
classes such that
(i) the root of either image of a nontrivial element of a cyclic
transition group belongs to D;
(ii) if a periodic axis is carried by p Lambda E and q Lambda E
for distinct cosets pE, qE, where E is I or J, its root
belongs to D;
(iii) if an axis root in a noncyclic transition group maps to a
proper power under either transition injection, the root of
that image belongs to D.
"Carried" means both endpoints lie in the indicated translate of the
limit set.
Why: for (i), include the images of one generator of each nontrivial
cyclic transition group under both injections: finitely many. For
(ii), strong inertia with test subgroup E makes every off-diagonal
intersection E intersected with xEx^{-1} (x not in E) at most cyclic,
and finite double-coset lists says only finitely many ExE have
nontrivial intersection; include the root of a generator of each such
intersection. For (iii), an axis root u whose image is r^k with
|k| > 1 has r outside E (else u would be a proper power), and the two
distinct cosets E, rE both carry the axis of r; case (ii) then applies.
The next lemma turns two repetitions of a root class into the very
obstruction the theorem forbids.
LEMMA (repetition gives a Baumslag-Solitar subgroup). Let (h_s e) be
a reduced edge path in T indexed by consecutive integers, and let r_s
in P be nontrivial. Suppose the cyclic subgroups h_s <r_s> h_s^{-1}
at consecutive positions are commensurable. If i < j have the same
traversal sign and <r_i>, <r_j> are conjugate in P, then G contains
a Baumslag-Solitar subgroup.
Why: choose p in P with p <r_i> p^{-1} = <r_j>, and set
g = h_j p h_i^{-1}. Since p fixes e, g carries the edge h_i e to
h_j e and the cyclic subgroup at i to a subgroup commensurable with
the one at j. Let sigma be the segment from the midpoint of h_i e to
the midpoint of h_j e along the given path. At each join of two
consecutive translates g^k sigma, the incoming and outgoing pieces use
opposite halves of the same edge, because g preserves the canonical
orientation and the two positions have the same traversal sign. So the
translates concatenate without backtracking to a bi-infinite geodesic,
and g translates it a positive distance: g is tree-hyperbolic. The
cyclic subgroup at i is infinite and elliptic, so elliptic
commensuration gives an embedded Baumslag-Solitar subgroup.
Now the payoff.
PROPOSITION (no atoms). Let mu_s be a bounded sequence of nonzero
currents with transition currents and compatible essential types, as
in the trajectory proposition. If G is Baumslag-Solitar-free, then
every mu_s is atomless.
Why: suppose a row has an atom, and index from there. At each forward
transition the coset formula for the induced current has a positive
contribution to this atom coming from an atom of the transition
current; choose one such atom and take its image in the next row.
Every chosen line is a periodic axis; let r_s be its root. Develop the
matching into an actual reduced tree path, which the repetition lemma
needs. Call a transition exceptional if its group is cyclic, or a
chosen image line is carried by more than one coset, or an upstairs
root has a proper-power image. By finite exceptional roots, at least
one adjacent root at every exceptional transition lies in D.
If infinitely many transitions are exceptional, some root class and
some traversal sign occur together at two positions, and repetition
gives a Baumslag-Solitar subgroup. If only finitely many are
exceptional, discard an initial segment; on the tail each push has a
single contributing coset, so both image atoms have exactly the mass
of their common upstairs atom. That mass is a fixed positive number,
and the norm bound together with the atomic-axis lemma bounds
|r_s|_cyc. There are finitely many conjugacy classes of roots of
bounded cyclic length, so again a root class repeats at two positions
with the same traversal sign, and repetition gives a Baumslag-Solitar
subgroup. Both cases contradict BS-freeness.
The atoms are gone. What remains is to turn the two-sided sequence into
a single current that is fixed by both edge inclusions.
PROPOSITION (a stationary current). There exist a nonzero current nu
on I and a nonzero atomless current mu on P such that
i_* nu = j_* nu = mu.
In particular, if the restricted HNN extension is not hyperbolic, the
trajectory proposition supplies such a stationary current.
Why: a nonzero current on an infinite cyclic group is necessarily
atomic, so an atomless trajectory has no cyclic transitions. Every
transition is therefore aligned, all traversal signs are the same, and
the maps are i, j or j, i according to that sign. Reverse the indexing
if needed; then there are currents nu_s on the same group I with
i_* nu_s = mu_s, j_* nu_s = mu_{s+1}.
In particular I is nontrivial. Fix a free basis of I and set
S_N = sum_{s=-N}^{N} nu_s, a_N = ||S_N|| = sum ||nu_s||,
the last equality using positivity. Telescoping gives
j_* S_N - i_* S_N = mu_{N+1} - mu_{-N}.
If a_N tends to infinity, pass to a subsequence on which S_N / a_N
converges to a current nu of norm 1. A uniform norm bound controls the
mass of any fixed compact cylinder, since such a cylinder is contained
in finitely many translates of the set of lines through the identity.
So the right-hand side above, divided by a_N, tends to zero on every
compact cylinder, and continuity gives i_* nu = j_* nu. The positive
lower norm bounds make this common current mu nonzero.
If instead a_N is bounded, the positive two-sided sum
nu = sum_{s in Z} nu_s
has finite positive norm and defines a Radon measure, because every
compact line set is covered by finitely many translates of the identity
line set. Monotone convergence lets the coset sums pass through, and
reindexing the positive series gives i_* nu = sum mu_s = sum mu_{s+1}
= j_* nu. These currents have finite norm by the upper bound and are
nonzero; equivalently, summability and the norm comparison make the
endpoint norms tend to zero.
Atomlessness needs one more step, because a weak limit of atomless
currents can gain atoms. Use the constant bilateral trajectory whose
rows are all mu and whose transition currents are all nu. Its
transitions are aligned and positive, it is bounded, and the no-atoms
proposition applies, so mu is atomless. Any atom of nu would push a
positive atomic mass through an injective boundary map into mu, so nu
is atomless too.
Fix the currents with i_* nu = j_* nu = mu, and let Omega be the closed
support of mu in line space. A boundary point is rational if a
nontrivial element of the free group fixes it, equivalently its ray
labels are eventually periodic. First remove all lines with a rational
endpoint.
LEMMA (irrational endpoints). For an atomless free-group current,
almost every line has two irrational endpoints. An injection between
finitely generated free groups preserves and reflects rationality of
boundary points.
Why: fix a rational endpoint xi and orient an axis toward it. Every
oriented line with positive endpoint xi, other than the axis itself,
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 pointed toward
xi, the sets r^{-n} D_v for n >= 0 are pairwise disjoint (they merge at
r^{-n}v) and have equal measure by invariance. Finiteness of the
measure on the compact set of lines through v gives mu(D_v) = 0. There
are countably many merge vertices and rational endpoints, and
atomlessness removes the exceptional axes themselves. Flip invariance
handles negative endpoints.
An injection preserves rationality by equivariance of its boundary map.
For reflection, identify the image with a finitely generated subgroup E
of a free group and suppose xi in Lambda E is fixed by a nontrivial
ambient element r. 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
(quasiconvexity, together with xi in Lambda E), so pick 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, at n > m; then
e_n e_m^{-1} = r^{n-m} lies in E, and its inverse image is a nontrivial
element fixing the original boundary point.
If E is I or J and an irrational endpoint lies in two translates
p Lambda E, q Lambda E with pE, qE distinct, the two conjugate
subgroups meet cyclically, so the endpoint would be rational. Thus an
irrational endpoint belongs to at most one translate of each.
PROPOSITION (transfers on a conull set). There is a translation- and
flip-invariant conull Borel set U contained in Omega with these
properties.
(i) Every line of U has two irrational endpoints and is carried by
a unique I-coset and a unique J-coset.
(ii) On lines carried by p Lambda I, translate by p^{-1} and apply
the boundary isomorphism from I to J: this forward transfer,
and the analogous backward transfer from q Lambda J, preserve
mu on their domains.
(iii) Every defined transfer, its inverse, and every output frame
translation take lines of U to lines of U, so all successive
forward and backward transfers exist on U.
Measure preservation in (ii) holds on the whole Borel domain of lines
with irrational endpoints, not only on U.
Why: let X be the set of lines with two irrational endpoints, a conull
Borel set by the lemma. 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 uniqueness observation proves (i) wherever carriers exist.
Let vartheta be the boundary isomorphism from I to J induced by j. On
the Borel domain D_p = X intersected with d^2(p Lambda I), define
F_p(L) = vartheta(p^{-1} L). Rationality preservation and reflection
make F_p(D_p) = X intersected with d^2 Lambda J, a Borel set. For every
Borel D inside D_p, each push formula has just one contributing coset,
so
mu(D) = nu(p^{-1} D) = mu(F_p D).
The maps are homeomorphisms between their domains and images, so the
images are Borel. The same holds for the backward branches, which
preserve measure as well.
Finally choose a single conull set on which everything iterates. Start
in X and remove the Borel null set consisting of X minus Omega together
with the lines having no I-carrier or no J-carrier. The complement of
the support is null because line space has a countable base. Saturate
this null set under the partial forward and backward branches, their
inverses, translations, and flip. There are countably many such maps
and countably many finite compositions, each taking null sets to null
sets on its domain, so the saturation is Borel and null. Its complement
in X is the required U. The set U is dense in Omega, but it is not
asserted to be closed; statements about successive transfers apply to
U until a separate support argument is given.
Every transferred line determines, edge by edge, a genuine path in the
Bass-Serre tree, and each finite piece of that path has a fat
stabilizer. This is the bridge to the finite core graphs.
PROPOSITION (actual paths and finite windows). Every line L of U
determines an actual aligned bilateral path of positively traversed
edges
..., E_{-1}, E_0 = e, E_1, ...
in the Bass-Serre tree T, with a transferred line in each edge frame.
Moreover:
(i) the path does not depend on the chosen coset representatives
or on compatible changes of frames and line coordinates;
(ii) for every finite window containing E_0, the stabilizer K in P
is finitely generated and noncyclic, and the endpoints of L
lie in Lambda K;
(iii) two distinct actual aligned windows of the same shape about e
have at most cyclic common stabilizer, so an irrational point
belongs to the limit set of at most one such stabilizer;
(iv) if an aligned one-sided window joins e to he, its stabilizer is
S = P intersected with h P h^{-1}, and the boundary map of
S -> h^{-1} S h, x -> h^{-1} x h, implements the line transfer;
on the irrational-endpoint domain it preserves mu and it takes
its domain in U into U.
In (iii), a shape fixes the numbers of edges before and after the
central edge, all traversed positively.
Why: in a frame h for the current edge, let pI be the unique forward
carrier coset. Choose the next edge hpt e and use the forward transfer
for its line coordinates. For the backward extension the unique coset
pJ gives hpt^{-1}e. The transfer proposition lets this iterate in both
directions, and the frame description in the neighboring-edges lemma
shows the result is an aligned positive path, hence reduced.
Changing p to p i_0 for i_0 in I does not change the next edge, since
p i_0 t = pt j(i_0); it changes the frame on the right by j(i_0) in P
and the line coordinates by the inverse translation, and all later
carrier cosets shift by the same amount, leaving later edges unchanged.
The backward analogue uses J. This proves (i), including compatibility
of forward and backward steps.
For (ii), build a finite window one end at a time. Its stabilizer
starts as P and, at each step, is intersected with p I p^{-1} (forward)
or p J p^{-1} (backward) in the current frame. An intersection of
finitely generated subgroups of a free group is finitely generated, and
the quasiconvex-intersection lemma says its limit set is the
intersection of limit sets. The current endpoints belong to both by
induction and the carrier choice, hence to the new stabilizer's limit
set. A trivial or cyclic subgroup cannot carry an irrational endpoint,
so the stabilizer is noncyclic.
For (iii), the distinct-paths lemma gives the cyclic common stabilizer
at the first divergence, and finitely generated limit sets intersect in
the limit set of the intersection.
For (iv), an element fixing the two endpoint edges of a tree segment
fixes every edge between them, so the stabilizer of the one-sided
window is exactly P intersected with h P h^{-1}; the successive
coordinate changes compose to conjugation by h^{-1}, and their boundary
maps agree with that conjugation.
LEMMA (window finiteness). Fix a finite positive window shape about e
and a finite ball in the Cayley tree of P. There are only finitely
many actual windows of that shape whose stabilizer is noncyclic and
whose stabilizer limit set has convex hull meeting that ball.
Why: the window rank bound gives finitely many P-orbits of such
windows; choose representatives W_i with stabilizers K_i. For each i the
quotient of the convex hull of 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 of pW_i meets the fixed
ball, pick a vertex x in the intersection; then p^{-1}x = k v_{i,j} for
some k in K_i, so pW_i = x v_{i,j}^{-1} W_i. There are finitely many
choices of x, i, j.
Each short window of the path fixed by a line of U has a stabilizer K that
is a finitely generated free group of bounded total reduced rank, but the
core graph of K can be huge. What follows tames that: after suppressing
degree-two vertices the graph shapes and sizes are bounded, the edges whose
labels stay bounded form a single fixed small graph, and its noncyclic
components give finitely many maximal subgroup carriers for all the
long-window stabilizers. A return along a carrier then produces an
automorphism to which Brinkmann's expansion theorem applies.
The immersed-core description is used for finitely generated free
subgroups, and the corresponding pullback description of intersections
(Stallings). For a finitely generated nontrivial subgroup K of P, let
C(K) = K minus Hull(Lambda K)
be its finite unbased core graph, an immersion into the basis rose R of
P. It has no degree-one vertices. A circuit is a locally injective map
from a circle; it need not be an embedded loop.
(local: Stallings, Topology of finite graphs, https://doi.org/10.1007/BF02095993)
For each n >= 1 choose one representative of every P-orbit of
positively traversed windows W = (E_{-n},...,E_n) with E_0 = e and
noncyclic stabilizer K_W, and form the finite graph
Gamma_n = disjoint union over W of C(K_W),
an immersion into R. There are finitely many summands by the window
rank bound. Set r_0 = rbar(P); since mu is nonzero and atomless, P has
rank at least two, so r_0 >= 1.
A state of a labeled core is one of its vertices. In a noncyclic core a
vertex is natural if its degree is at least three. A natural edge is a
maximal nonbacktracking path whose interior vertices have degree two;
suppressing those interiors gives the natural graph.
LEMMA (window cores).
(1) The sum of the reduced ranks of the components of Gamma_n is at
most r_0. After suppressing degree-two interiors, Gamma_n has at
most 3 r_0 unoriented edges and 2 r_0 vertices.
(2) Every line of Omega has a label-preserving lift to Gamma_n.
(3) A ray with irrational endpoint has at most one starting state
for a lift among all components of Gamma_n.
Why: (1) is the window rank bound. In a noncyclic connected core with
degree-two vertices suppressed, every vertex has degree at least three,
so 2E >= 3V and E - V = r - 1; hence E <= 3(r-1) and V <= 2(r-1).
Summing over components bounds the natural edges, not their subdivided
lengths.
The state description is exact. 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, independently of the coset representative; equality
x^{-1}W = y^{-1}W means yx^{-1} in K, that is Kx = Ky, so the
description is injective, and states in different cores cannot
represent the same window.
A ray starting at Kx lifts in the Cayley tree starting at x, staying in
Hull(Lambda K); its label, read from the identity, has endpoint in
Lambda(x^{-1}Kx). If that endpoint is irrational, the windows lemma
forbids two distinct represented windows carrying it, giving (3).
Continuation from a state is unique because the core map is an
immersion.
For (2): a line of U has endpoints in the limit set of its window
stabilizer, so its labels lift to Gamma_n. Liftability to a fixed finite
graph is closed in line space. Center a ray at a vertex on a limiting
line; close lines have compatible labels on each bounded segment; only
finitely many lift states exist, so a subsequence uses one state on
arbitrarily long segments, and the immersion extends it. Since U is
dense in Omega, every Omega-line lifts.
Now fix a subsequence and shrink each long edge to a fixed graph. Pass
to a subsequence on which the natural graph of Gamma_n has a fixed
abstract type, and identify those natural graphs with one fixed finite
graph. Pass to a further subsequence on which each natural edge has
either a fixed label or a length tending to infinity, and for a
divergent edge arrange convergence of all finite label prefixes in both
orientations. The small graph Gamma^sm consists of the natural vertices
and the natural edges with fixed labels, subdivided according to those
labels. Components of rank zero or one are allowed.
Each connected component D of Gamma^sm, with a chosen vertex, gives the
subgroup of P consisting of labels of its based loops. The map to R is
an immersion, so this subgroup is isomorphic to pi_1 D, and changing the
vertex only conjugates it. A bi-infinite reduced path entirely in D lies
in its unbased core, and its labels have endpoints in a translate of the
corresponding limit set.
Let C be the set of P-conjugacy classes of finitely generated
noncyclic subgroups Q of P that fix a positively traversed bilateral
path through e pointwise. Such a path is unique, or two windows would
eventually diverge, contradicting distinct windows. Define a preorder
[Q] <=_c [M] iff Q is contained in some conjugate a M a^{-1} of M.
LEMMA (conjugate containment). A nontrivial finitely generated
subgroup of a finite-rank free group cannot be properly contained in
a conjugate of itself. Consequently <=_c is a partial order on C.
Why: suppose a^{-1} M a is strictly smaller than M. No positive power
a^k lies in M, for otherwise the nested inclusions
M > a^{-1}Ma > ... > a^{-k}Ma^k = M
would force equality throughout. So the Schreier vertices M a^{-n},
n >= 0, are distinct. Fix a nontrivial reduced word w in M of length L.
At each of those vertices, w labels a closed reduced path, since
a^{-n} w a^n lies in M. The Schreier graph is a finite unbased core
with trees attached, and a closed reduced path of length L starts
within distance L/2 of the core. The L/2-neighborhood of the finite
core has only finitely many vertices and cannot contain all the
distinct M a^{-n}. So proper containment is impossible. Replacing a by
a^{-1} gives the reverse direction, and the antisymmetry of <=_c follows.
An elementary graph bound supplies the compactness step. Note that a
core can have bridges.
LEMMA (bounded circuits in a core). Let C be a finite connected graph
with no degree-one vertices and at least one edge. Every edge of C
lies on an immersed circuit of length at most 2|E(C)|, where |E(C)|
counts unoriented edges.
Why: 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 (a component without a cycle would be a finite tree
with at most one vertex of degree one, impossible). On each side take a
shortest path from the endpoint of a to a simple cycle, traverse the
cycle, and return. Combine the two with a and a^{-1}; the joins are
reduced because the paths stay on their own sides of a. The resulting
immersed circuit contains a and has length at most 2|E(C)|.
PROPOSITION (finite cofinal carriers). The noncyclic component
subgroup classes of Gamma^sm belong to C and form a finite cofinal
subset: every [Q] in C is contained, up to P-conjugacy, in one of
them. In particular C has only finitely many maximal elements, and
they are exactly the maximal members of this finite family.
Why: let M be a noncyclic subgroup given by loops at a vertex of a
small component. The same loop labels occur at the corresponding state
of every selected Gamma_n, so by the state description M fixes an
actual centered window of radius n. The windows for different n agree
on any common shape, because M is noncyclic, so their compatible
restrictions form a bilateral positive path fixed by M. Thus [M] is in C.
Now take [Q] in C with fixed unbased core C(Q), and put
D_Q = 2|E(C(Q))|. For every selected n, Q is contained in the full
stabilizer of its centered radius-n window, and translating to a
representative gives a label-preserving immersion C(Q) -> Gamma_n.
Choose n so large that every divergent natural edge of Gamma_n has
length greater than D_Q. If the image of some edge of C(Q) met the
interior of such a natural edge, then by the bounded-circuits lemma that
source edge lies on an immersed circuit of length at most D_Q. Its image
is immersed; at a degree-two interior of a natural edge there is no
reduced way to turn around, so the image circuit must traverse that
whole natural edge, cyclically, and is therefore longer than D_Q, a
contradiction. So the whole image of C(Q) lies in Gamma^sm, hence in
one small component with subgroup M. The induced inclusion gives
[Q] <=_c [M], and M is noncyclic because it contains Q. There are
finitely many small components, so a maximal one is maximal in all of
C, and conversely a maximal class in C equals a small class above it.
Moving one edge forward along the path fixed by Q, and re-expressing Q
in the new edge coordinates, defines a map
F: C -> C.
This moves subgroup coordinates, not individual lines, and is defined
for every class in C.
PROPOSITION (atoroidal return). The map F is an order-preserving
bijection with order-preserving inverse. It permutes the finite set of
maximal classes of C. For every representative M of such a maximal
class there are an integer p >= 1 and an element v in G of height p
with
v^{-1} M v = M,
and the automorphism alpha(m) = v^{-1} m v is atoroidal: no nontrivial
conjugacy class in M is periodic under alpha.
Why: choose a frame h for the next edge on the unique path fixed by Q.
Transfer gives h^{-1} Q h contained in P. Another frame has the form hq
with q in P, and conjugates this output inside P; conjugating Q inside P
translates its path and gives the same output class. So F is well
defined, and moving backward is its inverse. If Q is contained in M
after choosing representatives, both groups fix the same path by
noncyclic uniqueness, and the same partial conjugation preserves the
inclusion. This gives the order assertions.
An order automorphism permutes the maximal elements, and there are
finitely many of them, so F^p[M] = [M] for some p >= 1. Let h be a frame
for the p-th forward edge. Since h^{-1} M h and M are equal up to
conjugacy in P, there is c in P with h^{-1} M h = c M c^{-1}. Set
v = hc. Then v has height p and v^{-1} M v = M, so the return map is an
onto automorphism.
For atoroidality, suppose alpha^k(w) = u^{-1} w u for a nontrivial w in
M, an element u in M, and k >= 1. Then g = v^k u^{-1} commutes with w.
Its height is kp, nonzero, whereas w has height zero. A relation
w^a g^b = 1 forces b = 0 by height and then a = 0, so <w,g> is
Z^2 = BS(1,1), contrary to BS-freeness. So no conjugacy class is
periodic. (The element v preserves the bilateral path fixed by M:
normalizing M sends that path to another path fixed by M, both contain
v e, and uniqueness in that edge's coordinates identifies them.)
Brinkmann's theorem gives expansion. It states that an automorphism of a
finite-rank free group with no nontrivial periodic conjugacy class is
hyperbolic: there are b_0 >= 1 and lambda_0 > 1 with
lambda_0 |w| <= max(|alpha^{b_0}(w)|, |alpha^{-b_0}(w)|).
The argument needs a bigger expansion factor, and a consequence for
currents.
(local pdf: brinkmann.pdf; https://arxiv.org/abs/math/9906008)
LEMMA (expansion on elements and currents). Let M, v, alpha be as in
the return proposition, with a chosen free basis of M. There are
b >= 1, beta = alpha^b, and lambda > 4 such that, for all w in M,
lambda |w|_M <= max(|beta(w)|_M, |beta^{-1}(w)|_M),
lambda l_M(w) <= max(l_M(beta(w)), l_M(beta^{-1}(w))),
and for every M-current sigma,
lambda ||sigma||_M <= ||beta_* sigma||_M + ||(beta^{-1})_* sigma||_M.
The partial conjugations along the paths to v^b e and v^{-b} e have
domains
S_+ = P intersected with v^b P v^{-b},
S_- = P intersected with v^{-b} P v^b,
finitely generated full window stabilizers, both containing M. Their
boundary transfers restrict on d^2 Lambda M to beta and beta^{-1} and
preserve mu on their irrational-endpoint domains.
Why: apply Brinkmann to alpha and put gamma = alpha^{b_0}. If
|gamma(w)| >= lambda_0 |w|, applying the inequality to gamma(w) shows
the expanding direction must be forward, so induction gives
|gamma^k(w)| >= lambda_0^k |w|; the backward direction is symmetric.
Choose k with lambda_0^k > 4 and set b = b_0 k. This gives the based
inequality, which also holds for w = 1. Applying it to w^q, dividing by
q, and letting q tend to infinity gives the cyclic inequality (using
lim |w^q|/q = l_M(w)).
For counting currents, the cyclic inequality and ||eta_w|| = l_M(w)
give the current inequality with a maximum replaced by the sum. The sum
form is linear, so it holds for finite nonnegative combinations of
counting currents; a circulation approximation extends it to every
current. Concretely, decompose the length-k cylinder masses into
cyclically reduced loops, build a combination sigma_k with matching
lengths and the same norm, and pass to the limit.
Finally, the return paths are reduced, and fixing their first and last
edges is equivalent to fixing the intervening segment, so their full
stabilizers in P are S_+ and S_-. Since v normalizes M, both contain M.
Their partial conjugations x -> v^{-b} x v^b and x -> v^b x v^{-b}
restrict to beta and beta^{-1}, and the boundary and measure assertions
follow from the windows proposition.
PROPOSITION (zero mass on a maximal carrier). For every maximal
carrier M from the finite cofinal family,
mu(d^2 Lambda M) = 0,
and the same holds for every P-translate of this line set.
Why: pull back the restriction of mu to the relatively closed subset
d^2 Lambda M by the boundary embedding of M in P, getting an M-current
sigma. Both endpoints are irrational outside a null set, and on those
lines the return transfer Theta_+ is measure-preserving, maps
d^2 Lambda M onto itself, and agrees with beta. So beta_* sigma = sigma,
and the inverse has the same property. The current inequality gives
lambda ||sigma|| <= 2 ||sigma||,
and lambda > 4 forces sigma = 0. Translation invariance of mu gives the
last assertion.
This is a measure conclusion, not a support conclusion: a closed set of
measure zero can still contain lines of the support. The return expansion
excludes exactly those possible support lines.
So far mu is a nonzero stationary current with a conull transfer set U,
and every maximal carrier has zero mass. Zero mass does not remove a
carrier line from the closed support Omega: a line can sit in the
support with measure zero. What follows proves the stronger fact that
Omega contains no periodic line, and only then extends everything from
U to the whole support.
The obstruction is a possible carrier line in the support despite its
zero measure. Expanding a positive-measure cylinder near such a line
along a fixed return either keeps following the carrier, which would
concentrate mass on the zero-mass carrier and is impossible, or leaves
the return domain. In the second case two window stabilizers share a
long segment, and that shared segment is forced to contain a periodic
middle. The periodic middle then excludes periodic support lines
entirely.
Write C_Q for the unit-edge Cayley tree of a finitely generated free
group Q with a fixed basis. A homomorphism is extended over each edge by
its reduced image path. The image of a geodesic under a fixed injective
homomorphism is a uniform quasigeodesic, and its tightened line is at
uniformly bounded distance from it.
LEMMA (nearness and order under a fixed transfer). Let a: Q -> F and
b: Q -> F' be injective homomorphisms between finitely generated free
groups with fixed bases. For every D >= 0 there are E, T >= 0 with the
following property. Let L be an oriented line of C_Q, and L_a, L_b its
tightened image lines. If q in Q satisfies d(a(q), L_a) <= D, then
d(b(q), L_b) <= E. If two elements q_1, q_2 have projections to L_a
positively ordered and separated by more than T, then their projections
to L_b are in the same order, and their separation on L_b tends to
infinity uniformly over all L, q_1, q_2.
Why: choose a vertex u of L whose a-image is uniformly close to a point
of L_a nearest a(q); the lower quasi-isometry bound for a bounds d_Q(q,u);
the upper bound for b, then its line Hausdorff bound, gives the first
claim. For the order claim, prove it first for 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 far beyond v, so projecting to the tightened line shows the
projection of f(v) is not far behind f(u); the lower embedding bound
gives separation tending to infinity. Apply first to a, then to b.
Two facts about trees are used repeatedly. If x, y lie within D of a
geodesic line L, the part of [x,y] farther than D from both endpoints
lies on L. And the lines meeting a fixed finite ball form a compact
subset of line space, on which a current has finite mass.
LEMMA (expansion with fixed trimming). Fix a maximal carrier M and the
maps above. There are constants Q >= 0 and L_* > 0 with the following
property. Let L be an oriented M-Cayley-tree line, let x, y be vertices
on L with d_M(x,y) = l >= L_*, and let A be a Borel subset of U whose
lines lie within Q of both x and y in C_P. Suppose a sign epsilon in
{+1,-1} satisfies d_M(beta^epsilon x, beta^epsilon y) >= lambda l. If
every line of A belongs to d^2 Lambda S_epsilon, then its image
A' = Theta_epsilon(A) has the same measure, lies in U, and admits
vertices x', y' on the tightened line beta^epsilon L with
d_P(x', ell), d_P(y', ell) <= Q for all ell in A',
d_M(x', y') >= 2l,
and the same sign satisfies d_M(beta^epsilon x', beta^epsilon y') >=
lambda d_M(x', y'). The constants do not depend on L, x, y, A, or on
how many times this is applied.
Why: choose Q to bound the distance of every M-vertex on a segment to
its tightened P-image and back. The transport lemma with D = Q gives C_1
so that u = beta^epsilon x and z = beta^epsilon y lie within C_1 of every
line of A'. Let a > 0, b >= 0 be lower embedding constants and B a common
vertex-to-line bound for beta^{+-1}. Choose R with R > B and
a R - b - Q > C_1. Delete R edges at both ends of [u, z] in C_M and call
the new endpoints x', y'. They lie on the tightened line beta^epsilon L.
Their P-images lie within Q of [u,z]_P; the projection of either new
endpoint to that segment is at least a R - b - Q > C_1 from each end, so
the intervening part of [u,z]_P lies on every line of A' and x', y' are
still within Q of each such line. The new length l' satisfies
l' >= lambda l - 2R, and d_M(beta^{-epsilon}x', beta^{-epsilon}y') <=
l + 2 K R with K the Lipschitz constant of beta^{+-1}. Choose L_* so that
lambda l - 2R >= 2l and l + 2 K R < 2 lambda l; then the backward
direction at the new pair is too short, and the based inequality forces
expansion in the original direction. Measure preservation and membership
in U come from the invariant transfer construction.
Fix the carrier M, its return maps, and the trimming constants. The next
lemma is the common engine for both support arguments. Its reference line
is transferred in the boundary of M; only the competing line in its
conclusion is asserted to be in Omega.
LEMMA (finite failure of an expanding return). There are constants
L_0 >= L_*, c > 0, C_2 >= 0, and an integer D >= 1, depending only on
the splitting, carrier, and return maps, with the following property.
Let eta be in Omega intersected with d^2 Lambda M, let L be its
M-Cayley-tree line, and choose vertices x_0 < y_0 on L with
l_0 = d_M(x_0, y_0) >= L_0. Suppose epsilon in {+1,-1} satisfies
d_M(beta^epsilon x_0, beta^epsilon y_0) >= lambda l_0.
Then there are n >= 0, a line ell in U, and an oriented periodic P-line
zeta of primitive period at most D such that ell, zeta, and the
tightened P-image eta_n of beta^{epsilon n} L share a segment of length
at least
c 2^n l_0 - C_2.
The endpoints of zeta lie in the limit sets of the stabilizers of two
distinct actual windows of the shape defining Theta_epsilon; one of them
is the prescribed window, with stabilizer S_epsilon.
Why, part 1 (the process cannot run forever). Choose L_0 so that the
projections of x_0, y_0 to eta are distinct and occur in that order. The
cylinder prescribing the directed segment between those projections has
positive finite measure, since eta is in the support; intersecting with U
gives a Borel set A_0 of positive finite measure, every line of which
lies within Q of x_0 and y_0.
Try to apply Theta_epsilon to the whole set successively. At time n call
the attempt a failure if A_n is not contained in d^2 Lambda S_epsilon.
Otherwise set A_{n+1} = Theta_epsilon(A_n) and use the trimming lemma to
get the next segment. While 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, every
line of A_n lies within Q of both vertices, and the measure stays
delta = mu(A_0) > 0.
The process cannot continue forever. Pick an M-vertex a_n within one edge
of the midpoint of [x_n, y_n]_M and translate A_n by a_n^{-1}. The central
part of [x_n, y_n]_P lies on every line of A_n; the image of a_n is
uniformly close to that segment, and the lower embedding bound puts its
projection arbitrarily far from both ends. So the translated sets
eventually lie in one compact set of lines meeting a fixed ball, with
measure still delta. Every subsequential limit is in d^2 Lambda M, so
every relative neighborhood of the carrier-in-K set contains all
sufficiently late translated sets; outer regularity of the finite measure
then gives mu(K intersected with d^2 Lambda M) >= delta, contradicting
zero carrier mass.
Why, part 2 (a failure gives a bounded-period axis). Suppose failure
occurs at time n, and choose ell in A_n outside d^2 Lambda S_epsilon. The
line ell has an actual trajectory for the same finite one-sided window,
with noncyclic stabilizer K, and that path differs from the prescribed
path (equality would put both endpoints of ell in Lambda S_epsilon).
Both ell and eta_n pass within Q of x_n, y_n, and x_n, y_n lie in M, a
subgroup of S_epsilon. Let H_epsilon be the larger of Q and
d_P(1, Hull(Lambda S_epsilon)). Deleting H_epsilon from both ends of
[x_n, y_n]_P leaves a segment that lies in ell, in eta_n, and in the
prescribed stabilizer hull, hence in Hull(Lambda K) too. Its length is at
least a l_n - C, with a, C fixed.
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}, and they remain distinct; every loop in the
component therefore fixes both paths, so its fundamental group injects
into an at-most-cyclic group by distinct windows. That component has rank
at most one. Since the window shape is fixed, there are finitely many core
representatives, and the numbers of unoriented edges in all relevant
fiber products are bounded by one integer D.
Increase L_0 so that a L_0 - C > 2D + 1, taking the maximum over the two
signs. Then a 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 is one circuit with trees attached, so a
sufficiently long reduced path follows the circuit between initial and
terminal tree pieces of length at most D. The 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 of primitive
P-period at most D. Unfolding the circuit in each core puts the endpoints
of zeta in both actual window stabilizer limit sets. Since l_n >= 2^n l_0,
this gives the overlap bound.
PROPOSITION (a carrier line forces a periodic line). If Omega contains
a line carried, up to translation, by a maximal noncyclic carrier M,
then Omega contains a periodic line.
Why: translate the given line into d^2 Lambda M. Choose successively
longer segments on its M-line and, for each, a sign supplied by the based
inequality. Apply the failure lemma, passing to a subsequence with the
same sign. The support line and the 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, so there are only finitely many of them. Along a subsequence
the periodic axis is fixed; the translated lines of 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 lies in Omega.
The next lemma controls the number of periods in a common segment.
Controlling only its metric length would not suffice, because a period
can grow under an inverse automorphism.
LEMMA (bounded loss of periods). Let psi be an automorphism of a
finitely generated free group M with a fixed basis. There is an integer
C_4 >= 0, depending only on psi, with the following property. Suppose
two lines L_1, L_2 of C_M share a segment [u, z^k u], where u is a
vertex, z translates L_2, and k > C_4. Their tightened psi-images share
a segment consisting of k - C_4 whole psi(z)-periods. If z is primitive
as a translation in M, then so is psi(z).
Why: choose B >= 0 such that the image of any vertex on any line lies
within B of the tightened image line. Put q = 2B + 1 and C_4 = 2q. Write
L_i' = psi L_i 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 z'^k a, because projection onto an invariant
line commutes with the translation. Both endpoint images lie within B of
both output lines, so a and z'^k a lie within 2B of L_1'. The vertices
z'^q a and z'^{k-q} a therefore lie farther than 2B from both ends of
[a, z'^k a] and occur in that order, so the tree observation puts their
joining segment on L_1' while 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. Automorphisms preserve being a non-power, giving the last claim.
LEMMA (agreement of periodic labels). Two bi-infinite label sequences
with respective positive periods a, b that agree on a + b consecutive
positions agree everywhere. Consequently two oriented periodic lines in
a labeled Cayley tree with such a common segment are the same line.
Why: number the common positions 0, ..., a + b - 1. For 0 <= i < a, the
letter at i equals the letter at i + b by the second period, and by the
first period this equals the letter at (i + b) mod a. So the first period
pattern is invariant under adding b modulo a, hence periodic with period
d = gcd(a,b). The first b common letters then show the second pattern is
the same d-periodic pattern, and both entire sequences agree. A label
together with an initial vertex determines a unique path in a Cayley
tree, so equality of sequences and the shared segment identify the lines.
LEMMA (a bounded-period axis near a carrier). Let M be a finitely
generated nontrivial subgroup of P, with fixed bases, and fix D >= 1.
There are constants L_D, B_D such that if a periodic P-line zeta of
primitive period at most D shares a segment of length at least L_D with
an M-line, then zeta is itself an M-line. Its primitive translation in
M has cyclic M-length at most B_D.
Why: 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 Hull(Lambda M).
If r is the actual primitive P-translation of zeta, the segment contains
V+1 successive vertices r^i z_0 in one period phase. Two have the same
core state, M r^i z_0 = M r^j z_0 with 0 <= i < j <= V. Cancelling z_0
gives r^{j-i} in M, so a power of the translation of this very axis lies
in 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 in M (up to
orientation), so z = r^q with 1 <= q <= V. Choose embedding constants
|g|_M <= A |g|_P + B. Applying this to powers of z, dividing by the
exponent, and taking the limit gives l_M(z) <= A l_P(z) <= A V D, so
B_D = A V D works.
Periodic support points can now be ruled out, which in turn lets the
transfer arguments reach every support line.
PROPOSITION (no periodic support lines). The support Omega contains no
periodic line.
Why: suppose eta_0 in Omega is periodic.
Step 1: build a trajectory and locate a maximal carrier. Approximate
eta_0 by lines of U. For each fixed centered window, their stabilizer
hulls meet a common finite ball, so window finiteness allows passing to a
subsequence with a constant actual window path; doing this successively
gives a bilateral aligned path whose every finite window has noncyclic
stabilizer. The limiting endpoints lie in each such stabilizer's limit
set. Choose frames h_s on this path with h_0 = 1. Boundary transfer on a
fixed window is continuous, so the conull image lines converge in the
successive frames to lines eta_s in Omega. Each eta_s is periodic: the
core-state repetition argument in the carrier-period lemma puts a power
of the original axis translation in each fixed window stabilizer, and
that power transfers to a nontrivial element translating the image.
Choose the primitive P-translation r_s of eta_s toward its positive
endpoint, and 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_s I p_s^{-1}, and its image is r_{s+1}^{n_s} with n_s the least positive
integer with r_{s+1}^{n_s} in q_s^{-1} J q_s. The exponents are positive
because each translation is oriented toward the transferred positive
endpoint. The finite ambiguity dictionary of the finite-roots lemma says:
if either minimum exponent exceeds one, or an axis belongs to two distinct
domain cosets on its side, then 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, and the
repetition lemma would give a tree-hyperbolic element commensurating a
nontrivial elliptic cyclic group, contrary to BS-freeness. So only
finitely many indices are exceptional.
It follows that there are positive integers k_s with
m_s divides k_s, k_{s+1} = n_s k_s / m_s,
and only finitely many ratios differ from one. Choose k_0 divisible by all
the finitely many factors needed when propagating in both directions; then
h_s r_s^{k_s} h_s^{-1} is one fixed nontrivial element, independent of s.
Fix s. 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_s e: the edges h_s^{-1} E_{s+j} have
central edge e and belong to the same window family. Its length is fixed
as the window size tends to infinity. It cannot traverse a natural edge
whose length tends to infinity, since an immersed closed path that uses
the interior of such an edge traverses the whole edge cyclically. So 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 lie in finitely many conjugacy classes and repeat. So 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 d^2 Lambda M. (If there were no maximal
noncyclic carrier, this conclusion itself would be a contradiction.)
Step 2: follow the expanding return iterates. The forward path now 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, so
transfer and repeat give 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.
Let a_j >= 1 be the primitive cyclic M-length of its j-th positive
beta-iterate. At some time a_{j+1} >= lambda a_j; otherwise the cyclic
based inequality, applied at every j >= 1, would give a_{j-1} >= lambda
a_j, impossible for positive integer lengths. Once a forward expansion
occurs it continues, because 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, pick a
vertex x_0 on its axis, and set y_0 = z^k x_0 for a large integer k, so
l_0 = k d_0. The cyclic expansion gives l_M(beta z) >= lambda d_0, hence
d_M(beta x_0, beta y_0) = d_M(beta x_0, (beta z)^k beta x_0)
>= k l_M(beta z) >= lambda l_0,
even if beta x_0 is off the axis of beta z. So for l_0 >= L_0 the
finite-failure lemma supplies a failure time n, a support line, and a
bounded-period axis sharing the overlap segment.
Step 3: pull the failure axis back to the initial period. The periodic
P-line zeta has bounded period, allows both the prescribed and a competing
path, and shares with the current M-line a segment at least as long as the
overlap. Take l_0 large enough that, already at n = 0, this overlap exceeds
the threshold L_D of the carrier-period lemma. That lemma makes zeta itself
an M-line, with primitive M-period bounded by B_D.
The two corresponding M-lines now share a segment of length at least
c_0 2^n l_0 - C_0, for fixed c_0 > 0, C_0. Here is the 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,
and 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.
Choosing the pairs in oriented order, as the transport lemma allows, and
using that a tree is convex gives a common central segment in C_M after
deleting a fixed amount at both ends.
The uniformly bounded primitive period of zeta now shows the common segment
contains at least c' 2^n l_0 - C_3 whole primitive zeta-periods. Apply the
period-pullback lemma with psi = beta^{-1} successively n times. At time
zero at least
c' 2^n l_0 - C_3 - n C_4
whole periods remain. Choose the initial length so large that
c' l_0 > max(C_4, C_3 + d_0 + 1).
Since 2^n >= n + 1, the accumulated period count exceeds d_0 + 1 for every
possible failure time n, so every intermediate pullback has enough periods.
The primitive period of the second line after these pullbacks, call it b,
may be large, but what is controlled is the number of full periods, not
their metric length. More than d_0 + 1 periods give a common label segment
of length at least (d_0 + 2)b >= d_0 + b, and the periodic-labels lemma
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, contradicting the unique forward coset choices on the late
periodic trajectory. So failure is impossible too, completing the
contradiction.
Having excluded periodic support points, the transfer theory now reaches
every line of the support, not just those in U.
COROLLARY (support). The following hold for the closed support Omega.
(1) No line of Omega is carried by the small subgraph. There is an
integer B_0 such that every segment of a lift of an Omega-line to
any selected Gamma_n, if contained entirely in the small subgraph,
has length at most B_0.
(2) Every endpoint of every line of Omega is irrational.
(3) Every line of Omega lifts to each centered core union Gamma_n and
has a unique compatible bilateral trajectory. On each fixed window
the partial boundary transfer takes every support line in its
domain to a line of Omega.
Why: an immersed bi-infinite path in a finite cyclic small component is
periodic, excluded by no-periodic. A line carried by a noncyclic small
component is carried, up to conjugacy, by a maximal noncyclic carrier, and
carrier forces periodic would then produce a periodic support line, also
impossible. A tree component cannot carry an immersed bi-infinite path.
This proves the first statement 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 segments
converges to an immersed bi-infinite path in it, and the limit support line
would be carried by the small graph. This contradiction gives B_0.
A rational endpoint is represented by an eventually periodic ray.
Translating farther and farther along its periodic tail would give a
periodic line as a limit of support lines, and closedness together with
no-periodic excludes this. So both endpoints of every support line are
irrational.
The lifts to each Gamma_n already followed from the window-cores lemma,
using conull lifts, density, and closedness. Each such lift gives an actual
window path carrying the line's endpoints, and irrational uniqueness makes
these paths compatible when a window is shortened, hence a unique
bilateral trajectory.
For the transfers of support lines, approximate a given support line by
lines of U and fix one window. All approximants meet a common fixed ball,
so the hulls of their window stabilizers do too, and there are finitely many
possible actual paths. Any path occurring infinitely often carries the
limiting endpoints, since its limit set is closed, and irrational uniqueness
identifies it with the limiting line's own path. So the approximants
eventually have that same path. The boundary transfer on its fixed subgroup
is continuous, and their images lie in U, hence in Omega, so they converge
to the transferred limiting line, which is in Omega by closedness. This
proves the last statement without discarding any support point.
Almost everything is in place. If G were not hyperbolic, there would be a
nonzero stationary current mu whose closed support Omega has unique
bilateral trajectories, irrational endpoints, and transfers that map the
support into itself. This last part derives a contradiction by detecting
branching in the language of Omega. Bounded graph complexity leaves only
finitely many rays with more than one possible past. A periodic orbit among
their endpoints produces a partial conjugation that slides points
progressively farther along one ray, and no finite invariant measure on the
lines through a vertex can survive such a slide.
Fix the basis of P. Orient every line from its negative endpoint to its
positive endpoint. An infinite reduced word rho is occurring if it labels a
positive ray in some line of Omega. An occurring word rho is left-special
if there are distinct basis letters a, b such that both a rho and b rho
occur, both extensions being reduced. Translating the two rays to start at
the identity gives two lines of Omega with a common positive ray but
different incoming edges: two different pasts for the same future.
LEMMA (finitely many left-special words). The set of left-special
occurring words is finite and nonempty.
Why, finiteness: let rho be left-special. Both of its predecessor lines
lift to each Gamma_n by the window-cores lemma. Their common endpoint is
irrational, so uniqueness of the starting state for an irrational ray puts
both lifts at the same vertex. That vertex has two different incoming edges
as well as the outgoing ray edge, so it is a natural vertex.
Let B bound the lengths of small-only segments of support lines. Starting
at this natural vertex, the lift of rho follows at most B labeled 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, so on a subsequence both 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, proving
finiteness.
Why, nonempty: suppose no occurring ray were left-special. Then there is an
integer N such that every occurring word of length N has a unique
predecessor letter. Otherwise arbitrarily long finite words would have two
distinct predecessor letters, and passing to a fixed pair of letters and a
convergent subsequence of their full-line extensions would give a
left-special infinite word, using compactness of the space of lines through
a vertex and closedness of Omega. Now 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;
since the sum of the indegrees equals the number of vertices, every
outdegree is also one. The graph is a disjoint union of directed cycles, so
every line of Omega would have a periodic label, contradicting no-periodic.
Let E be the set of P-orbits of the positive endpoints of left-special
rays. It is finite and nonempty. The forward path transfer induces a
permutation of E. Take two predecessor lines of such a ray; their common
irrational endpoint puts them in the same unique domain coset. The
boundary embedding on that coset sends them to two support lines with the
same positive endpoint and distinct negative endpoints, which after their
positive rays merge exhibit a left-special ray. Changing the input
representative by an element of P only changes the output by an element of
P, and the backward transfer is the inverse construction, so this really
is a permutation.
Choose a left-special word based at 1 whose endpoint xi represents a
periodic element of E, of period p > 0. Compose the p forward transfers and
adjust the final frame by an element of P. The actual path construction
gives an element v in G of nonzero height p and a partial conjugation
Theta: S -> T, Theta(x) = v^{-1} x v,
S = P intersected with v P v^{-1},
T = P intersected with v^{-1} P v,
whose boundary map fixes xi. Here S and T are the full stabilizers,
expressed in initial and final frames, of the corresponding finite path, so
they are finitely generated subgroups of P, and the map and its inverse
preserve mu on their domains in U.
Write r_0 = 1, r_1, r_2, ... for the consecutive vertices of the ray to xi
in C_P. Since xi is in Lambda S and S is finitely generated, choose s_n in
S and a fixed K with d(s_n, r_n) <= K for every n. The embedding Theta is a
quasi-isometry onto T and fixes xi, so there is a sequence f(n) with, after
enlarging a fixed constant C,
d(Theta(s_n), r_{f(n)}) <= C, f(n) -> infinity,
|f(n+1) - f(n)| <= C.
These are the ray versions of the transport estimates.
LEMMA (positive lag). After replacing Theta by its inverse if necessary,
the choices above may be made so that
f(n) - n -> +infinity.
Why: if |f(n) - n| stayed bounded along an infinite subsequence, then
s_n^{-1} Theta(s_n) would take only finitely many values there, while the
s_n escape every finite ball. So there are distinct s_n, s_m and d in P
with Theta(s_n) = s_n d and Theta(s_m) = s_m d, giving
Theta(s_n s_m^{-1}) = (s_n d)(s_m d)^{-1} = s_n s_m^{-1}.
The nontrivial element s_n s_m^{-1} has infinite order and commutes with v.
Their heights are zero and p, nonzero, so together they generate Z^2,
contradicting BS-freeness.
So |f(n) - n| tends to infinity. The signed sequence f(n) - n has bounded
successive increments, 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) with Theta^{-1}(t_m) uniformly
close to r_{g(m)}. A first-crossing choice n(m) satisfies
|f(n(m)) - m| <= C and n(m) -> infinity. Then t_m and Theta(s_{n(m)}) are
uniformly close, and the inverse quasi-isometry gives g(m) = n(m) + O(1)
and g(m) - m -> +infinity, proving the assertion for the inverse map.
The positive-lag return iterates a measure-preserving transfer along the
ray. If a fixed positive-mass piece of the support could slide forward
indefinitely without shrinking, its translates would produce infinitely
many pairwise-disjoint copies of one fixed mass inside a single compact
set of finite measure.
LEMMA (forcing). Fix s >= n_0, and for N >= s let
U_N = { ell in U : ell contains the directed segment [r_s, r_N] }.
Every line of U_N contains r_{n_0} and lies in the domain of Theta.
Why: a line of U_N contains [r_s, r_N] with s >= n_0, so it contains
r_{n_0}. If a line of U_N exited the domain of Theta, one of the finitely
many actual window paths would produce a competing actual path for the
window defining Theta. Its stabilizer limit set contains the positive
endpoint xi. Uniqueness of the actual path for an irrational endpoint
then identifies the competing path with the path defining Theta,
contradicting the exit assumption.
LEMMA (nested cylinders). Choose s > max{n_0, f(n_0) + D} with D the
transport constant below. For every sufficiently large N the cylinder
U_N has finite positive measure and satisfies
Theta(U_N) is contained in U_{N+1} contained in U_N.
Why: U_N is compact, open, and meets the support, so it has finite
positive measure. The forcing lemma makes Theta(U_N) defined. Apply the
transport lemma (nearness and order) to the fixed inclusion S into P and
the fixed injection Theta: S -> P; it produces E such that Theta(s_{n_0})
and Theta(s_N) are within E of every line of Theta(U_N). The defining
property of f puts r_{f(n_0)} and r_{f(N)} within E + C of those image
lines, and the input projections are positively ordered with separation
at least N - n_0 - 2K. Choose D > E + C; then every line of Theta(U_N)
contains the trimmed segment [r_{f(n_0)+D}, r_{f(N)-D}] for all
sufficiently large N. The positive lag lemma also gives f(N) - D >= N + 1
for all large N. Hence the trimmed segment contains [r_s, r_{N+1}]
whenever s > f(n_0) + D, and every line of Theta(U_N) lies in U_{N+1}.
The containment U_{N+1} inside U_N is immediate from the definition of
U_N.
COROLLARY (the final contradiction). The hypotheses of the combination
theorem and the failure of hyperbolicity cannot both hold.
Why: by the nesting lemma, and by measure preservation of Theta on its
domain in U,
mu(U_N) = mu(Theta(U_N)) <= mu(U_{N+1}) <= mu(U_N)
for all sufficiently large N. So all sufficiently late U_N have the same
finite positive measure, call it delta. Continuity from above on the
decreasing sequence of compact-open cylinders gives
mu(U_infty) = delta > 0, U_infty = intersection of U_N,
and every line of U_infty contains the full directed ray [r_s, xi).
For each n >= s translate U_infty by r_n^{-1}. The translate consists of
lines through r_0 = 1 and so lies in the single compact set X_0 of lines
through 1; its measure is still delta. Its positive endpoints are
r_n^{-1} xi. For n != m these are distinct: an equality would give a
nontrivial element r_m r_n^{-1} in P fixing xi, making xi rational,
contrary to the fact that every support endpoint is irrational. So the
translated sets are pairwise disjoint subsets of the compact finite-
measure set X_0. Their total mass would be infinite, a contradiction.
Hence the combination theorem's hypotheses force G to be hyperbolic.
This completes the restricted combination theorem. The result shown:
a restricted HNN extension whose edge gluing satisfies the
intersection and strong-inertia hypotheses, and whose result contains no
Baumslag-Solitar subgroup, is word-hyperbolic.
With the combination theorem in hand, the main theorem now follows.
Every finitely generated one-relator group with a non-power relator is fed
to the combination theorem as its ambient group H, and the hypothesis that
this ambient group is hyperbolic is exactly what must be established, by a
separate induction. This is the classical Magnus induction, run
geometrically on a graph presentation: split the one-relator group over
free edge groups, notice that the vertex group is again a one-relator
group of smaller complexity, and induct down to free groups.
Two structural facts power the induction. First, Magnus subgroups of a
one-relator group inject into it. Second, in a hyperbolic one-relator
group they are quasiconvex (Linton's theorem). Together they supply the
combination theorem's hypotheses.
A one-relator group is best pictured as a graph with one disk glued in.
Let Gamma be a finite connected graph, and let lambda: S^1 -> Gamma be a
nontrivial immersed combinatorial circuit (no backtracking, including
around the closing point). Form the quotient space
X = Gamma glued along lambda to D^2, S_lambda = lambda(S^1),
so that pi_1 X = <pi_1 Gamma | lambda> is a one-relator group. A connected
subgraph Delta of Gamma is a Magnus subgraph if S_lambda is not contained
in Delta, i.e. Delta omits an edge traversed by lambda.
LEMMA (graph Magnus subgroups inject). If Delta is a Magnus subgraph,
there is a free basis Sigma of pi_1 Gamma in which pi_1 Delta is
generated by a subset omitting a letter occurring in the cyclically
reduced relator. Consequently pi_1 Delta -> pi_1 X is injective.
Why: extend a spanning tree of Delta to one of Gamma, and use them to
describe pi_1 as words in the edges outside the spanning tree. The relator
is r = u lambda u^{-1} for a path u to the basepoint. Its first letter lies
in the part of the edge left out by Delta, so r becomes a cyclically
reduced nonidentity word using that letter. A subgroup spanned by all but
one free generator that appears in the relator is a Magnus subgroup, and
Magnus subgroups of one-relator groups inject.
The relator r can be assumed cyclically reduced and a non-power, meaning
its primitive root is r itself. This is the setting of the main theorem:
the proper-power case is handled separately by a local spelling theorem
(via a Dehn algorithm), and the trivial-relator case is a free group.
The induction step replaces X by a cyclic cover and reads off an HNN
splitting. Complexity is lexicographic order on pairs of nonnegative
integers attached to a non-power graph presentation; each step strictly
decreases it, so the induction terminates.
Fix a nontrivial homomorphism chi: pi_1 Gamma -> Z and let Gamma_chi ->
Gamma be the connected cyclic cover with deck transformation tau generating
the deck group. Choose a spanning tree downstairs. Index lifted vertices by
(v, n) so that tree edges preserve n and each oriented non-tree edge e has
an integral increment d_e; the increments generate Z because chi is
nontrivial.
LEMMA (a tree domain in a cyclic cover). Every connected cyclic cover of
a finite connected graph has a domain D with D intersected with tau D a
nonempty tree.
Why: let D_N be the subgraph induced by all vertex types at levels
[-N, N]. For large N every base edge occurs and its occurrences form an
interval. The lifted spanning trees connect the different vertex types at
each level. Choose a word in the signed increments whose total is 1 and
whose partial sums are bounded by L; translates of it connect consecutive
levels away from the ends of [-N, N]. Choose also a nonzero increment a.
From near a boundary, repeated steps of size a in the positive direction
reach the interior, and repeated negative steps do the same from the top.
For N > L + a these steps stay in [-N, N], so D_N is connected; the same
argument applied to [-N+1, N] shows D_N intersected with tau D_N is a tree.
Now the split. Spread the domains tau^i D along the cover. Because cell
occurrences form intervals and D intersected with tau D is nonempty, the
union of an interval of translates is connected, and the intersection of
two adjacent unions is exactly the adjacent overlap.
LEMMA (the Magnus splitting). With the notation above there is a
non-power graph presentation
G = <pi_1 U, t | t A t^{-1} = B>,
where
- pi_1 U injects into pi_1 Gamma, so U is BS-free whenever G is;
- U_- = tau^0 D union ... union tau^{m-1} D and
U_+ = tau^1 D union ... union tau^m D are subgraphs whose
fundamental groups A = pi_1 U_- and B = pi_1 U_+ are finitely
generated free Magnus subgroups of pi_1 U, and U_- intersected with
U_+ is connected;
- the presentation of pi_1 U has strictly smaller complexity than that
of X.
When the overlap groups are trivial this is a free product with Z;
otherwise it is a genuine HNN extension.
(local pdf: linton-hierarchies.pdf; https://arxiv.org/abs/2202.11324)
Why: take a lift of the relator lambda to a closed circuit in the cover.
Either this lifted support fails to project injectively on vertices, or it
does.
Case 1 (it does not). Let m >= 0 be the least integer for which the union
U of tau^i D, 0 <= i <= m, contains such a closed lift. Cell occurrences in
U are the shifted intervals of those in D, so U is connected and each
adjacent overlap is a tree; when m = 0 the two overlaps are both D itself.
Minimality of m makes neither U_- nor U_+ contain a closed lift of lambda,
so both are Magnus subgraphs, and their intersection, a translate-union of
domains, is connected. The lifted circuit has the same length as lambda but
strictly more support vertices, which reduces the complexity of pi_1 U.
Case 2 (it does project injectively). Then the lifted support projects
isomorphically onto S_lambda, the character chi vanishes on pi_1 S_lambda,
and every lift of this support is a single copy of it lying at one level.
The support copies then tile the cover, and the rank of the ambient graph
strictly drops while the relator stays a non-power.
In both cases the quotient is the one-disk space Y = Gamma_chi glued along
a lift of lambda. Form a line of disjoint copies Z_i = tau^i Z of a chosen
piece Z, joining adjacent copies by the cylinder on their overlap; call the
result W, with its collapse map W -> Y.
Every loop and every homotopy uses finitely many copies, so W -> Y
induces an isomorphism on fundamental groups as well. The vertex group
pi_1 U embeds in pi_1 Gamma by inclusion into the cover and the covering
map, and a proper power in pi_1 U would remain one in pi_1 Gamma, giving
the non-power assertion.
Finally, the stable letter. With a point x in U_- and a path q in U_+ from
x to tau x, set t = [p(q)]; it has height one, and the edge isomorphism
sends a loop a in U_- to q (tau a) q^{-1} in U_+, projecting to
t a t^{-1}. When the overlaps are trivial there is no edge relation, and t
is chosen similarly. In either case q closes in the quotient by crossing
the cylinder at tau x, so it is a stable letter.
Now the induction concludes.
PROPOSITION (non-power hyperbolicity). Let Gamma be a finite connected
graph, and let lambda: S^1 -> Gamma be a nontrivial immersed circuit
representing a non-power in pi_1 Gamma. If the one-relator group
pi_1(Gamma glued to D^2 along lambda) is BS-free, then it is
word-hyperbolic.
Why: induct on the nonnegative lexicographic complexity. In rank one the
non-power relator generates the cyclic pi_1 Gamma and the quotient is
trivial; rank zero cannot support the circuit.
In higher rank apply the splitting lemma to get G = <H, t | t A t^{-1} = B>.
The vertex group H embeds in G, hence is BS-free, and has a non-power graph
presentation of smaller complexity, so H is hyperbolic by induction. The
one-relator torsion theorem (McCool and Schupp) makes H torsion-free. The
side groups A and B are finitely generated free Magnus subgroups of H, hence
quasiconvex (Linton). Collins's conjugate-intersection theorem, which
permits equal Magnus subgroups, gives: for M, N in {A, B} and g in H,
M intersected with g N g^{-1} noncyclic implies g in M N,
the first overlap hypothesis. Collins's direct-intersection theorem gives
I = A intersected with B = C or C * <u>,
where C is generated by the common basis letters and the second expression
is an internal free product with an infinite cyclic group. So I and its
transported copy J = t^{-1} I t are finitely generated.
If the edge group is nontrivial, the Magnus subgraphs U_- and U_+ have
connected intersection, and Linton's strong-inertia lemma says the
intersection of the fundamental groups of two such subgraphs is strongly
inert in either side group. Hence I is strongly inert in A and in B; since
isomorphisms preserve reduced ranks and the indexing double cosets,
transporting the latter assertion through the edge isomorphism makes J
strongly inert in P as well. These are exactly the two strong-inertia
inequalities the combination theorem asks for. When the edge group is
trivial, the same requirements hold directly. So G is a restricted HNN
extension whose edge gluing satisfies both hypotheses and whose result is
BS-free, and the combination theorem makes G hyperbolic.
MAIN THEOREM. Let X be a finite set and r an element of the free group
F(X). If G = <X | r> contains no subgroup isomorphic to BS(m,n) for any
nonzero integers m, n, then G is word-hyperbolic.
Why: freely and cyclically reduce r; conjugating or inverting r does not
change its normal closure. If the reduced relator is trivial, G is a
finitely generated free group and is hyperbolic. If it is a proper power
w^k with k >= 2, Newman's spelling theorem turns the presentation into a
Dehn algorithm: every nonempty freely reduced null word contains a piece
of a cyclic conjugate of w^{+-k} strictly longer than (k-1)|w|, i.e. more
than half the relator length; replacing that piece by the inverse
complementary piece strictly shortens the word and consumes one defining
relation. These shortening moves give a linear isoperimetric inequality,
so G is hyperbolic. In the remaining case r is a cyclically reduced
non-power; realize it as an immersed circuit in the basis rose and apply
the non-power proposition above.
The same machinery also proves the geometric subgroup statement that fuelled
the induction.
THEOREM (Magnus subgroups are quasiconvex). Let Gamma be finite connected,
lambda a nontrivial immersed circuit representing a non-power in
pi_1 Gamma, and Y the space obtained by gluing a D^2 along lambda. If
pi_1 Y is BS-free, then for every connected subgraph S of Gamma with
lambda(S^1) not contained in S, the map pi_1 S -> pi_1 Y is injective and
its image is quasiconvex.
Why: the proposition makes the ambient group hyperbolic; the graph Magnus
lemma identifies pi_1 S with an injective Magnus subgroup of a one-relator
presentation; Linton's theorem makes that subgroup quasiconvex, and a
basepoint change only conjugates it.
Hyperbolicity is a geometric condition. The main theorem converts it into
a structural one about virtually free-by-cyclic groups.
A group H is virtually free-by-cyclic if it has a finite-index subgroup H_0
fitting into a short exact sequence
1 -> F -> H_0 -> Z -> 1,
with F a finitely generated free group. Equivalently, H_0 is the mapping
torus of an automorphism of a free group: it is generated by F and a stable
letter t with t F t^{-1} contained in F, conjugating F by that
automorphism.
COROLLARY. Every finitely generated Baumslag-Solitar-free one-relator
group is virtually free-by-cyclic.
The route: BS-freeness gives hyperbolicity (main theorem); a separate
theorem makes the group virtually compact special. Since F(X) is the
fundamental group of a compact connected planar surface with |X|+1
boundary components, Kielak-Linton's Corollary 1.3 makes G virtually
free-by-cyclic, and their Theorem 1.1 sharpens this to a finite-index
embedding in F_d rtimes_alpha Z for some d >= 0 and some automorphism
alpha of F_d. The BS-free hypothesis enters only once, through the main
theorem; nothing downstream depends on BS-freeness.
(local pdf: kielak-linton.pdf; https://arxiv.org/abs/2302.11500)
The relator's primitivity rank controls a complementary story. The
primitivity rank of a relator r is the least rank of a subgroup of F(X)
containing r in which r is not part of a free basis, with value infinity if
no such subgroup exists; high primitivity rank means the relator lines up
with few subgroups, and "negative immersions" is the case of primitivity
rank greater than two. Louder and Wilton proved that groups with negative
immersions have no Baumslag-Solitar subgroups and are coherent, and Linton
proved that they are hyperbolic and virtually special. Linton's later
characterization: a finitely generated one-relator group is both hyperbolic
and locally quasiconvex exactly when its relator has primitivity rank
different from two. Local quasiconvexity means every finitely generated
subgroup is quasiconvex. That characterization does not by itself settle
hyperbolicity for BS-free groups whose relator has primitivity rank two.
The present argument does, for the whole BS-free class at once, by the
combination-and-induction argument; and the virtually free-by-cyclic
corollary extends the structural picture from the negative-immersion class
to all Baumslag-Solitar-free one-relator groups.
(local pdf: louder-wilton-neg.pdf, https://arxiv.org/abs/1803.02671; linton-hyperbolic.pdf, https://arxiv.org/abs/2211.04371)