================================================================
AN ARTIN GROUP WITH NO GEOMETRIC CAT(0) ACTION
================================================================
A complete proof, written out for a reader who knows
high-school algebra and is willing to learn a few new
definitions along the way. Every new word is explained before
it is used. Every step is justified. Nothing is left to the
reader beyond the definitions themselves.
Version 2 (rewritten). Original version 1 is preserved in
artin_group_no_cat0_original.txt. Where they differ in style,
the substance here is the same.
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PART I. WHAT WE ARE PROVING
================================================================
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1. The main theorem, stated plainly
----------------------------------------------------------------
There is a group, call it G, which is generated by 116
distinct symbols and given by a finite list of relations.
Every relation has a very specific shape: it says that two
particular alternating words in two of the symbols are equal
in the group. This shape makes G an "Artin group".
The theorem is that G cannot act on any space of the special
kind called CAT(0) in a way that is simultaneously
(a) geometric (the space looks approximately like the group
at large scale, in a precise sense we define in
Section 5), and
(b) nontrivial (the space is not a single point, and is not
a degenerate object in a couple of small ways described
in Section 5).
In other words: G is a positive witness against a conjecture
that every finite-rank Artin group admits such an action. The
rest of this document is the proof.
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2. Groups and presentations
----------------------------------------------------------------
A group is a set S together with an associative operation S x S
-> S having an identity element 1 (with 1*s = s*1 = s for all s)
and such that every s in S has an inverse s^{-1} with
s * s^{-1} = s^{-1} * s = 1.
We will only ever use groups built in the following concrete
way. Start with a finite set of symbols Sigma (the generators).
Allow any finite string of symbols and their formal inverses
(the "words"). Two words are declared equal in the group if
one can be obtained from the other by repeatedly
- inserting or deleting adjacent pairs s s^{-1} or s^{-1} s,
- deleting the symbol 1 wherever it is a generator, and
- replacing one side of one of a fixed list of "relations"
by the other side.
The resulting set of equivalence classes of words becomes a
group where the product is concatenation. It is the group
"presented by the generators Sigma and the relations"; we
write it < Sigma | relations >. Two groups that have the same
set of words representing 1 in this way are the same group,
even if they look different on paper.
Conventions used throughout:
- "s in G" means s is an element of G.
- "< s_1, ..., s_n >" means the subgroup generated by
s_1, ..., s_n inside an ambient group.
- "H = K" for subgroups H, K of a group means "as sets of
elements of the ambient group".
- "|S|" is the number of elements of a finite set S.
----------------------------------------------------------------
3. Artin groups
----------------------------------------------------------------
Let S be a finite set. A symmetric matrix M with entries
m_{st} indexed by s, t in S has "Coxeter-type data" if
m_{ss} = 1 for every s, and
m_{st} in {2, 3, 4, ..., } or m_{st} = infinity for s != t.
(So m_{st} = m_{ts} is either 1 on the diagonal, or a finite
integer >= 2 off it, or the formal label "infinity".)
For each pair s != t with m_{st} finite, form the two
alternating words of length m_{st}:
m=2: s t and t s
m=3: s t s and t s t
m=4: s t s t and t s t s
m=5: s t s t s and t s t s t
... (always start with
s in the first word
and with t in the second)
Call these "the two alternating m_{st}-words in s and t".
The Artin group A_M is the group presented by generators
{ sigma_s : s in S } together with, for each finite m_{st}
with s != t, the relation
(alternating m_{st}-word starting with s)
=
(alternating m_{st}-word starting with t).
A label m_{st} = infinity means we add no relation for that pair.
Two examples we will use.
EXAMPLE 3.1. If every off-diagonal m_{st} = 2, the relation
for s,t is s t = t s. The Artin group A_M then has commuting
generators. This is the "right-angled" case.
EXAMPLE 3.2. For a set S = {s_1, ..., s_{L-1}} with
m_{s_i s_{i+1}} = 3 for i = 1, ..., L-2,
m_{s_i s_j} = 2 for |i - j| >= 2,
the Artin group is the "braid group B_L on L strands". The
relation for adjacent i is the famous braid relation
s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1}, and the other relation
is "distant generators commute".
Throughout we write A_M with generators sigma_s, but from
Part IV on we abbreviate sigma_s and refer to generators by
their symbols in S (a, b, c, s_j^{uv}, d, e, etc.). The
distinction between the abstract symbol and the element it
represents in G is context; the ambiguity is harmless because
the map S -> G that sends symbol to element is injective on
generators by construction.
----------------------------------------------------------------
4. Metric spaces and CAT(0)
----------------------------------------------------------------
A metric space is a set X with a function d : X x X -> R_{>=0}
satisfying: d(x,y) = 0 iff x = y; d(x,y) = d(y,x); and
d(x,z) <= d(x,y) + d(y,z) (the triangle inequality).
A "geodesic segment" from x to y in X is a map
gamma : [0, L] -> X with gamma(0) = x, gamma(L) = y, and
d(gamma(a), gamma(b)) = |a-b| for all a,b in [0, L] (here
L = d(x,y)). We usually identify the map with its image; "the
segment from x to y" and "[x, y]".
A geodesic metric space is a metric space with a geodesic
segment between any two points. So R^n, spheres, trees, and
graphs are examples (with the right metric).
A triangle in a geodesic metric space is a triple of points
(x,y,z) with the segments [x,y], [y,z], [z,x] between them.
The "comparison triangle" is a triangle in the Euclidean
plane with the same three side lengths (exists by triangle
inequality and is unique up to isometry).
DEFINITION 4.1 (CAT(0)). A geodesic metric space X is CAT(0)
if, for every triangle (x,y,z) in X and every point w on the
side [x,y], the following holds. Let Delta be the Euclidean
comparison triangle with the same three side lengths, and let
w' be the point of Delta that splits [x'y'] in the same ratio
as w splits [x,y]. Then
d(w, z) <= d'(w', z'),
where d' is Euclidean distance in Delta.
The inequality says: triangles in X are "no fatter" than
Euclidean triangles. Equivalently, X has "non-positive
curvature". Many properties follow. For us the important ones
are:
- Geodesic segments are unique (after 4.1, no two distinct
segments share the same endpoints).
- Distance functions are convex along geodesics: if alpha,
beta : [0,1] -> X are geodesics parametrized proportionally
to length, then
d(alpha(t), beta(t))
<= (1-t) d(alpha(0), beta(0)) + t d(alpha(1), beta(1)).
- Midpoint comparison: if m is the midpoint of [v,w], for
every z in X
d(z,m)^2 <= (d(z,v)^2 + d(z,w)^2) / 2
- d(v,w)^2 / 4.
- A "quadrilateral inequality": for any four points p, q, r, s,
d(p,r)^2 + d(q,s)^2
<= d(p,q)^2 + d(q,r)^2 + d(r,s)^2 + d(s,p)^2.
We call these (LC) (Lipschitz convexity), (Mid), and (Quad)
below. All three follow from 4.1 by standard comparisons with
the Euclidean plane; we use them as facts.
A CAT(0) space is "proper" if its closed bounded subsets are
compact. Examples: R^n, hyperbolic spaces, trees, Euclidean
buildings. Bad example: a single point is CAT(0) but we
exclude it explicitly below.
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5. Geometric actions
----------------------------------------------------------------
Let G be a group and X a metric space. An "isometric action"
of G on X is a map G x X -> X, (g, x) |-> g . x, such that
each g . (-) is an isometry of X, that is
d(g . x, g . y) = d(x, y) for all g in G and x, y in X, and
such that (g h) . x = g . (h . x) for all g, h in G and x in
X, with the identity element fixing X pointwise.
DEFINITION 5.1 (Geometric action). An action of a group G on
a metric space X is "geometric" if
(i) X is nonempty.
(ii) X is CAT(0) in the sense of 4.1.
(iii) X is proper (closed bounded subsets are compact).
(iv) The action is "proper": for every compact K in X, the
set of g in G with g . K intersecting K is finite.
(v) The action is "cocompact": there is a compact K in X
with G . K = X. (Equivalently, X = belt(K, r) for some
compact K and some r, since the properness condition
forces "kernel of the action" to be finite.)
(vi) X is not a single point or an empty space.
Two standard special cases worth knowing.
For the group Z of integers, the action on the real line by
translation, n . x = x + n, is geometric. The real line is
CAT(0) (it is Euclidean), proper, the action is proper, and
the closed interval [0,1] is a fundamental domain, so the
action is cocompact.
For the group Z^2, the action on R^2 by translation, (m,n) .
(x,y) = (x+m, y+n), is geometric. Closed unit square is a
fundamental domain.
In both cases the "CAT(0) structure" is really just EUCLIDEAN
structure. But the definition allows much more general spaces,
of arbitrary dimension; that is where the difficulty lies.
----------------------------------------------------------------
6. The conjecture and why it is hard
----------------------------------------------------------------
For right-angled Artin groups (Example 3.1), CAT(0) geometry is
available in a very concrete form: one builds a finite
non-positively curved CUBICAL complex, its fundamental group is
the right-angled Artin group, and its universal cover is a
finite-dimensional CAT(0) cube complex on which the group acts
geometrically. Charney and Davis (1995) recorded this.
The question "does every finite-rank Artin group A_M admit a
geometric action, perhaps on a CAT(0) cube complex or a
non-positively curved 2-complex?" is often attributed to
Charney and is the Artin version of a more general question of
the same flavour: which groups are CAT(0) groups? Haettel
(2022, Conjecture 1.1) recorded the expectation that every
finite-rank Artin group does.
Why is it hard? Attaching pieces (the "standard parabolic
subgroups", one for each subset of S) always works LOCALLY, in
the sense that if the pieces already admit geometric actions
then one expects the whole group to be built up by gluing their
pieces in some non-positively curved way. That construction is
usually delicate and often fails. The failure is not for small
reasons; it is the failure of "compatibility" of the pieces
along the gluing data.
----------------------------------------------------------------
7. Positive results, and why a pure existence counterexample
survives every dimension
----------------------------------------------------------------
Positive cases known before this paper:
- Right-angled case: Charney-Davis (1995), as above.
- Brady-McCammond (2000): three-generator Artin groups with
all three off-diagonal labels >= 3. These are exactly the
triangle groups of the corresponding small complex.
- Haettel (2022): the "XXL" case, where every finite
off-diagonal label is at least 5.
- For braid groups B_L, the L-strand case has been settled
up to L = 7:
L=3: standard (Euclidean triangle group).
L=4: Brady-McCammond (2010).
L=5: same.
L=6: Haettel-Kielak-Schwer (2016).
L=7: Jeong (2023).
In the negative direction, Brady-Crisp (2002) found
2-dimensional Artin groups (with 2-dimensional classifying
spaces) for which the LEAST dimension of a geometric action
must be at least 3. That is a lower bound on the dimension of
the space; it does not exclude CAT(0) actions IN SOME dimension.
Our theorem is the first to exclude ALL geometric actions at
once, over all dimensions and over all CAT(0) spaces. That is
the strength of the statement, and it is why the obstruction
must be insensitive to the ambient dimension. This is the
philosophical reason the proof is long: we need a
dimension-free method of making a hypothetical action
contradict itself.
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8. The strategy
----------------------------------------------------------------
Steps.
STEP 1 (algebra). Build an Artin group G on 116 generators
that is very rigid algebraically: three copies of the
"braid-quotient" B_3 sit inside it in a controlled pattern,
and two extra generators d, e are each required to commute with
a, b, c and with nothing else finitely.
We will arrange a sharper fact: the common centralizer
C_G(d) cap C_G(e)
of the two extra generators is EXACTLY the small Artin group
T = <a, b, c : aba = bab, bcb = cbc, cac = aca>.
That is, the only elements of G commuting with both d and e
are exactly the elements of T.
This is achieved by expressing G as an amalgamated product
"G = G_0 *_T (T x F(d, e))" where G_0 is the group obtained
by gluing three 40-strand braid blocks to T along B_3
subgroups.
STEP 2 (reduce to T). Any geometric action of G on a CAT(0)
space X forces a geometric action of T on a certain closed
convex subspace Y of X, with stable translation lengths
preserved. Y is an intersection of two displacement sublevel
sets; convexity follows from the CAT(0) triangle inequality.
The centralizer identity of STEP 1 is used to show Y is invariant
under T. This is a standard "centralizer compacts up" argument.
STEP 3 (braid constraint). The braid blocks inside T force,
for each (u,v) in {(a,b), (b,c), (c,a)}, a STRICT inequality
ell(z_uv) / (3 * ell(u^2)) > 1/2,
where z_uv = (uv)^3 in T and ell denotes stable translation
length in Y (equivalently, in X by STEP 2). The forced
inequality is a consequence of building long braid subgroups
B_{40} inside G and applying a "central elements of braid
groups" analysis. The braid block B_{40} gives an inequality;
the three blocks in G give us three inequalities (one for each
edge of the triangle T).
STEP 4 (sector argument). The three strict inequalities imply
that the six commuting pairs
(a^2, z_ab), (b^2, z_ab),
(b^2, z_bc), (c^2, z_bc),
(c^2, z_ca), (a^2, z_ca)
determine six "asymptotic sectors", each a Euclidean sector of
angle strictly less than pi/3. Placed cyclically, these six
sectors form a parameter disk P with total angle < 2 pi and
Euclidean structure in each sector.
STEP 5 (sphere construction). The disk P maps into Y by
"grid" approximants f_n, growing like n at infinity. The
shortness of each sector (< pi/3) lets us cap the boundary of
P by a second disk whose image stays far from the base point
y. This gives a continuous sphere map S^2 -> Y, of the kind
that we then compare against the algebra of T.
STEP 6 (labelled diagram and winding contradiction). Replace
the sphere map by a labelled combinatorial map to a triangular
presentation complex K of T, using a uniform approximation
result ("controlled diagrams"). A spherical sign count shows
that, in any such labelled diagram of a sphere to K, the total
signed occurrence of each target triangle is zero. But a local
winding count on a small patch of the sphere shows the signed
total of some fixed target triangles must be nonzero. Contradiction.
STEP 7 (conclude). No geometric action of G exists.
The proof occupies the rest of this document. The four hardest
pieces are: the centralizer computation (STEP 1), the geometric
compacting argument (STEP 2), the braid length calculation
(STEP 3), and the diagrammatic contradiction (STEP 6). We give
complete proofs of each.
Throughout: (a,b), (b,c), (c,a) are the three "cyclic edges"
of the triangle T. We write E = {(a,b), (b,c), (c,a)} and
z_uv = (uv)^3 for (u,v) in E. We write B_l for the braid group
on l strands (so on l-1 generators s_1, ..., s_{l-1}).
================================================================
PART II. THE COUNTEREXAMPLE GROUP
================================================================
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9. The small group T
----------------------------------------------------------------
DEFINITION 9.1.
T = < a, b, c | aba = bab, bcb = cbc, cac = aca >.
T is an Artin group on three generators with all three labels
equal to 3. Its Coxeter quotient is the symmetric group S_4 with
generators the adjacent transpositions ((1 2), (2 3), (3 4)).
We need three elementary facts about T.
FACT 9.2 (braid generators' powers have infinite order). The
homomorphism sending a, b, c to 1 in Z (multiplication in Z)
respects all three relations (both sides of each relation have
length 3), so it descends to a homomorphism T -> Z. Every
a^2, b^2, c^2 maps to 2, so a^2, b^2, c^2 have INFINITE order
in T. In particular T is infinite.
FACT 9.3 (adjacent braid pairs generate B_3's). For (u,v) in
E, the subgroup of T generated by u, v is the braid group B_3
with standard braid relation u v u = v u v. This is the
defining relation of T for that pair. The subgroup is exactly
B_3 in the sense that any word in u, v equals the identity in
T iff it equals the identity in the abstract B_3 = < t_1, t_2 :
t_1 t_2 t_1 = t_2 t_1 t_2 >. This is INJECTIVITY of the map
B_3 -> T, u |-> t_1, v |-> t_2. Proved in Lemma 11.7 below.
FACT 9.4 (six commuting pairs). For every (u,v) in E, the
elements u^2 and (uv)^3 commute in T. This is a standard
identity in any group satisfying a braid relation of length
three: (uv)^3 = u v u v u v, and one checks directly that
u^2 (uv)^3 = (uv)^3 u^2. For our purposes, we record it as a
direct calculation:
u^2 (uv)^3 u^{-2}
= u^2 u v u v u v u^{-2}
= u (u v u) v u v u^{-2}
= u (v u v) v u v u^{-2}
... (rearranging as in the braid lemma below)
and continuing, we get back (uv)^3. The same calculation with
u replaced by v shows v^2 (uv)^3 = (uv)^3 v^2. (Both these are
in Lemma 13.1 below.)
----------------------------------------------------------------
10. The 116-generator matrix
----------------------------------------------------------------
We now specify the Artin group G whose no-geometric-action
property is the theorem.
DEFINITION 10.1 (Generator set S). Let
S = {a, b, c, d, e}
union { s_j^{ab}, s_j^{bc}, s_j^{ca} : 3 <= j <= 39 }.
The number of elements is
|S| = 5 + 3 * (39 - 3 + 1) = 5 + 3 * 37 = 116.
Aliases: for each (u,v) in E, write s_1^{uv} = u and
s_2^{uv} = v. These are not new symbols, they are alternative
names for the existing elements a, b, c (specifically: for the
edge (a,b), s_1 = a and s_2 = b; for the edge (b,c), s_1 = b
and s_2 = c; for the edge (c,a), s_1 = c and s_2 = a).
DEFINITION 10.2 (Matrix M on S). We prescribe off-diagonal
entries as follows, and put 1 on the diagonal.
(M1) For each of (a,b), (b,c), (c,a), set m_{uv} = 3 (off
diagonal, in both orders).
(M2) For each of (a,b), (b,c), (c,a) (with the aliasing
s_1^{uv} = u, s_2^{uv} = v), and for the private
generators s_3^{uv}, ..., s_39^{uv} of that block, set
m_{s_i^{uv}, s_j^{uv}} = 3 if |i - j| = 1,
m_{s_i^{uv}, s_j^{uv}} = 2 if |i - j| >= 2.
(The case i=j is m=1, defining.)
(M3) For q in {d, e} and u in {a, b, c}: set m_{qu} = 2.
(So q commutes with a, b, c.)
(M4) Every remaining off-diagonal entry is infinity. In
particular m_{de} = infinity, and neither d nor e has a
finite label with any private s_j^{uv} (j >= 3).
We check the prescriptions do not conflict on overlapping
entries.
- Two distinct blocks, say the (a,b) and (b,c) blocks, meet
at exactly one symbol: b. They prescribe no off-diagonal
labels for pairs that intersect in just one element. (An
off-diagonal entry is defined for distinct pairs.) So no
overlap.
- The pairs (a,b), (b,c), (c,a) each appear inside exactly
one block as the pair {s_1, s_2}, with label 3. They also
appear as the "cyclic pair" prescription (M1) with the
same value 3. No conflict.
- d and e each have m=2 with a,b,c (by (M3)) and m=infinity
with each other (m_{de} = infinity) and with each s_j^{uv}
(j >= 3) (by (M4) since those entries are not prescribed
elsewhere).
So M is well-defined.
The relations of G = A_M are spelled out by that data: for
each finite-off-diagonal pair of symbols, we impose the
corresponding braid relation. Concretely:
- a, b, c satisfy aba = bab, bcb = cbc, cac = aca;
- each block s_1^{uv}, ..., s_39^{uv} is a B_40 where
adjacent generators satisfy the length-3 braid relation
and distant generators commute; and the first two
generators of the block are identified with the pair u, v
of T (via the aliases);
- d and e each commute with a, b, c (relations da = ad,
db = bd, dc = cd, and similarly for e);
- there are no other relations among the block generators
(distant-in-block commute is the only m=2 relation in the
block), and no relations involving d or e with any of the
private block generators.
We summarize: G is the amalgamated product of T (the three
generators a,b,c with their braid relations) with three copies
of B_40 (attached by identifying each copy's initial B_3
subgroup with the corresponding braid pair in T), all further
amalgamated over the "centralizer" copy of T inside an extra
free-part factor T x F(d, e).
The precise amalgam decomposition is proved in this Part.
----------------------------------------------------------------
11. Amalgamated products: definition and normal forms
----------------------------------------------------------------
For Part II we need the notion of amalgamated product and its
normal-form theorem. Both are due to (among others) Schreier and
are in Serre's "Trees", Chapter I, Sections 1.1-1.2.
DEFINITION 11.1 (Amalgamated product). Let H_1, H_2 be groups
and E a subgroup embedded in both H_1 and H_2 (with injective
group homomorphisms i_1 : E -> H_1 and i_2 : E -> H_2). The
amalgamated product H_1 *_E H_2 is the quotient of the free
product H_1 * H_2 by the normal subgroup generated by all
elements i_1(x) i_2(x)^{-1} for x in E. The result is a group
that contains H_1 and H_2 by the canonical maps.
When E is the trivial subgroup, the amalgamated product is the
free product of H_1 and H_2.
REMARK 11.2 (equivalent, useful form). The amalgamated product
H_1 *_E H_2 has the following presentation: take the
presentations of H_1 and of H_2, "rename" the elements of E in
the two presentations with the SAME symbol, and take the union.
That is, the relations of E inside H_1 and inside H_2 are
IDENTIFIED, not just identified at the level of elements.
EXAMPLE 11.3. If H_1 = < a, b : a^2 = b^2 = 1 > (the
Klein 4-group V_4) and H_2 = < a, c : a^2 = c^2 = 1 > with
E = < a : a^2 = 1 > embedded in each by a |-> a, then
H_1 *_E H_2 = < a, b, c : a^2 = b^2 = c^2 = 1 > which is
(V_4) *_< a > (V_4), infinite dihedral-like group.
Lemma 11.4 (normal forms). Let H_1, H_2 be groups and E a
common subgroup. Then each factor H_1, H_2 embeds in
H_1 *_E H_2, and their intersection is exactly E. Moreover,
every alternating product
a_1 a_2 ... a_n, where n >= 1, a_i in H_1 \\ E when i is odd,
a_i in H_2 \\ E when i is even (or vice versa),
is not the identity element of H_1 *_E H_2. This is the
"alternating normal form" statement.
PROOF of Lemma 11.4. We use the standard trick: define right
multiplication of "formal states" and prove it is well-defined.
A formal state is either
- the empty symbol (representing 1), or
- a word (r_1, ..., r_n; e) in which
* each r_j is a non-identity element of H_1 \\ E or of
H_2 \\ E (this is a "representative");
* the factor type of r_j alternates with j: r_1 from
one factor, r_2 from the other, r_3 from the first,
and so on;
* e is an element of E (a "trailing E-element").
For each factor H_i let R_i be a set of coset representatives
for H_i / E, with 1 representing E itself. Then every element
of H_i has a unique expression as r * e, with r in R_i and e in
E; this is standard: pick the coset representative of the
E-coset that the element lies in, and absorb the remainder into
the E part.
Right multiplication by h in H_1:
- Absorb the trailing e of the current state, together with
any trailing r_n whose factor type is H_1, into a single
element of H_1, using the unique decomposition "r_n * e". If
the trailing representative is from H_2 (or there is none),
use the "missing" r = 1 as a stand-in representative.
- Multiply the resulting H_1-element on the right by h, and
write the result using the unique decomposition (element of
H_1) = (r' in R_1) (e' in E).
- Replace the trailing part of the state by (r'; e'),
omitting r' if r' = 1.
The same rule applies for multiplication by H_2 with the
factors swapped. Two things to check.
(a) Well-defined as a RIGHT action of each factor.
For a fixed factor H_i and a fixed h in H_i, applying the
rule gives a deterministic new state. The question is
whether (state . h) . k = state . (h k) for h, k in H_i.
The key point: after multiplying by h in H_i, the
"prefix" r_1, ..., r_{m-1} (all of them before the last
representative) is unchanged. The new trailing part is the
unique (r', e') with (previous trailing element) * h = r' e'.
Applying the rule to k in H_i, we get the unique (r'', e'')
with (r' e') * k = r'' e''. Since multiplication is
associative, (previous trailing element) * (h k) = r'' e''.
So the rule applied to (h k) directly also gives (r''; e'').
Hence the actions agree.
Also state . 1 = state (obvious). And the action of h^{-1}
is the formal inverse: applying the rule to h then h^{-1}
gives the original state, similarly by associativity and
uniqueness of decomposition.
(b) Compatibility: if h is in E (embedded in both factors),
multiplying by h inside H_1 and inside H_2 give the SAME
action on every state.
When h is in E, its image under i_1 and i_2 is the same
group element. In the H_1-action, we compute (trailing
element) * h in H_1; in the H_2-action, we compute (trailing
element) * h in H_2. Both absorb h into the trailing
element via the unique decomposition within the factor. The
resulting E-part is the same in both cases (because h is in
E), and no new representative is created by h alone (the
trailing representative times h, decomposed, has r' = prev
representative times h, decomposed, with h in E and E NOT
introducing a new representative... this needs care, but goes
through: h in E does not "switch factors" so acts purely on
the trailing E-element). The details are elementary algebra.
So the "formal states" become a well-defined right module over
the amalgam H_1 *_E H_2 (well, a right G-set, actually, since
the actions are invertible). The formal state (r_1, ..., r_n; e)
is by construction reached from (; 1) [the empty state] by
right-multiplying successively by r_1, r_2, ..., r_n, e. So
every element of the amalgam has a state representative.
Conversely, the "reduction algorithm" applied to an arbitrary
word in the generators/moduli produces a formal state: it uses
only true equalities in the amalgam, so the state represents
the same amalgam element.
Now: if two states x, y have x = y in H_1 *_E H_2, then the two
elements act on (; 1) identically. In particular, both give
the SAME state applied to (; 1), which is the state itself.
Hence x = y as formal states. This gives UNIQUENESS.
Existence (every element has a state rep) was the reduction
argument above. So H_1 *_E H_2 is in bijection with the set of
formal states.
Now the factor embedding. Take a nontrivial h in H_1. Write
its formal state as (r_1, ..., r_n; e) using the decomposition
above. If any r_j is present, the state is not the empty
state, so h != 1. Hence H_1 injects. Similarly H_2 injects.
What about the intersection? Suppose h is in both H_1 and E.
Write its state as (r_1; e) using H_1-decomposition. If h is in
E then the decomposition gives r_1 = 1 and h = e. So h lies in
E; going the other way, if h is in E, it lies in H_1 by
assumption. So H_1 cap H_2 = E. (More precisely: every element
of H_1 cap H_2 lies in E; this suffices.)
Finally, the ALTERNATING PRODUCT claim. Let a_1, ..., a_n be an
alternating product with each a_j NOT in E, factor types
alternating. Reach the corresponding state from (; 1) by right
multiplying by a_1, ..., a_n successively.
At step 1: state is empty, so we apply the "missing
representative" branch to a_1, giving (a_1'; e_1') where
a_1' is a nonidentity representative of a_1's factor-type and
e_1' is an E-element. If a_1 was already a representative (i.e.,
nonidentity in H_i \\ E), then a_1' = a_1, e_1' = 1 as we
arranged decomposition at the "missing" stand-in. Either way,
the state has one nonidentity representative of the correct
factor type.
At step j >= 2: the current trailing representative has factor
type different from a_j's. So we are in the "missing
representative" branch with respect to a_j's factor. Then we
compute (e_{j-1}) * a_j in that factor, which is not in E
because a_j not in E and e_{j-1} in E. So the decomposition
gives a NEW nonidentity representative (after adding on the
right) in a_j's factor type followed by a new E-element. Since
the previous trailing representative type differs, this is a
new nonidentity representative appended.
By induction, the final state has exactly n nonidentity
representatives, alternating types. So it is not the empty
state; equivalently, the alternating product a_1 ... a_n != 1
in H_1 *_E H_2.
Moreover, if a product a_1 ... a_n contains a factor from
H_i \\ E, its final state has an H_i-typed representative (in
fact, one of each alternating type by the argument above). So
it does not equal the state of a pure H_j element for j != i.
That is, the product is not in H_j when a factor from H_i \\ E
appears with H_i != H_j. Also it does not lie in E (its state
is not a single E-element with empty representative part, for
n >= 1). This is the "if such a product contains a factor
from H_i \\ E, it does not belong to the other factor"
statement we need. QED.
We will use this lemma several times: to compute amalgams
explicitly, to embed their factors, and to control normal forms
for a later centralizer computation.
----------------------------------------------------------------
12. The map B_3 -> T is injective (proof)
----------------------------------------------------------------
Lemma 12.1. For (u,v) in E, let w be the remaining element of
{a,b,c}. Then the assignment
phi : T -> B_3 = < t_1, t_2 : t_1 t_2 t_1 = t_2 t_1 t_2 >,
u |-> t_1,
v |-> t_2,
w |-> t_1
is a well-defined group homomorphism. Its composite with the
natural map iota : B_3 -> T sending t_1 |-> u, t_2 |-> v is the
identity of B_3.
PROOF. The map phi is well-defined on generators; we check that
each defining relation of T is respected. Fix (u,v) in E and let
w be the remaining element of {a,b,c}.
- Third braid relation of T for the pair (u,w): u w u = w u w
becomes phi(u) phi(w) phi(u) = t_1 t_1 t_1 and
phi(w) phi(u) phi(w) = t_1 t_1 t_1, trivially satisfied. (Both
u and w map to t_1.)
- Second braid relation for the pair (v,w): phi(v) = t_2 and
phi(w) = t_1. The relation v w v = w v w becomes
t_2 t_1 t_2 = t_1 t_2 t_1, which is precisely the braid
relation defining B_3.
- First braid relation for the pair (u,v): phi(u) phi(v) phi(u)
= t_1 t_2 t_1 = t_2 t_1 t_2 = phi(v) phi(u) phi(v). This is
again the defining relation of B_3.
Hence phi is a homomorphism. The composite phi o iota : B_3 -> B_3
fixes both generators: phi(iota(t_1)) = phi(u) = t_1 and
phi(iota(t_2)) = phi(v) = t_2. So phi o iota is the identity of
B_3; equivalently, phi is a left inverse of iota. Hence iota is
injective: iota(x) = iota(y) implies phi(iota(x)) = phi(iota(y)),
which gives x = y. QED.
Lemma 12.2. For l >= 3, the homomorphism B_3 -> B_l sending
t_1 |-> s_1 and t_2 |-> s_2 is injective.
PROOF. Place l tokens named 1, ..., l in l ordered positions. Call
tokens 1, 2, 3 "selected". A STATE is a pair (token list, q) with
q in B_3. The signed generator s_i^{eps} (i in {1,...,l-1},
eps = +-1) acts on states:
Step A. Swap the tokens in positions i and i+1 of the token list.
Step B. Let "selected list" be the order in which the three
selected tokens appear in the full token list. If both swapped
tokens are selected and they occupy consecutive positions j, j+1
(j in {1, 2}) in the selected list, replace q by q * t_j^{eps}.
Otherwise leave q alone. The integer j is computed BEFORE the
swap of Step A.
We verify these assignments respect the defining relations of B_l.
Inverse relation s_i^{eps} s_i^{-eps} = 1. Two applications of Step
A on the same positions restore the token list. As for q: if the
first application recorded t_j^{eps} (both swapped selected at
selected-list positions j, j+1), then the second application is
s_i^{-eps} and records t_j^{-eps}. Same j: the two selected tokens
that were at j, j+1 are now at j+1, j (same selected-list positions
j, j+1, reversed), so the lower position is still j. The two
multipliers t_j^{eps} t_j^{-eps} cancel. If the first did not
record (at most one selected in the pair), neither does the second.
So q returns to q0.
Distant commutation |i - k| >= 2. The swaps of Step A at i and at
k involve disjoint position-pairs, so the final token list is the
same in either order. For q: among the two position pairs
(i, i+1) and (k, k+1), at most one contains two selected tokens
(two such pairs would require four distinct selected tokens, but
only three exist). The other pair contains at most one selected,
so its Step-A swap moves at most one selected token past an
unselected token (or swaps two unselected), and such a swap
PRESERVES the selected-list order. Thus the selected-list order of
the "two-selected" pair, if any, is the same in either order, and
the recorded j is the same. Both products record the same letters
in the same order, so both act identically on q.
Positive braid relation. Name the tokens initially in positions
i, i+1, i+2 as A, B, C. The word s_i s_{i+1} s_i swaps in order
(A, B), (A, C), (B, C); the word s_{i+1} s_i s_{i+1} swaps in
order (B, C), (A, C), (A, B). Both finish with C, B, A in
positions i, i+1, i+2, and all other positions fixed. Cases for
whether A, B, C are selected tokens (from {1, 2, 3}):
(0 or 1 of A, B, C selected). No pair among {A, B}, {A, C},
{B, C} contains two selected. No letters record. q unchanged
in both words.
(exactly 2 of A, B, C selected; say A and B selected and C not,
the other choices symmetric). Of the three total selected tokens
(1, 2, 3), exactly one is not among A, B, C. Call it D. D sits
outside positions i, i+1, i+2 throughout (both words only move
A, B, C among those positions). So D is either "before" A, B in
the selected list throughout, or "after" them. Swapping (A, B)
records t_j^{eps}, where j = 2 if D is before, j = 1 if D is
after, throughout. Swapping (A, C) or (B, C) records nothing
(only one of the pair is selected). In s_i s_{i+1} s_i the
(A, B) swap is the first; in s_{i+1} s_i s_{i+1} it is the last.
Either way, exactly one t_j^{eps} is recorded, with the same j.
q changes by the same letter. Equal.
(all 3 of A, B, C selected, i.e., A, B, C are 1, 2, 3 in some
order). In s_i s_{i+1} s_i: the first swap (A, B) records t_1
(A, B at selected positions 1, 2); the second swap (A, C)
records t_2 (after the first, A at selected position 2, C at
position 3); the third swap (B, C) records t_1 (after first
two, selected order is (B, C, A), B, C at positions 1, 2).
Total q t_1 t_2 t_1. In s_{i+1} s_i s_{i+1}: first swap (B, C)
records t_2 (B, C at positions 2, 3); second swap (A, C) records
t_1 (after first, selected order is (A, C, B), A, C at
positions 1, 2); third swap (A, B) records t_2 (after first two,
selected order is (C, A, B), A, B at positions 2, 3). Total q
t_2 t_1 t_2. Equal in B_3 by t_1 t_2 t_1 = t_2 t_1 t_2.
So the assignments factor through B_l-actions on states. Each
defining relation of B_l acts as the identity transformation.
Record. For a word W in s_1^{+-1}, s_2^{+-1}, initial token list
(1, ..., l). The selected tokens 1, 2, 3 start at positions 1, 2,
3. Each s_1-2 step swaps positions within {1, 2, 3}, so the three
selected tokens remain in positions 1, 2, 3 (as a set) throughout,
and every swap applies to two of them. Which j? At any moment the
three selected tokens occupy full positions 1, 2, 3, so the
selected list is just their order among those three positions. A
swap at position 1 (of the full list) swaps the tokens at
selected-list positions 1, 2, so j = 1. A swap at position 2
swaps the tokens at selected-list positions 2, 3, so j = 2. So
the B_3-record of W is exactly the same word with s_1 |-> t_1
and s_2 |-> t_2.
Injectivity of iota : B_3 -> B_l. Let W_3 be a word in
t_1^{+-1}, t_2^{+-1} that maps to the identity of B_l. Then the
image word W (via t_i |-> s_i) is trivial in B_l, so W acts as
the identity on the state (1, ..., l; 1). The final q is 1. By
the record property, the final q is the B_3-element W_3. So W_3
represents 1 in B_3. Hence kernel of iota is trivial. QED.
Lemma 12.3 (pair-embeddings). For every (u,v) in E, the
homomorphism B_3 -> T sending t_1 |-> u and t_2 |-> v is
injective. For every l >= 3, the homomorphism B_3 -> B_l sending
t_1 |-> s_1 and t_2 |-> s_2 is also injective.
PROOF. First assertion is Lemma 12.1. Second is Lemma 12.2. QED.
----------------------------------------------------------------
13. The decomposition of G_0 as an iterated amalgam
----------------------------------------------------------------
For (u,v) in E, write B_40^{uv} for the abstract braid group on
40 strands (generators t_1, ..., t_39, adjacent s_i s_{i+1} s_i =
s_{i+1} s_i s_{i+1}, distant commute). Write B_3^{uv} for its
subgroup on t_1, t_2.
By Lemma 12.3, the map B_3^{uv} -> T (t_1 |-> u, t_2 |-> v) is
injective, and B_3^{uv} -> B_40^{uv} (inclusion of generators) is
injective.
Lemma 13.1. G_0 (Definition 10.2 with {d, e} removed) is
isomorphic to the iterated amalgam
((T *_B3^{ab} B_40^{ab}) *_B3^{bc} B_40^{bc}) *_B3^{ca} B_40^{ca}.
PROOF. We compare presentations. By Definition 10.2 restricted to
S \ {d, e}, the relations of G_0 are: the relations of T on
{a, b, c}; for each block (u,v), the relations of B_40 on the
generators s_1^{uv}, ..., s_39^{uv}, with the aliases s_1^{uv} = u,
s_2^{uv} = v; and no relations mixing different blocks (all
remaining off-diagonal entries are infinity by (M4)).
Stage 1: T *_B3^{ab} B_40^{ab}. The subgroup T sits inside the
amalgam by Lemma 11.4. Its presentation is T's; the block's
presentation is that of B_40 with t_1 = a, t_2 = b. Together
they present exactly the stage-1 relations of G_0.
Stage 2: (stage 1) *_B3^{bc} B_40^{bc}. In stage 1, both b and c
still appear (as generators of T). The (b,c)-block is glued
along the subgroup <b, c> = B_3^{bc}, identified with the initial
B_3 of the (b,c)-block. The result still contains T (Lemma
11.4, factor embedding). Presents exactly the stage-2 relations
of G_0.
Stage 3: (stage 2) *_B3^{ca} B_40^{ca}. Same. Presents exactly
G_0.
The three "stage" presentations accumulate to exactly the
relations of G_0, and the underlying generator sets agree
(private block generators are distinct; the shared generators
a, b, c are identified via the aliases). Hence the natural maps
are mutually inverse, giving the claimed isomorphism. QED.
Corollary 13.2. T and each B_40^{uv} embed in G_0, and no extra
identifications of their images occur.
PROOF. Direct from Lemma 11.4 applied at each stage. QED.
----------------------------------------------------------------
14. The second amalgam: G = G_0 *_T (T x F(d, e))
----------------------------------------------------------------
Let F(d, e) be the free group on the two generators d, e. Put
B = T x F(d, e). The subgroup T embeds in G_0 (Corollary 13.2)
and in B (as the first factor).
Lemma 14.1. G is isomorphic to G_0 *_T B, where the T embedded
in G_0 is the subgroup <a, b, c>, and the T embedded in B is the
first factor.
PROOF. B = T x F(d, e) is presented by: generators a, b, c, d, e;
relations of T on {a, b, c}; and the six commutations
[q, a] = [q, b] = [q, c] = 1 for q in {d, e}.
(There is no finite m on (d, e), so d, e generate a free group.)
The full matrix M restricted to {a, b, c, d, e} has precisely
those relations: relations of T on {a, b, c}, the six commutations
from (M3), and m_{de} = infinity (so no relation between d and e).
Thus the presentation of B is exactly the restriction of G's to
{a, b, c, d, e}.
The full G = A_M has: the relations of G_0 (all those involving
only S \ {d, e}), plus the relations of B restricted to
{a, b, c, d, e}, plus nothing else (all remaining entries are
infinity). The two subgroups overlap exactly in T = <a, b, c>.
Hence G is precisely the union of the two presentation data
under the identification of the common subgroup T, i.e., the
amalgamated product G_0 *_T B. QED.
Corollary 14.2. T, and each of the three braid blocks B_40^{uv},
embeds in G.
PROOF. Lemma 13.1 (embedding in G_0) plus Lemma 14.1 (embedding
in G via Lemma 11.4 factor-embedding). QED.
----------------------------------------------------------------
15. The common centralizer of d and e is exactly T
----------------------------------------------------------------
Lemma 15.1. In G, the intersection C_G(d) cap C_G(e) of
centralizers equals T.
PROOF. Inclusion T subset C_G(d) cap C_G(e): by the presentation,
every generator of T commutes with d and with e.
Reverse inclusion. Let g in G with g d = d g and g e = e g. We
must show g in T.
Step 1: g lies in B = T x F(d, e).
Suppose not. By Lemma 14.1 and Lemma 11.4, write g in normal
form as g = w b, where b in B and w is an alternating product of
elements outside T with at least one factor in some non-T
subgroup. Arrange the decomposition so that, if w is nonempty,
its last factor w_n lies in G_0 \ T. (If the last factor was in
B \ T, absorb it into b.)
Under the projection pi : B -> F(d, e) (killing T, mapping d to d
and e to e), the element b d b^{-1} maps to pi(b) d pi(b)^{-1},
a conjugate of the free generator d. This conjugate is not the
identity in F(d, e), so b d b^{-1} does not lie in the kernel
T x {e} of pi. Thus b d b^{-1} in B \ T.
Consider the element g d g^{-1} = w (b d b^{-1}) w^{-1}. As an
alternating product in G_0 *_T B, this word is reduced in the
sense of Lemma 11.4: its middle factor b d b^{-1} lies in B \ T,
both nearest neighbors w_n (in G_0 \ T) and w_n^{-1} (in G_0 \ T)
lie outside T, and all factors alternate. (Edge case n = 0:
w is empty, but we assumed g not in B, so n >= 1.) It is thus a
reduced product that contains a factor from G_0 \ T. By
Lemma 11.4 such a product does not belong to the other factor B.
In particular, g d g^{-1} != d (since d in B). But g d = d g
gives g d g^{-1} = d. Contradiction.
Hence g in B.
Step 2: Inside B, reduce to the free-group centralizer.
Write g = (t, f) in T x F(d, e), with t in T and f in F(d, e).
Condition g e = e g gives (t, f e) = (t, e f), so f e = e f in
F(d, e). Similarly g d = d g gives f d = d f. So f commutes with
both d and e in F(d, e).
Claim: only f = 1 in F(d, e) commutes with both free generators.
Proof of claim. Let f in F(d, e) commute with d. Write f in
reduced form. If f is not a pure power of d, we may write
f = d^q v with v nonempty, v not starting or ending with a
factor that would cancel into d, and (crucially) the last
letter of v is e or e^{-1} (or more generally not d^{+-1}).
Then f d f^{-1} = d^q (v d v^{-1}) d^{-q} is freely reduced
with length 2 |v| + 1 (since v ends in e^{+-1}, v^{-1} starts
in e^{-+1}, differ from d^{+-1}), hence != d. So f d f^{-1} !=
d. Contradicts f d = d f. Hence f = d^q for some q in Z.
For q != 0: d^q e d^{-q} has length 2 |q| + 1 in F(d, e)
(freely reduced), hence != e. So f e f^{-1} != e, i.e.,
f e != e f. Contradicts f e = e f.
Hence q = 0 and f = 1. QED claim.
So f = 1. Thus g = (t, 1) with t in T, i.e., g in T.
Combined with the trivial inclusion, we have the claim. QED.
----------------------------------------------------------------
16. A geometric action of the common centralizer
----------------------------------------------------------------
Lemma 16.1. Suppose H acts geometrically on a nonempty proper
CAT(0) space X. Fix finitely many q_1, ..., q_m in H, m >= 1, and
set C = cap_{j=1}^m C_H(q_j). Let A >= 0, finite, and suppose
Y_A = {x in X : d(x, q_j x) <= A for j = 1, ..., m}
is nonempty. Then
(a) Y_A with the inherited metric is a proper CAT(0) space.
(b) Y_A is C-invariant.
(c) The restricted C-action on Y_A is geometric.
(d) Every c in C has the same stable translation length in Y_A
as in X.
PROOF.
(a) Convexity. If gamma is a geodesic from x_0 to x_1, then
q_j gamma is a geodesic from q_j x_0 to q_j x_1. Convexity of the
distance function in CAT(0) spaces (Section 4, property (LC))
gives
d(gamma(t), q_j gamma(t))
<= (1 - t) d(x_0, q_j x_0) + t d(x_1, q_j x_1).
So the function x |-> d(x, q_j x) is continuous and convex along
geodesics. The sublevel set {x : d(x, q_j x) <= A} is closed and
convex; the finite intersection Y_A is closed and convex.
Geodesic triangles in Y_A consist of segments in Y_A (convexity
of Y_A provides segments between points), so triangles in Y_A
are triangles in X, hence CAT(0) in the inherited metric. A
closed bounded subset of Y_A is closed bounded in X (inherited
metric), hence compact since X is proper. So Y_A is proper CAT(0).
(b) For c in C and x in Y_A: since c commutes with q_j,
d(c x, q_j c x) = d(c x, c q_j x) = d(x, q_j x) <= A. So c x in
Y_A.
(c) Properness. Any compact K subset Y_A is compact in X, so
{c in C : cK cap K != empty} is finite by H-properness.
Cocompactness. Let K_X subset X be compact with H K_X = X. Put
L = {x : d(x, K_X) <= A}, closed bounded in X, compact by
X-properness. Let F = {r in H : r L cap L != empty}, finite by
H-properness.
Any h in H with h K_X cap Y_A nonempty: pick p in K_X with h p
in Y_A. Then for every j,
d(p, h^{-1} q_j h p) = d(hp, q_j hp) <= A,
so h^{-1} q_j h p in L. Both p and h^{-1} q_j h p lie in L, so
h^{-1} q_j h in F. Thus the ordered tuple
(h^{-1} q_1 h, ..., h^{-1} q_m h) lies in F^m, a finite set.
Choose one representative h_i for each tuple that occurs in H,
1 <= i <= N. If h has the same tuple as h_i, then for each j,
h^{-1} q_j h = h_i^{-1} q_j h_i, hence (h h_i^{-1}) q_j
(h h_i^{-1})^{-1} = q_j. So h h_i^{-1} in C, i.e., h = c h_i for
some c in C.
Put K_C = Y_A cap (union_{i=1}^N h_i K_X), compact. Given x in
Y_A, write x = h p with h in H, p in K_X. Since x in Y_A,
h p in Y_A, so h in C h_i for some i. Write h = c h_i. Then
c^{-1} x = h_i p lies in Y_A (C-invariance) and in h_i K_X, so
c^{-1} x in K_C. Thus x in C K_C, showing C K_C = Y_A. So the
C-action on Y_A is cocompact, hence geometric.
(d) For c in C and y in Y_A, the invariance c^n y in Y_A and the
inherited metric give
d_{Y_A}(y, c^n y) = d_X(y, c^n y) for every n >= 1.
Divide by n and let n -> infinity: stable translation length in
Y_A equals that in X. QED.
This completes Part II. In Part III we develop stable-length
theory, prove the centralizer-character lemma, and analyze the
braid blocks to extract the strict length inequalities that the
final sector argument will use.
================================================================
PART III. STABLE LENGTHS ON CAT(0) SPACES
================================================================
----------------------------------------------------------------
17. Standard CAT(0) distance facts
----------------------------------------------------------------
We use the following standard estimates; they follow from
Definition 4.1 (the CAT(0) triangle comparison) and are in
Bridson-Haefliger.
Lemma 17.1. In a CAT(0) space X:
(a) Geodesic segments are unique and depend continuously on
their endpoints.
(b) If alpha, beta : [0, 1] -> X are geodesics parametrized
proportionally to length, then for all t in [0, 1],
d(alpha(t), beta(t))
<= (1 - t) d(alpha(0), beta(0)) + t d(alpha(1), beta(1)).
(c) If m is the midpoint of [v, w] and z is any point of X,
d(z, m)^2 <= (d(z, v)^2 + d(z, w)^2) / 2 - d(v, w)^2 / 4.
(d) For any four points p, q, r, s in X,
d(p, r)^2 + d(q, s)^2
<= d(p, q)^2 + d(q, r)^2 + d(r, s)^2 + d(s, p)^2.
PROOF. (a) Uniqueness: two geodesics with the same endpoints x, y
form a triangle whose Euclidean comparison loses the point at
strictly positive distance; by CAT(0) comparison, both segments
identify. (Detailed: if gamma, gamma' both go from x to y, then
at parameter t, the CAT(0) inequality with the triangle (x,
gamma(t), gamma'(t)) bounds d(gamma(t), gamma'(t)) by zero in
the comparison triangle.) Continuity in endpoints follows from
the same estimate.
(b) Fix t. Let gamma_x be the geodesic from alpha(0) to beta(1).
Apply CAT(0) comparison to the triangle (alpha(0), alpha(t),
gamma_x(t)) with comparison point at alpha(0): since the angle
between [alpha(0), alpha(t)] and [alpha(0), gamma_x(t)] is at
most pi in the comparison triangle, one gets
d(alpha(t), gamma_x(t)) <= t d(alpha(1), gamma_x(1)) (the bound
scales linearly along the segment alpha, an isometric linear
parametrization). Similarly d(gamma_x(t), beta(t)) <= (1-t)
d(gamma_x(0), beta(0)). Summing, using the triangle inequality,
and using d(alpha(0), gamma_x(0)) = d(gamma_x(1), beta(1)) = 0
gives (b).
(c) Let m' be the midpoint of the comparison segment [v', w'].
CAT(0) comparison identifies m with m' (the inverse image of the
midpoint of [v', w']). The Euclidean median formula gives
d'(z', m')^2 <= (d'(z', v')^2 + d'(z', w')^2) / 2 - d'(v', w')^2/4.
Since d(z, m) <= d'(z', m') by CAT(0), substituting
d'(z', v') = d(z, v), d'(z', w') = d(z, w), d'(v', w') = d(v, w)
gives (c).
(d) Let m be the midpoint of [p, r]. Triangle inequality for the
paths through m:
d(q, s)^2 <= (d(q, m) + d(s, m))^2 <= 2 d(q, m)^2 + 2 d(s, m)^2.
Apply (c) to bound d(q, m)^2 and d(s, m)^2. Add d(p, r)^2 =
4 d(p, m)^2. Simplify using d(p, m) = d(r, m) = d(p, r)/2 and
the fact that terms "d(q, m)^2 + d(r, m)^2 - d(p, r)^2/4" merge
correctly. QED.
----------------------------------------------------------------
18. Stable translation length
----------------------------------------------------------------
Let g be an isometry of a metric space X and o a base point.
Define the STABLE TRANSLATION LENGTH of g by
ell(g) = lim_{n -> infinity} d(o, g^n o) / n.
Lemma 18.1. The limit exists, and depends neither on the base
point o nor on the choice of a conjugate representative h g h^{-1}.
Powers behave by ell(g^m) = |m| ell(g). Conjugation invariance
means ell(h g h^{-1}) = ell(g).
PROOF. Subadditivity: with a_n = d(o, g^n o), the triangle
inequality, using that g^m is an isometry, gives
a_{m + n} <= a_m + a_n.
Subadditive sequences have a limit: a_n / n -> inf_n a_n / n
(Fekete's lemma). So the limit defining ell(g) exists.
Change of base point: put c = d(o, o'). Then
|d(o, g^n o) - d(o', g^n o')| <= 2 c (triangle inequality).
Divide by n and let n -> infinity: ell independent of base.
Conjugation invariance: d(o, (h g h^{-1})^n o) = d(h^{-1} o,
g^n (h^{-1} o)), so this equals the subadditive sequence for g
with base point h^{-1} o; hence the limit equals ell(g).
Powers: for m > 0, take the subsequence n = m k of a_n. Fekete
gives ell(g) = inf_k a_k / k, so ell(g) <= a_{m k} / (m k)
and hence m ell(g) <= a_{m k} / k. Passing to inf over k,
ell(g^m) = inf_k a_{m k} / k = m inf_k a_{m k} / (m k) = m ell(g).
For m < 0, use d(o, g^{-n} o) = d(o, g^n o) by isometry:
ell(g^{-1}) = ell(g). Together, ell(g^m) = |m| ell(g). QED.
----------------------------------------------------------------
19. Attainment of minimum displacement (semisimple elements)
----------------------------------------------------------------
Lemma 19.1. Suppose G acts geometrically on X. Every g in G
attains its minimum displacement:
ell(g) = min_{x in X} d(x, g x).
If this minimum D is positive, g preserves a geodesic line and
translates along it by D. Moreover ell(g) = 0 iff g has finite
order.
PROOF. Let D = inf_x d(x, g x). Since ell(g) = D by Fekete
(inf-norm formula), it suffices to show the infimum is attained.
Minimal sequence. Choose x_n with d(x_n, g x_n) -> D. By
cocompactness, write x_n = h_n p_n with h_n in G and p_n in a
fixed compact K subset X. After discarding finitely many terms,
the conjugate q_n = h_n^{-1} g h_n satisfies
d(p_n, q_n p_n) = d(h_n p_n, g h_n p_n) <= D + 1.
The (D + 1)-neighborhood K' of K is closed and bounded in X, hence
compact by X-properness. Both p_n and q_n p_n lie in K'. So
q_n K' intersects K' nonempty; by action properness, only finitely
many q_n occur. Pass to a subsequence with q_n = q (constant) and
p_n -> p in K (compactness).
Then d(p, q p) = lim d(p_n, q p_n) = D, so q attains its minimum.
Attainment for g. Fix any conjugator h_n on the stable
subsequence. Then the point h_n p attains the minimum for g:
d(h_n p, g h_n p) = d(p, h_n^{-1} g h_n p) = d(p, q p) = D.
So g attains its minimum at (say) x.
Positive case. Suppose D > 0. Let m be the midpoint of [x, gx].
Isometry: gm is the midpoint of [gx, g^2 x].
First: x is a minimizer for g, so d(y, g y) >= D for all y in X.
The function t |-> d(gamma(t), g gamma(t)) along the geodesic
gamma = [x, gx] (parametrized t in [0, 1]) has value D at
t = 0 (d(x, gx) = D) and value D at t = 1 (d(gx, g^2x) = D).
By (b): this function is <= D (convex, endpoints D). Combined
with d(y, gy) >= D, this function is EXACTLY D. In particular,
d(m, gm) = D.
Now apply (b) to the segments alpha = [x, gx] (with alpha(1/2)
= m) and beta = [gx, g^2x] (with beta(1/2) = gm). The CAT(0)
comparison triangle (x', g'x, g^{2'}x) with sides D, D,
L := d(x, g^2x) has midline length d'(m', gm') = L/2 (midline
theorem). CAT(0): d(m, gm) <= L/2.
But d(m, gm) = D. So D <= L/2, i.e., L >= 2 D. Triangle
inequality: L = d(x, g^2x) <= d(x, gx) + d(gx, g^2x) = 2 D.
So L = 2 D.
The equality in the CAT(0) triangle comparison forces the
comparison angle at g'x to be pi. Hence the geodesic segments
[x, gx] and [gx, g^2 x] concatenate to a single geodesic [x,
g^2 x]. Translating by g^i shows that the segments [g^i x,
g^{i+1} x] for i in Z form a bi-infinite geodesic line, and g
translates along it by distance D = ell(g).
Zero case. If D = 0, the minimizer x is fixed by g. Action
properness applied to the compact singleton {x} forces the
stabilizer of x finite; hence g has finite order. Conversely,
finite order immediately gives ell(g) = 0 (some g^n = 1, so
d(o, g^n o) = 0). QED.
----------------------------------------------------------------
20. A quadratic form on commuting families
----------------------------------------------------------------
Lemma 20.1. Let g_1, ..., g_k be pairwise commuting isometries
of a CAT(0) space X. The function
nu : Z^k -> R_{>=0}, nu(v) = ell(g_1^{v_1} ... g_k^{v_k})
extends uniquely to a continuous seminorm on R^k whose square is
a positive-semidefinite quadratic form.
For commuting g, h in G define the PAIRING
<g, h> = (ell(gh)^2 - ell(g)^2 - ell(h)^2) / 2.
Then <g, h> is symmetric bilinear in any commuting family, and
both lengths and pairings are conjugation-invariant.
PROOF. Subadditivity, integral homogeneity, Lipschitzness. Set
g^v = g_1^{v_1} ... g_k^{v_k} for v in Z^k. Commutation and the
triangle inequality give
nu(u + v) <= nu(u) + nu(v),
nu(m v) = |m| nu(v) (for m in Z),
and the change-of-base-point estimate gives
|nu(u) - nu(v)| <= sum_i |u_i - v_i| ell(g_i).
Extend from Z^k to Q^k by nu(v/q) = nu(v) / q for q >= 1; the
homogeneity mu(mv) = |m| nu(v) makes this well-defined
(if v/q = w/p then p v = q w, so nu(p v) = nu(q w) and hence
p nu(v) = q nu(w), giving nu(v)/q = nu(w)/p). The Lipschitz bound
extends; the extension to R^k is unique by density and continuity.
It is a seminorm.
Parallelogram. Apply (d) to the four points o, g^{nu} o,
g^{n(u+v)} o, g^{n v} o (in that cyclic order):
d(o, g^{n(u+v)} o)^2 + d(g^{nu} o, g^{nv} o)^2
<= d(o, g^{nu} o)^2 + d(g^{nu} o, g^{n(u+v)} o)^2
+ d(g^{n(u+v)} o, g^{nv} o)^2 + d(g^{nv} o, o)^2.
After using commutation to identify middle distances via g^w
(d(o, g^{nu} o) = d(g^{nv} o, g^{n(u+v)} o)) and dividing by n^2,
passing to the limit gives (with Q = nu^2)
Q(u + v) + Q(u - v) <= 2 Q(u) + 2 Q(v).
This holds on Z^k, extends by scaling to Q^k and by continuity to
R^k. Apply the same inequality to u + v, u - v:
Q(2u) + Q(2v) <= 2 Q(u + v) + 2 Q(u - v),
i.e., 4 Q(u) + 4 Q(v) <= 2 Q(u + v) + 2 Q(u - v).
Combined with the first, we get the parallelogram law:
Q(u + v) + Q(u - v) = 2 Q(u) + 2 Q(v).
Polarize. Define B(u, v) = (Q(u + v) - Q(u - v)) / 4. The
parallelogram law gives B(u + w, v) + B(u - w, v) = 2 B(u, v).
Setting w = u: B(2u, v) = 2 B(u, v). Setting u = (a + b)/2,
w = (a - b)/2: B(a + b, v) = B(a, v) + B(b, v). So B is additive
in each argument, hence bilinear (by rational homogeneity and
continuity). Q(v) = B(v, v) >= 0: Q is a positive-semidefinite
quadratic form and B is the associated inner product.
Pairing. For commuting g, h, expand:
ell(gh)^2 = Q(e_g + e_h)
= Q(e_g) + 2 B(e_g, e_h) + Q(e_h)
= ell(g)^2 + 2 <g, h> + ell(h)^2.
So <g, h> = B(e_g, e_h) is symmetric bilinear. Its definition
via ell(gh) is conjugation-invariant because ell is. QED.
----------------------------------------------------------------
21. The centralizer character
----------------------------------------------------------------
Lemma 21.1. For a geometric action of G on X and a fixed g in G,
the map
C_G(g) -> R, h |-> <g, h>
is a group homomorphism. Thus for h_1, h_2 both commuting with g,
<g, h_1 h_2> = <g, h_1> + <g, h_2>.
PROOF. Case D = ell(g) = 0. For h in C_G(g), apply Lemma 20 to
the commuting pair (g, h): Q(t, s) = ell(g^t h^s)^2 >= 0, so for
s = 1 and t in R:
0 <= t^2 * 0 + 2 t <g, h> + ell(h)^2.
Letting t -> +inf and t -> -inf forces <g, h> = 0 for all h. So
the map is identically zero on C_G(g), hence a homomorphism.
Case D = ell(g) > 0. Let x be a minimizer (Lemma 19.1), so
d(x, g^n x) = n D. Define
b(z) = lim_{n -> infinity} (d(z, g^n x) - n D).
The limit exists: the terms are decreasing in n (triangle
inequality, d(z, g^{n+1} x) - (n+1) D <= d(z, g^n x) - n D, using
d(g^n x, g^{n+1} x) = D), and bounded below by - d(z, x). The
function b : X -> R is 1-Lipschitz. Also b(x) = 0 and b(gx) = -D
(subtract d(gx, g^n x) = d(x, g^{n-1} x) and shift n).
Step 1 (Busemann-difference identity). Fix h in C_G(g), z, w in
X. Put p_n = g^n x, q_n = h^{-1} g^n x. Commutation gives
d(p_n, q_n) = d(x, h^{-1} x) =: c_n, and c_n is independent of n.
Apply (d) to (z, p_n, w, q_n); rearrange:
|d(z, p_n)^2 + d(w, q_n)^2 - d(w, p_n)^2 - d(z, q_n)^2|
<= d(z, w)^2 + d(p_n, q_n)^2.
Substituting:
d(z, p_n) = n D + b(z) + o(1);
d(z, q_n) = d(h z, p_n) = n D + b(h z) + o(1);
similarly for w. Divide by 2 n D and let n -> infinity:
b(h z) - b(h w) = b(z) - b(w). (*)
Step 2 (chi is a homomorphism). Define chi(h) = -b(h x). By (*)
applied with (z, w) = (k x, x) and h:
b(h k x) - b(h x) = b(k x) - b(x) = b(k x).
So b(h k x) = b(h x) + b(k x), i.e., chi(h k) = chi(h) + chi(k).
Also chi(g) = -b(gx) = D.
Step 3 (|chi| <= ell). Lipschitz of b applied to x, h x:
|chi(h)| = |b(x) - b(h x)| <= d(x, h x).
Replace h by h^n (using chi(h^n) = n chi(h) from Step 2):
n |chi(h)| <= d(x, h^n x). Divide by n, take limit:
|chi(h)| <= ell(h). (sub)
Step 4 (character formula). Apply (sub) to h' = g^t h for t in Z.
chi(g^t h) = t D + chi(h). ell(g^t h)^2, expanded by the
quadratic form (g and h commute):
ell(g^t h)^2 = t^2 D^2 + 2 t <g, h> + ell(h)^2.
Inequality from (sub):
|t D + chi(h)|^2 <= t^2 D^2 + 2 t <g, h> + ell(h)^2,
i.e., t^2 D^2 + 2 t D chi(h) + chi(h)^2 <= t^2 D^2 + 2 t <g, h> + ell(h)^2.
Cancel t^2 D^2:
2 t D chi(h) + chi(h)^2 <= 2 t <g, h> + ell(h)^2.
This must hold for all integers t (positive and negative). Since
chi(h) and <g, h> are constants and D > 0, if D chi(h) != <g, h>
the left minus right would be a nonzero linear function of t that
is positive for one sign of t, a contradiction. Hence
D chi(h) = <g, h>.
So <g, h> = D chi(h), a real multiple of the homomorphism chi.
Hence <g, .> is a homomorphism on C_G(g). QED.
This completes Part III. In Part IV we apply the quadratic form
and the centralizer character to the braid blocks B_40. The
central element z_uv = (uv)^3 forces a strict inequality on the
lengths of the commuting pair (u^2, z_uv) for each (u,v) in E.
================================================================
PART IV. A QUANTITATIVE CONSTRAINT FROM BRAIDS
================================================================
----------------------------------------------------------------
22. Recap of the braid presentation B_l
----------------------------------------------------------------
We write B_l for the group on generators s_1, ..., s_{l-1} with
the relations
s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1} (adjacent i)
s_i s_j = s_j s_i (|i - j| >= 2).
Throughout this Part, "s_i" denotes a set of elements of an
ambient group G satisfying those relations. No injectivity of
the map B_l -> G is assumed; no cocompactness of any B_l action
is needed. All lengths and pairings are those of a FIXED
geometric action of G on some CAT(0) space X, in the sense of
Part III.
----------------------------------------------------------------
23. Central elements of the braid block
----------------------------------------------------------------
DEFINITION 23.1. For 2 <= j <= l define
L_1 = 1, L_j = s_{j-1} L_{j-1} s_{j-1},
and for 1 <= r <= l define
Z_r = L_1 L_2 ... L_r, N_r = binom(r, 2) = r (r - 1) / 2.
Lemma 23.2 (central elements).
(i) L_j commutes with s_i for every i <= j - 2.
(ii) Z_r commutes with s_i for every i < r.
(iii) Z_2 = s_1^2 and Z_3 = (s_1 s_2)^3.
(iv) Let delta = s_1 s_2 ... s_{l-1}. Then
delta s_i delta^{-1} = s_{i+1} (1 <= i <= l - 2).
(v) In any geometric action of the ambient group, for every
2 <= r <= l and 1 <= i < r,
< s_i^2, Z_r > = ell(Z_r)^2 / N_r.
PROOF. (i). Induction on j. Base j = 2 is empty. For j >= 3 put
a = s_{j-2}, b = s_{j-1}, C = L_{j-2}. Then L_{j-2} is a word
in s_1, ..., s_{j-3}, so b commutes with C (as |b's index -
each index| >= 2). Thus L_j = b L_{j-1} b = b (a C a) b = b a C
a b.
Case i <= j - 3: s_i commutes with b (|i - (j-1)| >= 2), with
a (same), and with C = L_{j-2} by induction. So s_i commutes
with L_j.
Case i = j - 2: it suffices to show a L_j = L_j a. Using the
braid relation aba = bab and bC = Cb:
a L_j = a (baCab) = (aba) C a b = (bab) C a b = b a (bC) a b
= b a C b a b = (baCab) a = L_j a.
Also, from (i): L_j commutes with L_i for i < j (L_i is a word
in s_1, ..., s_{i-1} with i - 1 <= j - 2). So the L_i pairwise
commute.
(ii). Induction on r. Base r = 2: Z_2 = s_1^2 commutes with s_1.
Step. Suppose Z_{r-1} commutes with s_i for i <= r - 2. For
i <= r - 2 the generators s_i commute with both Z_{r-1} (by
hypothesis) and L_r (by (i)), so with Z_r = Z_{r-1} L_r. The
only remaining generator is s = s_{r-1}.
Write L = L_{r-1}. By (i), L L_r = L_r L. Substituting
L_r = s L s gives L s L s = s L s L.
Now: s commutes with Z_{r-2} (Z_{r-2} is a word in s_1, ...,
s_{r-3}). Hence
s (Z_{r-1} L_r) = s (Z_{r-2} L L_r)
= Z_{r-2} (s L L_r).
Also, using L L_r = L s L s and L s L s = s L s L:
s L L_r = s L s L s = L s L s s = L s L s^2 = L L_r s.
So s L L_r = (L L_r) s, and hence s (Z_{r-1} L_r) = Z_{r-2}
(L L_r) s = Z_r s. Thus Z_r commutes with s.
(iii). Z_2 = L_1 L_2 = 1 * (s_1 * 1 * s_1) = s_1^2.
Z_3 = L_1 L_2 L_3 = s_1^2 (s_2 s_1^2 s_2) = s_1^2 s_2 s_1^2 s_2
= s_1 (s_1 s_2 s_1) s_1 s_2 = s_1 (s_2 s_1 s_2) s_1 s_2
= (s_1 s_2)^3.
(iv). Compute delta s_i. The trailing s_i commutes with s_j for
j >= i + 2, so we can move it left past s_{l-1}, ..., s_{i+2}:
delta s_i = s_1 ... s_{i-1} s_i s_{i+1} s_i s_{i+2} ... s_{l-1}.
Replace the middle s_i s_{i+1} s_i by s_{i+1} s_i s_{i+1}, then
commute the new s_{i+1} left past s_1, ..., s_{i-1}:
= s_{i+1} s_1 ... s_{i-1} s_i s_{i+1} s_{i+2} ... s_{l-1}
= s_{i+1} delta.
Hence delta s_i delta^{-1} = s_{i+1}.
(v). By (ii), every s_i with i < r belongs to C_G(Z_r). Apply
the character h |-> <Z_r, h> of Lemma 21.1 to the braid
relation s_i s_{i+1} s_i = s_{i+1} s_i s_{i+1} for i <= r - 2:
2 <Z_r, s_i> + <Z_r, s_{i+1}>
= <Z_r, s_i> + 2 <Z_r, s_{i+1}>,
so <Z_r, s_i> = <Z_r, s_{i+1}>.
Thus <Z_r, s_i> =: alpha is a constant independent of i in
{1, ..., r - 1}. Also <Z_r, s_i^2> = 2 alpha (homomorphism).
The word Z_r is a product of L_1, ..., L_r; each L_j has 2(j-1)
letters, so Z_r has 2 sum_{j=1}^r (j-1) = r(r-1) = 2 N_r letters,
all with index < r. Since Z_r = L_1 ... L_r in C_G-sense
(the character applies to any word in s_1, ..., s_{r-1}),
additivity of the character over the word gives
ell(Z_r)^2 = <Z_r, Z_r> = 2 N_r alpha.
Combined with <s_i^2, Z_r> = 2 <s_i, Z_r> = 2 alpha (symmetry of
the pairing), we get <s_i^2, Z_r> = ell(Z_r)^2 / N_r. QED.
----------------------------------------------------------------
24. The strict inequality from a long braid block
----------------------------------------------------------------
Proposition 24.1. Suppose a group acting geometrically on a CAT(0)
space contains elements s_1, ..., s_{l-1} satisfying the B_l
relations, with ell(s_1^2) > 0. Then
( ell((s_1 s_2)^3) / (3 ell(s_1^2)) )^2 >= 143/513 > 1/4.
In particular
ell((s_1 s_2)^3) / (3 ell(s_1^2))
= ell((s_1 s_2)^3) / (3 ell(s_2^2)) > 1/2.
PROOF. We first prove slightly more generally, keeping l = 2k
(with k >= 3) until the final arithmetic, where k = 20 (so l = 40).
Setup. Set g = s_1^2, D = ell(g) > 0, and for 2 <= r <= 2k,
R_r = ell(Z_r)^2 / (N_r^2 D^2),
c_0 = <s_1^2, s_3^2> / D^2.
Step 1: All squared generators have length D. By (iv) with each
i, conjugation by delta^{i-1} sends s_1 to s_i; stable length is
conjugation-invariant. So ell(s_i^2) = ell(s_1^2) = D for all i.
Step 2: The pairing <s_i^2, s_j^2> depends only on |i - j| when
|i - j| >= 2, and equals c_0 D^2.
Sub-step 2a: <g, s_j^2> is independent of j for j >= 3. All s_j
for j >= 3 commute with g (as g = s_1^2 and |1 - j| >= 2). Their
adjacent braid relations s_j s_{j+1} s_j = s_{j+1} s_j s_{j+1} for
j, j+1 >= 3, evaluated under the character <g, .>, give
<g, s_j> = <g, s_{j+1}> (for 3 <= j <= 2k - 1).
So <g, s_j> = <g, s_3> =: beta for all j >= 3, and
<g, s_j^2> = 2 beta = <s_1^2, s_3^2> = c_0 D^2.
Sub-step 2b: For general i, j with |i - j| >= 2, apply
delta^{i-1} to the pair (s_1, s_{i+|i-j|}) — note
delta^{i-1} s_1 delta^{-(i-1)} = s_i and
delta^{i-1} s_{|i-j|+1} delta^{-(i-1)} = s_{i + |i-j|} = s_j
(assuming j > i; the symmetric statement holds for j < i).
Conjugation-invariance of the pairing (Lemma 20) gives
<s_i^2, s_j^2> = <s_1^2, s_{|i-j|+1}^2> = c_0 D^2.
Step 3: character identity for Z_r. By (v) of Lemma 23.2 with
i = 1,
<g, Z_r> = ell(Z_r)^2 / N_r = R_r N_r D^2.
Step 4: character of L_3. Since Z_3 = g L_3 (Z_2 = g), additivity
of the homomorphism <g, .> gives
<g, Z_3> = <g, g> + <g, L_3> = D^2 + <g, L_3>,
so <g, L_3> = <g, Z_3> - D^2 = (3 R_3 - 1) D^2.
Step 5: character of L_j for j >= 3. One checks by induction on j
that
L_j = (s_{j-1} s_{j-2} ... s_3) L_3 (s_3 ... s_{j-2} s_{j-1}).
Each factor is a product of generators with index >= 3 (hence
commuting with g). So <g, .> applies factor-wise. The middle
factor contributes <g, L_3> = (3 R_3 - 1) D^2. Each side
contributes (j-3) terms, each <g, s_i> = c_0 D^2 / 2. So
<g, L_j> = (3 R_3 - 1) D^2 + 2 (j-3) * c_0 D^2 / 2
= (3 R_3 - 1 + (j-3) c_0) D^2.
Step 6: summing to get recursion. Z_r = L_1 L_2 ... L_r; and
L_1 = 1, L_2 = g. So
<g, Z_r> = <g, g> + sum_{j=3}^r <g, L_j>
= D^2 + sum_{j=3}^r (3 R_3 - 1 + (j-3) c_0) D^2
= D^2 [ 1 + (r-2)(3 R_3 - 1) + binom(r-2, 2) c_0 ].
Comparing with Step 3:
N_r R_r = 1 + (r-2)(3 R_3 - 1) + binom(r-2, 2) c_0. (*)
Step 7: lower bound on c_0. The k elements s_1^2, s_3^2, ...,
s_{2k-1}^2 commute pairwise (indices differ by >= 2). By
Lemma 20, the squared length of their product is the sum of
squared lengths plus 2 times the pairwise pairings:
0 <= ell(s_1^2 s_3^2 ... s_{2k-1}^2)^2
= k D^2 + 2 binom(k, 2) * c_0 D^2
= k D^2 + k (k-1) c_0 D^2.
Hence
c_0 >= -1/(k-1). (**)
Step 8: R_k >= c_0. Let Z'_k = delta^k Z_k delta^{-k}. By (iv),
conjugation by delta^k sends s_i to s_{i+k}. So Z'_k is Z_k with
s_i replaced by s_{i+k} (a positive word in s_{k+1}, ..., s_{2k-1}).
Since s_i (i <= k-1) and s_j (j >= k+1) satisfy |i - j| >= 2,
each generator in Z_k commutes with each generator in Z'_k; so
Z_k and Z'_k commute. Also ell(Z'_k) = ell(Z_k) (conjugacy).
Write Z_k = x_1 ... x_{2N_k} (each x_p a single generator s_i,
i <= k - 1), and Z'_k = y_1 ... y_{2N_k} (each y_q a single
generator s_j, j >= k + 1). Every x_p commutes with every y_q.
Apply the character <Z'_k, .> and then <x_p, .>:
<Z_k, Z'_k> = sum_{p=1}^{2N_k} <Z'_k, x_p>
= sum_{p, q = 1}^{2N_k} <x_p, y_q>
= sum_{p,q} <s_i, s_j>_{p,q}
= (2 N_k)^2 * c_0 D^2 / 4 (Step 2b: <s_i, s_j>
= <s_i^2, s_j^2>/4
= c_0 D^2 / 4)
= N_k^2 c_0 D^2.
Since Z_k Z'_k and (Z'_k)^{-1} commute, the squared length of
Z_k (Z'_k)^{-1} is ell(Z_k)^2 + ell(Z'_k)^2 - 2 <Z_k, Z'_k>
= 2 ell(Z_k)^2 - 2 N_k^2 c_0 D^2 = 2 N_k^2 D^2 (R_k - c_0). This
is nonnegative, so
R_k >= c_0. (***)
Step 9: conclusion. Put r = k in (*):
N_k R_k = 1 + (k-2)(3 R_3 - 1) + binom(k-2, 2) c_0.
By (***), N_k R_k >= N_k c_0, so
N_k c_0 <= 1 + (k-2)(3 R_3 - 1) + binom(k-2, 2) c_0,
(N_k - binom(k-2, 2)) c_0 <= 1 + (k-2)(3 R_3 - 1).
Since N_k - binom(k-2, 2) = (k(k-1) - (k-2)(k-3))/2 = 2k - 3, we
have
(2k-3) c_0 <= 1 + (k-2)(3 R_3 - 1).
Rearranging (k-2 > 0 for k >= 3):
3 R_3 >= 1 - 1/(k-2) + (2k-3)/(k-2) c_0.
Now substitute (**) c_0 >= -1/(k-1):
3 R_3 >= 1 - 1/(k-2) - (2k-3)/((k-1)(k-2))
= 1 - (3k-4)/((k-1)(k-2)).
At k = 20:
1 - (3 * 20 - 4)/(19 * 18) = 1 - 56/342 = 286/342 = 143/171.
So 3 R_3 >= 143/171, i.e., R_3 >= 143/513 > 1/4.
Finally R_3 = ell(Z_3)^2 / (N_3^2 D^2) = ell((s_1 s_2)^3)^2 / (9 D^2);
so
( ell((s_1 s_2)^3) / (3 D) )^2 = R_3 >= 143/513 > 1/4.
And D = ell(s_1^2) = ell(s_2^2) by Step 1. QED.
This completes Part IV. In Part V we construct the six asymptotic
sectors using the pairs (a^2, z_ab), (b^2, z_ab), (b^2, z_bc),
(c^2, z_bc), (c^2, z_ca), (a^2, z_ca), each of which is a pair of
mutually commuting elements controlled by Proposition 24.1.
================================================================
PART V. SIX ASYMPTOTIC SECTORS
================================================================
----------------------------------------------------------------
25. The obstruction for T: statement
----------------------------------------------------------------
Throughout this Part and the two following, set
T = < a, b, c | aba = bab, bcb = cbc, cac = aca >,
E = { (a, b), (b, c), (c, a) },
z_uv = (uv)^3 for (u, v) in E.
Proposition 25.1 (triangle obstruction). There is no geometric
action of T on a CAT(0) space Y on which, for every (u, v) in E,
ell(z_uv) / (3 ell(u^2)) = ell(z_uv) / (3 ell(v^2)) > 1/2. (R)
The two ratios in (R) are equal because u and v are conjugate in
T. Let delta = u v u = v u v (equality is the braid relation).
One checks: delta u delta^{-1} = u v u u u^{-1} v^{-1} u^{-1} =
u v u v^{-1} u^{-1} = (u v u) v^{-1} u^{-1} = (v u v) v^{-1} u^{-1}
= v u u^{-1} = v. So ell(u^2) = ell(v^2) by Lemma 18.1.
The denominators ell(u^2), ell(v^2) are positive in every geometric
action of T (the homomorphism T -> Z sending each generator to 1
shows a^2, b^2, c^2 have infinite order; Lemma 19.1).
Proposition 25.1 is proved in Section 34. The two intermediate
Parts (V and VI) build the two pieces: the asymptotic sectors
here, and the labelled diagram argument in Part VI.
Throughout this Part we work under the contrapositive hypothesis:
Assume T acts geometrically on a CAT(0) space Y and (R) holds
for all three edges (u, v) in E. Fix a base point y in Y.
We will derive a contradiction.
----------------------------------------------------------------
26. Orbit control (Svarc-Milnor)
----------------------------------------------------------------
We use the standard orbit-comparison result (Svarc-Milnor);
see Bridson-Haefliger I.8.18-19.
Lemma 26.1. Suppose T acts geometrically on Y with base point y.
There are R, C < infinity such that Ty is R-dense in Y, and for
all s, t in T,
d_T(s, t) <= C ( 1 + d_Y(sy, ty) ), (OC)
where d_T is the word metric for T with respect to the generating
set {a^{+-1}, b^{+-1}, c^{+-1}}.
PROOF. By cocompactness, there is a compact K subset Y with
T K = Y. Put R = max_{p in K} d_Y(y, p). Every t p is within R
of t y (isometry). So Ty is R-dense.
Now let L = d_Y(sy, ty). Divide the geodesic from sy to ty into
N = max(1, ceil(L)) pieces of length <= 1. At each division point,
pick a nearby orbit point: at the division point x_j (j = 0, ...,
N), choose t_j in T with d_Y(t_j y, x_j) <= R. Then t_0 = s
(forced: d(sy, sy) = 0 forces t_0 = s), t_N can be taken as t
(otherwise perturb). Each successive increment q_j = t_j^{-1}
t_{j+1} satisfies d_Y(y, q_j y) = d_Y(t_j y, t_{j+1} y) <= 2R + 1
(triangle inequality through x_j, x_{j+1}, with distance <= 1
between them). Let F = {q in T : d_Y(y, qy) <= 2R + 1}; by action
properness, F is finite (both y and qy lie in the compact ball of
radius 2R + 1). Let C' = max_{q in F} d_T(1, q), and C = max(1,
C').
Then d_T(s, t) <= sum_j d_T(1, q_j) <= N * C <= C (L + 1). QED.
----------------------------------------------------------------
27. Mirror coordinates (Digne's reflection action)
----------------------------------------------------------------
We need a way to "separate" disjoint regions of the parameter disk
in the CAT(0) space Y. The tool is an abstract affine action of T
on a linear space E via mirror reflections. This is the
equilateral-triangle specialization of Digne's reflection-coordinate
homomorphism; we give the concrete construction.
Take an equilateral triangle in the Euclidean plane. Let D_a,
D_b, D_c be its three side lines. Let r_u be reflection in line
D_u, u in {a, b, c}. Let D denote the set of all images of the
three lines D_a, D_b, D_c under the group generated by r_a, r_b,
r_c (the "triangle reflection group"). D is infinite but locally
finite.
Define E = direct sum over D of R e_D (formal basis vector e_D
indexed by D). E is a vector space with L1 norm
||sum_D xi_D e_D||_1 = sum_D |xi_D|. Every vector has finite
support (only finitely many xi_D nonzero).
The reflections r_u act linearly on E by permuting basis vectors:
r_u(xi) = sum_D xi_D e_{r_u(D)} (since r_u permutes D).
Lemma 27.1 (mirror action).
(a) The assignments
u . xi = e_{D_u} + r_u xi (u in {a, b, c})
define an affine isometric action of T on E.
(b) Let V(t) = t . 0 for t in T (the "translation part"). Then
||V(t) - V(s)||_1 <= d_T(s, t) (ML)
where d_T is the word metric.
(c) For each (u, v) in E, the elements u^2 and z_uv act by pure
translations on E, with vectors
L_u = 2 e_{D_u}, L_uv = 2 (e_{D_u} + e_{D_v} + e_{D_uv}),
where D_uv is the image of D_u under r_v (equivalently, the
third mirror through the vertex D_u intersect D_v of the
triangle).
(d) All six lines D_a, D_b, D_c, D_ab, D_bc, D_ca are distinct.
PROOF. (a) The maps xi |-> e_{D_u} + r_u xi are affine, with
linear part r_u and translation part e_{D_u}. We check the
defining relations of T. On E, both sides of a relation must
give the same affine map (same linear part and same translation
part).
The composition r_a r_b is rotation by 2 pi / 3 about the
intersection vertex D_a intersect D_b (composition of two
reflections in lines meeting at angle pi / 3 is a rotation by
2 pi / 3, i.e., the rotation that cyclically permutes the three
lines D_a, D_b, D_ab through the vertex). In particular
r_a r_b D_a = D_b, r_a r_b D_b = D_ab, r_a r_b D_ab = D_a.
The element aba = (e_{D_a} + r_a)(e_{D_b} + r_b)(e_{D_a} + r_a)
has translation part, evaluated at 0:
e_{D_a} + r_a(e_{D_b}) + r_a r_b(e_{D_a})
= e_{D_a} + e_{r_a D_b} + e_{r_a r_b D_a}
= e_{D_a} + e_{D_ab} + e_{D_b}
(using r_a D_b = D_ab and r_a r_b D_a = D_b).
Its linear part is r_a r_b r_a (a composition of reflections,
i.e., a reflection); it is exactly reflection in the third line
D_ab through the vertex, denoted r_{D_ab}.
The element bab = (e_{D_b} + r_b)(e_{D_a} + r_a)(e_{D_b} + r_b)
has translation part:
e_{D_b} + r_b(e_{D_a}) + r_b r_a(e_{D_b})
= e_{D_b} + e_{r_b D_a} + e_{r_b r_a D_b}
= e_{D_b} + e_{D_ab} + e_{D_a}
(since r_b D_a = D_ab and r_b r_a D_b = D_a).
Its linear part is r_b r_a r_b = reflection in the same line
D_ab.
So aba = bab as affine maps of E. The other two braid relations
bcb = cbc and cac = aca follow by the same computation with the
pair (a, b) replaced by (b, c) and (c, a). Hence (a).
(b) Each generator u or u^{-1} moves 0 to a vector of L1-norm 1
(e_{D_u} has L1-norm 1, and r_u preserves L1-norm). So a word
of length k moves 0 by at most k in E. That is:
||V(w)||_1 <= d_T(1, w) for every w in T. (first-est)
General bound. Use the cocycle identity for affine maps: for
any s, t in T,
V(s^{-1} t) = r_{s^{-1}} ( V(t) - V(s) ).
(This is the standard relation between translation parts of a
composition; r_{s^{-1}} is the linear part of s^{-1}, and
V(t) - V(s) is the affine difference.) Since r_{s^{-1}}
permutes basis vectors, it preserves L1-norm. Therefore:
||V(t) - V(s)||_1 = ||V(s^{-1} t)||_1 <= d_T(1, s^{-1} t) = d_T(s, t),
which is (ML).
(c) For u^2: applying u twice sends xi |-> e_{D_u} + r_u xi
|-> e_{D_u} + r_u(e_{D_u} + r_u xi)
= e_{D_u} + e_{r_u D_u} + r_u^2 xi = 2 e_{D_u} + xi
(using r_u D_u = D_u and r_u^2 = id). So u^2 acts by translation
by 2 e_{D_u} = L_u.
For z_uv = (uv)^3 = (uvu)^2: the map uvu has translation part
T_uvu = e_{D_u} + e_{D_uv} + e_{D_v} (by the computation of (a)
with u, v in place of a, b) and linear part r_u r_v r_u = r_{D_uv}
(reflection in the third line D_uv). Squaring an affine map
xi |-> T + L xi gives xi |-> T + L(T) + L^2 xi; with L = r_{D_uv}
(fixing D_uv and swapping D_u, D_v, so fixing T_uvu), we get
translation T_uvu + r_{D_uv}(T_uvu) = 2 T_uvu. So (uvu)^2 = z_uv
translates by 2(e_{D_u} + e_{D_v} + e_{D_uv}) = L_uv.
(d) The three initial lines D_a, D_b, D_c are the sides of an
equilateral triangle. The three lines D_ab, D_bc, D_ca are the
images of D_a, D_b, D_c under the reflections r_b, r_c, r_a
respectively. Concretely, D_ab is the reflection of D_a across
D_b, so it passes through the vertex D_a intersect D_b and is
parallel to D_c (the two lines D_a, D_b meet at angle pi / 3,
so reflecting D_a across D_b gives a line at pi / 3 angle from
D_b on the other side, i.e., parallel to D_c). Similarly
D_bc || D_a and D_ca || D_b. The six lines therefore consist of
the three "near" side lines (D_a, D_b, D_c) and the three
"far" side lines (each parallel to a distinct side of the
triangle), and are pairwise distinct. QED.
The important point of Lemma 27.1 is: the abstract "mirror" map
v_* : P -> E (constructed in Article 28) will SEPARATE disjoint
closed regions of the parameter disk, because V produces a
1-Lipschitz "model" of Y inside E. See Lemma 29.1 (sep) below.
----------------------------------------------------------------
28. The parameter disk P and its sectors
----------------------------------------------------------------
For each (u, v) in E, consider the two commuting pairs:
(U, H) = (u^2, z_uv) and (U, H) = (v^2, z_uv).
Both pairs consist of commuting elements (u^2 commutes with z_uv
by Fact 9.4; v^2 likewise). By Lemma 20, each pair determines a
seminorm on R^2. Because (R) holds, the six sectors have
total angle < 2 pi; this is the content of Lemma 28.1 below.
Lemma 28.1 (sector data). For each of the six pairs above:
(a) The restriction of the stable seminorm to R^2 is a positive-
definite Euclidean norm.
(b) The sector angle theta between the U-axis and the H-axis
satisfies 0 < theta < pi/3.
(c) Across the six sectors (in the cyclic order given below),
the sector angles sum to a total angle
0 < Omega < 2 pi,
and the sectors glue isometrically along their shared rays:
a^2, z_ab, b^2, z_bc, c^2, z_ca, a^2. (ro)
(d) The radius-one disk P of this cone gluing is a topological
disk, and the sextuple (v_*) below provides a continuous
injective map P -> E whose restriction to each sector is
the linear embedding
v_*(lambda, mu) = lambda L_u + mu L_uv (sv)
(and analogously with v in place of u for the other three
sectors).
PROOF. (a) and (b). For the commuting pair (U, H) = (u^2, z_uv),
the pure-translation formulas (Lemma 27.1(c)) give
V((U^i H^j)^m) = m (i L_u + j L_uv) for integers i, j.
Using (ML) and (OC):
m ||i L_u + j L_uv||_1 = ||V((U^i H^j)^m)||_1
<= d_T(1, (U^i H^j)^m) (by ML)
... and by OC applied with s = 1,
t = (U^i H^j)^m:
<= C ( 1 + m ||(i, j)||_{U,H} ) (approx)
Dividing by m and letting m -> infinity:
|| i L_u + j L_uv ||_1 <= C || (i, j) ||_{U, H}.
The same inequality holds for rational coordinates by homogeneity
and for real by continuity. In particular, the two vectors L_u and
L_uv must be linearly independent (else the norm ||(i,j)||_{U,H}
would fail for some (i, j) != 0 with i L_u + j L_uv = 0). This
gives positive-definiteness of ||.||_{U,H}.
The pairing <U, H> = <u^2, (uv)^3> by (v) of Lemma 23.2 with
r = 3 and i = 1: <U, H> = ell(H)^2 / N_3 = ell(H)^2 / 3. The
cosine of the sector angle:
cos theta = <U, H> / (ell(U) ell(H)) = ell(H) / (3 ell(U)).
The strict (R) inequality ell(H) / (3 ell(U)) > 1/2 gives
cos theta > 1/2, so theta < pi/3. Positive definiteness ensures
theta > 0.
(c) Around the six sectors, the shared rays are:
sector for (a^2, z_ab) meets sector for (b^2, z_ab) along z_ab;
sector for (b^2, z_ab) meets sector for (b^2, z_bc) along b^2;
...
Order (ro) lists the ray labels in sequence.
Along each shared ray for W (i.e., W = one of a^2, b^2, c^2, z_ab,
z_bc, z_ca), the ray parameter is t >= 0, and distance from the
origin in either adjacent sector is t ell(W) (by the ray formula
from Lemma 20). So the two sectors glue isometrically.
The total angle Omega is the sum of the six sector angles, each
strictly less than pi/3. Six times strictly less than pi/3 equals
strictly less than 2 pi. Also each angle > 0. So 0 < Omega < 2 pi.
(d) P is the part 0 <= r <= 1 of the cone disk parametrized by
radius r >= 0 and cyclic angle in [0, Omega] (angle 0 and Omega
identified along the ray for a^2). It is a topological disk (a
closed disk with a single point "cone apex" at r = 0).
Injectivity of v_* is verified by separating cases in the
E-coordinates: if the D_uv-coordinate of a nonzero image vector
is nonzero, it specifies (u, v) (which of the three sectors) and
the value mu (via the L_uv-coefficient). If D_uv-coefficient is
zero, the vector lies on the shared H-ray, which identifies the
two adjacent sectors. If additionally the entire "excess
coordinate" of D_u over 2 mu is zero (for the sector using u as
the squared generator), the vector is 0 or on the a^2, b^2, c^2
ray. All these cases are consistent with the gluing. QED.
We use only the Euclidean metric within each sector of P; we
do NOT claim that P is CAT(0). The metric on P is the (finite)
-piecewise Euclidean one obtained by gluing six Euclidean sectors
along their edges.
----------------------------------------------------------------
29. Uniform asymptotic maps f_n : P -> Y
----------------------------------------------------------------
We now build continuous maps f_n : P -> Y that grow like n and
whose metric structure within each sector approaches the Euclidean
one as n -> infinity. The "labels" t_n(p) are the group elements
whose orbit point t_n(p) y is close to f_n(p).
For (lambda, mu) in a sector (say the sector for U = u^2,
H = z_uv), define
t_n(lambda, mu) = U^{floor(n lambda)} H^{floor(n mu)}. (lab)
The two labels agree on the shared H-ray: for mu in R_{>=0},
floor(n * 0) = 0, so t_n(0, mu) = H^{floor(n mu)}, which is the
same as t_n in the adjacent sector (where the shared ray is the
H-axis with lambda-coordinate 0).
Lemma 29.1. There are continuous maps f_n : P -> Y with
f_n(0) = y, and a constant C_0 < infinity independent of n and
p, such that
d_Y(f_n(p), t_n(p) y) <= C_0. (grid)
For p, p' in the same sector,
d_Y(f_n(p), f_n(p')) / n -> || p - p' ||_{U, H} (sect)
uniformly. Additionally, uniformly on P,
d_Y(y, f_n(p)) / n -> r(p), V(t_n(p)) / n -> v_*(p). (rad)
Finally, if A_1, A_2 subset P are disjoint nonempty closed sets,
there is some delta > 0 with, for all large n,
d_Y(f_n(A_1), f_n(A_2)) >= delta n. (sep)
PROOF. Triangulation and construction of f_n. Triangulate the
entire exponent quadrant [0, inf)^2 in (lambda, mu) coordinates
by squares of side 1/n, each split along a diagonal. Assign to the
vertex (i/n, j/n) the point U^i H^j y (a group orbit point).
Parametrize edges by constant-speed geodesics in Y. On each
triangle, "cone" the opposite edge to the remaining vertex using
geodesics (an iterated geodesic interpolation). This gives a
continuous map f_n on the quadrant.
Edge consistency: all restrictions to shared edges are the same
geodesic parametrization, so the six sector-maps glue continuously
across the shared rays.
Grid bound (grid). Geodesic convexity of Y (property (LC))
controls the distance of any image point from any vertex: along a
triangle, the image lies within the convex hull (in the CAT(0)
sense) of the three vertex images. The vertex images are y_orbit
points whose distances from y are bounded by the vertex labels'
uniform displacements. Commutation bounds those: d_Y(y, U y) and
d_Y(y, H y) are finite constants. Let C_0 = max of these over the
two commuting pairs, plus rounding error. Then every image lies
within C_0 of some t_n(p) y, giving (grid).
Uniform convergence (sect) and (rad). For one commuting pair, set
F(i, j) = d_Y(y, U^i H^j y), (i, j) in Z^2.
Triangle inequality plus commutation gives the Lipschitz bound:
| F(i, j) - F(i', j') | <= d(y, U y) |i - i'|
+ d(y, H y) |j - j'|.
For a rational vector q = (a/m, b/m), compare floor(n q) to
floor(n/m) (a, b). The difference is bounded. The definition of
stable length gives:
F(floor(n q)) / n -> ell(U^a H^b) / m = || q ||_{U, H}.
This handles q with negative coordinates too. On a bounded region,
choose a finite rational net of arbitrarily small mesh; the common
Lipschitz bound plus rounding O(1/n) and continuity of the stable
norm give (sect) uniformly.
For (rad): the first limit d_Y(y, f_n(p))/n -> r(p) follows from
(sect) with p' = 0 and the definition of r(p) = ||p||_{U,H} (with
appropriate scaling). The second limit V(t_n(p))/n -> v_*(p) comes
from the mirror-translation formula; the error is O(1/n).
Region separation (sep). Compactness of A_1 x A_2 (both closed
and disjoint compact subsets of P) plus injectivity of v_* give a
minimum positive separation:
eta = min_{p in A_1, q in A_2} || v_*(p) - v_*(q) ||_1 > 0.
By (rad), the difference V(t_n(p)) - V(t_n(q)) has L1 norm at
least eta n / 2, uniformly in p, q, n. Then (ML), (OC), and (grid)
combine:
d_Y(f_n(p), f_n(q)) >= eta n / (2C) - 1 - 2 C_0.
For large n, this is >= delta n for any delta < eta / (2C).
QED.
----------------------------------------------------------------
30. The outer cap
----------------------------------------------------------------
The boundary of P is a closed curve of length proportional to the
total sector angle Omega. The shortness of Omega (Omega < 2 pi)
allows filling the boundary by a disk whose image under f_n stays
far from y.
Lemma 30.1. There is epsilon > 0 such that, for all sufficiently
large n, the map f_n|boundary P extends to a continuous map of a
second parameter disk whose entire image stays OUTSIDE the open
ball B_Y(y, epsilon n) around y. Gluing the two parameter disks
gives a continuous sphere map f_n : S^2 -> Y.
PROOF. Chord distance on boundary P. For any two points p, p'
on boundary P, choose the shorter angular arc between them.
Its angle phi is at most Omega / 2 < pi. Unfold its sectors to
a Euclidean wedge; the intersection of the wedge with the unit
disk is convex, so the chord between p, p' stays within it.
Chord length:
|p - p'| <= 2 sin(phi / 2) <= 2 sin(Omega / 4).
Each sector appears at most once in this development, except
that the sector containing both endpoints may appear in two end
pieces. Each such piece is convex, so the chord splits into at
most seven single-sector pieces (one for each of the six sectors
plus an extra piece if the arc crosses the sector containing
both endpoints). Apply (sect) to each piece; uniformly in
p, p' in boundary P:
d_Y(f_n(p), f_n(p')) <= 2 sin(Omega / 4) n + o(n). (B-short)
Coning the boundary. Fix p_0 in boundary P, and cone the boundary
map f_n|boundary P to the point f_n(p_0) using Y-geodesics. The
continuous dependence of geodesics on endpoints (Lemma 17.1(a))
makes this a continuous map of a parameter disk into Y.
Image stays far from y. Consider a point q on the geodesic segment
[A, B], where A = f_n(a), B = f_n(b) for a, b in boundary P.
Triangle inequality through q:
d(y, q) >= max{d(y, A) - d(A, q), d(y, B) - d(q, B)}
>= (d(y, A) + d(y, B) - d(A, B)) / 2.
(The max-min inequality: max{a - x, b - (d - x)} where d = d(A,B),
x = d(A,q), and d - x = d(q,B); the max is at least the average
of the two expressions, which is (a + b - d) / 2.)
Now d(y, A) and d(y, B) each are n + o(n) by (rad) (since
A = f_n(a) with a on boundary P, so r(a) = 1 in sector
coordinates). And d(A, B) <= 2 sin(Omega/4) n + o(n) by (B-short).
So:
d(y, q) >= ( 2n - 2 sin(Omega/4) n ) / 2 - o(n)
= ( 1 - sin(Omega/4) ) n - o(n).
Take epsilon = (1/2) (1 - sin(Omega/4)). Since Omega < 2 pi,
Omega / 4 < pi / 2, so sin(Omega/4) < 1 and epsilon > 0. For
large n, the image of the second disk stays outside B_Y(y,
epsilon n). Gluing gives the continuous sphere map. QED.
The two Lemmas 29.1 and 30.1 give exactly the geometric input we
needed: a continuous sphere map f_n : S^2 -> Y whose image
separates a small patch B (to be chosen in Part VI) from the
"outer" region far from y. In Part VI we compare this sphere map
with a labelled combinatorial diagram of the presentation of T,
to derive a contradiction with the algebraic structure of T.
This completes Part V. In Part VI we develop the labelled
diagram technology (Brady-McCammond triangular presentation,
link-girth lemma, spherical signed count, controlled approximation),
then in Part VII (localization) we combine it with the sector
geometry to prove Proposition 25.1.
================================================================
PART VI. LABELLED TRIANGULAR DIAGRAMS
================================================================
----------------------------------------------------------------
31. The labelled triangular complex K
----------------------------------------------------------------
The triangle group T has the following auxiliary presentation.
Lemma 31.1 (triangular presentation). For each (u, v) in E,
introduce two private generators x_uv and h_uv. Within that block,
write (x_1, x_2, x_3, h) = (u, v, x_uv, h_uv). The presentation
with generators {a, b, c} union {x_uv, h_uv : (u,v) in E} and,
in each block, the three relations
x_i x_{i+1} = h (i = 1, 2, 3 modulo 3), (tri)
presents T. Equivalently: eliminating the private h_uv and x_uv
recovers T itself.
PROOF. From the first two relations, h = x_1 x_2 and
x_3 = x_2^{-1} h = x_2^{-1} x_1 x_2. Substituting in the third
relation x_3 x_1 = h:
(x_2^{-1} x_1 x_2) x_1 = x_1 x_2,
x_1 x_2 x_1 = x_2 x_1 x_2 (multiply left by x_2).
This is exactly the (u, v) braid relation of T. Conversely, the
braid relation x_1 x_2 x_1 = x_2 x_1 x_2 with h := x_1 x_2 and
x_3 := x_2^{-1} x_1 x_2 satisfies:
x_2 x_3 = x_2 (x_2^{-1} x_1 x_2) = x_1 x_2 = h,
x_3 x_1 = x_2^{-1} x_1 x_2 x_1 = x_2^{-1} x_2 x_1 x_2 = h,
x_1 x_2 = h.
So the three relations (tri) in a block are equivalent to the
braid relation between u and v. Since the private generators are
introduced independently in each block, the presentation with all
these relations recovers T. QED.
Let A = {a, b, c} union {x_uv, h_uv : (u, v) in E}; this is a
set of 3 + 6 = 9 generators.
DEFINITION 31.2 (labelled Cayley complex K). K is a 2-dimensional
labelled combinatorial complex.
Vertices: the elements of T.
Edges: for every g in T and q in A, an oriented edge, carrying
the label (g, q), from vertex g to vertex g q. Two edges with
the same (g, q) data are still distinguished by their own cell
identity.
Triangles: for every relation x_i x_{i+1} = h in some block and
every g in T, attach a triangle T_{g, i} along its three sides
matching the edges of (g, x_i), (g x_i, x_{i+1}), (g, h).
Its vertices (as distinct combinatorial points) are g, g x_i,
g h.
Cell identities are permanent: edges and triangles keep their
identities even if two happen to share vertex/label data. The
homomorphism T -> Z (a, b, c |-> 1) sends x_uv |-> 1 and
h_uv |-> 2, so the three vertices g, g x_i, g h of every target
triangle are distinct.
Lemma 31.3 (link-girth). The link at every vertex of K has no
circuit of length < 6 (even with parallel edges).
PROOF. Fix a vertex v in T (i.e., an element of T as a vertex of
K). The link has vertices q^+ and q^- for each q in A (the
"outgoing" and "incoming" q-edges at v). For the relation
x_i x_{i+1} = h in one block, the initial, middle, and terminal
corners at v occur at translates g = v, g = v x_i^{-1}, and
g = v h^{-1}. The corresponding link edges are
{ x_i^+, h^+ }, { x_i^-, x_{i+1}^+ }, { x_{i+1}^-, h^- }.
So one block contributes a theta graph: three internally disjoint
paths of length 3 from h^+ to h^-:
h^+ - x_1^+ - x_3^- - h^-,
h^+ - x_2^+ - x_1^- - h^-,
h^+ - x_3^+ - x_2^- - h^-.
The 9 link edges of the block are distinct.
The three blocks meet only at the shared directions of a, b, c.
Across different blocks, the only link edges with both endpoints
shared directions are
{ a^-, b^+ } (in the ab block), { b^-, c^+ } (in the bc
block), { c^-, a^+ } (in the ca block).
These three are distinct.
Now bound a simple circuit. Split it into maximal nonempty
sub-paths within individual blocks; let k be their number. Their
endpoints are the "block-change" points, at shared a, b, c.
k = 1: fully inside a block; distance >= 6 in the theta graph.
k = 2: both changes at the intersection of the same two blocks
(say a). The endpoints are a^+ and a^- (the two signs of the
one shared generator), at distance >= 3 in either block (the
theta graph has a^+ - h^+ - ... - h^- - a^- of length 5, or
a^+ - x_3^- - h^- - a^- of length 3, or a^- - h^- - ... - h^+ -
a^+ symmetric). So total >= 6.
k = 3: circuit visits all three blocks, changing at one sign of
a, b, c. The six sign-pairs (σ_a, σ_b), (σ_b, σ_c), (σ_c, σ_a)
with σ_x in {+1, -1} have either all three equal, or exactly two
sign changes (in the cyclic order a -> b -> c -> a).
- All three signs equal: distances 2 + 2 + 2 = 6.
- Two sign changes: one is a "minus to plus" change (say from
- to + between consecutive pairs), one is a "plus to minus"
change. The minus-to-plus change might use the exceptional
adjacency { x_1^-, x_2^+ } of distance 1; the plus-to-minus
change is a non-exceptional opposite-sign pair of distance
>= 3. The third pair (unchanging sign) is distance 2. Lower
bound 1 + 3 + 2 = 6.
k >= 4: a sub-path of length 1 must be one of the exceptional
edges {a^-, b^+}, {b^-, c^+}, {c^-, a^+}. Each shared direction
is incident to at most one such edge, so two length-1 sub-paths
cannot be consecutive. So at most floor(k/2) sub-paths have
length 1, and the rest have length >= 2. Total >=
2(k - floor(k/2)) + floor(k/2) = 2k - floor(k/2) >= 6.
So every simple circuit in the link has length >= 6. QED.
----------------------------------------------------------------
32. Signed counts on spheres
----------------------------------------------------------------
DEFINITIONS. A COMBINATORIAL MAP from a finite triangulated oriented
2-sphere Sigma to K is: each edge of Sigma maps to a specified
target edge of K (possibly with reversed orientation), or to a
target vertex; each triangle either maps homeomorphically to a
specified target triangle T_{g, i} (respecting vertex
correspondence) or maps affinely into a specified target edge or
vertex. These choices agree on common sides. In the second case
we say the triangle is COLLAPSED and it contributes zero to all
counts.
Fix one orientation on every target triangle of K. On the oriented
source Sigma, a non-collapsed triangle has SIGN +1 or -1 according
to whether its map to the (oriented) target respects or reverses
the orientation.
Proposition 32.1 (sphere counts). For every combinatorial map from
a finite triangulated oriented 2-sphere Sigma to K, and for every
individual target triangle tau of K, the sum over all occurrences
of tau in Sigma of their signs is zero. Collapsed triangles
contribute zero and are ignored.
PROOF. We use the fold-deletion method of Gersten (1987, Def 3.1,
Cor 4.6), adapted here to keep cell identities and allow collapsed
parameter triangles.
Encoding Sigma as a labelled planar diagram. In each non-collapsed
source triangle, draw a central node with three short "legs" to
the three side midpoints. For a triangle collapsed to an edge,
its two non-constant sides share a labelled target edge; join
their midpoints by a single arc (both sides map to that target
edge). A triangle collapsed to a vertex contributes nothing.
Join across shared source sides: this produces an embedded
trivalent graph in Sigma, together with possibly some closed
curves without nodes; discard those curves. Each remaining node
records a particular occurrence of a target triangle, with its
sign. Each graph edge pairs two sides mapping to the same
labelled target edge.
Retaining endpoint data. Give the graph a thin regular
neighborhood: node disks + edge ribbons. Along either side of a
ribbon, the group label of the corresponding endpoint of its
target edge is constant: this is because the ribbon connects
sides that map to the same target edge, and the endpoint labels
of a given target edge are the same. At a node, both neighboring
ribbon sides of a given corner carry the same group label (the
corresponding corner of that target triangle). Consequently
every boundary circle of the regular neighbourhood carries one
group-vertex label throughout.
Capping. For each connected component containing nodes, cap its
regular neighbourhood by a disk on every boundary circle, using
the (constant) group label as the label of the dual vertex. This
produces a LABELLED TRIANGULAR CELL DECOMPOSITION OF A SPHERE:
graph nodes <-> triangles, graph edges <-> edges, boundary
circles <-> vertices. Multiple graph edges between the same two
nodes are allowed (the resulting cell decomposition need not be
a strict simplicial complex; incidences, not distinct vertex
pairs, are counted).
Excluded degeneracies. The labelling rules give:
(i) A graph edge cannot join two legs at the same node (those
legs represent different target sides; a ribbon has one
target edge).
(ii) No two corners of a face become the same dual vertex
(their actual target vertex labels are distinct).
(iii) A bridge of the graph (an edge whose removal disconnects)
would have both its ribbon sides on the same boundary
circle; constancy around that circle would force the two
distinct endpoint labels of the target edge to coincide.
So no bridges.
Deleting folds. A FOLD is a pair of faces sharing an edge and
mapping to the same target triangle across that same target side.
Their signs are opposite. Take a small disk D around the shared
edge and its two nodes, cutting the other four legs close to the
nodes. The chosen edge + endpoint disks form a disk (any other
parallel connecting edges between the two nodes are outside D).
Exposed legs occur in cyclic order i, j, j, i (up to reversal and
cyclic permutation). Reconnect i to i and j to j inside D without
crossing, and delete the two nodes. All labels match on each new
ribbon side. So both the label rules, the planarity, and the sum
of signs for each target triangle are preserved. This may
disconnect the graph or create closed curves; discard those
curves and re-cap the remaining components.
Euler contradiction. Suppose nodes remain and no fold is
available. Choose a component. Its DUAL spherical cell
decomposition has, at each dual vertex x (the label of some
target vertex of K), a closed walk in the link of x, obtained by
reading successive target corners of the faces incident at x.
A backtrack in this walk uses the same target corner on both
sides of a shared edge; by the exact corner description
({x_i^+, h^+}, {x_i^-, x_{i+1}^+}, {x_{i+1}^-, h^-}), that
identifies the same translated target triangle on both sides,
which is exactly a fold. These must be two distinct source
faces (one source triangle cannot use two different corners at
this dual vertex, nor have two of its sides joined by the same
ribbon). So a backtrack implies a fold, contradicting "no fold
available".
Thus every link walk is non-backtracking. Any finite nonempty
closed non-backtracking walk contains a simple circuit; by
Lemma 31.3, every such circuit has length >= 6. Let F, E, V
count faces, edges, vertices of this dual cell decomposition
(all incidences counted). Then:
3 F = 2 E (every face triangle; every edge shared by 2 faces),
2 E >= 6 V (every vertex lies on link walk of length >= 6, so
total vertex-edge incidences 2E >= 6V),
V - E + F = 2 (Euler characteristic of the sphere).
The first two imply V - E + F <= 0, contradicting the third. So
whenever nodes remain, a fold can be removed. Each removal
decreases their number by two, so the process terminates with no
nodes. All signed target-triangle counts are preserved by the
removals and are then zero. QED.
----------------------------------------------------------------
33. Controlled approximation by diagrams
----------------------------------------------------------------
Lemma 33.1 (controlled approximation). Suppose T acts
geometrically on Y and fix y in Y. There is a constant C_1,
depending only on this data and the fixed presentation, with the
following property. Let Sigma be a triangulated parameter sphere,
B a subcomplex, and f : Sigma -> Y continuous. Then there is a
refinement keeping B a subcomplex, plus a combinatorial map from
the refined sphere to K, such that, for every final parameter
triangle Q, every group-vertex label s of Q, and every p in Q,
d_Y(s y, f(p)) <= C_1. (all-pts)
In particular C_1 is independent of f, of its modulus of
continuity, and of the number of triangles required.
PROOF. Set up R and M. Choose R < infinity with Ty being R-dense
in Y (cocompactness + Lemma 26.1). Apply Lemma 26.1 (OC) to
choose an integer M with:
d_Y(s y, t y) <= 2 R + 1 => d_T(s, t) <= M.
Set D = max{d_Y(y, ay), d_Y(y, by), d_Y(y, cy)}.
Refine and label. Refine the initial triangulation, preserving B
as a subcomplex, until every face image has diameter <= 1 (by
uniform continuity of f). For each vertex v of the refined
triangulation, choose a group label s_v with d_Y(s_v y, f(v)) <=
R (using R-density).
Edge fill. For each edge of the refined triangulation, choose a
path in the Cayley graph of T from label(s_start) to
label(s_end), of word length <= M (which exists by the definition
of M: d_Y(s_start y, s_end y) <= 2 R + 1). Subdivide the
parameter edge into (M + 6) segments; add 3 constant steps at
each end (filling those by cones at a single vertex). Edge length
<= M + 6; every face boundary is a cyclic word of length at most
3(M + 6).
Finite boundary types. Relative to the label of any one base
vertex of a face, the boundary word has length at most 3(M+6)
and letters in {a^{plus-minus 1}, b^{plus-minus 1},
c^{plus-minus 1}, 1}. Finitely many possibilities. For each,
construct one finite filling (triangulated disk) with that
boundary, using only triangles from K.
Construction of a filling for a given null-loop boundary. Ignore
the 1-symbols (fill them by cones). The remaining word represents
the identity of T. By definition of a presented group, it lies in
the normal closure of the relators, which here are the triangular
relations (tri). So it can be reduced to the empty word by
finitely many:
- free insertions or deletions of q q^{-1} for q in A,
- insertions or deletions of a translated triangular loop
w (x_i x_{i+1} h^{-1}) w^{-1} or its reverse, at some vertex
of the current cycle,
- moves of "constant symbols" (treating 1 as a filler).
Each individual move takes place in one target edge, triangle, or
vertex.
Between successive intermediate cycles, realize the move as a
triangulated annulus:
- along an unchanged edge: a strip rectangle triangulated into
two (vertex- or edge-)collapsed triangles;
- a substituted arc: a polygonal region whose vertex labels lie
in the single target simplex used, triangulated with the
corresponding simplex maps.
For an EMPTY arc (deleting q q^{-1} at a vertex), use a fan of
triangles from the vertex rather than an identification of
parameter vertices (the parameter points p_0, p_k are distinct
even if their labels coincide).
Stack the finitely many annuli and cap the final three-step
constant loop by a cone. The result is a finite triangulated disk
with the original subdivided boundary. A boundary-preserving
homeomorphism to the original face puts all added triangles
inside that face. Do this independently for each face, gluing
along agreed edge subdivisions. The refinement of Sigma is
respecting B (which is preserved by uniform-continuity refinement
and only refined along edges not in B's interior).
Uniform estimate. Choose once and for all the finitely many
fillings constructed above. Normalize the base vertex of each
filling to 1 in T. Let L = largest word length d_T(1, s) over
all labels s appearing in any one of these fillings. Finite:
private generators (x_uv, h_uv) are words in a, b, c.
Now for a point p in an original face v_0 (with vertex label
s_0), and s any label appearing in the filling of that face:
d_Y(s y, f(p))
<= d_Y(s y, s_0 y) + d_Y(s_0 y, f(v_0)) + d_Y(f(v_0), f(p))
<= D L + R + 1.
(The first via d_T(s_0, s) <= L from the filling and the
correspondence between word metric and orbit distance giving
d_Y(s_0 y, s y) <= D * d_T(s_0, s) <= D L; the second because
f(v_0) is within R of s_0 y; the third because a face has
diameter <= 1 in Y by construction.)
Set C_1 = D L + R + 1. Every final triangle lies within some
face, so this bound holds at every point of every parameter
triangle. QED.
This completes Part VI. In Part VII (localization) we combine
the sector geometry of Part V with the diagram technology of
this Part to prove Proposition 25.1.