Criterion of Hurwitz equivalence for quasipositive factorizations of 3-braids

0.12 MB·English

by  Stepan Yu. Orevkov

CRITERION OF HURWITZ EQUIVALENCE FOR
QUASIPOSITIVE FACTORIZATIONS OF 3-BRAIDS
5
S. Yu. Orevkov
1
0
2 The problem of Hurwitz equivalence of n-tuples of braids appears in the study
y of braid monodromy of algebraic curves in C2. It was considered by many authors,
a
see [1, 3, 4] and references in [3]. We give an answer for the case mentioned in the
M
title.

0 Let B = hA | σ σ = σ σ = σ σ i, A = {σ ,σ ,σ } be the Birman–Ko–Lee
3 2 1 1 0 0 2 0 1 2
3
presentation (see [2]) of the group of braids with three strings. A quasipositive

] factorization of a braid X ∈ B is a collection (X ,...,X ) ∈ Bk such that X =
R 3 1 k 3
X X ...X andforeachi, thebraidX isconjugatetoσ . Notethatσ , σ , andσ
G 1 2 k i 1 0 1 2
are conjugate to each other. We denote the set of quasipositive factorizations of X
.
h by Q(X). A braid X is called quasipositive if Q(X) 6= ∅. The braid group B acts
t k
a on Q(X) by Σ : (X ,...,X ) 7→ (Y ,...,Y ) where (Y ,Y ) = (X X X−1,X )
m i 1 k 1 k i i+1 i i+1 i i
and Y = X for j 6∈ {i,i+1}. This action is called the Hurwitz action. Elements
j j
[
belonging to the same orbit are called Hurwitz-equivalent.

2
It is proven in [4] that each orbit of the Hurwitz action contains an element of a
v
9 certain explicitly specified finite set. The purpose of the present paper is to give an
6
easy criterion to decide if two given elements of this finite set belong to the same
2
0 orbit. To give precise statements, we need to introduce some notation with slightly
0 differsfromthatin[4]. LetusextendthealphabetAuptoAˆ= A∪{σˆ ,σˆ ,σˆ }. Let
0 1 2
.
1 A∗ and Aˆ∗ be the free monoids generated by A and Aˆ respectively. If U,V ∈ Aˆ∗,
0
5 then U ≡ V stands for equality in Aˆ∗ (i. e. letterwise coincidence of words) and
1 U = V (when U,V ∈ A∗) stands for equality of the corresponding elements of B .
3
:
v For U ∈ Aˆ∗, we denote the word obtained from U by erasing of all letters σˆ (resp.
i
i
X by replacing each σˆ with σ or by replacing each σ with σˆ ) by U¯ (resp. U′ or Uˆ).
i i i i
r For example, if U ≡ σ σˆ σ σˆ , then U′ ≡ σ σ , U¯ ≡ σ σ σ σ , and Uˆ ≡ σˆ σˆ σˆ σˆ .
a 0 1 2 1 0 2 0 1 2 1 0 1 2 1
We set δ = σ σ = σ σ = σ σ . It is easy to check that any X ∈ B can be
2 1 1 0 0 2 3
written as
X = Uδ−p, U ∈ A∗, p ∈ Z. (1)
If, moreover, U does not contain any subword which is equal (in B ) to δ, then
3
the presentation of X in the form (1) is unique and it is called the right Garside
normal form.
Let
W = W xˆ W xˆ ...W xˆ W ∈ Aˆ∗, W ∈ A∗, x ∈ A. (2)
1 1 2 2 k k k+1 i i
If W′ = δp, then [W] stands for the quasipositive factorization of W¯ δ−p of the form
(X ,...,X ) where X = A x A−1, A = W ...W .
1 k i i i i i 1 i
To each X ∈ B we associate a graph G (X) in the following way. Let (1) be
3 0
the right Garside normal form of X. We define the set of vertices of G (X) as
0
Typeset by AMS-TEX
1
2 S. YU. OREVKOV
V (X) = {W ∈ Aˆ∗ | W′ = δp,W¯ = U}. Two vertices are connected by an edge if
0
they are of the form AxˆByC and AxByˆC, x,y ∈ A where xB′ = B′y and, for some
q ≥ 0, either B′ = δq (an edge of type (h1)), or xB′ = δq (an edge of type (h2)).
Theorem 1. Suppose that the right Garside normal form of a braid X ∈ B is (1)
3
with p ≥ 0. Then each orbit of the Hurwitz action on Q(X) contains an element of
the form [W], W ∈ V (X). Two such elements belong to the same orbit if and only
0
if the corresponding vertices belong to the same connected component of G (X).
0
Remark 1. The condition p ≥ 0 in Theorem 1 is not very restrictive. Indeed, if
p < 0, then the Hurwitz action has a single orbit (see [4; Corollary 4]).
The rest of the paper is devoted to the proof of Theorem 1. Let τ : Aˆ∗ → Aˆ∗
be the monoid homomorphism defined on the generators by σ 7→ σ 7→ σ 7→ σ ,
0 1 2 0
σˆ 7→ σˆ 7→ σˆ 7→ σˆ . It induces the inner automorphism τ : X 7→ δ−1Xδ of B .
0 1 2 0 3
Lemma 1. If A,B ∈ A∗, A = B, and A 6≡ B, then A = Cδ = δτ(C), C ∈ A∗.
Proof. It is known (see [2; Theorem 2.7]) that if A = B, then B is obtained from
A by applying the relations without inserting the inverses of the generators. If
A 6≡ B, then a relation can be applied to A, hence A = DδE, D,E ∈ A∗. Then
A = Cδ = δτ(C) for C = τ(D)E. (cid:3)
Definition. Let W ≡ a ...a , a ∈ Aˆ, W′ = δq, q > 0. We say that letters a
1 n i i
and a (i < j) match each other in W if a ,a ∈ A, (a ...a )′ ≡ a Ba = δr+1 and
j i j i j i j
B = δr, r ≥ 0.
Formally speaking, the indices i and j rather than the letters a and a match
i j
each other in the above definition. However, abusing the language, we shall speak
about matching letters to avoid cumbersome notation.
Lemma 2. (follows from [4; Lemma 5]) Let W ∈ A∗, W = δq, q > 0. Then one
can associate parentheses to all letters of W so that the parentheses are balanced
and matching pairs of parentheses correspond to matching pairs of letters.
Lemma 3. Let W ≡ AuBCvD ∈ A∗, W = δq, and let v match u in W. Let
W ≡ CvDτq(AuB). Then W = δq and τq(u) matches v in W .
1 1 1
Proof. Let E = AuB, F = CvD. Then δq = EF, hence W = Fδ−qEδq =
1
F(F−1E−1)E(EF) = EF = δq. The same computation with E = Au, F = BCvD
yields BCvDτq(Au) = δq. Since BC = δr, we obtain vDτq(Au) = δq−r. Similarly,
by setting E = A and F = uBCvD, we obtain Dτq(A) = δq−r−1. (cid:3)
Lemma 4. Let W ≡ ABC ∈ A∗, W = δq, B = δr. Then, for any letter u in A
there is a letter in A or in C which matches u in W.
Proof. Follows from Lemma 3 combined with Lemma 2 applied to τq(C)A. (cid:3)
Let a braid X satisfies the hypothesis of Theorem 1. We extend the graph
G0(X) up to G(X) as follows. We define the set of vertices as V(X) = Sj≥0Vj
where V = {W ∈ Aˆ∗ | W′ = δp+j,W¯ δ−p−j = X}. The weight of a vertex
j
W ∈ V is defined as wt(W) = j. We define edges of types (h1) and (h2) in
j
the same way as in G (X) and we add more edges (W,V) in the following cases:
0
CRITERION FOR HURWITZ EQUIVALENCE FOR 3-BRAID 3
(h3) W ≡ ABˆ C, V ≡ ABˆ C, B ,B ∈ A∗, B = B ;
1 2 1 2 1 2
(v1) W ≡ APB, V ≡ Aτ−1(B), P ∈ A∗, P = δ;
(v2) either W ≡ APByC and V ≡ Aτ−1(ByˆC), or W ≡ AyBPC and V ≡
AyˆBτ−1(C) where P ∈ {uxˆ,xˆu}, P¯ = δ, and y matches u in W;
(v3) wt(V) = wt(W) − 1, W is connected by an edge of type (h3) to a vertex
which is connected to V by an edge of type (v2).
We call edges of types (h1–h3) horizontal and edges of types (v1–v3) vertical. The
notation (W,V) for vertical edges assumes that wt(W) > wt(V). If two vertices W
and V belong to the same connected component of G(W), then we write W ∼ V.
If either W ≡ V or there exists an edge (W,V) of type, say, (h1), then we write
W ∼ V. If wtW = wtV and W is connected to V by a path in G(X), which
h1
passes through vertices of weight ≤ wt(W) only, then we write W ∼ V.
h
We define f : (Aˆ∪ A−1)∗ → Aˆ∗ by f(1) = 1, f(xA) ≡ xf(A) for x ∈ Aˆ and
f(x−1A) ≡ τ(τ(x)f(A)) for x ∈ A. Note that the braid A−1f(A) is a power of δ.
Lemma 5. (a). Let W ∈ V(X). Then W ∼ f(Y ...Y ), Y ≡ A xˆ A−1 for some
1 k i i i i
A ∈ A∗, x ∈ A, i = 1,...,k. (b). If X is conjugate to σ , then V (X) contains a
i i 1 0
single element and it has the form f(AxA−1), A ∈ A∗, x ∈ A.
Proof. (a). Let W be as in (2) and set A ≡ W ...W . Then f(Y ...Y ) is
i 1 i 1 k
connected to W by a chain of edges of type (v1). (b). Follows from Lemma 2. (cid:3)
Lemma 6. [4; Lemmas 6 and 7]. If W ∈ V(X) then W ∼ V for some V ∈
V (X). (cid:3)
0
Lemma 7. Let W,V ∈ V(X). Then [W] ∼ [V] is equivalent to W ∼ V.
Proof. (⇒) Let [W] = (X ,...,X ). Lemma 5(a) combined with Lemmas 6 and
1 i
5(b) applied to Y ’s imply W ∼ f(Y ...Y ) where V (X ) = {f(Y )}. Thus [W] =
i 1 k 0 i i
[V] ⇒ W ∼ V. It remains to note that if W ≡ f(Y ...Y ), Y ≡ A xˆ A−1, and
1 k i i i i
V ≡ f(...Y Y¯ Y Y¯−1Y Y ...), W ≡ f(...Y Y Y Y¯−1Y¯ Y ...),
i−1 i i+1 i i i+2 1 i−1 i i+1 i i i+2
then Σ ([W]) = [V], W ∼ V, and W is connected to W by a chain of (v1)-edges.
i 1 h2 1
(⇐) (cp. [4; Lemma 6]). It suffices to consider the cases W ∼ V and
h1
W ∼ V. Let W be as in (2). Then, for some m, s, 1 ≤ m < s ≤ k + 1,
h2
we have W ≡ Axˆ ByC and V ≡ Ax ByˆC where A ≡ W xˆ W xˆ ...W , B ≡
m m 1 1 2 2 m
W xˆ ...W xˆ D, C ≡ Exˆ W xˆ ...xˆ W (if s = k + 1, then
m+1 m+1 s−1 s−1 s s+1 s+1 k k+1
C ≡ E), W ≡ DyE. Let [W] = (X ,...,X ) and [V] = (Y ,...,Y ).
s 1 k 1 k
It is clear that Y = X for i < m.
i i
If m ≤ i < s−1, then Y = B x B−1 where B = A′x W W ...W
i i i+1 i i m m+1 m+2 i+1
= (A′x (A′)−1)A′W ...W = X W ...W , whence Y = X X X−1.
m m+1 i+1 m 1 i+1 i m i+1 m
If i = s − 1, then Y = A′x B′y(A′x B′)−1. Using x B′ = B′y, we obtain
i m m m
Y = A′x (x B′)(A′x B′)−1 = A′x (A′)−1 = X .
i m m m m m
If i ≥ s, then Y = B x B−1 where B = A′x B′EF and F = W ...W .
i i i i i m s+1 i
Using x B′ = B′y, we obtain B = A′B′yEF = W W ...W , whence Y = X .
m i 1 2 i i i
Thus, [V] = (X ,...,X , X X X−1,...,X X X−1, X , X ,...,X )
1 m−1 m m+1 m m s−1 m m s k
= Σ ...Σ Σ [W]. (cid:3)
s−2 m+1 m
4 S. YU. OREVKOV
Let c : Aˆ∗ → Aˆ∗ be defined as aW 7→ Wτs(a), a ∈ Aˆ, W ∈ Aˆ∗.
s
Lemma 8. (follows from Lemma 3) (a). Let e = (W,V) ≡ (aA,bB), a,b ∈ Aˆ, be
an edge of G(X) of type (h1) or (h2). Let s = p+wt(W). Then e = (c (W),c (V))
1 s s
is an edge of G(c (W¯ )δ−s) of type (h1) or (h2). If a = b, then e and e are of the
s 1
same type. If a 6= b, then they are not.
(b). Let e = (W,V) be an edge of G(X) of type (v1) or (v2). Let s = p+wt(W).
Let e = (cm(W),cm (V)) where m = 2 if W starts with the word P used in the
1 s s−1
definition of e, and m = 1 otherwise. Then e is an edge of G(cm(W¯ )δ−s) of the
1 s
same type as e. (cid:3)
Lemma 9 (Diamond Lemma). Let e = (W,V ) and e = (W,V ) be vertical edges
1 1 2 2
in G(X). Then V ∼ V .
1 h 2
Proof. By Lemma 8, mutual arrangements of subwords of W used in the definition
of the edges can be considered up to cyclic permutations.
Case 1. Both e and e are of type (v1). The statement is evident.
1 2
Case 2. Both e and e are of type (v2). Let P ∈ {u xˆ ,xˆ u }, and y , i = 1,2,
1 2 i i i i i i
be the subwords of W used in the definition of e .
i
Case 2.1. P coincides with P , i. e., W ≡ Ay By CP, V ≡ Ayˆ By C, V ≡
1 2 1 2 1 1 2 2
Ay Byˆ C, P ∈ {xˆu,uxˆ}, P¯ = δ. By definition of edges of type (v2), we have
1 2
B′y C′ = δq, C′ = δr, y B′y C′u = δq+1, y C′u = δr+1. The former two identities
2 1 2 2
imply B′y = δq−r. The latter two identities imply y B′ = δq−r. Hence V ∼ V .
2 1 1 h2 2
Case 2.2. P and P have a common letter xˆ = xˆ = xˆ. Since u x = xu = δ,
1 2 1 2 1 2
it follows that u cannot match u . Hence y and y cannot coincide with them.
1 2 1 2
Case 2.2.1. y = y = y. W ≡ AyBu xˆu , V ≡ AyˆBτ−1(u ), V ≡ AyˆBu .
1 2 1 2 1 2 2 1
Then B′u = δq and B′ = δr, whence u = δq−r. Contradiction.
1 1
Case 2.2.2. W ≡ Ay By Cu xˆu , V ≡ Ay Byˆ Cτ−1(u ), V ≡ Ayˆ By Cu .
2 1 1 2 1 2 1 2 2 2 1 1
Since u δ = u xu = δu , we have u = τ(u ), whence V ≡ Ay Byˆ Cu . Since
1 1 2 2 2 1 1 2 1 1
B′y C′u = δq (e is of type (v2)) and y C′u = δr+1 (e is of type (v2)), it
1 1 2 1 1 1
follows that B′ = δq−r−1. To conclude that V ∼ V , it remains to check that
1 h1 2
y = τq−r−1(y ). Indeed, y δru = δr+1 implies τr(y )u = δ, and y δqu =
1 2 1 1 1 1 2 2
δq+1 implies τq(y )u = δ. Recall that u = τ(u ). So, we obtain δ = τ(δ) =
2 2 2 1
τ(τr(y )u ) = τr+1(y )u , whence τr+1(y ) = δu−1 = τq(y ).
1 1 1 2 1 2 2
Case 2.2.3. W ≡ Ay By Cu xˆu , V ≡ Ayˆ By Cτ−1(u ), V ≡ Ay Byˆ Cu .
1 2 1 2 1 1 2 2 2 1 2 1
Since e is of type (v2), we have C′u = δq. Hence, by Lemma 2, we have Cu ≡
2 1 1
C z C u where z matches u in Cu . Let V ≡ Ay By C zˆ C u . Then e =
1 1 2 1 1 1 1 3 1 2 1 1 2 1 3
(W,V ) is an edge of type (v2) and the pair (e ,e ) (resp. (e ,e )) satisfies the
3 1 3 3 2
conditions of Case 2.1 (resp. by Case 2.2.2).
Case 2.3. P and P have a common letter u = u = u, i. e., W ≡ Axˆ uxˆ .
1 2 1 2 1 2
Since x δ = x ux = δx , we have x = τ−1(x ). Hence the edge of type (v2)
1 1 2 2 2 1
defined by (P ,y ) coincides with e . The problem is reduced to Case 2.1.
1 2 2
Case 2.4. P and P are disjoint, y = u , and y = u . Then W ≡ AP BP ,
1 2 1 2 2 1 1 2
V ≡ Aτ−1(BPˆ ), V ≡ APˆ B, B′ = δq. If q = 0, then V ∼ V . Let q > 0.
1 2 2 1 1 h3 2
Let B ≡ CˆzD, z ∈ A and let xy = yz = δ. Since P C = τ−1(C)xy, we have
1
V ≡ Aτ−1(Cˆ)xˆyˆzD ∼ V . Then V ≡ Eτ−1(zDPˆ ) and V ≡ ExˆyˆzD for
3 h3 2 1 2 3
E ≡ Aτ−1(Cˆ). Since zD′ = (CˆzD)′ = B′ = δq, it follows from Lemma 2 that
zD includes a letter u which matches z. Let (V ,U ) be the edge of type (v2)
3 3
defined by (yˆz,u). Then D ≡ D uD and U ≡ Exˆτ−1(D uˆD ). We have D′ = δr
1 2 3 1 2 1
and zD′u = δr+1. Recall that zD′uD′ = δq which implies D′ = δq−r−1. Let
1 1 2 2
CRITERION FOR HURWITZ EQUIVALENCE FOR 3-BRAID 5
v = τq−r−1(u) and vw = δ. We may assume that P ≡ vw (otherwise we add an
2
edge of type (h3)). Then V ∼ V ≡ Eτ−1(zD uˆD vwˆ). Let us show that τ−1(z)
1 h1 4 1 2
matches τ−1(v) in V . Indeed, we have D′D′ = δrδq−r−1 = δq−1 and uD′ = D′v
4 1 2 2 2
(by the choice of v), whence zD′D′v = zD′uD′ = B′ = δq. Let (V ,U ) be the
1 2 1 2 4 4
edge of type (v2) defined by (τ−1(vwˆ),τ−1(z)). Then U ≡ Eτ−1(zˆD uˆD ). It
4 1 2
remains to note that xδ = xyz = δz, whence x = τ−1(z) and therefore U ≡ U .
3 4
Case 2.5. P , P , and y are pairwise disjoint: W ≡ D D where D ≡ AP B,
1 2 2 1 2 1 1
D ≡ y CP , C′ = δq, and D′ = δq+1. Then D′ = W′δ−q−1 is a power of δ. By
2 2 2 2 1
Lemma 2, this implies that D has a letter z which matches u . Let (W,V ) be
1 1 1 3
the edge of type (v2) defined by (P ,z ). Then we have V ∼ V (see Case 2.1),
1 1 1 h 3
V ≡ D E , and V ≡ E τ−1(D ) where D ∼ E in the corresponding graphs.
2 1 2 3 1 2 i v2 i
Hence V ∼ E τ−1(E ) ∼ V .
2 v2 1 2 v2 3
Case 3. e is of type (v3) and e is of type (v2) or (v3). Let W ∼ W ∼ V ,
1 2 h3 i v2 i
i = 1,2. Let P ∈ {xˆ u ,u xˆ } and y be the subwords of W used in the definition
i i i i i i i
of the edge (W ,V ). We may assume that W ≡ Eu xˆ Fˆ u and W ≡ Eu Fˆ xˆ u
i i 1 1 1 1 2 2 1 2 2 2
where x F = F x (otherwise the problem reduces to Case 2). By Lemma 1, we
1 1 2 2
have x F = δτ(D) = Dδ, D ∈ A∗. So, we may assume that x F ≡ x v τ(D) and
1 1 1 1 1 1
F x ≡ Dv x where x v = v x = δ. Without loss of generality we may assume
2 2 2 2 1 1 2 2
also that u x v ≡ σ σ σ .
1 1 1 2 1 0
Case 3.1. y = u and y = u . Then u (xˆ Fˆ )′u = u u = δ. Hence
1 2 2 1 1 1 1 2 1 2
W ≡ Eσ σˆ σˆ τ(Dˆ)σ and W ≡ Eσ Dˆσˆ σˆ σ , whence V ≡ Eτ−1(σˆ τ(Dˆ)σˆ ) ≡
1 2 1 0 1 2 2 0 2 1 1 0 1
Eσˆ Dˆσˆ ≡ V .
2 0 2
Case 3.2. E ≡ E y E . Then E′ = δq and y E′u = δq+1. Hence, by Lemma
2 1 1 1 1 1 1
4, there is a letter z in E which matches u . Let (W ,V ) be the edge of type
2 2 2 2 3
(v2) defined by (P ,z ). Then V ∼ V (see Case 2.1). By passing to a cyclic
2 2 2 h 3
permutation of W, we may assume that W ≡ Ay Bu xˆ vˆ τ(Dˆ)u Cz , W ≡
1 1 1 1 1 2 2 2
Ay Bu Dvˆ xˆ u Cz , B′ = δq, y Bu = δq+1, C′ = δr, u C′z = δr+1. For each
1 1 2 2 2 2 1 1 2 2
possible value of u , we find V such that V ∼ V ∼ V . (for u = σ ,σ we just
2 4 1 h 4 h 3 2 1 2
list the indices which should be inserted into the formulas for V , V , and V ).
1 3 4
u = σ :
2 0
V1 ≡ Aτ−q(σˆ0)Bτ−1(cid:0)σˆ0τ(Dˆ)σ0Cτr(σ2)(cid:1) ≡ Aτ−q(σˆ0)Bσˆ2Dˆσ2τ−1(C)τr(σ1),
V3 ≡ Aτ−q(σ0)Bσ2Dˆ σˆ2τ−1(cid:0)Cτr(σˆ2)(cid:1) ≡ Aτ−q(σ0)Bσ2Dˆσˆ2τ−1(C)τr(σˆ1),
V ≡ Aτ−q(σ )Bσˆ Dˆσ τ−1(C)τr(σˆ );
4 0 2 2 1
u = σ : V (ˆ0ˆ202), V (0ˆ2ˆ02), V (02ˆ0ˆ2); u = σ : V (ˆ0ˆ210), V (ˆ021ˆ0), V (02ˆ1ˆ0).
2 1 1 4 3 2 2 1 4 3
Case 4. e is of type (v1) and e is of type (v2). Let (W,V ) ≡ (AP ,A), P = δ.
1 2 1 1 1
Let P and y be the subwords of W used in the definition of e .
2
Case 4.1. P has one common letter with P: W ≡ AyBxˆuv, xu = uv = δ,
1
B′ = δq, yB′u = δq+1, V ≡ AyBxˆ, and V ≡ AyˆBτ−1(v). Then xδ = xuv = δv
1 2
implies τ−1(v) = x, and yδqu = δq+1 combined with xu = δ implies x = τq(y).
Hence V ∼ V .
1 v1 2
Case 4.2. P is disjoint from P. Lemma 4 implies A ≡ BzCPD where z matches
1
u. Let (W,V ) be the edge of type (v2) defined by (P,z). Then V ∼ V (see Case
3 2 h 3
2.1) and there exist vertical edges (V ,V ) and (V ,V ) where V ≡ BzˆCτ−1(D).
1 4 3 4 4
Case 5. e is of type (v1) and e is of type (v3). Let W ∼ W ∼ V . Then
1 2 h3 1 v2 2
we have W ∼ V which reduces the problem to Case 4. (cid:3)
1 v1 1
6 S. YU. OREVKOV
Lemma 10. Let e = (W ,W ) be a horizontal edge in G(X). If wt(W ) > 0, then
1 2 1
there exist vertical edges (W ,V ) and (W ,V ) such that V ∼ V .
1 1 2 2 1 h 2
Proof. The condition wt(W ) > 0 implies that W¯ has a subword P which is equal
1 1
to δ. By Lemma 8, it suffices to consider edges of type (h1) and (h3) only.
Case 1. e of type (h1). Let x and y be as in the definition of edges of type (h1).
It is clear that P 6≡ xy. If P, x, y are disjoint, then the statement is evident.
Case 1.1. W ≡ AyˆBux, W ≡ AyBuxˆ, ux = δ. Let V = AyˆB. Then
1 2 1
W ∼ V . Since e is an edge of type (h1), we have B′u = δq, x = τq(y). Hence,
1 v1 1
by Lemma 2, there is a letter z in Bu which matches u, i. e., B ≡ B zB , B′ = δr,
1 2 2
zB′u = δr+1. Hence W ∼ V ≡ AyB zˆB . Let us show that V ∼ V .
2 2 v2 2 1 2 1 h1 2
Indeed, B′zB′u = B′u = δq combined with zB′u = δr+1 imply B′ = δq−r−1, and
1 2 2 1
zδru = δr+1 combined with ux = δ imply x = τr+1(z). Since x = τq(y), we obtain
z = τq−r−1(y).
Case 1.2. W ≡ AyˆBxu, W ≡ AyBxˆu, xu = δ, B′ = δq, and yB′ = B′x. Then
1 2
yB′u = B′(xu) = δq+1, i. e., (W ,AyˆB) and (W ,AyˆB) are edges of type (v1) and
1 2
(v2) respectively.
Case 2. e of type (h3). Without loss of generality we may assume that W ≡
1
AuBˆ , W ≡ AuBˆ , u ∈ A, B ,B ∈ A∗, and B = B . Lemma 1 implies
1 2 2 1 2 1 2
B = δB. Let W ≡ AuxˆvˆBˆ where ux = xv = δ. Then W ∼ W ∼ W .
1 1 h3 h3 2
Let y matches u and let (W,V) be the edge of type (v2) defined by (uxˆ,y). Then
W ∼ V ∼ W . (cid:3)
1 v3 v3 2
Proof of Theorem 1. By Lemma 7, it is enough to show that: if W ∼ V and
wt(W) = wt(V), then W ∼ V. Indeed, if a path from W to V in G(X) passes
h
through vertices whose weight is greater than wt(W), then, by Lemmas 9 and 10,
we can modify the path so that either the number of vertices of the maximal weight
is reduced, or the number of horizontal edges incident to such vertices is reduced.
ACKNOWLEDGMENTS. This work was supported by the Russian Science
Foundation, project no. 14-21-00053.
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Steklov Math. Inst. (Moscow) and Univ. Paul Sabati´e (Toulouse-3)
E-mail address: [email protected]