On intersections of normal subgroups in free groups
Let N₁ (respectively N₂) be a normal closure of a set R₁ = {ui} (respectively R₂ = {vj}) of cyclically reduced words of the free group F(A). In the paper we consider geometric conditions on R₁ and R₂ for N₁ ∩ N₂ = [N₁, N₂]. In particular, it turns out that if a presentation < A | R₁, R₂ >...
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| Zitieren: | On intersections of normal subgroups in free groups / O.V. Kulikova // Algebra and Discrete Mathematics. — 2003. — Vol. 2, № 1. — С. 36–67. — Бібліогр.: 4 назв. — англ. |
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| citation_txt | On intersections of normal subgroups in free groups / O.V. Kulikova // Algebra and Discrete Mathematics. — 2003. — Vol. 2, № 1. — С. 36–67. — Бібліогр.: 4 назв. — англ. |
| collection | DSpace DC |
| container_title | Algebra and Discrete Mathematics |
| description | Let N₁ (respectively N₂) be a normal closure
of a set R₁ = {ui} (respectively R₂ = {vj}) of cyclically reduced
words of the free group F(A). In the paper we consider geometric
conditions on R₁ and R₂ for N₁ ∩ N₂ = [N₁, N₂]. In particular, it
turns out that if a presentation < A | R₁, R₂ > is aspherical (for
example, it satisfies small cancellation conditions C(p)&T(q) with
1/p + 1/q = 1/2), then the equality N₁ ∩ N₂ = [N₁, N₂] holds.
|
| first_indexed | 2025-12-01T08:15:38Z |
| format | Article |
| fulltext |
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h.Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 2. (2003). pp. 36–67
c© Journal “Algebra and Discrete Mathematics”
On intersections of normal subgroups in free
groups
O. V. Kulikova
Communicated by A. Yu. Olshanskii
Abstract. Let N1 (respectively N2) be a normal closure
of a set R1 = {ui} (respectively R2 = {vj}) of cyclically reduced
words of the free group F (A). In the paper we consider geometric
conditions on R1 and R2 for N1 ∩ N2 = [N1, N2]. In particular, it
turns out that if a presentation < A | R1, R2 > is aspherical (for
example, it satisfies small cancellation conditions C(p)&T (q) with
1/p + 1/q = 1/2), then the equality N1 ∩ N2 = [N1, N2] holds.
Introduction
Let F = F (A) be a free group generated by an alphabet A. Let N1
(respectively N2) be the normal closure of a set R1 = {ui} (respectively
R2 = {vj}) of cyclically reduced words of F . We will consider non-
intersecting symmetrized R1 and R2.
It is evident that the inclusion [N1, N2] ⊂ N1 ∩ N2 always holds.
But the reverse inclusion does not always hold (the simplest example is
R1 = {a1, a1
−1} and R2 = {a1, a2, a1
−1, a2
−1}). The aim of this paper is
to find necessary and sufficient conditions on R1 and R2 for
N1 ∩ N2 = [N1, N2]. (1)
These conditions are expressed in terms of certain geometric objects
called pictures (see, for example [4]).
2000 Mathematics Subject Classification: 20F05, 20F06.
Key words and phrases: normal closure of words in free groups, presentations
of groups, pictures, mutual commutants, intersection of groups, aspherisity, small can-
cellation conditions.
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h.O. V. Kulikova 37
In particular, it turns out that if a presentation < A | R1, R2 > is as-
pherical (for example, it satisfies small cancellation conditions C(p)&T (q)
with 1/p + 1/q = 1/2 ([4], theorem 2.2)), then the equality N1 ∩ N2 =
[N1, N2] holds.
The paper is divided into three sections, each of which is further
subdivided. In the first section we give main definitions, prove some
results concerning relations between them, formulate the main result of
the paper (Theorem 1) and prove corollaries of it. The second section is
devoted to the proof of Theorem 1. In the third section we prove some
simple corollaries of Theorem 1 in the case of free products.
It should be noted that a geometric test of the equality (1) obtained
in Theorem 1 is hard to verify, but its corollaries are useful. The following
question seems to be open: if the equality (1) holds if and only if there
exist sets of words R̃1 and R̃2 such that
(i) R̃F
1 = N1 and R̃F
2 = N2;
(ii) the presentation G =< A | R̃1 ∪ R̃2 > is strictly (R̃1, R̃2)-separable
(see Definition 1 below).
I am very grateful to my scientific advisor professor A. Yu. Ol’shanskii
for his constant attention to my work. Also I am grateful to A. A. Kly-
achko and D. V. Osin for conversations during preparation of this paper.
1. Formulation and corollaries of Theorem 1.
In the beginning we give some definitions and recall the definition of
pictures.
1.1 Main definitions. Relations between definitions.
Let N be the normal closure of a set R of cyclically reduced words of the
free group F (A).
A picture P over presentation G =< A | R > on a surface S ( see
details in [4] ) is a finite collection of ”vertices” V1, ..., Vn ∈ S, together
with a finite collection of simple pairwise disjoint connected oriented
”edges” E1, ..., Em ∈ S \ {{V1, ..., Vn} ∪ ∂S}, labelled by letters of A (in
[4], vertices are called discs and edges are called arcs). But these edges
need not all connect two vertices. An edge may connect a vertex to a
vertex (possibly coincident), a vertex to ∂S, or ∂S to ∂S. Moreover,
some edges need have no endpoints at all, but be circles disjoint from
the rest of P , such edges are called edges-circles. If an edge connects two
vertices, then one is the start of the edge, the other one is the end of the
edge.
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h.38 On intersections of normal subgroups in free groups
For each vertex V of P consider a circle C of a small radius with
center at V and a point p on C not lying on any edge of P . The labels
of edges intersected by C starting at p form a word r ∈ R. Changing of
the disposition of p on C and the direction of moving around C, we can
read any cyclic permutation of r and r−1.
Below we will consider pictures on S, where S is a sphere (spherical
pictures) or a disk (planar pictures).
In the case of a planar picture the labels of edges, intersected by a
circle C̄ near the boundary of the disk ∂S, starting at a point p̄ on C̄,
form a word W , which will be called the boundary label of the picture.
The following result is well-known (use Theorem 11.1 of [1] and du-
alise):
Lemma 1. Let W be a non-empty word on the alphabet A. Then W
represents the identity of the group G = F/N if and only if there is a
planar picture over the presentation G =< A | R > with the boundary
label W .
Let P be a picture over G =< A | R > and γ be a path on S not
passing through any vertex of P . If we travel around γ we encounter
a succession of edges. Reading the labels on these edges gives a word
called the word along the path γ and denoted by Lab(γ).
If γ is closed, consider a point p of γ not lying on any edge of P . The
word along γ read from p will be denoted by Labp
+(γ) or by Labp
−(γ)
(depending as the direction of reading is counterclockwise or not). It is
clear that Labp
+(γ)
−1
= Labp
−(γ). Changing the disposition of p we
obtain the same word up to cyclic permutation. Changing the direction
of reading we obtain the inverse word. We will denote the word along the
path γ by Lab(γ) when the disposition of p and the direction of reading
are not essential.
By 1 denote the identity element of the free group.
A dipole in a picture P over G =< A | R > is two vertices V1 and
V2 of P if there is a simple path ψ connecting points p1 and p2 lying on
circles C1 and C2 around these vertices such that the following conditions
hold:
(i) Lab(ψ) = 1;
(ii) Labp1
+(C1) = Labp2
−(C2).
A picture over G =< A | R > is reduced if it does not contain a
dipole.
A presentation G =< A | R > is aspherical if every connected spher-
ical picture over G =< A | R > contains a dipole.
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h.O. V. Kulikova 39
Definition 1. Let R1 and R2 be two sets of words in the free group F (A).
We say that a presentation G =< A | R1 ∪ R2 > is strictly (R1, R2)-
separable or satisfies the condition of strict (R1, R2)-separability if for
every reduced spherical picture P containing both R1-vertices and R2-
vertices there is a simple closed path γ dividing the sphere into two disks
such that the following conditions hold:
1) the both disks contain vertices;
2) Lab(γ) = 1.
Assertion 1. If in Definition 1 ”every reduced spherical picture” is re-
placed by ”every spherical picture”, then the class of presentations satis-
fying strict (R1, R2)-separability is not changed.
Proof. Let P be a non-reduced spherical picture over G =< A | R1 ∪
R2 > containing both R1-vertices and R2-vertices. Since P is not re-
duced, there is a dipole, i.e., there are two vertices V1 and V2 that inverse
words from R1 ∪ R2 correspond to, and V1 and V2 can be connected by
a simple path ψ such that Lab(ψ) = 1. It is easily seen that the simple
closed path γ from Definition 1 may be obtained going around V1 and
V2 and by-passing near ψ in the both directions.
Definition 2. Let R1 and R2 be two sets of words in F (A). We say
that a presentation G =< A | R1 ∪R2 > is weakly (R1, R2)-separable or
satisfies the condition of weak (R1, R2)-separability if for every reduced
spherical picture P containing both R1-vertices and R2-vertices there is
a simple closed path γ dividing the sphere into two disks such that the
following three conditions hold:
1) the both disks contain vertices;
2) Lab(γ) ∈ [N1, N2];
3) if one of the disks contains only R1-vertices, then the other one
contains only R2-vertices.
Assertion 2. If the condition 3) in Definition 2 is omitted, then the
class of presentations satisfying only 1) and 2) of Definition 2 will be
wider.
Proof. As a counterexample proving the assertion, one can consider a
presentation G =< A | R1 ∪ R2 >, where R1 is a symmetrized set of
words {a1, a3, [a1, a2]} and R2 is a symmetrized set of words
{a2, a3[a1, a2]}, where {ai} ∈ A. This presentation satisfies the condi-
tions 1), 2) of Definition 2, but it does not satisfy the condition 3).
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Indeed, let P denote a reduced spherical picture containing both both
R1-vertices and R2-vertices.
If there is an R1-vertex labelled by [a1, a2] ∈ [N1, N2] in P , then the
path γ is obtained going around this vertex.
If, in P , there is not any R1-vertex labelled by [a1, a2] and there is an
R1-vertex labelled by a3, then the edge starting at this a3-vertex must
have the end at an R2-vertex labelled by a3[a1, a2], since P is reduced.
Thus the path γ is obtained going around this (a3)-vertex, the edge
starting at this a3-vertex and the (a3[a1, a2])-vertex.
If in P there are neither R1-vertices labelled by [a1, a2] nor R1-vertices
labelled by a3, then P is non-reduced. Hence the conditions 1), 2) of
Definition 2 hold.
m
} }¢
¢
¢
¢
¢
¢
¢
¢
¢
¢̧
A
A
A
A
A
A
A
A
A
AK
A
A
A
A
A
A
A
A
A
AKA
A
A
A
A
A
A
A
AU
A
A
A
A
A
A
A
A
AAU
a3 [a1, a2]
a−1
3 [a1, a2]
−1
Fig. 1
But the presentation does not satisfy the condition 3), which is shown
by the picture containing only three vertices: an (a3)-vertex, an ([a1, a2])-
vertex and an (a3[a1, a2])-vertex (see Fig.1), since a3, a3[a1, a2] 6∈ [N1, N2].
Assertion 3. If a presentation G =< A | R1∪R2 > is strictly (R1, R2)-
separable, then it is weakly (R1, R2)-separable.
Proof. Let P be a reduced spherical picture over strictly (R1, R2)-
separable G =< A | R1 ∪ R2 > containing both R1-vertices and R2-
vertices. It is sufficient to find a simple closed path γ dividing the sphere
into two disks such that
1) the both disks contain vertices;
2) Lab(γ) = 1;
3) one of the disks contains only R1-vertices and the other one contains
only R2-vertices.
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h.O. V. Kulikova 41
It follows from the condition of strict (R1, R2)-separability that there
is a simple closed path γ1 in P dividing the sphere into two disks such
that the following conditions hold:
1) the both disks contain vertices;
2) Lab(γ1) = 1.
If at least one of these disks both contains R1-vertices and R2-vertices,
then it follows from the condition of strict (R1, R2)-separability that
there is another simple closed path γ2 in P non-crossing the path γ1 and
dividing such disk into two subdisks such that the following conditions
hold:
1) the both subdisks contain vertices;
2) Lab(γ2) = 1.
We continue in this fashion to obtain a finite set of simple closed
paths Γ = {γi} satisfying the following conditions:
1) these paths are pairwise disjoint;
2) Lab(γ) = 1 for each γi ∈ Γ;
3) the union of the paths γi ∈ Γ divides the sphere into parts {Dk}
each of which contains only R1-vertices or only R2-vertices.
Each path of Γ separates one part of {Dk} from another one. If
any path γi ∈ Γ separates one part of {Dk} from another one so that
the both parts contain either only R1-vertices or only R2-vertices, then
we will remove this path γi from Γ. We thus get that each path of Γ
separates a part of {Dk} containing only R1-vertices from another part
containing only R2-vertices.
Below we will transform each not simply connected part Di of {Dk}
in order to decrease the number of paths in Γ. Since Di is not simply
connected, there are at least two paths γi and γj of Γ bounding the part
Di. Join the paths as follows. Fix a point a+
i on the path γi and a point
a+
j on γj so that a+
i and a+
j do not belong to any edge of the picture
P . It is clear that the points a+
i and a+
j may be joined by a simple path
ψ+
(i,j) so that
1) the whole path ψ+
(i,j) lies in Di;
2) ψ+
(i,j) does not pass through any vertex of P ;
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h.42 On intersections of normal subgroups in free groups
3) ψ+
(i,j) does not intersect the paths of Γ except γi and γj at the
points a+
i and a+
j .
It is possible to draw a path ψ−
(i,j) through points a−i ∈ γi and a−j ∈ γj
close to the path ψ+
(i,j), in a parallel way with the properties similar to
ψ+
(i,j) so that ψ−
(i,j) intersects the same edges of P as ψ+
(i,j) does, and
that ψ+
(i,j) and ψ−
(i,j) are disjoint. Besides ψ−
(i,j) can be drawn so that the
segment [a+
i , a−i ] of γi and the segment [a+
j , a−j ] of γj do not intersect
the edges of P . Removing these segments gives a new simple closed path
γij = (γi/[a+
i , a−i ]) ∗ ψ+
(i,j) ∗ (γj/[a+
j , a−j ]) ∗ ψ−
(i,j) such that Lab(γij) = 1.
Replacing γi and γj by γij in Γ gives a set of paths satisfying the same
properties 1), 2), 3). Besides the number of paths in the resulting set of
paths becomes less than in the original set.
Consequently after a finite number similar transformations, all parts
from {Dk} become simply connected. Therefore the resulting set Γ con-
tains just one path, which is the desired path γ.
Assertion 4. The condition of weak (R1, R2)-separability is not equiva-
lent to the condition of strict (R1, R2)-separability.
Proof. As an example proving the assertion one can consider a presen-
tation G =< A | R1 ∪ R2 >, where R1 is a symmetrized set of words
{a1, [a1, a2]} and R2 = {a2, a2
−1}, where {ai} ∈ A.
It follows from Theorem 1 and Corollary 5 (see below) that this pre-
sentation is weakly (R1, R2)-separable since N1 ∩ N2 = [N1, N2].
Let us show that the existence of a spherical picture P containing
only an [a1, a2]-vertex and two a2-vertices contradicts the condition of
strict (R1, R2)-separability.
Indeed, suppose that there is a simple closed path γ dividing the
sphere into two disks. Then one of the disks must contain only one vertex.
Consequently Lab(γ) must be equal to the label of this vertex, which is
not equal to the identity element in the free group. This contradicts the
definition of strict (R1, R2)-separability.
Assertion 5. If every spherical picture over a presentation G =< A |
R1 ∪ R2 > containing both R1-vertices and R2-vertices is not reduced
(this condition will be called (R1, R2)-asphericity), then the presentation
is strictly (R1, R2)-separable.
Proof. is similar to the proof of Assertion 1.
Definition 3. Let R1 and R2 be two sets of words in F (A). We say
that a presentation G =< A | R1 ∪ R2 > is (R1, R2)-separable or sat-
isfies the condition of (R1, R2)-separability if for every spherical picture
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P containing both R1-vertices and R2-vertices there is a simple closed
path γ dividing the sphere into two disks such that the following three
conditions hold:
1) the both disks contain vertices;
2) Lab(γ) ∈ [N1, N2];
3) one of the disks contains only R1-vertices and the other one con-
tains only R2-vertices.
1.2 Formulation of Theorem 1 and corollaries from it.
Theorem 1. A presentation G =< A | R1 ∪ R2 > is weakly (R1, R2)-
separable if and only if N1 ∩ N2 = [N1, N2].
Corollary 1. The conditions of weak (R1, R2)-separability and (R1, R2)-
separability are equivalent.
Proof. It is easy to see that if a presentation is (R1, R2)-separable, then
it is weakly (R1, R2)-separable.
It remains to prove the converse statement. Let P be a spherical
picture over a weakly (R1, R2)-separable presentation G =< A | R1 ∪
R2 > containing both R1-vertices and R2-vertices. It is evident that
there is a simple closed path γ not passing through any vertex of P and
dividing the sphere into two disks one of which contains only R1-vertices
and the other one contains only R2-vertices. Hence Lab(γ) ∈ N1 ∩ N2.
Since G =< A | R1 ∪ R2 > is weakly (R1, R2)-separable, Theorem 1
leads to Lab(γ) ∈ [N1, N2]. Consequently γ is desired.
Corollary 2. The conditions of weak (R1, R2)-separability and weak
(R2, R1)-separability are equivalent. Moreover we get the equivalent con-
dition if in Definition 2 the item 3) is replaced by
3′) if one disk contains both R1- both R1- and R2-vertices, then the other
one both R1- and R2-vertices.
Corollary 3. Let a presentation G =< A | R1 ∪R2 > satisfy one of the
following conditions:
(i) strict (R1, R2)-separability;
(ii) (R1, R2)-asphericity;
(iii) asphericity.
Then N1 ∩ N2 = [N1, N2].
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Proof. (i) It follows directly from Theorem 1 and Assertion 3.
(ii) It follows directly from (i) and Assertion 4.
(iii) It follows from (ii).
Corollary 4. Let {R1, R2} be a set of words satisfying one of the fol-
lowing small cancellation conditions: either C(6), or C(4)&T (4), or
C(3)&T (6). Then N1 ∩ N2 = [N1, N2].
Proof. According to [4] (see also [3]), every small cancellation condition
described above is sufficient to asphericity of G =< A | R1 ∪ R2 >.
Corollary 5. Suppose that:
1) an alphabet A = X ⊔ Y ⊔ Z;
2) R1 is an arbitrary set of words on X and R2 is an arbitrary set of
words on Y .
Then N1 ∩ N2 = [N1, N2].
Proof. Without loss of generality, one can assume that R1 and R2 are
symmetrized.
By Corollary 3 it is sufficient to show that G =< A | R1 ∪ R2 > is
strictly (R1, R2)-separable.
Let P be a reduced spherical picture containing both both R1-vertices
and R2-vertices.
If there is an edge-circle in P dividing the sphere into two disks each
of which contains vertices, then Corollary 5 is proved (the simple closed
path can be drawn near this edge-circle).
If there is an edge-circle in P dividing the sphere into two disks one
of which does not contain vertices, then this edge-circle can be removed
from P .
Therefore we can suppose that there is no edge-circle, hence each
edge connects a vertex to a vertex.
Since R1 and R2 are written on the disjoint alphabets X and Y ,
we conclude that if any edge starts at an R1-vertex, then it ends at an
R1-vertex (similarly for R2-vertices).
For an R1-vertex, consider a connected component of P containing
this vertex. All vertices of this component are R1-vertices. Since P
also contains R2-vertices, the complement of this component also con-
tains vertices and falls into connected components. Among all these
connected components there is at least one, which can be covered by a
domain homeomorphic to a disk such that this domain does not inter-
sect the other connected components. Then the boundary of the domain
forms the desired simple closed path γ. Lab(γ) = 1, since no edge of P
intersects γ, and the statement follows.
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2. Proof of Theorem 1.
In the beginning we give several definitions, which we will use in the
proof of Theorem 1.
2.1 Additional definitions.
1) Picture P with equator Equ. Subpictures of P .
Let P be a picture on the sphere S2 over a presentation G =< A |
R1∪R2 >. Fix a simple closed path (denoted by Equ) on S2 not passing
through any vertex of P and dividing S2 into two parts so that one
part does not contain R1-vertices and the other one does not contain
R2-vertices. The path Equ is called an equator.
Lab(Equ) is denoted by W or W−1 (the sign depends on the direction
of reading). By the choice of the equator, it follows that W ∈ N1 ∩ N2.
Remark 1. The start point on Equ is not fixed, i.e., the equatorial
label W is considered as a cyclic word (if W belongs to some normal
subgroup, then all its cyclic permutations belong to the same subgroup).
For simplicity of notation all cyclic permutations of W will be denoted
again by W .
In the sequel, P denotes a picture with a fixed equator Equ.
Let P = P ′ ⊔ P ′′ be a disjoint union of two spherical pictures. P ′
(respectively, P ′′) will be called a subpicture of P . P ′ (respectively, P ′′)
may be both connected and disconnected.
2) Boundaries of vertices. North and south vertices.
Every vertex is labelled by a word of R1 (R1-word) or R2 (R2-word).
Consider a small disk with centre at a vertex such that the word along
its boundary coincides with the label of the vertex. The boundary of the
disk will be called the boundary of the vertex.
Equ divides the sphere into two hemispheres: so called north and
south. A vertex is called north (respectively, south) if it lies in the north
(respectively, south) hemisphere.
Suppose that R1-words correspond to the north vertices (so called
R1-vertices), R2-words correspond to the south vertices (so called R2-
vertices). Each word can be read along the boundary of the correspond-
ing vertex starting at some point on the boundary and choosing the
direction of reading.
3) Admissible moves.
In the proof of Theorem 1 we will transform a picture P with an equator
Equ on S2. A move (i.e., a transformation) is called admissible if, after
the move, the word W along Equ is replaced by a word W ′ equal to
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W to within an element of [N1, N2] (for simplicity of notation, we will
use the same letter W for the notation W ′). Moreover admissible moves
preserve the subdivision of P by Equ into the north and south vertices.
4) Map.
Each domain U ⊂ S2 homeomorphic to a square
{(x, y) ∈ R
2 | −1 < x < 1,−1 < y < 1 },
together with vertices and parts of edges lying in it is called a map.
5) Components.
Each spherical subpicture of P is called a component if it contains ver-
tices. A component is called reduced (respectively, non-reduced) if the
corresponding subpicture is reduced (respectively, non-reduced). A com-
ponent is called north (respectively, south) if the corresponding subpic-
ture contains only north (respectively, only south) vertices.
South and north components are called uniform. A component is
called mixed if the corresponding subpicture contains both south and
north vertices. Note that P is a component of itself.
6) States. Territories of states. Boundaries of states.
A component is called a state if it is uniform and can be covered by a
closed domain homeomorphic to a disk so that the domain does not inter-
sect the other components of P . The state is called north (respectively,
south) if the corresponding uniform component is north (respectively,
south).
The domain mentioned above is called a territory of the state. The
boundary of the territory is called the boundary of the state. Note that
the boundary of each state is homeomorphic to a circle. Without loss
of generality we can assume that the boundary of each state intersects
Equ in a finite set of points the number of which can not be decreased
by changing the territory.
7) Pieces of equator Equ.
Let T be a state. Equ intersects its boundary 2m times and is divided
into 2m connected parts by these intersection points. Those of parts
which lie on the territory of T are called pieces of the equator.
8) Regions of states.
Let T be a north state. Equ divides the territory of T into connected
parts, which are called regions north or south depending on what hemi-
sphere they belong to. A north region of a north state is called regular
if its boundary consists exactly of two parts: a connected part belonging
to the boundary of T and a piece of the equator. Respectively, a north
region of a north state is called irregular if it is not regular. Similarly,
one can define regions and regular regions of south states by replacing
the word ”north” by ”south”.
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9) σ-state.
A north state is called a north σ-state if all its north regions are regular.
(South σ-states are defined similarly by replacing the word ”north” by
”south”.)
10) σ-picture.
A picture with a fixed equator is called σ-picture if it can be reduced to a
picture containing only σ-states by a finite number of admissible moves.
2.2 Proof of Theorem 1.
The proof of a statement that weak (R1, R2)-separability implies N1 ∩
N2 = [N1, N2] is similar to the proof of Corollary 1.
Therefore it remains to prove the converse statement. Let a presen-
tation G =< A | R1 ∪ R2 > satisfy weak (R1, R2)-separability.
Since the inclusion [N1, N2] ⊂ N1 ∩ N2 always holds, it is sufficient
to prove the reverse inclusion.
Let W be an arbitrary word of the intersection N1 ∩N2. Then there
are two representations:
W =
∏
k
gkr1,kgk
−1, since W ∈ N1; (2)
and
W−1 =
∏
l
hlr2,lhl
−1, since W ∈ N2, , (3)
where gk, hl are words of F , each r1,k belongs to R1, and each r2,l belongs
to R2. To show that W ∈ [N1, N2], let us construct two planar pictures.
The word W in the form (2) is written on the boundary of the first
picture which contains only R1-vertices. The word W−1 in the form (3)
is written on the boundary of the second picture which contains only R2-
vertices. Pasting together the planar pictures by their boundaries gives
a picture P on S2 with a fixed equator Equ. Lab(Equ) is equal to W or
W−1 depending on the direction of moving along Equ.
Now the proof of Theorem 1 follows from the following Propositions
1 and 2, which will be proved in the subsections 2.4 and 2.5 below:
Proposition 1. Let a picture P with a fixed equator Equ be over a
presentation G =< A | R1 ∪ R2 >. If the presentation satisfies weak
(R1, R2)-separability, then P is a σ-picture.
Proposition 2. Let a picture P with a fixed equator Equ be over a
presentation G =< A | R1 ∪R2 >. If P is a σ-picture, then the word W
along Equ can be reduced to the identity element in the free group by a
finite number of admissible moves.
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Indeed, by Propositions 1, 2, there is a finite sequence of admissible
moves, which reduces the equatorial label W in P to the identity element
in the free group. Since admissible moves replace W by words equal
to W to within elements of [N1, N2], we have that W ∈ [N1, N2], as
claimed.
2.3 Some admissible moves. Auxiliary lemmas.
In the subsection 2.3 we describe admissible moves, which will be used
in the proof of Propositions 1 and 2.
1) Isotopy.
An isotopy of P is defined by replacing P by a picture F1(P ), where
Ft : S2 × [0, 1] → S2 × [0, 1] is a continuous isotopy of the sphere S2 such
that
(i) Ft leaves fixed all vertices, i.e., Ft(Vi) = Vi for t ∈ [0, 1] and for
each vertex Vi;
(ii) for each t and each edge Ej the intersection of an edge Ft(Ej) and
Equ consists of a finite number of points, moreover the edge F1(Ej)
intersects Equ transversally at every intersection point.
It is evident that an isotopy of P is an admissible move because either
it corresponds to an insertion or a cancellation of pairs of inverse letters
in the equatorial label W (see Fig.2) or it does not change W at all.
-Equ Equ?
?
6
6
½¼
½¼
a
a
a
a
Fig. 2
2) Bridge moves.
Assume that a map U contains only two edges
{x = −1/2,−1 < y < 1} and {x = 1/2,−1 < y < 1},
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which are contrariwise oriented and labelled by the same letter. A move
of P is called a bridge move (see also in [3]) if it does not change P out
of the map U and change P in U as is shown on Fig 3.
-
?
6
?
?
6
6
½¼¾»
a a
a a
a
a
Fig. 3
The bridge move is an admissible move because either it corresponds
to an insertion or a cancellation of pairs of inverse letters in the equatorial
label W or it does not change W at all.
3) Removing components and edge-circles of P not intersecting Equ.
If a component (in particular, a state) of P does not intersect Equ,
then it does not contribute to the equatorial label W . Therefore such
component can be removed. Similarly one can remove edge-circles not
intersecting Equ.
4) Removing superfluous loops.
Assume that Equ intersects any edge in two successive points which
divide Equ into two parts such that one of these two parts does not
intersect edges. Such part of the edge is called a superfluous loop. It is
evident that superfluous loops do not contribute to the equatorial label
W (considered as an element of the free group). Therefore superfluous
loops can be removed (see Fig. 2).
( This move is a special case of isotopy (see the move 1)) or a com-
position of the admissible moves 2) and 3).)
5) Uniting σ-states.
Let T1 and T2 be two distinguished north σ-states (the move of south
states is similar). Assume that there are points p1 on the boundary of
T1 and p2 on the boundary of T2 so that
(i) the points p1 and p2 lie in the south hemisphere;
(ii) it is possible to connect the points p1 and p2 by a simple path η
which does not intersect the territory of any state and lies in the
south hemisphere as a whole.
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Then the territories of the σ-states T1 and T2 can be united in one
by adding a small neighborhood of the path η (see Fig.4).
-Equ Equ
b bb bb b
¢¢ ¢¢AA AA
b bb b
Fig. 4
It is clear that the united territory is homeomorphic to a disc again.
The σ-states T1 and T2 lying in the united territory form one state, which
is a σ-state again.
Remark 2. By a deformation of territories of σ-states, we can exclude
the case when the territory of any σ-state successively intersects Equ in
two pieces I1 and I2 so that a part of Equ between I1 and I2 does not
intersect the territory of any state.
6) Pasting a map with a commutator subpicture in P .
In notation of the subsection 2.5, assume that there are two regular
regions: one of them belongs to a north state T1, the other one belongs
to a south state T2. By Lemma 2 below, the north region contains a
picture with a word w1 ∈ N1 written along a piece I2 of the equator,
the south region contains a picture with a word w2 ∈ N2 written along
a piece I3 of the equator.
We choose a map M1 so that the north region is contained in {0 ≤
y ≤ 1}, I2 ⊂ {y = 0}, and the intersection of T1 and {−1 < y <
0,−1 < x < 1} contains only parts of edges given in coordinates (x, y)
by {x = x′
1,−1 < y < 0}, ..., {x = x′
n′ ,−1 < y < 0}.
Similarly we choose a map M2 so that the south region is contained
in {0 ≥ y ≥ −1}, I3 ⊂ {y = 0}, and the intersection of T2 and {0 < y <
1,−1 < x < 1} contains only parts of edges given in coordinates (x, y)
by {x = x1, 0 < y < 1}, ..., {x = xn, 0 < y < 1}.
For the map M1 ( respectively, for M2) we construct a mirror-like map
M ′
1 (respectively, M ′
2) by reflecting M1 (respectively, M2) with respect
to the axis {x = 0} and by changing the orientations of the edges. The
map M ′
1 (respectively, M ′
2) contains the picture with the word w1
−1
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(respectively, w2
−1) written along the piece of the equator. The piece
of the equator in the map M ′
1 (respectively, M ′
2) will be denoted by I ′2
(respectively, I ′3).
M
M2 M1
M′
2 M′
1
bb bb bb
¢¢¢¢ AAAA
?
6
& %
rr rr
6 6
? ?
' $' $
Fig. 5
A map M containing a picture for the word w2w1w2
−1w1
−1 is con-
structed as follows:
(i) the equator passes through {y = 0};
(ii) a part corresponding to the south hemisphere is {0 ≥ y ≥ −1} and
a part corresponding to the north hemisphere is {0 ≤ y ≤ 1};
(iii) the map M2 is disposed in the rectangle {−1/5 < y < 1/5,−4/5 <
x < −3/5}; the map M1 is disposed in {−1/5 < y < 1/5,−2/5 <
x < −1/5}; the map M ′
2 is in {−1/5 < y < 1/5, 1/5 < x < 2/5};
the map M ′
1 is in {−1/5 < y < 1/5, 3/5 < x < 4/5};
(iv) in the part corresponding to the south hemisphere, we extend the
edges of the map M1 to join them with the corresponding edges of
M ′
1; the union of the pictures of these maps and the joining edges
forms a north σ-state which will be denoted by T comm
1 ;
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(v) in the part corresponding to the north hemisphere, we extend
the edges of the map M2 to join them with the corresponding
edges of M ′
2; the union of the pictures of these maps and the join-
ing edges forms a south σ-state which will be denoted by T comm
2 .
(See Fig. 5)
It is clear that the word along the equator of the map M
is w2w1w2
−1w1
−1.
A small map Ms in P containing nothing but a part {y = 0} of Equ
is replaced by the constructed map M . This move is admissible because
it corresponds to an insertion of the commutator w2w1w2
−1w1
−1 of the
elements from N1 and N2 in the equatorial label W .
Lemma 2. Let T be a state in P . Let I be an arbitrary piece of the
equator belonging to the territory of T . Then Lab(I) ∈ N1 if T is north,
and Lab(I) ∈ N2 if T is south.
Proof. Let T be north (the proof in the case of a south state is similar).
The piece I divides the territory of T into two parts T ′ and T ′′. We
consider one of them denoted by T ′. T ′ may contain only north vertices
(i.e., R1-vertices) and edges labelled by letters of the alphabet A. Hence
T ′ contains a planar picture over < A | R1 >. By Lemma 1, a word along
the boundary of T ′ belongs to N1. Since the edges intersect the boundary
of T ′ only in a part which coincides with I, Lemma 2 follows.
Lemma 3. Let T be a state in P . If Equ intersects the boundary of
T exactly two times, then the word along the piece of the equator lying
inside the territory of T is equal to the identity element in the free group.
Proof. Assume that T is north (the proof in the case of a south state
is similar). The given piece of the equator divides the territory of T
into two regions: north and south. Moreover all vertices lie in the north
region. Hence each edge intersects Equ even times. Clearly, there is an
edge a part of which forms a superfluous loop in the south hemisphere.
Removing superfluous loops (see the admissible move 4)) gives rise to
the case when no edge of T intersects Equ. Therefore the original word
along the piece of the equator lying inside the territory of T was equal
to the identity element in the free group.
2.4 The proof of Proposition 1.
The proof of Proposition 1 will be divided into several steps (lemmas).
In Step 1 we will show that the picture P with the fixed equator Equ
can be divided into a finite number of uniform components. In Step 2
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the uniform components will be transformed to states and edges-circles
not belonging to the states. In Step 3 we will get rid of the edges-circles
not belonging to the states. In Step 4 the states will be divided into σ-
states. In all steps we will use only a finite number of admissible moves.
Therefore we will get that P is a σ-picture.
All pictures obtained from P will be denoted by P again for simplicity
of notation.
Step 1. Reducing P to a picture containing only uniform components.
In Step 1 we will use the following two admissible moves.
Operation A: Transformations of reduced mixed components.
Let the presentation G =< A | R1 ∪R2 > be weakly (R1, R2)-separable.
By K denote a reduced mixed component.
Since K is a reduced spherical picture, the condition of weak (R1, R2)-
separability leads to the existence of a simple closed path γ dividing the
sphere into two parts so that
1) the both parts contains vertices;
2) U = Lab(γ) ∈ [N1, N2];
3) if one part contains only north vertices then the other one contains
only south vertices.
By the property 3) of γ, the following three cases are possible.
The first case: the path γ divides K into two parts one of which
contains only south vertices, the other one contains only north vertices.
The second case: γ divides K into two parts one of which contains only
south vertices, the other one contains both south and north vertices. In
these two cases we can assume that some segment ψ of the path γ lies
on Equ. The complement of ψ to γ will be denoted by ¬ψ. One of the
endpoints of ψ will be denoted by p.
The third case: the path γ divides K into two parts each of which
contains both south and north vertices. Consequently, γ is intersected by
Equ and divided by Equ into segments among which there are segments
lying in the north hemisphere wholly. Fix one of them. By ψ denote its
connected part not intersecting Equ. By p denote one of the endpoints
of ψ. By ¬ψ denote the complement of ψ to γ.
In each of these three cases one can assume that all edges intersecting
the path γ intersect it in the segment ψ, because otherwise all edges
intersecting ¬ψ can be moved by isotopy ( the admissible move 1) ) to
ψ along the path γ in the direction of the point p starting successively
at the nearest to p edge.
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In each of these three cases we select a map M on S2 with the follow-
ing properties: the map M contains the segment ψ of γ (coinciding with
the part of Equ in the first two cases) and parts of edges intersecting ψ:
more precisely,
1) {y = 0} is the segment ψ of the path γ;
2) {x = x1}, ..., {x = xn} correspond to the edges, which intersect the
segment ψ;
3) in the first two cases the rectangle {−1 < x < 1, y < 0} belongs
to the south hemisphere and the rectangle {−1 < x < 1, y >
0} belongs to the north hemisphere; in the third case the both
rectangles belong to the north hemisphere. (An example for the
third case is shown on Fig. 6 )
A new map M ′ will be constructed as follows. Since [N1, N2] ⊂
N1∩N2, the word U along ψ belongs to the both groups N1 and N2. We
construct planar pictures P1 and P2 with the boundary labels respectively
U and U−1. Moreover in the first two cases P1 is constructed over < A |
R2 > (using south vertices) and P2 is constructed over < A | R1 >
(using north vertices); in the third case the both pictures P1 and P2 are
constructed over < A | R1 > (using north vertices). Then these pictures
are disposed on the new map M ′ as follows:
1) P1 lies in the rectangle {−1 < y < −1/2,−1 < x < 1}; and the edges
corresponding to the boundary of P1 start at (x1,−1),...,(xn,−1);
2) P2 lies in the rectangle {1/2 < y < 1,−1 < x < 1}; and the edges
corresponding to the boundary of P2 start at (x1, 1),...,(xn, 1);
3) {y = 0} is the segment ψ of γ (coinciding with the part of Equ in the
first two cases).
The old map M is cut out from P and replaced by the new one M ′.
(An example for the third case is shown on Fig. 7.) By this replacing of
maps the equatorial label W is changed by the commutator from [N1, N2]
in the first two cases and it is not changed in the third case. After such
replacing all R1-vertices lie in the north hemisphere and all R2-vertices
lie in the south one. Therefore this move of P is admissible.
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Equ
' $γ
? ?
6
(−1, 1)
(−1,−1) (1,−1)
(1, 1)
x1 x2 x3
a
b
c
Fig. 6
Equ
' $γ
?
?
6
6
?
?
e
e
e
e
µ
¶ ¯
°(−1, 1)
(−1,−1) (1,−1)
(1, 1)
x1
x1
x2
x2
x3
x3
a
b
c
Fig. 7
By such replacing of maps the component K falls into two compo-
nents K1 and K2 separated from each other by the path γ. In the first
case each of the components Ki is uniform; in the second case one of the
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components (let it be denoted by K1) is mixed, the other one (K2) is
south uniform; in the third case each of the components Ki is mixed.
Remark 3. The components Ki can be non-reduced. The number of
south vertices in each mixed component Ki (K1 in the second case; K1
and K2 in the third case) is strictly less than the number of south vertices
in the original component K, since by Operation A we added only north
vertices to obtain the mixed components.
Operation B: Transformations of non-reduced mixed components.
Let K be a non-reduced component. Then there is a dipole in K, i.e.,
there are two vertices V ′ and V ′′ satisfying the following conditions:
(i) there is a simple path ψ joining some points p1 and p2, which lie
on the boundaries C1 and C2 of these vertices, so that Lab(ψ) = 1;
(ii) Labp1
+(C1) = Labp2
−(C2).
Remark 4. The both vertices of a dipole can be either north or south,
since the sets R1 and R2 are mutually disjoint.
Evidently, it is possible to surround V ′ and V ′′ by a simple closed
path γ′ passing along ψ such that Lab(γ′) = 1. Therefore if γ′ intersects
edges then among them there are two edges intersecting γ′ successively
and labelled by inverse letters. These edges can be removed from γ′ by
bridge moves. It is easily seen that no edge intersects γ′ after a finite
number of bridge moves.
Thus the component K falls into two components K1 and K2 not
connected with each other. The component K1 contains only the dipole
of V ′ and V ′′ and the edges joining them together with some edges-
circles. Hence K1 is uniform. The component K2 may be either uniform
or mixed, either reduced or non-reduced.
Remark 5. Operation B does not increase the number of vertices.
Therefore the number of vertices in each Ki is strictly less than the
number of vertices in the original component K. In particular, the num-
ber of south vertices in K2 is not more than the corresponding number
in the original component K.
Lemma 4. Let a presentation G =< A | R1 ∪R2 > be weakly (R1, R2)-
separable. Then a picture P with a fixed equator Equ falls into a finite
number of uniform components by a finite number of Operations A and
B (being admissible moves).
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Proof. Let P consist of n components {Ki}. In particular, n may be
equal to one. Using the admissible move 3) we have that all components
{Ki} intersect Equ.
To each mixed component Kl of P assign a number sl equal to the
number of south vertices in Kl. Let s be the maximum of {sl}.
The proof of Lemma 4 will be by induction on s.
If s = 0, then all components {Ki} are uniform, hence there is nothing
to prove.
Let s > 0. In this case we transform each mixed component K ∈ {Ki}
containing s south vertices as follows.
a) If K is not reduced, then by Operation B, the component K
falls into two components K ′
1 and K ′
2, at least one of which (say K ′
1)
contains only a dipole and possibly edges-circles. The component K ′
1 is
uniform. The component K ′
2 contains s′2 south vertices, where s′2 ≤ s. If
K ′
2 is mixed and non-reduced, then Operation B applies to it again. By
Remark 5, after applying a finite number of Operations B, the component
K falls into uniform components {K ′
i} containing dipoles and a reduced
component K ′′ containing s′′ south vertices, where s′′ ≤ s. If K ′′ is mixed
and s′′ = s, then we transform it in the way described in the item b)
below.
b) If K is reduced, then it follows from weak (R1, R2)-separability
that Operation A can be applied to K. By Operation A, the component
K falls into two components K ′′
1 and K ′′
2 . If any of K ′′
j is mixed, then by
Remark 3 it contains s′′j south vertices, where s′′j < s .
Thus after a finite number of Operations A and B the picture P falls
into a finite number of uniform components, since s and n are finite.
Step 2. Reducing a picture P containing only uniform components
to a picture containing only states and edges-circles not belonging to the
states.
Let {Ki} be a subdivision of P into uniform components.
Fix a point on S2 lying neither on any edge and any vertex of P nor
on Equ. Assume that this point lies in the north hemisphere. We will
call it the north pole.
Assume that a component K ∈ {Ki} is such that some edges of K
form a simple closed loop η. The loop η divides the sphere into two parts
homeomorphic to disks: a disk containing the north pole will be called
exterior and the other disk will be called interior.
If there is another component of {Ki} lying in the interior disk with
respect to the loop η, then this component is called interior for K. If
the interior disk of η contains at least one interior component for K but
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it does not contain other closed loops of K with the same property, then
η is called minimal.
The intersection of all exterior disks for all components contains the
north pole and is called the absolute exterior.
In the step 2 we will use the following two admissible moves.
Operation C: Uniting a component and its interior components belonging
to the same hemisphere.
Let η be a minimal loop of a component K. Assume that all interior com-
ponents being interior with respect to η belong to the same hemisphere
as the component K does. Then we unite these interior components and
K in one component which will be denoted again by K.
Operation D: Separating a component and its interior components be-
longing to the other hemisphere.
-
u uu u
u uu u
& %-
e ee ee eAA AA
¢¢ ¢¢
-± °#
"
Ã
!
¾
-
a
aa
a
Fig. 8
Let η be a minimal loop of a component K. If at least one of the interior
components being interior with respect to η does not belong to the same
hemisphere as K does, then we move all interior components of K into
the absolute exterior by bridge moves (see Fig. 8).
Remark 6. Operation D strictly decreases the summary number of in-
terior components in P and adds a finite number of edges-circles to P .
Lemma 5. Let {Ki} be a finite subdivision of a picture P into uniform
components. Then a finite number of Operation C and D (being admis-
sible moves) gives a subdivision of P into states and edges-circles not
belonging to the states.
Proof. To each component Ki of P assign a number si equal to the
number of minimal loops in the component Ki. Let s be the maximum
of {si}.
For s = 0, there is nothing to prove.
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Let s > 0. We transform each component K ∈ {Ki} containing s
minimal loops as follows.
Let η be one of s minimal loops in K. If all interior components
interior with respect to η belong to the same hemisphere as K does,
then by Operation C the loop η ceases to be minimal, hence the number
of minimal loops in K becomes strictly less.
If there are some interior components interior with respect to η such
that they belong to the other hemisphere, then by Operation D the
number of minimal loops in K becomes strictly less. Operation D in-
creases the number of minimal loops for none of components {Ki}, but
it increases the number of edges-circles (see Remark 6).
Decreasing the maximum number s of minimal loops by admissible
moves, we obtain that each component lies in exterior disks for the other
components, i.e., P consists only of the states and edges-circles not be-
longing to the states. This proves Lemma 5.
Step 3. Getting rid of edges-circles.
Assume that P consists not only of states but of some edges-circles not
belonging to the states. Each such edge-circle divides the sphere into
two parts homeomorphic to disks: an exterior disk containing the north
pole (see Step 2) and an interior disk not containing it.
Applying the admissible moves 4) and 3), we can assume that both
interior and exterior disks of each edge-circle contain states.
An edge-circle C is called minimal, if the interior disk of C does not
contain the other edges-circles.
In Step 3 we will use the following two admissible moves.
Operation E: Uniting an edge-circle and its interior states belonging to
the same hemisphere.
Let C be a minimal edge-circle such that its interior disk contains states
belonging to the same hemisphere only. Then we unite C and the states
interior to C in one state of the same hemisphere.
Remark 7. Operation E decreases the number of edges-circles in P .
An edge-circle, which was not minimal before Operation E, can become
minimal after Operation E.
Operation F: Cutting an edge-circle if its interior states belong to the
different hemispheres.
Let C be a minimal edge-circle such that its interior disk contains states
belonging to the different hemispheres. By bridge moves, C can be cut
into some edges-circles {Ci} so that for each Ci its interior states belong
only to the same hemisphere (see Fig. 9).
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Remark 8. Operation F is a composition of a finite number of bridge
moves. New edges-circles {Ci} are minimal again. An edge-circle, which
was not minimal before Operation F , remains not minimal after Opera-
tion F .
-
'
&
$
%
d dd dd dAA AA¢¢ ¢¢t tt tt tt t
'
&
$
%
'
&
$
%- -
¾ ¾
-
¾
a a
a a
a
a
Fig. 9
Lemma 6. Assume that there are a finite subdivision of P into states
{Ti} and edges-circles not belonging to the states. Then after a finite
number of bridge moves, P will consist of states only.
Proof. Assume that P contains s edge-circles not belonging to the states.
The proof is by induction on s. For s = 0, there is nothing to prove.
Let s > 0.
Let m be the number of minimal edges-circles (m > 0) and n be
the number of edges-circles not being minimal among all s edges-circles
(n = s − m).
If there are states from the different hemispheres interior to any mini-
mal edge-circle C, then by Operation F the edge-circle C can be cut into
a finite number of minimal edges-circles {Ci} such that for each Ci its
interior states belong only to the same hemisphere. Note that Operation
F does not change the number n of not minimal edges-circles.
Thus we can obtain that for each minimal edge-circle its interior
states are from the same hemisphere. By Operation E, we can unite each
minimal edges-circle and its interior states in one state. In addition, the
number of the edge-circles not belonging to the states becomes at most
n, where n < s.
Step 4. Reducing a picture P containing only states to a picture
containing only σ-states.
Let P contain only states.
On this step we will apply the following admissible move.
Operation G. Crushing a state into σ-states.
Let R be an irregular north region of a north state T ( the move of an
irregular south region of a south state is similar). The boundary of the
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region R contains at least two connected pieces of the equator. We fix any
of them and denote it by I. The following transformations are performed
near I. A map M is chosen in R so that it contains I together with edges
intersecting Equ at points of I: more precisely, {y = −1} coincides with
I; {x = x1}, ..., {x = xn} correspond to the edges intersecting Equ in I;
{x = −1} and {x = 1} coincide with parts of the boundary of T . ( See
an example on Fig. 10. )
The construction of a new map M ′ is similar to Step 1. Namely, by
Lemma 2, U = Lab(I) ∈ N1. Two planar pictures P1 and P2 with the
boundary labels respectively U and U−1 are constructed and disposed
on M ′ as follows:
1) P1 lies in the rectangle {−1 < y < −1/2,−1 < x < 1} and all edges
intersecting the boundary of P1 start at (x1,−1),...,(xn,−1);
2) P2 lies in the rectangle {1/2 < y < 1,−1 < x < 1} and all edges
intersecting the boundary of P2 start at (x1, 1),...,(xn, 1);
3) the territory of T divides into parts {y ≤ −1/2} and {y ≥ 1/2};
4) {y = −1} is the considered piece I of the equator.
The map M is cut out from P and replaced by the new map M ′ (see
Fig.11 ).
Equ
? ?
6
boundary of T boundary of T
(−1, 1)
(−1,−1) (1,−1)
(1, 1)
x1 x2 x3
a
b
c
Fig. 10
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Equ
?
?
6
6
?
?
e
e
e
e
µ
¶ ¯
°
boundary
(−1, 1)
(−1,−1) (1,−1)
(1, 1)
x1
x1
x2
x2
x3
x3
a
b
c
Fig. 11
This move does not change the equatorial label W , hence it is admis-
sible.
Remark 9. By this move the number of states increases by one and the
irregular region R divides into two parts: one is a regular region and
the other one is an irregular region such that the number of pieces of
the equator belonging to its boundary is strictly less than the number of
pieces belonging to the boundary of the original R.
Lemma 7. Let a picture P contain only a finite number of states. Then
P can be reduced to a picture containing only σ-states by a finite number
of Operations G (being admissible moves).
Proof. If all north regions of all north states and all south regions of all
south states are regular, then there is nothing to prove.
Otherwise, to each irregular region Ri assign the number si of pieces
of the equator lying on the boundary of Ri. Let s be the maximum of
{si} (s > 1).
Operation G applied to each irregular region decreases the number
s by one. Therefore all states become σ-states after a finite number of
Operations G.
Thus Proposition 1 follows from Lemmas 4 - 7.
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2.5 Proof of Proposition 2.
By admissible moves, P is reduced to a picture containing only σ-states.
We assume that P consists only of two σ-states: a north σ-state T1
and a south σ-state T2, since we can always unite all north σ-states with
each other and all south σ-states with each other (see the admissible
move 5)).
Equ can be divided into arcs so that each of these arcs is intersected
by one σ-state only. We fix the direction of moving along Equ (let
it be from the west to the east) and renumber these arcs J1, J2, ..., Jk
successively. We may assume that an arc Ji is intersected by the north
σ-state for even i and by the south σ-state for odd i. An arc will be
called south (respectively, north), if it is intersected only by the south
σ-state ( respectively only by the north σ-state ). According to Remark
2 (see the admissible move 5)), each arc Ji contains precisely one piece
Ii of the equator belonging to the territory of any σ-state (this piece Ii is
a part of the boundary of a regular region of the corresponding σ-state)
The proof of Proposition 2 is by induction on k equal to the number
of arcs.
If k ≤ 2, then Proposition 2 follows from Lemma 3.
Let k > 2. Then we do the following admissible move.
We consider the second north arc J2 and the third south arc J3. By
Lemma 2, a word w1 along the piece I2 of the north arc J2 belongs to
N1 and a word w2 along the piece I3 of the south arc J3 belongs to N2.
Construct a map M containing a subpicture with the part {y = 0} of
the equator corresponding to the word w2w1w2
−1w1
−1. Cut out a small
map Ms between I1 and I2 such that Ms contains nothing but a small
part {y = 0} of Equ. Paste M in place of Ms (see the admissible move
6)). This adds two new σ-states: north one T comm
1 and south one T comm
2 .
(See an example on Fig. 12) This move corresponds to the insertion of
the commutator w2w1w2
−1w1
−1 of w1 ∈ N1 and w2 ∈ N2 in the word
W = ...w1w2..., and replacing it by W ′ = ...(w2w1w2
−1w1
−1)w1w2....
By the admissible move 5), we unite the north σ-states T1 and T comm
1
in one σ-state Tn
1 , after that we use Remark 2. Let In
2 be a piece of the
equator in the north σ-state Tn
1 containing the piece of the equator I2
of the original σ-state T1 and the piece of the equator I ′2 of the original
σ-state T comm
1 ( the notation are from the description of the admissible
move 6)). Since Lab(In
2 ) = 1, after a finite number of bridge moves, no
edge of Tn
1 intersects In
2 (see the admissible move 2)).
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?
Equ
Equ
I1
I1
I2
I2
I3
I3I3
I2
I ′3
I ′2
T comm
2
T comm
1
b
b
b
b
b
b
¢¢
¢¢
AA
AA
6
6
r
r
r
r
?
?
?
?
bb bb bb
¢¢¢¢ AAAA
?
6
Á À
rr rr
6 6
? ?
º ·' $
Fig. 12
If a regular region (the boundary of which contains the piece In
2 ) con-
tains a subpicture not intersecting Equ, then this subpicture is removed
(see the admissible move 3)). Deformation of the territory of Tn
1 gives
that the territory of Tn
1 does not intersect Equ by the piece In
2 .
By the admissible move 5), we unite the south σ-states T2 and T comm
2
in one σ-state Tn
2 . By Remark 2, the piece In
3 of the south σ-state Tn
2
contains the piece I3 of the σ-state T2 and the piece I ′3 of the σ-state
T comm
2 . By similar transformations as above, we obtain that the σ-state
Tn
2 does not intersect Equ by the piece In
3 .
Note that as a result of the transformations described above the
equatorial label W ′ = ...(w2w1w2
−1w1
−1)w1w2... is reduced to the form
W ′ = ...w2w1....
Thus again P consists of the two σ-states Tn
1 and Tn
2 , which intersect
Equ by the same pieces as the σ-states T1 and T2 did, but Tn
1 and Tn
2
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intersect Equ in other order. More precisely, the south pieces of the
equator I1 and I2 turn out to be side by side (similarly the north pieces
I2 and I4 turn out to be side by side). Therefore by Remark 2, the
number of arcs in the subdivision of Equ becomes strictly less and the
number of states and vertices does not increase.
By induction, we get a subdivision of Equ into two arcs, i.e., the case
when the equatorial label W is equal to the identity element in the free
group, which is the desired conclusion.
3. Some corollaries in the case of free products.
Let G1 =< X | S1 > and G2 =< Y | S2 > represent some groups and
G = G1 ∗ G2 be a free product of them. We will assume that the sets
S1 and S2 are symmetrized. We consider the normal closure N1 of a set
of elements R1 = {ui} and the normal closure N2 of a set of elements
R2 = {vj} in G. We assume that R1 and R2 are mutually disjoint and
symmetrized.
Assertion 6. Let a presentation < X∪Y | S1∪S2∪R1∪R2 > be weakly
(T1, T2)-separable, where T1 and T2 are defined in one of the following
ways:
(i) T1 = R1 ∪ S1, T2 = R2 ∪ S2;
(ii) T1 = R1 ∪ S1 ∪ S2, T2 = R2.
Then N1 ∩ N2 = [N1, N2].
Proof. Let us show that Assertion 6 follows from the similar statement
in the case of free groups.
Let N1
′ =< T1 >F and N2
′ =< T2 >F denote the normal closures of
the sets of words respectively T1 and T2 in the free group F = F (X) ∗
F (Y ). By Theorem 1, weak (T1, T2)-separability leads to
N1
′ ∩ N2
′ = [N1
′, N2
′]. (∗)
Consider the canonical homomorphism ψ : F −→ G = F/ < S1 ∪S2 >F .
Let us show that the following equalities hold:
1) ψ(N1
′) = N1;
2) ψ(N2
′) = N2;
3) ψ([N1
′, N2
′]) = [N1, N2];
4) ψ(N1
′ ∩ N2
′) = N1 ∩ N2.
The first two equalities are immediate. It follows from them that the
third equality and the inclusion ψ(N1
′ ∩N2
′) ⊂ N1 ∩N2 hold. Therefore
it remains to prove only the inclusion N1 ∩ N2 ⊂ ψ(N1
′ ∩ N2
′).
Let c ∈ N1 ∩ N2 be an arbitrary element and n̄1 ∈ N1
′ and n̄2 ∈ N2
′
be some of its preimages. We have n̄1 = n ∗ n̄2, where n ∈ Ker ψ. If n
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is equal to the identity element in the free group, then there is nothing
to prove. Otherwise, we will look for such preimages of c that n will be
equal to the identity element in the free group.
In the case (ii) the element n−1 ∗ n̄1 belongs to N1
′. Thus in this
case preimages of c can be chosen so that n will be equal to the identity
element in the free group.
Consider the case (i). The element n can be represented as a product
of elements s1,i ∈< S1 >F and s2,j ∈< S2 >F . There are two possibili-
ties: the first factor belongs either to < S1 >F or to < S2 >F . We will
show that it is possible to replace n̄1 by ñ1 and n̄2 by ñ2 in each of these
cases so that ψ(n̄1) = ψ(ñ1), ψ(n̄2) = ψ(ñ2), and the number of factors
in the factorization of the product ñ1 ∗ ñ−1
2 will be strictly less than one
in the factorization of n̄1 ∗ n̄−1
2 .
If s2,1 ∈< S2 >F is the first factor of n, that is n = n̄1 ∗ n̄−1
2 = s2,1 ∗g,
then we conjugate it by the element s2,1 ∈ Kerψ: (s2,1
−1 ∗ n̄1 ∗ s2,1) ∗
(s2,1
−1 ∗ n̄−1
2 ∗ s2,1) = g ∗ s2,1 and denote the conjugated elements n̄1
and n̄2 by n̄1 and n̄2 again. We have that the number of factors in the
factorization of n does not increase.
Therefore we can assume that s1,1 ∈< S1 >F⊂ ker ψ is the first
factor of n. We multiply n by s1,1
−1 on the left and denote the product
s1,1
−1 ∗ n̄1 again by n̄1. Note that the new n̄1 also belongs to N1
′ and
the number of factors in the factorization of n decreases.
If the last factor of n is s2,k ∈< S2 >F⊂ kerψ, then we multiply n by
s2,k
−1 on the right and denote the product n̄2 ∗ s2,k
−1 again by n̄2. We
have that the new n̄2 ∈ N2
′ and the number of factors in the factorization
of n decreases.
Repeating as above, we can reduce n to the identity element in the
free group.
It follows now from (3), (4), (∗) that N1 ∩ N2 = [N1, N2].
Corollary 6. If R1 belongs to G1, and R2 belongs to G2, then N1∩N2 =
[N1, N2].
Proof. It follows from Corollary 5, Theorem 1, and (i) of Assertion 6.
Corollary 7. Under the notation of Assertion 6, let a presentation <
X ∪ Y | T1 ∪ T2 > satisfy one of the following conditions:
(i) strict (T1, T2)-separability;
(ii)(T1, T2)-asphericity;
(iii) asphericity.
Then N1 ∩ N2 = [N1, N2].
Proof. It follows from Corollary 3, Theorem 1, and Assertion 6.
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Corollary 8. Let a set {S1, S2, R1, R2} satisfy one of the following small
cancellation conditions: either C(6), or C(4)&T (4), or C(3)&T (6).
Then N1 ∩ N2 = [N1, N2].
Proof. It follows from Corollary 4, Theorem 1, and Assertion 6.
References
[1] A.Yu. Ol’shanskij. Geometry of defining relations in groups, Moscow, ”Nauka”,
1989 (in Russian).
[2] Ol’shanskij A.Yu. Geometry of defining relations in groups, Mathematics and Its
Applications. Soviet Series, 70, Dordrecht atc.: Kluwer Academic Publishers, 1991.
[3] R.S.Lindon, P.E.Schupp. Combinatorial group theory, Springer-Verlag, Berlin -
Heidelberg - NewYork, 1977.
[4] W.A. Bogley, S.J. Pride. Aspherical relative presentations, Proc. Edinburgh Math.
Soc., vol. 35 ( ser. II ), part 1, 1-40, 1992.
Contact information
O. V. Kulikova Department of Mechanics and Mathematics
Moscow State University,Vorobievy Gory 1
119992 Moscow, Russia
E-Mail: olga.kulikova@mail.ru
Received by the editors: 09.12.2002.
|
| id | nasplib_isofts_kiev_ua-123456789-154676 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1726-3255 |
| language | English |
| last_indexed | 2025-12-01T08:15:38Z |
| publishDate | 2003 |
| publisher | Інститут прикладної математики і механіки НАН України |
| record_format | dspace |
| spelling | Kulikova, O.V. 2019-06-15T17:41:20Z 2019-06-15T17:41:20Z 2003 On intersections of normal subgroups in free groups / O.V. Kulikova // Algebra and Discrete Mathematics. — 2003. — Vol. 2, № 1. — С. 36–67. — Бібліогр.: 4 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 20F05, 20F06. https://nasplib.isofts.kiev.ua/handle/123456789/154676 Let N₁ (respectively N₂) be a normal closure of a set R₁ = {ui} (respectively R₂ = {vj}) of cyclically reduced words of the free group F(A). In the paper we consider geometric conditions on R₁ and R₂ for N₁ ∩ N₂ = [N₁, N₂]. In particular, it turns out that if a presentation < A | R₁, R₂ > is aspherical (for example, it satisfies small cancellation conditions C(p)&T(q) with 1/p + 1/q = 1/2), then the equality N₁ ∩ N₂ = [N₁, N₂] holds. en Інститут прикладної математики і механіки НАН України Algebra and Discrete Mathematics On intersections of normal subgroups in free groups Article published earlier |
| spellingShingle | On intersections of normal subgroups in free groups Kulikova, O.V. |
| title | On intersections of normal subgroups in free groups |
| title_full | On intersections of normal subgroups in free groups |
| title_fullStr | On intersections of normal subgroups in free groups |
| title_full_unstemmed | On intersections of normal subgroups in free groups |
| title_short | On intersections of normal subgroups in free groups |
| title_sort | on intersections of normal subgroups in free groups |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/154676 |
| work_keys_str_mv | AT kulikovaov onintersectionsofnormalsubgroupsinfreegroups |