Structures of Fibers of Groups Actions on Graphs
ABDULLAH AL-HUSBAN1, DOAA AL-SHAROA2, MOHAMMAD AL-KASEASBEH3,
R. M. S. MAHMOOD4
1Department of Mathematics, Irbid National University,
P. O. Box 2600-Code 21110, Irbid, JORDAN
2Department of Applied Science, Ajloun College, Al-Balqa Applied University,
Ajloun 26816, JORDAN
3Department of Mathematics, Jerash University,
Jerash, JORDAN
4Department of Mathematics, Irbid National University,
P. O. Box 2600-Code 21110, Irbid, JORDAN
Abstract.- In this paper we prove that if G is a group acting on a tree X such that G is fixing no vertex of X, the
stabilizers of the edges of X are finite, and the stabilizers Gv of the vertices of X act on trees Xv where Xv X,
Xu Xv for all vertices u,v of X, where u v, and the stabilizer Ge of each edge contains no edge x of the tree Xo(x)
such that g(x) = for every edge gGx, then there exists a tree denoted
and is called the fiber of X such that G
acts on
.
Keywords: - Actions on trees with inversions, trees, groups acting on trees, stabilizers, orbits, and the orbit space.
Received: August 8, 2021. Revised: July 11, 2022. Accepted: August 12, 2022. Published: September 20, 2022.
1 Introduction
Let G be a group and HG such that H acts on the set
X. Define the relation on GX as follows. If f,gG
and u,vX let (f,u) (g,v) if there exists an element
hH such that f = gh, and u = h-1(v). It is clear that
is an equivalent relation on on GX. The equivalent
class containing the element (g,v)GH is denoted
by gHv. Thus,
gHv = {(gh, h-1(v)hH}(gH)H(v), where
gH = {ghhH} is the left coset of g and H(v) is the
orbit of v under the action of H on X.
Let
, and . We have the
following notation.
(1) = {
(2) = {
(4) = {
where
, the set of the orbits of the
action of H on X, and H(x) = {h(x)|h
H}, the orbit
that contains the element xX. It is clear that if
,
, and
, then = x and
= x).
The aim of this paper is to use above notation to
show that groups acting on trees with inversions,
fixing no vertex of the tree and of given trees on
which the stabilizers of the vertices act and of finite
edges stabilizers induce a new tree called the fiber
tree of the group.
2 Concepts of Graphs
A graph X is the disjoint union of vertices V(X) and
edges E(X). An edge e is called a loop if the initial
vertex o(e) equals its terminal vertex t(e). If all edges
in a graph are loop we call the graph a loop graph.
Moving on, if a graph has at least on loop then it is
called quasi-graph. A graph that all its initial vertices
and terminals and inverses in a graph X is called a
subgraph of X, say Y. Define
to be the set
where is the invers of e.
Let be edges in the graph X.
P = (e1, e2, ..., en) is called a path in X if
t(ei) = o(ei+1) for i = 1,2, ..., n-1. If u = o(e1) and
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v = t(en) then P is called a path in X joining (or
liking) the vertices u and v, or a path in X from u to
v. If o(e1) = t(en) then P is called a closed path in X.
If ei+1
for i = 1,2, ..., n, then P is called a reduced
path in X. o(P) = o(e1) is the initial of P,
t(P) = t(en) is the terminal of P,
= (
,
, ...,
,
) is the inverse of P. It is clear
that
is a path in X joining the vertices t(P) and
o(P). The edges in the path P = (e1, e2, ..., en) are
called the edges of P. If Q is a path in X such that
t(P) = o(Q), then PQ is a path in X such that o(PQ) =
o(P) and t(PQ) = t(Q). |P| = n is called the length of
P. It is clear that if eE(X) such that t(e) = o(P), then
(e, P) = (e, e1, e2, ..., en) is a path in X. A path P is
called a simple circuit if it is close and contains no
repeated edges. The set of all paths in the graph X is
denoted by Path(X). We recommend readers to [2, 9]
for the structures of groups acting on graphs without
inversions and [1, 4, 5, 6] for with inversions, when
an edge of the graph equals its inverse is allowed. For
further studies see [10, 11].
A group acts on a graph X if there exists a unique
element denoted by for every and every
. G acts on X with inversions if there exist an
element gG and an edge eE(X) such that
g(e) = .
Remark. We write (G;X) to mean that G is a group
acting on the tree X.
Definition 2.1. A subtree is called a tree of
representatives for the action of on if has a
unique vertex from each vertex orbit. A subtree is
called a transversal it has a unique edge y such that
move in different orbit than y. The pair (T;Y) is
called a cover or (a fundamental domain). See [3].
The following are some properties of the tree of the
representatives T and the transversal Y.
(1) For any vV(X), we have a unique vertex
denoted v* where v*V(T) and G(v) = G(v*). That
is, v = g(v*), gG, where G(v) = {g(v): gG} is the
orbit containing v.
(2) For every vV(X) we have gG where
g(v*) = v.
(3) v* = v for all vV(T).
(4) (v*)* = v* for all vV(X).
(5) (g(v))* = v* for all gG and all vV(X).
(6) If gG where g(u) = v then u = v.
(7) If gG and u,vV(X) where g(u) = v then
(g(u))* = u* = v*.
(8) If eE(Y) where o(e)V(T), then
(o(e))* = o(e), and if t(e)V(T), then (t(e))* = t(e).
(9) If eE(T), then (o(e))* = o(e), (t(e))* = t(e) and
o(e) t(e).
(10) For every aE(X) we have gG and bE(Y)
where a = g(b).
(11) If gG and a,bE(Y) on which g(a) = b, then
a = b or a =
.
For the rest of this section G will be a group acting
on a tree X of cover (T;Y).
The proofs of the following propositions are straight
forward.
Proposition 2.2.
The edges E(Y) of Y can be split in to the following
sets of edges, called the sets of splitting edges of Y.
(1) E0(Y) = {mE(Y): o(m), t(m)E(T)} = E(T).
(2) E1(Y) = {yE(Y): o(y)E(T), t(y)E(T),
G(y) G()}.
(3) E2(Y) = {xE(Y): o(x)E(T), t(x)E(T),
G(x) = G()}.
Proposition 2.3. For eE(Y), o(e)V(T), there
exists an element denoted [e]G where
[e]((t(e))*) = t(e). We choose [e] = 1 in case eE(T)
and [e](e) = if G() = G(e).
Proposition 2.4. Let mE0(Y), yE1(Y), and
xE2(Y). Then [m] = 1, [] = [y]-1, [] = [x], and
[x]2Gx.
Proposition 2.5. ([4]) The element gG, g 1 can
be written as a product
g = g0[e1]g1[e2]g2 ..., gn-1[en]gn, where e1, e2, ..., en are
edges of Y and g0, g1, g2 ..., gn-1, gn, are elements of
G such that (t(ei))* = (o(ei+1))* for i = 1, 2, ..., n-1,
g0 and gi for i = 1, 2, ..., n.
Definition 2.6. For eE(Y) define the sign +e of e
to be the edge +e = e if o(e)V(T) and +e = [e](e) if
t(e) V(T).
It is clear that if o(p)V(T) and t(p) V(T), then
pE(T), [p] = 1 and +p = p.
Proposition 2.7. (1) For eE(Y) we have the
following.
(i) o(+e) = (o(e))*, t(+e) = [e]((t(e))*), and
= .
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(ii) G+e G(o(e))* and [e]-1G[e] = = G+e.
(iii) If eV(T) or G(e) = G() then G+e = Ge.
(2) If p,qE(Y) on which +p = +q then p = q or
p = .
(3) If gG and, p,qE(Y) on which g(+p) = +q then
+p = +q.
(4) If mE0(Y), yE1(Y), xE2(Y) and gG, then
+m = m, +y = y, =
, +x = x,
and + = x.
Definition 2.8. Let gG and eE(X). The sum of g
and e is denoted by ge and is defined to be the pair
ge = (gG+e, +e).
Let X* be the set X* = {gegG, eE(Y)}.
We have the following facts. The proofs are clear.
(1) gm = (gGm,m), g[m] = (gGm,) = g
(2) gy = (gGy,y), g = )),
g[y] =
).
(3) gx = g[x] = = g[x]x.
(4) X* = {gm, gy, g, gxmE0(Y), yE1(Y),
xE2(Y)}.
(5) If f,gG and p,qE(Y) such that fp = gq,
then f = gh, where hG+p and +p = +q.
Proposition 2.9. X* E(X).
Proof. It is clear that the mapping :X*E(X)
giving by (ge) = g(+e) is one-one and onto.
3 Inversion Elements
Definition 3.1. If G is a group acting on a graph X,
gG and eE(X) where g(e) = , we say that g is an
inversion element of G and e is called an inversion
edge of X under g. It is clear that if X is a quasi-
graph on which G acts then we have eE(X) on
which = e. Then 1G(e) =. In this case 1G is an
inversion element of G and e is an inversion edge of
E(X) under 1G, the identity element of G
Proposition 3.2. Let X be a graph where the group
G acts. Then the following imply each other.
(1) The action of G on X is with inversions.
(2) E(X) has an inversion edge and G has an
inversion element.
(3) The orbit space G/X is a quasi-graph.
Proposition 3.3. Let X be a graph on which the
group G acts such that G has inversion element gG
and eE(X) be an inversion edge of X under g. Let
u{o(e), t(e)}. Then
(1) is an inversion under g, g2Ge and g2Gu.
(2) gGu if X is a tree.
Proof. Clear.
Lemma 3.4. Let (G;X) and HG. Then
(i) If H has an element that is an inversion, then H is
not contained in the stabilizer of any vertex of X.
(ii) If H is finite and contains no inversion element
then H is contained in a stabilizer of a vertex of X, H
fixes a vertex of X, and has a trivial orbit for the
action of H on X . Moreover, if u,vV(X) are two
vertices of X such that HGu and HGv, then
HeGe, where e is an edge of the reducing path in X
joining u and v.
Proof. (i) Let gH be an inversion element. Then
there exists an inversion edge eE(X) of X under g.
So g(e) = . Let u{o(e), t(e)}. Since X is a tree,
Proposition 3.3-(2) implies that gGu. If u = v we are
done. Now assume that u v. We need to show that
gGv. X being a tree implies that there exists a
unique reduced path P = ( e1, e2, ..., en)Path(X)
joining u and v. So the edges e1, e2, ..., en are distinct
and n 1. The properties of groups acting trees imply
that Q: g(e1), g(e2,), ..., g(en) Path(X) , where Q is
a unique reduced liking g(u) and g(v) of length n 1.
Assume that gGv. Then g(v) = v. Let u = o(e). We
consider the following cases.
Case 1. e = e1. So P is the path P = ( e, e2, ..., en). The
property g(e) = implies that Q is the reduced path
Q = (, g(e2,), ..., g(en)) Path(X) linking g(u) and
g(v) = v. Since t() = o(e) = u = o(e2), then o(g(e2,))
= g(u) and R = (g(e2,), ..., g(en))Path(X) such that
R is reduced and linking g(u) and g(v) = v and of
length n-1. Hence Q and R are two reduced paths in
Path(X) joining g(u) and g(v) = v of different lengths
n and n-1. This is impossible because X is a tree.
This implies that gGv.
Case 2. e e1. Then (, e1, e2, ..., en)Path(X) such
that it is reduced and linking o() = t(e) and v. Then
(g(e), e1, e2, ..., en)Path(X) is reduced and linking
t(e) and v. As X is a tree, S = (g2(e), g(e1), g(e2,), ...,
g(en))Path(X) is a unique and reduced linking
g(t(e)) and g(v) = v. Since u = o(e) and g() = g2(e),
therefore by Proposition 3.3-(2), g2Gu. So g2(u) =
u. So S is a reduced path in X joining u and v. Thus,
R and S are two distinct reduced paths in X joining u
and v. Since X is a tree, this contradicts a property of
a tree that two distinct vertices of a tree are joined by
exactly one reduced path. This implies that gGv.
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Let u = t(e). Then u = o() and by adjusting the cases
above yields that gGv. Hence H is not contained in
any stabilizer Gv. for any vertex vV(X) of X.
(ii) Since H is finite and contains no inversion
element of G, by [2, Theorem 8.1, p. 27], there exists
a vertex vV(X) such that HGv. Then the stabilizer
Hv of the vertex vV(X) is HGv The case HGv
implies that H = Hv. So H fixes the vertex v. Since H
is finite, the stabilizer Hv and the orbit H(v) = {h(v):
hH}of v under the action of H on X are finite. By
the Orbit-Stabilizer Theorem [8, Lemma 4.11, p. 72],
the orders of H, Hv, and H(v) satisfy the equation
H = HvH(v). The case H = Hv implies that H =
Hv. So H(v) = 1. So H has a trivial orbit for the
action of H on X. If u,vV(X) are two vertices of X
such that HGu and HGv then Gu = Gv or HGu
Gv and by Theorem 4.3 of [7], H is contained in the
intersection of the stabilizers of the edges of the
reduce path in X joining u and v.
Corollary 3.5. Let (G;X), yE(X), o(y) = v, Xv be a
tree where (Gv ;Xv) and is finite and contains no
inversion element of Gv. Then we have w(y)V(Xv )
of Xv on which Gy(Gv)w(y), (Gv)w(y) is the stabilizer
of the vertex w(y) under the action of Gv on Xv.
Proof. By Lemma 3.4-(ii).
4 Basics of the Fibers
For the rest of this section, we have (G;X) of a
cover (T;Y) of the following assumptions.
(a) For each vV(T) let Xv be a graph such that
XuXv = for all uV(T), u v, and the stabilizer
of v Gv acts on Xv.
(b) For gG, vV(T) let gXv = {gx xXv}
and GXv = {gxgG,
xXv}.
(c)Let
=
, and
= X*
,
where X* = {gegG,eE(Y)} and
ge = (G+e,+e) of Definition 2.8.
Definition 4.1. For eE(Y), let w(e)V(X(o(e))*) be
chosen so that G+e(G(o(e))*)w(e), where (G(o(e))*)w(e) is
the stabilizer of w(e). So w()V(X(t(e))*) and
.
Proposition 4.2. Let u,vV(T) and f,gG. Then
(1) If u1V(Xu) and v1V(Xv) where
fu1= gv1then u = v, Gu = Gv, and we have
hGu where f = gh and v1 = h(u1)Xu.
(2) If u v then [fXu][gXv] = and
[GXu][GXv] = .
Lemma 4.3. (1)
is a graph where
V(
) =
] and
E(
) =
], where the ends
of E(
) are
If X*, then = ge = (gG+e, +e), where gG, and
eE(Y). Let o() = o(ge) = gw(e),
t() = t(ge) = g[e]w(), and
. If
], then the ends of are
defined as follows. o() = o(ge) = go(e),
t() = t(ge) = gt(e), and,
= , where e and, o(e),
t(e), and are the initial, the terminal and the inverse
of the edge eE(Xv).
(2) GXv, vV(X), and
form subgraphs of
.
Proof. First we show that
forms a graph. Since Xv
is a graph, this implies that V(Xv)E(Xv) = . If
[GV(Xv)][GE(Xv)] , then there exists
an element a[GV(Xv)][GE(Xv)]. So
a = x = g, where f,gG, xV(Xv), and
eE(Xv). From we have hGv where f = gh and
e = h(x). The case h(x)V(Xv), because Gv acts on
Xv, implies that eV(Xv) which contradicts above
that V(Xv)E(Xv) = . So
[GV(Xv)][GE(Xv) = .
Since X*(
]) = , we have
(
])[X*
])] = . By taking the set of
vertices V(
) to be V(
) =
]
and the set of edges E(
) to be
E(
) = X*
]) we see that
V(
)E(
) = .
Now we show that for
we have o( t(),
t( o(), and = .
Let E(
).
Case 1. X*. Then = ge = (gG+e, +e), where
gG, and eE(Y).
Then o(
) = o( ) =
g[e]
w() = g[e]w() = t(),
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t(
) = t( ) =
) = gw(e) = o(),
because G+eG(o(e))* and = e, =
=
= g[e] = ge = .
Case 2.
], then
= ge, where gG and e and,
o(
) = o( = ) =
gt(e) = t(ge) = t(). Similarly, t(
) = t( = ) = go(e)
= o(ge) = o(). Furthermore, =
=
) = = ge = . Then
forms a
graph.
(2) From above we have V(GXv)E(GXv)
= . If aE(GXv) is an edge of GXv, then
a = ge , where gG, eE(Xv). It is clear that
o(a) = go(e), t(a) = gt(e), and
= are the ends of a, where
o(e), t(e), and are the ends of eE(Xv). So
GXv forms a subgraph of
. Since GXv
,
therefore GXv forms a subgraph of
. Since
V(
) = V(
) and
, this shows that
is a
subgraph of
.
Lemma 4.4. (G;
) where if f,gG, vV(T),
xV(Xv), pE(Xv), and eE(Y) then
f(g f(g and
f(ge) = fge. (G;
) is with inversions if the action
of (G;X) is with inversions.
Corollary 4.5. For each gxV(Xv), pE(Xv),
and eE(Y), the stabilizers of the elements
g
), g
), and ge
) are
the followings. = g(Gv)xg-1, =
g(Gv)pg-1, = gG+eg-1.
Proposition 4.6. If the stabilizer of every element of
X is finite and the stabilizer of every element of Xv
under the action of Gv on Xv is finite, then the
stabilizer of every element of
is finite.
Definition 4.7. For vV(T) and eE(Y), let
Lv = (G/Gv){v} and Le = (G/G+e){+e}.
Lemma 4.8. If gG, vV(T), xV(Xv), pE(Xv),
and eE(Y) then the orbits of g
),
g
), and ge
) are the following.
G(g) = GGv(x), and G(g) =
GGv(p), G(g) = GGv(p), and,
G(ge) = Le.
Corollary 4.9. G/
=
vV(T)[GGv/Xv)][eE(Y)Le], where
Gv/Xv = {Gv(a)aXv}, the orbit space
GGv/Xv) = {GGv(a)aXv}.
Corollary 4.10. If G/X, Gv/Xv, vV(T) , [G, Ga] are
finite aX, then G/
is finite.
Proof. It is clear that Lv and Le are finite, vV(T) ,
eE(Y). So G/
is finite.
Lemma 4.11. For vV(T) and Xv = {v} be the
trivial graph of one vertex v and no edges. Let
and
be the graphs defined above. Then
(1)
= {LvvV(T) and
= .
(2) For eE(Y), w(e) = (o(e))*, w() = (t(e))*.
(3) For gG, eE(Y), and ge , o(ge) =
g(o(e))* and t(ge) = g[e].
(4) V(
) =
= {G/GvvV(T)}and
E(
) = X* = {gegG,eE(Y)}, where
ge = (G+e,+e).
(5) For gvV(T) and eE(Y), the stabilizers, the
orbits of the vertex g
) and the edge
ge
) are = gGvg-1, a conjugate of Gv in
G, = gG+eg-1.
(6) The orbit space G/
is the set
G/
= {Lv, LevV(T), eE(Y)}.
(7) (G;
) is with inversions if (G;X) is with
inversions.
Proof. Xv = {v}is a trivial graph of one vertex v and
no edge for each vertex vV(T). That is, V(Xv) =
{v} and E(Xv) = . Gv acts on Xv trivially.
(1)
= )] =
{v}] = { gG, vV(T)}.
The case = {(gh,h-1(v)) hGv} = {(gh,v)
hGv} = [G/Gv]{v} = Lv implies that
= {LvvV(T)}. Since E(Xv) = E({v}) = ,
therefore E(
= )] =
{ gG, eE(Xv)}= { gG}= . So
is a null graph.
(2) w(e) = }and w() =
}. Therefore w(e) = (o(e))* and w() =
(t(e))*.
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(3) For gG, eE(Y), the initial and the terminal of
ge are o(ge) = gw(e) =
g(o(e))*, t(ge) = g[e]w() =
g[e].
(4) From (1), V(
) =
and,
E(
) =
) = X* = X* =
{gegG,eE(Y)}, where ge = (G+e,+e).
(5) By Corollary 4.5, = g(Gv)vg-1 = gGvg-1
because (Gv)v = Gv and = gG+eg-1, a conjugate
of G+e. By Lemma 4.8, G(g) = GGv(v) =
G{v}= {gvgG}= = (G/Gv){v} = Lv,
and G(ge) = {f(ge)fG}= {fgefG}=
= {(fgG+e ,+e)fG}= (G/G+e){+e} = Le.
(6) From above G/
= {G(g), G(ge)
vV(T), eE(Y)} = {Lv, LevV(T), eE(Y)}.
(7) By Lemma 4.4.
Corollary 4.12. For vV(T) let Xv = {v} such that
the index of the stabilizer Gv in G is of finite. Then
X.
5 Paths in the fiber graph
Again, in this section, G will be a group acting on a
connected graph X of fundamental domain (T;Y)
such that for each vV(T), Xv is a graph such that
XuXv = , uV(T), u v, and the stabilizer Gv
acts on Xv. Furthermore, for eE(Y), w(e) is a vertex
w(e)V(X(o(e))*) such that G+e(G(o(e))*)w(e) and
w()V(X(t(e))*) where (G(t(e))*)w(). Now we
state and prove relations in the graphs X and
.
Definition 5.1. Assume that gG and vV(T),
a,bV(Xv), e1, e2, ..., enE(Xv). Let
P = (e1, e2, ..., en). Define P = (ge1,
ge2, ..., gen).
Lemma 5.2. (1) PPath(Xv) if and only if
P ). If o(P) = a and t(P) = b,
then o( P) = a and t( P) = b.
(2) P is closed if and only if P is closed.
(3) P is reduced if and only if P reduced.
(4) P is a simple circuit if and only if P is a
simple circuit.
Proof. (1) By the definition of a,
bV( Xv), and, ge1, ge2, ...,
genE( Xv) = E(Xv). Let
PPath(Xv). Then for each i we have o(ei+1) = t(ei).
This implies that t(gei) = gei) =
gei+1) = o(gei+1). So
P ). Conversely, if
P ), then gei) =
gei+1). By the definition of we have
fGv on which g = gf and f-1(ei)) = ei+1). So
g = 1 and ei) = ei+1). This implies that
PPath(Xv). If o(P) = a then o(e1).
(2), (3), and (4) are clear.
Proposition 5.3. Let f, gG, vV(T), eE(Y),
PPath( Xv), and the edge a = ge of X*.
Then
(1) There exist two vertices denoted P and P of
V(Xv) such that the initial of P is o(P) = P
and the terminal of P is t(P) = P.
(2) If o(a) = t(P), then v = (o(e))* and we have
he on which g = fhe , he(w(e)) = P.
(3) If t(a) = o(P), then v = (t(e))* and there exists an
element ke such that g[e] = fke and
ke(w()) = P.
Proof. (1) Since Xv is a graph and PPath(
Xv), therefore o(P) and t(P) are in V( Xv)
= V(Xv). Then o(P) = P and t(P) =
P, where P,PV(Xv). (2) If o(a) = t(P)
then o(a) = g = t(P) = fP. Then
v = (o(e))* and = Gv and g =
fP. This implies we have he, g = fhe
and he(w(e)) = P.
(3) Similar to (2), v = (t(e))*, t(a) =
g[e] = o(P) = f P and we
have ke on which g[e] = fke and
ke(w()) = P.
Lemma 5.4. Let PPath(
). Then
(i) If vV(T) and gG such that PPath( ),
then we have the edges e1, e2, ..., enE(Xv) such that
P = (ge1, ge2, ..., gen),
o(P) = and t(P) = .
(ii) If PPath( ) for all vV(T), gG, then
there exist elements f1, f2, ..., fn, fn+1, g1, g2, ..., gn of
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G, vertices v1, v2, ..., vn, vn+1 of V(T), edges e1, e2, ...,
en of E(Y), and paths P1, P2, ..., Pn of ,
, ..., such that
P = (P1, g1e1, P2, g2e2, ..., Pn, gnen, Pn).
Furthermore, the following properties of P hold.
(1) vi = (o(ei))*, vi+1 = (t(ei))* and (t(ei))* = (o(ei+1))*.
(2) PiV()) and there exist
vertices , V() such that
o(Pi) = and t(Pi) = .
(3) o(P) = o(P1) = and t(P) = t(Pn)
= . So P joins the graphs
V() and
V(). That is, PPath(
) linking
the vertices and of
V(
).
(4) o(giei) = t(Pi) = , and t(giei) =
o(Pi+1) = .
(5) We have hi, ki and
gi = fihi, gi[ei] = fiki = hi,
w(ei) = hi( and
) = ki(.
(6) If P is closed, then (o(e1))* = (t(en))*, = h,
and g1 = gn[en]l-1hkwhere h,k,l
(7) If P is reduced, then Pi is reduced and +ei+1 +
Proof. (i) From Proposition 5.3-(1) and Lemma 5.2
where o(e1) = P and t(en) = P the result follows.
(ii) Since E(
) = ])X* =
)]X*, the edges of P consist
of edges of the forms and
edges of the form geX*. By (i) above, P consists
of edges from both of
] and
X*. So the edges of P consist of the edges of paths in
, vV(T), gG and edges of X*. This gives
the required structure of P introduced above. Now the
proofs of (1)-(7) of the lemma as follows.
(1) Follows from Proposition 5.3 -(2).
(2) Follows from Proposition 5.3-2.
(3) From (2) above.
(4) From (3) above.
(5) From Proposition 5.3-(3).
(6) Since P is closed, therefore o(P) = t(P). So
o(P1) = t(Pn). Since o(P1)
t(Pn) ,v1 = (o(e1))* and
vn+1 = (t(en))*, therefore =
and v1 = vn+1. Then
(o(e1))* = (t(en))*, = h and h( = ,
where h Since P is a path of
, therefore
t(P1) = o(g1e1) and t(gnen) = o(Pn), therefore
Then = w() and
gn[en]w(
) = . So
g1 = f1k and gn[en] = l, where k,l
From above, g1 = gn[en]l-1hk.
(7) Since P is a reduced path, no edge of P is the
inverse of its previous edge. So, if a,b are adjacent
edges of a path Pi, then b . So the path Pi is
reduced. If for some i, i = 1, 2, ..., n-1, we have +ei+1
= +
, then (t(ei))* = (o(ei+1))* and have gi+1 = gi[ei]h,
h. This implies thatgi+1ei+1 =
(gi+1) = (gi[ei]h
,+
) =
(gi[ei]
,+
), because h and
. This implies that gi+1ei+1 = gi[ei]
=
. Then P contains the edge giei and its
inverse
= gi[ei]
. Contradiction, because P
is reduced path.
Corollary 5.5. Let P = (P1, g1e1, P2, g2e2, ..., Pn,
gnen, Pn) be the path in Lemma 5.4. Then
(g1e1, g2e2, ..., gnen) is a path in the trivial fiber
graph
where Xv = {v} for all vV(T).
Proposition 5.6. Let P = (P1, g1e1, P2, g2e2, ..., Pn,
gnen, Pn) be the path of Lemma 5.4. Let
P* = (g1(+e1), g2(+e2), ..., gn(+en)). Then
(1) P*Path( X).
(2) If P is closed so P* is closed.
(3) If P is reduced so P* is reduced.
(4) If P is a simple circuit, so P* is a simple circuit.
Proof. Clear.
Lemma 5.7. Let g = g0[e1]g1[e2]g2 ..., gn-1[en]gn be a
product of the element g. For i = 1, 2, ..., n, let
fi = g0[e1]g1[e2]g2 ..., gi-2[ei-1]gi-1 with convention that
f1 = g0, let qi = fi(+ei), and pi = fiei. Then
(1) q = (q1, q2, ..., qn)Path(X ) linking (o(e1))* to
g((t(en))*).
(2) o(pi) and
t(pi) .
(3) For i = 1, 2, ..., n, assume that
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PiPath( such that t(P1) = o(p1),
o(Pi) = t(pi), t(Pi) = o(pi+1) for i = 2, ..., n-1. Then
P = (P1, p1, ..., pn-1, Pn) is a path in
joining the
vertices o(P1) and t(Pn).
(4) P* = q.
Proof. (1) We need to show that o(q) = (o(e1))*,
t(q) = g((t(en))*), t(qi) = o(qi+1), i = 1,2, ..., n-1.
Now o(q) = o(q1) = o(f1(+e1)) = fi(o(+ei)) =
f1((o(e1))*) = (o(e1))* because f1 = g0,
t(q) = t(qn) = t(fn(+en)) = t(g0[e1]g1[e2]g2 ..., gn-1[en]
t(+en ) = g0[e1]g1[e2]g2 ..., gn-1[en]gn((t(en))*) =
g((t(en))*), and t(qi) = t(fi(+ei) = fi(t(+ei)) =
fi[ei]((t(ei))*) = fi[ei]((o(ei+1))*) = o(qi+1).
(2) o(pi) = o(fiei) =
and,
t(pi) = t(fiei) =
because (t(ei))* = (o(ei+1))* and
gi+1. This shows that
t(pi) .
(3) By Lemma 5.4-(ii), the cases
o(pi) and
t(pi) of (2) above
implies that P is a path in
linking o(P1) and t(Pn).
(4) This follows from Proposition 5.4.
Definition 5. 8. The path P of Lemma 5.7 is called
the path in
obtained from the product of the
element g = g0[e1]g1[e2]g2 ..., gn-1[en]gn and q is the
path obtained by collapsing the vertices of P.
Lemma 5.9. (I) If X and Xv are connected, vV(T),
so
is connected.
(II) If X and Xv are trees, vV(T), so
is a tree.
Proof. (I) The following steps imply that X is
connected.
(1) Xv = { xxXv} is connected , fG
vV(T).
(2) Xu and Xv are linked by a path in
,
u,vV(T).
(3) Xv and Xv are linked by a path in
,
gG , vV(T).
(4) Xu and Xv are linked by a path in
,
f,gG, u,vV(T).
(1) Let p,qV( be
two distinct vertices of . By the definition of
, we have vertices a,bV(Xv) where
p = f and q = f. If a equals b, then p = q.
This contradicts the assumption that Since Xv
is a connected graph, we have PPath(Xv) on which
o(P) = a and t(P) = b. By Lemma 5.2,
P ), o( P) = a = p
and, t( P) = b = q. So Xv is
connected.
(2) For u,vV(T), there exist edges e1, e2, ...,
enE(T) such that p = (e1, e2, ..., en)Path(T),
o(p) = u, t(p) = v, t(ei) = o(ei+1), and w(
) = w(ei+1),
i = 1, 2, ..., n-1. Then for each e{e1, e2, ..., en},
[e] = 1, +e = e, + , (o(e))* = o(e), and
(t(e))* = t(e), w(e)V(Xo(e)), w()V(Xt(e)), G+e = Ge,
= = Ge. Consider the edges
1e1, 1e2, ..., 1en of X*. Then
o(1e1) = 1w(e) = 1w(e)
Xu,
t(1en) = [en]w(
) =1w(
)
Xv,
and, t(1ei) = [ei]w(
) = 1w(ei+1).
So Q = (1e1, 1e2, ..., 1en)Path(
) and liking
the subgraphs Xu and Xv of
.
(3) If g = 1, we have case (1). Assume that g 1. By
Proposition 2.5, the element g has the product
g = g0[e1]g1[e2]g2 ..., gn-1[en]gn where
(o(e1))* = (t(en))* = v. Then Pg = (P1, p1, ..., pn-1, Pn)
of Lemma 5.8, Pth(
and liking the subgraphs
Xv and Xv. Similarly, for fG, uV(T)
we have PfPath(
) liking the subgraphs Xv
and Xu.
(4) Let
be the converse of the path Pf of (3)
above. Then the composition
QPg of the paths
, Q and Pg,Path(
) liking the subgraphs
Xu and Xv of
. Consequently,
is a
connected graph.
(II) First we show that for gG, vV(T), the
subgraph Xv forms a subtree. If Xv
contains a loop, then there exists an edge
E( Xv) = E(Xv) such that o() = t().
Then = e where eE(Xv). For the case
o() = t() we have
o() = e) = o(e) = t() = e) =
t(e). The definition of implies that
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o(e) = t(e). So e is a loop in the tree Xv. This
contradicts the assumption that Xv is a tree. If
Xv contains a simple circuit
P = (P1, P2, ..., Pn)Path( Xv), then
o(P1) = t(Pn), t(Pi) = o(Pi+1) and Pi+1
for
i = 1, 2, ..., n-1. Xv being a subgraph of
implies that there exist edges e1, e2, ..., enE(Xv) such
that Pi = ei, i = 1, 2, ..., n. Then o(e1) = t(en),
t(ei) = o(ei+1), and ei+1
, i = 1, 2, ..., n. This implies
that (e1, e2, ..., en)Path(Xv) is a simple circuit. This
is a contradiction because Xv is a tree. Xv
is a subtree of
. If PPath(
) is a simple circuit,
then from above, PPath( Xv). Then P is the
path of the form of Lemma 5.4-(ii). Then Lemma 5.7
shows that the path P* obtained by collapsing the
vertices of P is a simple circuit in X. Since X is a
tree, we get contradiction because a tree contains no
simple circuits. Hence
is a tree.
6 The Main Result
Theorem 6.1. Assume (G;X) of a given cover (T;Y) where
Xv is a tree, XuXv = for all uV(T), u v.
Furthermore, for dE(Y), assume that Gd of d is finite and
containing no inversions of the tree X(o(d))*. Then
(1) There exists v(d)V(X(o(d))*) where G+d(G(o(d))*)v(d),
and v (d) = w(d), w(d) is the vertex of Definition 4.1.
(2) The fiber
is a tree.
(3) If (G;X) is with inversions or for vV(T), if (Gv;
Xv) is with inversions, then (G;
) is with inversions.
(4) The structures of the stabilizers of the elements of
are = f(Gv)xf-1, = f(Gv)pf-1, and =
fG+df-1 for all fzV(Xv), pE(Xv), and dE(Y).
(5) structures for the orbits of the elements of
are
G(f) = GGv(z),
G(f) = GGv(p), and, G(fd) = (G/G+d){+d},
fzV(Xv), pE(Xv), and dE(Y).
(6) The orbit space G/
has the form G/
=
vV(T)[GGv/Xv)][dE(Y) (G/G+d){+d}].
The edges of
have the properties that
o(fp) = fo(p), t(fp) = ft(p), and,
= , and o(fd) = fv(d), t(fd) =
f[d]v(
) and
for all f
pE(Xv), and dE(Y). Proof.
(1) Since the stabilizer of each edge eE(Y) is finite,
therefore G+e is finite. Since , and G+e
contains no inverter edges of the tree , therefore by
Corollary 2.5, there exists a vertex denoted v(e) where
G+e((. Since w(e) is arbitrary, we take w(e) =
v(e). (2) Th assumptions that X and Xv,
vV(Y) are trees, Lemma 5.9-(II) implies that the fiber
is a tree. (3) Lemma 4.4. (4) Corollary 4.5.
(5) Lemma 4.8. (6) Corollary 4.9.
Corollary 6.2. If (G;X) is without inversions and
Gd , dE(Y) is finite, then
forms a tree.
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