747
Internat. J. Math. & Math. Sci.
Vol. 8 NO. 4 (1985) 747-754
ON ORDERABILITY OF TOPOLOGICAL GROUPS
G. RANGAN
The Ramanujan Institute
University of Madras
Madras-5, India
(Received October 15, ]984)
ABSTRACT.
A necessary and sufficient condition for a topological group whose topology
can be induced by a total order compatible
with the group structure is given and such
groups are called ordered or orderable topological groups.
A separable totally-
disconnected ordered topological group is proved to be non-archimedean metrizable while
the converse is shown to be false by means of an example.
A necessary and sufficient
condition for a no-totally disconnected locally compact abelian group to be orderable
is also given.
KEY WORDS AND PHRASES. Topological groups, orderable or ordered topological groups,
non-archimedean metrizable, totally disconnected, locally compact abelian groups.
1980 AMS SUBJECT CLASSIFICATION CODES. 22A05, 54F05.
I.
INTRODUCTION.
Topological groups whose topology can be induced by a total order are called top-
ologically orderable groups and they are studied by Nyikos and Reichel [i] and
Venkataraman, Rajagopalan and Soundarajan [2].
In this paper we give a necessary and
sufficient condition so that topological group is orderable in the sense that it admits
a total order which ind,,ces the topology of the topological group and which is also
compatible
with its group structure.
We call such groups ordered topological groups.
As a first step to this we give a necessary and sufficient condition that a group be
orderable so that it admits a totoal order compatible with the group structure and such
groups are called ordered groups.
shown (Theorem 2.6, [2]
Venkataraman, Rajagopalan, and Soundararajan have
that a separable totally-disconnected topological group is a
topologically orderable group if and only if it is metrizable and zero dimensional.
It
turns out that a separable totally-disconnected ordered topological group is non-
archimedean metrizable (in the sense of Rangan
[3]) while the converse fails
to be true.
It is interesting to note that the totally-disconnected non-archimedean metrizable
locally compact abelian topological group
(Qp,+)
of p-adic numbers under addition
admits an order that is compatible with the group structure (+) alone and another order
that is compatible with its topology alone but admits no order that is compatible with
G. RANGAN
748
+
both its group structure
and sufficient condition for a non-totally disconnected
locally compact abelian group to be orderable.
2.
ORERABILITY OF A TOPOLOGICAL GROUP.
A group
THEOREM 2.1.
homomorphism
f
x
G
corresponding to each
and a
into the additive group of real numbers can be found
(e
e
x
C(x)
can be ordered if and only if a subgrodp
C(x)
from
G) which satisfies the fol-
being the identity of
lowing conditions
and f (x)
x
xC(x)
(i)
(ii) for
y
(iii) for
C(x),
in
x,y
C(y)
C(x)
G
in
0
for every
C(y)
C(x)
e,
e, y
x
aC(x)a
-I
(iv) for every
a
in
G
for every
t
in
C(x) i.e. if
C(x)
and
x
taining
Let
PROOF.
G
ker f
x
C(x)
group of
C(x)
a
-I
:x/axa
then
for each
be an ordered group.
x
C(x)/C(x)*
that
and
C(y) and f
where
f
f
x
or
y
x
axa
z
-I
f (t)
x
and
is the smallest convex subgroup con-
C(x)*
x
e
G
in
x.
C(x) denote the
let
be the largest convex subgroup not
[5]) that
C(x)*
is a normal sub-
is order isomorphic to a subgroup of the reals
and we identify this quotient group with the corresponding subgroup of the reals.
fx
C(x)
be the natural map of
C(x)*
and
C(y)*
respectively
C(x)
C(y)
C(y)
C(x)
to
C(x)/C(x)*
C(y)
implies
C(x)*
C(y)*
and hence
groups
f
Hence (i), (ii)
C(axa-1) *
C(x)/C(x)*
I
x
f
aC(x)a
C(z)/C(z)*
and
where
z
axa
-I
-I
C(y)
and
If
y
C(x)
I*a
C(axa
-!
Let
Since
C(y)
and (iii) of the theorem are easily verified.
induces an order isomorphism
a
0
C(x)
then being convex subgroups of an ordered group either
automorphism being order preserving it follows that
aC(x)*a -I
fx(X)
Then clearly
are the largest convex subgroups contained in
C(x)
-i
f oI
z a
For each
and
f (ata
z
is the inner automorphism deter-
x
Then it i well known (see [4] and
x
e
is the largest convex subgroup not containing
smallest convex subgourp containing
containing
I
x
O,
C(z)
G
in
C(x)
G
Further for the order on
a
G
in
either
fy(X)
in which case
mined by the element
x
or
Inner
and
between the quotient
If we identify these two sub-
groups of the reals forgetting the order isomorphism between them we get that
f
f oI
z a
x
which proves (iv) of the theorem.
Conversely if
G
f (x) > 0
of all elements such that
on
For
G
4=>f x -l(x
x e
x
C(x
G C(x)
O<x
-I)
serves as the strict positive part of an ordering
by (ii) and hence
P.
f
x
f -1
x
x
P>f x (x)
0
P
We now observe a useful property of the subgroups
perty
P
satisfies the conditions of the theorem we show that the set
C(x)’s
which we call the pro-
ORDERABILITY OF tOPOLOGICAL GROUPS
PROPERTY P.
If
a, b
G, a
ab
C(a)
e, b
C(a)
then
= C(a)
and
C(b)
and
e
749
C(a)
C(ab)=
C(ba).
tion to (i).
a
If
+
0
C(y)
+
fy(y)
C(z)
0.
x e P
Let
where
belongs to
Similarly it can be seen that
C(x)
C(y)
f
f
For
x
axa
_
y
xy
C(x)
-1
implies that
f (x) > 0
Then
f (axa
z
-1
y e C(x),
f
C(x)
t c
xy
f (x) >
x
G
C(x)
that
C(yt
-I)
C(yt
C(x)
Suppose
C(yt
impossible.
For
C(t)
fyt-l(yt -I)
f
ft(t)
0
yt
-1)
-l(y) + f
and
e
C(t)
C(t
-I
CASE 2.
Then
f (x)
X
f
xt
f
-l(t
f (xt
-I
f (x)
+
e
X
x
t
C(x)
C(t)
C(xt
-I)
X
-i
e
(xy)
aC(x)a
-I
axa
i.e. z
f
x
f
t
f (t
X
-i
yt
is a convex
C(x)
and hence
thereby proving
C(yt
fty-I
fyt -I
ft(t
-I
y
C(x)
).
Property P
_
and
0
and so
fyt-1
since
C(yt
Thus we get-that
(y)
-I
0
i.e.
C(x)
x
For this
x
and let us assume that
C(t) c_ C(x)
f (x)
C(xt
0
The convexity of
-i
and so
since
C
fx fxt-!
f (t
-i
X
implies that
We now discuss the two possibilities
separately.
by
C(x).
C(x)
X
-1
We will now show that this is
ft
This implies
e
and Property P imply that
Hence
f (t)
C(t)
-1
t
xy
is an ordered group.
C(x)
-I
C(y) c_ C(x)
-1)
0
is the smallest convex subgroup containing
x, t
t
f
C(x)
C(x).
-I
yt
y
From the assumption
implies
yt
C(x) can be
and
is impossible
is any conves subgroup containing
X
-1
C(ty
-l(t
C(x)
C
C(x),
t e
C(t)
-I
C(t
fxt-l(xt
C(t
In other words
We now prove that
let us suppose that
CASE I.
yt
-I)
a contradiction to the choice of
(iii) of the theorem.
x
C(y-lyt -I)
C(yt
C(yt -I)
C(y
condition (ii) of the theorem we get that
now implies that
implies that
C(x)
This is done by proving that
f
If
xy e P.
f ol (x)
z a
t < y
e
C(x), C(xy)
C(y)
(the case
and so
To prove the second part of the theorem we first show that
subgroup by showing that
fxy(Xy)=
and
O, a contradiction.
Now by Theorem 2, p. 13, [4] it follows that
P.
C(ba).
C(y)
By (iv) of the theorem we have
X
f (z)
z
and
C(y)
C(y)
a contradic-
C(a)
xy
oherwise since
C(xy)
0
Now
0
a
C(x)
f (x) + f (y)
y
x
f (xy)
x
0
C(xy)
then
f (b)
f a (b)
since
and
x,y e P
i.e.
0
C(ab).
By Property P,
a e G
z
+ f a (b)f a (a)
let us suppose that
similarly dealt with).
fy(X)
f
since
C(ab)
0
a
0
and hence by (iii)
C(x)
however
f (y)
y
0
fy(y)
C(y)
C(x)
C(a)
Hence
f (x)
x
f (b)
and
f a (a)
f (ab)
C(a) 0
C(ab)
fx(X)
C(a)
C(b)
For
f (x)
x
0
t
Now
in view of
C.
> fx(t)
and
G. RANGAN
750
fx (x)
e
fx(t) ==fxt-l(xt
f (x)
x
C(x)
fx(t)
and
f
C(xt
f (xt
x
Then
== fx(t)
f n -I
x t
x
In either case
CASE 3.
f (x) > 0
x
for some integer
=> f xn t -l(xnt -I)
=
0
x
n
> t
C(x)
0
n
=>f x (xnt -I)
e
=
f
-I
fx(t).
f
x
C(x)
z,y
C(x)
0.
x
i.e.
2 >
t
_
fx2t-l(x2t-l)
e
The convexity of
C(x) c_ C.
x
= fx (y)
C(x)
z > y
C(zy
Then
-I
C(x)
fx (z)
=
f
-1
x
f
x
C(x)
C.
C(x)
taining
Ker f
Then since
x
x
4 C,
C(x)*
C(t)
C.
i.e.
-1
==
0
f
-1
zy
C
C(x)
fx(Xt -I)
now implies
-I
C(x)
which
x.
=
-I
For let
f (zy
x
fx(Zy
-1
0
-I
fx(Z)
Ker
This proves that
fx
G.
is the largest convex subgroup not contain-
C
is any convex subgroup not containing
and so
C(t)
C
x
x
C(x
C(x)
-l(zy
+
f (x)
to the reals.
fx(Z) _> fx(y)
C c_ C(x)
by (i) of the theorem and
t, C(x)
zy
C(x)*
For this let us suppose that
Let
C(zy
and hence also of
It remains for us to show that
x
C
e
C(x)
C(x)
In either case
is a convex subgroup of
C(t)
For
is the smallest convex subgroup containing
C(zy
f (y)
f (z)
x
fx (y)
and
is convex.
C(x2t -I)
Now
is an order preserving homomorphism from
x
C
since
From Property P we get that
fx2t-1
t
Thus we see that in all cases
f
C
t
=>c(xnt -I)
0
C(t).
i.e. f (x)
x
f (xt
x
since
proves that
i.e.
C.
t
t e C.
that
t
we conclude that
C
x.
t
C.
t
-I
-I
nf (x)
x
which in turn implies that
ing
by (i) of the THeorem since
> 0
Again from the convexity of
x.
t
-I
C(x).
t
If
f (t)
x
0
x
then
being the smallest convex subgroup con-
a contradiction.
Hence
f (t)
x
0
i.e.
This completes the proof of the theorem
Even though possibly one can prove Theorem 2.1 in a different way using Theorem 11,
p.51 of [4] or Theorem 2
of [5] we have prefered the above proof since it is elementary.
The above form of formulating the theorem gives rise in a natural way for a criterion
of orderability of a topological group as well.
e is the first convex subgroup of
REMARK. C, the intersection of all C(x), x e G, x
G
Hence when C
(e) the above theorem gives a necessary and sufficient condition
that
G
may be made into an ordered group without first convex subgroup.
THEOREM 2.2.
A non-discrete topological group
G
can be ordered so that the in-
terval topology of the order coincides with the given topology and also simultaneously
this order is compatible with the group structure of
G
if and only if
G
satisfies
the following criterion (v) in addition to those of Theorem 2.1.
(v)
either the intersection of all
which case the collection
the C(x), x
{C(x)}
x
G, x
e is the singleton
(e) in
G form a neighbouhod base at the
ORDERABILITY OF TOPOLOGICAL GROUPS
C
of
e
identity
or the intersection
G
is an open subgroup of
and for each
751
C
of all the
x
in
C
f
C(x), x
e
G,
x
e
is an open continuous
x
homomorphism into the reals.
In view of Theorem 2.1, it is enough to show that condition (v) above is
PROOF.
equivalent to the coincidence of the interval topology and the given topology of the
G
group
Let now
G
be an ordered group.
group (see 4.19 [6]).
CASE I.
Let
a
such that
t
I
e
in
i.e.
C
f
y
-I
e,
t
C(x) E I
G,
x
x
as in Theorem 2.1.
e
I
are in
C(x)
t
G
Then since
e
and this proves that
nC(x)
Since
t N y
implies
is not discrete there exists
C(x), x
(e)
G
there exists
C(x)
since
is convex.
G
x
Hence
form a neighbourhood base of
C z (e)
oC(x)
C
is a convex subgroup and hence open in
C(x)
Hence for
C ==f
x,y
for
G
CASE 2.
Then
and
y
y # C(x)
such that
C(x)*
C(x) and
We define
be an open interval containing
y > e
with the interval topology is a topological
(e)
nC(x)
I
G
f
f
x
C(x)/C(x)*
f
y
C
is one-to-one on
C,
x,y
xy
f
-i
C(x)
C
=
0
f
C(y)
xy
-l(xy
For
G
and so
-1
x e G
f
f(x)
x
f(y)
being the cannonical map from
x
f
C(x)
f (say).
y
==> f(x)
C(x)
being order isomorphic to a subgroup of the reals,
ous homomorphism from
it is clear that
f
to
x
x
y,
f(y)
i.e.
C(x)/C(x)
and
is an open continu-
to the reals.
To prove the converse also we consider two cases.
CASE I.
nC(x)
From Case
(e)
{C(x)}, x e G
{C(x)},
of the necessity part it follows that
bourhood base for the interval topology.
neighbourhood base at
e
By hypothesis
x
e, form a neigh-
x e G, x
e
form a
for the given topology and hence the two topologies coincide.
CASE 2. oC(x)
C z (e)
For x,y
C, it is clear from the convexity of C(x) and (ii) of Theorem 2.1 that
f
f
f (say)
Then f is a one-to-one open continuous homomorphism from C to
y
x
the reals, the topology of
of
G
Hence
C
C
being the relative topology got from the order topology
with respect to this (order) topology is homeomorphic to a subgroup
of the reals with respect to its relative topology.
continuous homomorphism from
topology on
G
onto this subgroup of the reals.
the two topologies on
on
G
the
two
C
conincide.
topologies on
COROLLARY 2.3.
Let
G
But by hypothesis
G
C
e
is an open
Now
f
is one-to-one implies that
being an open subgroup in both the topologies
themselves coincide.
be an ordered topological group (i.e. the topology of
is giben by a total order compatible with the group structure of
ponent of the identity
f
C, taken with the relative topology got from the given
of
G
is either the trivial subgroup
subgroup topologically isomorphic to the additive group of reals.
G).
(e)
G
Then the comor it is an open
G. RANGAN
752
If
(e) by Theorem 2.2 above G becomes a zero dimensional group.
PROOF. If oC(x)
(e) then C is the component of identity. For, each C(x) being an
nC(x)
C
open and closed subgroup, contains the component of
C
e and so
itself contains the
But the component being a connected subgroup is also a convex subgroup (see
component.
Proposition 1.3(2), [2]) and hence contains
C
From Theorem 2.2 it follows easily that
the component.
C
which proves that
C
is the same as
is topologically isomorphic
to the additive group of reals.
3.
TOTALLY DISCONNECTED ORDERED GROUPS.
A metric
X
on a set
d
is said to be a non-archimedean metric if
d
satisfies
the stronger triangle inequality
d(x,y)
for
X.
x,y,z
max(d(x,z), d(z,y)
A topological group is said to be non-archimedean metrizable if there
exists a right (or left) invariant metric on
G (see
G
which induces the topology of
G
is such that its topology is given by
Rangan [3]).
LEMMA 3.1.
Suppose a topological group
d
a non-archimedean metric
invariant metric
then there is an equivalent non-archimedean right (or left)
G (i.e. a non-archimedean metric
on
0
PROOF.
When
d
z
varies in
easy to check that
G
Put
G
Then
positive integers
d’(x,y)
V
where
n
0
p
P(x,e)
l/n}
<
k.
W
{x
I/m
d’
and choose
z z o)
<
I/m
z
in
Then
with respect to the metric
existence of a positive integer
W
is in
U
k
V
Hence
and
U
U
n
d’(x,e)
x
n
k
Wz
V
M
for all
n
d’.
contains a
l/n}.
To prove
U
k
for
O, choose positive integers n,m such that
O(x,e)
d’(XZo,Z o + I/n. Let
such that
0
is the open sphere with centre
The continuity of
such that
d’(XZo,Z o)
+ I/n
x
WZo
UkZ
I/m + I/n
xz
o
Let
I/M
x
at
x
E
z
e
U k.
O
and radius
implies the
Clearly
x
proving thereby that
This completes the proof of the lemma.
M
THEOREM 3.2.
Then
group.
0(x,e)
d’.
G
We will now
G
is finer than the topology induced by
Corresponding to the Integer M
1/M
V
is clear that
that the two topologies are the same we will show that each
1/n + 1/m
It is
is bounded.
coincide.
G, it
x,y
{x
n
d’
and
for all
i.e. the topology induced by
some
d’
d/l+d
where supremum is
is a right invaPiant non-archimedean metric on
0
Ox,y)
sup d’(xz,yz)
0(x,y)
is well-defined since
0
show that the topologies induced by
Since
then the metric d’
G
is a non-archimedean metric on
also defines the same topology on
taken as
which gives the same
0
d).
topology as
PROOF.
O
Let
G
be a separable totally disconnected ordered topological
is non-archimedean metrizable.
By Theorem 2.6 of [2], it follows that
G
is topologically orderable,
metrizable, zero-dimensional group and so carries a non-archimedean metric inducing the
topology of
that
G
G (see proof of Theorem 7 of [I] or [7]).
is non-archimedean metrizable.
Now from Lemma 3.1 it follows
ORDERABILITY OF TOPOLOGICAL GROUPS
Vnkataraman,
753
Rajagopalan and Soundararajan proved (see Theorem 2.6, [2]) that a
separable totally disconnected group is topologically orderable (i.e. the topology is
given by a total order) if and only if it is metrizable and zero-dimensional.
Non-
archimedean metrizable groups are totally disconnected groups (see Rangan [3]).
Hence
in view of Theorem 3.2 above it is natural to ask whether the converse of Theorem 3.2
is also true.
Is it possible to total order a non-archimedean separable topo-
i.e.
logical group in such a way that the order is compatible with both the group structure
That the answer to this question is in the negative is shown by
and the topology?
Theorem 3.4 given below.
For the counter example given in Theorem 3.4 we require the
following definition.
Each
Let
Q
x
in
x’
I/p
Ix-yl P
and d(x,y)
for
and
x,y
Then
Qp
The completion of
archimedean metric.
where
of
Q
d
Qp
does not divide
Xlp
is easily seen to be a non-
with this metric (which can be made
is called the field of
Q
into a field extending the addition and multiplication in
We consider the additive group of this field
p-adic numbers.
p
We define
is an integer.
s
Q
in
be a fixed prime number.
p
pSx’
can be written uniquely in the form
Q
the numerator and denominator of
s
Let
denote the field of rational numbers.
Qp
and denote it by
It is a separable totally disconnected group (see section 2, [8] for
itself.
details).
(see van Rooij [8]).
LEMMA 3.3.
PROOF.
ln!l P
Since
En!
follows that
A
B
which proves that
THEOREM 3.4.
Qp
PROOF.
topology.
In.n,l P
En.n!
and
tend to zero as
B
Then
There is no order
A + B
Qp
Suppose
Let
0
Qp
(Qp,+)
on
En! + En.n!
E(n+l)!
which is compatible with both
From the compatibility of the group structure
under this order.
s
n
k=l
0
k.k!
and from the compatibility of the order
-i
lim s
This proves that
simultaneously with the group structure
Since
+
+ alone.
+
0
which is a contradiction.
Qp
and the topology.
admits no order compatible
However by Theorem 2.6
of [2] it follows that it admits an order which induces the topology of
Qp
Qp
(see also
is a torsion-free abelian group it admits a total order
compatible with the group structure
4.
A
admits a total order compatible with the group structure and
is also impossible.
Remark 5.7 of [2]).
-I.
even though it admits an order compatible with the topo-
and topology it follows by Lemma 3.3 that
0
converges to
tends to infinity, it
n
alone and an order compatible with the group structure
and order it follows that
Similarly
Z n.n!
the series
-i.
and the usual topology of
logy of
and
Qp
In
+ alone (Corollary 5, p.36, [4]).
NON-TOTALLY DISCONNECTED ORDERED GROUPS.
In this section we consider only abelian non-totally disconnected groups. In what
follows we refer the reader to Wright [9] for the deinitions of radical-free, maximal
G. RANGAN
754
radical free etc.
A non-totally disconnected locally compact ableian topological group
THEOREM 4.1.
is orderable if and only if it is maximally radical free and its component is an open
subgroup of
PROOF.
topologically isomorphic to the reals.
G
Suppose
is a mon-totally disconnected locally compact abelian group
G
whose topology is given by a total order which is compatible with the group structure
of
G
Then by Corollary 2.3 its component
C
is an open subgroup topologically
Now by Theorem 5.1 of Wright [9] it follows that
isomorphic with the reals.
G
is
maximally radical-free in its interval topology.
Let now
G
satisfy the conditions of the Theorem.
Theore 5.2 of Wright [9]
orderable.
G
G
is connected then by
G
is
If not since the component of the identity is open and topologically iso-
morphic to the reals
G
is one-dimensional
free by Theorem 5.3 of Wright
By Theorem
If
is topologically isomorphic to the reals and so
of Isiwata
[I0],
G/C
is discrete and maximally radical
[9] and hence torsion-free by Theorem 4.1 of Wright [9].
G
can be totally ordered so that the order topology
coincides with the given topology of
G
and this order is compatible with the group
structure as well.
ACKNOWLEDGEMENT.
I thank Professors M. Rajagopalan and R. Venkataman for the many use-
ful discussions I had with them while preparing this paper.
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I.
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