ON ORDERABILITY OF TOPOLOGICAL GROUPS

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.

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.Venkataraman, Rajagopalan, and Soundararajan have shown (Theorem 2.6, [2]   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 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.THEOREM 2.1.A group G can be ordered if and only if a subgrodp C(x) and a homomorphism f from C(x) into the additive group of real numbers can be found -I for every t in C(x) i.e. if I :x/axa is the inner automorphism deter- a mined by the element a in G then f f oI x z a Further for the order on G C(x) for each x is the smallest convex subgroup con- taining x and C(x) ker f is the largest convex subgroup not containing x.
x PROOF.Let G be an ordered group.For each x e in G let C(x) denote the smallest convex subgourp containing x and C(x)* be the largest convex subgroup not containing x Then it i well known (see [4] and [5]) that C(x)* is a normal sub- group of C(x) and that C(x)/C(x)* is order isomorphic to a subgroup of the reals and we identify this quotient group with the corresponding subgroup of the reals.Let fx be the natural map of C(x) to C(x)/C(x)* Then clearly fx(X) 0 Since If we identify these two sub- groups of the reals forgetting the order isomorphism between them we get that f f oI which proves (iv) of the theorem.
x z a Conversely if G satisfies the conditions of the theorem we show that the set P of all elements such that f (x) > 0 serves as the strict positive part of an ordering x on G For x e G C(x) C(x-I) by (ii) and hence f We now observe a useful property of the subgroups C(x)'s which we call the pro- perty P. fy(X) + fy(y) 0. i.e. x,y e P implies that xy e P. -I Let x e P a e G Then f (x) > 0 By (iv) of the theorem we have aC(x)a i.e. z axa z z z a x belongs to P. Now by Theorem 2, p. 13, [4] it follows that G is an ordered group.
To prove the second part of the theorem we first show that C(x) is a convex -I subgroup by showing that y e C(x), e t < y implies that yt C(x) and hence This is done by proving that C(x) that C(x) _ C(yt -I) Suppose C(x) C(yt -I) From the assumption y C(x) and condition (ii) of the theorem we get that C(y C(y) c_ C(x) C(yt ).Property P now implies that C(yt-1) C(y-lyt-I) C(t-1) C(t) We will now show that this is impossible.For C(t) C(yt C(ty implies ft fyt -I fty -I and so 0 fyt-l(yt -I) f -l(y) + f -l(t The convexity of C implies that t C. f (x) fx(t) ==fxt-l(xt > 0 by (i) of the THeorem since t x. i.e.
x e t x. Again from the convexity of C we conclude that t C.
In either case t C. - -I Then f (xt 0 i.e. f (x) fx(t).From Property P we get that C(x2t-I) C(x) x x which in turn implies that f f fx2t-1 Now fx2t-l(x2t-l) f (x) + fx(Xt -I) The convexity of C now implies x x that t e C.
-I Thus we see that in all cases C(x) c_ C. proves that C(x) is the smallest convex subgroup containing x.
f is an order preserving homomorphism from C(x) to the reals.For let x -1 -1 -1 z,y C(x) and z > y Then C(zy C(x) C(zy zy zy fx (y) = fx (y) fx (z) In either case fx(Z) _> fx(y) This proves that Ker fx C(x) is a convex subgroup of C(x) and hence also of G.
It remains for us to show that C(x)* is the largest convex subgroup not contain- ing x For this let us suppose that C is any convex subgroup not containing x Let C. Then since x 4 C, C c_ C(x) and so t C(x).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.
REMARK.C, the intersection of all C(x), x e G, x e is the first convex subgroup of 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.PROOF.In view of Theorem 2.1, it is enough to show that condition (v) above is equivalent to the coincidence of the interval topology and the given topology of the group G Let now G be an ordered group.G with the interval topology is a topological group (see 4.19 [6]).We define C(x) and C(x)* for x G, x e as in Theorem 2.1.To prove the converse also we consider two cases.

CASE I. nC(x) (e)
From Case of the necessity part it follows that {C(x)}, x e G x e, form a neigh- bourhood base for the interval topology.By hypothesis {C(x)}, x e G, x e form a neighbourhood base at e 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 x y the reals, the topology of C being the relative topology got from the order topology of G Hence C with respect to this (order) topology is homeomorphic to a subgroup of the reals with respect to its relative topology.But by hypothesis f is an open continuous homomorphism from C, taken with the relative topology got from the given topology on G onto this subgroup of the reals.Now f is one-to-one implies that the two topologies on C conincide.C being an open subgroup in both the topologies on G the two topologies on G themselves coincide.
COROLLARY 2.3.Let G be an ordered topological group (i.e. the topology of G is giben by a total order compatible with the group structure of G).  open and closed subgroup, contains the component of e and so C itself contains the component.But the component being a connected subgroup is also a convex subgroup (see Proposition 1.3(2), [2]) and hence contains C which proves that C is the same as the component.From Theorem 2.2 it follows easily that C is topologically isomorphic to the additive group of reals.

TOTALLY DISCONNECTED ORDERED GROUPS.
A metric d on a set X is said to be a non-archimedean metric if d satisfies the stronger triangle inequality d(x,y) max(d(x,z), d(z,y) for x,y,z X.A topological group is said to be non-archimedean metrizable if there exists a right (or left) invariant metric on G which induces the topology of G (see Rangan [3]).
LEMMA 3.1.Suppose a topological group G is such that its topology is given by a non-archimedean metric d then there is an equivalent non-archimedean right (or left) invariant metric 0 on G (i.e. a non-archimedean metric 0 which gives the same topology as d).
PROOF.When d is a non-archimedean metric on G then the metric d' d/l+d also defines the same topology on G Put 0(x,y) sup d'(xz,yz) where supremum is taken as z varies in G Then 0 is well-defined since d' is bounded.It is easy to check that 0 is a right invaPiant non-archimedean metric on G We will now show that the topologies induced by 0 and d' coincide.
Since Ox,y) PROOF.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 G (see proof of Theorem 7 of [I] or [7]).Now from Lemma 3.1 it follows that G is non-archimedean metrizable.

C
(x)* and C(y)* are the largest convex subgroups contained in C(x) and C(y) respectively C(x) C(y) implies C(x)* C(y)* and hence f ) then being convex subgroups of an ordered group either C(x) C(y) or C(y) C(x) Hence (i), (ii) and (iii) of the theorem are easily verified.Inner -I -! automorphism being order preserving it follows that aC(x)a C(axa and aC(x)*a -I C(axa-1) * I induces an order isomorphism I* between the quotient a a -I groups C(x)/C(x)* and C(z)/C(z)* where z axa to (i).Hence C(a) C(ab).Similarly it can be seen that C(a) C(ba).If f (x) 0 f (y) 0 and C(x) C(y) then C(xy) C(x) C(y) and fxy(Xy)= x y fx(X) + fy(y) 0 since f f f For oherwise since xy C(x), C(xy) x y xy C(x) C(y) and hence by (iii) 0 f (xy) f (x) + f (y) O, a contradiction.If x x y C(x) can be however C(x) C(y) let us suppose that C(x) C(y) (the case C(y) similarly dealt with).By Property P, C(xy) C(y) and so f f and f (xy) xy y xy

CASE 2 .
C(x) C(t) C(xt -I) Then f -I f fWe now discuss the two possibilities f (x) > fx(t) (v) either the intersection of all the C(x), x G, x e is the singleton (e) in which case the collection {C(x)} x G form a neighbouhod base at the identity e of C or the intersection C of all the C(x), x e G, x e is an open subgroup of G and for each x in C f is an open continuous x homomorphism into the reals.
implies t N y since C(x) is convex.Hence t I i.e.C(x) E I and this proves that C(x), x G form a neighbourhood base of e in G CASE 2. oC(x) C z (e) Then C is a convex subgroup and hence open in G For x e G it is clear that C C(x) Hence for x,y C, C C(x) C(y) and so f f f -to-one on C f being the cannonical map from C(x) to C(x)/C(x) and x C(x)/C(x)* being order isomorphic to a subgroup of the reals, f is an open continu- x ous homomorphism from C(x) to the reals.
oC(x) (e) by Theorem 2.2 above G becomes a zero dimensional group.If nC(x) C(e) then C is the component of identity.For, each C(x) being an . the topology induced by p is finer than the topology induced by d'.To prove that the two topologies are the same we will show that each V M contains a U k for some k.Corresponding to the Integer M O, choose positive integers n,m such that 1/n + 1/m 1/M and choose z in G such that O(x,e) d'(XZo,Z + I/n. with respect to the metric d'.The continuity of x xz at x e implies the o existence of a positive integer k such that UkZ WZo Let x E U k. Clearly x is in W Hence 0(x,e)d'(XZo,Zo) + I/n I/m + I/n I/M proving thereby that U k V M This completes the proof of the lemma.THEOREM 3.2.Let G be a separable totally disconnected ordered topological group.Then O is non-archimedean metrizable.
Then the com- ponent of the identity e of G is either the trivial subgroup (e) or it is an open subgroup topologically isomorphic to the additive group of reals.