Equal column sum and equal row sum dimension group realizations

0.41 MB·English

by  David Handelman

Equal column sum and equal row sum dimension group realizations
Abstract Motivated by connections between minimal actions,
especiallyTplitz,onCantorsets,anddimensiongroups,wefind
realizationsofclassesofdimensiongroupsaslimits ofprimitive
matrices all of which have equal column sums, or equal row
sums.
David Handelman1
All groups are abelian, free means free as an abelian group, all partially ordered groups are un-
3 perforated and torsion-free. Equivalence classes representing elements of the direct limit, limA :
1 n
F F are expressed as [v,n], where v belongs to the nth free abelian group.
0 n n+1
→
2 Suppose U is a noncyclic subgroup of the rationals, and let τ : G U R be an onto group
n homomorphism from a torsion-free group G to U. We may impose an→order⊂ed group structure on
a G simply by declaring g G+ 0 iff τ(g) > 0. This makes G into a simple dimension group with
J ∈ \{ }
unique trace, and the trace is rational-valued; all such simple dimension groups are constructed in
3
this manner. Thatis, G is an extension(in thecategoryof abeliangroups)of a torsion-freeabelian
1
group kerτ by the subgroup U of the rationals.

] As G is a countable dimension group, by [EHS], it has a representation as ordered groups,
A
G limA : Zf(n) Zf(n+1) where f : N N is a function, we take the usual coordinatewise
F ≃ n → →
ordering on each Zf(n) and impose the usual direct limit ordering. The A have only nonnegative
. n
h
entries (and, by telescoping, can be made strictly positive when G is simple). However, [EHS]
t
a does not give specific representations, that is, the matrices A cannot be constructed from the
m n
argument, except by extremely complicated machinations. Here we consider the case that G be of
[
rankk+1(sokerτ is rankk), andprovide explicit realizationsfor G with theECS property(equal

1 column sums: each of the nonnegative(or strictly positive) matrices A has all of its column sums
n
v
equal).
9
9 With an ECS realization, there is a canonical choice of trace, namely (up to scalar multiple),
7 the sequence of normalized multiples of constant rows; in this case, we say the trace has an ECS
2
realization. We show that for general dimension groups with order unit, a trace admits an ECS
.
1 realization iff it is faithful, rational-valued, and good (in the sense of Akin, after translation to
0
dimension groups as in [BeH]). In this case, there is no control on the matrix sizes, but we do not
3
1 require simplicity.
: WhenwetakethetransposesofthematricesusedforECSrealizations,andthusobtainERS—
v
i equalrow sum—realizations, theresultingdimension groupsrun over allpossible simple dimension
X
groups of finite rank with unique trace (not generally rational-valued) which could admit an ERS
r
a realization. These are very closely related to Tplitz systems (pairs (X,T) consisting of a Cantor
set and a minimal self-homeomorphism, which is an almost everywhere one-to-one extension of an
odometer), as explained to me by Chris Skau, whose question about ERS realizations motivated
this paper.
An ERS realization of a simple dimension group G with respect to a (noncyclic rank one
subgroup)H suchthatτ(H) = 0andG/H istorsion-freeisanorderedgroupisomorphismφ : G
6 →
limA : Zf(n) Zf(n+1) where f : N N is some function,A are nonnegativeinteger matrices
n n
→ →
each having equal row sums, the direct limit ordering is imposed, such that φ(H) = [111 ,n]Z,
n f(n)
∪
with 111 the column consisting of ones. If f(n)= s for all n, the realization is of size s.
f(n)
If G and H are as in the prevous paragraph, and there exists an ERS realization of G with
respect to H that is also ECS, then we refer to this as an ECRS realization of G with respect to
H.
1 Supported in part by an NSERC Discovery Grant.
1
We establish following results on ECS, ERS, and ECRS realizations.
Let G be a dimension group.
(i) If G is simple, of rank k + 1 with unique trace τ, and τ(G) is a subgroup of the
rationals, then there exists an ECS realization of G of size k+2. (Theorem 4.1)
(ii) Let τ be a trace on G with τ(G) Q. Then there exists an ECS realization of G
⊆
representing τ if and only if τ is good (in the sense of Akin, as translated to the
dimension group setting [BeH]) and faithful (that is, kerτ G+ = 0 . (Theorem
∩ { }
6.1(b))
(iii) Suppose G is simple and has unique trace τ, and H is a noncyclic rank one subgroup
of G such that τ(H) = 0 and G/H is torsion-free.
6
(a) If rankG = k+1, then there exists a size k+2 ERS realization of G with respect
to H. (Theorem 7.1(a))
(b) There exists an ERS realization of G with respect to H. (Theorem 7.1(b))
(iv) Suppose G is as in (iii), and in addition, τ(G) Q.
⊆
(a) If τ(G) has no primes of infinite multiplicity (that is, τ(G) is not p-divisible for
any prime p), then G admits an ECRS realization with respect to H if and only
if λ := τ(G)/τ(H) rankG; when λ< , there is an ECRS realization of size λ
| |≥ ∞
but none of other sizes. (Theorem 11.9)
(b) If τ(G) has a prime of infinite multiplicity (that is, τ(G) is p-divisible for some
prime p), then G admits an ECRS realization with respect to H; this can be
constructed to be bounded if τ(G)/τ(H) < . (Theorem 11.13)
| | ∞
Part(ii)aboveappliestoalldimensiongroups(withanorderunit),buttheotherpartsrequire
simplicity and unique trace.
Much of the time, we work in the category of abelian groups with group homomorphism to
the reals: a torsion-free abelian group G together with a group homomorphism t : G R such
→
that t(G) is dense in R; we denote this (G,t). Isomorphism in this category is the obvious one,
(G,t) (G′,t′) if there exists a group isomorphism φ : G G′ such that t′φ is a nonzero scalar
≃ →
multiple of τ. Automatically, this induces an isomorphism kert kert′.
≃
Suppose that G and G′ are noncyclic simple dimension groups with unique trace, τ and τ′
respectively. Then G G′ as ordered groups if and only if (G,τ) (G′,τ′) as abelian groups with
≃ ≃
real-valued group homomorphism. One way is trivial. Conversely, suppose φ : G G′ is a group
→
isomorphism such that τ′φ = λτ for some nonzero real λ. By replacing φ by φ if necessary, we
−
may assume λ > 0. Then φ is an isomorphism of ordered groups.
To see this, we note that g G+ 0 iff τ(g) > 0; this is equivalent to τ′(φ(g))> 0, which is
∈ \{ }
equivalent to φ(g) (G′)+ 0 . As φ is a group isomorphism, this says both φ and φ−1 are order
∈ \{ }
preserving, hence φ is an order isomorphism.
Hence to decide if two simple dimension groups with unique trace are order isomorphic, it
is sufficient to find a group isomorphism between them that scales the trace(s). This makes life
simple, at least when the dimension group has unique trace.
Thedimension groupswe will beconsideringforECS realizationshave anadditionalproperty,
that the range of their trace is (up to nonzero scalar multiple) a subgroup of the rationals. So we
consider them as groups with real-valued group homomorphism, (G,t) such that t(G) is rank one
and dense in R (so up to scalar multiple, t(G)= U Q).
⊆
Although we will often be talking about extensions of abelian groups, 0 C G U 0,
→ → → →
itistoorestrictivetodealwiththeclassificationasextensions(thatis,withinExt1(C,U)); instead,
we are dealing with the coarser classification, isomorphism for maps G U, where we are allowed
→
2
to multiply by 1 (and if U is p-divisible, by powers of p). There are still generically uncountably
±
many isomorphism classes of these, since Aut(C) and Aut(U) are usually countable.
As usual, if a groupor orderedgroupis given as a directlimit, limM :F F (typically,
n n n+1
→
F will be free abelian groups, and if the ordered direct limit is required, the entries of M will be
n n
nonnegative), then elements of the direct limit can be written as equivalence classes, [a,n] where
a F , and the equivalence relation is generated by [a,n] = [M a,n+1].
n n
∈
1 Via subsemigroups
For this section, G need only be a partially ordered group with positive cone G+. Let P de-
note the set of nonnegative integers. If a G+, we denote by a P, the set of sums
i i
{ } ⊆
a n(i) n(i) P , the semigroup (or subsemigroup) generated by a .
i i
{ | ∈ } { P}
Let S be a collection of subsemigroups of G+ with S S S ... such that
P { n}n∈N 1 ⊆ 2 ⊆ 3 ⊆
f(n)
G+ = S . SupposeS isgeneratedby a(n) . SinceS S ,wecanfindanf(n+1) f(n)
∪ n n i i=1 n ⊆ n+1 ×
matrix A (called a transition matrix)nwith eontries from P such that for all i,
n
f(n+1)
(n) (n+1)
(*) a = (A ) a ;
i n ji j
j=1
X
there is usually a great deal of choice available for the matrix entries, since there is no assumption
of any sort of unique decomposition. Note the subscript ji, not ij.
Let F = Zf(n), the free abelian group on f(n) generators (denoted e(n), but when su-
n i
perscripted (n) is understood, it is removed), equipped with the usual coordinatewise ordering.
Now form the direct limit dimension group from the A s, H = limA : F F . Define
n n n n+1
→
(n)
ψ : F G via ψ (e ) = a . This is a well-defined positive group homomorphism from F
n n → n i i n
to G. The condition in (*) is precisely what we need in order that ψ A = ψ . Hence the
n+1 n n
◦
family ψ induces a positive group homomorphism Ψ : H G (explicitly, Ψ[v,n] = ψ (v)
n n
{ } →
where v F ).
n
∈
Since G+ = S , Ψ(H+) = G+; if we assume G is directed (as we may as well), that is,
n
∪
G = G+ G+, then Ψ is onto. If Ψ is one to one, then it is an isomorphism of ordered groups
−
(in particular, G is a dimension group), and we have a realization for it using the matrices A .
n
If rankH rankG < , then Ψ is automatically an isomorphism (since an onto homomorphism
≤ ∞
from a torsion-free abelian group of finite rank to a torsion-free group of the same or more rank is
automatically one to one), then Ψ is an isomorphism.
The construction of Ψ depends on the choice(s) of the generators for the semigroups S , and
n
then on the matrices A ; different choices for the matrices (even fixing the generators of all the
n
S ) can result in different Ψ functions, some of which may be one to one while others need not be.
n
We summarize this in one gigantic statement.
LEMMA1.1 Suppose that G is a partially ordered abelian group with an increasing set of
subsemigroups, S S ... such that G+ = S , and suppose that A is the transition
1 2 n n
⊆ ⊂ ∪
matrix associated to a choice of generators for S S , with each S generated by
n n+1 n
⊂
a(n) . Form the dimension group H = limA :F F .
i n n → n+1
n(a) Tohere is a unique positive group homomorphism Ψ : H G such that [e(n),n] a(n);
→ i 7→ i
moreover, Ψ(H+) = G+.
(b) If Ψ is one to one, then it is an isomorphism of ordered abelian groups, and thus G
is a dimension group.
3
(c) If G is torsion-free such that rankH rankG and rankH < , then Ψ is an ordered
≤ ∞
group isomorphism.
2 Realizing G as ECS (free kernel that splits)
Over this and the next few sections, we deal with the simple dimension group G of rank k + 1
having unique trace τ, and in addition, τ(G) = U is a rank one (necessarily noncyclic) subgroup
of the reals. For expository reasons, we proceed in three steps.
This section deals with a rather special case, that kerτ be free of rank k and the extension
splits. In the next section , we drop the splitting property (but maintain freeness of the kernel);
finally, we deal with the general case, wherein kerτ is an arbitrary rank k torsion-free abelian
group, and the extension by U is arbitrary. We could go straight to the general case, but this
would have resulted in a very complicated argument. Instead, as we proceed through the cases,
we find the extra complications can be dealt with in a relatively smooth manner.
Herewedealwiththeeasiestcase,G = U Zk whereG+ 0 = (u,w) u> 0 . Althoughwe
⊕ \{ } { | }
know that G = U Zk is a dimension group, and G+ 0 = (u,w) u Q++ and w Zk , and
⊕ \{ } ∈ ∈
thus is a limit of free abelian groups with their coordinatewise limit by [EHS], the latter does not
(cid:8) (cid:12) (cid:9)
give an explicit form. Here we obtain from a natural (but not the mo(cid:12)st natural) subsemigroups of
G+, an explicit realization with all the matrices being size k+2 and column stochastic(all column
sums equal for each matrix; this is abbreviated ECS). The following is the result of this section.
PROPOSITION 2.1 Let G = U Zk where U is a noncyclic subgroup of the rationals, and
⊕
G is the simple dimension group obtained from the map G U. Then G can be realized
→
as a direct limit (in the category of ordered abelian groups) limA : Zk+2 Zk+2 where
n
→
A are primitive and ECS.
n
First, we finda suitablerepresentationof Zk as a unionof k+1subsemigroups. For 1 i k,
let ǫ denote the standard basis vector of Zk, and set ǫ = ǫ . Obviously k+1ǫ ≤=000≤and
i k+1 − i i=1 i
it is easy to verify that k+1ǫ P = Zk.
j=1 i P P
Now let the supernatural number of U be given. We may block (telescope) all the primes
P
and their powers that appear, so that we have a sequence of positive integers p ,p ,..., with
1 2
p > (k+1)2 for all n and U lim p : Z Z. Let q = n p . Now define the elements, for
n ≃ × i → n i=1 i
i = 0,1,2,...,k+1,
Q
1 ,000 if i = 0
(n) qn
a =
i  (cid:16) 1 ,ǫ(cid:17) if 1 i k+1
 qn i ≤ ≤
k+1 (n) (cid:16) (cid:17)
Set Sn = i=0 ai P, so that f(n)is constant in n with value k + 2. Now we can write (in
(n) (n+1) (n)
lots of different ways) a as a nonnegative linear combination of the a , for example, a =
P i i 0
(n+1) (n) (n+1) (n+1) (n+1)
p a and a = a +2a +(p k 1)a (this exploits the facts that
n+1 0 i j6=0,2 j i n+1− − 0
k+1ǫ = 000 and p > k +1; in fact, we assumed p > (k +1)2 which we will need later).
i=1 i n+1P n+1
This yields that S S ; the matrices resulting from these representations are not suitable for
n n+1
P ⊆
our purposes, as the resulting map Ψ is not one to one.
ItiselementarythatG+ = S ;anarbitraryelementofG+ isoftheformx = (b/q ,v)whereb
n n
∪
isapositiveintegerandv Zk. Letdbethemaximumabsolutevalueofthecoordinatesofv (asan
∈
elementofZk,i.e.,theusualcoordinates),andfindlsothatp p ... p > (k+1)d(b+1). We
n+1 n+2 n+l
· · ·
k+1
can find positive integers r(1),r(2),...,r(k+1) with r(i)ǫ = v such that r(i) < d(k+1).
i=1 i
Then
P P
k+1 k
b 1 1
,v = r(i) ,ǫ + b(p p ... p ) r(i) ,0
i n+1 n+2 n+l
q q · · · − q
(cid:18) n (cid:19) i=1 (cid:18) n+l (cid:19) i=1 !(cid:18) n+l (cid:19)
X X
4
expresses x as an element of S .
n+l
Now we make a very particular choice of the transition matrices, A ; not only do they have
n
to satisfy (*), but they have to be rank k+1 (or less, but strictly less is not possible, except for
finitely many n). Since the matrices are all square of size k +2, the rank condition turns out to
be not so onerous, especially since imposing the obvious constraint on the trace will force the rank
condition to apply.
Temporarilydropthesubscriptn on some of thevariables, so we will obtaina matrixA whose
large eigenvalue is p (corresponding to p ); we insist that p > (k+1)2. We write,
n
k+1
(n) (n+1) (n+1)
a = (p k 1)a + a
0 − − 0 i
i=1
X
(n) (n+1) (n+1)
a = (p 1)a +a for 1 i k
i − 0 i ≤ ≤
k
a(n) = (p k2 k 1)a(n+1)+ ka(n+1)+(k+1)a(n+1).
k+1 − − − 0 i k+1
i=1
X
These relations are trivially verified by using k+1ǫ =000. The correspondingmatrix (we act from
i=1 i
the left, so each equation gives rise to a column), is rather simple to describe (but really tedious
P
to TEX). There is a k k identity matrix occupying most of the space.
×
p k 1 p 1 p 1 p 1 ... p 1 p 1 p k2 k 1
− − − − − − − − − −
1 k
 
1 k
. .
A =  .. I .. .
 k 
 1 k 
 
 1 k 
 
 1 0 0 0 ... 0 0 k+1 
 
 
Aconsequenceoftheequationsisthatthecolumnsumsareallp. Ifwesumallbutthefirstcolumn,
theresultisk+1timesthefirstcolumn(asfollowsfromp k2 k 1+k(p 1)= (k+1)(p k 1)).
− − − − − −
Hence rankA < k+2, that is, rankA k+1.
≤
Now restore the subscript n; we obtain square matrices A of size k + 2 with nonnegative
n
entries, such that rankA k+1, each with large eigenvalue p (this latter is not needed now).
n n
≤
ThentherankofH = limA : Zk+2 Zk+2 is atmostliminfrankA k+1,andsothepositive
n n
→ ≤
map Ψ : H G obtained from this sequence of relations is an isomorphism of ordered groups.
→
This yields an ECS realization of G.
•
3 Arbitrary extensions by free abelian groups
Now wetrytofindexplicit realizationsof dimensiongroupsG thataregiven as arbitaryextensions
ofZk byU asubgroupoftherationals. Explicitly, wehaveashortexactsequenceofabeliangroups
0 Zk G U 0; regarding U as a subgroup of the reals, the quotient map τ : G U
→ → → → →
yields the ordering: for nonzero g in G, g G+ iff τ(g) > 0. This describes all dimension groups
∈
with unique trace, whose values lie in the rationals, and whose kernel is free abelian of finite rank.
The previous constructions of course dealt with the case wherein τ is split.
PROPOSITION 3.1 Let G be a noncyclic simple dimension group with unique trace t
such that t(G) := U is a subgroup of the rationals, and such that kert is free of rank k.
Then G admits an ECS realization by primitive matrices of size k+2.
5
Given the data (G,τ,U,kerτ) we can construct semigroups S S ... of G+ with the
n n+1
property that G+ = S . To begin with, write U = lim p : Z ⊆Z; for⊂m q = n p , and
∪ n × n → n i=1 i
make an initial selection,one for eachn, of g τ−1(1/q ). Theng p g kerτ, so we can
n n n n+1 n+1
∈ − ∈ Q
write g = p g +vn for unique vn kerτ = Zk (obviously, vn depends on the selection of
n n+1 n+1
∈
the sequence (g ). The sequence (v ,p ) determines the isomorphism class of G, but by itself,
n n n+1
this is not that useful.
The obvious candidate for the subsemigroup S is the semigroup generated by
n
g ,g +ǫ ,...,g +ǫ ;g +ǫ ,
n n 1 n k n k+1
{ }
k
where ǫ have their usual meaning: standard basis elements if i < k +1 and ǫ = ǫ ;
i k+1 − j=1 j
k+1
it is convenient to define ǫ = 0, so we can write S = (g +ǫ )P. Unfortunately, there is
0 n i=0 n i P
no guarantee that S S (in other words, that the matrix entries be nonnegative), largely
n n+1
⊆ P
because at this stage, we have no control over vn.
The idea is to make a better choice of g , and then a telescoping (amounting to telescoping
n
the p ), and if we are careful, we will obtain S S for all n, and the correspondingtransition
n n n+1
⊆
matricescanbe writtendown; in fact, we will writedown the transitionmatrices, verify theentries
are all nonnegative, from which it follows that the S are increasing.
n
Letusseewhatweneedtoobtainthis;wewillwritedowntherelationsbetweenthegenerators,
and hope for the best. Fix n and order (g + ǫ ) according to the subscript of the ǫ , with
n i i
0 i k +1. The relations are given by g +ǫ = A (g +ǫ ), where A are integers,
≤ ≤ n i j ji n+1 j ji
hopefully nonnegative, and this forces various equations.
P
Since τ(g +ǫ ) are all equal to 1/q = p /q and τ(g +ǫ ) = 1/q , we deduce that
n i n n+1 n+1 n+1 i n+1
for all j, A = p , that is, the row sums of AT are all p , so that the column sums of A
i ji n+1 n+1
are all p . (This is a useful way of calibrating the matrix—I am always confused as to whether
n+1
P
it should be A or AT, and keeping in mind that the column sums must be equal determines which
it is.)
Now fix i; we have the equation, g +ǫ = p g + A ǫ . Using g = p g +vn,
n i n+1 n+1 j ji j n n+1 n+1
we have, for all i = 0,1,...,k + 1 (suppressing the subscript n on A , as otherwise it gets too
n
P
crowded),
k+1
vn +ǫ = A ǫ .
i ji j
j=0
X
When i = 0 (so ǫ = 0), we obtain
i
A A = (vn) if j > 0
j,0 k+1,0 j
−
k
A = p A = p (k+1)A (vn) .
00 n+1 i,0 n k+1,0 i
− − −
i>0 i=1
X X
Already we see a problem; the coefficients are suppose to be nonnegative, and so we require
p (k+1)A + k (vn) withA 0(wehavenocontrolyetonthesumofthecoefficients
n ≥ k+1,0 i=1 i k+1,0 ≥
of vn). Anyway, we continue; for 1 i k,
P ≤ ≤
A = A +(vn) +1
i,i k+1,i i
A = A +(vn) if j = i and 1 j k
j,i k+1,i j
6 ≤ ≤
k+1 k
A = p = A = p (k+1)A (vn) 1.
0,i n+1 j,i n+1 k+1,i i
− − −
j=1 i=1
X X
6
Finally, with i= k+1,
A = A +(vn) 1 if k+1 > j > 0
j,k+1 k+1,k+1 j
−
k
A = p A k = p (k+1)A (vn) k.
0,k+1 n+1 i,k+1 n k+1,k+1 i
− − − − −
i>0 i=1
X X
Now set a = A (obviously this depends on n, but for now we suppress the sub/superscript);
i k+1,i
then all the entries are linear in the choice of a . If the entries do happen to be nonnegative, then
i
the resulting matrix A = (A ) (order of the subscripts reversed) will implement the embedding
n ij
S S . The resulting matrix is similar to the preceding ones, in that the interior k k matrix
n n+1
⊆ ×
is v 111T +111 (a ,...,a )+I where 111 is the column of size k consisting of ones, we regard v as
1 k k
· ·
a column, and represents the usual product of matrices. Notice that v 111T is k k but vT111 is
just the sum of·the coefficients of v, k (vn) . We sometimes suppress th·e sub/su×perscripts n or
i=1 i
n+1 in vn and p , and the implicit superscripts in a(n).
n+1 P i
(1)
p (vn)T111 (k+1)a (p (vn)T111)111T (k+1)(a ,a ,...,a )
n+1 0 n+1 1 2 k
− − − − ∗
A = vn+a 111 vn111T +111(a ,...,a )+I vn+(a 1)111 ,
n 0 1 k k k+1
 − 
a a ,a ,...,a a
0 1 2 k k+1
 
wherethe(0,k+1)entry(theupperright;leftblank,becauseofoverflow)isp vT111 (k+1)a +
n+1 0
− −
k. The column sums are all p , as follows from the choice of generators of the subsemigroups.
n+1
Withoutyetworryingaboutpositivityorrank,wecancalculatetheeigenvaluesandtheirgeometric
multiplicities, by explicitly computing the left eigenvectors.
First, 111T is the left eigenvector for p. Next, define ⊥v := w Z1×k wv = 0 (we use Zm
k+2 ∈
tomeanZm×1, thatis, thedefaultis columns). For eachu ⊥v, therowof sizek+2, (0,u, uT111),
∈ (cid:8) (cid:12) (cid:9) −
is a left eigenvector for the eigenvalue 1. If v =000 (as we are assuming implic(cid:12)itly anyway), then ⊥v
6
is rank k 1, and thus even the geometric multiplicity of 1 as an eigenvalue is at least k 1.
− −
This leaves two eigenvalues. We may find u Q1×k such that u v = 1 p + vT111 ; then
0 0 k
∈ −
(1,u ,k+1 u 111 ) is another left eigenvector for the eigenvalue 1 (we may multiply by an integer
0 0 k
−
and so obtain an integer eigenvector if desired), and since its first coordinate is not zero, it is not
in the R-span of the previous eigenvectors for 1; hence the multiplicity of 1 is at least k.
There is one remaining eigenvalue, in addition to p,1k, and it is easily determined from the
k+1
trace; the trace of the matrix is p+k+ a (k+1)a , hence the last remaining eigenvalue
i=1 i − 0
k+1
is a (k +1)a . Since we want the rank of the matrix to be k +1, we are free to choose
i=1 i − 0 P
k+1
any selection of integers a such that a = (k + 1)a (that is, a is the average of all the
P i i=1 i 0 0
others). Whenthisisimposed,weseequicklythatthecorrespondingrelationholdsforthecolumns,
P
that is, the sum of all but the first column is k + 1 times the first. In other words, if we set
z = (k+1, 1, 1,..., 1)T Zk+2, then Az =000. Moreover, z is independent of the choice of n
− − − ∈
(that is, A z =000 for all n).
n
A particular consequence is that W := ⊥z = w Z1×(k+2) wz = 0 is a common A -
n
∈
invariant subgroup(on the left, of course,meaning WA W for all n); moreover, the eigenvalues
n
(cid:8) ⊆ (cid:12) (cid:9)
of A restricted to this subgroup are exactly p,1k (the zero eigenv(cid:12)ector has conveniently been
n
eliminated, since z spans, as a real vector space, the right zero-eigenspace of all the A ).
n
Now we modifythe sequence(g ) and corresponding(vn) to permit a selection of integersa
n i,n
k+1
(and with a = (k+1)a ) so that the matrix A has only nonnegative entries.
i=1 i 0 n
Let G be given by the sequence (p ,vn). Let E GL(k,Z) and W Zk; then the group
n+1
P ∈ ∈
extension given by the sequence (p ,Evn +(p 1)W) is equivalent. To see this, form the
n+1 n+1
−
7
square matrices of size k+1, A = pn+1 000 and F = 1 000 . Then F−1 = 1 000 , and
n vn Ik W E −E−1W E−1
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
p 000
D := FA F−1 = n+1 .
n n Evn +(p 1)W I
n+1
(cid:18) − (cid:19)
Now G limA : Zk+1 Zk+1 as abelian groups, and the map on each copy of Zk+1 given
n
≃ →
by F induces a group isomorphism from G to G′ := limD : Zk+1 Zk+1. The corresponding
n
→
data for the sequence of D s is (p ,Evn +(p 1)W) and these maps preserve the map to
n n+1 n+1
−
U. Of course the drawback with this equivalence relation is that it applies to all the vns at once.
LEMMA 3.2 Suppose V (R+)k. There exist E GL(k,Z) and W Zk such that the
∈ ∈ ∈
coefficients of EV W are all nonnegative, and sum to less than one.
−
Proof. Define W0 Zk via (W0) = V (the floor function). Then V := V W0 has all its
i i 0
∈ ⌊ ⌋ −
coefficients nonnegative and strictly less than one. If either all or all but one of the entries of V is
0
zero, we are done.
Otherwise, let s = max V and t = max V s (the notation is not very clear, but if
i i
{ } { \{ }}
there is a tie for maximum, then t = s). We apply the division algorithm to s and t; there
exists an integer m > 1 such that s = mt + s′ where 0 s′ < t; this is implemented by an
≤
elementary transformation, hence by an element of GL(k,Z), and the new vector (replacing s in
one its positions by the smaller s′) either has strictly smaller maximal entry, or the multiplicity of
its maximal entry has been reduced. If in the resulting vector, there are still more than one entry,
we can continue the process.
Itiseasytoseethatiteitherterminatesinasinglenonzeroentry(whichoccurspreciselywhen
all the nonzero entries of V are rational multiples of each other), or we can make the maximal
entry as small as we like, say less than 1/k. Either way, we have constructed E as a product of
elementarytransformations(hencein GL(k,Z)) such thatE(V W0) has onlynonnegativeentries
−
and whose entries sum to less than 1. Now set W = EW0 Zk.
∈ •
LEMMA 3.3 Suppose that the extension G of Zk by U is implementable by (p ,vn)
n+1
such that vn/p converges (in Rk) and p . Then the corresponding dimension
n+1 n+1
→ ∞
group G with unique trace being the map to U is realizable as a limit A : Zk+2 Zk+2
n
→
with A of the form (1) primitive.
n
Proof. Set V = limvn/p . By the preceding, there exists E GL(k,Z) and W Zk such
n+1
∈ ∈
that EV W has only nonnegative entries adding to λ < 1. Now the extension corresponding
−
to (p ,vn) is equivalent to (p ,(vn)′ := Evn (p 1)W), so it suffices to show that
n+1 n+1 n+1
− −
(p ,(vn)′) can be realized by a sequence of primitive matrices of the form (1).
n+1
We observe that (vn)′/p = Evn/p W(p 1)/p , and this sequence converges
n+1 n+1 n+1 n+1
− −
to EV W. Thus, given ǫ < min (1 λ)/3(k+1),λ/3k , for all sufficiently large n, we have
− { − }
ǫ111 (vn)′/p and the sum of the entries is less than λ+ǫ. Thus
n+1
− ≤
ǫp (vn)′ and (vn)′ < λ+ǫp .
− n+1 ≤ i i n+1
X
Set a = a = = a to be 1 if min(vn)′ 0 and equal to 1 min(vn)′ otherwise.
0 1 ··· k+1 i ≥ − i
If min(vn)′ 0, then we note that from (vn)′ (λ+ǫ)p , we obtain an upper bound
i ≥ i ≤ n+1
on the sum, p µ, where µ = 1 (1 λ)(1 1/3k) (what is important is that the coefficient is
n+1
− − −P
boundedaboveawayfromone,uniformlyinsufficientlylargen). Thenp (vn)′ (k+1) 1
n+1− i− − ≥
p (1 µ) (k+2). Sincep , forallfurthersufficientlylargen,thisexpressionispositive;
n+1 n+1
− − → ∞ P
thus the matrix A in (1) has only nonnegative entries.
n
8
If min(vn)′ < 0, then for any j,
i
p (vn)′ (k+1)a 1 p p (λ+ǫ) (k+1) (k+1)p ǫ 1
n+1− i− j − ≥ n+1 − n+1 − − n+1 −
X p (1 λ (k+2)ǫ) (k+1).
n+1
≥ − − −
Now 1 λ (k+1)ǫ> 1 λ (1 λ)/3= 2(1 λ/3)> 0. Hence by further increasing n, we have
− − − − − −
that for all sufficiently large n, the matrix entries of A are nonnegative. We can always delete a
n
k+1
finite number of the matrices at the outset. Because (k+1)a = a , the rank of each A is
0 i=1 i n
k+1, so that Lemma 1.1 applies.
P •
LEMMA 3.4 Let G be a group extension of Zk by noncyclic U Q with data (p ,vn).
n+1
⊆
Then there is an equivalent representation, (q ,(vn)′), such that q is increasing,
n+1 n+1
q , and (vn)′/q converges.
n+1 n+1
→ ∞
Proof. First, we may make an initial telescoping, and thus may assume that p are increasing
n+1
to infinity at the outset. Now we perform the following substitution transform, to ensure that the
resulting v entries are all between 0 and p 1. Suppose we have done this up to n = m 1;
n n+1
− −
thatis, g = p g +vi for1 i m 1. Now set g′ = g +um, the um to bedetermined.
i i+1 i+1 ≤ ≤ − m+1 m+1
Then we have g = p g′ +vm p um (and the subsequent relations, for larger m, are
m m+1 m+1 − m+1
also affected, but we come to themby the induction argument). We can obviously choose um Zk
∈
so that all the entries of vm p um lie in the set 0,1,2,...,p 1 . This completes the
m+1 m+1
− { − }
induction, and allows us to assume that the newly relabelled vn satisfy 000 vn (p 1)111.
n+1
≤ ≤ −
In particular, with the current notation, vn/p is a bounded sequence in [0,1]k. Hence
n+1
{ }
there exists a subsequence indexed by n(i) N such that vn(i)/p converges, say to V
n(i)+1
∈ ∈
[0,1]k. The integers n(1) < n(2) < n(3) < suggest a telescoping; set M = pj+1 000 ,
··· → ∞ (cid:8) (cid:9) j vj Ik
discard the M for j < n(1), and define (cid:16) (cid:17)
j
Mn(i) = M ... M M ;
n(i+1)−1 n(i)+1 n(i)
· · ·
n(i+1)−n(i)
the upper left entry is q = p , and of course the lower right k k is the identity
i j=1 n(i)+j ×
matrix. The column of size k to the left of the identity is obtained by an easy induction argument.
Q
This yields (and probably better is to prove this by induction):
v(i) vn(i) vn(i)+1 vn(i)+2
= + + +...
q p p p p p p
n(i)+1 n(i)+1 n(i)+2 n(i)+1 n(i)+2 n(i)+3
v(i) vn(i) vj 1 1
sup + +...
q − p ≤ p p p p
(cid:13) n(i)+1(cid:13)∞ (cid:13) j+1(cid:13)(cid:18) n(i)+2 n(i)+2 n(i)+3 (cid:19)
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) 1 (cid:13) 1 1 2 1 3
(cid:13) (cid:13) δ (cid:13) (cid:13) 1+ + + +...
≤ p p p p
n(i)+2 n(i)+2 (cid:18) n(i)+2(cid:19) (cid:18) n(i)+2(cid:19) !!
δ
= .
p 1
n(i)+2
−
The δ was obtained from vn/p being a bounded sequence. We have used p p
n+1 n+2 n+1
{ } ≥
to convert the estimate into a geometric series. Finally, since p , the sequence v(i)
n+1 → ∞ qi
converges.
n o
•
The previous two results yield Theorem 3.1.
9
4 Arbitrary extensions
In this section, we deal with arbitrary extensions of U by arbitrary finite rank torsion-free abelian
groups, instead of Zk; that is, let C be a rankk torsion-freeabelian group, U a noncyclic subgroup
of therationals, and considerextensions0 C G U. We realize thedimension groupG with
→ → →
strict ordering induced by G U R as a limit of ECS matrices of size k +2. The arguments
→ ⊂
are rather similar to those of the previous section, but involve a couple of extra features.
Let C = kert; this has rank k, so we can write C as a limit (as abelian groups) B : Zk Zk
n
→
for some choice of B with detB = 0. We can incorporatethe identity as many times as we wish,
n n
6
say B ,I ,...,I ,B ,I ,...,I ,B ,..., and this gives the same abelian group (the idea is that
1 k k 2 k k 3
we will be telescoping the k + 1 size matrices, and we want to ensure that the absolute column
sums of the B are ooo(p )). Re-indexing, we can obtain G (the extension) as the abelian group
n n+1
direct limit arising from the square matrices of size k + 1, pn+1 000 , and we can assume that
vn Bn
Bn ∞,∞ =ooo √pn+1 (or anything reasonable). (cid:16) (cid:17)
k k
When we realize the corresponding semigroup coming from the relations, we obtain rather
(cid:0) (cid:1)
similar matrices to those previously encountered. Let ǫn be the standard basis elements for Zk at
i
the nth level, and define ǫn =000 and ǫn = k ǫn.
0 k+1 − i=1 i
We can express G (as an abelian group with real-valued homomorphism) as the limit, G =
P
limM := pn+1 000 : Zk+1 Zk+1, where the common left eigenvector x = (1,0,0,...,0)
n vn Bn →
induces the(cid:16)map to U(cid:17). Call the nth copy of Zk+1, F , so the elements of G are the equivalence
n
classes[a,n] = [M a,n+1]wherea F , andthemap toU is given by t : [a,n] xa/p ...p . In
n n 1 n
∈ 7→
particular,thecorrespondingpositivecone(ofthedimensiongroup,onceweputthestrictordering
arising from t) will be t−1(1/p ...p )P.
1 n
∪
Set g = [xT := (1,0,...,0)T,n]; so t(g ) = 1/p p . Then g p g = [M xT
n n 1 n n n+1 n+1 n
··· − −
xT,n+1], and this is simply the column whose top entry is 0 and the rest of which is vn, which
we write as k vnǫn+1.
i=1 i i
Take as our generatorsfor the positive cone at the nth level, g +ǫn (now i = 0,1,...,k+1 to
P n i
incorporatethetwo extraelementsrequired);werewritethisasg + k (B ) ǫn+1 (arisingfrom
n j=1 n ji j
the effect of B ; note that the coefficients are transposed). From the equation g = p g +
n n n+1 n+1
P
k vnǫn+1, we want to find a (k+2) (k+2) matrix An := (An) (0 i,j k+1; this is not
i=1 i i × ij ≤ ≤
supposed to represent the nth power of some matrix A, but is merely superscripting the indexthe
P
previous notation was A ; we frequently drop the super/subscript n for simplicity) such that for
n
all i,
k k+1
g + (B ) ǫn+1 = (An) (g +ǫn+1).
n n ji j ji n+1 j
j=1 j=0
X X
We are free to choose the entries of An (subject of course to positivity constraints) so long as this
setofequationsholds. First,fromt(g +ǫn) = 1/p ...p ,thecolumnsumsofAn mustallbep .
n j 1 n n+1
Hence we require k+1An = p for all i. Using the relation g p g = k vnǫn+1, the
j=0 ji n+1 n− n+1 n+1 i=1 i i
previous equations become
P P
0 if i = 0
k+1 k
Anjiǫnj+1 = vlnǫnl+1+ k1(Bn)jiǫnj+1 if 1 ≤ i≤ k
Xj=0 Xl=1  −P kl=1 kj=1(Bn)jlǫnj+1 if i = k+1.
 P P
Restrict to the case j > 0 (each entry in the top row is determined by the remaining ones in its
column, as the column sums are all p ); taking coordinates, we obtain the following equations,
n+1
10