This paper characterizes the rings of continuous linear operators on certain topological spaces of formal sums, extending to modules over ordered groups, thus advancing the understanding of operator structures in topological algebra.
Contribution
It describes the rings of continuous linear operators on topological spaces of formal sums associated with filters and involutions, generalizing previous results to modules over ordered groups.
Findings
01
Characterization of rings of continuous operators on formal sum spaces
02
Application to modules over left ordered groups
03
Extension of operator theory in topological algebra
Abstract
The rings of linear continuous operators on the topological spaces of G-zero maps were described, where G is a filter on a set with an involution. This applies to modules of formal series with well ordered support over left ordered groups.
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Taxonomy
TopicsRings, Modules, and Algebras · Advanced Topology and Set Theory · Homotopy and Cohomology in Algebraic Topology
Full text
Topological linear spaces of formal linear sums and continuous linear operators
Nikolay I. Dubrovin
Department of Algebra and geometry, Vladimir State University,
Gorky Street 87, Vladimir, 600000, Russia
The rings of linear continuous operators on the topological spaces
of G-zero maps were described, where G is a filter on a set with an involution.
This applies to modules of formal series with well ordered support over left
ordered groups.
Key words and phrases:
Filter, operator topology, involution, left ordered groups
2000 Mathematics Subject Classification:
06F15, 46K05, 54A20
The author is supported by the EPSRC grant no. EP/D077907/1.
He also thanks Manchester University for kind hospitality.
1. Introduction
If X is a topological space, then the ring C(X) of continuous functions from
X to R is a classical object in topology and analysis. For instance, one can be
interested in properties of C(X) as a ring, and from this point of view the situation
is well understood (see [7]).
If X has an additional structure, for instance, if X is a linear topological space,
one can consider the properties of the ring of linear continuous functions from X
to R endowing it with different ‘natural’ topologies (see [1]). However
sometimes we have to deal with the case, when the target, K, of maps from X is a
noncommutative skew field. For instance, this situation occurs trying to embed a group ring
FG of a (torsion-free) group G over a field F into a skew field.
For example, let G be the universal covering of the group SL2(R) and
U=\bigl{\{}\left(\begin{smallmatrix}a&b\\
0&a^{-1}\end{smallmatrix}\right)\mid a,b\in\mathbb{R},\,a>0\bigl{\}} is a subgroup of
SL2(R). Since U is metabelian and torsion-free, the group ring FU (over any field F)
is an Ore domain, therefore its classical quotient field K is a (noncommutative) skew field.
Trying to extend this
embedding to an embedding of KG into a skew field, the first author developed the
following approach (see [3]). He considered the space K{CoDccG} of all formal
series on G with well ordered support, and the group ring KG acting on this space by left
multiplication. Then one can invert the elements of KG as linear maps forming a
rational closure D of KG in K{CoDccG}. The behavior of elements of D is
quite complicated, and the aforementioned paper contains a series of algebraic conditions on
elements of D. It is quite difficult and tedious to verify that these conditions
respect basic operations. Later (see [6]) Dubrovin noticed that an essential part of the
proof can be simplified by endowing K{CoDccG} with a structure of a linear topological
space such that elements of D become linear continuous maps. Thus the aforementioned
algebraic conditions can be understood as well known properties of continuous maps.
In this paper we develop a very general approach to tackling this situation. Namely,
with each filter G on a set G we connect a linear space K{G} of all maps from
G to K whose zero set belongs to G. We endow this space with a linear Hausdorff
topology making it into linear topological space.
The examples of such topologies
include Tychonoff topology on the product of spaces, but also the adic topology on the space
of Laurent power series. However, the example of our main interest will be the space
of formal series K{CoDccG} with well ordered support over a left ordered group G.
One of the main result of the paper describes linear continuous maps between topological
spaces K{G} and K{H}, where G and H are filters on sets G and H
with involution (see Theorem 9.6). In particular, we completely characterize
such maps in terms of zero sets of their (infinite) matrices. Namely, these zero sets must
belong to a special filter on the direct product of H and G, which were
introduced and investigated in [5].
As a corollary we give a matrix description of the ring of continuous operators of the
space K{G} (see Theorem 10.7).
There is a different approach how to embed a group ring of a countable torsion-free group
into a skew field, based on the theory of C∗-algebras and operators on Hilbert spaces
(see [9]). From this point of view this paper is a first step in developing a similar
machinery in a more general and abstract situation. For instance, in Section 5
we introduce the operation of pairing on formal sums which resembles scalar product in
Hilbert spaces.
2. Filters
In this section we recall some basic facts and definitions, and also
some results from [5].
Let G be a set. A nonempty collection G of subsets of G is
said to be a filter, if it is closed with respect to finite
intersections and supersets. For instance, we allow the set of all
subsets of G, P(G), to be a filter. Clearly, if G is a
filter, then G=P(G) iff ∅∈G. If G=P(G),
then G is said to be a proper filter on G.
Let L be a collection of subsets of G with the following
property: for all A,B∈L there is C∈L such that C⊆A∩B. Then G={B∣A⊆B for some A∈L} is a
filter generated by L, and L is a filter base
for G.
We say that a subset A of G is cofinite, if its
complement A is a finite set. The Frechet filter on
G, Cof(G), consists of all cofinite subsets of G. Clearly
Cof(G) is a proper filter iff G is an infinite set.
Let G be a linearly ordered set. A subset Δ⊆G is
said to be well ordered, if every nonempty subset of Δ
has a minimal element. This is the same as Δ has a
descending chain condition (d.c.c.): every descending chain
of elements a1≥a2≥… of Δ stabilizes. Clearly
Δ has a d.c.c. iff it contains no (strictly) descending chain
a1>a2>a3>….
Similarly, Δ⊆G has an ascending chain
condition (a.c.c.), if every ascending chain a1≤a2≤…
of its elements stabilizes. Thus Δ has an a.c.c. iff it
contains no (strictly) ascending chain a1<a2<… iff G
is well ordered in the dual ordering.
Suppose that (G,≤) is a linearly ordered set. Let CoDcc(G)
denote the collection of all subsets of G whose complement has a
d.c.c. Since the union of two well ordered subsets of G is well
ordered, CoDcc(G) is a filter on G, and it is a proper filter iff
G is not well ordered.
Similarly let CoAcc(G) be a collection of all subsets of G whose
complement has an a.c.c. Then CoAcc(G) is a filter on G and this
filter is proper iff G contains a strictly ascending chain.
We can order the filters on G by inclusion: G1≤G2 if
G1⊆G2. It is easily checked that with respect to this
ordering the set of all filters on G forms a lattice, that is, for
any filters G1 and G2 there is a least filter G1∨G2 containing G1 and G2, and there is a largest
filter G1∧G2 which is contained in both
G1 and G2.
The following remark describes the operations in this lattice.
Remark 2.1**.**
Let G1, G2 be filters on a set G. Then G1∧G2
is given by the intersection of filters: G1∩G2={A⊆G∣A∈G1 and A∈G2}. Furthermore, G1∨G2
is the filter generated by all intersections A∩B, A∈G1, B∈G2.
We define a new operation on filters. In ring theory this operation
corresponds to the quotient of ideals. Suppose that G1 and
G2 are filters on G. Define G1:G2={A⊆G∣A∪A′∈G1 for every A′∈G2}.
G1:G2* is a filter on G. Furthermore, G1:G2 is
the largest filter F on G with the property F∩G2⊆G1.*
The following remark is straightforward.
Remark 2.3**.**
1) If F1⊆F2 then F1:G⊆F2:G.
2) If G1⊆G2 then F:G1⊇F:G2.
3) F:G=(F∩G):G.
4) F⊆F:G.
If G is a filter, then define
[TABLE]
For instance,
Cof(G)⊥=P(G) and P(G)⊥=Cof(G). It follows
from the definition that G⊥=Cof(G):G, hence
G⊥ is a filter by Fact 2.2. Furthermore,
Remark 2.3, 4) implies that Cof(G)⊆G⊥.
Lemma 2.4**.**
G⊆G⊥⊥* and G⊥=G⊥⊥⊥
for any filter G.*
Proof.
We prove that G⊆G⊥⊥. Fix A∈G and choose
any A′∈G⊥. Then (by the definition of G⊥)
A∪A′ is cofinite, hence A∈G⊥⊥.
Remark 2.3 2) applied to G⊆G⊥⊥ yields
G⊥⊇G⊥⊥⊥, and the reverse inclusion
follows from what we have just proved.
∎
A filter G is said to be balanced, if
G⊥⊥=G. For instance, Cof(G) and P(G) are
balanced filters. Furthermore, every balanced filter on G contains
Cof(G).
Let G be a linearly ordered set. Then CoDcc(G)⊥=CoAcc(G) and
CoAcc(G)⊥=CoDcc(G), therefore CoDcc(G) and CoAcc(G) are
balanced filters.
3. Space of G-zero functions
Most results of this paper can be proven for normed skew fields K.
But to avoid technicalities, in what follows K will always denote
a skew field with a discrete topology.
A left (right) K-linear space L with a topology T is
said to be a linear topological space, if the addition of
elements of L defines a continuous function L×L→L,
where L×L is taken with product topology; and the same is
true for any function k×L→L given by multiplication by k∈K.
Since K is discrete the last condition can be
replaced by the following: for each open set U⊆L and every
0=k∈K the product kU is open.
Suppose that Ui, i∈I is a collection of subspaces of L
such that for all i,j∈I there exists k∈I with Uk⊆Ui∩Uj. A subset V of L is defined to be open, if for
every a∈V there exists i∈I such that a+Ui⊆V. This
defines a linear topologyT on L, therefore L is a
topological space with a linear topology. Note that T is
Hausdorff iff ∩i∈IUi={0}. In this paper we will
consider only Hausdorff linear topologies.
For instance, let G be a set and let L=Map(G,K) be a left
(right) vector space of all maps from G to K. Let I be the
collection of all finite subsets of G, and we consider I as a
set of indices. For each i∈I define a subspace Ui of L
consisting of all maps f:G→K such that f(g)=0 for every
g∈i. Then the family Ui, i∈I defines a linear topology
on L called the Tychonoff topology. For instance, if K is
a finite field (with discrete topology), then L is a compact
space (Tychonoff theorem).
Suppose that f:G→K is a map. Then the support of f,
supp(f), is the following subset of G: supp(f)={g∈G∣f(g)=0}. Similarly the zero-set of f, Z(f), is
defined as Z(f)={g∈G∣f(g)=0}. Clearly G=supp(f)∪Z(f) is a partition of G. Furthermore, if f,h:G→K and
0=k∈K then Z(f+h)⊇Z(f)∩Z(h), Z(k⋅f)=Z(f) and Z(0)=G, where [math] stands for the zero function.
If G is a filter on G, then Funct(G)={f∈Map(G,K)∣Z(f)∈G} will denote the space of G-zero
functions. Clearly Funct(G) is a left and right subspace of the
(linear) topological space Map(G,K). For instance,
Funct(P(G))=Map(G,K).
The following remark shows that operations on linear spaces
Funct(G) correspond to operations on the lattice of filters (see
Remark 2.1).
Lemma 3.1**.**
Let G1 and G2 be filters on G. Then Funct(G1)∩Funct(G2)=Funct(G1∩G2) and Funct(G1)+Funct(G2)=Funct(G1∨G2). Furthermore, if G1⊆G2
then Funct(G1) is a subspace of Funct(G2).
Proof.
Clearly f∈Funct(G1)∩Funct(G2) iff Z(f)∈G1∩G2 iff f∈Funct(G1∩G2), which proves that
Funct(G1)∩Funct(G2)=Funct(G1∩G2).
To prove the inclusion Funct(G1)+Funct(G2)⊆Funct(G1∨G2) let f∈Funct(G1)+Funct(G2). Then f=h1+h2, where
Z(hi)∈Gi. It follows that Z(f)⊇Z(h1)∩Z(h2)∈G1∨G2, hence f∈Funct(G1∨G2).
For the reverse inclusion suppose that f∈Funct(G1∨G2),
that is, Z(f)⊇A1∩A2 for some Ai∈Gi, i=1,2. Define
h∈Map(G,K) as follows:
[TABLE]
Then f=h1+h2 and Z(h1)⊇A1, Z(h2)⊇A2,
therefore hi∈Funct(Gi).
∎
Given A⊆G, we set U(A,G)={f∈Funct(G)∣Z(f)⊇A}. Clearly this is the same as supp(f)⊆A.
The proof of the following lemma is straightforward.
Lemma 3.2**.**
1) U(A,G) is a left (right) subspace of Funct(G).
2) If A1,A2⊆G then U(A1∩A2,G)=U(A1,G)∩U(A2,G).
Now we are in a position to construct a linear topology on
Funct(G).
1) The family of subspaces {U(A,G)∣A∈G⊥} form a
base of zero neighborhoods of a linear topology T(G) on the
spaces Map(G,K) and Funct(G), and this topology is Hausdorff.
2) If G is a balanced filter, then Funct(G) is complete in
this topology.
3) If Cof(G)⊆G then the set of all maps from Map(G,K)
with finite support is dense in Funct(G).
Let us consider some examples of topologies T(G).
Example 3.4**.**
1) If G=Cof(G), then Funct(G) consists of all functions with
finite support and G⊥=P(G). Then ∅∈G⊥, hence
U(∅,G)={0} is an open set. It follows that every subset of
Funct(G) is open and closed, hence T(G) is a discrete
topology.
2) If G=P(G), then G⊥=Cof(G), Funct(G)=Map(G,K) and
A runs over all cofinite subsets of G. Thus we obtain the
Tychonoff topology whose subbase is given by the subspaces
Ug={f∈Map(G,K)∣f(g)=0}.
3) Suppose that (G,≤) is a linearly ordered set and
G=CoAcc(G). Then G⊥=CoDcc(G), therefore the base of zero
neighborhoods is given by UD={f∈Funct(G)∣f(g)=0 for every
g∈D}, where D is a well ordered subset of G.
Note that, if G=(Q,≤), then the space Map(Q,K) with
Tychonoff topology is metrizable and separable. This is not longer
true for the topology T(CoAccG). Indeed, suppose that D1,D2,… are well ordered subsets of Q such that UD1,UD2,… form a basis of zero neighborhoods. This means that for every
well ordered D⊆Q there is k such that UD1∩…∩UDk⊆UD. It follows easily that D⊆D1∪…∪Dk.
Clearly there exists an ascending sequence d1<d2<… such that
dk∈/D1∪⋯∪Dk for every k. If D={d1,d2,…},
then D has a d.c.c, therefore D⊆D1∪…∪Dk for some k,
and then dk∈D1∪…∪Dk, a contradiction.
This shows that in the space (Funct(CoAccQ),T(CoAccQ)) no point has a
countable base of neighborhoods, in particular, this space is not separable.
4) Let G=⟨t⟩ be an infinite cyclic group with the usual
linear ordering: tn≥tm iff n≥m. Let G=CoDcc(G),
that is, G is generated by the following collection of sets:
{tn∣n<l}, l∈Z. Then G⊥ is generated by
{tn∣n>m}, m∈Z, hence Funct(G) is the space of
Laurent power series ∑i≥lkiti and T(G) is the
t-adic topology.
4. Direct sum decompositions
Let C be a subset of G. We identify Map(C,K) with a subspace
of Map(G,K) consisting of all maps f:G→K such that the
restriction of f to C is zero. If G is a filter on G,
then the family GC={A∩C∣A∈G} will be a filter on
G called an induced filter.
Note that with respect to the above identification, Funct(GC)⊆Funct(G). Indeed, if f∈Map(C,K), then f∈Funct(GC) means
that Z(f)=A∩C for some A∈G. If we consider f as a
map from G to K, then Z(f)=(A∩C)∪C⊇A, hence
f∈Funct(G).
Proposition 4.1**.**
Let C⊆G. Then the topology T(GC) on the space
Funct(GC) coincides with the topology induced by T(G).
Proof.
Note that A∩C∈GC⊥ for every A∈G⊥. Indeed,
every set from GC can be written in the form A′∩C for
some A′∈G. Since A∪A′ is cofinite in G, it follows
that (A∩C)∪(A′∩C)=(A∪A′)∩C is cofinite in C.
Thus A∩C∈GC⊥ and clearly U(A,G)∩Funct(GC)⊆U(A∩C,GC).
Now take any A1∈GC⊥. We claim that B=A1∪C∈G⊥ which would imply the inclusion U(A1,GC)⊆U(B,G)∩Funct(GC). Indeed, for every A′∈G the set A1∪(A′∩C) is cofinite in C. Then (A1∪C)∪A′ is
cofinite in G because it contains A1∪(A′∩C) (cofinite
in C) and C.
∎
Let A be a subset of G. For every map f:G→K we have
f=f∣A+f∣A, where supp(f∣A)⊆A and
supp(f∣A)⊆A. This yields a decomposition of
linear spaces: Funct(G)=Funct(GA)⊕Funct(GA), where
Funct(GA)⊆Map(A,K) and Funct(GA)⊆Map(A,K).
In the following proposition we will single out two important
particular cases.
Proposition 4.2**.**
1) If A∈G then Funct(G)=Funct(GA)⊕Map(A,K).
Furthermore, T(G) induces the Tychonoff topology on Map(A,K).
2) If G⊇Cof(G) and A∈G⊥, then Funct(G)=U(A,G)⊕Funct(CofA). Furthermore, T(G) induces the
discrete topology on Funct(CofA).
Proof.
Since A∩A=∅∈Funct(GA), the induced filter on A coincides
with P(A), hence Funct(GA)=Map(A,K). Thus, by
Proposition 4.1 and Example 3.4 2), the topology
induced on Map(A,K) by T(G) will be Tychonoff.
It remains to apply the equality Funct(G)=Funct(GA)⊕Funct(GA).
By the definition of ⊥-operation, we obtain GA=Cof(A).
Again, by Proposition 4.1 and Example 3.4 1),
T(G) induces discrete topology on Funct(GA). Now the
result follows from the same equality.
∎
5. Involution and pairing
Now assume that G is a set with an involution ∗, that is, with a
map G→G such that g∗∗=g. If A is a subset of G then
we define A∗={a∗∣a∈A}, and we put G∗={A∗∣A∈G}, if G is a filter on G.
The following remark is obvious.
Remark 5.1**.**
If G is a filter, then G∗ is also a filter. Furthermore,
the map G→G∗ defines an automorphism of the lattice of
filters on G. For instance, G⊥∗=G∗⊥ for every
filter G.
A filter G⊥∗ is said to be adjoint to the filter
G. Thus a filter G is self-adjoint, if G⊥∗=G. Note that every self-adjoined filter is balanced. Indeed,
G⊥⊥=G⊥⊥∗∗=G⊥∗⊥∗=(G⊥∗)⊥∗=G⊥∗=G. For instance, let
G be a linearly orderer group, G=CoDcc(G) and the involution ∗
is given by g∗=g−1. Then CoDcc(G)⊥∗=CoAcc(G)∗. Since
taking the inverse in the linearly ordered group reverses the
ordering, CoAcc(G)∗=CoDcc(G), therefore CoDcc(G) is self-adjoint, and
the same is true for CoAcc(G).
From now on each group G will be considered as a group with the
involution ∗:g→g−1.
Suppose that G is a group with a linear ordering ≤. We say that G is
left (right) ordered, if g1≤g2 implies hg1≤hg2 (g1h≤g2h)
for every g1,g2,h∈G. A group G is said to be linearly ordered,
if it is left and right ordered with respect to ≤.
Proposition 5.2**.**
A left ordered group (G,≤) is linearly ordered iff the filter
CoDcc(G) (or CoAcc(G)) is self-adjoint.
Proof.
We have already proved that for a linearly ordered group G,
both CoDcc(G) and CoAcc(G) are self-adjoint filters.
Suppose that G is a left ordered group and CoDcc(G) is a
self-adjoint filter. This means that CoDcc(G)⊥∗=CoDcc(G), which
is the same as CoDcc(G)⊥=CoDcc(G)∗ or CoAcc(G)=CoDcc(G)∗.
To prove that G is linearly ordered it suffices to check that the
cone P={g∈G∣g≥e} is invariant, that is, a−1Pa=P
for every a∈G. Moreover it is enough to verify that a−1Pa⊆P for every a∈G. Assuming otherwise we will find a,b∈G such that b>e and a−1ba<e. Multiplying ba<a on
the left by b we obtain a descending chain Δ={a>ba>b2a>…}. Since Δ has an a.c.c., Δ∈CoAcc(G). From CoAcc(G)=CoDcc(G)∗
it follows that Δ∗∈CoDcc(G) where Δ∗={a−1b−n∣n∈ω},
hence Δ∗ has a smallest element. If a−1b−n is
such, then a−1b−n≤a−1b−n−1. Multiplying by bn+1a on the left
we obtain b≤e, a contradiction.
∎
Now we define a pairing on the space Map(G,K) as the
following partially defined non-degenerate bilinear form. If f,h∈Map(G,K) then
[TABLE]
and the result ⟨f,h⟩ is defined, if supp(f)∩(supph)∗ is a finite set.
Clearly this is the same as
(suppf)∗∩supp(h) is finite. In particular, this is the case
when f∈Funct(G) and h∈Funct(G⊥∗) or vice versa. If
K is a field, this form is symmetric.
We say that the equality (1) defines a pairing between
subspaces L and L′ of Map(G,K), if the following holds true:
the product ⟨f,h⟩ is defined for all f∈L, h∈L′;
if f∈Map(G,K) and the product ⟨f,h⟩ is defined
for every h∈L′, then f∈L;
if h∈Map(G,K) and the product ⟨f,h⟩ is defined
for every f∈L, then h∈L′.
In this case (L,L′) is said to be a dual pair.
Proposition 5.3**.**
If G is a balanced filter on a set G with an involution, then
(Funct(G⊥∗),Funct(G)) is a dual pair and the same is true
for the pair (Funct(G),Funct(G⊥∗)).
Proof.
is satisfied by the definition of G⊥.
Let f∈Map(G,K) is such that ⟨f,h⟩ is defined for
every h∈Funct(G). Then (suppf)∗∩supp(h) is finite for
every h∈Funct(G), therefore Z(f)∗∪Z(h) is a cofinite
set for every such h. Take any B∈G and let hB be the
characteristic function of G∖B (that is, hB(g)=0 if g∈B, and hB(g)=1 otherwise). Clearly hB∈Funct(G). It follows
that Z(f)∗∪B is a cofinite set, therefore, by the
definition of ⊥, we obtain Z(f)∗∈G⊥. It follows
that Z(f)∈G⊥∗ hence f∈Funct(G⊥∗).
Suppose that h∈Map(G,K) and the result ⟨f,h⟩ is
defined for every f∈Funct(G⊥∗). As in the proof of 2) it
follows that Z(h)∈(G⊥∗)⊥∗=G⊥⊥=G,
since G is balanced.
∎
Thus we obtain the following diagram of pairing:
[TABLE]
where ⟷ stands for pairing, and G1⊆G2 are balanced filters.
Because Funct(G) and Funct(G⊥∗) are paired, it follows
that Funct(G⊥∗) is isomorphic to the space of linear
forms on Funct(G), that is to the space of linear continuous maps from
Funct(G) to K. We will derive this fact later from a more
general description of continuous linear operators on spaces of
G-zero functions.
6. Filters on direct products
Let G be a filter on a set G, and let H be a filter on
H. In this section we consider different extensions of these
filters to a filter on the direct product H×G. One of these
extensions is well known.
Fact 6.1**.**
(see [1, Sec. 6.7])
The family {B×A∣B∈H,A∈G} is a base of a filter
H×G on H×G. This filter is proper iff both G and
H are proper.
But this particular filter bears no significance for topological
spaces of formal sums. The following filter is more useful.
We introduce a new filter Cof(H,G) whose subbase is given by
the following collections of sets: {h×A∣h∈H,A∈G} and {B×g∣B∈H,g∈G}. Thus
Cof(H,G) consists of subsets of H×G whose complement is
a subset of a finite union of sets {h}×A and B×{g}. But every subset of {h}×A, A∈G is of the
form {h}×A′ for some A′∈G, and every subset of
B×{g}, B∈H is of the form B′×{g} for
some B′∈H. It follows that Cof(H,G) consists of
complements to the sets (⋃i=1n{hi}×Ai)∪(⋃j=1mBj×{gj}), where gj∈G,
hi∈H and Ai∈G, Bj∈H.
Suppose that G contains the Frechet filter on G, and H
contains the Frechet filter on H. Then
1) Cof(H×G)⊆Cof(H,G)⊆H⊗G.
2) Cof(H,G)=(CofH×CofG)∩(H⊗G).
Now we are in a position to introduce the main construction of this
section.
Definition 6.6**.**
Suppose that H is a filter on H and G is a filter on G.
Let ⟨H,G⟩ consists of all subsets X of H×G
with the following properties:
a) for every A∈G⊥ there exists B′∈H such that
B′×A⊆X;
b) for every B∈H⊥ there exists A′∈G such that B×A′⊆X.
If t∈G, then define Ht(X)={h∈H∣(h,t)∈X}. Similarly,
if s∈H, then set Gs(X)={g∈G∣(s,g)∈X}.
It is easily seen that X∈⟨H,G⟩ iff the following
holds:
a)′⋂t∈AHt(X)∈H for every A∈G⊥;
b)′⋂s∈BGs(X)∈G for every B∈H⊥.
For instance, the equivalence of a) and a)′ can be seen as
follows. If a) holds then B′⊆⋂t∈AHt, hence
this intersection is in H; and if a)′ holds then we take B′
to be equal to this intersection.
If H and G contain Frechet filters, then (H⊗G)∩(H⊥⊗G⊥)=Cof(H×G).
The following remark is obvious.
Remark 6.9**.**
If H⊆H′ and G⊆G′, then H⊗G⊆H′⊗G′.
Proposition 6.10**.**
If H and G are balanced filters, then ⟨H,G⟩⊇Cof(H,G):(H⊥⊗G⊥)=(H⊥⊗G⊥)⊥.
Proof.
First we prove the equality. We claim that Cof(H,G)∩(H⊥⊗G⊥)=Cof(H×G). Indeed, by Fact 6.5 2),
we obtain
[TABLE]
By Fact 6.8 this is the same as (CofH×CofG)∩Cof(H×G)=Cof(H×G), as
desired.
Then by Remark 2.3 3) and the definition of ⊥ we obtain
Cof(H,G):(H⊥⊗G⊥)=(Cof(H,G)∩(H⊥⊗G⊥)):(H⊥⊗G⊥)=Cof(H×G):(H⊥⊗G⊥)=(H⊥⊗G⊥)⊥.
Now we prove the inclusion. Suppose that Z∈(H⊥⊗G⊥)⊥ and we have to show that Z∈⟨H,G⟩.
Take any B∈H⊥ and define A′⊆G by the following rule:
A′={g∈G∣(h,g)∈Z for some h∈B}=πG[(B×G)∩Z], where πG is a projection on G.
Suppose that A′∈/G. Then A′∈/(G⊥)⊥ because
G is balanced. It follows that there exists A∈G⊥ such
that the union A′∪A is not cofinite in G, that is, the
intersection A′∩A is infinite. Since B×A∈H⊥⊗G⊥, it follows that Z∪B×A is cofinite in H×G, hence Z∩(B×A) is a
finite set. On the other hand for every g∈A′∩A
(there are infinitely many of them) there exists h∈B such
that (h,g)∈Z∩(B×A), hence this set must be
infinite, a contradiction. Thus A′∈G.
We prove that B×A′⊆Z. Indeed, otherwise (h,g)∈Z for some (h,g)∈B×A′. Since g∈/A′, by the
construction of A′ we obtain h∈/B, a contradiction.
Thus for every B∈H⊥ there exists A′∈G such that
B×A′⊆Z. Similarly for every A∈G⊥ there exists
B′∈H such that B′×A⊆Z. It follows that Z∈⟨H,G⟩.
∎
Before proving the next lemma, let us recall a useful equality: if
B⊆H and A⊆G, then (B×A)=(B×A)∪(B×A)∪(B×A).
Lemma 6.11**.**
H⊥⊗G⊥⊆⟨H,G⟩⊥* for all filters
H and G.*
Proof.
Suppose that Z∈H⊥⊗G⊥, hence Z⊆B×A for some B∈H⊥, A∈G⊥. We have to prove
that Z∈⟨H,G⟩⊥, that is, Z∪X is a cofinite
set for every X∈⟨H,G⟩. Since X∈⟨H,G⟩ there are B′∈H and A′∈G such that B′×A,B×A′⊆X. Then
[TABLE]
[TABLE]
Intersecting this with Z⊆B×A and taking into account that the intersections of B×A with B′×A, B′×A, B×A′ and B×A′ are empty, we obtain X∩Z⊆(B′∩B)×(A∩A′). Since B∈H⊥ and B′∈H, therefore
B∪B′ is a cofinite set, hence B′∩B is finite.
Similarly A′∩A is a finite set, hence X∩Z is finite, as desired.
∎
Remark 6.12**.**
H⊗G⊆⟨H⊥,G⊥⟩⊥* for all filter
H and G.*
Proof.
Indeed, by Lemma 2.4 we have G⊆G⊥⊥ and
H⊆H⊥⊥, therefore H⊗G⊆H⊥⊥⊗G⊥⊥ by Remark 6.9. Furthermore
H⊥⊥⊗G⊥⊥⊆⟨H⊥,G⊥⟩⊥
by Lemma 6.11.
∎
Theorem 6.13**.**
Suppose that H and G are balanced filters. Then ⟨H,G⟩=(H⊥⊗G⊥)⊥. In
particular, ⟨H,G⟩ is a balanced filter.
Proof.
By Proposition 6.10 we have ⟨H,G⟩⊇⟨H⊥⊗G⊥⟩⊥. On the other hand applying ⊥ to
the inclusion in Lemma 6.11, we obtain (H⊥⊗G⊥)⊥⊇⟨H,G⟩⊥⊥. Thus ⟨H,G⟩⊇⟨H⊥,G⊥⟩⊥⊇⟨H,G⟩⊥⊥, hence ⟨H,G⟩⊇⟨H,G⟩⊥⊥. Then Lemma 2.4 yields the desired.
∎
7. G-sums
Suppose that (X,T) is a typological abelian group and xi,
i∈I is a family of elements of X. An element x∈X is said to be
a sum of this family with respect to T, written
x=∑i∈ITxi, if the following holds. For every
neighbourhood U of x there is a finite subset Δ⊆I such
that ∑i∈Δ′xi∈U for every finite set Δ′⊆I
containing Δ. Clearly, if X is Hausdorff, then the sum is
unique. It is easily seen that, if the family xi, i∈I is
summable, then the limit of the xi with respect to the Frechet
filter on I is equal to zero. This means that for every zero
neighbourhood U there exists a finite subset Δ of I such
that xi∈U for every i∈I∖Δ.
Suppose that (X,TX) and (Y,TY) are topological abelian
groups. If φ is a continuous morphism from X to Y then
φ preserves topological sums. This means that, if x=∑i∈ITXxi in X, then the sum ∑i∈ITYφ(xi) exists and equal to φ(x).
For more on sums in topological abelian groups see [2]
Now we define a sum with respect to a filter.
Definition 7.1**.**
Let hj∈Map(G,K), j∈J be a family of maps and let G
be a filter. This family is said to be G-summable, and
the map f:G→K is a G-sum of this family,
f=∑j∈JGhj, if the following holds:
1) for every g∈G there are only finitely many j∈J such
that g∈supp(hj), and f(g)=∑j∈Jhj(g);
2) ⋂j∈JZ(hj)∈G.
Note that the condition 1) of this definition means that the family
{hj}, j∈J is summable with respect to Tychonoff topology
on Map(G,K). Furthermore, 2) implies that hj∈Funct(G) for
every j; and Z(f)⊇⋂j∈JZ(hj) yields that
f∈Funct(G).
In the following theorem we compare these two types of summability.
Theorem 7.2**.**
1) If the family hj, j∈J is G-summable, then it is
summable with respect to topology T(G). Furthermore, ∑j∈JGhj=∑j∈JT(G)hj.
2) If the family hj∈Funct(G), j∈J is summable with
respect to T(G) and G is balanced, then it is G-summable
and ∑j∈JGhj=∑j∈JT(G)hj again.
Proof.
Suppose that f=∑j∈JGhj and A=⋂j∈JZ(hj)∈G. Choose any A′∈G⊥. Then A∪A′ is a
cofinite set, hence A∪A′=G∖{g1,…,gn}, gi∈G. Let J0 consist of all j∈J such that gt∈supp(hj)
for some t=1,…,n. By the assumption J0 is a finite subset
of J.
Suppose that J′ is any finite subset of J containing J0. Then
Z(f−∑j∈J′hj)⊇A∪{g1,…,gn}⊇A′, therefore f−∑j∈J′hj∈U(A′,G). This proves
that f=∑j∈JT(G)hj.
Because the family hj, j∈J is summable with respect to
T(G), by what we have already noticed, hj converges to [math]
with respect to the Frechet filter on I. It follows that for every
g∈G there exist only finitely many j∈J such that
hj(g)=0, therefore 1) holds true.
Thus it remains to prove that A=⋂j∈JZ(hj)∈G.
Let f=∑j∈JT(G)hj. As above the hj converge to
zero with respect to the Frechet filter on J. Thus for every
A′∈G⊥ there is a finite set of indices F(A′) such that
for every j∈J∖F(A′) we have hj∈U(A′,G). Then
Z(hj)⊇A′ yields ⋂j∈J∖F(A′)Z(hj)⊇A′. Since hj∈Funct(G) and F(A′) is finite, it follows
that A1=⋂j∈F(A′)Z(hj)∈G. Then
[TABLE]
is a cofinite set. Since this is true for any A′∈G⊥, we conclude that A∈G⊥⊥=G.
∎
8. Matrix notations
Suppose that G is a set and H is a set with an involution ∗.
Each map Ψ:H×G→K can be consider as an H×G
matrix over K whose (h,g)-entry, Ψhg, is Ψ(h,g).
These notations resemble the notations in tensor calculus and, as we
will see below, they are quite advantageous, when we consider
multiplication of matrices.
The set of all such maps form a (left and right) vector space over
K and will be denoted by HKG. If H consists of one
element, a map Ψ:H×G→K is said to be a row,
and we use small Greek letters α,β,… to denote rows.
Similarly, if G consists of one element, then a map Ψ:H→K
is a said to be a column, and we use small boldfaced letters
a,b, … to denote columns. In case, when
H consists of one element h, we simplify notations:
KG={h}KG; and similarly HK means HK{g},
when G consists of one element g.
Let δ be the Kronecker symbol on G, that is, δ is a map
from G×G to K such that δhg=1 if g=h and
δhg=0 otherwise. If Ψ∈HKG and h∈H, then
Ψh will denote the row of Ψ with number h, therefore
(Ψh)g=Ψhg for every g∈G. Similarly, Ψg will
denote the column of Ψ with number g, therefore
(Ψg)h=Ψhg for every h∈H. In particular, δg is
a row whose gth entry is 1 and all remaining entries are zero,
and similarly for the column δg.
Suppose that Φ∈JKH and Ψ∈HKG. We say the the
product Θ=Φ⋅Ψ∈JKG is defined if, for
every pair (j,g)∈J×G, we have Φjh∗⋅Ψhg=0 only for finitely many h∈H and
Θjg=∑h∈HΦjh∗⋅Ψhg. This defines
a partial operation JKH×HKG→JKG.
Note that ∑h∈HΦjh∗⋅Ψhg=∑h∈HΦjh⋅Ψh∗g. More precisely, the left and right
parts are defined simultaneously and, if they are defined, they are
equal. Immediately from the definition it follows that Θj=Φj⋅Ψ and Θg=Φ⋅Ψg.
If G is a finite set, then the multiplication on GKG is
defined everywhere, therefore GKG is a ring isomorphic to the
ring of ∣G∣×∣G∣ matrices over K. But the unity of this
ring is given by the map E:G×G→G such that Egg∗=1
for every g∈G and zero otherwise. For instance, if G consists
of one element g, then the ring {g}K{g} is isomorphic
to K. If G is infinite, the partial multiplication we have just
defined is usually not associative. Indeed, suppose that G=N with
the identical involution ∗, and let Φ, Ψ and Θ be
the following matrices:
[TABLE]
Then (Φ⋅Ψ)⋅Θ=1 and Φ⋅(Ψ⋅Θ)=0.
The following lemma claims distributivity and can be easily verified
by direct calculations.
Proposition 8.1**.**
Suppose that Φ∈JKH and Ψ,Θ∈HKG. If both
products Φ⋅Ψ and Φ⋅Θ are defined, then
Φ⋅(Ψ+Θ) is also defined and equal to Φ⋅Ψ+Φ⋅Θ.
An arbitrary row γ∈KG is uniquely determined by its
coordinates γg∈K, g∈G. Thus γ is a topological sum
(see Section 7) with respect to Tychonoff topology on
Map(G,K): γ=∑g∈GTychγgδg. Furthermore,
if γ∈Funct(G), then clearly γ=∑g∈GGγgδg. Similarly, each column a∈HK is uniquely
determined by its coordinates ah∈K, h∈H, hence we can
write a=∑h∈HTychδhah; and if a∈Funct(H), then a:=∑h∈HHδhah.
9. Continuous linear maps
Recall that G is a filter on a set G and H is a filter on
a set H. In what follows we will always assume that G and
H contain Frechet filters. Thus Funct(CofG)⊆Funct(G) and
Funct(CofH)⊆Funct(H). In this section we describe continuous
linear maps of topological linear spaces Funct(G)→Funct(H).
We will consider Funct(G) as a left or right K-vector space. To
specify the side, we will use K{G} to denote this space
considered as a left vector space over K, and call it a
space of G-zero rows. Similarly {G}K will denote
Funct(G) considered as a right vector space over K, and will be
called a space of G-zero columns. We use K[G] to denote
K{CofG}, and [G]K to denote {CofG}K. If G is a
group, then K[G] with the operations defined in
Section 8 is isomorphic to the usual group ring. In
what follows we will always assume that G and H are endowed with
an involution.
Recall that we agreed to denote rows with small Greek letters α,β…, and columns with small boldfaced letters a,b,…. Suppose that φ is a map from Funct(G) to Map(H,K).
First we consider Funct(G) as a right vector space {G}K and
Map(H,K) as the space of columns HK. The image of a column
a∈{G}K will be denoted by φ[a]. Then we can
assign to φ an H×G matrix Φ, whose (h,g∗)-entry
is equal to φ[δg]h:
[TABLE]
(Recall that δg is a column whose gth coordinate is 1 and
all the remaining coordinates are zero, and φ[δg]h is the
hth coordinate of the column φ[δg]). We say that Φ
is a matrix of φ. For instance,
Φg∗=φ[δg] is the g∗th column of Φ.
The zero set of Φ, Z(Φ), is a collection of all
(h,g)∈H×G such that Φhg=0. This is in accordance
with our previous definition of the zero set of a map.
Now we restrict ourselves to the case, when φ is a map from
{G}K to {H}K⊆Map(H,K).
Definition 9.1**.**
A right (left) linear map φ:Funct(G)→Funct(H) is said to be
G-linear, if for any G-summable family hj, j∈J, where hj∈Funct(G), the family φ[hj], j∈J is
H-summable and
[TABLE]
Apparently this condition bears no connection with topology. But below
(see Theorem 9.6) we will see that G-linearity is the same
as continuity. The following lemma shows that both G-linear or continuous
linear operators are uniquely determined by their matrices.
Lemma 9.2**.**
Suppose that φ:{G}K→{H}K is either G-linear or a
linear continuous map. If Φ is a matrix of φ and a∈{G}K,
then φ[a]=Φ⋅a.
Proof.
We will prove this lemma only when φ is continuous. The proof
in the case, when φ is G-linear, is similar.
As we have already noticed (see a remark after Lemma 8.1),
a=g∈G∑Gδgag, ag∈K, hence
a=g∈G∑T(G)δgag by Theorem 7.2.
Since φ is continuous and linear, we obtain
φ[a]=g∈G∑T(H)φ[δg]ag. Invoking
Theorem 7.2 again, we get φ[a]=g∈G∑Hφ[δg]ag=g∈G∑HΦg∗ag. On the
other hand, by the definition of product of matrices, (Φ⋅a)h=g∈G∑Φhg∗ag, therefore
φ[a]=Φ⋅a, as desired.
∎
Our next objective is to decide when the left multiplication by an
H×G matrix Φ defines a continuous linear map φ
from {G}K to {H}K. As a first approximation we
consider the following condition:
b) the rule a→Φ⋅a, where a∈{G}K, defines a
K-linear map φ:{G}K→HK.
Proof.
a) If B=H∖{h}, then B={h}. Since B∈H,
(2) yields Z(Φh)∈G⊥∗.
b) By what we have just proved, Z(Φh)∈G⊥∗ for
every row Φh of Φ. Since a∈{G}K and
Z(Φh)∈G⊥∗, the product Φh⋅a is
defined (see Proposition 5.3) and belongs to K. Then
Φ⋅a is defined and belongs to HK.
∎
We need to put one extra condition on Φ to ensure that the
image of φ is contained in {H}K.
[TABLE]
The next proposition shows that (2) and (3)
together imply that φ is continuous.
Proposition 9.4**.**
Let Φ be an H×G-matrix satisfying (2) and
(3). Then the rule a→Φ⋅a defines a
continuous linear map φ:{G}K→{H}K.
Proof.
By Remark 9.3, Φ⋅a∈HK. We prove that
Φ⋅a∈{H}K. Indeed, from a∈{G}K it
follows that A=Z(a)∈G. Furthermore, (3) implies
that B=⋂t∈AZ(Φt∗)∈H, therefore
B×A∗⊆Z(Φ). To show that Φ⋅a∈{H}K, take any h∈B. Then
[TABLE]
It follows that Z(Φ⋅a)⊇B, therefore
Z(Φ⋅a)∈H, as desired.
As we have already noticed (see Lemma 8.1),
Φ⋅(a+b)=Φ⋅a+Φ⋅b for all columns
a,b∈{G}K. Furthermore, clearly (Φ⋅a)⋅k=Φ⋅(a⋅k) for every k∈K. It follows easily that
φ is linear.
To prove that φ is continuous, it suffices to check that φ
is continuous at zero. This means that for every zero neighbourhood
U(B,H), B∈H⊥ in {H}K, there exists a zero
neighbourhood U(A,G), A∈G⊥ in {G}K such that
[TABLE]
From (2) we obtain C=⋂s∈BZ(Φs)∈G⊥∗. If A=C∗, then A∈G⊥, and we prove that A
is as required. By the definition of A, we have B×A∗⊆Z(Φ), that is, Φsg∗=0 for any s∈B and g∈A. Take any a∈U(A,G), hence ag=0 for every
g∈A. Then for any s∈B we obtain:
[TABLE]
It follows that Z(Φ⋅a)⊇B, therefore
Φ⋅a∈U(B,H).
∎
Below (see Theorem 9.6) we will show the the converse is
also true: if φ:{G}K→{H}K is a continuous linear map,
then its matrix Φ satisfies both (2) and
(3). But first we connect these conditions with filters on
direct products.
Lemma 9.5**.**
A matrix Φ∈HKG satisfies (2) and (3)
iff Z(Φ)∈⟨H,G⊥∗⟩.
Proof.
Let us rewrite a)′ from Definition 6.6 replacing
G by G⊥∗ and X by Z(Φ). Then G⊥
should be replaced by G∗.
[TABLE]
where Ht={h∈H∣(h,t)∈Z(Φ)} and Gs={g∈G∣(s,g)∈Z(Φ)}.
But clearly Ht=Z(Φt) and Gs=Z(Φs). Then
(4) can be rewritten as follows:
[TABLE]
Applying the involution, we see that the first condition in
(5) is equivalent to (3). Furthermore, the
second condition in (5) coincides with (2).
∎
The following theorem characterizes continuous linear maps between
spaces {G}K and {H}K in terms of filters on H×G.
Theorem 9.6**.**
Suppose that H and G are balanced filters on H and G and
Φ is an H×G-matrix. Then the following are equivalent.
a) The left multiplication by Φ defines a continuous linear map
φ from {G}K to {H}K.
b) The zero set Z(Φ) belongs to the filter ⟨H,G⊥∗⟩;
c) Each row Φh belongs to K{G⊥∗}. Furthermore, for every A∈G
the collection Φt∗, t∈A is H-summable and
[TABLE]
for any set of coefficients kt∈K.
d) the left multiplication by Φ defines a G-linear map φ
from {G}K to {H}K.
If φ is a linear continuous map from {G}K to {H}K, then
its matrix Φ satisfies these equivalent conditions.
Proof.
b) ⇒ a) follows from Proposition 9.4 and Lemma 9.5.
a) ⇒ c). Since the product Φ⋅a is defined for every a∈{G}K,
by the definition of product of matrices we obtain Φh∈K{G⊥∗}.
If t∈A, then Z(δt)⊇A, hence
Z(δtkt)⊇A for any kt∈K, and therefore
⋂t∈AZ(δtkt)⊇A. It follows that
⋂t∈AZ(δtkt)∈G, hence the family
δtkt, t∈A is summable and clearly
[TABLE]
Furthermore, by Theorem 7.2 we have
t∈A∑Gδtkt=t∈A∑T(G)δtkt.
Then
[TABLE]
Since φ is continuous, the family Φt∗kt, t∈A is summable.
By Theorem 7.2 and the previous equality, we obtain
[TABLE]
as desired.
c) ⇒ d). Because Φh∈K{G⊥∗} for every h∈H,
the product Φ⋅a is defined for every a∈{G}K.
Furthermore, (6) yields that Φ⋅a∈{H}K.
Indeed, if a∈{G}K, then Z(a)∈G. If A=Z(a),
then a=t∈A∑Tychδtat. By the assumption,
[TABLE]
Since the family Φt∗, t∈A is H summable, then clearly the family
Φt∗at, t∈A is H-summable, therefore
Φ⋅a=t∈A∑HΦt∗at∈H (see a remark after
Definition 7.1).
Suppose that bj, j∈J, where bj∈{G}K, is a G-summable family.
Then
A=⋂j∈JZ(bj)∈G and a=j∈J∑Gbj∈{G}K.
From Z(bj)⊇A it follows that
[TABLE]
where j∈J and ktj=(bj)t∈K. Since the family bj,
j∈J is G-summable, for every t∈A there are only finitely
many j∈J such that ktj=0. Thus we can set kt=∑j∈Jktj and then a=t∈A∑Tychδtkt.
First we check that the family Φ⋅bj, j∈J is
H-summable. Indeed, by the assumption,
Φ⋅bj=t∈A∑HΦt∗ktj,
therefore ⋂j∈JZ(Φ⋅bj)⊇⋂t∈AZ(Φt∗). Because the family Φt∗, t∈A is
H-summable, it follows that ⋂t∈AZ(Φt∗)∈H, hence
⋂j∈JZ(Φ⋅bj)∈H.
It remains to prove that for every h∈H there are only finitely
many j∈J such that (Φ⋅bj)h=0. Indeed, the family
Φt∗, t∈A
is H-summable, hence Δ={t∈A∣Φht∗=0} is
a finite set. Furthermore, for any t∈A the set Jt={j∈J∣ktj=0} is also finite. Hence the set
J0:=⋃t∈ΔJt is finite being a finite union of
finite sets. If j∈/J0, then j∈/Jt for any
t∈Δ, and thus ktj=0. Therefore,
[TABLE]
Now we prove the equality
Φ⋅j∈J∑Gbj=j∈J∑HΦ⋅bj. Indeed,
by the assumption we have
Φ⋅j∈J∑Gbj=Φ⋅a=t∈A∑HΦt∗kt. On the other hand
[TABLE]
Indeed, the first equality follows from (7) and the
assumption; and the last equality follows from kt=∑jktj.
The second equality will be checked coordinate-wise. Take any h∈H. Then
[TABLE]
where we omitted the superscript H in the right hand part of the
equality, because the sum by J in this part is finite. By the
definition of multiplication of matrices, this is equal to
j∈J∑t∈A∑Φht∗ktj. We have already
proved that Φht∗ktj=0 iff t∈Δ and
ktj=0, where Δ is a finite set. It follows that the set
of pairs (j,t) such that Φht∗ktj=0 is also
finite. Therefore me can change the summation order to get
[TABLE]
d) ⇒ b). Let a=g∈G∑Tychδgkg∈{G}K.
Since φ is G-linear, we obtain
[TABLE]
We will prove that Z(Φ)∈⟨H,G⊥∗⟩.
First we check condition (3): B′=⋂t∈AZ(Φt∗)∈H
for every A∈G.
Because the family δt, t∈A is
G-summable, by the assumption, the family of columns
Φt∗, t∈A is H-summable. By Theorem 7.2
and the necessary condition of convergency (see
Section 7) we obtain
[TABLE]
that is, the limit of the map t→Φt∗ from A to
HK with respect to the Frechet filter on A is equal
to zero. This means that for any zero neighborhood U(B,H),
B∈H⊥ in {H}K there exists a cofinite subset A0′
of A such that Φt∗∈U(B,H) for any t∈A0′.
In other words, if t∈A0′, then Z(Φt∗)⊇B,
that is, Z(Φt∗)∪B=H.
Then
[TABLE]
[TABLE]
[TABLE]
Note that the last set is cofinite in H. Indeed, Φt∗∈{H}K
and B∈H⊥ implies that each Z(Φt∗)∪B is cofinite, and
A∪A0′=A∖A0′ is a finite set.
Thus we proved that the union B′∪B is cofinite for any
B∈H⊥. This means that B′∈H⊥⊥=H, as
required.
This is the same as the
union A∗∪A′ is cofinite for any A∈G. We have the
following sequence of equivalences:
[TABLE]
[TABLE]
Hence:
[TABLE]
We have already proved that there is a cofinite subset A0′ of
A such that Z(Φt∗)∪B=H for all t∈A0′.
Then A0′∗⊆A′. Since A∪A0′ is cofinite in H,
the same is true for A∗∪A0′∗. From
A0′∗⊆A′ it follows that A∗∪A′ is also
cofinite in H, as desired.
∎
If X is a K-linear topological space, then a map
φ:X→K is said to be a linear form, if φ is linear
and continuous. This space can be endowed with a weak topology:
a net of linear forms φi:X→K, i∈I converges to φ,
if φi(x) converges to φ(x) for every x∈X.
Corollary 9.7**.**
Suppose that G is a balanced filter on G. Then the space of
continuous linear forms of the topological linear space of columns
({G}K,T(G)) is isomorphic to the topological linear space of
rows (K{G⊥∗},T(G⊥∗))
Proof.
Applying Theorem 9.6 to the case when H consists of
one element, we obtain that every continuous linear map
from {G}K to K is given by a left multiplication by a row
from K{G⊥∗}.
∎
For further applications we need a dual variant of Theorem 9.6.
Remark 9.8**.**
Suppose that Ψ is an H×G matrix.
Then the following are equivalent:
a) the right multiplication by Ψ defines a linear continuous map
from K{G} to K{H};
b) Z(Ψ)∈⟨G⊥∗,H⟩.
If ψ is a continuous linear map from K{G} to K{H}, then its matrix
Ψ satisfies these equivalent conditions.
10. The ring of continuous operators
Suppose that Φ is an H×G matrix, γ is a row and a is a
column such that the products γ⋅Φ and Φ⋅a are defined.
Remark 10.1**.**
[TABLE]
Proof.
Indeed, if h∈⋂t∈G∖Z(a)Z(Φt∗), then
[TABLE]
The second inclusion has a similar proof.
∎
Now we prove an auxiliary lemma.
Lemma 10.2**.**
For any β∈K{G⊥∗} and b∈{H}K the matrix
Φ=b⋅β∈HKG satisfies (2) and
(3). Therefore (by Proposition 9.4) the left multiplication
by Φ defines a continuous linear map from {G}K to {H}K.
Proof.
First we check (2). Take any B∈H⊥. If s∈H, then
(b⋅β)s=bsβ,
where bs∈K is the sth coordinate of b. If bs=0, then
Z((b⋅β)s)=G, and otherwise
Z((b⋅β)s)=Z(β). Therefore,
[TABLE]
which implies ⋂s∈HZ((b⋅β)s)∈G⊥∗, as desired. The proof of
(3) is similar.
∎
Recall that in Section 8 we gave an example that a product of
matrices is not associative. It will become associative if we put
some extra restrictions.
Proposition 10.3**.**
Suppose that Υ, Φ and Ψ are matrices of the
following size: Υ∈JKH, Φ∈HKG,
Ψ∈GKI. Further assume that
a) for any pair (j,g) there exist only finitely many h∈H
such that Υjh∗Φhg=0;
b) for any pair (h,i) there exist only finitely many g∈G
such that Φhg∗Ψgi=0;
c) for any pair (j,i)∈J×I the exist only finitely many
(h,g)∈H×G such that
Υjh∗Φhg∗Ψgi=0;
Then the products Υ⋅(Φ⋅Ψ) and
(Υ⋅Φ)⋅Ψ are defined and equal.
Proof.
Clearly a) and b) is nothing more that the existence of the products
Υ⋅Φ and Φ⋅Ψ. From c) we obtain the
following equality:
[TABLE]
which implies
associativity.
∎
This proposition applies in the following situation.
Proposition 10.4**.**
Suppose that b1, b2, … are columns from {G}K, and
β1,β2,… are rows from K{G⊥∗}. Then
any (finite) alternating product
b1⋅β1⋅b2⋅β2⋅… or
β1⋅b1⋅β2⋅b2⋅… does not
depend on the way we put brackets on it.
Proof.
We prove only associativity for short products:
[TABLE]
for any columns b,c and any rows β,γ. The general case is derived by
induction as in [8, Sec. 1].
Clearly the conditions a) and b) from Proposition 10.3 are satisfied.
Furthermore, c) of this proposition means the following: for every
j∈G there are only finitely many g∗∈G such that
bjβg∗cg=0. But this follows from the fact that g
belong to (suppβ)∗∩suppc, which is a finite set.
The proof of the second equality is similar.
∎
Now we define a ring M0(G) of ‘finitary’ G×G-matrices.
Later we will show that this ring can be considered as a
dense subring in the algebra of all continuous linear operators on
{G}K.
Let M0(G) denote the set of all G×G-matrices
[TABLE]
where ai∈{G}K and γi∈K{G⊥∗}, i=1,…,n.
Proposition 10.5**.**
The set (M0(G),+,⋅) is an associative ring.
Proof.
Clearly this set is closed with respect to addition.
We show that the product of elements of (M0(G),+,⋅) is an
element of (M0(G),+,⋅). Clearly it suffices to check this for
elements of the form a⋅γ and b⋅β, where
a,b are columns and γ,β are rows. Indeed,
γ⋅b=k∈K. Applying Proposition 10.4, we obtain
[TABLE]
It remains to use distributivity (see Lemma 8.1).
∎
The following proposition also claims an associativity.
Proposition 10.6**.**
Suppose that G and H are balanced filters on G and H.
Further assume that γ∈K{H⊥∗} is a row,
a∈{G}K is a column, and Φ∈HKG is an H×G-matrix such that Z(Φ)∈⟨H,G⊥∗⟩. Then
(γ⋅Φ)⋅a=γ⋅(Φ⋅a).
Proof.
From Theorem 3.3 3) it follows that
a=ilimT(G)ai, where ai are columns with
finite support. If i is fixed, then (ai)g=0 for finitely
many g∈G. For any such g, there exists only finitely many h
such that γh∗Φhg∗=0. Thus there are only
finitely many pairs (g,h) such that γh∗Φhg∗(ai)g=0. This means that hypotheses of
Proposition 10.3 are satisfied.
Now we have
[TABLE]
Furthermore, by the assumption (and the dual variant of
Proposition 9.4) the right multiplication by Φ acts
as a linear continuous operator from K{H⊥∗} to K{G⊥∗}. In
particular, γ⋅Φ∈K{G⊥∗}. Then, by
Corollary 9.7, the left multiplication by γ⋅Φ is a
linear continuous map from {G}K to K. Thus
[TABLE]
where Disc denotes the discrete topology on K. By
Proposition 10.5 we obtain
(γ⋅Φ)⋅ai=γ⋅(Φ⋅ai), therefore
[TABLE]
Since left multiplications by γ and Φ are continuous operators
(see Theorem 9.6), we conclude that
[TABLE]
Because a=limiT(G)ai, this is equal to γ⋅(Φ⋅a), as desired.
∎
Now we give a matrix description of the algebra of linear continuous operators of
the column space {G}K.
Theorem 10.7**.**
Let G be a balanced filter on a set G with involution, and
let M(G) be a set of all G×G-matrixes Φ such that
Z(Φ)∈⟨G,G⊥∗⟩. Then
a) M(G) is a ring with respect the above defined addition
and multiplication;
b) If Φ∈M(G), then the map Φ:a→Φ⋅a
is a continuous endomorphism of the column space {G}K.
c) The map Φ→Φ^ is an isomorphism from M(G) onto the
ring of all continuous endomorphisms of the space {G}K.
Proof.
b) follows from the equivalence of a) and b) in Theorem
9.6.
a), c) First we prove that M(G) is closed with respect to
multiplication. Suppose that Φ,Ψ∈M(G). Since Z(Φ),Z(Ψ)∈⟨G,G⊥∗⟩, by Remark 9.3 a) and its dual variant,
we obtain that Z(Φg)∈G⊥∗ and Z(Ψh∗)∈G for all
g,h∈G. It follows that the product Φ⋅Ψ is defined.
Take any g∈G. By Proposition 10.6 we obtain
(Φg⋅Ψ)⋅a=Φg⋅(Ψ⋅a).
Then by the definition of matrix multiplication we derive
[TABLE]
and Φg⋅(Ψ⋅a)=(Φ⋅(Ψ⋅a))g, therefore (Φ⋅Ψ)⋅a=Φ⋅(Ψ⋅a). Because the left
multiplication by Φ and Ψ is continuous, this equality
implies that the left multiplication by Φ⋅Ψ is
continuous. By Theorem 9.6 we conclude that Φ⋅Ψ∈M(G).
Furthermore, writing (Φ⋅Ψ)⋅a=Φ⋅(Ψ⋅a) at the level of endomorphisms, we obtain Φ⋅Ψ=Φ∘Ψ, therefore the map Φ→Φ preserves multiplication. Clearly this map preserves an
addition, hence it is a morphism of rings (associative or not).
By Theorem 9.6, if φ is a continuous linear operator
of the space {G}K, then there is a unique matrix Φ∈M(G) such that φ[a]=Φ⋅a. It follows that the map Φ→Φ
is an isomorphism of rings, in particular M(G) is an associative ring.
∎
Suppose that G, H are balanced filters on sets with involution G and H.
Let L(H,G) be the space of all linear continuous maps {G}K→{H}K.
By Theorem 9.6 we may identify L(H,G) with the space of
H×G-matrices Φ such that Z(Φ)∈⟨H,G⊥∗⟩. Therefore
we may consider L(H,G) as a topological space with the topology
T(⟨H,G⊥∗⟩) defined by the filter
⟨H,G⊥∗⟩ on H×G.
It is well known that for normed linear spaces, the convergency
of a net of continuous operators with respect to the operator norm
implies their strong convergency. The following proposition gives a version of this
result in our situation.
Proposition 10.8**.**
Suppose that a net of H×G-matrices Φi∈L(H,G) converges
to Φ with respect to T(⟨H,G⊥∗⟩). Then for any column
a∈{G}K and for any row γ∈K{H⊥∗}
[TABLE]
Proof.
Clearly we may assume that Φ=0, hence Φ⋅a=0. Thus for the
first equation we have to prove that ilimT(G)Φi⋅a=0.
Recall that by Theorem 6.13 we have
⟨H,G⊥∗⟩⊥=H⊥⊗G∗.
Furthermore (see Section 3) the basis of the filter
H⊥⊗G∗ is
given by the sets B1×A1, where B1∈H and A1∈G∗.
Let U(B,H), B∈H⊥ be a zero neighborhood of the space {H}K.
We have to find a zero neighborhood
U=U(B1×A1,T(⟨H,G⊥∗⟩)), of the space L(H,G)
such that Φi∈U implies Φi⋅a∈U(B,H).
Take B1=B and A1=Z(a)∗. If Φi∈U then, by the definition U, we have
[TABLE]
If h is an arbitrary element of B, then
[TABLE]
It follows that Z(Φi⋅a)⊇B, that is,
Φi⋅a∈U(B,H), as desired.
The second equality can be verified similarly.
∎
By Lemma 10.2, M0(G) is a subring of M(G). The following remark
shows that this subring is dense.
Remark 10.9**.**
M0(G)* is a dense subring of M(G) with respect to topology T(⟨G,G⊥∗⟩).*
Proof.
The set K[H×G] of all matrices with finite support forms a subalgebra of the
algebra M0(G). By Theorem 3.3, this algebra is dense in M(G). It remains
to notice that K[H×G]⊆M0(G), because every finite matrix is a linear
combination of matrices δg⋅δh∈M0(G).
∎
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