Topological linear spaces of formal linear sums and continuous linear operators

TL;DR

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

Characterization of rings of continuous operators on formal sum spaces

Application to modules over left ordered groups

Extension of operator theory in topological algebra

Abstract

The rings of linear continuous operators on the topological spaces of $\mathfrak{G}$-zero maps were described, where $\mathfrak{G}$ is a filter on a set with an involution. This applies to modules of formal series with well ordered support over left ordered groups.

Equations

$${\mathfrak{G}}^{\perp }=\{A\subseteq G\mid A\cup A^{\prime }\text{ is cofinite for every }{A}^{\prime }\in \mathfrak{G}\}.$$

$$h_1(g)=\{\begin{array}{ll}f(g),& \text{if g∈A1}\\ 0,& \text{if g∈A1,}\end{array}{h}_2(g)=\{\begin{array}{ll}f(g),& \text{if g∈A1∖A2}\\ 0,& \text{if g∈/A1∖A2.}\end{array}$$

$$⟨f,h⟩=\sum _{x\in G}f(x)h(x^{\ast })=\sum _{x\in G}f(x^{\ast })h(x)$$

$$\mathrm{Funct}(\mathrm{Cof}G)\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\mathrm{Cof}(\mathfrak{H},\mathfrak{G})\cap ({\mathfrak{H}}^{\perp }\otimes {\mathfrak{G}}^{\perp })=(\mathrm{Cof}H\times \mathrm{Cof}G)\cap (\mathfrak{H}\otimes \mathfrak{G})\cap ({\mathfrak{H}}^{\perp }\otimes {\mathfrak{G}}^{\perp }).$$

$$\overline{X}\subseteq \overline{(B^{\prime }\times \overline{A})\cup (\overline{B}\times A^{\prime })}=\overline{B^{\prime }\times \overline{A}}\cap \overline{\overline{B}\times A^{\prime }}=$$

$$(({\overline{B}}^{\prime }\times A)\cup ({\overline{B}}^{\prime }\times \overline{A})\cup (B^{\prime }\times A))\cap ((B\times {\overline{A}}^{\prime })\cup (B\times A^{\prime })\cup (\overline{B}\times {\overline{A}}^{\prime })).$$

$$A\cup A^{\prime}=\bigl{(}\bigcap_{j\in J\setminus F(A^{\prime})}\mathcal{Z}(h_{j})\cup A^{\prime}\bigl{)}\cap\bigl{(}\bigcap_{j\in F(A^{\prime})}\mathcal{Z}(h_{j})\cup A^{\prime}\bigl{)}=G\cap(A_{1}\cup A^{\prime})=A_{1}\cup A^{\prime}$$

$$\Phi =\left(\begin{array}{cccc}1& 1& 1& \dots \end{array}\right),\Psi =\left(\begin{array}{ccccc}1& -1& 0& 0& \dots \\ 0& 1& -1& 0& \dots \\ 0& 0& 1& -1& \dots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right)\text{mbox}and\Theta =\left(\begin{array}{c}1\\ 1\\ 1\\ ⋮\end{array}\right).$$

$${\Phi }_h^{g^{\ast }}=\phi [{\delta }^g{]}_h.$$

$$\varphi\bigl{[}\sum\limits_{j\in J}{}^{\mathfrak{G}}{h}_{j}\bigl{]}=\sum\limits_{j\in J}{}^{\mathfrak{H}}\varphi[{h}_{j}].$$

$$\bigcap _{s\in \overline{B}}\mathcal{Z}({\Phi }_s)\in {\mathfrak{G}}^{\perp \ast }\text{ for every }B\in {\mathfrak{H}}^{\perp }.$$

$$\bigcap _{t\in \overline{A}}\mathcal{Z}({\Phi }^{t^{\ast }})\in \mathfrak{H}\text{ for every }A\in \mathfrak{G}.$$

$$(\Phi \cdot \mathbf{a}{)}_h=\sum _{t\in G}{\Phi }_h^{t^{\ast }}{a}_t=\sum _{t\in A}{\Phi }_h^{t^{\ast }}\cdot 0+\sum _{t\in \overline{A}}0\cdot a_t=0.$$

$$\Phi \cdot U(A,\mathfrak{G})\subseteq U(B,\mathfrak{H}).$$

$$(\Phi \cdot \mathbf{a}{)}_s=\sum _{g\in G}{\Phi }_s^{g^{\ast }}{a}_g=\sum _{g\in \overline{A}}{\Phi }_s^{g^{\ast }}\cdot 0+\sum _{g\in A}0\cdot a_g=0.$$

$$\mathcal{Z}(\Phi )\in ⟨\mathfrak{H},{\mathfrak{G}}^{\perp \ast }⟩\iff \{\begin{array}{l}\bigcap _{t\in \overline{A}}{H}_t\in \mathfrak{H}\text{ for every }A\in {\mathfrak{G}}^{\ast };\\ \bigcap _{s\in \overline{B}}{G}_s\in {\mathfrak{G}}^{\perp \ast }\text{ for every }B\in {\mathfrak{H}}^{\perp },\end{array}$$

$$\mathcal{Z}(\Phi )\in ⟨\mathfrak{H},{\mathfrak{G}}^{\perp \ast }⟩\iff \{\begin{array}{l}\bigcap _{t\in \overline{A}}\mathcal{Z}({\Phi }^t)\in \mathfrak{H}\text{ for every }A\in {\mathfrak{G}}^{\ast };\\ \bigcap _{s\in \overline{B}}\mathcal{Z}({\Phi }_s)\in {\mathfrak{G}}^{\perp \ast }\text{ for every }B\in {\mathfrak{H}}^{\perp }.\end{array}$$

$$\Phi\cdot\bigl{(}\sum\limits_{t\in\overline{A}}{}^{\mathrm{Tych}}\delta^{t}k_{t}\bigl{)}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}k_{t}$$

$$\sum _{t\in \overline{A}}{}^{\mathrm{Tych}}{\delta }^t{k}_t=\sum _{t\in \overline{A}}{}^{\mathfrak{G}}{\delta }^t{k}_t.$$

$$\varphi\bigl{[}\sum\limits_{t\in\overline{A}}{}^{{\mathrm{Tych}}}\delta^{t}k_{t}\bigl{]}=\varphi\bigl{[}\sum\limits_{t\in\overline{A}}{}^{T(\mathfrak{G})}\delta^{t}k_{t}\bigl{]}=\sum\limits_{t\in\overline{A}}{}^{T(\mathfrak{H})}\varphi[\delta^{t}]k_{t}\,.$$

$$\Phi\cdot\bigl{(}\sum\limits_{t\in\overline{A}}{}^{\mathrm{Tych}}\delta^{t}k_{t}\bigl{)}=\varphi\bigl{[}\sum\limits_{t\in\overline{A}}{}^{\mathrm{Tych}}\delta^{t}k_{t}\bigl{]}=\sum\limits_{t\in\overline{A}}{}^{T(\mathfrak{H})}\varphi[\delta^{t}]k_{t}=\sum\limits_{t\in\overline{A}}{}^{T(\mathfrak{H})}\Phi^{t^{*}}k_{t}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}k_{t},$$

$$\Phi\cdot\mathbf{a}=\Phi\cdot\bigl{(}\sum\limits_{t\in\overline{A}}{}^{\mathrm{Tych}}\delta^{t}a_{t}\bigl{)}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}a_{t}\,.$$

$${\mathbf{b}}_j=\sum _{t\in \overline{A}}{}^{\mathrm{Tych}}{\delta }^t{k}_{tj},$$

$$(\Phi \cdot {\mathbf{b}}_j{)}_h=\sum _{t\in \overline{A}}{\Phi }_h^{t^{\ast }}{k}_{tj}=\sum _{t\in \overline{A}\setminus \Delta }0\cdot k_{tj}+\sum _{t\in \Delta }{\Phi }_h^{t^{\ast }}\cdot 0=0.$$

$$\sum _{j\in J}{}^{\mathfrak{H}}\Phi \cdot {\mathbf{b}}_j=\sum _{j\in J}{}^{\mathfrak{H}}\sum _{t\in \overline{A}}{}^{\mathfrak{H}}{\Phi }^{t^{\ast }}{k}_{tj}=\sum _{t\in \overline{A}}{}^{\mathfrak{H}}{\Phi }^{t^{\ast }}\sum _{j\in J}{k}_{tj}=\sum _{t\in \overline{A}}{}^{\mathfrak{H}}{\Phi }^{t^{\ast }}{k}_t.$$

$$\bigl{(}\sum\limits_{j\in J}{}^{\mathfrak{H}}\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}k_{tj}\bigl{)}_{h}=\sum\limits_{j\in J}\Bigl{(}\sum\limits_{t\in\overline{A}}{}\Phi^{t^{*}}k_{tj}\Bigl{)}_{h}\,,$$

$$\sum\limits_{j\in J}\Bigl{(}\sum\limits_{t\in\overline{A}}{}\Phi^{t^{*}}k_{tj}\Bigl{)}_{h}=\sum\limits_{t\in\overline{A}}\Phi_{h}^{t^{*}}\sum\limits_{j\in J}k_{tj}=\Bigl{(}\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}\sum\limits_{j\in J}k_{tj}\Bigl{)}_{h}$$

$$\phi [\mathbf{a}]=\sum _{g\in G}{}^{\mathfrak{H}}\phi [{\delta }^g]k_g=\sum _{g\in G}{}^{\mathfrak{H}}{\Phi }^{g^{\ast }}{k}_g=\Phi \cdot \mathbf{a}.$$

$$\underset{\mathrm{Cof}(\overline{A})}{\lim }{}^{T(\mathfrak{H})}{\Phi }^{t^{\ast }}=0,$$

Full text

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 $\mathbb{R}$ 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 $\mathbb{R}$ 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 $\mathbb{G}$ be the universal covering of the group $SL_{2}(\mathbb{R})$ 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 $SL_{2}(\mathbb{R})$. 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 $K\mathbb{G}$ into a skew field, the first author developed the following approach (see [3]). He considered the space $K\{\operatorname{CoDcc}\mathbb{G}\}$ of all formal series on $\mathbb{G}$ with well ordered support, and the group ring $K\mathbb{G}$ acting on this space by left multiplication. Then one can invert the elements of $K\mathbb{G}$ as linear maps forming a rational closure $D$ of $K\mathbb{G}$ in $K\{\operatorname{CoDcc}\mathbb{G}\}$. 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 $K\{\operatorname{CoDcc}\mathbb{G}\}$ 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 $\mathfrak{G}$ on a set $G$ we connect a linear space $K\{\mathfrak{G}\}$ of all maps from $G$ to $K$ whose zero set belongs to $\mathfrak{G}$. 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 $K\{\operatorname{CoDcc}G\}$ 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 $K\{\mathfrak{G}\}$ and $K\{H\}$, where $\mathfrak{G}$ and $\mathfrak{H}$ 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 $K\{\mathfrak{G}\}$ (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 $\mathfrak{G}$ 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 $G$, $\mathcal{P}(G)$, to be a filter. Clearly, if $\mathfrak{G}$ is a filter, then $\mathfrak{G}=\mathcal{P}(G)$ iff $\emptyset\in\mathfrak{G}$. If $\mathfrak{G}\neq\mathcal{P}(G)$, then $\mathfrak{G}$ is said to be a proper filter on $G$.

Let $\mathcal{L}$ be a collection of subsets of $G$ with the following property: for all $A,B\in\mathcal{L}$ there is $C\in\mathcal{L}$ such that $C\subseteq A\cap B$. Then $\mathfrak{G}=\{B\mid A\subseteq B$ for some $A\in\mathcal{L}\}$ is a filter generated by $\mathcal{L}$, and $\mathcal{L}$ is a filter base for $\mathfrak{G}$.

We say that a subset $A$ of $G$ is cofinite, if its complement $\overline{A}$ is a finite set. The Frechet filter on $G$, $\mathrm{Cof}\,(G)$, consists of all cofinite subsets of $G$. Clearly $\mathrm{Cof}\,(G)$ is a proper filter iff $G$ is an infinite set.

Let $G$ be a linearly ordered set. A subset $\Delta\subseteq G$ is said to be well ordered, if every nonempty subset of $\Delta$ has a minimal element. This is the same as $\Delta$ has a descending chain condition (d.c.c.): every descending chain of elements $a_{1}\geq a_{2}\geq\dots$ of $\Delta$ stabilizes. Clearly $\Delta$ has a d.c.c. iff it contains no (strictly) descending chain $a_{1}>a_{2}>a_{3}>\dots$.

Similarly, $\Delta\subseteq G$ has an ascending chain condition (a.c.c.), if every ascending chain $a_{1}\leq a_{2}\leq\dots$ of its elements stabilizes. Thus $\Delta$ has an a.c.c. iff it contains no (strictly) ascending chain $a_{1}<a_{2}<\dots$ iff $G$ is well ordered in the dual ordering.

Suppose that $(G,\leq)$ is a linearly ordered set. Let $\operatorname{CoDcc}(G)$ 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, $\operatorname{CoDcc}(G)$ is a filter on $G$, and it is a proper filter iff $G$ is not well ordered.

Similarly let $\operatorname{CoAcc}(G)$ be a collection of all subsets of $G$ whose complement has an a.c.c. Then $\operatorname{CoAcc}(G)$ 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: $\mathfrak{G}_{1}\leq\mathfrak{G}_{2}$ if $\mathfrak{G}_{1}\subseteq\mathfrak{G}_{2}$. 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 $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$ there is a least filter $\mathfrak{G}_{1}\vee\mathfrak{G}_{2}$ containing $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$, and there is a largest filter $\mathfrak{G}_{1}\wedge\mathfrak{G}_{2}$ which is contained in both $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$.

The following remark describes the operations in this lattice.

Remark 2.1**.**

Let $\mathfrak{G}_{1}$, $\mathfrak{G}_{2}$ be filters on a set $G$. Then $\mathfrak{G}_{1}\wedge\mathfrak{G}_{2}$ is given by the intersection of filters: $\mathfrak{G}_{1}\cap\mathfrak{G}_{2}=\{A\subseteq G\mid A\in\mathfrak{G}_{1}$ and $A\in\mathfrak{G}_{2}\}$. Furthermore, $\mathfrak{G}_{1}\vee\mathfrak{G}_{2}$ is the filter generated by all intersections $A\cap B$, $A\in\mathfrak{G}_{1}$, $B\in\mathfrak{G}_{2}$.

We define a new operation on filters. In ring theory this operation corresponds to the quotient of ideals. Suppose that $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$ are filters on $G$. Define $\mathfrak{G}_{1}:\mathfrak{G}_{2}=\{A\subseteq G\mid A\cup A^{\prime}\in\mathfrak{G}_{1}$ for every $A^{\prime}\in\mathfrak{G}_{2}\}$.

Fact 2.2** ([5], property 6).**

$\mathfrak{G}_{1}:\mathfrak{G}_{2}$* is a filter on $G$. Furthermore, $\mathfrak{G}_{1}:\mathfrak{G}_{2}$ is the largest filter $\mathcal{F}$ on $G$ with the property $\mathcal{F}\cap\mathfrak{G}_{2}\subseteq\mathfrak{G}_{1}$.*

The following remark is straightforward.

Remark 2.3**.**

1) If $\mathfrak{F}_{1}\subseteq\mathfrak{F}_{2}$ then $\mathfrak{F}_{1}:\mathfrak{G}\subseteq\mathfrak{F}_{2}:\mathfrak{G}$.

2) If $\mathfrak{G}_{1}\subseteq\mathfrak{G}_{2}$ then $\mathfrak{F}:\mathfrak{G}_{1}\supseteq\mathfrak{F}:\mathfrak{G}_{2}$.

3) $\mathfrak{F}:\mathfrak{G}=(\mathfrak{F}\cap\mathfrak{G}):\mathfrak{G}$.

4) $\mathfrak{F}\subseteq\mathfrak{F}:\mathfrak{G}$.

If $\mathfrak{G}$ is a filter, then define

[TABLE]

For instance, $\mathrm{Cof}\,(G)^{\perp}=\mathcal{P}(G)$ and $\mathcal{P}(G)^{\perp}=\mathrm{Cof}\,(G)$. It follows from the definition that $\mathfrak{G}^{\perp}=\mathrm{Cof}\,(G):\mathfrak{G}$, hence $\mathfrak{G}^{\perp}$ is a filter by Fact 2.2. Furthermore, Remark 2.3, 4) implies that $\mathrm{Cof}\,(G)\subseteq\mathfrak{G}^{\perp}$.

Lemma 2.4**.**

$\mathfrak{G}\subseteq\mathfrak{G}^{\perp\perp}$* and $\mathfrak{G}^{\perp}=\mathfrak{G}^{\perp\perp\perp}$ for any filter $\mathfrak{G}$.*

Proof.

We prove that $\mathfrak{G}\subseteq\mathfrak{G}^{\perp\perp}$. Fix $A\in\mathfrak{G}$ and choose any $A^{\prime}\in\mathfrak{G}^{\perp}$. Then (by the definition of $\mathfrak{G}^{\perp}$) $A\cup A^{\prime}$ is cofinite, hence $A\in\mathfrak{G}^{\perp\perp}$.

Remark 2.3 2) applied to $\mathfrak{G}\subseteq\mathfrak{G}^{\perp\perp}$ yields $\mathfrak{G}^{\perp}\supseteq\mathfrak{G}^{\perp\perp\perp}$, and the reverse inclusion follows from what we have just proved. ∎

A filter $\mathfrak{G}$ is said to be balanced, if $\mathfrak{G}^{\perp\perp}=\mathfrak{G}$. For instance, $\mathrm{Cof}\,(G)$ and $\mathcal{P}(G)$ are balanced filters. Furthermore, every balanced filter on $G$ contains $\mathrm{Cof}\,(G)$.

Fact 2.5** ([5], Thm. 14).**

Let $G$ be a linearly ordered set. Then $\operatorname{CoDcc}(G)^{\perp}=\operatorname{CoAcc}(G)$ and $\operatorname{CoAcc}(G)^{\perp}=\operatorname{CoDcc}(G)$, therefore $\operatorname{CoDcc}(G)$ and $\operatorname{CoAcc}(G)$ are balanced filters.

3. Space of $\mathfrak{G}$-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 $\mathcal{T}$ is said to be a linear topological space, if the addition of elements of $L$ defines a continuous function $L\times L\to L$, where $L\times L$ is taken with product topology; and the same is true for any function $k\times L\to L$ given by multiplication by $k\in K$. Since $K$ is discrete the last condition can be replaced by the following: for each open set $U\subseteq L$ and every $0\neq k\in K$ the product $kU$ is open.

Suppose that $U_{i}$, $i\in I$ is a collection of subspaces of $L$ such that for all $i,j\in I$ there exists $k\in I$ with $U_{k}\subseteq U_{i}\cap U_{j}$. A subset $V$ of $L$ is defined to be open, if for every $a\in V$ there exists $i\in I$ such that $a+U_{i}\subseteq V$. This defines a linear topology $\mathcal{T}$ on $L$, therefore $L$ is a topological space with a linear topology. Note that $\mathcal{T}$ is Hausdorff iff $\cap_{i\in I}U_{i}=\{0\}$. In this paper we will consider only Hausdorff linear topologies.

For instance, let $G$ be a set and let $L=\operatorname{Map}(G,K)$ 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 $i\in I$ define a subspace $U_{i}$ of $L$ consisting of all maps $f:G\to K$ such that $f(g)=0$ for every $g\in i$. Then the family $U_{i}$, $i\in I$ 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 $f:G\to K$ is a map. Then the support of $f$, $\operatorname{supp}(f)$, is the following subset of $G$: $\operatorname{supp}(f)=\{g\in G\mid f(g)\neq 0\}$. Similarly the zero-set of $f$, $\mathcal{Z}(f)$, is defined as $\mathcal{Z}(f)=\{g\in G\mid f(g)=0\}$. Clearly $G=\operatorname{supp}(f)\cup\mathcal{Z}(f)$ is a partition of $G$. Furthermore, if $f,h:G\to K$ and $0\neq k\in K$ then $\mathcal{Z}(f+h)\supseteq\mathcal{Z}(f)\cap\mathcal{Z}(h)$, $\mathcal{Z}(k\cdot f)=\mathcal{Z}(f)$ and $\mathcal{Z}(0)=G$, where [math] stands for the zero function.

If $\mathfrak{G}$ is a filter on $G$, then $\operatorname{Funct}(\mathfrak{G})=\{f\in\operatorname{Map}(G,K)\mid\mathcal{Z}(f)\in\mathfrak{G}\}$ will denote the space of $\mathfrak{G}$-zero functions. Clearly $\operatorname{Funct}(\mathfrak{G})$ is a left and right subspace of the (linear) topological space $\operatorname{Map}(G,K)$. For instance, $\operatorname{Funct}(\mathcal{P}(G))=\operatorname{Map}(G,K)$.

The following remark shows that operations on linear spaces $\operatorname{Funct}(\mathfrak{G})$ correspond to operations on the lattice of filters (see Remark 2.1).

Lemma 3.1**.**

Let $\mathfrak{G}_{1}$ and $\mathfrak{G}_{2}$ be filters on $G$. Then $\operatorname{Funct}(\mathfrak{G}_{1})\cap\operatorname{Funct}(\mathfrak{G}_{2})=\operatorname{Funct}(\mathfrak{G}_{1}\cap\mathfrak{G}_{2})$ and $\operatorname{Funct}(\mathfrak{G}_{1})+\operatorname{Funct}(\mathfrak{G}_{2})=\operatorname{Funct}(\mathfrak{G}_{1}\vee\mathfrak{G}_{2})$. Furthermore, if $\mathfrak{G}_{1}\subseteq\mathfrak{G}_{2}$ then $\operatorname{Funct}(\mathfrak{G}_{1})$ is a subspace of $\operatorname{Funct}(\mathfrak{G}_{2})$.

Proof.

Clearly $f\in\operatorname{Funct}(\mathfrak{G}_{1})\cap\operatorname{Funct}(\mathfrak{G}_{2})$ iff $\mathcal{Z}(f)\in\mathfrak{G}_{1}\cap\mathfrak{G}_{2}$ iff $f\in\operatorname{Funct}(\mathfrak{G}_{1}\cap\mathfrak{G}_{2})$, which proves that $\operatorname{Funct}(\mathfrak{G}_{1})\cap\operatorname{Funct}(\mathfrak{G}_{2})=\operatorname{Funct}(\mathfrak{G}_{1}\cap\mathfrak{G}_{2})$.

To prove the inclusion $\operatorname{Funct}(\mathfrak{G}_{1})+\operatorname{Funct}(\mathfrak{G}_{2})\subseteq\operatorname{Funct}(\mathfrak{G}_{1}\vee\mathfrak{G}_{2})$ let $f\in\operatorname{Funct}(\mathfrak{G}_{1})+\operatorname{Funct}(\mathfrak{G}_{2})$. Then $f=h_{1}+h_{2}$, where $\mathcal{Z}(h_{i})\in\mathfrak{G}_{i}$. It follows that $\mathcal{Z}(f)\supseteq\mathcal{Z}(h_{1})\cap\mathcal{Z}(h_{2})\in\mathfrak{G}_{1}\vee\mathfrak{G}_{2}$, hence $f\in\operatorname{Funct}(\mathfrak{G}_{1}\vee\mathfrak{G}_{2})$.

For the reverse inclusion suppose that $f\in\operatorname{Funct}(\mathfrak{G}_{1}\vee\mathfrak{G}_{2})$, that is, $\mathcal{Z}(f)\supseteq A_{1}\cap A_{2}$ for some $A_{i}\in\mathfrak{G}_{i}$, $i=1,2$. Define $h\in\operatorname{Map}(G,K)$ as follows:

[TABLE]

Then $f=h_{1}+h_{2}$ and $\mathcal{Z}(h_{1})\supseteq A_{1}$, $\mathcal{Z}(h_{2})\supseteq A_{2}$, therefore $h_{i}\in\operatorname{Funct}(\mathfrak{G}_{i})$. ∎

Given $A\subseteq G$, we set $U(A,\mathfrak{G})=\{f\in\operatorname{Funct}(\mathfrak{G})\mid\mathcal{Z}(f)\supseteq\overline{A}\}$. Clearly this is the same as $\operatorname{supp}(f)\subseteq A$.

The proof of the following lemma is straightforward.

Lemma 3.2**.**

1) $U(A,\mathfrak{G})$ is a left (right) subspace of $\operatorname{Funct}(\mathfrak{G})$.

2) If $A_{1},A_{2}\subseteq G$ then $U(A_{1}\cap A_{2},\mathfrak{G})=U(A_{1},\mathfrak{G})\cap U(A_{2},\mathfrak{G})$.

Now we are in a position to construct a linear topology on $\operatorname{Funct}(\mathfrak{G})$.

Theorem 3.3** ([6], Thm. 1).**

1) The family of subspaces $\{U(A,\mathfrak{G})\mid A\in\mathfrak{G}^{\perp}\}$ form a base of zero neighborhoods of a linear topology $T(\mathfrak{G})$ on the spaces $\operatorname{Map}(G,K)$ and $\operatorname{Funct}(\mathfrak{G})$, and this topology is Hausdorff.

2) If $\mathfrak{G}$ is a balanced filter, then $\operatorname{Funct}(\mathfrak{G})$ is complete in this topology.

3) If $\mathrm{Cof}\,(G)\subseteq\mathfrak{G}$ then the set of all maps from $\operatorname{Map}(G,K)$ with finite support is dense in $\operatorname{Funct}(\mathfrak{G})$.

Let us consider some examples of topologies $T(\mathfrak{G})$.

Example 3.4**.**

1) If $\mathfrak{G}=\mathrm{Cof}\,(G)$, then $\operatorname{Funct}(\mathfrak{G})$ consists of all functions with finite support and $\mathfrak{G}^{\perp}=\mathcal{P}(G)$. Then $\emptyset\in\mathfrak{G}^{\perp}$, hence $U(\emptyset,G)=\{0\}$ is an open set. It follows that every subset of $\operatorname{Funct}(\mathfrak{G})$ is open and closed, hence $T(\mathfrak{G})$ is a discrete topology.

2) If $\mathfrak{G}=\mathcal{P}(G)$, then $\mathfrak{G}^{\perp}=\mathrm{Cof}\,(G)$, $\operatorname{Funct}(\mathfrak{G})=\operatorname{Map}(G,K)$ and $A$ runs over all cofinite subsets of $G$. Thus we obtain the Tychonoff topology whose subbase is given by the subspaces $U_{g}=\{f\in\operatorname{Map}(G,K)\mid f(g)=0\}$.

3) Suppose that $(G,\leq)$ is a linearly ordered set and $\mathfrak{G}=\operatorname{CoAcc}(G)$. Then $\mathfrak{G}^{\perp}=\operatorname{CoDcc}(G)$, therefore the base of zero neighborhoods is given by $U_{D}=\{f\in\operatorname{Funct}(\mathfrak{G})\mid f(g)=0$ for every $g\in D\}$, where $D$ is a well ordered subset of $G$.

Note that, if $G=(\mathbb{Q},\leq)$, then the space $\operatorname{Map}(\mathbb{Q},K)$ with Tychonoff topology is metrizable and separable. This is not longer true for the topology $T(\operatorname{CoAcc}G)$. Indeed, suppose that $D_{1},D_{2},\dots$ are well ordered subsets of $\mathbb{Q}$ such that $U_{D_{1}},U_{D_{2}},\dots$ form a basis of zero neighborhoods. This means that for every well ordered $D\subseteq\mathbb{Q}$ there is $k$ such that $U_{D_{1}}\cap\ldots\cap U_{D_{k}}\subseteq U_{D}$. It follows easily that $D\subseteq D_{1}\cup\ldots\cup D_{k}$.

Clearly there exists an ascending sequence $d_{1}<d_{2}<\dots$ such that $d_{k}\notin D_{1}\cup\dots\cup D_{k}$ for every $k$. If $D=\{d_{1},d_{2},\dots\}$, then $D$ has a d.c.c, therefore $D\subseteq D_{1}\cup\ldots\cup D_{k}$ for some $k$, and then $d_{k}\in D_{1}\cup\ldots\cup D_{k}$, a contradiction.

This shows that in the space $(\operatorname{Funct}(\operatorname{CoAcc}\mathbb{Q}),T(\operatorname{CoAcc}\mathbb{Q}))$ no point has a countable base of neighborhoods, in particular, this space is not separable.

4) Let $G=\langle t\rangle$ be an infinite cyclic group with the usual linear ordering: $t^{n}\geq t^{m}$ iff $n\geq m$. Let $\mathfrak{G}=\operatorname{CoDcc}(G)$, that is, $\mathfrak{G}$ is generated by the following collection of sets: $\{t^{n}\mid n<l\}$, $l\in\mathbb{Z}$. Then $\mathfrak{G}^{\perp}$ is generated by $\{t^{n}\mid n>m\}$, $m\in\mathbb{Z}$, hence $\operatorname{Funct}(\mathfrak{G})$ is the space of Laurent power series $\sum_{i\geq l}k_{i}t^{i}$ and $T(\mathfrak{G})$ is the $t$-adic topology.

4. Direct sum decompositions

Let $C$ be a subset of $G$. We identify $\operatorname{Map}(C,K)$ with a subspace of $\operatorname{Map}(G,K)$ consisting of all maps $f:G\to K$ such that the restriction of $f$ to $\overline{C}$ is zero. If $\mathfrak{G}$ is a filter on $G$, then the family $\mathfrak{G}_{C}=\{A\cap C\mid A\in\mathfrak{G}\}$ will be a filter on $G$ called an induced filter. Note that with respect to the above identification, $\operatorname{Funct}(\mathfrak{G}_{C})\subseteq\operatorname{Funct}(\mathfrak{G})$. Indeed, if $f\in\operatorname{Map}(C,K)$, then $f\in\operatorname{Funct}(\mathfrak{G}_{C})$ means that $\mathcal{Z}(f)=A\cap C$ for some $A\in\mathfrak{G}$. If we consider $f$ as a map from $G$ to $K$, then $\mathcal{Z}(f)=(A\cap C)\cup\overline{C}\supseteq A$, hence $f\in\operatorname{Funct}(\mathfrak{G})$.

Proposition 4.1**.**

Let $C\subseteq G$. Then the topology $T(\mathfrak{G}_{C})$ on the space $\operatorname{Funct}(\mathfrak{G}_{C})$ coincides with the topology induced by $T(\mathfrak{G})$.

Proof.

Note that $A\cap C\in\mathfrak{G}^{\perp}_{C}$ for every $A\in\mathfrak{G}^{\perp}$. Indeed, every set from $\mathfrak{G}_{C}$ can be written in the form $A^{\prime}\cap C$ for some $A^{\prime}\in\mathfrak{G}$. Since $A\cup A^{\prime}$ is cofinite in $G$, it follows that $(A\cap C)\cup(A^{\prime}\cap C)=(A\cup A^{\prime})\cap C$ is cofinite in $C$. Thus $A\cap C\in\mathfrak{G}_{C}^{\perp}$ and clearly $U(A,\mathfrak{G})\cap\operatorname{Funct}(\mathfrak{G}_{C})\subseteq U(A\cap C,\mathfrak{G}_{C})$.

Now take any $A_{1}\in\mathfrak{G}_{C}^{\perp}$. We claim that $B=A_{1}\cup\overline{C}\in\mathfrak{G}^{\perp}$ which would imply the inclusion $U(A_{1},\mathfrak{G}_{C})\subseteq U(B,\mathfrak{G})\cap\operatorname{Funct}(\mathfrak{G}_{C})$. Indeed, for every $A^{\prime}\in\mathfrak{G}$ the set $A_{1}\cup(A^{\prime}\cap C)$ is cofinite in $C$. Then $(A_{1}\cup\overline{C})\cup A^{\prime}$ is cofinite in $G$ because it contains $A_{1}\cup(A^{\prime}\cap C)$ (cofinite in $C$) and $\overline{C}$. ∎

Let $A$ be a subset of $G$. For every map $f:G\to K$ we have $f=f|_{A}+f|_{\overline{A}}$, where $\operatorname{supp}(f|_{A})\subseteq A$ and $\operatorname{supp}(f|_{\overline{A}})\subseteq\overline{A}$. This yields a decomposition of linear spaces: $\operatorname{Funct}(\mathfrak{G})=\operatorname{Funct}(\mathfrak{G}_{A})\oplus\operatorname{Funct}(\mathfrak{G}_{\overline{A}})$, where $\operatorname{Funct}(\mathfrak{G}_{A})\subseteq\operatorname{Map}(A,K)$ and $\operatorname{Funct}(\mathfrak{G}_{\overline{A}})\subseteq\operatorname{Map}(\overline{A},K)$.

In the following proposition we will single out two important particular cases.

Proposition 4.2**.**

1) If $A\in\mathfrak{G}$ then $\operatorname{Funct}(\mathfrak{G})=\operatorname{Funct}(\mathfrak{G}_{A})\oplus\operatorname{Map}(\overline{A},K)$. Furthermore, $T(\mathfrak{G})$ induces the Tychonoff topology on $\operatorname{Map}(\overline{A},K)$.

2) If $\mathfrak{G}\supseteq\mathrm{Cof}\,(G)$ and $A\in\mathfrak{G}^{\perp}$, then $\operatorname{Funct}(\mathfrak{G})=U(A,\mathfrak{G})\oplus\operatorname{Funct}(\mathrm{Cof}\,\overline{A})$. Furthermore, $T(\mathfrak{G})$ induces the discrete topology on $\operatorname{Funct}(\mathrm{Cof}\,\overline{A})$.

Proof.

1. Since $A\cap\overline{A}=\emptyset\in\operatorname{Funct}(\mathfrak{G}_{\overline{A}})$, the induced filter on $\overline{A}$ coincides with $\mathcal{P}(\overline{A})$, hence $\operatorname{Funct}(\mathfrak{G}_{\overline{A}})=\operatorname{Map}(\overline{A},K)$. Thus, by Proposition 4.1 and Example 3.4 2), the topology induced on $\operatorname{Map}(\overline{A},K)$ by $T(\mathfrak{G})$ will be Tychonoff. It remains to apply the equality $\operatorname{Funct}(\mathfrak{G})=\operatorname{Funct}(\mathfrak{G}_{A})\oplus\operatorname{Funct}(\mathfrak{G}_{\overline{A}})$.

2. By the definition of $\perp$-operation, we obtain $\mathfrak{G}_{\overline{A}}=\mathrm{Cof}\,(\overline{A})$. Again, by Proposition 4.1 and Example 3.4 1), $T(\mathfrak{G})$ induces discrete topology on $\operatorname{Funct}(\mathfrak{G}_{\overline{A}})$. 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 $G\to G$ such that $g^{**}=g$. If $A$ is a subset of $G$ then we define $A^{*}=\{a^{*}\mid a\in A\}$, and we put $\mathfrak{G}^{*}=\{A^{*}\mid A\in\mathfrak{G}\}$, if $\mathfrak{G}$ is a filter on $G$.

The following remark is obvious.

Remark 5.1**.**

If $\mathfrak{G}$ is a filter, then $\mathfrak{G}^{*}$ is also a filter. Furthermore, the map $\mathfrak{G}\to\mathfrak{G}^{*}$ defines an automorphism of the lattice of filters on $G$. For instance, $\mathfrak{G}^{\perp*}=\mathfrak{G}^{*\perp}$ for every filter $G$.

A filter $\mathfrak{G}^{\perp*}$ is said to be adjoint to the filter $\mathfrak{G}$. Thus a filter $\mathfrak{G}$ is self-adjoint, if $\mathfrak{G}^{\perp*}=\mathfrak{G}$. Note that every self-adjoined filter is balanced. Indeed, $\mathfrak{G}^{\perp\perp}=\mathfrak{G}^{\perp\perp**}=\mathfrak{G}^{\perp*\perp*}=(\mathfrak{G}^{\perp*})^{\perp*}=\mathfrak{G}^{\perp*}=\mathfrak{G}$. For instance, let $G$ be a linearly orderer group, $\mathfrak{G}=\operatorname{CoDcc}(G)$ and the involution $*$ is given by $g^{*}=g^{-1}$. Then $\operatorname{CoDcc}(G)^{\perp*}=\operatorname{CoAcc}(G)^{*}$. Since taking the inverse in the linearly ordered group reverses the ordering, $\operatorname{CoAcc}(G)^{*}=\operatorname{CoDcc}(G)$, therefore $\operatorname{CoDcc}(G)$ is self-adjoint, and the same is true for $\operatorname{CoAcc}(G)$.

From now on each group $G$ will be considered as a group with the involution $*:g\to g^{-1}$.

Suppose that $G$ is a group with a linear ordering $\leq$. We say that $G$ is left (right) ordered, if $g_{1}\leq g_{2}$ implies $hg_{1}\leq hg_{2}$ ($g_{1}h\leq g_{2}h$) for every $g_{1},g_{2},h\in G$. A group $G$ is said to be linearly ordered, if it is left and right ordered with respect to $\leq$.

Proposition 5.2**.**

A left ordered group $(G,\leq)$ is linearly ordered iff the filter $\operatorname{CoDcc}(G)$ (or $\operatorname{CoAcc}(G)$) is self-adjoint.

Proof.

We have already proved that for a linearly ordered group $G$, both $\operatorname{CoDcc}(G)$ and $\operatorname{CoAcc}(G)$ are self-adjoint filters.

Suppose that $G$ is a left ordered group and $\operatorname{CoDcc}(G)$ is a self-adjoint filter. This means that $\operatorname{CoDcc}(G)^{\perp*}=\operatorname{CoDcc}(G)$, which is the same as $\operatorname{CoDcc}(G)^{\perp}=\operatorname{CoDcc}(G)^{*}$ or $\operatorname{CoAcc}(G)=\operatorname{CoDcc}(G)^{*}$.

To prove that $G$ is linearly ordered it suffices to check that the cone $P=\{g\in G\mid g\geq e\}$ is invariant, that is, $a^{-1}Pa=P$ for every $a\in G$. Moreover it is enough to verify that $a^{-1}Pa\subseteq P$ for every $a\in G$. Assuming otherwise we will find $a,b\in G$ such that $b>e$ and $a^{-1}ba<e$. Multiplying $ba<a$ on the left by $b$ we obtain a descending chain $\Delta=\{a>ba>b^{2}a>\dots\}$. Since $\Delta$ has an a.c.c., $\overline{\Delta}\in\operatorname{CoAcc}(G)$. From $\operatorname{CoAcc}(G)=\operatorname{CoDcc}(G)^{*}$ it follows that $\overline{\Delta}^{*}\in\operatorname{CoDcc}(G)$ where $\Delta^{*}=\{a^{-1}b^{-n}\mid n\in\omega\}$, hence $\Delta^{*}$ has a smallest element. If $a^{-1}b^{-n}$ is such, then $a^{-1}b^{-n}\leq a^{-1}b^{-n-1}$. Multiplying by $b^{n+1}a$ on the left we obtain $b\leq e$, a contradiction. ∎

Now we define a pairing on the space $\operatorname{Map}(G,K)$ as the following partially defined non-degenerate bilinear form. If $f,h\in\operatorname{Map}(G,K)$ then

[TABLE]

and the result $\langle f,h\rangle$ is defined, if $\operatorname{supp}(f)\cap(\operatorname{supp}h)^{*}$ is a finite set. Clearly this is the same as $(\operatorname{supp}f)^{*}\cap\operatorname{supp}(h)$ is finite. In particular, this is the case when $f\in\operatorname{Funct}(\mathfrak{G})$ and $h\in\operatorname{Funct}(\mathfrak{G}^{\perp*})$ 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 $L^{\prime}$ of $\operatorname{Map}(G,K)$, if the following holds true:

1. the product $\langle f,h\rangle$ is defined for all $f\in L$, $h\in L^{\prime}$;

2. if $f\in\operatorname{Map}(G,K)$ and the product $\langle f,h\rangle$ is defined for every $h\in L^{\prime}$, then $f\in L$;

3. if $h\in\operatorname{Map}(G,K)$ and the product $\langle f,h\rangle$ is defined for every $f\in L$, then $h\in L^{\prime}$.

In this case $(L,L^{\prime})$ is said to be a dual pair.

Proposition 5.3**.**

If $\mathfrak{G}$ is a balanced filter on a set $G$ with an involution, then $(\operatorname{Funct}(\mathfrak{G}^{\perp*}),\operatorname{Funct}(\mathfrak{G}))$ is a dual pair and the same is true for the pair $(\operatorname{Funct}(\mathfrak{G}),\operatorname{Funct}(\mathfrak{G}^{\perp*}))$.

Proof.

1. is satisfied by the definition of $\mathfrak{G}^{\perp}$.

2. Let $f\in\operatorname{Map}(G,K)$ is such that $\langle f,h\rangle$ is defined for every $h\in\operatorname{Funct}(\mathfrak{G})$. Then $(\operatorname{supp}f)^{*}\cap\operatorname{supp}(h)$ is finite for every $h\in\operatorname{Funct}(\mathfrak{G})$, therefore $\mathcal{Z}(f)^{*}\cup\mathcal{Z}(h)$ is a cofinite set for every such $h$. Take any $B\in\mathfrak{G}$ and let $h_{B}$ be the characteristic function of $G\setminus B$ (that is, $h_{B}(g)=0$ if $g\in B$, and $h_{B}(g)=1$ otherwise). Clearly $h_{B}\in\operatorname{Funct}(\mathfrak{G})$. It follows that $\mathcal{Z}(f)^{*}\cup B$ is a cofinite set, therefore, by the definition of $\perp$, we obtain $\mathcal{Z}(f)^{*}\in\mathfrak{G}^{\perp}$. It follows that $\mathcal{Z}(f)\in\mathfrak{G}^{\perp*}$ hence $f\in\operatorname{Funct}(\mathfrak{G}^{\perp*})$.

3. Suppose that $h\in\operatorname{Map}(G,K)$ and the result $\langle f,h\rangle$ is defined for every $f\in\operatorname{Funct}(\mathfrak{G}^{\perp*})$. As in the proof of 2) it follows that $\mathcal{Z}(h)\in(\mathfrak{G}^{\perp*})^{\perp*}=\mathfrak{G}^{\perp\perp}=\mathfrak{G}$, since $\mathfrak{G}$ is balanced. ∎

Thus we obtain the following diagram of pairing:

[TABLE]

where $\longleftrightarrow$ stands for pairing, and $\mathfrak{G}_{1}\subseteq\mathfrak{G}_{2}$ are balanced filters.

Because $\operatorname{Funct}(\mathfrak{G})$ and $\operatorname{Funct}(\mathfrak{G}^{\perp*})$ are paired, it follows that $\operatorname{Funct}(\mathfrak{G}^{\perp*})$ is isomorphic to the space of linear forms on $\operatorname{Funct}(\mathfrak{G})$, that is to the space of linear continuous maps from $\operatorname{Funct}(\mathfrak{G})$ to $K$. We will derive this fact later from a more general description of continuous linear operators on spaces of $\mathfrak{G}$-zero functions.

6. Filters on direct products

Let $\mathfrak{G}$ be a filter on a set $G$, and let $\mathfrak{H}$ be a filter on $H$. In this section we consider different extensions of these filters to a filter on the direct product $H\times G$. One of these extensions is well known.

Fact 6.1**.**

(see [1, Sec. 6.7]) The family $\{B\times A\mid B\in\mathfrak{H},A\in\mathfrak{G}\}$ is a base of a filter $\mathfrak{H}\times\mathfrak{G}$ on $H\times G$. This filter is proper iff both $\mathfrak{G}$ and $\mathfrak{H}$ are proper.

But this particular filter bears no significance for topological spaces of formal sums. The following filter is more useful.

Fact 6.2** (see [5], Prop. 16).**

The family $\{\overline{\overline{B}\times\overline{A}}\mid B\in\mathfrak{H},A\in\mathfrak{G}\}$ form a base of a filter $\mathfrak{H}\otimes\mathfrak{G}$ on $H\times G$. The filter $\mathfrak{H}\otimes\mathfrak{G}$ is proper iff $\mathfrak{G}$ or $\mathfrak{H}$ is proper. Furthermore, $\mathfrak{H}\otimes\mathfrak{G}\subseteq\mathfrak{H}\times\mathfrak{G}$.

Note that, if $X$ is a subset of $H\times G$, then $X\in\mathfrak{H}\otimes\mathfrak{G}$ iff $X\supseteq\overline{\overline{B}\times\overline{A}}$ iff $\overline{X}\subseteq\overline{B}\times\overline{A}$ for some $B\in\mathfrak{H}$ and $A\in\mathfrak{G}$.

Fact 6.3** (see [5], (6)).**

$\mathrm{Cof}\,(H)\otimes\mathrm{Cof}\,(G)=\mathrm{Cof}\,(H\times G)$.

We introduce a new filter $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})$ whose subbase is given by the following collections of sets: $\{\overline{h\times\overline{A}}\mid h\in H,A\in\mathfrak{G}\}$ and $\{\overline{\overline{B}\times g}\mid B\in\mathfrak{H},g\in G\}$. Thus $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})$ consists of subsets of $H\times G$ whose complement is a subset of a finite union of sets $\{h\}\times\overline{A}$ and $\overline{B}\times\{g\}$. But every subset of $\{h\}\times\overline{A}$, $A\in\mathfrak{G}$ is of the form $\{h\}\times\overline{A}^{\prime}$ for some $A^{\prime}\in\mathfrak{G}$, and every subset of $\overline{B}\times\{g\}$, $B\in\mathfrak{H}$ is of the form $\overline{B}^{\prime}\times\{g\}$ for some $B^{\prime}\in\mathfrak{H}$. It follows that $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})$ consists of complements to the sets $\left(\bigcup_{i=1}^{n}\{h_{i}\}\times\overline{A}_{i}\right)\cup\left(\bigcup_{j=1}^{m}\overline{B}_{j}\times\{g_{j}\}\right)$, where $g_{j}\in G$, $h_{i}\in H$ and $A_{i}\in\mathfrak{G}$, $B_{j}\in\mathfrak{H}$.

Fact 6.4** (see [5], Prop. 18).**

$\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})=(\mathrm{Cof}\,H\otimes\mathfrak{G})\vee(H\otimes\mathrm{Cof}\,G)$.

Fact 6.5** (see [5], Prop. 18, 20).**

Suppose that $\mathfrak{G}$ contains the Frechet filter on $G$, and $\mathfrak{H}$ contains the Frechet filter on $H$. Then

1) $\mathrm{Cof}\,(H\times G)\subseteq\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})\subseteq\mathfrak{H}\otimes\mathfrak{G}$.

2) $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})=(\mathrm{Cof}\,H\times\mathrm{Cof}\,G)\cap(\mathfrak{H}\otimes\mathfrak{G})$.

Now we are in a position to introduce the main construction of this section.

Definition 6.6**.**

Suppose that $\mathfrak{H}$ is a filter on $H$ and $\mathfrak{G}$ is a filter on $G$. Let $\langle\mathfrak{H},\mathfrak{G}\rangle$ consists of all subsets $X$ of $H\times G$ with the following properties:

a) for every $A\in\mathfrak{G}^{\perp}$ there exists $B^{\prime}\in\mathfrak{H}$ such that $B^{\prime}\times\overline{A}\subseteq X$;

b) for every $B\in\mathfrak{H}^{\perp}$ there exists $A^{\prime}\in\mathfrak{G}$ such that $\overline{B}\times A^{\prime}\subseteq X$.

If $t\in G$, then define $H_{t}(X)=\{h\in H\mid(h,t)\in X\}$. Similarly, if $s\in H$, then set $G_{s}(X)=\{g\in G\mid(s,g)\in X\}$.

It is easily seen that $X\in\langle\mathfrak{H},\mathfrak{G}\rangle$ iff the following holds:

a)′  $\bigcap_{t\in\overline{A}}H_{t}(X)\in\mathfrak{H}$ for every $A\in\mathfrak{G}^{\perp}$;

b)′  $\bigcap_{s\in\overline{B}}G_{s}(X)\in\mathfrak{G}$ for every $B\in\mathfrak{H}^{\perp}$.

For instance, the equivalence of a) and a)′ can be seen as follows. If a) holds then $B^{\prime}\subseteq\bigcap_{t\in\overline{A}}H_{t}$, hence this intersection is in $\mathfrak{H}$; and if a)′ holds then we take $B^{\prime}$ to be equal to this intersection.

Fact 6.7** (see [5], Thm. 23).**

$\langle\mathfrak{H},\mathfrak{G}\rangle$* is a filter on $H\times G$ containing $\mathfrak{H}\otimes\mathfrak{G}$.*

Fact 6.8** (see [5], L. 24).**

If $\mathfrak{H}$ and $\mathfrak{G}$ contain Frechet filters, then $(\mathfrak{H}\otimes\mathfrak{G})\cap(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=\mathrm{Cof}\,(H\times G)$.

The following remark is obvious.

Remark 6.9**.**

If $\mathfrak{H}\subseteq\mathfrak{H}^{\prime}$ and $\mathfrak{G}\subseteq\mathfrak{G}^{\prime}$, then $\mathfrak{H}\otimes\mathfrak{G}\subseteq\mathfrak{H}^{\prime}\otimes\mathfrak{G}^{\prime}$.

Proposition 6.10**.**

If $\mathfrak{H}$ and $\mathfrak{G}$ are balanced filters, then $\langle\mathfrak{H},\mathfrak{G}\rangle\supseteq\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G}):(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})^{\perp}$.

Proof.

First we prove the equality. We claim that $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})\cap(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=\mathrm{Cof}\,(H\times G)$. Indeed, by Fact 6.5 2), we obtain

[TABLE]

By Fact 6.8 this is the same as $(\mathrm{Cof}\,H\times\mathrm{Cof}\,G)\cap\mathrm{Cof}\,(H\times G)=\mathrm{Cof}\,(H\times G)$, as desired.

Then by Remark 2.3 3) and the definition of $\perp$ we obtain $\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G}):(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=(\mathrm{Cof}\,(\mathfrak{H},\mathfrak{G})\cap(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})):(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=\mathrm{Cof}\,(H\times G):(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})=(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})^{\perp}$.

Now we prove the inclusion. Suppose that $Z\in(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})^{\perp}$ and we have to show that $Z\in\langle\mathfrak{H},\mathfrak{G}\rangle$. Take any $B\in\mathfrak{H}^{\perp}$ and define $A^{\prime}\subseteq G$ by the following rule: $\overline{A}^{\prime}=\{g\in G\mid(h,g)\in\overline{Z}$ for some $h\in\overline{B}\}=\pi_{G}[(\overline{B}\times G)\cap\overline{Z}]$, where $\pi_{G}$ is a projection on $G$.

Suppose that $A^{\prime}\notin\mathfrak{G}$. Then $A^{\prime}\notin(\mathfrak{G}^{\perp})^{\perp}$ because $\mathfrak{G}$ is balanced. It follows that there exists $A\in\mathfrak{G}^{\perp}$ such that the union $A^{\prime}\cup A$ is not cofinite in $G$, that is, the intersection $\overline{A}^{\prime}\cap\overline{A}$ is infinite. Since $\overline{\overline{B}\times\overline{A}}\in\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp}$, it follows that $Z\cup\overline{\overline{B}\times\overline{A}}$ is cofinite in $H\times G$, hence $\overline{Z}\cap(\overline{B}\times\overline{A})$ is a finite set. On the other hand for every $g\in\overline{A}^{\prime}\cap\overline{A}$ (there are infinitely many of them) there exists $h\in\overline{B}$ such that $(h,g)\in\overline{Z}\cap(\overline{B}\times\overline{A})$, hence this set must be infinite, a contradiction. Thus $A^{\prime}\in\mathfrak{G}$.

We prove that $\overline{B}\times A^{\prime}\subseteq Z$. Indeed, otherwise $(h,g)\in\overline{Z}$ for some $(h,g)\in\overline{B}\times A^{\prime}$. Since $g\notin\overline{A}^{\prime}$, by the construction of $\overline{A}^{\prime}$ we obtain $h\notin\overline{B}$, a contradiction.

Thus for every $B\in\mathfrak{H}^{\perp}$ there exists $A^{\prime}\in\mathfrak{G}$ such that $\overline{B}\times A^{\prime}\subseteq Z$. Similarly for every $A\in\mathfrak{G}^{\perp}$ there exists $B^{\prime}\in\mathfrak{H}$ such that $B^{\prime}\times\overline{A}\subseteq Z$. It follows that $Z\in\langle\mathfrak{H},\mathfrak{G}\rangle$. ∎

Before proving the next lemma, let us recall a useful equality: if $B\subseteq H$ and $A\subseteq G$, then $\overline{(\overline{B}\times\overline{A})}=(B\times A)\cup(B\times\overline{A})\cup(\overline{B}\times A)$.

Lemma 6.11**.**

$\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp}\subseteq\langle\mathfrak{H},\mathfrak{G}\rangle^{\perp}$* for all filters $\mathfrak{H}$ and $\mathfrak{G}$.*

Proof.

Suppose that $Z\in\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp}$, hence $\overline{Z}\subseteq\overline{B}\times\overline{A}$ for some $B\in\mathfrak{H}^{\perp}$, $A\in\mathfrak{G}^{\perp}$. We have to prove that $Z\in\langle\mathfrak{H},\mathfrak{G}\rangle^{\perp}$, that is, $Z\cup X$ is a cofinite set for every $X\in\langle\mathfrak{H},\mathfrak{G}\rangle$. Since $X\in\langle\mathfrak{H},\mathfrak{G}\rangle$ there are $B^{\prime}\in\mathfrak{H}$ and $A^{\prime}\in\mathfrak{G}$ such that $B^{\prime}\times\overline{A},\overline{B}\times A^{\prime}\subseteq X$. Then

[TABLE]

[TABLE]

Intersecting this with $\overline{Z}\subseteq\overline{B}\times\overline{A}$ and taking into account that the intersections of $\overline{B}\times\overline{A}$ with $\overline{B}^{\prime}\times A$, $B^{\prime}\times A$, $B\times\overline{A}^{\prime}$ and $B\times A^{\prime}$ are empty, we obtain $\overline{X}\cap\overline{Z}\subseteq(\overline{B}^{\prime}\cap\overline{B})\times(\overline{A}\cap\overline{A}^{\prime})$. Since $B\in\mathfrak{H}^{\perp}$ and $B^{\prime}\in\mathfrak{H}$, therefore $B\cup B^{\prime}$ is a cofinite set, hence $\overline{B}^{\prime}\cap\overline{B}$ is finite. Similarly $\overline{A}^{\prime}\cap\overline{A}$ is a finite set, hence $\overline{X}\cap\overline{Z}$ is finite, as desired. ∎

Remark 6.12**.**

$\mathfrak{H}\otimes\mathfrak{G}\subseteq\langle\mathfrak{H}^{\perp},\mathfrak{G}^{\perp}\rangle^{\perp}$* for all filter $\mathfrak{H}$ and $\mathfrak{G}.$*

Proof.

Indeed, by Lemma 2.4 we have $\mathfrak{G}\subseteq\mathfrak{G}^{\perp\perp}$ and $\mathfrak{H}\subseteq\mathfrak{H}^{\perp\perp}$, therefore $\mathfrak{H}\otimes\mathfrak{G}\subseteq\mathfrak{H}^{\perp\perp}\otimes\mathfrak{G}^{\perp\perp}$ by Remark 6.9. Furthermore $\mathfrak{H}^{\perp\perp}\otimes\mathfrak{G}^{\perp\perp}\subseteq\langle\mathfrak{H}^{\perp},\mathfrak{G}^{\perp}\rangle^{\perp}$ by Lemma 6.11. ∎

Theorem 6.13**.**

Suppose that $\mathfrak{H}$ and $\mathfrak{G}$ are balanced filters. Then $\langle\mathfrak{H},\mathfrak{G}\rangle=(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})^{\perp}$. In particular, $\langle\mathfrak{H},\mathfrak{G}\rangle$ is a balanced filter.

Proof.

By Proposition 6.10 we have $\langle\mathfrak{H},\mathfrak{G}\rangle\supseteq\langle\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp}\rangle^{\perp}$. On the other hand applying $\perp$ to the inclusion in Lemma 6.11, we obtain $(\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{\perp})^{\perp}\supseteq\langle\mathfrak{H},\mathfrak{G}\rangle^{\perp\perp}$. Thus $\langle\mathfrak{H},\mathfrak{G}\rangle\supseteq\langle\mathfrak{H}^{\perp},\mathfrak{G}^{\perp}\rangle^{\perp}\supseteq\langle\mathfrak{H},\mathfrak{G}\rangle^{\perp\perp}$, hence $\langle\mathfrak{H},\mathfrak{G}\rangle\supseteq\langle\mathfrak{H},\mathfrak{G}\rangle^{\perp\perp}$. Then Lemma 2.4 yields the desired. ∎

7. $\mathfrak{G}$-sums

Suppose that $(X,\mathcal{T})$ is a typological abelian group and $x_{i}$, $i\in I$ is a family of elements of $X$. An element $x\in X$ is said to be a sum of this family with respect to $\mathcal{T}$, written $x=\sum_{i\in I}^{\mathcal{T}}x_{i}$, if the following holds. For every neighbourhood $U$ of $x$ there is a finite subset $\Delta\subseteq I$ such that $\sum_{i\in\Delta^{\prime}}x_{i}\in U$ for every finite set $\Delta^{\prime}\subseteq I$ containing $\Delta$. Clearly, if $X$ is Hausdorff, then the sum is unique. It is easily seen that, if the family $x_{i}$, $i\in I$ is summable, then the limit of the $x_{i}$ 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 $\Delta$ of $I$ such that $x_{i}\in U$ for every $i\in I\setminus\Delta$.

Suppose that $(X,\mathcal{T}_{X})$ and $(Y,\mathcal{T}_{Y})$ are topological abelian groups. If $\varphi$ is a continuous morphism from $X$ to $Y$ then $\varphi$ preserves topological sums. This means that, if $x=\sum_{i\in I}^{\mathcal{T}_{X}}x_{i}$ in $X$, then the sum $\sum_{i\in I}^{\mathcal{T}_{Y}}\varphi(x_{i})$ exists and equal to $\varphi(x)$.

For more on sums in topological abelian groups see [2]

Now we define a sum with respect to a filter.

Definition 7.1**.**

Let $h_{j}\in\operatorname{Map}(G,K)$, $j\in J$ be a family of maps and let $\mathfrak{G}$ be a filter. This family is said to be $\mathfrak{G}$-summable, and the map $f:G\to K$ is a $\mathfrak{G}$-sum of this family, $f=\sum_{j\in J}^{\mathfrak{G}}h_{j}$, if the following holds:

1) for every $g\in G$ there are only finitely many $j\in J$ such that $g\in\operatorname{supp}(h_{j})$, and $f(g)=\sum_{j\in J}h_{j}(g)$;

2) $\bigcap_{j\in J}\mathcal{Z}(h_{j})\in\mathfrak{G}$.

Note that the condition 1) of this definition means that the family $\{h_{j}\}$, $j\in J$ is summable with respect to Tychonoff topology on $\operatorname{Map}(G,K)$. Furthermore, 2) implies that $h_{j}\in\operatorname{Funct}(\mathfrak{G})$ for every $j$; and $\mathcal{Z}(f)\supseteq\bigcap_{j\in J}\mathcal{Z}(h_{j})$ yields that $f\in\operatorname{Funct}(\mathfrak{G})$.

In the following theorem we compare these two types of summability.

Theorem 7.2**.**

1) If the family $h_{j}$, $j\in J$ is $\mathfrak{G}$-summable, then it is summable with respect to topology $T(\mathfrak{G})$. Furthermore, $\sum_{j\in J}^{\mathfrak{G}}h_{j}=\sum_{j\in J}^{T(\mathfrak{G})}h_{j}$.

2) If the family $h_{j}\in\operatorname{Funct}(\mathfrak{G})$, $j\in J$ is summable with respect to $T(\mathfrak{G})$ and $\mathfrak{G}$ is balanced, then it is $\mathfrak{G}$-summable and $\sum_{j\in J}^{\mathfrak{G}}h_{j}=\sum_{j\in J}^{T(\mathfrak{G})}h_{j}$ again.

Proof.

1. Suppose that $f=\sum_{j\in J}^{\mathfrak{G}}h_{j}$ and $A=\bigcap_{j\in J}\mathcal{Z}(h_{j})\in\mathfrak{G}$. Choose any $A^{\prime}\in\mathfrak{G}^{\perp}$. Then $A\cup A^{\prime}$ is a cofinite set, hence $A\cup A^{\prime}=G\setminus\{g_{1},\dots,g_{n}\}$, $g_{i}\in G$. Let $J_{0}$ consist of all $j\in J$ such that $g_{t}\in\operatorname{supp}(h_{j})$ for some $t=1,\dots,n$. By the assumption $J_{0}$ is a finite subset of $J$.

Suppose that $J^{\prime}$ is any finite subset of $J$ containing $J_{0}$. Then $\mathcal{Z}(f-\sum_{j\in J^{\prime}}h_{j})\supseteq A\cup\{g_{1},\dots,g_{n}\}\supseteq\overline{A}^{\prime}$, therefore $f-\sum_{j\in J^{\prime}}h_{j}\in U(A^{\prime},\mathfrak{G})$. This proves that $f=\sum_{j\in J}^{T(\mathfrak{G})}h_{j}$.

2. Because the family $h_{j}$, $j\in J$ is summable with respect to $T(\mathfrak{G})$, by what we have already noticed, $h_{j}$ converges to [math] with respect to the Frechet filter on $I$. It follows that for every $g\in G$ there exist only finitely many $j\in J$ such that $h_{j}(g)\neq 0$, therefore 1) holds true.

Thus it remains to prove that $A=\bigcap_{j\in J}\mathcal{Z}(h_{j})\in\mathfrak{G}$. Let $f=\sum_{j\in J}^{T(\mathfrak{G})}h_{j}$. As above the $h_{j}$ converge to zero with respect to the Frechet filter on $J$. Thus for every $A^{\prime}\in\mathfrak{G}^{\perp}$ there is a finite set of indices $F(A^{\prime})$ such that for every $j\in J\setminus F(A^{\prime})$ we have $h_{j}\in U(A^{\prime},\mathfrak{G})$. Then $\mathcal{Z}(h_{j})\supseteq\overline{A}^{\prime}$ yields $\bigcap_{j\in J\setminus F(A^{\prime})}\mathcal{Z}(h_{j})\supseteq\overline{A}^{\prime}$. Since $h_{j}\in\operatorname{Funct}(\mathfrak{G})$ and $F(A^{\prime})$ is finite, it follows that $A_{1}=\bigcap_{j\in F(A^{\prime})}\mathcal{Z}(h_{j})\in\mathfrak{G}$. Then

[TABLE]

is a cofinite set. Since this is true for any $A^{\prime}\in\mathfrak{G}^{\perp}$, we conclude that $A\in\mathfrak{G}^{\perp\perp}=\mathfrak{G}$. ∎

8. Matrix notations

Suppose that $G$ is a set and $H$ is a set with an involution $*$. Each map $\Psi:H\times G\to K$ can be consider as an $H\times G$ matrix over $K$ whose $(h,g)$-entry, $\Psi^{g}_{h}$, is $\Psi(h,g)$. 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 ${}^{H}K^{G}$. If $H$ consists of one element, a map $\Psi:H\times G\to K$ is said to be a row, and we use small Greek letters $\alpha,\beta,\dots$ to denote rows. Similarly, if $G$ consists of one element, then a map $\Psi:H\to K$ is a said to be a column, and we use small boldfaced letters $\mathbf{a},\mathbf{b}$, $\dots$ to denote columns. In case, when $H$ consists of one element $h$, we simplify notations: $K^{G}={}^{\{h\}}K^{G}$; and similarly ${}^{H}K$ means ${}^{H}K^{\{g\}}$, when $G$ consists of one element $g$.

Let $\delta$ be the Kronecker symbol on $G$, that is, $\delta$ is a map from $G\times G$ to $K$ such that $\delta^{g}_{h}=1$ if $g=h$ and $\delta^{g}_{h}=0$ otherwise. If $\Psi\in{}^{H}K^{G}$ and $h\in H$, then $\Psi_{h}$ will denote the row of $\Psi$ with number $h$, therefore $(\Psi_{h})^{g}=\Psi^{g}_{h}$ for every $g\in G$. Similarly, $\Psi^{g}$ will denote the column of $\Psi$ with number $g$, therefore $(\Psi^{g})_{h}=\Psi^{g}_{h}$ for every $h\in H$. In particular, $\delta_{g}$ is a row whose $g$th entry is $1$ and all remaining entries are zero, and similarly for the column $\delta^{g}$.

Suppose that $\Phi\in{}^{J}K^{H}$ and $\Psi\in{}^{H}K^{G}$. We say the the product $\Theta=\Phi\cdot\Psi\in{}^{J}K^{G}$ is defined if, for every pair $(j,g)\in J\times G$, we have $\Phi^{h^{*}}_{j}\cdot\Psi^{g}_{h}\neq 0$ only for finitely many $h\in H$ and $\Theta^{g}_{j}=\sum_{h\in H}\Phi^{h^{*}}_{j}\cdot\Psi^{g}_{h}$. This defines a partial operation ${}^{J}K^{H}\times{}^{H}K^{G}\to{}^{J}K^{G}$.

Note that $\sum_{h\in H}\Phi^{h^{*}}_{j}\cdot\Psi^{g}_{h}=\sum_{h\in H}\Phi^{h}_{j}\cdot\Psi^{g}_{h*}$. 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 $\Theta_{j}=\Phi_{j}\cdot\Psi$ and $\Theta^{g}=\Phi\cdot\Psi^{g}$.

If $G$ is a finite set, then the multiplication on ${}^{G}K^{G}$ is defined everywhere, therefore ${}^{G}K^{G}$ is a ring isomorphic to the ring of $|G|\times|G|$ matrices over $K$. But the unity of this ring is given by the map $E:G\times G\to G$ such that $E^{g^{*}}_{g}=1$ for every $g\in G$ 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 $G=\mathbb{N}$ with the identical involution $*$, and let $\Phi$, $\Psi$ and $\Theta$ be the following matrices:

[TABLE]

Then $(\Phi\cdot\Psi)\cdot\Theta=1$ and $\Phi\cdot(\Psi\cdot\Theta)=0$.

The following lemma claims distributivity and can be easily verified by direct calculations.

Proposition 8.1**.**

Suppose that $\Phi\in{}^{J}K^{H}$ and $\Psi,\Theta\in{}^{H}K^{G}$. If both products $\Phi\cdot\Psi$ and $\Phi\cdot\Theta$ are defined, then $\Phi\cdot(\Psi+\Theta)$ is also defined and equal to $\Phi\cdot\Psi+\Phi\cdot\Theta$.

An arbitrary row $\gamma\in K^{G}$ is uniquely determined by its coordinates $\gamma^{g}\in K$, $g\in G$. Thus $\gamma$ is a topological sum (see Section 7) with respect to Tychonoff topology on $\operatorname{Map}(G,K)$: $\gamma=\sum_{g\in G}^{\mathop{{\rm Tych}}\nolimits}\gamma^{g}\delta_{g}$. Furthermore, if $\gamma\in\operatorname{Funct}(\mathfrak{G})$, then clearly $\gamma=\sum_{g\in G}^{\mathfrak{G}}\gamma^{g}\delta_{g}$. Similarly, each column $\mathbf{a}\in{}^{H}K$ is uniquely determined by its coordinates $a_{h}\in K$, $h\in H$, hence we can write $\mathbf{a}=\sum_{h\in H}^{\mathop{{\rm Tych}}\nolimits}\delta^{h}a_{h}$; and if $\mathbf{a}\in\operatorname{Funct}(\mathfrak{H})$, then $\mathbf{a}:=\sum_{h\in H}^{\mathfrak{H}}\delta^{h}a_{h}$.

9. Continuous linear maps

Recall that $\mathfrak{G}$ is a filter on a set $G$ and $\mathfrak{H}$ is a filter on a set $H$. In what follows we will always assume that $\mathfrak{G}$ and $\mathfrak{H}$ contain Frechet filters. Thus $\operatorname{Funct}(\mathrm{Cof}\,G)\subseteq\operatorname{Funct}(\mathfrak{G})$ and $\operatorname{Funct}(\mathrm{Cof}\,H)\subseteq\operatorname{Funct}(\mathfrak{H})$. In this section we describe continuous linear maps of topological linear spaces $\operatorname{Funct}(\mathfrak{G})\to\operatorname{Funct}(\mathfrak{H})$.

We will consider $\operatorname{Funct}(\mathfrak{G})$ as a left or right $K$-vector space. To specify the side, we will use $K\{\mathfrak{G}\}$ to denote this space considered as a left vector space over $K$, and call it a space of $\mathfrak{G}$-zero rows. Similarly $\{\mathfrak{G}\}K$ will denote $\operatorname{Funct}(\mathfrak{G})$ considered as a right vector space over $K$, and will be called a space of $\mathfrak{G}$-zero columns. We use $K[G]$ to denote $K\{\mathrm{Cof}\,G\}$, and $[G]K$ to denote $\{\mathrm{Cof}\,G\}K$. 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 $\alpha,\beta\dots$, and columns with small boldfaced letters $\mathbf{a},\mathbf{b},\dots$. Suppose that $\varphi$ is a map from $\operatorname{Funct}(\mathfrak{G})$ to $\operatorname{Map}(H,K)$. First we consider $\operatorname{Funct}(\mathfrak{G})$ as a right vector space $\{\mathfrak{G}\}K$ and $\operatorname{Map}(H,K)$ as the space of columns ${}^{H}K$. The image of a column $\mathbf{a}\in\{\mathfrak{G}\}K$ will be denoted by $\varphi[\mathbf{a}]$. Then we can assign to $\varphi$ an $H\times G$ matrix $\Phi$, whose $(h,g^{*})$-entry is equal to $\varphi[\delta^{g}]_{h}:$

[TABLE]

(Recall that $\delta^{g}$ is a column whose $g$th coordinate is $1$ and all the remaining coordinates are zero, and $\varphi[\delta^{g}]_{h}$ is the $h$th coordinate of the column $\varphi[\delta^{g}]$). We say that $\Phi$ is a matrix of $\varphi$. For instance, $\Phi^{g^{*}}=\varphi[\delta^{g}]$ is the $g^{*}$th column of $\Phi$.

The zero set of $\Phi$, $\mathcal{Z}(\Phi)$, is a collection of all $(h,g)\in H\times G$ such that $\Phi^{g}_{h}=0$. This is in accordance with our previous definition of the zero set of a map.

Now we restrict ourselves to the case, when $\varphi$ is a map from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K\subseteq\operatorname{Map}(H,K)$.

Definition 9.1**.**

A right (left) linear map $\varphi:\operatorname{Funct}(\mathfrak{G})\to\operatorname{Funct}(\mathfrak{H})$ is said to be $\mathfrak{G}$-linear, if for any $\mathfrak{G}$-summable family $h_{j}$, $j\in J$, where $h_{j}\in\operatorname{Funct}(\mathfrak{G})$, the family $\varphi[h_{j}]$, $j\in J$ is $\mathfrak{H}$-summable and

[TABLE]

Apparently this condition bears no connection with topology. But below (see Theorem 9.6) we will see that $\mathfrak{G}$-linearity is the same as continuity. The following lemma shows that both $\mathfrak{G}$-linear or continuous linear operators are uniquely determined by their matrices.

Lemma 9.2**.**

Suppose that $\varphi:\{\mathfrak{G}\}K\to\{\mathfrak{H}\}K$ is either $\mathfrak{G}$-linear or a linear continuous map. If $\Phi$ is a matrix of $\varphi$ and $\mathbf{a}\in\{\mathfrak{G}\}K$, then $\varphi[\mathbf{a}]=\Phi\cdot\mathbf{a}$.

Proof.

We will prove this lemma only when $\varphi$ is continuous. The proof in the case, when $\varphi$ is $\mathfrak{G}$-linear, is similar.

As we have already noticed (see a remark after Lemma 8.1), $\mathbf{a}=\sum\limits_{g\in G}{}^{\mathfrak{G}}\delta^{g}a_{g}$, $a_{g}\in K$, hence $\mathbf{a}=\sum\limits_{g\in G}{}^{T(\mathfrak{G})}\delta^{g}a_{g}$ by Theorem 7.2. Since $\varphi$ is continuous and linear, we obtain $\varphi[\mathbf{a}]=\sum\limits_{g\in G}{}^{T(\mathfrak{H})}\varphi[\delta^{g}]a_{g}$. Invoking Theorem 7.2 again, we get $\varphi[\mathbf{a}]=\sum\limits_{g\in G}{}^{\mathfrak{H}}\varphi[\delta^{g}]a_{g}=\sum\limits_{g\in G}{}^{\mathfrak{H}}\Phi^{g^{*}}a_{g}$. On the other hand, by the definition of product of matrices, $(\Phi\cdot\mathbf{a})_{h}=\sum\limits_{g\in G}\Phi^{g^{*}}_{h}a_{g}$, therefore $\varphi[\mathbf{a}]=\Phi\cdot\mathbf{a}$, as desired. ∎

Our next objective is to decide when the left multiplication by an $H\times G$ matrix $\Phi$ defines a continuous linear map $\varphi$ from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K$. As a first approximation we consider the following condition:

[TABLE]

It guarantees that $\varphi$ is a linear map.

Remark 9.3**.**

Let $\Phi$ be an $H\times G$-matrix satisfying (2). Then

a) The rows of $\Phi$ belong to the space $K\{\mathfrak{G}^{\bot*}\}$.

b) the rule $\mathbf{a}\to\Phi\cdot\mathbf{a}$, where $\mathbf{a}\in\{\mathfrak{G}\}K$, defines a $K$-linear map $\varphi:\{\mathfrak{G}\}K\rightarrow{}^{H}K$.

Proof.

a) If $B=H\setminus\{h\}$, then $\overline{B}=\{h\}$. Since $B\in\mathfrak{H}$, (2) yields $\mathcal{Z}(\Phi_{h})\in\mathfrak{G}^{\perp*}$.

b) By what we have just proved, $\mathcal{Z}(\Phi_{h})\in\mathfrak{G}^{\perp*}$ for every row $\Phi_{h}$ of $\Phi$. Since $\mathbf{a}\in\{\mathfrak{G}\}K$ and $\mathcal{Z}(\Phi_{h})\in\mathfrak{G}^{\perp*}$, the product $\Phi_{h}\cdot\mathbf{a}$ is defined (see Proposition 5.3) and belongs to $K$. Then $\Phi\cdot\mathbf{a}$ is defined and belongs to ${}^{H}K$. ∎

We need to put one extra condition on $\Phi$ to ensure that the image of $\varphi$ is contained in $\{\mathfrak{H}\}K$.

[TABLE]

The next proposition shows that (2) and (3) together imply that $\varphi$ is continuous.

Proposition 9.4**.**

Let $\Phi$ be an $H\times G$-matrix satisfying (2) and (3). Then the rule $\mathbf{a}\to\Phi\cdot\mathbf{a}$ defines a continuous linear map $\varphi:\{\mathfrak{G}\}K\to\{\mathfrak{H}\}K$.

Proof.

By Remark 9.3, $\Phi\cdot\mathbf{a}\in{}^{H}K$. We prove that $\Phi\cdot\mathbf{a}\in\{\mathfrak{H}\}K$. Indeed, from $\mathbf{a}\in\{\mathfrak{G}\}K$ it follows that $A=\mathcal{Z}(\mathbf{a})\in\mathfrak{G}$. Furthermore, (3) implies that $B=\bigcap_{t\in\overline{A}}\mathcal{Z}(\Phi^{t^{*}})\in\mathfrak{H}$, therefore $B\times\overline{A^{*}}\subseteq\mathcal{Z}(\Phi)$. To show that $\Phi\cdot\mathbf{a}\in\{\mathfrak{H}\}K$, take any $h\in B$. Then

[TABLE]

It follows that $\mathcal{Z}(\Phi\cdot\mathbf{a})\supseteq B$, therefore $\mathcal{Z}(\Phi\cdot\mathbf{a})\in\mathfrak{H}$, as desired.

As we have already noticed (see Lemma 8.1), $\Phi\cdot(\mathbf{a}+\mathbf{b})=\Phi\cdot\mathbf{a}+\Phi\cdot\mathbf{b}$ for all columns $\mathbf{a},\mathbf{b}\in\{\mathfrak{G}\}K$. Furthermore, clearly $(\Phi\cdot\mathbf{a})\cdot k=\Phi\cdot(\mathbf{a}\cdot k)$ for every $k\in K$. It follows easily that $\varphi$ is linear.

To prove that $\varphi$ is continuous, it suffices to check that $\varphi$ is continuous at zero. This means that for every zero neighbourhood $U(B,\mathfrak{H})$, $B\in\mathfrak{H}^{\bot}$ in $\{\mathfrak{H}\}K$, there exists a zero neighbourhood $U(A,\mathfrak{G})$, $A\in\mathfrak{G}^{\bot}$ in $\{\mathfrak{G}\}K$ such that

[TABLE]

From (2) we obtain $C=\bigcap_{s\in\overline{B}}\mathcal{Z}(\Phi_{s})\in\mathfrak{G}^{\bot*}$. If $A=C^{*}$, then $A\in\mathfrak{G}^{\bot}$, and we prove that $A$ is as required. By the definition of $A$, we have $\overline{B}\times A^{*}\subseteq\mathcal{Z}(\Phi)$, that is, $\Phi_{s}^{g^{*}}=0$ for any $s\in\overline{B}$ and $g\in A$. Take any $\mathbf{a}\in U(A,\mathfrak{G})$, hence $a_{g}=0$ for every $g\in\overline{A}$. Then for any $s\in\overline{B}$ we obtain:

[TABLE]

It follows that $\mathcal{Z}(\Phi\cdot\mathbf{a})\supseteq\overline{B}$, therefore $\Phi\cdot\mathbf{a}\in U(B,\mathfrak{H})$. ∎

Below (see Theorem 9.6) we will show the the converse is also true: if $\varphi:\{\mathfrak{G}\}K\to\{\mathfrak{H}\}K$ is a continuous linear map, then its matrix $\Phi$ satisfies both (2) and (3). But first we connect these conditions with filters on direct products.

Lemma 9.5**.**

A matrix $\Phi\in{}^{H}K^{G}$ satisfies (2) and (3) iff $\mathcal{Z}(\Phi)\in\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle$.

Proof.

Let us rewrite a)′ from Definition 6.6 replacing $\mathfrak{G}$ by $\mathfrak{G}^{\bot*}$ and $X$ by $\mathcal{Z}(\Phi)$. Then $\mathfrak{G}^{\bot}$ should be replaced by $\mathfrak{G}^{*}$.

[TABLE]

where $H_{t}=\left\{h\in H\mid(h,t)\in\mathcal{Z}(\Phi)\right\}$ and $G_{s}=\left\{g\in G\mid(s,g)\in\mathcal{Z}(\Phi)\right\}$.

But clearly $H_{t}=\mathcal{Z}(\Phi^{t})$ and $G_{s}=\mathcal{Z}(\Phi_{s})$. 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 $\{\mathfrak{G}\}K$ and $\{\mathfrak{H}\}K$ in terms of filters on $H\times G$.

Theorem 9.6**.**

Suppose that $\mathfrak{H}$ and $\mathfrak{G}$ are balanced filters on $H$ and $G$ and $\Phi$ is an $H\times G$-matrix. Then the following are equivalent.

a) The left multiplication by $\Phi$ defines a continuous linear map $\varphi$ from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K$.

b) The zero set $\mathcal{Z}(\Phi)$ belongs to the filter $\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle$;

c) Each row $\Phi_{h}$ belongs to $K\{\mathfrak{G}^{\perp*}\}$. Furthermore, for every $A\in\mathfrak{G}$ the collection $\Phi^{t^{*}}$, $t\in\overline{A}$ is $\mathfrak{H}$-summable and

[TABLE]

for any set of coefficients $k_{t}\in K$.

d) the left multiplication by $\Phi$ defines a $\mathfrak{G}$-linear map $\varphi$ from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K$.

If $\varphi$ is a linear continuous map from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K$, then its matrix $\Phi$ satisfies these equivalent conditions.

Proof.

b) $\Rightarrow$ a) follows from Proposition 9.4 and Lemma 9.5.

a) $\Rightarrow$ c). Since the product $\Phi\cdot\mathbf{a}$ is defined for every $\mathbf{a}\in\{\mathfrak{G}\}K$, by the definition of product of matrices we obtain $\Phi_{h}\in K\{\mathfrak{G}^{\perp*}\}$.

If $t\in\overline{A}$, then $\mathcal{Z}(\delta^{t})\supseteq A$, hence $\mathcal{Z}(\delta^{t}k_{t})\supseteq A$ for any $k_{t}\in K$, and therefore $\bigcap_{t\in\overline{A}}\mathcal{Z}(\delta^{t}k_{t})\supseteq A$. It follows that $\bigcap_{t\in\overline{A}}\mathcal{Z}(\delta^{t}k_{t})\in\mathfrak{G}$, hence the family $\delta^{t}k_{t}$, $t\in\overline{A}$ is summable and clearly

[TABLE]

Furthermore, by Theorem 7.2 we have $\sum\limits_{t\in\overline{A}}{}^{\mathfrak{G}}\delta^{t}k_{t}=\sum\limits_{t\in\overline{A}}{}^{T(\mathfrak{G})}\delta^{t}k_{t}$. Then

[TABLE]

Since $\varphi$ is continuous, the family $\Phi^{t^{*}}k_{t}$, $t\in\overline{A}$ is summable. By Theorem 7.2 and the previous equality, we obtain

[TABLE]

as desired.

c) $\Rightarrow$ d). Because $\Phi_{h}\in K\{\mathfrak{G}^{\perp*}\}$ for every $h\in H$, the product $\Phi\cdot\mathbf{a}$ is defined for every $\mathbf{a}\in\{\mathfrak{G}\}K$. Furthermore, (6) yields that $\Phi\cdot\mathbf{a}\in\{\mathfrak{H}\}K$. Indeed, if $\mathbf{a}\in\{\mathfrak{G}\}K$, then $\mathcal{Z}(\mathbf{a})\in\mathfrak{G}$. If $A=\mathcal{Z}(\mathbf{a})$, then $\mathbf{a}=\sum\limits_{t\in\overline{A}}{}^{\mathop{{\rm Tych}}\nolimits}\delta^{t}a_{t}$. By the assumption,

[TABLE]

Since the family $\Phi^{t^{*}}$, $t\in\overline{A}$ is $\mathfrak{H}$ summable, then clearly the family $\Phi^{t^{*}}a_{t}$, $t\in\overline{A}$ is $\mathfrak{H}$-summable, therefore $\Phi\cdot\mathbf{a}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}a_{t}\in\mathfrak{H}$ (see a remark after Definition 7.1).

Suppose that $\mathbf{b}_{j}$, $j\in J$, where $\mathbf{b}_{j}\in\{\mathfrak{G}\}K$, is a $\mathfrak{G}$-summable family. Then $A=\bigcap_{j\in J}\mathcal{Z}({\mathbf{b}}_{j})\in\mathfrak{G}$ and $\mathbf{a}=\sum\limits_{j\in J}{}^{\mathfrak{G}}{\mathbf{b}}_{j}\in\{\mathfrak{G}\}K$.

From $\mathcal{Z}(\mathbf{b}_{j})\supseteq A$ it follows that

[TABLE]

where $j\in J$ and $k_{tj}=({\mathbf{b}}_{j})_{t}\in K$. Since the family $\mathbf{b}_{j}$, $j\in J$ is $\mathfrak{G}$-summable, for every $t\in\overline{A}$ there are only finitely many $j\in J$ such that $k_{tj}\neq 0$. Thus we can set $k_{t}=\sum_{j\in J}k_{tj}$ and then ${\mathbf{a}}=\sum\limits_{t\in\overline{A}}{}^{\mathrm{Tych}}\,\delta^{t}k_{t}$.

First we check that the family $\Phi\cdot{\mathbf{b}}_{j}$, $j\in J$ is $\mathfrak{H}$-summable. Indeed, by the assumption, $\Phi\cdot{\mathbf{b}}_{j}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}k_{tj}$, therefore $\bigcap_{j\in J}\mathcal{Z}(\Phi\cdot{\mathbf{b}}_{j})\supseteq\bigcap_{t\in\overline{A}}\mathcal{Z}(\Phi^{t^{*}})$. Because the family $\Phi^{t^{*}}$, $t\in\overline{A}$ is $\mathfrak{H}$-summable, it follows that $\bigcap_{t\in\overline{A}}\mathcal{Z}(\Phi^{t^{*}})\in\mathfrak{H}$, hence $\bigcap_{j\in J}\mathcal{Z}(\Phi\cdot{\mathbf{b}}_{j})\in\mathfrak{H}$.

It remains to prove that for every $h\in H$ there are only finitely many $j\in J$ such that $(\Phi\cdot\mathbf{b}_{j})_{h}\neq 0$. Indeed, the family $\Phi^{t^{*}}$, $t\in\overline{A}$ is $\mathfrak{H}$-summable, hence $\Delta=\{t\in\overline{A}\mid\Phi^{t^{*}}_{h}\neq 0\}$ is a finite set. Furthermore, for any $t\in\overline{A}$ the set $J_{t}=\{j\in J\mid k_{tj}\neq 0\}$ is also finite. Hence the set $J_{0}:=\bigcup_{t\in\Delta}J_{t}$ is finite being a finite union of finite sets. If $j\notin J_{0}$, then $j\notin J_{t}$ for any $t\in\Delta$, and thus $k_{tj}=0$. Therefore,

[TABLE]

Now we prove the equality $\Phi\cdot\sum\limits_{j\in J}{}^{\mathfrak{G}}{\mathbf{b}}_{j}=\sum\limits_{j\in J}{}^{\mathfrak{H}}\Phi\cdot\mathbf{b}_{j}.$ Indeed, by the assumption we have $\Phi\cdot\sum\limits_{j\in J}{}^{\mathfrak{G}}{\mathbf{b}}_{j}=\Phi\cdot\mathbf{a}=\sum\limits_{t\in\overline{A}}{}^{\mathfrak{H}}\Phi^{t^{*}}k_{t}$. On the other hand

[TABLE]

Indeed, the first equality follows from (7) and the assumption; and the last equality follows from $k_{t}=\sum_{j}k_{tj}$. The second equality will be checked coordinate-wise. Take any $h\in H$. Then

[TABLE]

where we omitted the superscript $\mathfrak{H}$ 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 $\sum\limits_{j\in J}\sum\limits_{t\in\overline{A}}\Phi^{t^{*}}_{h}k_{tj}$. We have already proved that $\Phi^{t^{*}}_{h}k_{tj}\neq 0$ iff $t\in\Delta$ and $k_{tj}\neq 0$, where $\Delta$ is a finite set. It follows that the set of pairs $(j,t)$ such that $\Phi^{t^{*}}_{h}k_{tj}\neq 0$ is also finite. Therefore me can change the summation order to get

[TABLE]

d) $\Rightarrow$ b). Let $\mathbf{a}=\sum\limits_{g\in G}{}^{\mathrm{Tych}}\delta^{g}k_{g}\in\{\mathfrak{G}\}K$. Since $\varphi$ is $\mathfrak{G}$-linear, we obtain

[TABLE]

We will prove that $\mathcal{Z}(\Phi)\in\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle$. First we check condition (3): $B^{\prime}=\bigcap_{t\in\overline{A}}\mathcal{Z}(\Phi^{t^{*}})\in\mathfrak{H}$ for every $A\in\mathfrak{G}$.

Because the family $\delta^{t}$, $t\in\overline{A}$ is $\mathfrak{G}$-summable, by the assumption, the family of columns $\Phi^{t^{*}}$, $t\in\overline{A}$ is $\mathfrak{H}$-summable. By Theorem 7.2 and the necessary condition of convergency (see Section 7) we obtain

[TABLE]

that is, the limit of the map $t\to\Phi^{t^{*}}$ from $\overline{A}$ to ${}^{H}K$ with respect to the Frechet filter on $\overline{A}$ is equal to zero. This means that for any zero neighborhood $U(B,\mathfrak{H})$, $B\in\mathfrak{H}^{\bot}$ in $\{\mathfrak{H}\}K$ there exists a cofinite subset $A_{0}^{\prime}$ of $\overline{A}$ such that $\Phi^{t^{*}}\in U(B,\mathfrak{H})$ for any $t\in A_{0}^{\prime}$. In other words, if $t\in A_{0}^{\prime}$, then $\mathcal{Z}(\Phi^{t^{*}})\supseteq\overline{B}$, that is, $\mathcal{Z}(\Phi^{t^{*}})\cup B=H$.

Then

[TABLE]

[TABLE]

[TABLE]

Note that the last set is cofinite in $H$. Indeed, $\Phi^{t^{*}}\in\{\mathfrak{H}\}K$ and $B\in\mathfrak{H}^{\perp}$ implies that each $\mathcal{Z}(\Phi^{t^{*}})\cup B$ is cofinite, and $\overline{A\cup A^{\prime}_{0}}=\overline{A}\setminus A_{0}^{\prime}$ is a finite set.

Thus we proved that the union $B^{\prime}\cup B$ is cofinite for any $B\in\mathfrak{H}^{\bot}$. This means that $B^{\prime}\in\mathfrak{H}^{\bot\bot}=\mathfrak{H}$, as required.

Now we have to check (2):

[TABLE]

This is the same as the union $A^{*}\cup A^{\prime}$ is cofinite for any $A\in\mathfrak{G}$. We have the following sequence of equivalences:

[TABLE]

[TABLE]

Hence:

[TABLE]

We have already proved that there is a cofinite subset $A_{0}^{\prime}$ of $\overline{A}$ such that $\mathcal{Z}(\Phi^{t^{*}})\cup B=H$ for all $t\in A_{0}^{\prime}$. Then $A^{\prime*}_{0}\subseteq A^{\prime}$. Since $A\cup A^{\prime}_{0}$ is cofinite in $H$, the same is true for $A^{*}\cup A^{\prime*}_{0}$. From $A^{\prime*}_{0}\subseteq A^{\prime}$ it follows that $A^{*}\cup A^{\prime}$ is also cofinite in $H$, as desired. ∎

If $X$ is a $K$-linear topological space, then a map $\varphi:X\to K$ is said to be a linear form, if $\varphi$ is linear and continuous. This space can be endowed with a weak topology: a net of linear forms $\varphi_{i}:X\to K$, $i\in I$ converges to $\varphi$, if $\varphi_{i}(x)$ converges to $\varphi(x)$ for every $x\in X$.

Corollary 9.7**.**

Suppose that $\mathfrak{G}$ is a balanced filter on $G$. Then the space of continuous linear forms of the topological linear space of columns $(\{\mathfrak{G}\}K,T(\mathfrak{G}))$ is isomorphic to the topological linear space of rows $(K\{\mathfrak{G}^{\bot*}\},T(\mathfrak{G}^{\bot*}))$

Proof.

Applying Theorem 9.6 to the case when $H$ consists of one element, we obtain that every continuous linear map from $\{\mathfrak{G}\}K$ to $K$ is given by a left multiplication by a row from $K\{\mathfrak{G}^{\bot*}\}$. ∎

For further applications we need a dual variant of Theorem 9.6.

Remark 9.8**.**

Suppose that $\Psi$ is an $H\times G$ matrix. Then the following are equivalent:

a) the right multiplication by $\Psi$ defines a linear continuous map from $K\{\mathfrak{G}\}$ to $K\{\mathfrak{H}\}$;

b) $\mathcal{Z}(\Psi)\in\langle\mathfrak{G}^{\perp*},\mathfrak{H}\rangle$.

If $\psi$ is a continuous linear map from $K\{\mathfrak{G}\}$ to $K\{\mathfrak{H}\}$, then its matrix $\Psi$ satisfies these equivalent conditions.

10. The ring of continuous operators

Suppose that $\Phi$ is an $H\times G$ matrix, $\gamma$ is a row and $\mathbf{a}$ is a column such that the products $\gamma\cdot\Phi$ and $\Phi\cdot\mathbf{a}$ are defined.

Remark 10.1**.**

[TABLE]

Proof.

Indeed, if $h\in\bigcap_{t\in G\setminus\mathcal{Z}(\mathbf{a})}\mathcal{Z}(\Phi^{t^{*}})$, then

[TABLE]

The second inclusion has a similar proof. ∎

Now we prove an auxiliary lemma.

Lemma 10.2**.**

For any $\beta\in K\{\mathfrak{G}^{\bot*}\}$ and $\mathbf{b}\in\{\mathfrak{H}\}K$ the matrix $\Phi=\mathbf{b}\cdot\beta\in{}^{H}K^{G}$ satisfies (2) and (3). Therefore (by Proposition 9.4) the left multiplication by $\Phi$ defines a continuous linear map from $\{\mathfrak{G}\}K$ to $\{\mathfrak{H}\}K$.

Proof.

First we check (2). Take any $B\in\mathfrak{H}^{\bot}$. If $s\in H$, then $(\mathbf{b}\cdot\beta)_{s}=b_{s}\beta$, where $b_{s}\in K$ is the $s$th coordinate of $\mathbf{b}$. If $b_{s}=0$, then $\mathcal{Z}((\mathbf{b}\cdot\beta)_{s})=G$, and otherwise $\mathcal{Z}((\mathbf{b}\cdot\beta)_{s})=\mathcal{Z}(\beta)$. Therefore,

[TABLE]

which implies $\bigcap_{s\in\overline{H}}\mathcal{Z}((\mathbf{b}\cdot\beta)_{s})\in\mathfrak{G}^{\bot*}$, 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 $\Upsilon$, $\Phi$ and $\Psi$ are matrices of the following size: $\Upsilon\in{}^{J}K^{H}$, $\Phi\in{}^{H}K^{G}$, $\Psi\in{}^{G}K^{I}$. Further assume that

a) for any pair $(j,g)$ there exist only finitely many $h\in H$ such that $\Upsilon_{j}^{h^{*}}\Phi_{h}^{g}\neq 0$;

b) for any pair $(h,i)$ there exist only finitely many $g\in G$ such that $\Phi_{h}^{g^{*}}\Psi_{g}^{i}\neq 0$;

c) for any pair $(j,i)\in J\times I$ the exist only finitely many $(h,g)\in H\times G$ such that $\Upsilon_{j}^{h^{*}}\Phi_{h}^{g^{*}}\Psi_{g}^{i}\neq 0$;

Then the products $\Upsilon\cdot(\Phi\cdot\Psi)$ and $(\Upsilon\cdot\Phi)\cdot\Psi$ are defined and equal.

Proof.

Clearly a) and b) is nothing more that the existence of the products $\Upsilon\cdot\Phi$ and $\Phi\cdot\Psi$. From c) we obtain the following equality:

[TABLE]

which implies associativity. ∎

This proposition applies in the following situation.

Proposition 10.4**.**

Suppose that $\mathbf{b}_{1}$, $\mathbf{b}_{2}$, $\ldots$ are columns from $\{\mathfrak{G}\}K$, and $\beta_{1},\beta_{2},\ldots$ are rows from $K\{\mathfrak{G}^{\bot*}\}$. Then any (finite) alternating product $\mathbf{b}_{1}\cdot\beta_{1}\cdot\mathbf{b}_{2}\cdot\beta_{2}\cdot\ldots$ or $\beta_{1}\cdot\mathbf{b}_{1}\cdot\beta_{2}\cdot\mathbf{b}_{2}\cdot\dots$ does not depend on the way we put brackets on it.

Proof.

We prove only associativity for short products:

[TABLE]

for any columns $\mathbf{b},\mathbf{c}$ and any rows $\beta,\gamma$. 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 $j\in G$ there are only finitely many $g^{*}\in G$ such that $b_{j}\beta^{g^{*}}c_{g}\neq 0$. But this follows from the fact that $g$ belong to $(\operatorname{supp}\beta)^{*}\cap\operatorname{supp}\mathbf{c}$, which is a finite set.

The proof of the second equality is similar. ∎

Now we define a ring $\mathcal{M}_{0}(\mathfrak{G})$ of ‘finitary’ $G\times G$-matrices. Later we will show that this ring can be considered as a dense subring in the algebra of all continuous linear operators on $\{\mathfrak{G}\}K$.

Let $\mathcal{M}_{0}(\mathfrak{G})$ denote the set of all $G\times G$-matrices

[TABLE]

where $\mathbf{a}_{i}\in\{\mathfrak{G}\}K$ and $\gamma_{i}\in K\{\mathfrak{G}^{\bot*}\}$, $i=1,\dots,n$.

Proposition 10.5**.**

The set $(\mathcal{M}_{0}(\mathfrak{G}),+,\cdot)$ is an associative ring.

Proof.

Clearly this set is closed with respect to addition.

We show that the product of elements of $(\mathcal{M}_{0}(\mathfrak{G}),+,\cdot)$ is an element of $(\mathcal{M}_{0}(\mathfrak{G}),+,\cdot)$. Clearly it suffices to check this for elements of the form $\mathbf{a}\cdot\gamma$ and $\mathbf{b}\cdot\beta$, where $\mathbf{a},\mathbf{b}$ are columns and $\gamma,\beta$ are rows. Indeed, $\gamma\cdot\mathbf{b}=k\in K$. 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 $\mathfrak{G}$ and $\mathfrak{H}$ are balanced filters on $G$ and $H$. Further assume that $\gamma\in K\{\mathfrak{H}^{\bot*}\}$ is a row, $\mathbf{a}\in\{\mathfrak{G}\}K$ is a column, and $\Phi\in{}^{H}K^{G}$ is an $H\times G$-matrix such that $\mathcal{Z}(\Phi)\in\langle\mathfrak{H},\mathfrak{G}^{\perp*}\rangle$. Then $(\gamma\cdot\Phi)\cdot\mathbf{a}=\gamma\cdot(\Phi\cdot\mathbf{a})$.

Proof.

From Theorem 3.3 3) it follows that $\mathbf{a}=\lim\limits_{i}{}^{T(\mathfrak{G})}\,\mathbf{a}_{i}$, where $\mathbf{a}_{i}$ are columns with finite support. If $i$ is fixed, then $(\mathbf{a}_{i})_{g}\neq 0$ for finitely many $g\in G$. For any such $g$, there exists only finitely many $h$ such that $\gamma^{h^{*}}\Phi^{g^{*}}_{h}\neq 0$. Thus there are only finitely many pairs $(g,h)$ such that $\gamma^{h^{*}}\Phi^{g^{*}}_{h}(\mathbf{a}_{i})_{g}\neq 0$. 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 $\Phi$ acts as a linear continuous operator from $K\{\mathfrak{H}^{\perp*}\}$ to $K\{\mathfrak{G}^{\perp*}\}$. In particular, $\gamma\cdot\Phi\in K\{\mathfrak{G}^{\perp*}\}$. Then, by Corollary 9.7, the left multiplication by $\gamma\cdot\Phi$ is a linear continuous map from $\{\mathfrak{G}\}K$ to $K$. Thus

[TABLE]

where $\mathop{{\rm Disc}}\nolimits$ denotes the discrete topology on $K$. By Proposition 10.5 we obtain $(\gamma\cdot\Phi)\cdot\mathbf{a}_{i}=\gamma\cdot(\Phi\cdot\mathbf{a}_{i})$, therefore

[TABLE]

Since left multiplications by $\gamma$ and $\Phi$ are continuous operators (see Theorem 9.6), we conclude that

[TABLE]

Because $\mathbf{a}=\lim_{i}{}^{T(\mathfrak{G})}\mathbf{a}_{i}$, this is equal to $\gamma\cdot(\Phi\cdot\mathbf{a})$, as desired. ∎

Now we give a matrix description of the algebra of linear continuous operators of the column space $\{\mathfrak{G}\}K$.

Theorem 10.7**.**

Let $\mathfrak{G}$ be a balanced filter on a set $G$ with involution, and let $\mathcal{M}(\mathfrak{G})$ be a set of all $G\times G$-matrixes $\Phi$ such that $\mathcal{Z}(\Phi)\in\langle\mathfrak{G},\mathfrak{G}^{\bot*}\rangle$. Then

a) $\mathcal{M}(\mathfrak{G})$ is a ring with respect the above defined addition and multiplication;

b) If $\Phi\in\mathcal{M}(\mathfrak{G})$, then the map $\widehat{\Phi}:\mathbf{a}\to\Phi\cdot\mathbf{a}$ is a continuous endomorphism of the column space $\{\mathfrak{G}\}K$.

c) The map $\Phi\to\hat{\Phi}$ is an isomorphism from $\mathcal{M}(\mathfrak{G})$ onto the ring of all continuous endomorphisms of the space $\{\mathfrak{G}\}K$.

Proof.

b) follows from the equivalence of a) and b) in Theorem 9.6.

a), c) First we prove that $\mathcal{M}(\mathfrak{G})$ is closed with respect to multiplication. Suppose that $\Phi,\Psi\in\mathcal{M}(\mathfrak{G})$. Since $\mathcal{Z}(\Phi),\mathcal{Z}(\Psi)\in\langle\mathfrak{G},\mathfrak{G}^{\bot*}\rangle$, by Remark 9.3 a) and its dual variant, we obtain that $\mathcal{Z}(\Phi_{g})\in\mathfrak{G}^{\perp*}$ and $\mathcal{Z}(\Psi^{h^{*}})\in\mathfrak{G}$ for all $g,h\in G$. It follows that the product $\Phi\cdot\Psi$ is defined.

Take any $g\in G$. By Proposition 10.6 we obtain $(\Phi_{g}\cdot\Psi)\cdot\mathbf{a}=\Phi_{g}\cdot(\Psi\cdot\mathbf{a})$. Then by the definition of matrix multiplication we derive

[TABLE]

and $\Phi_{g}\cdot(\Psi\cdot\mathbf{a})=(\Phi\cdot(\Psi\cdot\mathbf{a}))_{g}$, therefore $(\Phi\cdot\Psi)\cdot\mathbf{a}=\Phi\cdot(\Psi\cdot\mathbf{a})$. Because the left multiplication by $\Phi$ and $\Psi$ is continuous, this equality implies that the left multiplication by $\Phi\cdot\Psi$ is continuous. By Theorem 9.6 we conclude that $\Phi\cdot\Psi\in\mathcal{M}(\mathfrak{G})$.

Furthermore, writing $(\Phi\cdot\Psi)\cdot\mathbf{a}=\Phi\cdot(\Psi\cdot\mathbf{a})$ at the level of endomorphisms, we obtain $\widehat{\Phi\cdot\Psi}=\widehat{\Phi}\circ\widehat{\Psi}$, therefore the map $\Phi\to\widehat{\Phi}$ preserves multiplication. Clearly this map preserves an addition, hence it is a morphism of rings (associative or not).

By Theorem 9.6, if $\varphi$ is a continuous linear operator of the space $\{\mathfrak{G}\}K$, then there is a unique matrix $\Phi\in\mathcal{M}(\mathfrak{G})$ such that $\varphi[\mathbf{a}]=\Phi\cdot\mathbf{a}$. It follows that the map $\Phi\to\widehat{\Phi}$ is an isomorphism of rings, in particular $\mathcal{M}(\mathfrak{G})$ is an associative ring. ∎

Suppose that $\mathfrak{G}$, $\mathfrak{H}$ are balanced filters on sets with involution $G$ and $H$. Let $\mathcal{L}(\mathfrak{H},\mathfrak{G})$ be the space of all linear continuous maps $\{\mathfrak{G}\}K\to\{\mathfrak{H}\}K$. By Theorem 9.6 we may identify $\mathcal{L}(\mathfrak{H},\mathfrak{G})$ with the space of $H\times G$-matrices $\Phi$ such that $\mathcal{Z}(\Phi)\in\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle$. Therefore we may consider $\mathcal{L}(\mathfrak{H},\mathfrak{G})$ as a topological space with the topology $T(\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle)$ defined by the filter $\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle$ on $H\times G$.

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 $H\times G$-matrices $\Phi_{i}\in\mathcal{L}(\mathfrak{H},\mathfrak{G})$ converges to $\Phi$ with respect to $T(\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle)$. Then for any column $\mathbf{a}\in\{\mathfrak{G}\}K$ and for any row $\gamma\in K\{\mathfrak{H}^{\bot*}\}$

[TABLE]

Proof.

Clearly we may assume that $\Phi=0$, hence $\Phi\cdot\mathbf{a}=\mathbf{0}$. Thus for the first equation we have to prove that $\lim\limits_{i}{}^{T(\mathfrak{G})}\Phi_{i}\cdot\mathbf{a}=0$.

Recall that by Theorem 6.13 we have $\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle^{\bot}=\mathfrak{H}^{\bot}\otimes\mathfrak{G}^{*}$. Furthermore (see Section 3) the basis of the filter $\mathfrak{H}^{\perp}\otimes\mathfrak{G}^{*}$ is given by the sets $\overline{\overline{B}_{1}\times\overline{A}_{1}}$, where $B_{1}\in\mathfrak{H}$ and $A_{1}\in\mathfrak{G}^{*}$.

Let $U(B,\mathfrak{H})$, $B\in\mathfrak{H}^{\bot}$ be a zero neighborhood of the space $\{\mathfrak{H}\}K$. We have to find a zero neighborhood $U=U(\overline{\overline{B}_{1}\times\overline{A}_{1}},T(\langle\mathfrak{H},\mathfrak{G}^{\bot*}\rangle))$, of the space $\mathcal{L}(\mathfrak{H},\mathfrak{G})$ such that $\Phi_{i}\in U$ implies $\Phi_{i}\cdot\mathbf{a}\in U(B,\mathfrak{H})$.

Take $B_{1}=B$ and $A_{1}=\mathcal{Z}(\mathbf{a})^{*}$. If $\Phi_{i}\in U$ then, by the definition $U$, we have

[TABLE]

If $h$ is an arbitrary element of $\overline{B}$, then

[TABLE]

It follows that $\mathcal{Z}(\Phi_{i}\cdot\mathbf{a})\supseteq\overline{B}$, that is, $\Phi_{i}\cdot\mathbf{a}\in U(B,\mathfrak{H})$, as desired.

The second equality can be verified similarly. ∎

By Lemma 10.2, $\mathcal{M}_{0}(\mathfrak{G})$ is a subring of $\mathcal{M}(\mathfrak{G})$. The following remark shows that this subring is dense.

Remark 10.9**.**

$\mathcal{M}_{0}(\mathfrak{G})$* is a dense subring of $\mathcal{M}(\mathfrak{G})$ with respect to topology $T(\langle\mathfrak{G},\mathfrak{G}^{\perp*}\rangle)$.*

Proof.

The set $K[H\times G]$ of all matrices with finite support forms a subalgebra of the algebra $\mathcal{M}_{0}(\mathfrak{G})$. By Theorem 3.3, this algebra is dense in $\mathcal{M}(\mathfrak{G})$. It remains to notice that $K[H\times G]\subseteq\mathcal{M}_{0}(\mathfrak{G})$, because every finite matrix is a linear combination of matrices $\delta^{g}\cdot\delta_{h}\in\mathcal{M}_{0}(\mathfrak{G})$. ∎