Description of unitary representations of the group of infinite $p$-adic integer matrices

TL;DR

This paper classifies all irreducible unitary representations of the group of infinite p-adic integer matrices, revealing their structure via finite-dimensional representations of related groups over residue rings.

Contribution

It provides a complete classification of irreducible unitary representations for infinite p-adic integer matrix groups, a new result in representation theory of infinite p-adic groups.

Findings

Irreducible representations factor through groups over residue rings.

Representations are induced from finite-dimensional representations of open subgroups.

The classification applies specifically for p ≠ 2.

Abstract

We classify irreducible unitary representations of the group of all infinite matrices over a $p$-adic field ($p\ne 2$) with integer elements equipped with a natural topology. Any irreducible representation passes through a group $GL$ of infinite matrices over a residue ring modulo $p^k$. Irreducible representations of the latter group are induced from finite-dimensional representations of certain open subgroups.

Equations

$$p>2.$$

$${\mathbb{O}}_p=\underset{⟵n}{\lim }{\mathbb{Z}}_{p^n}$$

$$\cdots ⟵{\mathbb{Z}}_{p^{n-1}}⟵{\mathbb{Z}}_{p^n}⟵{\mathbb{Z}}_{p^{n+1}}⟵\dots$$

$$(v,w)\mapsto (vg,w{g}^{t-1}).$$

$$\mathfrak{l}({\mathbb{O}}_p)=\underset{⟵n}{\lim }\mathfrak{l}({\mathbb{Z}}_{p^n}),$$

$$\mathrm{GL}(\infty ,{\mathbb{O}}_p)=\underset{⟵n}{\lim }\mathrm{GL}(\infty ,{\mathbb{Z}}_{p^n})$$

$$\sum _{x\in Q\setminus G}\Vert f(x){\Vert }^2<\infty .$$

$$⟨f_1,f_2{⟩}_H:=\sum _{x\in Q\setminus G}⟨f_1(x),f_2(x){⟩}_V.$$

$$\rho (g)f(x)=U(g,x)f(xg)$$

$$U(x,g_1{g}_2)=U(x,g_1)U(x{g}_1,g_2).$$

$$U(g,x)\sim A(gx{)}^{-1}U(g,x)A(x),$$

$$\mathbb{G}:=\mathrm{GL}(\infty ,{\mathbb{Z}}_{p^{\mu }}).$$

$$\mathfrak{l}({\mathbb{Z}}_{p^{\mu }})\times \mathfrak{l}({\mathbb{Z}}_{p^{\mu }})\to {\mathbb{Z}}_{p^{\mu }}$$

$$\{v,w\}:=\sum v_j{w}_j=v{w}^t,$$

$$\{vg,v{g}^{t-1}\}=\{v,w\}.$$

$${\mathrm{Ind}}_{\hat{\mathbb{G}}[L;M]}^{\mathbb{G}}(\hat{\tau })$$

$${\mathrm{Ind}}_{\hat{\mathbb{G}}[L_1;M_1]}^{\mathbb{G}}({\tau }_1)\sim {\mathrm{Ind}}_{\hat{\mathbb{G}}[L_2;M_2]}^{\mathbb{G}}({\tau }_2)$$

$$g=\left(\begin{array}{cc}a& bv\\ wc& z\end{array}\right),$$

$$a=1-bS,a=1-Tc.$$

$$g=\left(\begin{array}{cc}1-buc& bv\\ wc& z\end{array}\right).$$

$$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\sim \left(\begin{array}{cc}{1}_{\alpha }& 0\\ 0& u\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}{1}_{\beta }& 0\\ 0& v\end{array}\right)=\left(\begin{array}{cc}a& bv\\ uc& udv\end{array}\right),$$

$$\left[\begin{array}[]{c|c}a&b\\ \hline\cr c&d\end{array}\right]_{\alpha\beta},$$

$$\mathbb{G}(\alpha )\setminus \mathbb{G}/\mathbb{G}(\beta )\times \mathbb{G}(\beta )\setminus \mathbb{G}/\mathbb{G}(\gamma )\to \mathbb{G}(\alpha )\setminus \mathbb{G}/\mathbb{G}(\gamma ).$$

$$g_{1}=\left[\begin{array}[]{c|cc}a&b&\\ \hline\cr c&d&\\ &&1_{\infty}\end{array}\right]_{\alpha\beta},\,\,g_{1}=\left[\begin{array}[]{c|cc}p&q&\\ \hline\cr r&t&\\ &&1_{\infty}\end{array}\right]_{\beta\gamma}.$$

$$\theta^{\beta}(j):=\left(\begin{array}[]{c|ccc}1_{\beta}&&&\\ \hline\cr&0&1_{j}&\\ &1_{j}&0&\\ &&&1_{\infty}\end{array}\right)\in\mathbb{G}(\beta).$$

$$\mathbb{G}(\alpha )\cdot g_1{\theta }^{\beta }(j)g_2\cdot \mathbb{G}(\gamma )\in \mathbb{G}(\alpha )\setminus \mathbb{G}/\mathbb{G}(\gamma ).$$

$${\mathfrak{g}}_1\circ {\mathfrak{g}}_2=$$

$$\displaystyle=\left[\left(\begin{array}[]{c|ccc}a&b&&\\ \hline\cr c&d&&\\ &&1_{L}&\\ &&&1_{\infty}\end{array}\right)\left(\begin{array}[]{c|ccc}1_{\beta}&&&\\ \hline\cr&0&1_{L}&\\ &1_{L}&0&\\ &&&1_{\infty}\end{array}\right)\left(\begin{array}[]{c|ccc}p&q&&\\ \hline\cr r&t&&\\ &&1_{L}&\\ &&&1_{\infty}\end{array}\right)\right]_{\alpha\gamma},$$

$$\mathfrak{g}_{1}\circ\mathfrak{g}_{2}=\left[\begin{array}[]{c|ccc}ap&aq&b&\\ \hline\cr cp&cq&d&\\ r&t&0&\\ &&&1_{\infty}\end{array}\right]_{\alpha\gamma}\sim\,\,\left[\begin{array}[]{c|ccc}ap&b&aq&\\ \hline\cr cp&d&cq&\\ r&0&t&\\ &&&1_{\infty}\end{array}\right]_{\alpha\gamma}.$$

$${\mathfrak{g}}_1\in \mathbb{G}(\alpha )\setminus \mathbb{G}/\mathbb{G}(\beta ),{\mathfrak{g}}_2\in \mathbb{G}(\beta )\setminus \mathbb{G}/\mathbb{G}(\gamma ),{\mathfrak{g}}_3\in \mathbb{G}(\gamma )\setminus \mathbb{G}/\mathbb{G}(\delta ),$$

$$({\mathfrak{g}}_1\circ {\mathfrak{g}}_2)\circ {\mathfrak{g}}_3={\mathfrak{g}}_1\circ ({\mathfrak{g}}_2\circ {\mathfrak{g}}_3).$$

Full text

Description of unitary representations

of the group of infinite $p$-adic integer matrices

Yury A. Neretin111Supported by the grants FWF, P28421, P31591.

We classify irreducible unitary representations of the group of all infinite matrices over a $p$-adic field ($p\neq 2$) with integer elements equipped with a natural topology. Any irreducible representation passes through a group $\operatorname{GL}$ of infinite matrices over a residue ring modulo $p^{k}$. Irreducible representations of the latter group are induced from finite-dimensional representations of certain open subgroups.

1 Introduction

1.1. Notations and definitions. a) Rings. Let $p$ be a prime,

[TABLE]

Let ${\mathbb{Z}}_{p^{n}}:={\mathbb{Z}}/p^{n}{\mathbb{Z}}$ be a residue ring, $\mathbb{F}_{p}:={\mathbb{Z}}_{p}$ be the field with $p$ elements. The ring of $p$-adic integers ${\mathbb{O}}_{p}$ is the projective limit

[TABLE]

of the following chain (see, e.g., [32]:

[TABLE]

we have ${\mathbb{Z}}_{p^{n}}={\mathbb{O}}_{p}/p^{n}{\mathbb{O}}_{p}$. Denote by ${\mathbb{Q}}_{p}$ the field of $p$-adic numbers.

b) The infinite symmetric group and oligomorphic groups. Let $\Omega$ be a countable set. Denote by $S(\Omega)$ the group of all permutations of $\Omega$, denote $S_{\infty}:=S({\mathbb{N}})$. The topology on the infinite symmetric group $S(\Omega)$ is determined by the condition: stabilizers of finite subsets are open subgroups and these subgroups form a fundamental system of neighborhoods of the unit222Thus we get a structure of a Polish group, moreover this topology is a unique separable topology on the infinite symmetric group, see [13]. In particular, this means that a unitary representation of $S_{\infty}$ in a separable Hilbert space is automatically continuous.. Equivalently, a sequence $g^{(\alpha)}$ converges to $g$ if for each $\omega\in\Omega$ we have $\omega g^{(\alpha)}=\omega g$ for sufficiently large $\alpha$.

A closed subgroup $G$ of $S(\Omega)$ is called oligomorphic if for each $k$ it has only a finite number of orbits on the product $\Omega\times\dots\times\Omega$ of $k$ copies of $\Omega$, see [5].

c) Modules $\mathfrak{l}({\mathbb{Z}}_{p^{n}})$ and groups $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$. Define the module $\mathfrak{l}({\mathbb{Z}}_{p^{n}})$ as the set of all sequences $v=(v_{1},v_{2},\dots)$, where $v_{j}\in{\mathbb{Z}}_{p^{n}}$ and $v_{j}=0$ for sufficiently large $j$. The set $\mathfrak{l}({\mathbb{Z}}_{p^{n}})$ is countable, we equip it with a discrete topology. Denote by $e_{j}$ the standard basis elements, i.e., $e_{j}$ has a unit on $j$-th place, other elements are [math].

Define groups $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ as groups of infinite invertible matrices $g$ over ${\mathbb{Z}}_{p^{n}}$ such that:

$\bullet$ each row of $g$ contains only a finite number of nonzero elements;

$\bullet$ each column contains only a finite number of nonzero elements;

$\bullet$ the inverse matrix $g^{-1}$ satisfies the same conditions.

Notice that rows of a matrix $g$ are precisely vectors $e_{i}g$, and columns are $e_{j}g^{t}$ (we denote by $g^{t}$ a transposed matrix).

Actually, the topic of this paper are representations of $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$.

This group is continual and we must define a topology on $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$. A sequence $g^{(\alpha)}\in\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ converges to $g$ if all sequences $e_{i}g^{(\alpha)}$ and $e_{i}(g^{(\alpha)})^{t}$ are eventually constant and their limits are $e_{i}g$ and $e_{j}g^{t}$ respectively. Thus we get a structure of a totally disconnected topological group.

The group $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ acts on the countable set $\mathfrak{l}({\mathbb{Z}}_{p^{n}})\oplus\mathfrak{l}({\mathbb{Z}}_{p^{n}})$ by transformations

[TABLE]

In particular, this define an embedding of $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ to a symmetric group S\bigl{(}\mathfrak{l}({\mathbb{Z}}_{p^{n}})\oplus\mathfrak{l}({\mathbb{Z}}_{p^{n}})\bigr{)}. The image of the group $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ is a closed subgroup of S\bigl{(}\mathfrak{l}({\mathbb{Z}}_{p^{n}})\oplus\mathfrak{l}({\mathbb{Z}}_{p^{n}})\bigr{)} and the induced topology coincides with the natural topology on $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$. By [27], Lemma 3.7, the group $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ is oligomorphic.

d) Modules $\mathfrak{l}({\mathbb{O}}_{p})$ and groups $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$. Denote by $\mathfrak{l}({\mathbb{O}}_{p})$ the set of all sequences $r=(r_{1},r_{2},\dots)$, where $r_{j}\in{\mathbb{O}}_{p}$ and $|r_{j}|\to 0$ as $j\to\infty$. The space $\mathfrak{l}({\mathbb{O}}_{p})$ is a projective limit,

[TABLE]

we equip it with the topology of the projective limit. In other words, a sequence $r^{(j)}\in\mathfrak{l}({\mathbb{O}}_{p})$ converges if for any $p^{n}$ the reduction of $r^{(j)}$ modulo $p^{n}$ is eventually constant in ${\mathbb{Z}}_{p^{n}}$.

We define $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ as the group of all infinite matrices $g$ over ${\mathbb{O}}_{p}$ such that:

$\bullet$ each row of $g$ is an element of $\mathfrak{l}({\mathbb{O}}_{p})$;

$\bullet$ each column of $g$ is an element of $\mathfrak{l}({\mathbb{O}}_{p})$;

$\bullet$ the matrix $g$ has an inverse and $g^{-1}$ satisfies the same conditions.

We say that a sequence $g^{(\alpha)}\in\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ converges to $g$ if for any $i$ the sequence $e_{i}g^{(\alpha)}$ converges to $e_{i}g$ and for any $j$ the sequence $e_{i}(g^{(\alpha)})^{t}$ converges to $e_{j}g^{t}$. This determines a structure of a totally disconnected topological group on $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$.

We have obvious homomorphisms $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})\to\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n-1}})$, the group $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ is the projective limit

[TABLE]

and its topology is the topology of projective limit.

1.2. Preliminary remarks. A priori we know the following statement:

Theorem 1.1

a)* The group $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ is a type $I$ group, it has a countable number of irreducible unitary representations. Any unitary representation $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ is a sum of irreducible representations. Any irreducible unitary representation of $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ is in fact a representation of some group $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$.*

b)* Each irreducible representation of $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ is induced from a finite-dimensional representation of an open subgroup. More precisely, for any irreducible unitary representation of $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$ there exists an open subgroup $\widehat{Q}\subset\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{n}})$, a normal subgroup $Q\subset\widehat{Q}$ of finite index and an irreducible representation $\nu$ of $\widehat{Q}$, which is trivial on $Q$, such that $\rho$ is induced from $\nu$.*

This is a special case of a theorem of Tsankov about unitary representations of oligomorphic groups and projective limits of holomorphic groups, see [34], Theorem 1.3333A reduction of representations of $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ to representations of quotients $\operatorname{GL}(\infty,\mathbb{Z}_{p^{\mu}})$ easily follows from [20], Proposition VII.1.3, see [27], Corollary 3.5. In our proof of Theorem 1.5 Tsankov’s theorem is used in the proof of Proposition 2.1, which was done in [27].. It seems that [34], [2] is not sufficient to give a precise answer in our case.

Let us give a definition of an induced representation (see, e.g., [33], Sect. 7, Kirillov, Sect. 13), which is appropriate in our case. Let $G$ be a totally disconnected separable group, $Q$ its open subgroup. Let $\nu$ be a unitary representation of $Q$ in a Hilbert space $V$. Consider the space $H$ of $V$-valued functions $f$ on a countable homogeneous space $Q\setminus G$ such that

[TABLE]

Equip $H$ with the inner product

[TABLE]

Let $U$ be a function on $G\times(Q\setminus G)$ taking values in the group of unitary operators in $V$ such that:

$\bullet$ Formula

[TABLE]

determines a representation of $G$ in $H$.

$\bullet$ Let $x_{0}$ be the initial point of $Q\setminus G$, i.e., $x_{0}Q=x_{0}$. Then for $q\in Q$ we have $U(q,x_{0})=\nu(q)$.

The first condition implies that the function $U(g,x)$ satisfies the functional equation

[TABLE]

It can be shown that $U(g,x)$ is uniquely defined up to a natural calibration

[TABLE]

where $A$ is a function on $Q\setminus G$ taking values in the unitary group of $V$ (see, e.g., [15], Sect 13.1). For this reason, an induced representation $\rho(g)=\operatorname{Ind}^{G}_{Q}(\nu)$ is canonically defined up to a unitary equivalence.

We also can choose $U(g,x)$ in the following way. For any $x\in Q\setminus G$ we choose an element $s(x)\in G$ such that $x_{0}s(x)=x$. Then $U(g,x)=\nu(q)$, where $q$ is determined from the condition $s(x)g=q\,s(xg)$.

1.3. The statement. Thus we fix a ring ${\mathbb{Z}}_{p^{\mu}}$ and examine the group

[TABLE]

We consider two right actions of $\mathbb{G}$ on $\mathfrak{l}(\mathbb{Z}_{p^{\mu}}),g:v\mapsto vg$, $g:v\mapsto vg^{t-1}$. Define a pairing

[TABLE]

by

[TABLE]

our action preserves this pairing, i.e.,

[TABLE]

Let $L\subset\mathfrak{l}(\mathbb{Z}_{p^{\mu}})$, $M\subset\mathfrak{l}(\mathbb{Z}_{p^{\mu}})$ be finitely generated $\mathbb{Z}_{p^{\mu}}$-submodules. Denote by $\widehat{\mathbb{G}}[L;M]$ the subgroup of $\mathbb{G}$ consisting of $g$ sending $L$ to $L$ and $M$ to $M$. By $\mathbb{G}^{\circ}[L;M]\subset\widehat{\mathbb{G}}[L;M]$ we denote group of matrices fixing $L$ and $M$ pointwise. Obviously, the quotient group $\widehat{\mathbb{G}}[L;M]/\mathbb{G}^{\circ}[L;M]$ is finite, it acts on the direct sum $L\oplus M$ preserving the pairing $\{f,g\}$. Any irreducible representation $\tau$ of $\widehat{\mathbb{G}}[L;M]/\mathbb{G}^{\circ}[L;M]$ can be regarded as a representation $\widehat{\tau}$ the group $\widehat{\mathbb{G}}[L;M]$, which is trivial on $\mathbb{G}^{\circ}[L;M]$. For given $L$, $M$, $\tau$ we consider the representation

[TABLE]

of $\mathbb{G}$ induced from the representation $\widehat{\tau}$ of the group $\widehat{\mathbb{G}}[L;M]$. Olshanski [30] obtained the following statement444A proof in [30] is only sketched, other proofs were given by Dudko [8] and Tsankov [34]. for the group $\operatorname{GL}(\infty,\mathbb{F}_{p})=\operatorname{GL}(\infty,{\mathbb{Z}}_{p})$.

Theorem 1.2

a)* Any irreducible unitary representation of the group $\operatorname{GL}(\infty,\mathbb{F}_{p})$ has this form.*

b)* Two irreducible representations can be equivalent only for a trivial reason, i.e.,*

[TABLE]

if and only if there exists $h\in\mathbb{G}$ such that $L_{1}h=L_{2}h$, $M_{1}h^{t-1}=M_{2}h^{t-1}$ and $\tau_{2}(q)=\tau_{1}(hqh^{-1})$.

For groups $\operatorname{GL}(\infty,\mathbb{Z}_{p^{\mu}})$ with $\mu>1$ the situation is more delicate. Let $L$, $M$ actually be contained in $(\mathbb{Z}_{p^{\mu}})^{m}\subset\mathfrak{l}(\mathbb{Z}_{p^{\mu}})$. Fix a matrix $b$ such that555We assume that each row of $b$ and each column of $c$ contains only a finite number of nonzero elements. $\ker b=L$ and a matrix $c$ such that $\ker c^{t}=M$.

Lemma 1.3

The group $\mathbb{G}^{\circ}[L;M]$ consists of all invertible matrices admitting the following representation as a block matrix of size $m+\infty$:

[TABLE]

where the block ’$a$’ can be written in both forms

[TABLE]

Next, define a subgroup $\mathbb{G}^{\bullet}[L;M]\subset\mathbb{G}^{\circ}[L;M]$ consisting of matrices having the form

[TABLE]

Proposition 1.4

The group $\mathbb{G}^{\bullet}[L;M]$ is the minimal subgroup of finite index in $\widehat{\mathbb{G}}[L;M]$, i.e., it is contained in any subgroup of finite index in $\widehat{\mathbb{G}}[L;M]$.

Theorem 1.5

a)* Any irreducible unitary representation of $\mathbb{G}$ is induced from a representation $\tau$ of some group $\widehat{\mathbb{G}}[L;M]$ that is trivial on the subgroup $\mathbb{G}^{\bullet}[L;M]$.*

b)* Two irreducible representations of this kind can be equivalent only for the trivial reason as in Theorem 1.2.*

Remark. Recall that $p\neq 2$. In several places of our proof we divide elements of residue rings $\mathbb{Z}_{p^{\mu}}$ by 2. Usually, this division can be replaced by longer considerations. But in Lemma 6.8 this seems crucial. $\boxtimes$

Remark. Let $L$, $M\subset p\cdot\mathfrak{l}(\mathbb{Z}_{p^{\mu}})$. Then $\mathbb{G}[L;M]$ contains a congruence subgroup $N$ consisting of elements of $\mathbb{G}$ that are equal $1$ modulo $p^{\mu-1}$. Since $N$ is a normal subgroup in $\mathbb{G}$, it is normal in $\widehat{\mathbb{G}}[L;M]$. Let $\tau$ be trivial on $N$. Then the induced representation $\operatorname{Ind}^{\mathbb{G}}_{\mathbb{G}[L;M]}(\widehat{\tau})$ is trivial on the congruence subgroup $N$ and actually we get representations of $\operatorname{GL}(\infty,{\mathbb{Z}}_{p^{\mu-1}})$. $\boxtimes$

Remark. The statement b) is a general fact for oligomorphic groups, see [34], Proposition 4.1(ii). So we omit a proof (in our case this can be easily established by examination of intertwining operators). $\boxtimes$

1.4. Remarks. Infinite-dimensional $p$-adic groups. Now there exists a well-developed representation theory of infinite symmetric groups and of infinite-dimensional real classical groups. Parallel development in the $p$-adic case meet some difficulties. However, infinite dimensional $p$-adic group were a topic of sporadic attacks since late 1980s, see [19], [36], [18]. We indicate some works on $p$-adic groups and their parallels with nontrivial constructions for real and symmetric groups.

a) An extension of the Weil representation of the infinite-dimensional symplectic group $\mathrm{Sp}(2\infty,{\mathbb{C}})$ to the semigroup of lattices (Nazarov [19], [18], see a partial exposition in [22], Sect. 11.1-11.2).

b) A construction of projective limits of $p$-adic Grassmannians and quasiinvariant actions of $p$-adic $\operatorname{GL}(\infty)$ on these Grassmannians [24]. This is an analog of virtual permutations (or Chinese restaurant process, see, e.g., [1], 11.19, they are a base of harmonic analysis related to infinite symmetric group, see [14]), and of projective limits of compact symmetric spaces (see [31], [21]), they are a standpoint for a harmonic analysis related to infinite-dimensional classical groups, see [3]).

c) An attempt to describe a multiplication of double cosets (see the next section) for $p$-adic classical groups in [25]. In any case this leads to a strange geometric construction, namely to simplicial maps of Bruhat–Tits buildings whose boundary values are rational maps of $p$-adic Grassmannians.

d) The work [4] contain a $p$-adic construction in the spirit of exchangeability666i.e., of higher analogs of the de Finetti theorem, see, [1], namely, descriptions of invariant ergodic measures on spaces of infinite $p$-adic matrices. By the Wigner–Mackey trick (see, e.g., [15], Sect. 13.3), such kind of statements can be translated to a description of spherical functions on certain groups.

So during last years new elements of a nontrivial picture related to infinite-dimensional $p$-adic groups appeared. For this reason, understanding of representations $\operatorname{GL}(\infty,{\mathbb{O}}_{p})$ becomes necessary.

1.5. Another completion of a group of infinite matrices over ${\mathbb{Z}}_{p^{n}}$. Define a group $\mathcal{G}$ consisting of infinite matrices $g$ over ${\mathbb{Z}}_{p^{n}}$ such that:

$\bullet$ $g$ contains only a finite number of elements in each column;

$\bullet$ $g^{-1}$ exists and satisfies the same property.

A sequence $g^{(\alpha)}$ converges to $g$ if for each $j$ we have a convergence of $e_{j}g^{(\alpha)}$.

Clearly, $\mathcal{G}\supset\mathbb{G}$. Classification of irreducible unitary representations of $\mathcal{G}$ is the following. For each finitely generated submodule in $\mathfrak{l}({\mathbb{Z}}_{p^{n}})$ we consider the subgroup $\widehat{\mathcal{G}}[L]$ consisting of transformations sending $L$ to itself and the subgroup $\mathcal{G}^{\circ}[L]$ fixing $L$ pointwise.

Proposition 1.6

Any irreducible unitary representation of $\mathcal{G}$ is induced from a representation of some group $\widehat{\mathcal{G}}[L]$ trivial on $\mathcal{G}^{\circ}[L]$.

This follows from Theorem 1.5, on the other hand this can be deduced in a straightforward way from Tsankov’s result [34].

2 Preliminaries: the category of double cosets

2.1. Multiplication of double cosets and the category $\mathcal{K}$. Here we discuss a version of a general construction of multiplication of double cosets (see [29], [30], [20], [26], [27]).

Denote by $\mathbb{G}_{\mathrm{fin}}\subset\mathbb{G}$ the subgroup of finitary matrices, i.e., matrices $g$ such that $g-1$ has only a finite number of nonzero elements. For $\alpha=0$, 1, …denote by $\mathbb{G}(\alpha)\subset\mathbb{G}$ the subgroups consisting of matrices having the form $\begin{pmatrix}1_{\alpha}&0\\ 0&u\end{pmatrix}$, where $1_{\alpha}$ denotes the unit matrix of size $\alpha$ and $u$ is an arbitrary invertible matrix over $\mathbb{Z}_{p^{\mu}}$. Obviously, $\mathbb{G}(\alpha)$ is isomorphic to $\mathbb{G}$. Consider double cosets spaces $\mathbb{G}(\alpha)\setminus\mathbb{G}/\mathbb{G}(\beta)$, their elements are matrices determined up to the equivalence

[TABLE]

where a matrix $g$ is represented as a block matrix of size $(\alpha+\infty)\times(\beta+\infty)$. For a matrix $g$ we write the corresponding double coset as

[TABLE]

we will omit subscripts $\alpha\beta$ if it is not necessary to indicate a size. We wish to define a natural multiplication

[TABLE]

Let $\mathfrak{g}_{1}\in\mathbb{G}(\alpha)\setminus\mathbb{G}/\mathbb{G}(\beta)$, $\mathfrak{g}_{2}\in\mathbb{G}(\beta)\setminus\mathbb{G}/\mathbb{G}(\gamma)$ be double cosets. By [27], Lemma 4.1, any double coset has a representative in $\mathbb{G}_{\mathrm{fin}}$. Choose such representatives $g_{1}$ and $g_{2}$ for $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}$,

[TABLE]

let sizes of submatrices $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$, $\begin{pmatrix}p&q\\ r&t\end{pmatrix},$ be $N\times N$. Denote by $\theta^{\beta}(j)$ the following matrix

[TABLE]

Consider the sequence

[TABLE]

It is more or less obvious that this sequence is eventually constant and its limit is

[TABLE]

where $L\geqslant N-\beta$. The final expression is

[TABLE]

In calculations below we use the last expression for $\circ$-product.

It is is easy to verify that this multiplication is associative, i.e., for any

[TABLE]

we have

[TABLE]

In other words, we get a category. Objects of this category are numbers $\alpha=0$, 1, 2, …. Sets of morphisms are

[TABLE]

The multiplication is given by formula (2.17). Denote this category by $\mathcal{K}$.

The group of automorphisms $\operatorname{Aut}_{\mathcal{K}}(\alpha)$ is $\operatorname{GL}(\alpha,\mathbb{Z}_{p^{\mu}})$, it consists of double cosets of the form \left[\begin{array}[]{c|c}a&0\\ \hline\cr 0&1_{\infty}\end{array}\right].

Next, the map $g\mapsto g^{-1}$ induces maps

[TABLE]

denote these maps by $\mathfrak{g}\mapsto\mathfrak{g}^{*}$. It is easy to see that we get an involution in the category $\mathcal{K}$, i.e.,

[TABLE]

The map $g\mapsto g^{t-1}$ determines an automorphism of the category $\mathcal{K}$, denote it by $\mathfrak{g}\mapsto\mathfrak{g}^{\bigstar}$. It sends objects to themselves and

[TABLE]

Remarks on notation.

1. In formulas (2.2), (2.3), (2.17), the last columns, the last rows, and the blocks $1_{\infty}$ contain no information and only enlarge sizes of matrices. For this reason, below we will omit them. Precisely, for a matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ of finite size we denote

[TABLE]

2. We will denote a multiplication of $[g]$ by an automorphism $A$ as $A\cdot[g]$,

[TABLE]

2.2. The multiplicativity theorem. Consider a unitary representation $\rho$ of the group $\mathbb{G}$ in a Hilbert space $H$. Denote by $H_{\alpha}\subset H$ the space of $\mathbb{G}(\alpha)$-fixed vectors. Denote by $P_{\alpha}$ the operator of orthogonal projection to $H_{\alpha}$.

Proposition 2.1

a)* For any $\beta$ the sequence \rho\bigl{(}\theta^{\beta}(j)\bigr{)} converges to $P_{\beta}$ in the weak operator topology.*

b)* The space $\cup H_{\alpha}$ is dense in $H$.*

The first statement is Lemma 1.1 from [27], the claim b) is a special case of Proposition VII.1.3 from [20]. $\square$

Let $g\in\mathbb{G}$, $\alpha$, $\beta\in{\mathbb{Z}}_{+}$. Consider the operator

[TABLE]

given by

[TABLE]

It is easy to see that for $h_{1}\in\mathbb{G}_{\alpha}$, $h_{2}\in\mathbb{G}_{\beta}$ we have

[TABLE]

i.e., $\widetilde{\rho}_{\alpha\beta}(g)$ actually depends on the double coset $\mathfrak{g}$ containing $g$.

Theorem 2.2

a)* The map $g\mapsto\widetilde{\rho}_{\alpha\beta}(\mathfrak{g})$ is a representation of the category $\mathcal{K}$, i.e., for any $\alpha$, $\beta$, $\gamma$ for any $\mathfrak{g}_{1}\in\operatorname{Mor}(\beta,\alpha)$, $\mathfrak{g}_{2}\in\operatorname{Mor}(\gamma,\beta)$ we have*

[TABLE]

b)* $\widetilde{\rho}$ is a $*$-representation, i.e.,*

[TABLE]

The statement a) is an automatic corollary of Proposition 2.1, see [27], Theorem 2.1, the statement b) is obvious. $\square$

Remark. The considerations of Subsections 2, 2 are one-to-one repetitions of similar statements for real classical groups and symmetric groups, see [30], [28], [23], [26]. Further considerations drastically differ from these theories. $\boxtimes$

2.3. Structure of the paper. We derive the classification of unitary representations of $\mathbb{G}$ from the multiplicativity theorem and the following argumentation. The semigroups $\Gamma(m):=\operatorname{End}_{\mathcal{K}}(m)$ are finite. It is known that a finite semigroup with an involution has a faithful $*$-representation in a Hilbert space if and only if it is an inverse semigroup (see discussion below, Subs. 3). More generally, if a category having finite sets of morphisms acts faithfully in Hilbert spaces, then it must be an inverse category, see [12]. However, semigroups $\operatorname{End}_{\mathcal{K}}(\alpha)$ are not inverse777This was observed by Olshanski [30] for $\operatorname{GL}(\infty,\mathbb{F}_{p})$., and $*$-representations of $\mathcal{K}$ pass through a smaller category.

Section 3 contains preliminary remarks on inverse semigroup and construction of an inverse category $\mathcal{L}$, which is a quotient of $\mathcal{K}$. This provides us lower estimate of maximal inverse semigroup quotients of semigroups $\Gamma(m)$.

In Section 4 we examine idempotents in maximal inverse semigroup quotients $\operatorname{inv}(\Gamma(m))$ of $\Gamma(m)$. In Section 5 we show that some of idempotents of $\operatorname{inv}(\Gamma(m))$ act by the same operators in all representations of $\mathbb{G}$. Next, for any representation of $\mathbb{G}$ there is a minimal $m$ such that $H_{m}\neq 0$. In Section 6 we examine the image of $\Gamma(m)$ in such representation.

In Section 7 we discuss properties of the groups $\mathbb{G}^{\circ}[L;M]$ and $\mathbb{G}^{\bullet}[L;M]$.

The final part of the proof is contained in Section 8.

3 The reduced category and inverse semigroups

3.1. Notation. Below we work only with the group $\mathbb{G}:=\operatorname{GL}(\infty,\mathbb{Z}_{p^{\mu}})$. To simplify notation, we write

[TABLE]

For a unitary representation $\rho$ of a $\mathbb{G}$ we define the height $h(\rho)$ as the minimum of $\alpha$ such that $H_{\alpha}\neq 0$.

By $x(\operatorname{mod\,}p)$ we denote a reduction of an object (a scalar, a vector, a matrix) defined over $\mathbb{Z}_{p^{\mu}}$ modulo $p$, i.e. to the field $\mathbb{F}_{p}$. Notice that a square matrix $A$ of finite size over $\mathbb{Z}_{p^{\mu}}$ is invertible if and only if $A(\operatorname{mod\,}p)$ is invertible. A matrix $B$ is nilpotent (i.e., $B^{N}=0$ for sufficiently large $N$) if and only if $B(\operatorname{mod\,}p)$ is nilpotent.

We use several symbols for equivalences in $\operatorname{Mor}_{\mathcal{K}}(\beta,\alpha)$, the $\sim$ was defined by (2.1), the symbols

[TABLE]

are defined in the next two subsections.

3.2. The reduced category $\operatorname{red}(\mathcal{K})$. Let $\mathfrak{g}_{1}$, $\mathfrak{g}_{2}\in\operatorname{Mor}(\beta,\alpha)$. We say that they are $\thickapprox$-equivalent if for any unitary representation of $\mathbb{G}$ we have $\widetilde{\rho}_{\alpha\beta}(\mathfrak{g}_{1})=\widetilde{\rho}_{\alpha\beta}(\mathfrak{g}_{2})$. The reduced category $\operatorname{red}(\mathcal{K})$ is the category, whose objects are nonnegative integers and morphisms $\beta\to\alpha$ are $\thickapprox$-equivalence classes of $\operatorname{Mor}(\beta,\alpha)$. Denote by $\operatorname{red}(\Gamma(m))$ semigroups of endomorphisms of $\operatorname{red}(\mathcal{K})$.

Also we define a weaker equivalence, $\mathfrak{g}_{1}\thickapprox_{m}\mathfrak{g}_{2}$ if $\widetilde{\rho}_{\alpha\beta}(\mathfrak{g}_{1})=\widetilde{\rho}_{\alpha\beta}(\mathfrak{g}_{1})$ for all $\rho$ of height $\geqslant m$. Denote by $\operatorname{red}_{m}(\mathcal{K})$ the corresponding $m$-reduced category.

Our proof of Theorem 1.5 is based on an examination of the categories $\operatorname{red}(\mathcal{K})$ and $\operatorname{red}_{m}(\mathcal{K})$. We obtain an information sufficient for a classification of representations of $\mathbb{G}$. However, the author does not know an answer to the following question.

Question 3.1

Find a transparent description of the category $\operatorname{red}(\mathcal{K})$.

3.3. Inverse semigroups. Let $\mathcal{P}$ be a finite semigroup with an involution $x\mapsto x^{*}$. Then the following conditions are equivalent.

A) $\mathcal{P}$ admits a faithful representation in a Hilbert space.

B) $\mathcal{P}$ admits an embedding to a semigroup of partial bijections888Recall that a partial bijection $\sigma$ from a set $A$ to a set $B$ is a bijection from a subset $S$ of $A$ to a subset $T$ of $B$, see e.g., [17] or [20], Sect. VIII.1. The adjoint partial bijection $\sigma^{*}:B\to A$ is the inverse bijection $T$ to $S$. of a finite set compatible with the involutions in $\mathcal{P}$ and in partial bijections.

C) $\mathcal{P}$ is an inverse semigroup (see [6], [17], [16]), i.e., for any $x$ we have

[TABLE]

and any two idempotents in $\mathcal{P}$ commute.

Discuss briefly some properties if inverse semigroups. Any idempotent in $\mathcal{P}$ is self-adjoint, and for any $x$, the element $x^{*}x$ is an idempotent. Since idempotents commute, a product of idempotents is an idempotent. The semigroup of idempotents has a natural partial order,

[TABLE]

We have $xy\preceq x$. If $x\preceq y$ and $u\preceq v$, then $xu\preceq yv$. Since our semigroup is finite, the product of all idempotents is a minimal idempotent $\boldsymbol{0}$, we have $\boldsymbol{0}x=x\boldsymbol{0}=\boldsymbol{0}$ for any $x$.

Let $\mathcal{R}$ be a finite semigroup with involution. Then there exists an inverse semigroup $\operatorname{inv}(\mathcal{R})$ and epimorphism $\pi:\mathcal{R}\to\operatorname{inv}(\mathcal{R})$ such that any homomorphism $\psi$ from $\mathcal{R}$ to an inverse semigroup $\mathcal{Q}$ has the form $\psi=\varkappa\pi$ for some homomorphism $\varkappa:\operatorname{inv}(\mathcal{R})\to\mathcal{Q}$. We say that $\operatorname{inv}(\mathcal{R})$ is the maximal inverse semigroup quotient of $\mathcal{R}$.

Lemma 3.1

The semigroups $\Gamma(m)$ are finite.

This is a corollary of the following statement, see [27], Lemma 4.1.a.

Lemma 3.2

Any double coset in $\mathbb{G}(m)\setminus\mathbb{G}/\mathbb{G}(m)$ has a representative in $\operatorname{GL}(3m)$.

We consider the following quotients of $\Gamma(m)$:

1. $\operatorname{inv}(\Gamma(m))$ is the maximal inverse semigroup quotient of $\Gamma(m)$;

2. $\operatorname{red}(\Gamma(m)):=\operatorname{End}_{\operatorname{red}\mathcal{K}}(m)$;

3. $\operatorname{red}_{m}(\Gamma(m)):=\operatorname{End}_{\operatorname{red}_{m}(\mathcal{K})}(m)$.

We have the following sequence of epimorphisms999All these semigroups are different.:

[TABLE]

For $g\in\mathbb{G}_{\mathrm{fin}}$ we denote by $[g]_{mm}$ the corresponding element of $\Gamma(m)$ and by $[[g]]_{mm}$ the corresponding element of \operatorname{inv}\bigl{(}\Gamma(m)\bigr{)}. The equality in $\Gamma(m)$ we denote by $\sim$, in $\operatorname{inv}\Gamma((m))$ by $\equiv$, in $\operatorname{red}(\Gamma(m))$ by $\thickapprox$, in $\operatorname{red}_{m}(\Gamma(m))$ by $\thickapprox_{m}$. Denote by $[[g_{1}]]\diamond[[g_{2}]]$ the product in $\operatorname{inv}(\Gamma(m))$.

Our next purpose is to present some (non-maximal) inverse semigroup quotient of $\Gamma(m)$.

3.4. The category $\mathcal{L}$ of partial isomorphisms. Let $V$, $W$ be modules over $\mathbb{Z}_{p^{\mu}}$. A partial isomorphism $p:V\to W$ is an isomorphism of a submodule $A\subset V$ to a submodule $B\subset W$. We denote $\operatorname{dom}p:=A$, $\operatorname{im}p:=B$. By $p^{*}$ we denote the inverse map $B\to A$. Let $p:V\to W$, $q:W\to Y$ be partial isomorphisms. Then the product $pq$ is defined in the following way:

[TABLE]

for $v\in\operatorname{dom}pq$ we define $v(pq)=(vp)q$.

A partial isomorphism $p$ is an idempotent if $\operatorname{dom}p=\operatorname{im}p$ and $p$ is an identical map.

Objects of the category $\mathcal{L}$ are modules

[TABLE]

equipped with the following pairing

[TABLE]

where $v_{\pm}\in\mathfrak{l}_{\pm}^{\alpha}$. We say that two partial isomorphisms

[TABLE]

are compatible if for any $y_{+}\in\operatorname{dom}\xi_{+}$ and $y_{-}\in\operatorname{dom}\xi_{-}$, we have

[TABLE]

Next, we define a category $\mathcal{L}$. Its objects are spaces $\mathfrak{l}_{+}^{\alpha}\oplus\mathfrak{l}_{-}^{\alpha}$ and morphisms are pairs of compatible partial isomorphisms $\xi_{+}:\mathfrak{l}_{+}^{\alpha}\to\mathfrak{l}_{+}^{\beta}$, $\xi_{-}:\mathfrak{l}_{-}^{\alpha}\to\mathfrak{l}_{-}^{\beta}$.

The category $\mathcal{L}$ is equipped with an involution

[TABLE]

and an automorphism

[TABLE]

Lemma 3.3

The semigroups $\operatorname{End}_{\mathcal{L}}(m)$ are inverse.

Indeed, $\operatorname{End}_{\mathcal{L}}(m)$ is a semigroup of partial bijections of a finite set $\mathfrak{l}^{m}_{+}\oplus\mathfrak{l}^{m}_{-}$. The whole category $\mathcal{L}$ is inverse for the same reason. $\square$

3.5. The functor $\Pi:\mathcal{K}\to\mathcal{L}$. Consider $g\in\mathbb{G}_{\mathrm{fin}}$. Let actually $g$ be contained in $\operatorname{GL}(N)$. Represent $g$ as a block $(\beta+(N-\beta))\times(\alpha+(N-\alpha))$ matrix and $g^{-1}$ as an $(\alpha+(N-\alpha))\times(\beta+(N-\beta))$-matrix,

[TABLE]

Define maps $\xi_{\pm}:\mathfrak{l}_{\pm}^{\alpha}\to\mathfrak{l}_{\pm}^{\beta}$ by:

$\bullet$ $\operatorname{dom}\xi_{+}:=\ker b$ and $\xi_{+}$ is the restriction of $a$ to $\ker b$;

$\bullet$ $\operatorname{dom}\xi_{-}:=\ker C^{t}$ and $\xi_{-}$ is the restriction of $A^{t}$ to $\ker C^{t}$.

Proposition 3.4

a)* The pair $\xi_{+}$, $\xi_{-}$ depends only on the double coset containing $\mathfrak{g}$.*

b)* Partial isomorphisms $\xi_{+}$, $\xi_{-}$ are compatible.*

c)* The map $\mathfrak{g}\mapsto(\xi_{+},\xi_{-})$ determines a functor from the category $\mathcal{K}$ to the category $\mathcal{L}$.*

Denote this functor by $\Pi$. By $\Pi(\mathfrak{g})$ we denote the morphism of $\mathcal{L}$ corresponding to $\mathfrak{g}$. We have

[TABLE]

Proof. For any invertible matrix $v$ we have, $\ker b=\ker bv$. Therefore $\xi_{+}$ depends only on a double coset. For $\xi_{-}$ we apply (3.2).

b) Let $v\in\ker b$, $w\in\ker C^{t}$. Then

[TABLE]

c) We look to formula (2.17) for a product in $\mathcal{K}$. The new $\xi_{+}$ is a restriction of $ap$ to $\ker b\cap\ker aq$. This is the product of two $\xi$-es. $\square$

Remark. According Olshanski [30], for the case $\operatorname{GL}(\infty,\mathbb{F}_{p})$ the functor $\Pi:\mathcal{K}\to\mathcal{L}$ determines an isomorphism of categories $\operatorname{red}(\mathcal{K})\to\mathcal{L}$. However, for $\mu>1$ the maps $\Pi:\operatorname{red}(\Gamma(m))\to\operatorname{Mor}_{\mathcal{L}}(m)$ are neither surjective, nor injective. However we will observe, that $\Pi$ induce isomorphisms of semigroups of idempotents; this provides us an important argument for the proof of Proposition 6.1. $\boxtimes$

4 Idempotents in $\operatorname{inv}(\Gamma(m))$

Here we examine idempotents in the semigroup $\operatorname{inv}(\Gamma(m))$. The main statement of the section is Proposition 4.10.

4.1. Projectors101010This subsection contains generalities, $\mathcal{K}$ is an ordered category in the sense of [20], Sect. III.4, this implies all statements of the subsection. $P_{\alpha}$. Consider an irreducible representation $\rho$ of $\mathbb{G}$, let subspaces $H_{m}\subset H$ and orthogonal projectors $P_{m}:H\to H_{m}$ be as above.

Lemma 4.1

a)* The projector*

[TABLE]

is given by the operator \widetilde{\rho}_{mm}\bigl{(}\Theta_{[m]}^{\alpha}\bigr{)}, where

[TABLE]

b)* The tautological embedding $H_{\alpha}\to H_{m}$ is defined by the operator \widetilde{\rho}_{m\alpha}\bigl{(}\Lambda^{\alpha}_{[m]}\bigr{)}, there*

[TABLE]

c)* The orthogonal projector $H_{m}\to H_{\alpha}$ is given by \widetilde{\rho}_{\alpha m}\bigl{(}(\Lambda_{[m]}^{\alpha})^{*}\bigr{)}*

[TABLE]

Proof. a) We apply Proposition 2.1.a. For $j>m-\alpha$ we have $[\theta^{\alpha}(j)]_{mm}=\Theta_{[m]}^{\alpha}$. The same argument proves b) and c). $\square$

Lemma 4.2

a)* The map*

[TABLE]

is a homomorphism $\Gamma(\alpha)\to\Gamma(m)$.

b)* We have*

[TABLE]

This follows from a straightforward calculation. $\square$

Corollary 4.3

The map $\iota_{m}^{\alpha}$ is compatible with representations $\widetilde{\rho}$ of $\Gamma(\alpha)$ and $\Gamma(m)$. Namely, operators $\widetilde{\rho}_{mm}(\iota^{\alpha}_{m}(\mathfrak{g}))$ have the following block structure with respect to the decomposition $H_{m}=H_{\alpha}\oplus(H_{m}\ominus H_{\alpha})$:

[TABLE]

4.2. Idempotents in $\operatorname{inv}(\Gamma(m))$. Here we formulate several lemmas (their proofs occupy Subsections 4–4), as a corollary we get Proposition 4.10.

Lemma 4.4

Let for

[TABLE]

one of the blocks $a$, $d$ be degenerate. Then $[[g]]_{mm}\in\operatorname{inv}(\Gamma(m))$ has a representative $[g^{\prime}]$, for which both blocks $a$, $d$ are degenerate.

Denote by

[TABLE]

the subsemigroup in $\Gamma(m)$ consisting of all $[g]$, for which both blocks $a$, $d$ are nondegenerate.

Lemma 4.5

Any idempotent in $\operatorname{inv}(\Gamma(m))$ has a representative of the form $q\cdot[[R]]\cdot q^{-1}$ with $q$ ranging $\operatorname{GL}(m)$ and $R$ having the form

[TABLE]

where

[TABLE]

represents an idempotent in $\operatorname{inv}(\Gamma^{\circ}(\alpha))$. The parameter $\alpha$ ranges in the set [math], $1$, $2$, …, $m$.

Remark. Denote

[TABLE]

Then the following elements of $\Gamma(m)$ coincide:

[TABLE]

Denote

[TABLE]

Lemma 4.6

Elements of the form $[X(b,c)]$ are idempotents in $\Gamma^{\circ}(m)$. They depend only on $\ker b$ and $\ker c^{t}\subset\mathfrak{l}^{m}$.

Let $L:=\ker b$ and $M:=\ker c^{t}$. Denote

[TABLE]

Lemma 4.7

We have

[TABLE]

Lemma 4.8

Any idempotent in $\operatorname{inv}(\Gamma^{\circ}(m))$ has the form $\mathcal{X}[L,M]$.

Corollary 4.9

*Idempotents $\mathcal{X}[L,M]$ are pairwise distinct in $\operatorname{inv}(\Gamma^{\circ}(m))$. *

Proof. Indeed, $\operatorname{End}_{\mathcal{L}}(m)$ is an inverse semigroup, therefore we have a chain of maps

[TABLE]

The image of $X(b,c)$ in $\operatorname{Mor}_{\mathcal{L}}(m)$ is precisely the pair of identical partial isomorphisms $M\to M$, $L\to L$. Therefore for nonequivalent $X(b,c)$ we have different images. $\square$

Proposition 4.10

Any idempotent in $\operatorname{inv}(\Gamma(m))$ has a representative of the form

[TABLE]

where $q\in\operatorname{GL}(m)=\operatorname{Aut}_{\mathcal{K}}(m)$.

Proof. Lemma 4.2 defines a canonical embedding $i^{\beta}_{m}:\Gamma(\beta)\to\Gamma(m)$ for $\alpha<m$. By Lemma 4.5 any idempotent in $\operatorname{inv}(\Gamma(m))$ is equivalent to and idempotent lying in some $i^{\alpha}_{m}(\Gamma^{\circ}(\alpha))$. Lemma 4.8 gives us a canonical form of this idempotent. $\square$

Now we start proofs of Lemmas 4.4–4.8,

4.3. Proof of Lemma 4.4. Clearly $\Gamma(m)\setminus\Gamma^{\circ}(m)$ is a two-side ideal in $\Gamma(m)$. Since $[[g\circ(g^{-1}\circ g)]]_{mm}=[[g]]_{mm}$, it is sufficient to prove the statement for idempotents.

Let

[TABLE]

Then

[TABLE]

If $a$ is degenerate, then $aA$ is degenerate. Now let $a$ be non-degenerate, $d$ degenerate. Since the matrices (4.5) are inverse one to another, we have

[TABLE]

We see that $(1-Cb)(\operatorname{mod\,}p)$ is degenerate, $(1-bC)(\operatorname{mod\,}p)$ also is degenerate, and therefore $aA$ is degenerate.

4.4. Proof of Lemma 4.5. Step 1.

Lemma 4.11

Let $x$ be an idempotent in $\operatorname{inv}(\Gamma(m))$. Then it can be represented as $[[u]]$, where $u=u^{-1}$.

Proof. Let $x=[[g]]$. Then

[TABLE]

for sufficiently large $j$. We set $u:=g\theta^{m}(j)g^{-1}$. $\square$

Lemma 4.12

Let $g=g^{-1}\in\mathbb{G}_{\mathrm{fin}}$. For any $N>0$ there exists a representative $r\in\mathbb{G}_{\mathrm{fin}}$ of $[g]^{\circ 2N}$ such that $r=r^{-1}$.

Proof. Let actually $g\in\operatorname{GL}(m+l)$. Then we choose the following representative of $[g]^{\circ 8}$:

[TABLE]

Step 2.

Lemma 4.13

Let $g=g^{-1}=\begin{pmatrix}a&b\\ c&d_{\star}\end{pmatrix}\in\mathbb{G}$. Then there exists a matrix

[TABLE]

and $N$ such that

[TABLE]

has a form

[TABLE]

where $k$ is the rank of the reduced matrix $a^{m}(\operatorname{mod\,}p)$.

Clearly our lemma is a corollary of the following statement:

Lemma 4.14

For any $m\times m$ matrix $a$ over $\mathbb{Z}_{p^{\mu}}$ there exists $\zeta\in\operatorname{GL}(m)$ and $N$ such that

[TABLE]

Proof. We split the operator $a(\operatorname{mod\,}p)$ over the field $\mathbb{F}_{p}$ as a direct sum of a nilpotent part $S$ and a invertible part $T$. For sufficiently large $M$ the matrix $\begin{pmatrix}S&0\\ 0&T\end{pmatrix}^{M}$ has the form $\begin{pmatrix}0&0\\ 0&P\end{pmatrix}$ with a nondegenerate $P$. Since the group $\operatorname{GL}(k,\mathbb{F}_{p})$ is finite, $P^{L}=1_{k}$ for some $L$.

Thus without loss of generality, we can assume that $a$ has a form

[TABLE]

where $\alpha$, $\beta$, $\gamma$, $\delta$ are matrices over $\mathbb{Z}_{p^{\mu}}$. We conjugate it as follows

[TABLE]

We wish to choose $u$ to make zero in the boxed block. It is sufficient to find a matrix $u$ satisfying the following equation:

[TABLE]

We look for a solution in the form

[TABLE]

First, we consider $S_{k}$ as formal noncommutative variables. Then we get a system of equations of the form

[TABLE]

where $F_{k}$ are polynomial expressions with integer coefficients. These equations can be regarded as recurrence formulas for $S_{k}$. In this way we get a solution $u$.

Thus without a lose of generality we can assume that $a$ has the form

[TABLE]

Raising it to $\mu$-th power, we come to a matrix of the form

[TABLE]

We conjugate it as

[TABLE]

Taking $v=(1+p\delta^{\prime\prime})^{-1}\gamma^{\prime\prime}$ we kill the left lower block and come to a matrix of the form $\begin{pmatrix}0&0\\ 0&1+p\delta^{\prime\prime\prime}\end{pmatrix}$. Raising it in $p^{\mu-1}$-th power we come to $\begin{pmatrix}0&0\\ 0&1\end{pmatrix}$. $\square$

Step 3. Thus the element $[[g]]^{\diamond 2N}$ from Lemma 4.13 has a representative of the following block $(m-k)+k+(m-k)+\infty$ form:

[TABLE]

Lemma 4.15

There is a matrix U=\left(\begin{array}[]{c|c}1_{m}&0\\ \hline\cr 0&u_{\star}\end{array}\right) such that $UrU^{-1}$ has the form

[TABLE]

Recall that $[r]\sim[UrU^{-1}]$.

Proof. Since the matrix $\begin{pmatrix}\beta_{11}&\beta_{12}\end{pmatrix}$ is nondegenerate (otherwise $r$ is degenerate), we can choose a conjugation of $r$ by matrices $U=\begin{pmatrix}1_{m}&0\\ 0&*\end{pmatrix}$ reducing this block to the form $\begin{pmatrix}1&0\end{pmatrix}$. We have $r^{2}=1$, evaluating $r^{2}$ we get $\gamma_{11}$ in the left upper block. Therefore $\gamma_{11}=1$. Thus we come to new $r$,

[TABLE]

with new $\beta$, $\gamma$, $\delta$. Next, we conjugate this matrix by

[TABLE]

and kill $\gamma_{21}$. Thus we come to new $r$,

[TABLE]

But $(r^{\sim\sim})^{2}=1$. Looking to third row and third column of $(r^{\sim\sim})^{2}$ we observe that

[TABLE]

Thus, $r^{\sim\sim}$ has the desired form. $\square$

4.5. Proof of Lemma 4.6. Denote

[TABLE]

We can conjugate this matrix by $\begin{pmatrix}1&0\\ 0&u_{\star}\end{pmatrix}$. Therefore a matrix $A$ is defined up to multiplications $A\sim Au$, where $u$ is an invertible matrix. The invariant of this action is $\ker A$ (this is more or less clear, formally we can refer to Lemma 7.3 proved below).

Next,

[TABLE]

We have $\ker\begin{pmatrix}A&A\end{pmatrix}=\ker A$ and therefore $[X_{+}(A)]$ is an idempotent. In the same way, [X_{-}(B)]:=\left[\begin{array}[]{cc}1&0\\ B&1_{\star}\end{array}\right] is an idempotent. It remains to notice that

[TABLE]

Thus $[X(A,B)]$ is an idempotent.

4.6. Proof of Lemma 4.6. In notation of the previous subsection

[TABLE]

i.e.,

[TABLE]

or

[TABLE]

On the other hand, we have

[TABLE]

and now the statement becomes obvious.

4.7. Proof of Lemma 4.7. Indeed,

[TABLE]

and $\ker\begin{pmatrix}b_{1}&b_{2}\end{pmatrix}=\ker b_{1}\cap\ker b_{2}$. $\square$

4.8. Proof of Lemma 4.8.

Step 1. Any idempotent $[[g]]\in\operatorname{inv}(\Gamma^{\circ}(m))$ has a representative of the form $\begin{pmatrix}1&a\\ b&1_{\star}\end{pmatrix}$, where $ab=0$.

Let [[g]]=\left[\left[\begin{array}[]{cc}\alpha&\beta\\ \gamma&\delta_{\star}\end{array}\right]\right] be an idempotent, let $\alpha$, $\delta$ be nondegenerate. By Lemma 4.11 without loss of generality we can assume $g=g^{-1}$. Taking an appropriate power $[r]=[g]^{\circ 2N}$, we can achieve $\alpha=1$. By Lemma 4.12, we can assume $r=r^{-1}$.

Set $r=\begin{pmatrix}1&-a\\ b&c_{\star}\end{pmatrix}$. Evaluating $r^{2}=1$ we get the following collection of conditions

[TABLE]

We replace $r$ by an equivalent matrix

[TABLE]

here we used the identity $-ac^{-1}=a$.

Step 2. We evaluate $[r]^{\circ 2}$,

[TABLE]

But $ab=0$ and therefore $aba=0$. Repeating the same reasoning, we get

[TABLE]

Step 3. Next, we set $N=p^{\mu}$ in formula (4.8). Consider the following block matrix $u$ of size $p^{\mu}$,

[TABLE]

We conjugate the matrix $q$ defined by (4.8) as

[TABLE]

We have

[TABLE]

and we get a matrix of the form $X(A,B)$.

5 Idempotents in $\operatorname{red}(\Gamma(m))$

Here the main statement is Proposition 5.1, which shows that all idempotents in $\operatorname{red}(\Gamma(m))$ have representatives in $\operatorname{red}(\Gamma^{\circ}(m))$, therefore they have the form $\mathcal{X}[L,M]$), where $L\subset\mathfrak{l}^{m}$, $M\subset\mathfrak{l}^{m}$. The second fact (Proposition 5.3), which is important for the proof below, is a coherence of elements $\mathcal{X}[L,M]$ in different semigroups $\operatorname{red}(\Gamma(n))$.

5.1. Coincidence of idempotents.

Proposition 5.1

The following idempotents in $\operatorname{inv}(\Gamma(m))$ coincide as elements of $\operatorname{red}(\Gamma(m))$:

[TABLE]

and

[TABLE]

Corollary 5.2

Any idempotent in $\operatorname{red}(\Gamma(m))$ has the form $\mathcal{X}[L,M]$.

Proof of corollary. The semigroup $\operatorname{red}\Gamma(m))$ is a quotient of $\operatorname{inv}(\Gamma(m))$, the semigroup of idempotents also is a quotient of the semigroup of idempotents. By Proposition 4.10 all idempotents in $\operatorname{inv}(\Gamma(m))$ have $[[X_{\alpha}^{\bigcirc}[b,c]]$. By Proposition 5.1, they also can be written as $[[X_{\alpha}^{\square}[b,c]]$ . $\square$

Proposition will be proved below in Subsection 5.

Remarks. a) The idempotents $[[X_{\alpha}^{\bigcirc}(b,c)]]$ and $[[X_{\alpha}^{\square}(b,c)]]$ are different in $\operatorname{inv}(\Gamma(m))$. Indeed, we have the following homomorphism from $\Gamma(m)$ to the inverse semigroup $\operatorname{End}_{\mathcal{L}}(m)$. On $\Gamma^{\circ}(m)$ we define it as the map $\Pi$ described in Subs. 3. On the other hand, we send $\Gamma(m)\setminus\Gamma^{\circ}(m)$ to $\boldsymbol{0}$, i.e., to a pair of partial bijections with empty domains of definiteness. This map separates our idempotents.

b) Idempotents $\mathcal{X}[L,M]$ are pairwise different in $\operatorname{red}(\Gamma(m))$. To verify this, consider the representation of $\mathbb{G}$ in $\ell^{2}(\mathbb{G}/\mathbb{G}[L;M])$. It is easy to show that $\mathcal{X}(L,M)$ is the minimal idempotent of $\operatorname{red}(\Gamma(m))$ acting in this representation nontrivially. $\boxtimes$

5.2. Coherence. Let $L$, $M\subset\mathfrak{l}^{m}$ be submodules. Formula (4.4) defines the idempotent $\mathcal{X}[L,M]=X(b,c)$ as an element of $\Gamma(m)$, recall that $L=\ker b$, $M=\ker c^{t}$. However, for $n>m$ we can regard $L$, $M\subset\mathfrak{l}^{m}$ as submodules $L$ in $\mathfrak{l}^{n}\supset\mathfrak{l}^{m}$. In the larger space we have

[TABLE]

Consider a unitary representation $\rho$ of $\mathbb{G}$ in a Hilbert space $H$. For any $n\geqslant m$ we have an operator

[TABLE]

We claim that these operators as operators $H\to H$ depend only on $L$, $M$ and not on $n$. Precisely, we have the following statement.

Proposition 5.3

a)* Let $n\geqslant m$. Then a block matrix structure of the operator (5.1) with respect to the orthogonal decomposition $H_{n}=H_{m}\oplus(H_{n}\ominus H_{m})$ is*

[TABLE]

b)* For any $L$, $M\subset\mathfrak{l}^{m}$ we have a well-defined operator $\widetilde{\rho}(\mathcal{X}[L,M])$ in $H$, which sends $H_{m}$ to $H_{m}$ as $\widetilde{\rho}_{mm}(\mathcal{X}[L,M])$ and is zero on the orthocomplement $H\ominus H_{m}$.*

Proof. According Corollary 4.3, the right hand side of (5.2) is $\widetilde{\rho}_{nn}(X_{m}^{\square}(b,c))$. By Proposition 5.1, this operator coincides with $\widetilde{\rho}_{nn}(X_{m}^{\bigcirc}(b,c))$. $\square$

5.3. Proof of Proposition 5.1.

Lemma 5.4

Let $\mathfrak{g}\in\operatorname{red}(\Gamma(m))$ be an idempotent. Let $g$ be a representative of $\mathfrak{g}$ in $\mathbb{G}_{\mathrm{fin}}$. Then for any unitary representation $\rho$ of $\mathbb{G}$ in a Hilbert space $H$ the image of the orthogonal projector $\widetilde{\rho}_{mm}(\mathfrak{g})$ coincides with the space of fixed points of the subgroup in $\mathbb{G}$ generated by $\mathbb{G}(m)$ and $g$.

Proof. Let $v\in\operatorname{im}\widetilde{\rho}_{mm}(\mathfrak{g})$, i.e.,

[TABLE]

This happens if and only if $P_{m}v=v$, $\rho(g)v=v$. The condition $P_{m}v=v$ means that $\rho(h)v=v$ for all $h\in\mathbb{G}(m)$. $\square$

Therefore, it is sufficient to show that the group generated by $\mathbb{G}(m)$ and $X^{\bigcirc}(b,c)$ coincides with the group generated by $\mathbb{G}(m)$ and $X^{\square}(b,c)$.

Lemma 5.5

The group generated by the subgroup $\mathbb{G}(\beta)$ and the matrix

[TABLE]

coincides with $\mathbb{G}$.

Proof. Denote by $G$ the group generated by $X(1,1)$ and $\mathbb{G}(\beta)$. Conjugating $X(1,1)$ by block diagonal matrices we can get any matrix of the form $X(A,B)$ with nondegenerate $A$, $B$. Multiplying such matrices we observe that elements of the form $X(A_{1}+A_{2},B_{1}+B_{2})$ are contained in $G$. In particular, $X(0,2)\in G$. Since $p\neq 2$, conjugating $X(0,2)$ by a block scalar matrix we come to $X(0,1)\in G$. In the same way $X(1,0)\in G$. Now the statement became more-or-less obvious. $\square$

Lemma 5.6

The group generated by $\mathbb{G}(\beta)$ and the matrix

[TABLE]

coincides with $\mathbb{G}$.

Proof. Denote this group by $G$. Denote $S_{\infty}(\beta):=S_{\infty}\cap\mathbb{G}(\beta)$. Multiplying the matrix (5.3) from the left and right by elements of $S_{\infty}(\beta)$ we can get an arbitrary matrix of the form \left(\begin{array}[]{c|c}0&\sigma_{1}\\ \hline\cr\sigma_{2}&0_{\star}\end{array}\right) with $\sigma_{1}$, $\sigma_{2}\in S_{\beta}$. Multiplying two matrices of this type we can get any matrix \left(\begin{array}[]{c|c}\sigma&1\\ \hline\cr 0&1_{\star}\end{array}\right), where $\sigma\in S_{\beta}$. Therefore our group contains the subgroup $S_{\beta}\times S_{\infty}(\beta)$, which is maximal in $S_{\infty}$. Therefore $G\supset S_{\infty}$. But $S_{\infty}$ and $\mathbb{G}(\beta)$ generate $\mathbb{G}$, see [27], Lemma 3.6. $\square$

Proof of Proposition 5.1. Denote by

— $G^{\bigcirc}$ the group generated by $\mathbb{G}(m)$ and $X^{\bigcirc}_{\alpha}(b,c)$;

— $G^{\square}$ the group generated by $\mathbb{G}(m)$ and $X^{\square}_{\alpha}(b,c)$;

— $G$ the group generated by $\mathbb{G}(\alpha)$ and the matrix $X_{\diamond}(b,c)$ defined by

[TABLE]

Obviously, $G\supset G^{\bigcirc}$, $G\supset G_{\square}$. Let us verify the opposite inclusions.

The inclusion $G^{\bigcirc}\supset G$. Clearly $X_{\alpha}(-b,-c)\in G^{\bigcirc}$. Therefore $G^{\bigcirc}$ contains

[TABLE]

By Lemma 5.5, the group generated by $Y$ and $\mathbb{G}(m)$ is $\mathbb{G}(\alpha)$. On the other hand, $Y^{-1}X^{\bigcirc}(b,c)\sim X_{\diamond}(b,c)$.

The inclusion $G^{\square}\supset G$. We have

[TABLE]

Next, $X_{\diamond}(b,c)^{-1}X^{\square}_{\alpha}(b,c)\sim X_{\alpha}(0,0)$ and we refer to Lemma 5.6.

Thus, $G^{\bigcirc}=G^{\square}$. By Lemma 5.4, for any unitary representation $\rho$ of $\mathbb{G}$ we have

[TABLE]

and this completes the proof of Proposition 5.1. $\square$

6 The semigroup $\operatorname{red}_{m}(\Gamma(m))$

6.1. Structure of the semigroup $\operatorname{red}_{m}(\Gamma(m))$. Denote by $\boldsymbol{0}$ the minimal idempotent of the semigroup $\operatorname{red}_{m}(\Gamma(m))$.

Proposition 6.1

Any element $\neq\boldsymbol{0}$ in $\operatorname{red}_{m}(\Gamma(m))$ has a representative of a form $aX(b,c)$, where $a\in\operatorname{GL}(m)$.

The proof occupies the rest of the section. As a byproduct of Lemma 6.3 we will get the following statement.

Lemma 6.2

Any idempotent $[X(b,c)]$ by a conjugation by $a\in\operatorname{GL}(m)$ can be reduced to a form

[TABLE]

where $\gamma\beta=0(\operatorname{mod\,}p)$, $\beta\gamma=0(\operatorname{mod\,}p)$.

6.2. Proof of Proposition 6.1. Step 1.

Lemma 6.3

a)* Let $B$ be an $m\times N$ matrix over $\mathbb{Z}_{p^{\mu}}$, $C$ an $N\times m$ matrix. Then transformations*

[TABLE]

allow to reduce them to the form

[TABLE]

where $b_{12}$, $c_{21}$ are square nondegenerate matrices of the same size, products $b_{21}c_{12}$, $c_{12}b_{21}$ are nilpotent and $b_{22}=0(\operatorname{mod\,}p)$, $c_{22}=0(\operatorname{mod\,}p)$.

b)* The transformations*

[TABLE]

where $u$, $v$, $w$ are invertible, allow to reduce a pair $(B,C)$ to the form

[TABLE]

where $c_{21}b_{12}=0(\operatorname{mod\,}p)$, $b_{12}c_{21}=0(\operatorname{mod\,}p)$.

Proof. a) Reduce our matrices modulo $p$. A canonical form of a pair of counter operators $P:\mathbb{F}_{p}^{m}\to\mathbb{F}_{p}^{N}$ and $Q:\mathbb{F}_{p}^{N}\to\mathbb{F}_{p}^{m}$ is a standard problem of linear algebra, see, e.g., [7], [11]. In particular, such operators in some bases admit block decompositions $P=\begin{pmatrix}P_{r}&0\\ 0&P_{n}\end{pmatrix}$, $Q=\begin{pmatrix}Q_{r}&0\\ 0&Q_{n}\end{pmatrix}$, where $P_{r}Q_{r}$, $Q_{r}P_{r}$ are nondegenerate and $P_{n}Q_{n}$, $Q_{n}P_{n}$ are nilpotent.

Thus the matrices $B$, $C$ can be reduced to the form

[TABLE]

where

1. $b_{21}$, $c_{12}$ are invertible matrices of the same size;

2. products $b_{12}c_{21}$, $c_{21}b_{12}$ are nilpotent;

3. the matrices $b_{11}$, $b_{22}$, $c_{11}$, $c_{22}$ reduced $(\operatorname{mod\,}p)$ are zero.

Set

[TABLE]

notice that $u_{1}$ ($\operatorname{mod\,}p)$ is 1. We pass to new matrices

[TABLE]

For new $B$ the block $b_{11}=0$, other properties 1)–3) of matrices $B$, $C$ are preserved. Next, we take a unique matrix of the form $u_{2}=\begin{pmatrix}1&0\\ *&1\end{pmatrix}$ such that $C^{\prime\prime}u_{2}$ has zero block $c_{12}$. On the other hand the block $b_{11}$ of $u_{2}^{-1}B^{\prime\prime}$ is zero. We come to a desired form.

b) We apply statement a) and reduce $(B,C)$ to the form (6.1). Next, we multiply $\widetilde{B}$ from right by $\begin{pmatrix}b_{21}&\\ &1\end{pmatrix}^{-1}$ and get 1 on the place of $b_{21}$. After this, we multiply new $B$ from right by $\begin{pmatrix}1&-b_{22}\\ 0&1\end{pmatrix}$ and kill $b_{22}$. Finally, we repeat the same transformations with $\widetilde{C}$.

Now the problem is reduced to the same question for a pair $b_{12}$, $c_{21}$. If $c_{21}b_{12}\neq 0(\operatorname{mod\,}p)$, then we choose an invertible matrix $U$ such that $b_{12}Uc_{21}$ is not nilpotent and again repeat a). Etc. $\square$

Step 2.

Lemma 6.4

Let $[g]\in\Gamma(m)$ have the form

[TABLE]

and $[[g]]\not\thickapprox_{m}\boldsymbol{0}$. Then $bc$ and $cb$ are nilpotent.

Proof. We apply the previous lemma and represent $[g]$ as

[TABLE]

Set

[TABLE]

Let us show that

[TABLE]

Indeed,

[TABLE]

to establish the equivalence we multiply $[g]\circ[h_{m}^{\alpha}]$ from the left by

[TABLE]

Next, denote

[TABLE]

We have

[TABLE]

the latter matrix is obtained from $r$, see (6.3), by removing two boxed blocks $\boxed{1_{m-\alpha}}$, all other blocks are the same. Thus $[r]\sim[g]$, i.e., we established (6.2).

Suppose that $\alpha\neq m$. Then by Proposition 5.1,

[TABLE]

But $\Theta_{[m]}^{\alpha}\thickapprox_{m}\boldsymbol{0}$, therefore $[[g]]\approxeq_{m}\boldsymbol{0}$. $\square$

Step 3. Thus it is sufficient to prove Proposition 6.1 for $[g]$ having the form

[TABLE]

Lemma 6.5

Let [g]=\left[\begin{array}[]{c|c}1&b\\ \hline\cr c&1_{\star}\end{array}\right]_{mm} be invertible111111This is equivalent to invertibility of $(1-bc)^{-1}$ or invertibility of $(1-cb)^{-1}$. Here we do not need a nilpotency of $bc$.. Then

[TABLE]

Proof. By (3.1),

[TABLE]

We have (see, e.g., [9], Sect. 2.5)

[TABLE]

We also keep in mind the identity

[TABLE]

to establish it, we multiply both sides from the left by $(1-cb)$ and from the right by $(1-bc)$.

Next,

[TABLE]

This matrix defines an idempotent in $\operatorname{inv}(\Gamma^{\circ}(m))$. We must verify the following statement:

Lemma 6.6

Under our conditions,

[TABLE]

Proof. By Corollary 4.9 we can identify an idempotent in $\operatorname{inv}(\Gamma^{\circ})$ evaluating its image in $\operatorname{Mor}_{\mathcal{L}}(m)$. So we get

[TABLE]

where

[TABLE]

We have $\ker B=\ker b$, $\ker C^{t}=\ker c^{t}$, therefore by Lemma 4.6 we have $[[X(B,C)]]\equiv[[X(b,c)]]$. $\square$

Corollary 6.7

Let

[TABLE]

be invertible and $u$ also be invertible. Then

[TABLE]

Proof. Indeed, $\ker bu=\ker b$. So both sides are $[[X(b,c)]]$. $\square$

Step 4.

Lemma 6.8

Let [g]=\left[\begin{array}[]{c|c}1&b\\ \hline\cr c&1_{\star}\end{array}\right]_{mm}, let $bc$ and $cb$ be nilpotent. Then there exists $u$ having the form

[TABLE]

such that

[TABLE]

Proof. The product is

[TABLE]

here

[TABLE]

We claim that there exists a unique $u$ such that $rq=0$. A straightforward calculation shows that

[TABLE]

Since $cb$ is nilpotent, we can write the equation $rq=0$ as

[TABLE]

the sum actually is finite. Clearly we can find a solution in the form $u=-1/2+\sum_{j>0}s_{j}(cb)^{j}$, where $s_{j}$ are dyadic rationals, for coefficients $s_{j}$ we have a system of recurrent equations. This $u$ is invertible (since we can write a finite series for $u^{-1}$).

Next, we must show that the matrix \left(\begin{array}[]{c|c}1&bu\\ \hline\cr c&1_{\star}\end{array}\right) is invertible. Indeed, this is equivalent to existence of $(1-cbu)^{-1}$ and this is clear since by (6.5) $cbu$ is nilpotent.

Next we wish to simplify the matrix \left(\begin{array}[]{c|c}A&br\\ \hline\cr qc&1_{\star}\end{array}\right) by conjugations by matrices of the form $\begin{pmatrix}1&0\\ 0&D_{\star}\end{pmatrix}$. In fact, we have transformations

[TABLE]

For such transformations we have $r^{\prime}q^{\prime}=rq$. Set

[TABLE]

Then $r^{\prime}=\begin{pmatrix}u&0&0\end{pmatrix}$. But $u$ is invertible and $r^{\prime}q^{\prime}=0$. Therefore $q^{\prime}$ has the form $\begin{pmatrix}0\\ *\\ *\end{pmatrix}$, on the other hand multiplication $q\mapsto Dq$ does not change the second and third elements of the column $q$. Thus we came to the matrix

[TABLE]

Consider the following matrices:

[TABLE]

The conjugation $R\mapsto TRT^{-1}$ kills boxed elements of $R$. The conjugation $R\mapsto STRT^{-1}S^{-1}$ reduces the matrix to the desired form. $\square$

Proof of Proposition 6.1. Thus we have

[TABLE]

The second factor is

[TABLE]

Passing to adjoint elements we get

[TABLE]

It remains to notice that

[TABLE]

6.3. Proof of Lemma 6.2. We refer to Lemma 6.3.

7 The groups $\mathbb{G}^{\bullet}[L;M]$

In this section we examine subgroups $\mathbb{G}^{\circ}[L;M]$, $\mathbb{G}^{\bullet}[L;M]\subset\mathbb{G}$ defined in Subsection 1. We prove that $\mathbb{G}^{\bullet}[L;M]$ is well-defined (Lemma 7.5, shows that it is generated by $\mathbb{G}(m)$ and the idempotent $X(b,c)$ (Proposition 1.4). Also we prove that it is a minimal subgroup of finite index in $\mathbb{G}[L;M]$ (equivalently, $\mathbb{G}^{\bullet}[L;M]$ has no subgroups of finite index, Proposition 7.11).

7.1. Several remarks on submodules in $\mathfrak{l}^{k}$.

Lemma 7.1

Let $L\subset\mathfrak{l}^{k}$ be a submodule. Then there exists a basis $e_{j}\in\mathfrak{l}^{k}$ such that $M:=\oplus p^{s_{j}}\mathbb{Z}_{p^{\mu}}e_{j}$. The collection $s_{1}$, $s_{2}$, …is a unique $\operatorname{GL}(m)$-invariant of a submodule $L$.

This is equivalent to a classification of sublattices in $({\mathbb{O}}_{p})^{k}$ under the action of $\operatorname{GL}(k,{\mathbb{O}}_{p})$ or equivalently to a classification of pairs of lattices in ${\mathbb{Q}}_{p}^{k}$ under $\operatorname{GL}(k,{\mathbb{Q}}_{p})$, the latter question is standard, see, e.g., [35], Theorem I.2.2. $\square$

Corollary 7.2

Any submodule $L\subset\mathfrak{l}^{k}$ is a kernel of some endomorphism $\mathfrak{l}^{k}\to\mathfrak{l}^{k}$.

Indeed, we pass to a canonical basis $e_{j}$ as in the lemma and consider the map sending $e_{j}$ to $p^{\mu-s_{j}}e_{j}$. $\square$

Lemma 7.3

a)* Let $L$ be a submodule in $\mathfrak{l}^{m}$. Let $b$, $b^{\prime}:\mathfrak{l}^{m}\to\mathfrak{l}^{N}$ be morphisms of modules such that $L=\ker b=\ker b^{\prime}$. Then there is a transformation $u\in\operatorname{GL}(N)$ such that $b^{\prime}=bu$.*

b)* Let $\ker b=L$, $\ker b^{\prime}=L^{\prime}\supset L$. Then there is an endomorphism $u:\mathfrak{l}^{N}\to\mathfrak{l}^{N}$ such that $b^{\prime}=bu$.*

Proof. a) The modules $\operatorname{im}b\simeq\operatorname{im}b^{\prime}\simeq\mathfrak{l}^{m}/L$ are isomorphic. By the previous lemma there is an automorphism of $\mathfrak{l}^{N}$ identifying these submodules.

b) $L$ is a submodule of $L^{\prime}$, therefore $\operatorname{im}b^{\prime}$ is a quotient module of $\operatorname{im}b$. Therefore there is a projection map $\pi:\operatorname{im}b\to\operatorname{im}b^{\prime}$, orders of elements do not increase under this map. By Lemma 7.1 we have a basis $e_{j}\in\mathfrak{l}^{N}$ such that $p^{s_{j}}e_{j}$, where $j=1$, …, $m$, is the system of generators of $\operatorname{im}b$. Choose arbitrary vectors $v_{j}$ such that $p^{s_{j}}v_{j}=\pi(p^{s_{j}}e_{j})$ and consider the map sending $e_{j}$ to $v_{j}$. $\square$.

7.2. The group $\mathbb{G}^{\bullet}$. Here we show that $\mathbb{G}^{\bullet}[L;M]$ is a group, and its definition does not depend on the choice of matrices $b$, $c$.

Lemma 7.4

a)* Fix a matrix $B$ of size $l\times N$. Then the set of invertible matrices $g$ of the form $1-BS$, where $S$ ranges in the set of $N\times l$ matrices, is a group.*

b)* Fix matrices $B$, $C$ of sizes $l\times N$ and $N\times l$ respectively. Then the set of invertible matrices $g$ of the form $g=1-BuC$ is a group.*

Proof. Clearly, both sets are closed with respect to multiplication. We must show that $g^{-1}$ satisfies the same property. In the first case,

[TABLE]

In the second case,

[TABLE]

Lemma 7.5

Fix matrices $b$, $c$ of sizes $m\times N$ and $N\times m$ respectively.

a)* The set of invertible matrices $g=\begin{pmatrix}a&bv\\ wc&z\end{pmatrix}$ such that the block ’$a$’ admits representations $a=1-bS$, $a=1-Tc$ is a group*

b)* The set $\mathbb{G}^{\bullet}[L;M]$, i.e., the set of all invertible matrices of the form $g=\begin{pmatrix}1-buc&bv\\ wc&z\end{pmatrix}$, is a group.*

Proof. In the first case we write

[TABLE]

and reduce the statement to the previous lemma.

In the second case we write

[TABLE]

and again we apply the previous lemma. $\square$

7.3. The group $\mathbb{G}^{\circ}[L;M]$.

Proof of Lemma 1.3. Let $g=\begin{pmatrix}\alpha&\beta\\ \gamma&\delta_{\star}\end{pmatrix}\in\mathbb{G}^{\circ}[L,M]$, i.e., $g$ fix pointwise $L\subset\mathfrak{l}^{m}$ and $g^{t}$ fix pointwise of $M\subset\mathfrak{l}^{m}$. Then $L\subset\ker\beta$ and by Lemma 7.3.b we have $\beta=bv$ for some matrix $v$. Also $L\subset\ker(1-\alpha)$ and therefore $\alpha=1-bS$ for some $S$. $\square$

7.4. Changes of coordinates.

Lemma 7.6

Let $L$, $M\subset\mathfrak{l}^{m}$. Let $a\in\operatorname{GL}(m)$. Then

[TABLE]

The first statement is an immediate consequence of the definition, the second is straightforward. $\square$

7.5. Generators of $\mathbb{G}^{\bullet}[L;M]$. Let $m$, $b$, $c$ be as in Subsection 1, i.e., $L=\ker b$, $M=\ker c^{t}\subset\mathfrak{l}^{m}$.

Proposition 7.7

The group $\mathbb{G}^{\bullet}[L;M]$ is generated by $\mathbb{G}(m)$ and the matrix $X(b,c)$.

Proof. Consider the group $G$ generated by $\mathbb{G}(m)$ and $X(b,c)$. Clearly, $\mathbb{G}^{\bullet}[L,M]\supset G$. Let us prove the converse.

1. Conjugating $X(b,c)$ by block diagonal matrices $\in\mathbb{G}(m)$ we get arbitrary matrices of the form $X(bv,wc)$, where $v$, $w$ are invertible matrices. Consider products

[TABLE]

We set $b=-b^{\prime}$, for any matrix $\sigma$ we can find invertible matrices $c$, $c^{\prime}$ such that121212It is sufficient to verify this statement for matrices over $\mathbb{F}_{p}$. Without loss of generality we can assume that $\sigma$ is diagonal. For $p\neq 2$ any element of $\mathbb{F}_{p}$ is a sum of two nonzero elements, where $\sigma$ can be represented as a sum of two diagonal matrices. $c+c^{\prime}=\sigma$. Thus $G$ contains all matrices of the form

[TABLE]

where $v$, $w$ are arbitrary matrices.

2. In virtue of Lemma 6.2, conjugating the matrices (7.2) by elements of $\operatorname{GL}(m)$ and multiplying from the left and the right by elements of $\mathbb{G}(m)$ we can reduce the matrices (7.2) to the forms

[TABLE]

where $\gamma\beta=0(\operatorname{mod\,}p)$, $\beta\gamma=0(\operatorname{mod\,}p)$. Multiplying $Y[\beta]$ from right by elements of $\mathbb{G}(m+\alpha)$ we can get that any matrix $Y[\beta r]$ with invertible $r$. The condition of invertibility of $r$ can be removed, because

[TABLE]

and we can represent any matrix $r$ as a sum of 3 invertible matrices.

In the same way we get that $G$ contains all elements of the form $Z[q\gamma]$.

Take $r=0$, $q=0$. Then the matrices $Y[0]=Y[\beta\cdot 0]$, $Z[0]=Z[0\cdot\gamma]$ together with $\mathbb{G}(m)$ generate the group $\mathbb{G}(m-\alpha)$.

Next, $G$ contains matrices $Y[\beta]Y[0]^{-1}$ and $Z[\gamma]Z[0]^{-1}$. They are matrices of the form (7.3), where boxed blocks are replaced by zeroes.

Therefore our problem is reduced to a description of the subgroup generated by $\mathbb{G}(m-\alpha)$ and $X(\beta,\gamma)$.

Thus, without loss of generality, we can assume that $\alpha=0$ and $cb=0(\operatorname{mod\,}p)$, $bc=0(\operatorname{mod\,}p)$.

3. Multiplying the matrices (7.2), we get

[TABLE]

Since $cb=0(\operatorname{mod\,}p)$, the $bvwc$ is nilpotent, and therefore $1-bvwc$ is invertible. We represent our matrix as

[TABLE]

Since the whole product and three factors are contained in $G$, the fourth factor also is contained in $G$,

[TABLE]

for any $v$, $w$.

4. Now consider an arbitrary element of $\mathbb{G}^{\bullet}[L;M]$,

[TABLE]

All factors of the right hand side are contained in $G$, and therefore $\mathbb{G}^{\bullet}[L;M]$ is contained in $G$. $\square$

Corollary 7.8

The group $\mathbb{G}^{\bullet}[L;M]$ does not depend on a choice of $m$.

Proof. Let $L$, $M\subset\mathfrak{l}^{m}$, let $L=\ker b$, $M=\ker c^{t}$. Let us regard $L$, $M$ as submodules $L^{\prime}$, $M^{\prime}$ of $\mathfrak{l}^{m}\oplus\mathfrak{l}^{k}$. Then

[TABLE]

Clearly the subgroup $G_{m}$ generated by $\mathbb{G}(m)$ and $X(b,c)$ and the subgroup $G_{m+k}$ generated by $\mathbb{G}(m+k)$ and $X(b^{\prime},c^{\prime})$ coincide. Formally, we must repeat the first two steps of the previous proof. $\square$

7.6. The quotient $\mathbb{G}^{\circ}/\mathbb{G}^{\bullet}$.

Lemma 7.9

A group $\mathbb{G}^{\bullet}[L;M]$ has finite index in $\mathbb{G}^{\circ}[L;M]$.

Proof. Without loss of generality we can assume that $cb=0(\operatorname{mod\,}p)$, $bc=0(\operatorname{mod\,}p)$. Denote by $A^{\circ}\subset\operatorname{GL}(m)$ the subgroup consisting of matrices $a$ admitting representations $a=1-bS$, $a=1-Tc$. Notice that $1-a$ is a nilpotent, since $TcbS=0(\operatorname{mod\,}p)$. Therefore $a$ is invertible. Denote by $A^{\bullet}$ the subgroup consisting of elements of the form $1-buc$.

The subgroup $A^{\bullet}$ is normal in $A^{\circ}$. Indeed, let $a\in A^{\circ}$, $a=1-bS$, $a^{-1}=1-Tc$. Then

[TABLE]

Let $g=\begin{pmatrix}a&bv\\ wc&z\end{pmatrix}\in\mathbb{G}^{\circ}[L;M]$. Let us show that the map $g\mapsto a$ induces a homomorphism from $\mathbb{G}^{\circ}[L;M]\to A^{\circ}/A^{\bullet}$. Indeed,

[TABLE]

In the left upper block we have

[TABLE]

We represent $a_{1}^{-1}=1-bS_{1}$, $a_{2}^{-1}=1-bS_{2}$ and get

[TABLE]

The expression in the curly brackets is contained in $A^{\bullet}$.

Clearly, the kernel of the homomorphism is $\mathbb{G}^{\bullet}[L,M]$. Thus we have an isomorphism of quotient groups,

[TABLE]

The group in the right-hand side is finite. $\square$

7.7. Absence of subgroups of finite index.

Lemma 7.10

The group $\mathbb{G}$ has not proper open subgroups of finite index.

Proof. Let $P$ be a proper open subgroup. Then it contains some group $\mathbb{G}(\nu)$. On the other hand $\mathbb{G}$ contains a complete infinite symmetric group $S_{\infty}$, and $S_{\infty}$ has no subgroups of finite index. Therefore $P$ contains $S_{\infty}$. But the subgroup in $\mathbb{G}$ generated by $\mathbb{G}(\nu)$ and $S_{\infty}$ is the whole group $\mathbb{G}$, see [27], Lemma 3.6. $\square$

Proposition 7.11

The subgroup $\mathbb{G}^{\bullet}[L;M]$ has no proper open subgroups of finite index.

Proof. Let $Q$ be such subgroup. By the previous lemma, $\mathbb{G}(m)$ has not open subgroups of finite index, we have $Q\supset\mathbb{G}(m)$. Hence $Q$ contains a minimal normal subgroup $R$ containing $\mathbb{G}(m)$. The quotient $Q/R$ is generated by the image $\xi$ of $X(b,c)$, therefore $Q/R$ is a cyclic group. But

[TABLE]

Since $p\neq 2$ the elements $X(b,c)^{2}$ and $X(b,c)$ have the same images in $Q/R$. Therefore the image of $X[b,c]$ is 1. $\square$

Corollary 7.12

Any subgroup of finite index in $\mathbb{G}[L,M]$ contains $\mathbb{G}^{\bullet}[L,M]$.

8 End of the proof

This section contains the end of the proof of Theorem 1.5. We know that all idempotents in semigroups $\operatorname{red}(\Gamma(n))$ have the form $\mathcal{X}[L,M]$, see Corollary 5.2, for different $n$ they can be identified in a natural way, see Proposition 5.3. We also know that any non-zero element of $\operatorname{red}_{m}(\Gamma(m))$ is a product of an invertible element and an idempotent $\mathcal{X}[L,M]$, see Proposition 6.1. This implies that all irreducible representations of $\mathbb{G}$ are induced from representations $\tau$ of groups $\mathbb{G}[L;M]$. Proposition 7.11 implies that such $\tau$ must be trivial on $\mathbb{G}^{\bullet}[L;M]$.

8.1. A preliminary remark.

Lemma 8.1

Consider an irreducible $*$-representation $\sigma$ of the category $\mathcal{K}$ in a sequence of Hilbert spaces $H_{j}$. Let $\xi\in H_{m}$ be a nonzero vector. Then the matrix element

[TABLE]

determines $\sigma$ up to equivalence.

This is a general statement on $*$-representations of categories (and a copy of a similar statement for unitary representations of groups), we give a proof for completeness.

Proof. For each $\mathfrak{g}\in\operatorname{Mor}_{\mathcal{K}}(m,\alpha)$ we define a vector

[TABLE]

Since $\sigma$ is irreducible, vectors $\omega_{\mathfrak{g}}^{\alpha}$, where $\mathfrak{g}$ ranges in $\operatorname{Mor}_{\mathcal{K}}(m,\alpha)$, generate the space $H_{\alpha}$. Their inner products are determined by the function $c$:

[TABLE]

Next, let $\mathfrak{h}\in\operatorname{Mor}_{\mathcal{K}}(\alpha,\beta)$. Let $\mathfrak{g}$, $\mathfrak{f}$ range respectively in $\operatorname{Mor}_{\mathcal{K}}(m,\alpha)$, $\operatorname{Mor}_{\mathcal{K}}(m,\beta)$. Then

[TABLE]

Clearly an operator $\sigma(\mathfrak{h})$ is uniquely determined by such inner products. $\square$

8.2. Representations of the semigroup $\operatorname{red}_{m}(\Gamma(m))$. Consider an irreducible representation of $\mathcal{K}$ of height $m$ and the corresponding representation $\lambda$ of the semigroup $\operatorname{End}_{\mathcal{K}}(m)$ in $H_{m}$. Recall that $\tau$ passes through semigroup $\operatorname{red}_{m}(\Gamma(m))$. By Proposition 6.1, any nonzero element of the latter semigroup can be represented as $a\cdot\mathcal{X}[L,M]$, where $a\in\operatorname{GL}(m)$. Denote

[TABLE]

The following lemma is a special case of general description of representations of finite inverse semigroups, see, e.g., [10]. However, due to Proposition 6.1 our case is simpler than general inverse semigroups. We show that the representation of $\operatorname{GL}(m)$ in $H_{m}$ is induced from an irreducible representation of some subgroup $\widehat{\mathbb{G}}_{m}[L,M]$ and idempotents $\mathcal{X}[N,K]$ act in the induced representation as multiplications by indicator functions of certain sets. Precisely,

Lemma 8.2

Let $\mathcal{X}[L,M]$ be the minimal idempotent in $\operatorname{red}_{m}(\Gamma(m))$ such that $\lambda(\mathcal{X}[L,M])\neq 0$. Then there is an irreducible representation $\tau_{m}$ of $\widehat{\mathbb{G}}_{m}[L,M]$ in a space $V$ such that $H_{m}$ can be identified with the space $\ell_{2}$ of $V$-valued functions on the homogeneous space $\widehat{\mathbb{G}}_{m}[L,M]\setminus\operatorname{GL}(m)$ and

1)* The group $\operatorname{GL}(m)$ acts by transformations of the form*

[TABLE]

and for $q\in\widehat{\mathbb{G}}_{m}[L,M]$ we have $R(p,x_{0})=\tau_{m}(q)$ (where $x_{0}$ denotes the initial point of $\widehat{\mathbb{G}}_{m}[L,M]\setminus\operatorname{GL}(m)$).

2)* The semigroup of idempotents acts by multiplications by indicator functions. Namely $\mathcal{X}[K,N]$ acts by multiplication by the function*

[TABLE]

Proof. Consider the image $V$ of the projector \lambda\bigl{(}\mathcal{X}[L,M]\bigr{)}. The idempotent $\mathcal{X}[L,M]$ commutes with $\widehat{\mathbb{G}}_{m}[L,M]$. Indeed, for $q\in\widehat{\mathbb{G}}_{m}[L,M]$ we have

[TABLE]

Therefore the subspace $V$ is $\widehat{\mathbb{G}}_{m}[L,M]$-invariant. Denote by $\tau_{m}$ the representation of the group $\widehat{\mathbb{G}}_{m}[L,M]$ in $V$. We need the following lemma:

Lemma 8.3

For any $\mathfrak{g}\in\operatorname{red}_{m}(\Gamma(m))$ we have $\lambda(\mathfrak{g})V=V$ or $\lambda(\mathfrak{g})V\bot V$.

Proof of Lemma 8.3. Let us apply an arbitrary element of $\operatorname{red}_{m}(\Gamma(m))$ to $v\in V$,

[TABLE]

We have the following cases:

If $K\not\supset L$ or $N\not\supset M$, then by our choice of $\mathcal{X}[L,M]$, we have

[TABLE]

2. Otherwise we come to $\lambda(a)\lambda\left(\mathcal{X}[L,M]\right)v=\lambda(a)v$.

2.1) If $a\in\widehat{\mathbb{G}}_{m}[L,M]$, we get $\lambda(a)v\in V$.

2.2) Let $a\notin\widehat{\mathbb{G}}_{m}[L,M]$. Then

[TABLE]

Since an idempotent $\mathcal{X}[a^{-1}L\cap L,a^{t}M\cap M]$ is strictly smaller than $\mathcal{X}[L,M]$, the $\lambda(\mathcal{X}[\dots])=0$. $\square$

End of proof of Lemma 8.2. Thus $H_{m}$ is an orthogonal direct sum of spaces $V_{x}$, where $x$ ranges in the homogeneous space $\widehat{\mathbb{G}}_{m}[L,M]\setminus\operatorname{GL}(m)$, and $\lambda(a)$ sends each $V_{x}$ to $V_{xa}$. This means that the representation $\lambda$ of $\operatorname{GL}(m)$ is induced from the representation of $\widehat{\mathbb{G}}_{m}[L,M]$ in $V$, see., e.g., [33], Sect.7.1.

Operators

[TABLE]

act as orthogonal projectors to $V_{x_{0}a}$. A projector \lambda\bigl{(}\mathcal{X}[K,N]\bigr{)} is identical on $V_{x_{0}a}$ if and only if $\mathcal{X}[K,N]\,\mathcal{X}[La^{-1},Ma^{t}]=\mathcal{X}[La^{-1},Ma^{t}]$ and this give us the action of the semigroup of idempotents.

It remains to show the representation of $\widehat{\mathbb{G}}_{m}[L,M]$ in $V$ is irreducible. Assume that it contains a $\widehat{\mathbb{G}}_{m}[L,M]$-invariant subspace $W$, then each $V_{x}$ contains a copy $W_{x}$ of $W$ and $\oplus_{x}W_{x}$ is a $\operatorname{GL}(m)$-invariant subspace in the whole $H_{m}$. $\square$

Corollary 8.4

Let $\lambda(\mathfrak{g})$ be a nonzero operator leaving $V$ invariant. Then there is $b\in\widehat{\mathbb{G}}_{m}[L,M]$ such that

[TABLE]

Proof. This operator can be represented as $\lambda(a)\lambda(\mathcal{X}[N,K])$ An operator $\lambda(\mathcal{X}[N,K])$ restricted to $V$ is zero 0 or 1. Let this operator be 1. Then $\lambda(a)$ preserves $V$ only if $a\in\widehat{\mathbb{G}}_{m}[L,M]$. In this case we set $b=a$. $\square$

Keeping in mind Lemma 8.1 we get the following statement:

Corollary 8.5

An irreducible $*$-representation of the category $\mathcal{K}$ is determined by its height $m$, a minimal idempotent $\mathcal{X}[L,M]$ acting nontrivially in $H_{m}$ and an irreducible representation $\tau$ of the group $\widehat{\mathbb{G}}_{m}[L,M]$.

We do not claim an existence of representation corresponding to given data of this kind.

8.3. End of proof. Let $\rho$ be an irreducible unitary representation of $\mathbb{G}$ of height $m$ in a Hilbert space $H$. Then we have a chain of subspaces in $H$:

[TABLE]

Lemma 4.2 defines a chain of semigroups

[TABLE]

Each semigroup $\Gamma(n)$ acts in $H$ as follows: in $H_{n}$ it acts by operators $\widetilde{\rho}_{nn}(\cdot)$, on $H_{n}^{\bot}$ these operators are zero (see Lemma 4.1).

On the other hand, we have a chain of groups

[TABLE]

acting by unitary operators, their inductive limit is the group $\mathbb{G}_{\mathrm{fin}}$. Each group $\operatorname{GL}(n)$ preserves the subspace $H_{n}$, on this subspace the action of $\operatorname{GL}(n)$ coincides with the action of the group $\operatorname{Aut}_{\mathcal{K}}(n)=\operatorname{GL}(n)$.

Consider the data listed in Corollary 8.5. We regard the subspace $V=\operatorname{im}\widetilde{\rho}_{mm}(\mathcal{X}[L,M])\subset H_{m}$ as a subspace in $H$. Denote the $\operatorname{GL}(n)$-cyclic of $V$ by $W_{n}$, it is a subspace in $H_{n}$.

Lemma 8.6

Let $g\in\operatorname{GL}(n)$. If $g\in\widetilde{\mathbb{G}}_{n}[L;M]$, then $\rho(g)V=V$. Otherwise, $\rho(g)V\bot V$.

Proof. In the first case, we have

[TABLE]

and therefore the image $V$ of $\widetilde{\rho}_{nn}(\mathcal{X}[L,M])$ is invariant with respect to $\rho(g)$.

In the second case we repeat the line (8.1). $\square$

Thus the representation of $\operatorname{GL}(n)$ in $W_{n}$ is induced from the subgroup $\widehat{\mathbb{G}}_{n}[L,M]$. If $k>n$, then we have embeddings

[TABLE]

and therefore the map of homogeneous spaces

[TABLE]

On the other hand, we have an embedding $W_{n}\to W_{k}$ regarding the orthogonal decompositions of these spaces into copies of $V$, therefore the map $\Xi_{n,k}$ is an embedding.

Finally, we get a representation of $\mathbb{G}_{\mathrm{fin}}$ induced from the subgroup $\widehat{\mathbb{G}}_{\mathrm{fin}}[L,M]$. By continuity, $\mathbb{G}$ acts regarding the same orthogonal decomposition $\oplus V_{xa}$. Hence a representation of $\mathbb{G}$ is induced from closure131313This closure contains $\mathbb{G}(m)$ and we refer to Lemma 3.2. of $\widehat{\mathbb{G}}_{\mathrm{fin}}[L,M]$, i.e., $\widehat{\mathbb{G}}[L,M]$.

Lemma 8.7

The image of $\widehat{\mathbb{G}}_{\mathrm{fin}}[L,M]$ in the group of operators in $V$ coincides with the image of $\widehat{\mathbb{G}}_{m}[L,M]$.

Proof. Let $u\in\widehat{\mathbb{G}}_{n}[L,M]$. Then $\rho(u)$ preserves $V\subset H_{m}$. Therefore

[TABLE]

By Corollary 8.4, this operator has the form \rho(u^{\prime})\Bigr{|}_{V}, where $u^{\prime}\in\widehat{\mathbb{G}}_{m}[L,M]$. $\square$

Thus the representation $\tau$ of $\widehat{\mathbb{G}}_{\mathrm{fin}}[L;M]$ in $V$ has a finite image. It continuous extension to $\mathbb{G}[L;M]$ has the same image. The kernel of the representation $\tau$ is a closed subgroup. Since it has a finite index, it is open. By Proposition 7.11, $\tau$ is trivial on the subgroup $\mathbb{G}^{\bullet}[L;M]$.