Direct finiteness of representable regular rings with involution

TL;DR

This paper investigates conditions under which elements in certain regular rings with involution are units, aiming to establish criteria for the rings' direct finiteness, though a complete proof remains open.

Contribution

It introduces a new condition based on finite-dimensional subspace actions for elements to be units in von Neumann *-regular rings with involution.

Findings

Provides a condition for invertibility of elements in regular rings with involution

Highlights an open problem regarding the direct finiteness of such rings

Clarifies the gap in previous claims about ring properties

Abstract

For von Neumann *-regular rings R of endomorphisms (the involution given by taking adjoints) of inner product spaces we provide a condition on r in R (in terms of action of r on finite dimensional subspaces) for r being a unit. It remains open whether this result can be used to prove direct finiteness of R. This was claimed in earlier versions but no proof was given.

Equations

$$\begin{array}[]{lcl}B_{U}&=&\{\varepsilon_{U}\varphi\pi_{U}+\lambda\operatorname{{\sf id}}_{V}\mid\varphi\in\operatorname{{\sf End}^{\ast}}(U_{F}),\ \lambda\in C(F)\}\\ &=&\{\psi\in\operatorname{{\sf End}}(V_{F})\mid\psi(U)\subseteq U\;\&\;\exists\lambda\in C(F)\,\psi|U^{\perp}=\lambda\operatorname{{\sf id}}_{U^{\perp}}\}\end{array}$$

$$\begin{array}[]{lcl}J(V_{F})&=&\{\varphi\in\operatorname{{\sf End}^{\ast}}(V_{F})\mid\dim\operatorname{{\sf im}}\varphi<\omega\}\\ \hat{J}(V_{F})&=&\{\varphi+\lambda\operatorname{{\sf id}}_{V}\mid\varphi\in J(V_{F}),\,\lambda\in C(F)\}\end{array}$$

$$\Sigma(x)=\bigl{\{}[xe=re]\ \&.\ [x^{\ast}e=r^{\ast}e]\&[\exists y.\,yx=1]\mid e\in J_{0}\bigr{\}}$$

Full text

1. Introduction

A $*$-ring, that is ring with involution, is called finite if $rr^{*}=1$ implies $r^{*}r=1$. This is a basic notion in the classification of von Neumann algebras; in particular, as shown by Murray and von Neumann, a finite von Neumann algebra admits a finite $*$-ring of quotients. This ring is also $*$-regular and directly finite, that is $rs=1$ implies $sr=1$. As Ara and Menal [1] have shown, any $*$-regular ring is at least finite, while direct finiteness remains an open question, as stated by Handelman [5, Problem 48]. The present note gives a positive answer for certain [von Neumann] regular rings with involution.

For $\ast$-rings, there is a natural and well established concept of [faithful] representation in a vector space $V_{F}$ endowed with a non-degenerate orthosymmetric sesquilinear form: an embedding into the $*$-ring $\operatorname{{\sf End}^{\ast}}(V_{F})$ of those endomorphisms of $V_{F}$ which admit an adjoint. Famous examples are due to Gel’fand-Naimark-Segal ($C^{\ast}$-algebras in Hilbert space) and Kaplansky (primitive $\ast$-rings with a minimal right ideal). For $\ast$-regular rings of classical quotients of finite Rickart $C^{\ast}$-algebras existence of representations has been established in [8], jointly with M. Semenova. N. Niemann [11, 6] has shown that a subdirectly irredcucible $*$-regular ring is representable if and only its ortholattice lattice of principal right ideals is representable within the ortholattice of closed subspaces of some $V_{F}$.

According to joint work with Susan M. Roddy [7], representability of modular ortholattices is equivalent to membership in a variety generated by finite height members. Using ideas from Tyukavkin [12], the analogue for $*$-regular rings was obtained by F. Micol [10]. Here, we rely on the presentation given in [9]: A regular $*$-ring can be represented within $V_{F}$, respectively some ultrapower thereof, if and only if it can be obtained via formation of ultraproducts, regular $*$-subrings, and homomorphic images from the class of the $\operatorname{{\sf End}^{\ast}}(U_{F})$, $U_{F}$ ranging over finite dimensional non-degenerate subspaces of $V_{F}$. It will be shown that direct finiteness is inherited under the particular construction of [9] which proves the reduction to finite dimensions.

Thanks are due to Ken Goodearl for the hint to reference [4].

2. Preliminaries

When mentioning rings, we always mean associative rings $R$ with unit $1_{R}$, considered as a constant. In any ring, if $a$ has a left inverse $x$, that is $xa=1$, then $a$ is left cancellable, that is $ay=az$ implies $y=z$. $R$ is directly finite if, for all $r,s\in R$, $sr=1$ implies $rs=1$. In such ring, $a$ has a left inverse if and only if $a$ is a unit (and $x=a^{-1}$). The endomorphism ring $\operatorname{{\sf End}}(V_{F})$ of a vector space is directly finite if and only if $\dim V_{F}<\omega$. A $\ast$-ring is a ring endowed with an involution $r\mapsto r^{*}$; an element $e$ of such a ring is a projection, if $e=e^{2}=e^{\ast}$.

A ring $R$ is [von Neumann] regular if for any $a\in R$, there is an element $x\in R$ such that $axa=a$; such an element is called a quasi-inverse of $a$. If, for all $a$, $x$ can be chosen a unit, then $R$ is unit regular. Examples of such are the $\operatorname{{\sf End}}(V_{F})$, $\dim V_{F}<\omega$. A detailed discussion of direct finiteness in regular rings is given in Goodearl [5]. A regular $*$-ring is $*$-regular if $xx^{*}=0$ only for $x=0$.

In a regular ring, any left cancellable $a$ has a left inverse; indeed $axa=a$ implies $xa=1$. If $xa=1$ and $aua=a$ with a unit $u$ then $a=u^{-1}$ and $x=a^{-1}$. It follows

Fact 1**.**

In a directly finite regular ring every left cancellable element is a unit – similarly on the right. Every unit regular ring is directly finite.

We recall some basic concepts and facts from [9] (here, $\Lambda$ can be taken the $\ast$-ring of integers). In the sequel, $F$ will be a division ring endowed with an involution and $V_{F}$ a [right] $F$-vector space of $\dim V_{F}>1$ endowed with a non-degenerate sesquilinear form $\langle{.}\mid{.}\rangle$ which is orthosymmetric, that is $\langle{v}\mid{w}\rangle=0$ iff $\langle{w}\mid{v}\rangle=0$. Such space will be called pre-hermitean and denoted by $V_{F}$, too. Within such space, any endomorphism $\varphi$ has at most one adjoint $\varphi^{*}$; and these $\varphi$ form a subring of $\operatorname{{\sf End}}(V_{F})$ which is a $\ast$-ring $\operatorname{{\sf End}^{\ast}}(V_{F})$ under the involution $\varphi\mapsto\varphi^{*}$. For $\dim V_{F}<\omega$, $\operatorname{{\sf End}^{\ast}}(V_{F})$ contains all of $\operatorname{{\sf End}}(V_{F})$. A [faithful] representation of a $\ast$-ring $R$ is an embedding of $R$ into some $\operatorname{{\sf End}^{\ast}}(V_{F})$.

Consider a linear subspace $U$ of $V_{F}$, $1<\dim U_{F}<\omega$. With the induced sesquilinear form, $U_{F}$ is pre-hermitean if and only if $V=U\oplus U^{\perp}$; in particular, there is a projection $\pi_{U}\in\operatorname{{\sf End}^{\ast}}(V_{F})$ such that $U=\operatorname{{\sf im}}\pi_{U}$ and such that the inclusion map $\varepsilon_{U}:U\to V$ is the adjoint of $\pi_{U}$ (here, considerd as a map $V\to U$). We write in this case $U\in\mathbb{O}(V_{F})$ and say that $U$ is a finite-dimensional orthogonal summand. A crucial fact is that $V_{F}$ is the directed union of the $U_{F}$, $U\in\mathbb{O}(V_{F})$. Let $C(F)$ denote the center of $F$ and, for $U\in\mathbb{O}(V_{F})$,

[TABLE]

($\varphi$ and $\psi$ are related via $\varphi(v)=\psi(v)+\lambda v$). Thus, $B_{U}$ is a $\ast$-subring of $\operatorname{{\sf End}^{\ast}}(V_{F})$ and embeds into $\operatorname{{\sf End}^{\ast}}(W_{F})$ for any $W\in\mathbb{O}(V_{F})$, $U\subset W\neq U$. In particular, $B_{U}$ is directly finite. Moreover, $B_{U}$ is unit regular; indeed, $\chi\in B_{U}$ is a unit quasi-inverse of $\psi\in B_{U}$ if $\chi|U$ is one of $\psi|U$ and $\chi|U^{\perp}=(\psi|U^{\perp})^{-1}$ (considering these as endomorphisms of $U$ and $U^{\perp}$, respectively). We put

[TABLE]

According to [9, Proposition 4.4] $\hat{J}(V_{F})$ is a $\ast$-subring of $\operatorname{{\sf End}^{\ast}}(V_{F})$ and $J(V_{F})$ is an ideal of $\operatorname{{\sf End}^{\ast}}(V_{F})$ closed under the involution. Also, the following holds.

• ($*$)

For any finite $\Phi\subseteq J(V_{F})$ there is $U\in\mathbb{O}(V_{F})$ such that $\varphi=\pi_{U}\varphi=\varphi\pi_{U}$ for all $\varphi\in\Phi$.

Thus, $\hat{J}(V_{F})$ is the directed union of the $B_{U},\,U\in\mathbb{O}(V_{F})$, whence unit-regular.

Lemma 2**.**

Every regular $\ast$-subring $R$ of $\operatorname{{\sf End}^{\ast}}(V_{F})$ extends to a regular $\ast$-subring $\hat{R}$ of $\operatorname{{\sf End}^{\ast}}(V_{F})$ containing $\hat{J}(V_{F})$ and such that $J(V_{F})$ is an ideal of $\hat{R}$.

Proof.

$\{\lambda\varphi\mid\varphi\in R,\,\lambda\in C(F)\}$ is a regular $\ast$-subring $R^{\prime}$ of $\operatorname{{\sf End}^{\ast}}(V_{F})$ and [9, Proposition 4.5] applies to $R^{\prime}$. ∎

Recall that a [faithful] representation of a $\ast$-ring $R$ within a pre-hermitian space $V_{F}$ is an embedding $\varepsilon\colon R\to\operatorname{{\sf End}^{\ast}}(V_{F})$. It is convenient to consider representations as unitary $R$-$F$-bimodules ${}_{R}V_{F}$ (where the action of $R$ is given as $rv=\varepsilon(r)(v)$) with sesquilinear form on $V_{F}$; that is, a $3$-sorted structure with sorts $V$, $R$, and $F$. Considering a $*$-subring $A$ of $R$ we may add a fourth sort, $A$, and the embedding map. to obtain $(_{R}V_{F};A)$. Any elementary extension $(_{\tilde{R}}\tilde{V}_{\tilde{F}};\tilde{A})$ is again such a structure, that is, a representation of $\tilde{R}$ and a $*$-ring $\tilde{A}$ which may be considered as $*$-subring of $\tilde{R}$. It is a modestly saturated extension if, for each set $\Sigma(\bar{x})$ of first order formulas in finitely many [sorted] variables and with parameters from $(_{R}V_{F};A)$, one has $(_{\tilde{R}}\tilde{V}_{\tilde{F}};\tilde{A})\models\exists\bar{x}.\Sigma(\bar{x})$, provided that $\Sigma(\bar{x})$ is finitely realized in $(_{R}V_{F};A)$, that is $(_{R}V_{F};A)\models\exists\bar{x}.\Psi(\bar{x})$ for every finite subset $\Psi(\bar{x})$ of $\Sigma(\bar{x})$. Such extension always exists, cf. [2, Corollary 4.3.1.4].

3. Main result

Theorem 3**.**

Every representable regular $\ast$-ring is directly finite.

Proof.

We recall the relevant steps of the proof of [9, Theorem 10.1]. Given a representation ${}_{R}V_{F}$ of the regular $*$-ring $R$, we may assume that $\dim V_{F}\geqslant\omega$. In view of Proposition 2, we also may assume that $R$ is a $*$-subring of $\operatorname{{\sf End}^{\ast}}(V_{F})$ containing $A=\hat{J}(V_{F})$ and having ideal $J(V_{F})$. Choose $(_{\tilde{R}}\tilde{V}_{\tilde{F}};\tilde{A})$ a modestly saturated elementary extension of $(_{R}V_{F};A)$.

Let $J_{0}$ denote the set of projections in $J(V_{F})$. For $a\in\tilde{A}$ and $r\in R$, we put $a\sim r$ if $ae=re$ and $a^{\ast}e=r^{\ast}e$ for all $e\in J_{0}$. According to Claims 1–4 in the proof of [9, Theorem 10.1], $S=\{a\in\hat{A}\mid a\sim r\ \text{for some}\ r\in R\}$ is a regular $\ast$-subring of $\hat{A}$ and there is a surjective homomorphism $g:S\to R$ such that $g(a)=r$ if and only if $a\sim r$.

Being an elementary extension of $A$, $\tilde{A}$ is directly finite and so is its subring $S$. Now, assume $sr=1$ in $R$. Consider a finite set $E\subseteq J_{0}$. According to $(*)$, there is $e\in J_{0}$ such that $ef=f$ and $er^{\ast}f=r^{\ast}f$ for all $f\in E$. Take $a=re$ and observe that $af=ref=rf$ and $a^{\ast}f=er^{\ast}f=r^{\ast}f$ for all $f\in E$. Thus, the set

[TABLE]

of formulas with a free variables $x,y$ of type $A$ and $R$, respectively, is finitely realized in $(_{R}V_{F};A)$. Indeed, given a finite subset $\Psi$ of $\Sigma(x)$ there is finite $E\subseteq J_{0}$ containing all $f\in J_{0}$ which occur in $\Sigma(x)$; choose $e$ for $E$ as above, $g,t,u$ according to Lemma 4 below, and $x=t+1-g$, $y=u+1-g$.

By saturation, there are $a\in\tilde{A}$ and $b\in\tilde{R}$ with $ba=1$ and $a\sim r$, whence $a\in S$ and $g(a)=r$. Moreover, $a$ is left cancellable in $\tilde{R}$ whence in the subring $S$ and so a unit of $S$ by regularity. Hence, $r=g(a)$ is a unit of $R$ and $s=r^{-1}$ whence $rs=1$. ∎

Lemma 4**.**

Consider a regular ring $R$ with ideal $I$ such that each $eRe$, $e\in I$, is unit-regular. Then for any $r,s\in R$ with $sr=1$ and idempotent $e\in I$ there are an idempotent $g\in I$, $e\in gRg$, and $t,u\in gRg$ such that $ut=g$, $te=re$, and $et=er$.

Proof.

Following [3] we consider $R$ the endomorphism ring of a (right) $R$-module, namely $M_{R}=R_{R}$. Observe that $r$ is an injective endomorphism of $M_{R}$. Let $U=\operatorname{{\sf im}}e$, $W_{1}=U+r^{-1}(U)$, $W_{2}=r(W_{1})$; in a particular, these are submodules of $M_{R}$ and $r|W_{1}$ is an isomorphism of $W_{1}$ onto $W_{2}$. By (the proof of) [4, Lemma 2] there is an idempotent $g\in I$ such that $e,re,se\in S:=gRg$. Put $W=\operatorname{{\sf im}}g$ which is a submodule of $M_{R}$, and an $S$-module under the induced action of $S$, so that $S=\operatorname{{\sf End}}(W_{S})=\operatorname{{\sf End}}(W_{R})$.

By hypothesis, $S$ is unit-regular whence, in particular, directly finite. Due to regularity of $S$, for any $h\in S$ and $S$-linear map $\phi:hS\to W$ there is an extension $\bar{\phi}\in S$, namely $\bar{\phi}|(g-h)S=0$. Due to direct finiteness, any injective such $\phi$ has an inverse in $S$. Also, by regularity, the submodules $W_{1}=\operatorname{{\sf im}}e+\operatorname{{\sf im}}se$ and $W_{2}=\operatorname{{\sf im}}e+\operatorname{{\sf im}}re$ are of the form $W_{i}=\operatorname{{\sf im}}g_{i}$ with idempotents $g_{i}\in S$.

Let $X_{i}=\operatorname{{\sf im}}(g-g_{i})$ whence $W=W_{i}\oplus X_{i}$. Since $r|W_{1}:W_{1}\to W_{2}$ is an $S$-linear isomorphism, according to [3, Theorem 3] there is an $S$-linear isomorphism $\varepsilon:X_{1}\to X_{2}$. Put $\delta(v)=\varepsilon(v)+g_{2}(r(v))$ for $v\in X_{1}$. If $\delta(v)=w\in W_{2}$ then $\varepsilon(v)\in W_{2}\cap X_{2}$ whence $\varepsilon(v)=0$ and $v=0$; it follows that $\delta$ is an $S$-linear isomorphism of $X_{1}$ onto $Y\subseteq W$ where $Y\cap W_{2}=0$. Also, $g_{2}(\delta(v))=g_{2}(r(v))$ since $g_{2}(X_{2})=0$. Define $t\in S$ as $t(v+w)=r(v)+\delta(w)$ for $v\in W_{1}$ and $w\in X_{1}$. $t$ is injective whence it has inverse $u$ in $S$. ∎

An example of a simple regular $*$-ring which is not finite is obtained as follows: Let $V_{F}$ a vector space of countably infinite dimension, and $R=\operatorname{{\sf End}}(V_{F})/J(V_{F})$. Of course, $R$ is not directly finite. Define the involution on the direct product $R\times R^{op}$ by exchange: $(r,s)^{*}=(s,r)$ to obtain the $*$-ring $S$. Now, if $rs=1$ but $sr\neq 1$ then $xx^{*}=1$ but $x^{*}x\neq 1$ in $S$ for $x=(r,s)$.

Problem 1**.**

Is every simple directly finite $*$-regular ring representable?