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AW*-algebras added
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Bert Lindenhovius
Bert Lindenhovius
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I am not sure whether your definition of a Baer *-ring is correct. In the book of Berberian on Baer *-rings a ring $A$ is called a Baer *-ring if

  1. $A$ is a *-ring (not necessarily an algebra), i.e., it admits an involution $a\mapsto a^*$ satisfying $a^{**}=a$, $(a+b)^*=a^*+b^*$, and $(ab)^*=b^*a^*$.
  2. For each non-empty subset $S\subseteq A$ the right annihilator $R(S)=\{a\in A: sa=0\text{ for all }s\in S\}$ equals $pA=\{pa:a\in A\}$ for some projection $p\in A$.

The C*-algebra $C([0,1])$ of continuous functions from the unit interval to the complex numbers has only two projections, hence its projections form a complete lattice, but is certainly not a Baer *-algebra.

Nevertheless, also with the correct definition of aThe Baer *$\phantom{.}^*$-ringrings that are C-algebras are called AW$^*$-algebras. The most prominent examples are of course von Neumann algebras, but there are many AW$^*$-algebras that are not von Neumann algebras, see for instance the book of Saito and Wright on monotone complete C-algebras, therea subclass of AW$^*$-algebras.

There are also examples of Baer *$\phantom{.}^*$-rings that are neither WAW$^*$-algebras nor locally W$^*$-algebras. Let $F$ be a finite field. For cardinality reasons, it cannot be isomorphic to an AW$^*$-algebra nor a locally W$^*$-algebra. Since its multiplication is commutative, the identity is an involution. If $S\subseteq F$ is non-empty, then it is easy to see that $R(S)$ is an ideal. Since $F$ has only trivial ideals, we have either $R(S)=(0)$, whence $R(S)=0F$, or $R(S)=F$, whence $R(S)=1R$. So $F$ is a Baer *-ring.

I am not sure whether your definition of a Baer *-ring is correct. In the book of Berberian on Baer *-rings a ring $A$ is called a Baer *-ring if

  1. $A$ is a *-ring (not necessarily an algebra), i.e., it admits an involution $a\mapsto a^*$ satisfying $a^{**}=a$, $(a+b)^*=a^*+b^*$, and $(ab)^*=b^*a^*$.
  2. For each non-empty subset $S\subseteq A$ the right annihilator $R(S)=\{a\in A: sa=0\text{ for all }s\in S\}$ equals $pA=\{pa:a\in A\}$ for some projection $p\in A$.

The C*-algebra $C([0,1])$ of continuous functions from the unit interval to the complex numbers has only two projections, hence its projections form a complete lattice, but is certainly not a Baer *-algebra.

Nevertheless, also with the correct definition of a Baer *-ring, there are examples of Baer *-rings that are neither W$^*$-algebras nor locally W$^*$-algebras. Let $F$ be a finite field. Since its multiplication is commutative, the identity is an involution. If $S\subseteq F$ is non-empty, then it is easy to see that $R(S)$ is an ideal. Since $F$ has only trivial ideals, we have either $R(S)=(0)$, whence $R(S)=0F$, or $R(S)=F$, whence $R(S)=1R$. So $F$ is a Baer *-ring.

I am not sure whether your definition of a Baer *-ring is correct. In the book of Berberian on Baer *-rings a ring $A$ is called a Baer *-ring if

  1. $A$ is a *-ring (not necessarily an algebra), i.e., it admits an involution $a\mapsto a^*$ satisfying $a^{**}=a$, $(a+b)^*=a^*+b^*$, and $(ab)^*=b^*a^*$.
  2. For each non-empty subset $S\subseteq A$ the right annihilator $R(S)=\{a\in A: sa=0\text{ for all }s\in S\}$ equals $pA=\{pa:a\in A\}$ for some projection $p\in A$.

The C*-algebra $C([0,1])$ of continuous functions from the unit interval to the complex numbers has only two projections, hence its projections form a complete lattice, but is certainly not a Baer *-algebra.

The Baer$\phantom{.}^*$-rings that are C-algebras are called AW$^*$-algebras. The most prominent examples are of course von Neumann algebras, but there are many AW$^*$-algebras that are not von Neumann algebras, see for instance the book of Saito and Wright on monotone complete C-algebras, a subclass of AW$^*$-algebras.

There are also examples of Baer$\phantom{.}^*$-rings that are neither AW$^*$-algebras nor locally W$^*$-algebras. Let $F$ be a finite field. For cardinality reasons, it cannot be isomorphic to an AW$^*$-algebra nor a locally W$^*$-algebra. Since its multiplication is commutative, the identity is an involution. If $S\subseteq F$ is non-empty, then it is easy to see that $R(S)$ is an ideal. Since $F$ has only trivial ideals, we have either $R(S)=(0)$, whence $R(S)=0F$, or $R(S)=F$, whence $R(S)=1R$. So $F$ is a Baer *-ring.

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Bert Lindenhovius
Bert Lindenhovius
  • 505
  • 4
  • 10

I am not sure whether your definition of a Baer *-ring is correct. In the book of Berberian on Baer *-rings a ring $A$ is called a Baer *-ring if

  1. $A$ is a *-ring (not necessarily an algebra), i.e., it admits an involution $a\mapsto a^*$ satisfying $a^{**}=a$, $(a+b)^*=a^*+b^*$, and $(ab)^*=b^*a^*$.
  2. For each non-empty subset $S\subseteq A$ the right annihilator $R(S)=\{a\in A: sa=0\text{ for all }s\in S\}$ equals $pA=\{pa:a\in A\}$ for some projection $p\in A$.

The C*-algebra $C([0,1])$ of continuous functions from the unit interval to the complex numbers has only two projections, hence its projections form a complete lattice, but is certainly not a Baer *-algebra.

Nevertheless, also with the correct definition of a Baer *-ring, there are examples of Baer *-rings that are neither W$^*$-algebras nor locally W$^*$-algebras. Let $F$ be a finite field. Since its multiplication is commutative, the identity is an involution. If $S\subseteq F$ is non-empty, then it is easy to see that $R(S)$ is an ideal. Since $F$ has only trivial ideals, we have either $R(S)=(0)$, whence $R(S)=0F$, or $R(S)=F$, whence $R(S)=1R$. So $F$ is a Baer *-ring.