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No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

Edit Here is an easier example, this time exploiting the nonemptiness of the spectrum (rather than boundedness). Let $M = {\bf C}(x)$, the rational functions in one variable over the complexes, with involution given by $x \mapsto x$ and complex conjugate on the complexes. This determines a positive definite involution, and $M$ and all matrix rings over it are Baer *. However, $M$ cannot be embedded in a Banach algebra, since $M$ is (or unitally contains in the case of matrix rings) a field over than ${\bf C}$: any nonscalar element, $b$, of $M$ satisfies $b - \lambda$ is invertible in $M$ and thus is invertible in any Banach algebra unitally containing it, whence its spectrum (in the Banach algebra) is empty, a contradiction.

No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

Edit Here is an easier example, this time exploiting the nonemptiness of the spectrum (rather than boundedness). Let $M = {\bf C}(x)$, the rational functions in one variable over the complexes, with involution given by $x \mapsto x$ and complex conjugate on the complexes. This determines a positive definite involution, and $M$ and all matrix rings over it are Baer *. However, $M$ cannot be embedded in a Banach algebra, since $M$ is (or unitally contains in the case of matrix rings) a field strictly containing ${\bf C}$: any nonscalar element, $b$, of $M$ satisfies $b - \lambda$ is invertible in $M$ and thus is invertible in any Banach algebra unitally containing it, whence its spectrum (in the Banach algebra) is empty, a contradiction.

No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

Edit Here is an easier example, this time exploiting the nonemptiness of the spectrum. Let $M = {\bf C}(x)$, the rational functions in one variable over the complexes, with involution given by $x \mapsto x$ and complex conjugate on the complexes. This determines a positive definite involution, and $M$ and all matrix rings over it are Baer *. However, $M$ cannot be embedded in a Banach algebra, since $M$ is (or unitally contains in the case of matrix rings) a field over than ${\bf C}$: any nonscalar element, $b$, of $M$ satisfies $b - \lambda$ is invertible in $M$ and thus is invertible in any Banach algebra unitally containing it, whence its spectrum (in the Banach algebra) is empty, a contradiction.

No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

Edit Here is an easier example, this time exploiting the nonemptiness of the spectrum (rather than boundedness). Let $M = {\bf C}(x)$, the rational functions in one variable over the complexes, with involution given by $x \mapsto x$ and complex conjugate on the complexes. This determines a positive definite involution, and $M$ and all matrix rings over it are Baer *. However, $M$ cannot be embedded in a Banach algebra, since $M$ is (or unitally contains in the case of matrix rings) a field over than ${\bf C}$: any nonscalar element, $b$, of $M$ satisfies $b - \lambda$ is invertible in $M$ and thus is invertible in any Banach algebra unitally containing it, whence its spectrum (in the Banach algebra) is empty, a contradiction.

No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

No. Begin with a faithful non-atomic probability measure on a space, $X,\mu$ (e.g., $[0,1]$ with Lebesgue measure). Then $L^{\infty}(X,\mu)$ is of course a vN algebra. Form $M(X,\mu)$, the algebra of measurable functions modulo zero ae. As is well-known and easy to verify, $M \equiv M(X,\mu)$ is the classical ring of quotients of $L^{\infty}(X,\mu)$, and is itself a vN regular Baer *-ring. We will show $M$ cannot be unitally embedded in any Banach algebra.

There exists a sequence of nonzero orthogonal projections $p_n \in L^{\infty}$ whose supremum is one, and if $\lambda_n$ is a sequence of complex numbers, there exists $b \in M(X,\mu)$ such that $bp_n =\lambda p_n$ (formally, $b = \sum \lambda_n p_n$). In particular, $b - \lambda_n$ is a zero divisor in $M$. Hence if $b$ belonged to a Banach algebra containing $L^{\infty}(X,\mu)$, it would follow that all the $\lambda_n$ belong to the spectrum of $b$ (as an element of the Banach algebra). We can simply choose the sequence $(\lambda_n)$ to be unbounded, contradicting the fact that the spectrum is bounded.

This yields a commutative example. For a simple example, replace $L^{\infty}$ by a type II vN factor, $N$, and let $M$ be its associated regular ring; it is again a Baer *-ring. It is also true (but more difficult to prove) that $M$ is a (the) classical ring of quotients of $N$. We can certainly embed $L^{\infty}$ as a masa in $N$, and the same element $b$ can be used.

A necessary condition for embeddability is that $*$ be definite, that is, $\sum a_i a_i* = 0$ implies all $a_i$ are $0$, which also follows from all matrix rings themselves being definite. Sufficient and almost necessary for there to be a unital $*$-homomorphism (of a definite Baer $*$-ring) to a $B(H)$ is that for all $a \in A$, there should exist finitely many $a_j \in A$ such that $aa*+\sum a_j a_j* $ is a (complex) scalar multiple of the identity.

In general, the bounded subalgebra of a Baer $*$-algebra (the set of elements $a$ for which such $a_j$ exist; it is a $*$-subalgebra) admits a seminorm, which after its kernel is factored out, can be completed to a C*-algebra; hence sufficiency of the criterion—however, in general, we don't get that the image in a $B(H)$ is a Baer * subring.

In the finite-dimensional case, assuming the involution is definite, the algebra must be semisimple, and then the involution can be converted to a standard one.

Edit Here is an easier example, this time exploiting the nonemptiness of the spectrum. Let $M = {\bf C}(x)$, the rational functions in one variable over the complexes, with involution given by $x \mapsto x$ and complex conjugate on the complexes. This determines a positive definite involution, and $M$ and all matrix rings over it are Baer *. However, $M$ cannot be embedded in a Banach algebra, since $M$ is (or unitally contains in the case of matrix rings) a field over than ${\bf C}$: any nonscalar element, $b$, of $M$ satisfies $b - \lambda$ is invertible in $M$ and thus is invertible in any Banach algebra unitally containing it, whence its spectrum (in the Banach algebra) is empty, a contradiction.