Questions tagged [von-neumann-algebras]

A von Neumann algebra is a unital -subalgebra of the algebra of bounded operators on a Hilbert space, closed in the weak operator topology. Also called a $W^$-algebra, and may be regarded as a non-commutative generalization of $L^\infty$ space. These algebras are extensively used in knot theory, non-commutative geometry, and quantum field theory.

A von Neumann algebra is a unital *-subalgebra of the algebra of bounded operators on a Hilbert space, closed in the weak operator topology. Also called a $W^*$-algebra, and may be regarded as a non-commutative generalization of $L^\infty$ space. These algebras are extensively used in knot theory, non-commutative geometry, and quantum field theory.

1137 questions
5
votes
2 answers

von Neumann algebra factors

I have two basic questions on von Neumann algebras : 1) If a von Neumann algebra $M$ is simple (only trivial ideals), is it a factor (i.e. $M\cap M'=\mathbb C \cdot 1_M$ ?). 2) If the reduced group algebra of a discrete group $C_r(\Gamma)$ (i.e. the…
vonN
  • 51
4
votes
1 answer

Radon- Nikodym Theorem For von Neumann algebras

Sakai's Radon Nikodym Theorem for von Neumann algebra goes as follows: Let $\phi$ and $\psi$ be normal forms on a $von$ $Neumann$ $algebra$ $M$ such that $\phi$ $\leq$ $\psi$. Then $\exists$ $a$ $\in$ $M$ uniquely determined by the following…
Ester
  • 3,057
  • 2
  • 23
  • 34
3
votes
0 answers

von Neumann algebra isomorphism

Let $\pi_1$ and $\pi_2$ be two representations of a unital $C^*$-algebra $A$ on Hilbert spaces $H_1$ and $H_2$ respectively. Assume that $\pi_1$ and $\pi_2$ are unitarily equivalent. So there exists a unitary $U\in B(H_1,H_2)$ such that…
budi
  • 1,780
3
votes
1 answer

Isomorphism of Von Neumann algebras

In general what is it meant to say two Von Neumann Algebras are isomorphic? Is it an injective and surjective *-homomorphism which is also strongly continuous or one which is also normal? I would suppose the correct definition would be normal as…
sirjoe
  • 875
3
votes
2 answers

Does weak convergence imply strong convergence in a von Neumann Algebra

Let $H$ be an Hilbert space, and $A$ a von Neumann subalgebra of $B(H)$. It is made abundantly clear that this is equivalent to: $A$ is a s-o closed *-subalgebra, and also $A$ is w-o closed *-subalgebra. What is not clear (to me), is whether the…
RS8
  • 41
3
votes
1 answer

Is $\mathcal{B}(H)$ amenable for any Hilbert space H?

This question is motivated by the following, given $M$ an amenable Von Neumann algebra and a Hilbert space H, is $M \overline{\otimes} \mathcal{B}(H)$ amenable? I have two main questions: 1) Given any finite dimensional Hilbert space H, is…
ADA
  • 1,653
3
votes
1 answer

Equivalent projections and 2-sided ideals in von Neumann algebras

Let $M$ be a von Neumann algebra. For given $x$ in $M$, we put $I_x=MxM$ (the algebraic ideal generated by $x$). We know that if two projections $p$ and $q$ are equivalent (that is $\exists u\in M$ with $p=u^*u,q=uu^*$) then $I_p=I_q$. What about…
ABB
  • 1,998
2
votes
1 answer

Is the double commutant of a unital $\ast$ algebra equal to its double dual?

The Sherman Takeda theorem says that the double dual $A^{\ast \ast}$ of a $C^{\ast}$ algebra $A$ can be identified as a Banach space with $A''$, the enveloping von Neumann algebra of $A$, i.e. the weak operator topology closure of $\pi_{U}(A)$ in…
Arundhathi
  • 1,453
2
votes
1 answer

Compression of a von Neumann algebra

Let $M$ be a von Neumann algebra on $B(H)$. and let $p \in M$ be a projection How can we prove that the compression $pMp$ is a von Neumann algebra on $B(pH)$? I saw a result that if $p \in M$ is a projection, then $(pMp)'=pM'p$. Using this can we…
budi
  • 1,780
2
votes
0 answers

is $L^2(M\ltimes G)\cong \oplus_G L^2(M)$ as Hilbert spaces

Let $M$ be a $II_1$ factor then I want to show that if $G$ is a finite group acting on $M$ then considering the GNS of the crossed product with respect to the trace one has the unitary isomorphism $L^2(M\ltimes G)\cong \oplus_G L^2(M)$. It seems…
sirjoe
  • 875
2
votes
1 answer

Generated von neumann algebra and commutant

Let $(\mathfrak{M}_i, \mathcal{H})_i$ a family of Von neumann algebras acting on the same Hilbert space. The von Neumann algebra $$\bigcup_{i}\mathfrak{M}_i = \bigcap\{\mathfrak{M} \, : \, \mathfrak{M} \supset \mathfrak{M}_i \, \, \, \forall…
James Arten
  • 1,953
  • 1
  • 8
  • 20
2
votes
1 answer

Image of a normal $*$-homomorphism

Let $\mathcal M$ be a von Neumann algebra. Let $\pi:\mathcal M\to\mathcal M$ be a normal $*$-homomorphism Is $\pi(\mathcal M)$ again a von Neuman algebra? By [J. Dixmier, Les algebres d’operateurs dans l’Espace Hilbertien, 2nd ed., Gauthier-Vallars,…
2
votes
1 answer

Definition of von Neumann subalgebra

Let $\mathcal M\subseteq B(\mathcal H)$ be a von Neumann algebra. Then, what should be natural definition of a von Neumann subalgebra?
2
votes
1 answer

Spectral theorem for von Neumann algebras

What is the motivation of studying spectral theorem? Is it just to know when two normal operators are unitary equivalent or not? Further, what is the viewpoint of seeing spectral theorem in terms of von Neumann algebras??
user548061
2
votes
1 answer

Zhu's definition of type 1 von Neumann algebras

in Zhu's Book on operator algebras he defines a von Neumann algebra $M$ to be of type 1 iff for every central projection $0\neq p \in M$ there is an abelian projection $0\neq q\leq p$ I'm told this is equivalent to $M$ is of type 1 iff for every…
user412810
1
2 3