Abstract Von Neumann Algebras

I have just read this question Is a von Neumann algebra just a C*-algebra which is generated by its projections? and am wondering about Robert Israel's answer when he says that a subalgebra of $C(X)$ is not a Von Neumann algebra. Some abstract definition must be being used, but he also says weak closure (by which I assume he means weak operator closure) and that only exists in $B(H)$. When someone says that a C* algebra $A$ is a von Neumann algebra, is he saying that $A$ is $*$-isomorphic (i.e. a bijective map preserving the 3 algebraic opera...Read more

Do all *-isomorphisms between von Neumann algebras preserve strong operator topology?

Do all $*$-isomorphisms between von Neumann algebras preserve the strong operator topology?Seems clearly true for $*$-isomorphisms with a unitary implementation, but I don't see the answer for other cases ... perhaps there is an easy argument from the fact that von Neumann algebras are closed in this topology, but I've spent a while looking for one and don't see it....Read more

operator algebras - the semigroup $D(\Bbb C)$

We define $D(\Bbb C)=\cup_n P(M_n(\Bbb C))/\sim$,where $P(M_n(\Bbb C)) $ is the set of projections in $M_n(\Bbb C)$.$\sim$ is the equivalence relation as follows:suppose $p$ is a projection in $P(M_n(\Bbb C))$,$q$ is a projection in $P(M_m(\Bbb C))$, $p\sim q$ if there is an element $v $ in $M_{m,n}(\Bbb C^n)$ with $p=v^*v,q=vv^*$.How to show that $D(\Bbb C)\cong \Bbb Z^{+}$?...Read more

There are 3 points in the spectrum of some self-adjoint element of a non-unital C*-algebra.

Let $A$ be a non-unital C*-algebra. I would like to know a simple way to show that $A$ contains a self-adjoint element whose spectrum has at least $3$ elements. Note that the spectrum of an element of $A$ is by definition the spectrum inside the unitization of $A$, and $0$ is in each spectrum. Thus the problem is to have 2 nonzero elements of the spectrum of some self-adjoint element. A self-adjoint element with only 1 nonzero element in its spectrum is a scalar multiple of a projection, so the problem is equivalent to showing that there ...Read more

c star algebras - operator norm of completely bounded map

Definition: An operator space $X$ is a closed subspace of a C*-algebra $A$. For each $n\in\mathbb{N}$, $\mathbb{M}_{n}(X)$ inherits a norm from $\mathbb{M}_{n}(A)$. Let $\phi$ be a linear map from an operator space $X\subset A$ into an operator space $Y\subset B$, $\phi$ is completely bounded (cb) if $$\|\phi\|_{cb}=sup_{n\in\mathbb N}\|\phi_{n}: \mathbb{M}_{n}(X)\to\mathbb{M}_{n}(Y)\|,$$ where $\phi_{n}$ is defined by $\phi_{n}([x_{i,j}])=[\phi(x_{i,j})]$ for $x=[x_{i,j}]\in\mathbb{M}_{n}(X)$.My question is how to comprehend $\|\phi\|_{cb}=\|\...Read more

$B$ stably finite simple $C^*$-algebra $\Rightarrow$ $B\otimes \mathcal{K}$ contains no infinite projections

A unital $C^*$-Algebra $B$ is called stably finite, if, for all $n\in\mathbb{N}$, $1_{M_n}\in M_n(B)$ is a finite projection.Claim: If $B$ is a unital, simple, stably finite $C^*$-Algebra, then $B\otimes \mathcal{K}$ contains no infinite projections. I want to prove this, but I don't really know, how ( I only know the definition) Here, $\mathcal{K}$ denotes the compact operators on a separable, infinite dimensional Hilbert space. Write $\mathcal{K}=\varinjlim M_n$ and $M_n(B)=B\otimes M_n$, then $B\otimes \mathbb{K}=B\otimes \varinjlim M_n=\var...Read more

minimal projections in finite dimensional von Neumann algebras

The algebras I'm working with are defined as follows Let $\mathcal{H}$ be a Hilbert space of finite dimension and denote by $\mathcal{B}(\mathcal{H})$ the bounded operators on $\mathcal{H}$. A matrix algebra $\mathcal{M}$ is a *-subalgebra of $\mathcal{B}(\mathcal{H})$.Am I correct in saying that these are exactly the finite-dimensional von Neumann-algebras? I need to prove the following: The set of minimal projections in a commutative matrix algebra $\mathcal{M}$ is a finite set $\{p_1,\dots,p_k \}$ of pairwise orthogonal projections.Note th...Read more

Suspension of Cuntz algebra is traceless

I saw a conclusion in a reference book: the suspension of the Cuntz algebra $C_0((0,1))\otimes \mathcal O_2$ has no tracial states.My thought: there are many tracial states on $C_0((0,1))$. We take a tracial state $\tau$ on $C_0((0,1))$, then we can define a tracial state $\tilde{\tau}$ on $C_0((0,1))\otimes \mathcal O_2$ as follows:$$\tilde{\tau} (x\otimes y)=\tau(x),\forall x\in C_0((0,1)),y\in \mathcal O_2.$$Can anyone point out the mistake, thanks!...Read more

Isometries in type III von Neumann algebra

Consider a type III von Neumann algebra and an isometry $W$. How does one show that there exists a sequence of unitaries $U_n$ that converge strongly to $W$?Connes-Stormer in their "Homogeneity of the state space of Factor of type III$_1$" last page say: "Since $M$ is of type of III standard arguments show that we can find a sequence of unitaries in $M$ converging strongly to $W$."...Read more

operator algebras - Infinite tensor product and partial isometries

Consider a countably infinite tensor product of the algebra of complex $2\times 2$ matrices:$\otimes_{k=1}^\infty M^{(k)}_2 (\mathbb{C})$, and two different states: 1) $\phi$ that is the tensor product of the density matrix $\frac{1}{2}\left(\begin{matrix} 1 & 0 \\0 & 1 \end{matrix}\right)$ at each site.2) $\psi$ that is the tensor product of the density matrix $\frac{1}{1+\lambda}\left(\begin{matrix} 1 & 0 \\0 & \lambda \end{matrix}\right)$ with$\lambda\neq 0$ at each site.The von Neumann algebra one obtains using $\psi$ is the...Read more

Von Neumann algebra generated by a non-self-adjoint operator

Let $A$ be a bounded linear operator on a separable Hilbert space ${\cal H}$, and suppose that $A$ is distinct from its adjoint $A^*$. Question: Can the double commutant of $A$ be distinct from the double commutant of $\{A,A^*\}$? If so, is there a simple example?This question was inspired by the following statement from section 3.3 in Vaughan Jones (2009), Von Neumann Algebras (https://math.berkeley.edu/~vfr/VonNeumann2009.pdf): If $S \subseteq {\cal B(H)}$, we call $(S \cup S^*)''$ the von Neumann algebra generated by $S$.I don't know if thi...Read more

quantum mechanics - Prove: the density operator of a pure state has exactly one non-zero eigenvalue equal to unity

What is the proper way of proving : the density operator $\hat{\rho}$ of a pure state has exactly one non-zero eigenvalue and it is unity, i.e,the density matrix takes the form (after diagonalizing):\begin{equation}\hat{\rho}={\begin{bmatrix}1 & 0 & \cdots & 0 \\0 & 0 & \cdots & 0 \\\vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & 0 \\\end{bmatrix}}\end{equation}For mixed state: $\hat{\rho}=\sum \limits_{i}P_{i}|\psi_{i}\rangle\langle\psi_{i}|$ For any state: $Tr(\hat{\rho})=\sum\limits_{i}P_{...Read more

spectral theory - Brown measure of left shift operator

Let $L$ be the left shift operator on $\ell^2(\mathbb{Z})$ with trace $\tau(T) := \langle T \delta_0, \delta_0 \rangle$. How can I show that the Brown measure of $L$ is the uniform measure on the unit circle?The Brown measure is defined as follows: For each $z \in \mathbb{C}$ define $\nu_{z}$ to be the spectral measure of the (self-adjoint) operator $(L - z)^*(L-z)$. Then let $$f(z) = \frac{1}{2} \int_0^\infty \log x \,d\nu_z(x).$$ Then the Brown measure is defined as $$\mu_L := \frac{1}{2\pi} \Delta f,$$ where the Laplacian is taken in the...Read more