0
votes
2answers
27 views

Some doubts concerning spectral theory.

Probably I'm saying something wrong (that's why the conclusions are strange) so please correct me! There is the continuous functional calculus for a normal element $N$ of a C*-Algebra. This means ...
1
vote
1answer
21 views

If $N$ is normal, $W^\ast(N)=\{N\}''$

I want to prove that if $N$ is normal, then $W^\ast(N)=\{N\}''$, where $W^\ast(N)$ is the Von Neumann algebra generated by $N$ and $\{N\}''$ is the bicommutant of $N$. for the inclusion $W^\ast(N) ...
0
votes
1answer
29 views

norm of a matrix that its entries are operators in B(H).

Let S is a subset of B(H). Define $M_2(S)=\{T= \left( \begin{array}{ccc} A & B \\ C & D \\ \end{array} \right) : A,B,C,D \in S\}$. what is the relationship between $||T||$ and ...
0
votes
1answer
25 views

polar decomposition of multiplicative operator on L^2 induced by identity function.

We know that every operator in B(H) has a polar decomposition. $T=VP$ that $P=|T|$ and V is a partial isometry with initial space closure of ImP and final space ImT. How can i obtain polar ...
0
votes
0answers
49 views

Closed unit ball of B(H) is not compact in strong operator topology of B(H).

In operator theory we prove that closed unit ball of B(H) is compact in weak operator topology and is closed in strong operator topology. But a book of operator theory states that closed unit ball of ...
0
votes
1answer
39 views

The map $T\longmapsto \|T\|$ is not continuous in the strong operator topology of $\mathscr B(H)$

In the context of Strong and Weak operator topologies on $\mathscr B(H)$ there is an statement that says: the map on $\mathscr B(H)$ that $T\longmapsto \|T\|$ is not continuous in the strong operator ...
1
vote
1answer
23 views

Why is this statement true for two equivalent projections in $B(H)$?

In a book of operator theory it is stated that two projections $P$ and $Q$ in a von Neumann algebra $A$ are equivalent if there exist $V$ in $A$ that $V^*V=P$ and $VV^*=Q$. After this definition, it ...
2
votes
1answer
55 views

Why is the weak operator closure of a commutative $\boldsymbol{C^*\!\!\!\!-}$algebra also commutative?

In a book on Operator Theory there is the following statement: If $\mathscr A$ is a commutative $C^*$-subalgebra of $\mathscr B(\mathcal H)$, where $\mathcal H$ is a Hilbert space, then the weak ...
1
vote
2answers
33 views

Strong closure of a C*-algebra of operators.

In Arveson's book, the Kaplansky density theorem is proved in order to have this corollary: "Let $A$ be a self-adjoint algebra of operators on a separable Hilbert space $H$. Then for every operator ...
1
vote
1answer
26 views

About what happens to eigenspace under functional calculus for Unbounded Operator

Let $T$ be an unbounded self adjoint positive operator on a Hilbert Space $\mathcal{H}$. Let $x \in \mathcal{H}$ be a vector such that $Tx = x$. Is it true that $T^{\frac{1}{2}} x = x$. For what $f$ ...
2
votes
1answer
35 views

Zhou operator theory book, Kaplanskys formula

In Zhou's operator theory book, Kaplanskys formula has stated that if $P$ and $Q$ are projection in a von neumann algebra $A$ acting on $H$, then $P\vee Q-Q\sim P-P\wedge Q$. In the proof, it says ...
2
votes
1answer
37 views

A question about $C^\ast$-algebra

In Kadinson's book Fundamentals of The Theory of Operator Algebra, when the author proved the Theorem 7.2.1, he let $V$ be an extreme point of the unit ball of $C^\ast$-algebra $\cal{U}$, $h$ be a ...
1
vote
2answers
40 views

The unitary implementation of $*$-isomorphism of $B(H)$

Is it possible to construct $*$-isomorphism of (factor von Neumann) algebra $B(H)$ which is not unitary implementable?
1
vote
1answer
47 views

Automorphism of $W^*$ algebra

Let $\mathfrak{A}$ be von Neumann algebra. It is in particular $C^*$ algebra. Is it true that every $*$-isomorphism of $\mathfrak{A}$ is also $W^*-$isomorphism? (Note that every $*$-isomorphism of ...
1
vote
1answer
22 views

A separating set which is not cyclic

Let $H=L^2[0,1]$ , $T_g$ be the multiplication operator on $H$, i.e. $f\to fg$ . Let $A$ be the set of the $T_g$ as $g$ runs through the set of polynomials with complex coefficients. Let $h$ be te ...
0
votes
1answer
24 views

Can we find an invertible projection in an arbitrary von Neumann algebra?

I am looking for an answer for this question: Let $\mathcal{A}$ be an arbitrary von Neumann algebra, can we say there is an invertible projection ($P\neq I$) in $\mathcal{A}$? I think, if there is ...
0
votes
0answers
32 views

spectral projections

Let $A$ be a von Neumann algebra, and $T$ be an hermitian element of $A$. Show that the spectral projections of $ T $ belong to $A$. Proof: the spectral projections of $ T $ commute with every ...
0
votes
1answer
33 views

Abelian von Neumann algebra

Let $M$ be a family of commutative normal operators which is closed under adjoint. clearly $M\subset M'$, but I do not know why $M'\subset M^{''}$? and how can conclude that $M^{''}$ is abelian?
3
votes
1answer
48 views

A simple question about completely positive linear maps

Let $A$ be the C*-algebra and $M_{n}(A)$ be the C*-algebra of $n\times n$ matrices with entries in $A$. We use $(a_{ij})$ to denote the element of $M_{n}(A)$. My question is: For every $a\in A$, ...
3
votes
1answer
112 views

Question about projections on Hilbert space

Let $P_i$ be projections from a Hilbert space $\cal{H}$ to its closed subspace $\cal{H}_i$, $i=1,2,\cdots,n$, such that $\sum^n_{i=1} P_i$ is also a projection. And let $P$ be a projection from ...
0
votes
1answer
34 views

$L(H)$ and functions belong to predual of this space

Does every W*-continuous linear functional belong to $L^1(H)$? is it true? I cannot understand about it. Please regard me.
3
votes
1answer
167 views

Does an irreducible operator generate an exact $C^{*}$-algebra?

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Definition : An operator $T \in B(H)$ is irreducible if $W^{*}(T)=B(H)$. Definition : A ...
2
votes
1answer
123 views

Does an irreducible operator generate a nuclear $C^{*}$-algebra?

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Definition : An operator $T \in B(H)$ is irreducible (Halmos 1968) if its commutant $\{ T\}'$ ...
2
votes
1answer
87 views

Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous?

Let $H$ be an infinite dimensional separable Hilbert space. Definition: The commutant $\mathcal{S}'$ of a subset $\mathcal{S} \subset B(H)$ is $ \{A \in B(H) : AB=BA \ , \ \forall B \in \mathcal{S} ...
3
votes
1answer
135 views

On the use of nets when defining operator topologies

Let's consider the strong operator topology and the weak operator topology on bounded operators of a infinite-dimensional Hilbert space $H$. When they define these operator topologies, some authors ...
3
votes
1answer
161 views

The span of the orthorgonal projections is norm dense in $B(H)$

This is a question in my functional analysis book. How to use the spectral theorem to prove that the span of the orthogonal projections is norm dense in $B(H)$?
3
votes
1answer
99 views

Counterexample for a polar decomposition in von Neumann and $C^\ast$ algebras

For a von Neumann algebra, we have that partial isometry and positive operator of an operator in its polar decomposition belongs to the algebra, but in a $C^\ast$ algebra this may not be true. Can ...
9
votes
1answer
173 views

Do we have Maximal Abelian Algebras (MAAs)?

Let $\mathcal{H}$ be a Hilbert space and $B(\mathcal{H})$ the algebra of bounded linear operators on $\mathcal{H}$. A MASA $\mathcal{M}$ is a subalgebra of $B(\mathcal{H})$ that is abelian and ...
1
vote
1answer
187 views

Two questions from Dixmier's book on Von Neumann algebras

It seems something is going wrong with the preview I linked in some of my previous questions, so I will just type out the question. I am having trouble with Dixmier's proof of Corollary 5 on p. 46. ...
0
votes
0answers
90 views

weak closures of ideals [duplicate]

Possible Duplicate: Two questions from Dixmier's book on Von Neumann algebras On p. 46-47 in Dixmier's book on Von Neumann Algebras, which I just realized can be accessed through this ...
1
vote
0answers
93 views

When the ultrastrong closure of a *-algebra contains the double commutant

As lemma 6 on p.44 of Dixmier's book on Von Neumann algebras, he states that if $A$ is a *-algebra (i.e. possibly without identity, not necessarily closed in any topology) of operators in $B(H)$ such ...
11
votes
1answer
411 views

Renorming $\mathcal{B}(\mathcal{H})$?

Consider the Banach space of all bounded operators $\mathcal{B}(\mathcal{H})$ on a (separable if you wish) Hilbert space $\mathcal{H}$ with the operator norm. Can we renorm this space to a strictly ...
1
vote
1answer
272 views

Complementability of von Neumann algebras

Is every von Neumann algebra complemented in its bidual? It is certainly true for commutative von Neumann algebras as their spectrum is hyperstonian. Is it 1-complemented?