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17 votes
5 answers
2k views

If two projections are close, then they are unitarily equivalent

Given two projections $p,q\in B(H)$, it is well-known that if $\|p-q\|<1$, then there exists a unitary $u\in B(H)$ with $q=upu^*$. The proof that immediately occurs to me uses comparison of ...
1 vote
1 answer
180 views

Conditioning a $\mathrm{C}^*$-algebra state with infinite precision

This question (and a second part) have been asked at MSE and gone through two bounties without an answer. I have been beating my head at it for a while without success. Let $\mathcal{A}$ be a unital $\...
8 votes
0 answers
952 views

About generator and isomorphism problems for free groups operator algebras

Let $H$ be an infinite dimensional separable Hilbert space. The $C^{*}$-algebras and von Neumann here unital and subalgebras of $B(H)$. Definition : Let $\mathcal{A}$ be $C^{*}$-algebra (resp. a von ...
6 votes
1 answer
643 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} \}...
1 vote
1 answer
129 views

Which elements live in the image of the canonical map $X \otimes_\mathcal{F} M \to B(M_*, X)$?

Let $X\subseteq B(H)$ be an operator system and let $M\subseteq B(K)$ be a von Neumann algebra. We form the Fubini-tensor product $$X \otimes_\mathcal{F} M := \{z \in B(H\otimes K): (\sigma\otimes \...
3 votes
1 answer
255 views

Takesaki: Lemma about enveloping von Neumann algebra

Consider the following lemma with proof from Takesaki's book "Theory of operator algebra I" (p121): It appears to me that Takesaki claims at the end of the proof that $\pi(A)_1$ is $\sigma$-...
5 votes
1 answer
322 views

Takesaki's proof of the Kaplansky density theorem

Consider the following fragment from Takesaki's book "Theory of operator algebra I": Why is the boxed sentence true? It looks like they replace $A$ by its strong$^*$-closure. Is this ...
3 votes
1 answer
306 views

Opposite $C^*$ algebras

$\DeclareMathOperator\op{op}$Let $A$ be a $C^*$-algebra. We know that $A$ admits a natural operator space structure, namely the operator space structure induced by any faithful $*$-representation of $...
2 votes
0 answers
135 views

On crossed product of L^{P} spaces

Let $M$ be a von Neumann algebra with faithful normal state $\varphi$, and $G$ be a group action on $M$ preserving $\varphi$. Then is it true \begin{align*} L^{p}(M\rtimes G, \varphi^{M\rtimes G})\...
3 votes
0 answers
178 views

A point concerning absolute value of functionals

Let $M$ be a von Neumann sub-algebra in $B(H)$. Let $\phi$ be a normal functional on $M$. Assume $\psi$ is a normal functional on $B(H)$ with $\psi_{|_M}=\phi$ (note that $\phi$ and $\psi$ may have ...
5 votes
2 answers
988 views

Projections in a W*-algebra as a continuous lattice?

A continuous lattice is a complete lattice $L$ in which every element $y$ is equal to $\bigvee${$x \in L \mid x \ll y$} where $x \ll y$ ("x approximates y" or "x is way below y") if for any directed ...
3 votes
1 answer
481 views

Question on structure of von Neumann algebras, clarification in Conway's "A course in operator theory"

I was reading the section on the structure of type I von Neumann algebras in John B. Conway's "A course in operator theory" and a few questions about certain definitions and references arose, I was ...
12 votes
1 answer
901 views

Is there a proof that the $C^{*}$-algebras don't see the invariant subspace problem?

This post is an appendix of this one. Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Invariant subspace problem: Let $T \in B(H)$. Is ...
6 votes
1 answer
680 views

Is there an operator algebraic reformulation of the invariant subspace problem?

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant ...