All Questions
Tagged with oa.operator-algebras fa.functional-analysis
778 questions
2
votes
1
answer
362
views
Quasinilpotent example [duplicate]
Possible Duplicate:
Quasinilpotent operator
Do you know any example of a quasinilpotent operator such that every its power is non-compact?
Of course direct sum of nilpotent operators(or Volterra ...
- 21
33
votes
3
answers
3k
views
Reference request for translating from Top to C*-alg
Some recent questions on MO (for example, Do subalgebras of C(X) admit a description in terms of the compact Hausdorff space X?) have been about Gelfand duality — namely, that the categories of ...
- 18.7k
7
votes
1
answer
682
views
$c_0$-direct sum of $\mathcal{K}(\mathcal{H})$
Let $\mathcal{K}(\mathcal{H})$ be the C*-algebra of compact operators on a Hilbert space $\mathcal{H}$. I am interested in the ($c_0$-)sum
$A=\sum \mathcal{K}(\mathcal{H})$
of countably many ...
- 113
4
votes
1
answer
773
views
Algebraically simple Banach algebras
There are plenty of semi-simple Banach algebras - this broad class includes C*-algebras and algebras of bounded operators on a given Banach space. On the other hand, it seems unlikely to me that there ...
- 113
10
votes
1
answer
783
views
When do tensor products of C*-algebras commute with colimits?
Let $I$ be a filtered poset, which you should think of as being huge. Let $A_i$ be an $I$-diagram of $C^{\star}$-algebras and let $A$ be the colimit of this diagram; if necessary, we can also assume ...
- 976
20
votes
2
answers
1k
views
P-adic C* algebras
I understand that there is a definition of p-adic Banach algebras and that a significant amount of functional analysis can be developed in the non-archimedean setting. Is there a p-adic version of C*-...
- 373
8
votes
1
answer
749
views
Is $SU(\infty)$ amenable?
We can write the finitary special unitary group $SU(\infty)$ as the direct limit
$\varinjlim SU(n)$ of ordinary special unitary groups. These groups $SU(n)$ are compact, thus amenable. In other ...
- 606
2
votes
1
answer
199
views
Uniqueness of free complements
Let $A,B$ be subfactors of a II$_1$ factor $M$ with $A*B\simeq M$. That is, $A$ and $B$ are freely independent with respect to the trace and $M\simeq A\vee B$. We'll call $B$ a free complement for $A$ ...
- 1,411
12
votes
2
answers
547
views
Balls in spaces of operators
I am interested in some geometrical aspects of spaces $L(E)$, of bounded operators on a given Banach space $E$. I am unable to estimate if my problem deserves to be asked at MO, but let me try.
Is ...
- 121
35
votes
2
answers
9k
views
tr(ab) = tr(ba)?
It is well known that given two Hilbert-Schmidt operators $a$ and $b$ on a Hilbert space $H$, their product is trace class and $tr(ab)=tr(ba)$. A similar result holds for $a$ bounded and $b$ trace ...
- 43.2k
3
votes
0
answers
455
views
Morphism of von Neumann Algebras
Hello,
Is there a counterexample to the following statement:
let $A,B$ two von Neumann algebras, every morphism $A \rightarrow B$ of $C^* $-algebras is a $W^*$-homomorphism ?
( a $W^* $-...
- 663
6
votes
4
answers
8k
views
Characterization of the non-negative definite functions $f(x,y)$
The common definition of the non-negative definite functions is as follows:
Definition 1: A continuous complex-valued function $f(x)$ is called non-negative definite, if for any real numbers $x_1,\...
- 1,649
2
votes
1
answer
535
views
about decomposition of a non-negative definite operators
Hello,
Many years before, I had the following problem.
We first give a definition. Given a non-negative definite real-valued definite matrix $n^2\times n^2$ matrix $M$, it is called separable if it ...
- 1,649
5
votes
1
answer
673
views
Unbounded representations of Banach algebras
Can a representation of a Banach algebra be unbounded?
To clarify, I'm not asking about a representation as unbounded operators, but
rather a homomorphism $\pi: A \to B(H)$ for some Hilbert space $H$,...
- 275
22
votes
1
answer
745
views
The Mackey Topology on a Von Neumann Algebra
Every von Neumann algebra $\mathcal M$ is the dual of a unique Banach space $\mathcal M_* $. The Mackey topology on $\mathcal M$ is the topology of uniform convergence on weakly compact subsets of $\...
- 1,199
3
votes
1
answer
1k
views
Self-adjoint bounded operator, resolution of the identity, def. of the diagonal
Let $A$ be a self adjoint bounded linear operator with a continuous spectrum
$\sigma(A)=[a,b]$ which acts on a separable Hilbert space. Let
$E_\lambda$ be its resolution of the identity.
For ...
3
votes
1
answer
785
views
Maximal ideals of some algebras
This question comes up after I read the chapter 7 : Banach algebras and spectrum theory from Conway's book. As we have known that if $X$ is compact space, then all the maximal ideals of $C(X)=\{ f : X\...
- 281
1
vote
2
answers
177
views
Restriction on the coefficients for an operator in the free group factor $ L(\mathbb{F}_2) $
Let $\mathbb{F}_2$ denotes the free group generated by a,b, denote this group by $G$. Then consider the von Neumann algebra $L(G)$ generated by the family
$\{L_{x_g} : g \in G\}$, here, with $g \in G$...
- 1,528
23
votes
4
answers
2k
views
Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)?
I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{\mathrm{Tr}(A^* A)/n}$.
My question is whether a $k$-uple of hermitian matrices that are almost ...
- 9,486
4
votes
1
answer
874
views
equality in noncommutative Hölder inequality
Let $1\leq p,q,r\leq \infty$ such that $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$. Let $S_p$ denote the Schatten space. For any $x\in S_p$ and any $y\in S_q$ we have
$$
||xy||_{S_r} \leq ||x||_{S_p}||y||_{...
- 1,222
7
votes
1
answer
592
views
topologies on U(H)
There are many topologies on the algebra $B(H)$ of bounded operators on Hilbert space:
the weak, strong, ultraweak (also called σ-weak), ultrastrong (also called σ-strong), and some more......
- 43.2k
5
votes
2
answers
491
views
Is independence meaningful for commutative $C^*$-algebras?
I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate.
Let's say I have two self-adjoint operators on a Hilbert space and ...
- 2,243
8
votes
1
answer
2k
views
Limits of Nilpotent and Quasi-nilpotent Operators in a $\mathrm{II}_1$-factor
A bounded operator $A$ in a Hilbert space is called nilpotent if there exists $n$ such that $A^{n}=0$. An operator is called quasi-nilpotent iff
$$
\limsup_{n\to\infty}{ \|A^{n}\|^{1/n}}=0.
$$
Every ...
- 3,626
9
votes
2
answers
928
views
Property (T) for pseudogroups
Let $H$ be a Hilbert space, $S(H)$ be the inverse semigroup (pseudogroup) of linear maps between (closed) subspaces of $H$ preserving the dot product (the operation is composition of partial maps). ...
user6976
9
votes
2
answers
1k
views
Which Banach algebras are group algebras?
Given a locally compact Hausdorff group $G$, one can construct several Banach star-algebras using $G$ (and its associated Haar measure): $L^1 (G)$, $M(G)$ (regular complex measures on $G$), $L^{\infty}...
- 4,874
0
votes
3
answers
752
views
center of the algebra of bounded operators [closed]
Suppose that $X$ is a Banach space. How to prove that the center of the algebra $B(X)$ (the algebra of bounded operators on $X$) consists only of operators of the form $aI$, where $a$ is scalar and $I$...
- 33
12
votes
1
answer
329
views
Ideals in smooth subalgebras of C*-algebras
Let $B$ be a $C^{*}$-algebra and $\mathcal{B}$ a dense *-subalgebra stable under holomorphic functional calculus and $C^{1}$-functional calculus for selfadjoint elements. Also, $\mathcal{B}$ is a ...
- 213
5
votes
1
answer
410
views
Is the unitary group of $l^2(A)$ with the strict topology contractible?
Let $A$ be a $C^*$-algebra with countable approximate unit. Let $\mathbb{K}$ denote the compact operators on a separable Hilbert space. Mingo and later Cuntz and Higson have shown that the unitary ...
- 7,637
13
votes
1
answer
404
views
Self map of unitary group
Let $H$ be a Hilbert space and let $u_1 \in U(H)$ be a unitary operator on $H$. Consider the self-map $w: U(H) \to U(H)$ which is given by
$$w(v) := v^2 u_1 v^{-1}.$$
Since $U(H)$ is connected, there ...
- 25.5k
18
votes
1
answer
1k
views
Commuting unitaries
Is the following true:
For every unit vectors $x_1,..., x_n$, $y_1,..., y_n$ in $\mathbb{C}^k$
there exist a Hilbert space $H$, unitary operators $U_1,...,U_n$ and $V_1,...,V_n$ in $B(H)$ and unit ...
- 4,705
11
votes
1
answer
8k
views
Double Orthogonal Complement
Let $V$ be a complex inner product space. If $W$ is a closed subspace of $V$, we may define $W^\perp$ to be the subspace of all vectors $v \in V$ such that $\langle v | w\rangle =0$ for all $w \in W$....
- 1,199
4
votes
2
answers
452
views
Is every bounded representation of Z unitarisable when all sets are measurable?
For the purpose of this question, a group is amenable iff there exists a Følner sequence.
Dixmier unitarisability problem asks whether a (countable discrete) group G is amenable iff every bounded ...
- 3,897
2
votes
0
answers
200
views
Fredholmness and invertibility in a C* algebra generated convolution-type operators
Let $PC$ be the algebra of complex-valued, piecewise-continuous functions from $[-\infty,+\infty]$, $SO$ be the algebra of bounded, continuous, complex-valued functions on $\mathbb R$ which are slowly ...
- 21
37
votes
2
answers
2k
views
Moving one family of commuting self-adjoint operators to another without losing commutativity on the way
This is actually not a question of mine, so I'll be short on motivation and say nothing beyond that if this were true, a few fancy harmonic analysis techniques that a colleague of mine used in proving ...
- 61.9k
34
votes
2
answers
3k
views
Can we recover a von Neumann algebra from its predual?
By definition, a von Neumann algebra is a C*‑algebra A
that admits a predual, i.e., a Banach space Z such that
Z* is isomorphic to the underlying Banach space of A.
(We require that isomorphisms in ...
- 37.8k
2
votes
2
answers
864
views
Decomposition of an abelian von Neumann algebra
Hi, I came across the statement below and I couldn't figure out why it is true. I was hoping someone could explain it or give me a good reference. Thank you in advance.
"Let $\pi$ be a non-degenerate ...
14
votes
2
answers
926
views
"Explicit" embedding of $\ell^1$ as a closed subalgebra of a direct sum of matrix algebras
For sake of brevity let $A$ denote the Banach algebra formed by equipping $\ell^1({\mathbb N})$ with pointwise multiplication. This algebra is clearly not isomorphic as a Banach algebra to any uniform ...
- 25.8k
3
votes
0
answers
130
views
Positive block matrices over tensor algebras
Let $A$ be a unital C*-algebra. A positive block matrix in $M_2(A)$ must have the form
$$ \begin{pmatrix} a & a^{1/2} x b^{1/2} \\ b^{1/2} x^* a^{1/2} & b \end{pmatrix}, $$
where $a,b$ are ...
- 18.7k
4
votes
0
answers
256
views
A matrix minimisation problem
Feel free to edit the title!
Suppose A is a C*-algebra and $a,b\in A$ are self-adjoint. I'd be very happy with A being just $n\times n$ matrices.
Question: If there are $t\in\mathbb R$ and $\...
- 18.7k
1
vote
1
answer
2k
views
Square root of integral operator
Consider the 1-torus $\mathbb{T}$. Let $k$ be a smooth function on $\mathbb{T}^2$ and $K$ be the integral operator on $L^2(\mathbb{T})$ with kernel $k$. One can show that $K$ is of trace class, hence $...
- 1,702
8
votes
1
answer
431
views
Injectivity for bimodules and Hochschild cohomology
Let $A$ be a Banach algebra and let $X$ be an $A$-bimodule. Is there a notion of (relative) injectivity for $X$ which would imply that $\mathcal{H}^n(A,X)$ vanishes for all $n\ge 1$? Here $\mathcal{H}^...
user2412
0
votes
1
answer
1k
views
Linear Mapping and integration
I have been reading the paper - "Introduction to Quantum Fisher Information".
In section 1.2 the author talks about the linear map $\mathbb{J}_D$, which he defines as follows:
Let $D \in M_n$ be a ...
- 145
15
votes
2
answers
2k
views
Range of completely positive projection
Let $A$ be a C*-algebra. Suppose that $P:A \rightarrow A$ is a contractive completely positive projection. Does the range $P(A)$ is completely order isomorphic to a $C^*$-algebra?
In the case where ...
- 1,222
11
votes
2
answers
2k
views
How "generalized eigenvalues" combine into producing the spectral measure?
Hi... I am wondering how 'eigenvalues' that don't lie in my Hilbert space combine into producing the spectral measure. I study probability and I am quite ignorant in the field of spectral analysis of ...
- 333
7
votes
1
answer
577
views
Are the compact and Haagerup approximation properties equivalent?
The following essentially implies the equivalence of Anantharaman-Delaroche's compact approximation property (page 337 of Link) and the Haagerup approximation property.
Let $M$ be a type ${II}_{1}$ ...
- 7,047
2
votes
2
answers
709
views
Are there good inequalities on the norm?
It's well known that in a Hilbert space, good inequalities exist concerning the norm due to the existence of inner product.Now let X be a general Banach algebra, are there good inequalities concerning ...
- 1,528
0
votes
2
answers
377
views
"Frobenius-finite" linear operators on a Hilbert Space
Let $H = L_2(S)$ be the complex Hilbert space over $S$ with the counting measure. (There might be another term for this concept, but) I define a continuous linear operator $L$ on $H$ with matrix ...
user5810
6
votes
0
answers
354
views
Ordering of completely bounded maps
Let A be a C*-algebra, let H be a Hilbert space, and let $T:A\rightarrow B(H)$ be a completely bounded (cb) map (that is, the dilations to maps $M_n(A)\rightarrow M_n(B(H))$ are uniformly bounded). ...
- 18.7k
0
votes
0
answers
320
views
A result about Fredholm operator
When I read the article "Index Theory" in Handbook of global analysis, I meet a result as below(Corollary 2.13):
If every $F_0\in \mathcal {F}(H_1,H_2)$, there is an open neighborhood $U_0\subseteq \...
- 381
0
votes
2
answers
337
views
Is there a general notion of entropy for the states of a C*algebra?
I've seen some definition of the relative entropy between two states of a C*algebra. However this definitions work only for finite dimensional C*algebras and I don't know if there is a correspondent ...
