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7 votes
0 answers
164 views

Nontrivial examples of locally compact quantum groups

What are some families of locally compact quantum groups that are neither groups, duals of groups, compact, nor discrete?
Cameron Zwarich's user avatar
1 vote
1 answer
295 views

An example of non-invertible operator $F$ such that $P_nF$ is invertible on $\operatorname{Im}P_n$ or proving that It is impossible

Given: $X$ - any Banach space $F : X \to X$ (linear bounded and non-invertible) $P_n$, which is projector that strongly converges to the identity operator $I$ as $n \to\infty$ Can you help me come ...
TorteDeline's user avatar
5 votes
1 answer
208 views

Hyperfinite factors and increasing fatorization of states

If a factor $R$ contains a matrix algebra $M\subset R$ (i.e., a $M$ is a type $I_n$ factor), then $R \cong M \otimes M^c$ where $M^c=R\cap M'$ is the relative commutant. Each state $\omega$ on $R$ ...
Lau's user avatar
  • 759
4 votes
1 answer
270 views

Kaplansky inverse element theorem on group C-star algebra

In a class talking about $C^*$ algebra and (higher) index theory, I heard a theorem (related to Kaplansky, proved?), that is Suppose $\Gamma$ is a group (admitting Haar measure if necessary) while $\...
YOTAL's user avatar
  • 193
4 votes
1 answer
203 views

weights of projections and norms of operators in a von Neumann algebra

Let $M$ be an atomless von Neumann algebra equipped with a (semifinite faithful normal) weight $w$. Let $x\in M$ and let $\varepsilon>0$. Can we find a constant $\delta>0$ such that whenever a ...
user92646's user avatar
  • 617
5 votes
1 answer
221 views

Arens regularity of $\mathrm{BV}(\mathbb{R})$

$\DeclareMathOperator\BV{BV}$A Banach algebra $A$ is called Arens regular if the two canonical multiplications on the double dual $A^{**}$ coincide. Let $\BV(\mathbb{R})$ denote the Banach algebra of ...
Tobias Fritz's user avatar
  • 6,406
4 votes
0 answers
220 views

Bochner theorem for (non-abelian) discrete groups

I am interested in Pontryagin duality-like theories for discrete groups, more particularly, whether an analogue to Bochner's theorem for abelian groups exists in the discrete non-finite and non-...
Tomás Pacheco's user avatar
1 vote
0 answers
144 views

Reconstructing the manifold from space of functions in quantum mechanics

Due to Banach–Mazur, every separable Banach space is isomorphic to a subspace of $C([0,1])$. But some spaces, like $C([0,1]^n)$ and generally $C(M)$ for $M$ a manifold, allow one to reason about the ...
0x11111's user avatar
  • 593
6 votes
0 answers
98 views

Conditions for completely positive maps to act homomorphically across multiple subalgebras

For a completely positive (CP) map $u: A \to A'$ of $C^*$-algebras $A, A'$, the concept of multiplicative domains characterizes the largest subalgebra of $A$ on which $u$ behaves as a $*$-homomorphism....
BiPolarBear's user avatar
2 votes
1 answer
232 views

Completely contractive Banach algebra structure on the dual of a Hopf $C^*$-algebra

Let $(A, \Delta)$ be a Hopf $C^*$-algebra, i.e. $A$ is a $C^*$-algebra, and $\Delta: A \to M(A\otimes A)$ is an injective non-degenerate $*$-homomorphism that is coassociative: $$(\iota \otimes \Delta)...
Andromeda's user avatar
  • 175
1 vote
1 answer
211 views

Tensor product of faithful normal states is faithful

I know that given C*-algebras $A, B$ with faithful states $\omega,\varpi$, the tensor product state $\omega\otimes\varpi$ on the minimal tensor product $A\otimes_{\text{min}}B$ is faithful. I also ...
J_P's user avatar
  • 439
5 votes
1 answer
165 views

Approximation from below of positive elements in tensor product of von Neumann algebras

Let $\mathcal M$ and $\mathcal N$ be two von Neumann algebras. If $x$ is a positive element of $\mathcal M$ and $y$ is a positive element of $\mathcal N$, it is known that $x\otimes y$ is a positive ...
P. P. Tuong's user avatar
2 votes
0 answers
232 views

Show that $\mathbb{K}\cong M_{n}(\mathbb{K})$ [closed]

I would like to show the following isomorphy but not sure how to go about this: $\mathbb{K}\cong M_{n}(\mathbb{K})$ Also in Blackadar (Operator Algebras, page 171) he states that this isomorphism ...
craaaft's user avatar
  • 129
14 votes
2 answers
871 views

Which finite dimensional Banach spaces can be represented isometrically as spaces of bounded operators on a finite dimensional Hilbert space?

Background: It is known that every Banach space $X$ can be embedded isometrically as a subspace in the space $C(K)$ of continuous functions on a compact Hausdorff space $K$. Indeed, one can take $K$ ...
Orr Shalit's user avatar
3 votes
1 answer
244 views

inclusion of von Neumann algebras implies reversing inequality of its modular operators

I'm working with von Neumann algebras and I stumbled with this statement in a work of Borchers (1999) Since $\mathcal N \subseteq \mathcal M$, it follows by standard arguments that $\Delta_{\mathcal ...
Gabriel Palau's user avatar
1 vote
1 answer
256 views

Intersection of two intermediate subalgebras

Suppose $B\subset A$ is an inclusion of simple $C^*$-algebras with a conditional expectation of (Watatani) index-finite type and $B^{\prime}\cap A=\mathbb{C}$. Then we know $B^{\prime}\cap A_1$ is ...
Keshab Bakshi's user avatar
3 votes
1 answer
142 views

$K_0$ group of an infinite factor

The following question was already posted in this link but I could not understand hints given in this post. Let $\mathcal{M}$ be an infinite factor and my question is how to prove that $K_0(\mathcal{M}...
Sanae Kochiya's user avatar
0 votes
0 answers
157 views

Dependence of functional integral on the function space

In physics, the following functional integral is considered \begin{gather} Z[J]= \int Df \exp(-\int d^dx( f\Box f+\lambda f^4 +Jf )) \end{gather} It is usually said that the integration is performed ...
0x11111's user avatar
  • 593
1 vote
1 answer
209 views

Borel functions in C*-algebras

Is there a way of defining representations of separable $C^*$-algebras, say $\Phi$, so that $\Phi(A)$ is faithful representation of $A$ on a separable Hilbert space. There is a closure operation $A\...
user52345435's user avatar
28 votes
6 answers
6k views

Any real contribution of functional analysis to quantum theory as a branch of physics?

In the last paragraph of this last paper of Klaas Landsman, you can read: Finally, let me note that this was a winner's (or "whig") history, full of hero-worship: following in the footsteps of ...
2 votes
0 answers
121 views

Invariant weights associated to algebraic quantum groups

Consider an algebraic quantum group $(A, \Delta)$ in the sense of Van Daele, i.e. a regular multiplier Hopf $^*$-algebra with a positive left integral $\varphi$ and a positive right integral $\psi$. ...
Andromeda's user avatar
  • 175
3 votes
1 answer
155 views

Is a compact set of extreme points contained in a compact face?

I have run into the following question in convex analysis, which I haven't found answered in the literature: Suppose that $K$ is a "nice-enough" non-compact convex subset of a Hausdorff ...
Sean's user avatar
  • 135
2 votes
0 answers
177 views

What is meant by saying that the Shilov boundary of the polydisc $\mathbb D^n$ is $\mathbb T^n\ $?

Let $A$ be a complex Banach algebra and $M_A$ be the space of all non-zero multiplicative linear functionals on $A$ equipped with the weak$^*$-topology. Let $\widehat A$ be the image of $A$ under the ...
Anacardium's user avatar
13 votes
0 answers
174 views

Existence of more than two C*-norms on algebraic tensor product of C*-algebras

Let $A$ and $B$ be two C*-algebras. Then $(A,B)$ is called is a nuclear pair if there is a unique $C^*$-norm on the algebraic tensor product $A\odot B$. If $A$ or $B$ is nuclear, then all pairs $(A,B)$...
Alcides Buss's user avatar
5 votes
1 answer
158 views

Backwards stable factors

A factor $R$ is called stable if $M_n(R)\cong R$ for all $n>0$. For the sake of this question, we call a factor backwards stable if $R\cong M_n(S)$ implies $S\cong R$ where $S$ is allowed to be any ...
Lau's user avatar
  • 759
-2 votes
1 answer
143 views

Relationship between noncommutative torus for different values of theta [closed]

Let $u,v\in B(L_2(\mathbb T))$ defined as $u(f)(z)=zf(z)$ and $v(f)(z)=f(ze^{-2\pi i\theta})$ for $z\in\mathbb T$ where $\theta\in\mathbb R\setminus\mathbb{Q}$. Denote the $C^*$ algebra generated by $...
A beginner mathmatician's user avatar
1 vote
1 answer
286 views

A subalgebra of $B(H)$ which does not contain a commutator element

Is there a commutative subalgebra $A\subset B(H)$ containing the 1 dimensional scalars with the following property: The algebra $A$ has trivial intersection with the set of commutator ...
Ali Taghavi's user avatar
0 votes
2 answers
381 views

What are the properties of umbra with moments $\{1,1/2,1/3,1/4,1/5,...\}$?

If we apply operator $D\Delta^{-1}$ to a function, we will get the (Bernoulli) umbral analog of the function. Particularly, applying it to $x^n$ we will get the Bernoulli polynomials $B_n(x)$. ...
Anixx's user avatar
  • 10.1k
6 votes
0 answers
253 views

Are bounded groups of thin operators on Hilbert space similar to groups of unitaries?

QUESTION. Let $G$ be a group of bounded operators on $\ell^2$, satisfying $\sup_{x\in G} \lVert x\rVert <\infty$, whose elements are all of the form "identity+compact" (sometimes called &...
Yemon Choi's user avatar
  • 25.8k
14 votes
0 answers
220 views

Unitary group of a von Neumann algebra: is it a retract of $U(H)$?

Let $M\subset B(H)$ be a properly infinite von Neumann algebra (the case I care about is $M=$ hyperfinite $\mathrm{III}_1$). Consider the unitary groups $U(M)$ and $U(H)$ in their strong operator ...
André Henriques's user avatar
1 vote
0 answers
125 views

Transforming nilpotency into diagonalizability [closed]

We designate the $k$-th standard vector as $e_k$ in $\mathbb{C}^n$. We consider the backward shift operator, denoted as $T: \mathbb{C}^n \to \mathbb{C}^n$, which is defined as follows: $Te_1=0$ and $...
ABB's user avatar
  • 4,058
6 votes
1 answer
195 views

Is there an element in an infinite tensor power of a C*-algebra that is invariant under finite permutations?

Let $A$ be a C*-algebra, and consider the infinite tensor power $A^{\otimes {J}}$, where $J$ is infinite (we consider the minimal or maximal tensor product). To any finite permutation, which is a ...
Antonio Lorenzin's user avatar
4 votes
0 answers
242 views

On the Dunford-Pettis property and multiplier algebras

I am not an expert in operator algebras, so if the answer to this question might be trivial, that might be one reason for that: Let $\mathcal{A}$ be a $C^\ast$-algebra. Then $\mathcal{A}^{\ast \ast}$ ...
Alexander Dobrick's user avatar
0 votes
1 answer
152 views

Unitary representation of a group of automorphism on an abelian algebra

Given an abelian C*-algebra $\mathcal{A}$, a state $\omega$, a strongly continuous group of *-automorphism $\{\tau_t : t \in \mathcal{R}\}$, and given a representation $ (\pi(\mathcal{A}), \mid \...
MBlrd's user avatar
  • 33
69 votes
4 answers
13k views

What is a foliation and why should I care?

The title says everything but while it is a little bit provocative let me elaborate a bit about my question. First time when I met the foliation it was just an isolated example in the differential ...
truebaran's user avatar
  • 9,330
6 votes
1 answer
574 views

Integration in Banach algebra

Let $\mu$ be a Borel measure on the real line $\mathbb{R}$ taking values in a separable Banach algebra $A$. Assume that $\mu$ is such that the total variation measure $|\mu|$ is finite. Let $f$ be a ...
user72829's user avatar
  • 552
2 votes
0 answers
174 views

Zeta zeros and prolate wave operators

Recently, Connes, Consani and Moscovici in https://arxiv.org/abs/2310.18423 have blended two of their results on zeta zeros and the prolate wave operators, which, they say, "suggests the ...
Jon23's user avatar
  • 1,139
2 votes
0 answers
209 views

Literature on Lyndon words and the Lie commutator

Since I lost my paper notes in a domestic conflagration in Japan some ten years ago, I've occasionally tried to recall one particular author who wrote in the 1900s about Lyndon words / strings, or ...
Tom Copeland's user avatar
  • 10.5k
2 votes
0 answers
157 views

Why do von Neumann algebras possess identity?

My starting point is that a von Neumann algebra is a $C^*$-algebra with a predual. The usual approaches to showing the existence of identity involve spectral theory (for approximate identity), but ...
MrPajeet's user avatar
  • 433
6 votes
1 answer
175 views

Lifting a semicommutative von Neumann algebra into an algebra of operator-valued functions

Given a complete probability space $\Omega$ and a von Neumann algebra $\mathcal M$, it is well-known that the tensor product von Neumann algebra $\mathrm L^\infty(\Omega)\overline\otimes\mathcal M$, ...
P. P. Tuong's user avatar
5 votes
0 answers
114 views

Realize a $K_0$-group homomorphism by a unital $\ast$-homomorphism

This question is inspired by Exercise $7.7$ in *An Introduction to $K$-theory for $C^*$-algebras (available here). Given a unital AF-algebra $A$ and another unital $C^*$-algebra $B$ that has ...
Sanae Kochiya's user avatar
6 votes
1 answer
287 views

Sigma-weakly dense *-subalgebra of von Neumann algebra has increasing net of positive elements convergent to the identity

Let $M$ be a von Neumann algebra and $A\subseteq M$ a $\sigma$-weakly dense $*$-subalgebra of $M$. Does there exist an increasing net $\{a_i\}_{i\in I}\subseteq A\cap M^+$ such that $a_i\to 1$ in the $...
Andromeda's user avatar
  • 175
5 votes
1 answer
413 views

Is the sigma-strong topology generated by bounded sets?

Let $V$ be a Banach space and $B(V)$ be the bounded operators on $V$. Now, the space $B(V)$ has a norm, the strong topology (which is the topology of pointwise convergence) and it has a stronger ...
Sebastian Meyer's user avatar
2 votes
1 answer
170 views

Ultralimit of $w^*$-continuous maps

Let $\omega$ be a free ultrafilter on $\mathbb N.$ Let $(\mathcal M_n)$ be a sequence of finite von Neumann algebras. Let $\mathcal N$ be another finite von Neumann algebra and we have maps $\phi_n:\...
A beginner mathmatician's user avatar
0 votes
0 answers
35 views

Operator-form correspondence without lower semiboundedness

When dealing with a normal unbounded operator $A$, it is often useful to frame questions about the operator in terms of questions about the associated form $\omega,$ which has domain $D(|A|^{1/2})$ ...
user_35's user avatar
  • 109
9 votes
2 answers
516 views

Why operator systems?

A $\mathrm{C}^*$-algebra $\mathcal{A}\subset B(\mathsf{H})$ is a norm-closed, self-adjoint subalgebra of bounded operators on a Hilbert space. If we then take a unital self-adjoint (possibly closed) ...
JP McCarthy's user avatar
  • 1,027
4 votes
0 answers
211 views

Irreducible representations of $\mathrm{UHF}_n$

I have a question about the irreducible representations of the $C^*$-algebra $\mathrm{UHF}_n = \bigotimes_{k=1}^\infty M_n$. For every sequence of unit vectors $(\xi_k)$ in $\mathbb C^n$ there is a ...
Lau's user avatar
  • 759
5 votes
1 answer
199 views

Is the unit ball of $B(H)$ a Baire space (with the SOT)?

Let $H$ be a Hilbert space, and let $B(H)$ be the set of bounded linear operators $t \colon H \to H$. Recall that we say $t_i \to t$ in the strong operator topology if $t_i \xi \to t \xi$ for every $\...
Diego Martinez's user avatar
1 vote
1 answer
128 views

Compare the weight of $p\vee q$ and that of $p+q$

Let $M$ be a von Neumann algebra. If it has a semifinite faithful normal trace $\tau$, then we have $\tau(p\vee q)\le \tau(p)+\tau(q)$. However, for the weight (even a faithful normal state) $\omega$ ...
user92646's user avatar
  • 617
1 vote
1 answer
207 views

Weak-star convergence implies trace-norm convergence

By definition, if bounded operators $a_i$ converge to $0$ in the weak*-star topology, then $\operatorname{tr} a_it \to 0$ for any trace-class $t$. Does this also hold for the trace-norm instead of the ...
Dominique Unruh's user avatar

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