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3 votes
0 answers
134 views

What are all the possible indices for the finite depth subfactors?

Jones index theorem (1983) states that the set of all possible (finite) indices of subfactors is exactly $$\mathrm{Ind}=\{ 4 \cos(\pi/n)^2 \ | \ n \ge 3 \} \cup [4, \infty),$$ but if we restrict to ...
Sebastien Palcoux's user avatar
3 votes
0 answers
160 views

Non-emptiness of spectrum $\sigma(a)$ in non-Archimedean Banach algebras

I'm trying to determine whether or not the standard proof that the spectrum of a point in a unital Banach algebra is non-empty can be adapted to prove the same thing over certain non-Archimedean ...
Very Forgetful Functor's user avatar
3 votes
2 answers
361 views

${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?

Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. Let $\mathcal{A}$, $\mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}'$, $...
Sebastien Palcoux's user avatar
3 votes
0 answers
229 views

The convolution on a semisimple finite quantum groupoid

Let $\mathbb{A}$ be a finite dim. weak Hopf $C^*$-algebra (or semisimple finite quantum groupoid) and $\hat{\mathbb{A}}$ its dual. Let the Fourier transform $\mathcal{F}: \mathbb{A} \to \hat{\mathbb{...
Sebastien Palcoux's user avatar
3 votes
2 answers
207 views

Commutative direct summands of C*-algebras

I have a question about commutative direct summands of $C$*-algebras. Let $A$ be a $C$*-algebra (with unit) and suppose that its bidual $A^{**}$ has a commutative direct summand, that is, $A^{**}=B\...
Mark Roelands's user avatar
3 votes
1 answer
274 views

Finite codimensional subvector space of $C^{*}$ algebras which contains no invertible elements

Assume that $A$ is a unital $C^{*}$ algebra. Is there a subvector space $Y\subset A$ of finite codimension which does not contain any invertible element? Let $n(A)$ be the infimum of such ...
Ali Taghavi's user avatar
3 votes
1 answer
325 views

Reference on completely positive maps which are isometries

Let $\Phi:\mathcal{L}(H)\rightarrow \mathcal{L}(K)$ be a completely positive map sending positive self-adoint operators on a finite-dimensional Hilbert space $H$ to positive self-adoint operators on a ...
Stefano Gogioso's user avatar
3 votes
1 answer
161 views

Simple $Z^{*}$ algebra

What is an example of a simple $C^{*}$ algebra which all elements are (two sided or equivalently one sided) zero divisor?
Ali Taghavi's user avatar
3 votes
3 answers
1k views

Are the finite dimensional von Neumann algebras, singly generated?

Let $\mathcal{M}$ be a finite dimensional von Neumann algebra, then : $$\mathcal{M} \simeq \bigoplus_i M_{n_i}(\mathbb{C})$$ Question : Is it singly generated (as von Neumann algebra)? how ? ...
Sebastien Palcoux's user avatar
3 votes
1 answer
252 views

What is the story behind this Hilbert space in the definition of Hilbert Modules

Here is Deflnition 1.5 of Hilbert module in "L^2-invariants: theory and applications to geometry and K-theory", Springer-Verlag, 2002, by W. Lück: A Hilbert $\mathcal N(G)$-module $V$ is a Hilbert ...
MSMalekan's user avatar
  • 2,118
3 votes
0 answers
331 views

Textbook covering superoperators and tensor products

I am looking for a textbook to cover the following tensor product (and, of course, the theory around it): Let $\otimes_1$ denote the tensor product on Hilbert spaces. Let $\otimes_2$ denote the ...
Dominique Unruh's user avatar
3 votes
0 answers
146 views

Closed containment of open projections in C*-algebras

For a C*-algebra $A$ and open projections $p,q\in A^{**}$, consider the following statements. $\overline{p}\leq q$ $p\leq q$ and there exists open $r\in A^{**}$ with $rp=0$ and $r\vee q=1$ $p\leq q$ ...
Tristan Bice's user avatar
  • 1,307
3 votes
0 answers
168 views

The left regular representation of the Jacobi groups over local fields of characteristic >2 is type I?

Let $K$ be a non-archimedean local field of characteristic $>2$. Consider the Jacobi group $G=H_{2n+1}(K)\rtimes Sp_{2n}(K)$, which is the semidirect product of the Heisenberg group $H_{2n+1}(K)$ ...
m07kl's user avatar
  • 1,702
3 votes
2 answers
797 views

Banach algebraic proof of the Borsuk Ulam theorem

I am wondering whether there exists a proof of the classical Borsuk Ulam theorem for the Euclidean n-sphere, $n>2$ that is based only on the theory of Banach algebras. I checked on MR but had no ...
ray's user avatar
  • 687
3 votes
1 answer
335 views

What's the natural equivalence of subfactors in general?

Let $A$ be a factor and $\mathcal{C}_{A}$ be the category of all the subfactors $(M \subset N)$ such that $M$ and $N$ are isomorphic to $A$. The most famous of them is perhaps $\mathcal{C}_{R}$ with $...
Sebastien Palcoux's user avatar
3 votes
2 answers
397 views

Is the ideal property of $X^{**}$ inheritable to $X$?

Let $X$ be an operator space such that there is a weak$^*$-continuous complete isometry $\phi$ from its second dual $X^{**}$ into a $W^*$-algebra $M$ in which $\phi(X^{**})$ is a (necessarily weak$^*$-...
Masayoshi Kaneda's user avatar
2 votes
0 answers
208 views

A functor on the category of rings, algebras or compact Hausdorff topological space

Assume that $R$ is a unital ring or a complex or real (Banach or $C^{*}$) algebra. We define a relation $M$ on $R$ as follows: $$a\;M b \;\;\; \text{iff}\;\; a=xy,\;b=yx \;\; \text{for ...
Ali Taghavi's user avatar
2 votes
0 answers
101 views

Are there infinitely many amenable Hadamard-Petrescu subfactors?

The complex Hadamard matrices of dimension $n$ are used to build index $n$ subfactors through the commuting square construction. For more details, see the paper Subfactors and Hadamard Matrices by W....
Sebastien Palcoux's user avatar
2 votes
2 answers
865 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 ...
Wishiwere Smarter's user avatar
2 votes
0 answers
89 views

On the set of indices of irreducible depth 3 subfactors

Let $I_n$ be the set of indices of (finite index) irreducible depth $n$ subfactors. Then $I_2 = \mathbb{Z}_{>0}$. Question 1: Is it true that $I_3$ has no accumulation point? If so: ...
Sebastien Palcoux's user avatar
2 votes
1 answer
757 views

Minimal Tensor product of von Neumann algebras

Let $M$ and $N$ be two von Neumann algebras. A natural way to define $M\overline{\otimes}N$ is the von Neumann algebra generated by the algebraic tensor product $M\otimes N$ in $B(H\otimes K)$ where $...
Mathbuff's user avatar
  • 455
2 votes
1 answer
199 views

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 ...
keej's user avatar
  • 123
2 votes
0 answers
156 views

Extension of a theorem of Bisch to cyclotomic integers of fixed degree

Theorem 3.2 in this paper (Bisch, 1994) states that for a finite depth ${\rm II}_1$ subfactor of integral index, every intermediate subfactor are also of integral index. As an application, every such ...
Sebastien Palcoux's user avatar
2 votes
0 answers
105 views

Multiplicativity of the analytic index (or of kernel bundle)

What I want to ask is the multiplicativity of the analytic index of a family of Dirac operators. In the single operator case the analytic index of elliptic operator is multiplicative. This is proved ...
Ho Man-Ho's user avatar
  • 1,173
2 votes
1 answer
143 views

Inequality for hermitian matrices

Let $p_1$ and $p_2$ be a complete system of orthogonal projections on $\mathbf R^n, n \geq 2$ (i.e., $p_i^2=p_i=p_i^*$ and $p_1+p_2=\bf{1}$) and $S_1, S_2$ be two hermitian operators such that $S_i \...
Svata's user avatar
  • 73
2 votes
1 answer
164 views

Cocompact lattices in $\mathrm{Sp}(n, 1)$

This is a continuation from my previous question. I am reading the following paper of Cowling-Haagerup, and I was wondering whether there are uniform lattices in $\mathrm{Sp}(n, 1)$. Is there some way ...
Y. Paka's user avatar
  • 131
2 votes
1 answer
471 views

Takesaki II proposition 3.15 proof about modular automorphism groups: mistake in book?

Consider the following fragment from Takesaki's second volume "Theory of operator algebras" in chapter VIII "Modular automorphism groups" p122-123. Why is it possible to choose an ...
Andromeda's user avatar
  • 175
2 votes
2 answers
190 views

Results which are known about ideals of spatial tensor product

I am studying about ideals of spatial (minimal) tensor product of $C^{\ast}$-algebras but I did not find any book/paper in which all the results are given. What are some results or folklore which ...
Math Lover's user avatar
  • 1,115
2 votes
1 answer
352 views

K-Theory of $C^{*}(X)$

I'm new to K-Theory for $C^{*}$-algebra and $C^{*}$-algebra of groups. If $X$ is the group of finite support bijections of natural numbers then what is the K-Theory of $C^{*}(X)$? I was planning to ...
Peg Leg Jonathan's user avatar
2 votes
1 answer
462 views

Direct integral decomposition relative to a given measure space

It is well known that a separable Hilbert space $H$ decomposes as a direct integral in the presence of an abelian von Neumann algebra $\mathscr A\subseteq B(H)$. More precisely, and quoting from ...
Black's user avatar
  • 483
2 votes
1 answer
132 views

a nuclear $C^*$-subalgebra in $\prod_n M_n(\Bbb C)$

Does there exist a non-unital nuclear $C^*$ algebra $A$ of $\prod_nM_n(\Bbb C)$ such that $A$ properly contains $\oplus_n M_n(\Bbb C)$ and each element $(x_n)\not \in \oplus_n M_n(\Bbb C)$ we have $\...
math112358's user avatar
2 votes
1 answer
416 views

Questions about Maharam's classification theorem

I am studying von Neumann algebras. In the wiki article abelian von Neumann algebras, it mentions that every abelian von Neumann algebras acting on a separable Hilbert space is *-isomorphic to $L^{\...
Ken.Wong's user avatar
  • 523
2 votes
1 answer
368 views

On diagonal part of tensor product of $C^*$-algebras

Suppose we have a $C^*$-algebra $\mathcal{U}$, Consider the $C^*$-subalgebra generated by elements of the form $a\otimes a$, what is it isomorphic to? Is it isomorphic to $\mathcal{U}$ itself?
user136400's user avatar
2 votes
1 answer
131 views

Closeness of points in the irreducible decomposition of a C$^{*}$-algebra representation

Suppose $X$ and $Y$ are compact metric spaces. Let $\varphi\colon C(X)\to M_{n}(C(Y))$ be any $*$-homomorphism. If $\pi$ is an irreducible representation of $M_{n}(C(Y))$, then $\pi$ is unitarily ...
ervx's user avatar
  • 267
2 votes
1 answer
244 views

$R$ is a right multiplier and $R(a)b=a\overset{?}{\implies} A$ is unital

Let $A$ be a $C^*$-algebra, and $R:A\to A$ its right multilplier. Is it true that $$ \exists b\in A\quad \forall a\in A \quad R(a)b=a\qquad $$ implies $A$ is unital. I know this is true if A is a weak$...
Norbert's user avatar
  • 1,697
1 vote
1 answer
114 views

Is a $1_A \otimes U$ invariant subspace of $\mathcal{H}_A \otimes \mathcal{H}_B$ a product $V_A \otimes \mathcal{H}_B$?

Consider a subspace $V$ of $\mathcal{H}_A \otimes \mathcal{H}_B$, with $\mathcal{H}_A$ and $\mathcal{H}_B$ finite-dimensional Hilbert spaces, that is $1_A \otimes U$ invariant for all unitary ...
Matias Heikkilä's user avatar
1 vote
0 answers
81 views

A consecutive resolution of continum algebras to a simple continum algebra

Motivated by classical Gelfand Naimark duality, the correspondence between the category of commutative $C^{*}$ algebras and the category of locally compact Hausdorff spaces, we ...
Ali Taghavi's user avatar
1 vote
1 answer
181 views

Existence of a third intermediate if there are two intermediate subfactors of index 2

Let $(N \subset M)$ be an irreducible finite index unital inclusion of hyperfinite ${\rm II}_1$ factors. Let $K_1$ and $K_2$ be two distinct intermediate subfactors $N \subset K_i \subset M$, such ...
Sebastien Palcoux's user avatar
1 vote
0 answers
182 views

Characterization of exact groups via the existence of amenable actions on unital C*-algebras

It is known that a discrete group $G$ is exact if and only if it admits an amenable action on a compact topological space. Would it also be true that $G$ is necessarily exact when it admits an ...
Ruy's user avatar
  • 2,263
1 vote
0 answers
345 views

Duality of maps on bounded vs trace-class operators (Schrödinger-Heisenberg dual)

$\newcommand\calH{\mathcal H} \newcommand\calK{\mathcal K} \newcommand\tr{\operatorname{Tr}}$I am looking for a (citable) reference for the following fact: Bounded linear maps $g:T(\calH)\to T(\calK)$...
Dominique Unruh's user avatar
1 vote
1 answer
322 views

A particular commutator of the discrete Fourier matrix

For $N$ be a fixed natural number, define $w=e^{\frac{2\pi i}{N}}$ and $z=e^{\frac{\pi i}{N}}$, so that $z^2=w$. Let $D$ be the diagonal matrix $D=\operatorname{diag}(1,z,z^2,\ldots,z^{N-1})$ and $F$ ...
ABB's user avatar
  • 4,058
1 vote
1 answer
132 views

Is the coproduct of central operators, also central?

Let $\mathcal{P}$ be an irreducible finite index-depth subfactor planar algebra. The $2$-boxes space $\mathcal{P}_{2,+}$ is equipped with the coproduct $(a,b) \mapsto a*b = \mathcal{F}(\mathcal{F}^{-...
Sebastien Palcoux's user avatar
1 vote
0 answers
164 views

When a finite codimensional subalgebra contains a finite codimension ideal?

What is a classification of all algebras $A$ (purely algebraic algebras, Banach or $C^*$ algebras or Lie algebras) with the following property: Every finite codimensional subalgebra $B$ of $A$ ...
Ali Taghavi's user avatar
1 vote
1 answer
683 views

About separability of von Neumann algebras [closed]

Is a von Neumann algebra always separable in the $\sigma$-weak topology? If not, give a counterexample. Under what conditions will it be separable?
mathlover's user avatar
  • 227
1 vote
1 answer
306 views

Simple $C^*$ algebras whose all commutator elements have scalar square

Is there a simple $C^*$ algebra $A$, not isomorphic to $M_2(\mathbb{C})$, such that for every commutator element $x=ab-ba$, $x^2$ is an scalar element?
Ali Taghavi's user avatar
1 vote
0 answers
178 views

A locally convex $C^*$ algebra without zero divisor

Let we have a locally convex $C^*$ algebra $A$. That is $A$ is a TVS equipped with an algebra and an involution structure such that all operations are continuous. Moreover the topology on $A$ ...
Ali Taghavi's user avatar
1 vote
1 answer
259 views

Existence of homogeneous single chain compositions of a given maximal subfactor?

All the subfactors here are irreducible inclusion of hyperfinite II$_1$ factors. A subfactor $(N \subset M)$ is Homogeneous Single Chain ($HSC$) if its lattice of intermediate subfactors is a single ...
Sebastien Palcoux's user avatar
1 vote
0 answers
81 views

Why does one only consider one-parameter groups in Borchers-Arveson theorem?

(question from math.stackexchange) The theorem (Operator algebras and Quantum statistical mechanics vol. 1, Bratteli, Robinson, Thm. 3.2.46 p.261) roughly says that if one has a one parameter ...
Noix07's user avatar
  • 189
1 vote
0 answers
174 views

The planar algebra generated by the biprojections

Let $(N_1 \subset M_1)$ and $(N_2 \subset M_2)$ be two irreducible finite index subfactors. Let $\mathcal{B}_i$ be the set of all the biprojections of $\mathcal{P}_{2+}(N_i \subset M_i)$. Let $\...
Sebastien Palcoux's user avatar
1 vote
1 answer
79 views

Are the two-side TLJ subfactors maximal?

Let $(N \subset M)$ be a unital inclusion of ${\rm II}_1$ factors, with the following principal graph (called two-side TLJ) Question: Is $(N \subset M)$ a maximal subfactor?
Sebastien Palcoux's user avatar