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23 votes
2 answers
3k views

States in C*-algebras and their origin in physics?

in $C^*-$algebras with unit element, there is the definition of a state, as a functional $\omega$ with $\omega(e)=||\omega||=1.$ Now, of course there is also in classical physics and quantum ...
Acuriousmind's user avatar
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 ...
Matthew Daws's user avatar
  • 18.7k
15 votes
4 answers
3k views

Universal $C^*$-algebra with generators and relations

We say that the $C^*$-algebra $A$ generated by $a_1,...,a_n$ is universal subject to relations $R_1,...,R_m$ if for every $C^*$-algebra $B$ with elements $b_1,...,b_n$ satisfying relations $R_1,...,...
truebaran's user avatar
  • 9,330
14 votes
1 answer
2k views

Infinite tensor product of states

Tensor products of finite number of different objects are always well described in the literature. However, the situation of infinite tensor products seems to be much tougher. Even in the simplest ...
Glacier's user avatar
  • 143
13 votes
2 answers
1k views

Calkin Algebra and the embedding

Let $H$ be a separable, infinite dimensional Hilbert Space and $Calk(H):=B(H)/K(H)$ denotes the Calkin algebra. There is obvious surjection $\pi: B(H) \to Calk(H)$ but I'm interested in somehow ...
truebaran's user avatar
  • 9,330
9 votes
3 answers
2k views

Generalizations and relative applications of Fekete's subadditive lemma

Fekete's (subadditive) lemma takes its name from a 1923 paper by the Hungarian mathematician Michael Fekete [1]. A historical overview and references to (a couple of) generalizations and applications ...
Salvo Tringali's user avatar
8 votes
1 answer
716 views

A non-hyperfinite type III factor from an action of the free group on the circle

We define below a von Neumann algebra $\mathcal{M}$ from an action of the free group on the circle, and we prove that $\mathcal{M}$ is a non-hyperfinite type ${\rm III}$ factor. Question : Is $\...
Sebastien Palcoux's user avatar
81 votes
3 answers
9k views

Norms of commutators

If an $n$ by $n$ complex matrix $A$ has trace zero, then it is a commutator, which means that there are $n$ by $n$ matrices $B$ and $C$ so that $A= BC-CB$. What is the order of the best constant $\...
Bill Johnson's user avatar
  • 31.5k
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 ...
André Henriques's user avatar
26 votes
3 answers
2k views

About the category of von neumann algebras

I am looking for one (or more) reference about properties of the category of von Neumann algebra. More precisely, in an answer of a previous question, Dmitri Pavlov mentions that the $W^*$ category ...
Oliver's user avatar
  • 357
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 ...
Mikael de la Salle's user avatar
17 votes
3 answers
905 views

Existence of translation-invariant basis on $C_c(\mathbb R)$

Consider the space $C_c(\mathbb R)$ of complex-valued continuous functions of compact support. This is a vector space over $\mathbb C$, and I am not considering any topology, so the question is ...
Nick S's user avatar
  • 2,071
10 votes
1 answer
492 views

Which W*-algebras are the duals of C*-coalgebras?

A Banach algebra (assumed associative and unital) is precisely a monoid object in the monoidal category of Banach spaces, short linear maps, and the projective tensor product. A Banach coalgebra is ...
Toby Bartels's user avatar
  • 2,754
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 ...
Fabian Lenhardt's user avatar
10 votes
2 answers
926 views

Continuity of the product map

Let $A$ be a $C^*$-algebra. Is it possible to characterize $A$ for which the product map defined by $$\sum\limits_{i=1}^n a_i\otimes b_i \mapsto \sum\limits_{i=1}^n a_i b_i$$ is continuous with ...
Kate Juschenko's user avatar
9 votes
1 answer
956 views

A problem in functional calculus

This is embarrassing, I think it must work, but I can't see how to prove it works. If anyone knows enough functional calculus of operators on a Hilbert space to tell me how to do it, I would be very ...
Edwin Beggs's user avatar
  • 1,143
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 ...
Sebastien Palcoux's user avatar
6 votes
1 answer
765 views

An equivalence relation on the space of polynomials in one complex variable

Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$ where these integers are ...
Ali Taghavi's user avatar
4 votes
1 answer
414 views

A question on an argument in Woronowicz’s paper on the compact quantum group $ {\text{SU}_{q}}(2) $

Let $ q \in [0,1) $. The compact quantum group $ {\text{SU}_{q}}(2) $ is defined to be the universal unital $ C^{*} $-algebra that is generated by two elements $ \alpha $ and $ \beta $ satisfying the ...
Transcendental's user avatar
3 votes
0 answers
200 views

What are the first non-maximal non-group-subgroup simple irreducible subfactors?

Definition: For an irreducible (finite index) subfactor $(\mathcal{N} \subset \mathcal{M})$, an intermediate $(\mathcal{N} \subset \mathcal{P} \subset \mathcal{M})$ is normal if the biprojections $e_{\...
Sebastien Palcoux's user avatar
3 votes
1 answer
6k views

About eigen-functions of the Gaussian kernel

If I look at the Guassian kernel function $e^{- \frac {\vert x - y\vert_2^2 }{2 w^2 } }$ for $x, y \in \mathbb{R}$. Then w.r.t the Gaussian measure $N(\mu,\sigma)$ I believe it is true that this has a ...
gradstudent's user avatar
  • 2,246
2 votes
1 answer
233 views

${\rm II}_1$-factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?

Let $\mathcal{A} , \mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}', \mathcal{B}' $ are also a ${\rm II}_1$-factors. Question: $\mathcal{A} \cap \mathcal{B} = \mathbb{...
Sebastien Palcoux's user avatar
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 ...
fedja's user avatar
  • 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 ...
Dmitri Pavlov's user avatar
27 votes
0 answers
1k views

Unital $C^{*}$ algebras whose all elements have path connected spectrum

A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$. What is an example of a non commutative ...
Ali Taghavi's user avatar
25 votes
2 answers
1k views

Can nuclearity be determined by tensoring with a single C*-algebra?

A C*-algebra is nuclear if the algebraic tensor product $A\odot B$ ($B$ is any other C*-algebra) admits a unique C*-norm. This definition requires testing the condition for nuclearity with `all' C*-...
Lech Roch's user avatar
  • 505
22 votes
5 answers
1k views

Rigorous justification for this formal solution to $f(x+1)+f(x)=g(x)$

Let $g\in C(\Bbb R)$ be given, we want to find a solution $f\in C(\Bbb R)$ of the equation $$ f(x+1) + f(x) = g(x). $$ We may rewrite the equation using the right-shift operator $(Tf)(x) = f(x+1)$...
BigbearZzz's user avatar
  • 1,245
20 votes
2 answers
870 views

C$^*$-algebras isomorphic after tensoring with $M_n(\mathbb C)$

In 1977, Joan Plastiras gave a striking example of two non $*$-isomorphic C$^*$-algebras $\mathcal A$ and $\mathcal B$ such that $$\mathcal A \otimes M_2(\mathbb C) \simeq \mathcal B\otimes M_2(\...
Chris Ramsey's user avatar
  • 3,984
17 votes
4 answers
4k views

How much does the absolute value of an operator behave like an absolute value?

Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the ...
Iian Smythe's user avatar
  • 3,115
16 votes
1 answer
526 views

Equivariant Fredholm operators classify equivariant K-theory

Let $\mathcal{F}$ be the space of Fredholm operators on a separable Hilbert space $H$ with the topology induced by the operator norm. If $X$ is compact, Atiyah-Jänich proved that $$[X,\mathcal{F}]\...
Bo Liu's user avatar
  • 673
14 votes
2 answers
723 views

Why do the projections in the Calkin algebra not form a lattice?

Let $H$ be an infinite dimensional separable complex Hilbert space. Denote by $\mathcal{B}(H)$ the C*-algebra of bounded operators on $H$, $\mathcal{K}(H)$ the ideal of compact operators on $H$, and $\...
Iian Smythe's user avatar
  • 3,115
15 votes
1 answer
1k views

Convolution algebras for double groupoids?

There is a lot of work of course on convolution algebras of measured groupoids, and this gives "Noncommutative geometry". However there is a lot of interest in algebraically structured groupoids, for ...
Ronnie Brown's user avatar
  • 12.3k
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 ...
BigBill's user avatar
  • 1,222
14 votes
4 answers
550 views

About the existence of characters on $B(X)$

Let $X$ be a Banach space. Let $B(X)$ be the space of all bounded linear operators on $X$. Does $B(X)$ have an empty character space for any $X$? I know the proof of the fact that $M_n(\mathbb{C})$ ...
User93709's user avatar
  • 355
13 votes
1 answer
807 views

Inner and extendible automorphisms of C*-algebras

If an automorphism $\alpha$ of a C*-algebra $A$ is inner then whenever $A$ is a subalgebra of another C*-algebra $B$, $\alpha$ obviously extends to $B$. Is the converse true: if an automorphism $\...
Aaron Tikuisis's user avatar
13 votes
1 answer
1k views

A generalization of the Powers-Stormer inequality

The well-known Powers-Stormer inequality says the following: for positive semidefinite operators $A, B$, we have that $\mathrm{Tr}((A - B)(A - B)) \leq \| A^2 - B^2 \|_1$, where $\| \cdot \|_1$ ...
Henry Yuen's user avatar
  • 2,019
13 votes
2 answers
775 views

Properties of orthogonality-preserving c.p. maps between $C^*$-algebras

Suppose that $A,C$ are $C^*$-algebras and $\phi:A \to C$ is a completely positive, orthogonality-preserving linear map. (Orthogonality preserving means: if $a,b \in A$ satisfy $ab=0$ then $\phi(a)\phi(...
Aaron Tikuisis's user avatar
12 votes
1 answer
2k views

Making sense of "every non-commutative algebra has its own internal time evolution (aka a one-parameter group)"?

I've listened to many interviews and lectures of Alain Connes, in which he says something which goes roughly as follows "Every non-commutative algebra has its own time (evolution of), by which I ...
dohmatob's user avatar
  • 6,853
12 votes
0 answers
373 views

Does Thompson's group $V$ have property AP?

Property AP: A discrete group $\Gamma$ has property AP (Approximation Property) if there exists a net $(\phi_i)_{i \in I}$ of finitely supported functions on $\Gamma$ such that $\phi_i \to 1 $ weak$^*$...
tattwamasi amrutam's user avatar
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 ...
Sebastien Palcoux's user avatar
11 votes
2 answers
545 views

Is $\mathcal{B}^{\mathbb{Z}}(l^\infty(\mathbb{Z}))$ a commutative algebra?

Consider $l^\infty(\mathbb{Z})$ the Banach space of bounded complex valued functions on the abelian group $\mathbb{Z}$ with the supremum norm. It has a natural action by $\mathbb{Z}$ given by $(zf)(g):...
Werner Thumann's user avatar
11 votes
2 answers
2k views

Schur's Lemma for Hilbert spaces

Let $H$ be a complex Hilbert space and let a group $G$ act on $H$ such that there are no invariant closed subspaces besides $H$ and $(0)$. Let $D$ be the ring of bounded operators which commute with ...
David E Speyer's user avatar
10 votes
1 answer
1k views

Separating vectors for C$^*$-algebras

(I asked this on math.stackexchange, without response). Let $A$ be a C$^*$-algebra, concretely acting on a Hilbert space $H$. Suppose that $\xi_0\in H$ is cyclic and separating for $A$ (that is, the ...
Matthew Daws's user avatar
  • 18.7k
9 votes
1 answer
338 views

Commuting nets for commuting projections

I think this should not be too difficult, but I am not an expert. I did not get an answer on stackexchange. Let $A$ be a $C$*-algebra and let $p,q\in A^{**}$ be two commuting projections. Then there ...
Mark Roelands's user avatar
9 votes
1 answer
596 views

Why is the Berkovich spectrum of a C*-Algebra the same as the Gelfand spectrum?

Let $A = \mathcal{C}(X)$ be a commutative (unital) C*-Algebra. Let $Spec(A)$ denote its Gelfand spectrum $$ Spec(A) = \{A \rightarrow \mathbb{C} : \text{non-zero *-homomorphism} \} \simeq X. $$ Now ...
Niki's user avatar
  • 335
9 votes
5 answers
870 views

Abelianization of GL(H)

This is related to Theo's question about the abelianizations of finite dimensionsal Lie groups. I am interested in a specific (infinite-dimensional) case of the above question. Let H be an infinite-...
Mike Hartglass's user avatar
8 votes
1 answer
390 views

Order bounded version of monotone complete $C^*$-algebras

Let $A$ be a $C^*$-algebra with self-adjoint part $A_{\operatorname{sa}}$. Then $A$ is called monotone complete if every increasing norm bounded net in $A_{\operatorname{sa}}$ has a supremum (with ...
Jochen Glueck's user avatar
8 votes
3 answers
691 views

Commutant of the conjugations by unitary matrices

Let $\mathcal{L}(\mathbb{C}^{n \times n})$ denote the algebra of all linear mappings from $\mathbb{C}^{n \times n}$ to $\mathbb{C}^{n \times n}$ and let $\mathcal{C} \subseteq \mathcal{L}(\mathbb{C}^{...
Jochen Glueck's user avatar
8 votes
1 answer
340 views

characterization of commutative Banach algebras

Let $A$ be a Banach algebra with the following property: For every two nets $ x_{\alpha}$ and $y_{\alpha}$ in $A$, $x_{\alpha}y_{\alpha}$ converges if and only if $y_{\alpha}x_{\alpha}$ converges. ...
Ali Taghavi's user avatar
7 votes
1 answer
220 views

Is $C^{\infty}(E)$ a projective Frechet $C^{\infty}(M)$-module for a $C^{\infty}$-fiber bundle $E\to M$ with compact fiber?

The question is a special case of a previous question. Let $M$ be a compact smooth manifold, then it is clear that $C^{\infty}(M)$ is a Frechet algebra with pointwise multiplication and a collection ...
Zhaoting Wei's user avatar
  • 9,019