All Questions
82 questions
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 ...
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 ...
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,...,...
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 ...
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 ...
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 ...
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 $\...
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 $\...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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_{\...
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 ...
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{...
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 ...
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 ...
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 ...
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*-...
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 $(Tf)(x) = f(x+1)$...
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 $$\mathcal A \otimes M_2(\mathbb C) \simeq \mathcal B\otimes M_2(\...
17
votes
4
answers
4k
views
How much does the absolute value of an operator behave like an absolute value?
Recall that the absolute value of a bounded operator $T$ on a Hilbert space $H$ is the unique positive operator $|T|$ such that $$\||T|x\|=\|Tx\|$$ for all $x\in H$. It can be defined using the ...
16
votes
1
answer
526
views
Equivariant Fredholm operators classify equivariant K-theory
Let $\mathcal{F}$ be the space of Fredholm operators on a separable Hilbert space $H$ with the topology induced by the operator norm.
If $X$ is compact,
Atiyah-Jänich proved that
$$[X,\mathcal{F}]\...
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 $\...
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 ...
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 ...
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 $M_n(\mathbb{C})$ ...
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 $\...
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 $\| \cdot \|_1$ ...
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(...
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 ...
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$^*$...
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 ...
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):...
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 ...
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 ...
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 ...
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 ...
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-...
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 ...
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}^{...
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.
...
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 ...