All Questions
Tagged with oa.operator-algebras gn.general-topology
38 questions
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 $\...
4
votes
0
answers
152
views
Maximally fine topologies on $B(H)$ making the unit ball compact
Let $H$ be a Hilbert space, and $B(H)$ its algebra of bounded operators. One of the reasons the Ultraweak topology is (in a way) more useful than the weak operator topology is that the Ultraweak ...
1
vote
1
answer
114
views
Continuous surjection between spectra of commutative von Neumann algebras
Suppose that $V_1,V_2$ are two commutative von Neumann algebras and $V_1 \subset V_2$. Being in particular commutative $C^*$-algebras we have that $V_1 \cong C(X_1), V_2 \cong C(X_2)$ for some ...
2
votes
0
answers
198
views
A generalisation of closed and bounded subsets of non-Archimedean fields to topological spaces
The definition of compactness in topological spaces generalises the notion of a subset of $\mathbb{R}^n$ being closed and bounded, as expressed by the Heine-Borel Theorem.
In finite-dimensional vector ...
9
votes
1
answer
550
views
Is the unit sphere of a Banach space dense in the unit sphere of its second dual with respect to the weak-$\ast$ topology
To be a bit more precise and fix notations, let $X$ be a Banach space (over $\mathbb{R}$ or $\mathbb{C}$), $X^{\ast\ast}$ its second dual (as a Banach space). Here and in the following we identify $X$ ...
6
votes
0
answers
124
views
Meagre sets of bounded operators
Let $H$ be a separable, infinite-dimensional Hilbert space and let $\mathbb{B}(H)$ be the algebra of bounded operators on $H$. The norm topolology on $\mathbb{B}(H)$ is stricly finer, hence the ...
6
votes
1
answer
212
views
Embedding of $C(X)$ into $B(H)$ where $H$ is separable
I would like to ask a question which may look strange at the first sight nevertheless I find it interesting. Let $H$ be a separable Hilbert space: for any separable $C^*$-algebra $A$ one can embed $A$ ...
8
votes
0
answers
181
views
Continuous functions on a compact $T_1$ space
Let $X$ be a compact $T_1$ topological space consisting of more than one point, and suppose that $X$ is locally compact (i.e. every point has a local base of compact neighbourhoods), second countable, ...
0
votes
1
answer
495
views
Separability of an algebra is equivalent to separability of its spectrum
Let $A$ be a commutative C*-algebra.
I would like to show that $A$ is separable (i.e. has a countable dense subset) if and only if the spectrum of $A$ (denoted by $\Omega(A)$) is separable.
Notes ...
0
votes
0
answers
147
views
Approximation of Inductive Tensor Product $C(X) \bar{\otimes} C(Y)$
The following question is from Banach Algebra Techniques in Operator Theory written by Ronald G. Douglas.
Assume both $X, Y$ are Banach spaces and $X \otimes Y$ is the algebraic tensor product. Let ${...
6
votes
1
answer
353
views
Equivalence of $\sigma$-weak topology to another topology
Let $\mathcal H$ be a Hilbert space. Define a topology $\tau_1$ on $B(\mathcal H)$ by the family of seminorms $x\mapsto |Tr(xa)|,$ $a\in L^1(B(\mathcal H)).$ Here $B(\mathcal H)$ denotes the set of ...
1
vote
0
answers
253
views
When is the weak topology generated by a family of functions Baire?
Suppose we are given a locally compact space $X$ with $C_b(X)$ denoting the continuous bounded complex or real functions on $X$.
Now, if $A\subset C_b(X)$ is given, I am trying to figure out when the ...
3
votes
2
answers
816
views
Is the ring of $p$-adic integers extremally disconnected?
We call a topological space $X$ extremally disconnected if the closures of its open sets remain open. Obviously, Hausdorff extremally disconnected spaces are totally disconnected in the sense that ...
3
votes
2
answers
264
views
Ultraweak topology in abelian von Neumann algebras
Let $A$ be an abelian von Neumann algebra acting on the (not necessarily separable) Hilbert space $\mathcal{H}$ (with identity $I$). From the Gelfand-Neumark theorem, there is a compact Hausdorff ...
0
votes
1
answer
208
views
strict topology on multiplier algebras
Suppose $A$ is a $C^*$ algebra,$M(A)$ is the multiplier algebra.If $S$ is a subset of $M(A)$ which is compact for the strict topology on $M(A)$,is $S$ also a subset of $M(M(A))$ which is compact for ...
2
votes
1
answer
324
views
Direct proof a property of hyperstonean spaces
First, let me state some basic facts and definitions for my question. I believe these are well-known among experts working on von Neumann algebras, but let me state them anyway since my question is ...
9
votes
1
answer
384
views
Comparing two $\sigma$-algebras on $B(\ell^1)$
Let us consider $B(\ell^1)$, bounded linear operators on $\ell^1$. We recall the weak operator topology, denoted by $w$, on $B(\ell^1)$ is determined as follow
$$w-\lim T_i=T \Longleftrightarrow \...
0
votes
1
answer
142
views
A property of compact topological space via certain $C^*$ embedding in operator algebras
Assume that $A$ is a unital $C^*$ algebra. Is there a $C^*$ embedding of $A$ in some $B(H)$ whose image is a hereditary $C^*$ subalgebra of $B(H)$?
If not, is the answer affirmative when $A$ is ...
3
votes
1
answer
199
views
What can be said about the algebra of continuous functions on compact countable ordinals?
Let $X$ be a compact countable Hausdorff space. By Sierpinski-Mazurkiewicz Theorem we know that $X$ is a compact countable ordinal, i.e.
$$
X \simeq \omega ^{\alpha} \cdot n + 1
$$
where $\alpha$ is ...
1
vote
0
answers
81
views
A consecutive resolution of continum algebras to a simple continum algebra
Motivated by classical Gelfand Naimark duality, the correspondence between the category of commutative $C^{*}$ algebras and the category of locally compact Hausdorff spaces, we ...
2
votes
0
answers
208
views
A functor on the category of rings, algebras or compact Hausdorff topological space
Assume that $R$ is a unital ring or a complex or real (Banach or $C^{*}$) algebra.
We define a relation $M$ on $R$ as follows: $$a\;M b \;\;\; \text{iff}\;\; a=xy,\;b=yx \;\; \text{for ...
3
votes
2
answers
796
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 ...
2
votes
1
answer
320
views
Totally non hereditary $C^{*}$-subalgebras
Assume that $B$ is a $C^{*}$ subalgebra of $A$. We say $B$ is totally non hereditary subalgebra of $A$ if not only $B$ is not a hereditary subalgebra but also it is not isomorphic to any ...
17
votes
2
answers
2k
views
The letters of the word "ART"
Edit: According to the Gelfand duality between topological spaces and commutative $C^{*}$algebras, I add some new tags. So the question is that what is the structure of $ Ext (A,A)$ where $A$ is $...
4
votes
0
answers
146
views
A question on extension of $Z^{*}$ algebras
A $Z^{*}$ algebra is a $C^{*}$ algebra which all elements are(two sided or equivalently one sided) zero divisor.
Are there two $Z^{*}$ algebras $A,B$ such that for every short exact sequence of ...
4
votes
0
answers
171
views
$S^{3}$-valued harmonic analysis
Edit:
Note that $S^{3}$ with the quaternion operation is a group. For a locally compact Abelian group $\Gamma$ we consider
$$\tilde{\tilde{\Gamma}}=\{\phi:\Gamma \to S^{3} \mid \phi(xy)=\phi(...
4
votes
1
answer
312
views
A question on $Z^{*}$ algebras
A $Z^{*}$ algebra is a $C^{*}$ algebra which satisfies each of the following equivalent conditions:
All elements of $A$ are left zero divisor.
All elements are right zero divisor.
All elements are ...
4
votes
1
answer
271
views
Strange (?) definition of the spectrum
Suppose that $A$ is a commutative, unital $C^*$-algebra. Then it is isomorphic to $C(X)$ for some compact Hausdorff topological space $X$. $X$ can be identified as the space of all unital ...
7
votes
1
answer
730
views
Formal series convergence in deformation quantization and $C^*$-condition
A link between formal series convergence in deformation quantization (strict deformation quantization) and producing $C^*$-algebras instead of mere $*$-algebras (which $(\mathcal{C}^{\infty}(M)[[t]],\...
5
votes
1
answer
897
views
Folium in GNS construction and von Neumann algebras
The GNS construction allows one to represent a $C^*$-algebra as the algebra of bounded operators on a Hilbert space when a state is fixed, this state being represented as a vector on the Hilbert space....
3
votes
0
answers
431
views
Bohr topos and quantization
Bohrification is a natural way to construct a quantum "phase space" (with some nice insights on foundational problems like non-contextuality through Kochen-Specker etc). I was wondering, since we get ...
8
votes
2
answers
2k
views
von neumann algebras and measurable spaces
I've read some pages on links between von neumann (VN) algebras and measurable spaces (Spectra of $C^*$ algebras and Non-commutative geometry from von Neumann algebras?), but I can't get the following:...
5
votes
1
answer
494
views
When is a $*$-homomorphism between multiplier algebras strictly continuous?
(This question was posted on MSE here but didn't get any answers.)
The strict topology on the multiplier algebra M(A) of a C*-algebra A is that generated by the seminorms
$$ x\mapsto \|ax\|\quad x\...
12
votes
2
answers
479
views
C*-algebras with no nontrivial endomorphisms
Pick a C*-algebra $A$ and call a (*-)endomorphism $\alpha:A\to A$ nontrivial if it is injective and $\alpha(A)\neq A$.
Question: Do there exist infinite dimensional C*-algebras with no nontrivial ...
2
votes
1
answer
245
views
Assumptions on a commutative C*-algebra to get a nice C(X) - space
I have the following question,
Is it possible to get somehow a compact Hausdorff space $X$ which is second-countable from a unital commutative C*-algebra. If it is possible, what should we assume ...
0
votes
1
answer
386
views
The functor of continuous functions from compact CW-spaces to the reals
The contravariant functor $C(-)$ given by
$$
\hom_{Top}(-,\mathbb{R}):cCW\to Rng
$$
where $cCW$ is the category of compact CW complexes is injective on objects. What is known about surjectivity, ...
37
votes
5
answers
4k
views
Reference for the Gelfand duality theorem for commutative von Neumann algebras
The Gelfand duality theorem for commutative von Neumann algebras states that the following three categories are equivalent:
(1) The opposite category of the category of commutative von Neumann ...
38
votes
7
answers
5k
views
Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?)
In this question, Harry Gindi states:
The fact that a commutative ring has a natural topological space associated with it is a really interesting coincidence.
Moreover, in the answers, Pete L. ...