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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
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 ...
Aareyan Manzoor's user avatar
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 ...
truebaran's user avatar
  • 9,330
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 ...
Very Forgetful Functor's user avatar
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 $X$ ...
Rick Sternbach's user avatar
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 ...
Matthias Ludewig's user avatar
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 $A$ ...
truebaran's user avatar
  • 9,330
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, ...
Douglas Somerset's user avatar
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 ...
ned grekerzberg's user avatar
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 ${...
Sanae Kochiya's user avatar
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 ...
A beginner mathmatician's user avatar
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 ...
Merry's user avatar
  • 173
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 ...
Rick Sternbach's user avatar
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 ...
condexp's user avatar
  • 159
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 ...
math112358's user avatar
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 ...
Rick Sternbach's user avatar
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 \...
ABB's user avatar
  • 4,058
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 ...
Ali Taghavi's user avatar
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 ...
Niki's user avatar
  • 335
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
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
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 ...
ray's user avatar
  • 687
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 ...
Ali Taghavi's user avatar
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 $...
Ali Taghavi's user avatar
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 ...
Ali Taghavi's user avatar
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(...
Ali Taghavi's user avatar
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 ...
Ali Taghavi's user avatar
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 ...
truebaran's user avatar
  • 9,330
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]],\...
Issam Ibnouhsein's user avatar
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....
Issam Ibnouhsein's user avatar
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 ...
Issam Ibnouhsein's user avatar
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:...
Issam Ibnouhsein's user avatar
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\...
Paul McKenney's user avatar
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 ...
Ollie's user avatar
  • 1,411
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 ...
Celeban's user avatar
  • 145
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, ...
roger123's user avatar
  • 2,782
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 ...
Dmitri Pavlov's user avatar
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. ...
Kevin H. Lin's user avatar