All Questions
Tagged with oa.operator-algebras ra.rings-and-algebras
66 questions
40
votes
9
answers
10k
views
Simplest examples of rings that are not isomorphic to their opposites
What are the simplest examples of
rings that are not isomorphic to their
opposite rings? Is there a science to constructing them?
The only simple example known to me:
In Jacobson's Basic Algebra (...
- 5,654
26
votes
7
answers
5k
views
Commutative subalgebras of M_n
For a given $n$, is there any characterization for the commutative subalgebras of $M_n(\Bbb{C})$? I would like to know how many commutative subalgebras there are for each possible dimension.
In view ...
- 397
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 ...
- 2,071
17
votes
1
answer
710
views
Standard polynomials applied to matrices
The standard polynomial in $r$ non-commuting indeterminates $x_1,\ldots,x_r$ is defined by
$${\mathcal S}_r(x_1,\ldots,x_r):=\sum_{\sigma\in S_r}\epsilon(\sigma)x_{\sigma(1)}x_{\sigma(2)}\cdots x_{\...
- 52.3k
16
votes
3
answers
3k
views
Are there other semidirect product/crossed products in other areas
Suppose $(O, G, \alpha)$ is a triple where $O$ is some mathematical object, $G$ is a group and $\alpha : G \rightarrow Aut(O)$. Many different areas of mathematics study such triples. However, I only ...
- 3,984
14
votes
2
answers
549
views
$\mathbb{Z}G$ (left) Noetherian$\Rightarrow$ $\ell^1(G)$ is a flat (right) $\mathbb{Z}G$-module?
Let $G$ be a countable discrete group (not necessarily abelian), and suppose the group ring $\mathbb{Z}G$ is a left-Noetherian ring, for example, when $G$ is a polycyclic-by-finite group.
Denote the ...
- 1,528
14
votes
1
answer
616
views
How "nondegenerate" are amalgamated free products of C*-algebras?
In the following, I assume all algebras are unital. Let $A$ and $B$ be C*-algebras that each contain (isomorphic copies of) a common C*-subalgebra $C$. Let $A *_C B$ denote the amalgamated free ...
- 5,407
13
votes
0
answers
573
views
Classical (i.e. commutative) spaces with quantum symmetry but no classical symmetry
In a recent preprint (arXiv:2311.04889), my coauthors and I constructed a sequence of graphs with no classical symmetry which nevertheless have quantum symmetry.
For graphs this had been an open ...
- 2,339
12
votes
5
answers
2k
views
Group ring and left zero divisor
Let $K$ be a finite field and $G$ be a discrete group.
Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?
It does not seem to be related to zero divisor problem, any ...
- 4,705
9
votes
1
answer
521
views
Which group algebras in analysis are "true group algebras"?
Let $G$ be a group, $A$ a unital associative algebra over ${\mathbb C}$, and let us call a representation of $G$ in $A$ an arbitrary map $\pi:G\to A$ such that
$$
\pi(1)=1,\qquad \pi(a\cdot b)=\pi(a)\...
- 7,426
8
votes
1
answer
809
views
Strong Morita equivalence and representation theory
In the context of pure algebra we say that two algebras (in general: rings) $A,B$ are Morita equivalence when there are bimodules $_AP_B,_BQ_A$ such that $P \otimes_B Q \cong _AA_A$ and $Q \otimes_A P ...
- 9,330
8
votes
1
answer
740
views
A question regarding Kadison-Kaplansky idempotent conjecture (A nearest group element $g$ to a nontrivial self adjoint unitary element u )
Edit: According to answer and comments by Prof. Valette we edite the question.
The Kadison Kaplansky conjecture says:
Kadison-Kaplansky conjecture: If $G$ is a torsion-free discrete group then $C^*_{\...
- 356
8
votes
1
answer
366
views
Does the following $ C^{*} $-algebraic result have a purely algebraic proof?
While studying the proof of Bott periodicity for operator $ K $-theory in this set of notes, I learned this fact:
Theorem. Let $ A $ and $ B $ be $ C^{*} $-algebras. Let $ f,g: A \to B $ be $ * $-...
- 1,396
8
votes
1
answer
281
views
Factor traces of the Temperley-Lieb algebra
Given $\delta\in\mathbb C$, let $A(\delta)$ denote the complex unital $*$-algebra generated by an identity $1$ and selfadjoint elements $e_k$, $k\in\mathbb N$, satisfying $e_k^2=\delta e_k$, $e_ke_l=...
- 1,017
8
votes
0
answers
251
views
When does a semisimple $\mathbb{C}$-algebra come from a group?
Let $\mathcal{A}$ be a semisimple $\mathbb{C}$-algebra. By the Artin-Wedderburn theorem, it is isomorphic to a direct product of matrix algebras:
$$ \mathcal{A} = \prod_{i=1}^m M_{n_i}(\mathbb{C})$$
...
- 297
7
votes
3
answers
409
views
Are nearby subalgebras of matrix algebras conjugate?
Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M_n(k)$ are ...
- 9,853
7
votes
1
answer
479
views
Characterisation of finite dimensional C*-algebras?
$\DeclareMathOperator\Spec{Spec}$Let $A$ be a finite dimensional $*$-algebra over $\mathbb C$.
(Namely, an associate algebra equipped with an involution $*:A\to A$ satisfying $(ab)^*=b^*a^*$ and $(\...
- 43.2k
7
votes
1
answer
250
views
$K_0$ group of graph underlying an approximately finite (AF) C* algebra
Say we have an AF C* algebra $A$ described by some Bratteli diagram $E$. If $M_\infty (A)=\displaystyle{\lim_\rightarrow M_n(A)}$ and $P(A)$ are the projections in this algebra, we know that $K_0(A)^+=...
- 125
7
votes
1
answer
339
views
Why is the definition of von Neumann trace independent of the choice of the Hilbert space?
A Hilbert module defined 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 space $V$ ...
- 2,118
7
votes
1
answer
1k
views
The saturation of Murray von Neumann relation
Edit: According to comment of Pace Nielsen, I remove question 2 of the previous version:
Let $R$ be a unital ring. We define Murray Von Neumann relation $M$ on $R$ as follows:
We say $a M b$ iff $...
- 356
6
votes
2
answers
1k
views
Are the banded versions of a positive definite matrix positive definite?
Consider $M$, a positive definite matrix. Let $M^{(1)}$ be the diagonal matrix which agrees with $M$ on the diagonal ($M_{ii}=M^{(1)}_{ii}$). We have that $M^{(1)}$ is positive definite because it is ...
- 375
6
votes
1
answer
550
views
A generalized K- theory via generalized idempotents
Edit After the answer by Neil Strickland, I add the word "a ring" in this new version.
In the literature, there is a concept of generalized idempotent: an n- idempotent is an element $a$ of a Banach ...
- 356
6
votes
1
answer
322
views
Pairwise orthogonality for partitions of unity in a *-algebra
Let $\mathcal{A}$ be a $*$-algebra with unit $1_{\mathcal{A}}$. As in the $\mathrm{C}^*$-setting, a projection is an element $p\in\mathcal{A}$ such that $p=p^2=p^*$. A partition of unity is a finite ...
- 1,027
6
votes
0
answers
106
views
Automorphisms of algebraic Clifford algebra of a Hilbert space
Let $H$ be a real separable, infinite-dimensional Hilbert space and let
$$\mathrm{Cl}(H) = \mathcal{T}(H_{\mathbb{C}}) / \{v\otimes w + w\otimes w - 2\langle v, w\rangle \cdot \mathbf{1} ~|~ v, w \in ...
- 9,853
5
votes
1
answer
254
views
Well defined Tensoring of spectral triples
Hi,
I have a misunderstanding that I am hoping is really quite trivial. I will give my question directly and context below for those that need/want it.
Question: In connes standard model he takes ...
- 133
5
votes
1
answer
342
views
Can the trivial module be stably free for a monoid ring?
Let $M$ be a non-trivial monoid and $\mathbb ZM$ its monoid ring. All modules are left modules in what follows. Suppose that $M$ contains a zero element (or absorbing element) $z$. That is $mz=z=zm$ ...
- 38.6k
5
votes
0
answers
296
views
Are these element in a group algebra of a torsion-free group zero divisors?
Let $G$ be an arbitrary torsion-free group. For $x,y\in G$, which of these elements can be decided immediately not to be zero divisors in $\mathbb ZG$ (or in $\mathbb CG$)?
$$1+x+y,\quad 4+x+x^{-1}+y+...
- 2,118
4
votes
2
answers
499
views
rank of fin gen projective modules over C* algebras
Apologies - a better explanation than I started with - thanks to people for helping. It is obvious that there are many bad cases for rank - the problem is are there a reasonable number of good cases?
...
- 1,143
4
votes
1
answer
199
views
Groups for which all projections of $C^*_{\text{red}}G$ belong to $\mathbb{C}G$
Revision: According to comment of Wojowu we give a complete revise for this post.
A group $G$ is a pr-group if all projections of $C^*_{\text{red}} G$ are contained in its dense subalgebra $\mathbb{...
- 356
4
votes
0
answers
81
views
Question about terminology for a class of "self-modular" mappings between rings
(In the scenario I have in mind, rings need not be unital.)
The following notion has come up in some joint work that is being written up. Let $R$ and $S$ be rings, and let $D$ be a subring of $R$. Is ...
- 25.8k
4
votes
0
answers
133
views
Contractible Banach algebras
A Banach algebra $A$ is contractible if $H^1(A, X)=0$ for all Banach $A$-bimodules $X$. Now to my question
Let $A$ be Banach algebra and $I$ be closed ideal of $A$. If $I$ and $A/I$ are both ...
- 823
4
votes
0
answers
185
views
ring structure of $KK_*(A,A)$ for a separable $C^*$-algebra $A$
Motivation:
For a topological space $X$ one can consider under certain circumstances the cohomology ring of suitable cohomology theories, for example:
1) The cohomology ring $H^*(X;R)=\oplus_{i\ge ...
- 1,188
4
votes
0
answers
290
views
C*-algebras and bounded relations
I'm trying to get used to the language of generators and relations for C*-algebras through Loring's "Lifting Solutions to Perturbing Problems in C*-Algebras". So far this is what I got from the first ...
- 417
4
votes
0
answers
172
views
reference for direct finiteness of the ring of affiliated operators
Let $\Gamma$ be a group, $N(\Gamma)$ its group von Neumann algebra,
$\newcommand{\cUG}{{\mathcal U}(\Gamma)}$
and $\cUG$ the ring of all densely-defined, closed operators $\ell^2(\Gamma)\to\ell^2(\...
- 25.8k
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 ...
- 2,118
3
votes
1
answer
393
views
Kaplansky Idempotent conjecture and Extension theory
We consider the Idempotent Kaplansky conjecture with $\mathbb{C}$- coefficients, that is the problem of nontrivial idempotents for group algebra $\mathbb{C}\Gamma$ where $\Gamma$ is a torsion free ...
- 356
3
votes
1
answer
309
views
Metrics and completions on the direct limit of matrices of all sizes over arbitrary fields
Let $M_n$ denote the $n$ by $n$ matrices.
Consider the homomorphisms
$$\phi_{n,kn}: M_n \rightarrow M_{kn}$$
which takes a matrix $A \in M_n$ to $A \otimes I_k \in M_{kn}.$
This gives a sensible way ...
- 1,429
3
votes
0
answers
109
views
Does this element belong to $\mathbb CG$?
Let $G$ be a torsion-free group. Let $\alpha$ be a symmetric element of $\mathbb CG$, i.e. $\alpha^*=\alpha$, with $\|\alpha\|_1=\sum|\alpha(g)|<1$, so $\beta:=\sum_{n\ge 0}(-1)^n\alpha^n$ is an ...
- 2,118
3
votes
0
answers
183
views
Is the construction of ring C*-algebra functorial?
Cunz and Li defined defined C*-algebras for arbitrary rings with of course some condition. One can look at their article (http://arxiv.org/abs/0905.4861). My question: is the construction functorial? ...
- 95
2
votes
2
answers
207
views
Are all automorphisms of Lin(V) given by similarity transforms?
Let $V$ be a vector space with dimension greater than 1 over the field $F$ and $Sim = \{(f\in \operatorname{Lin}(V))\mapsto gf(g^{-1}) : g\in \operatorname{GL}(V)\}$, ie $Sim$ is the set of all ...
user5810
2
votes
1
answer
171
views
algebraic version and polar decomposition
I have been thinking about polar decomposition of $C^*$-algebras. I could not find a proper reference where it says: Let $A$ be a $C^*$-algebra, and $b$ an invertible element of $A$ with modulus $\...
2
votes
1
answer
438
views
Reference request (or otherwise): Adjoint action
I am interested in clearing up a confusion of mine. I will try to make my question as clear as I can but I apologize in advance if this is not the case.
Given a unitary group of some unital ...
- 133
2
votes
0
answers
88
views
Non-existence of idempotent via evaluation of higher order cocycle on a tuple of idempotents
The Kaplansky conjecture and Kaplanski-Kadison conjecture are classical conjectures about non existence of non-trivial idempotents in group algebra $\mathbb{C}\Gamma $ or the reduced group algebra $...
- 356
2
votes
0
answers
145
views
Solvability and nilpotency for Banach algebras
Do we have topological counterparts of solvability and nilpotency, which are central concepts for (finite-dimensional) Lie algebras, for infinite dimensional Banach algebras with the commutator ...
- 2,605
2
votes
0
answers
95
views
Algebraic language of fundamental results in operator algebras
Ignoring topological structures of von Neumann algebras, the larger category of Baer $*$-rings was emerged. In the unique text written by Sterling K. Berberian (1), some concepts and results in von ...
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 ...
- 356
1
vote
1
answer
286
views
A subalgebra of $B(H)$ which does not contain a commutator element
Is there a commutative subalgebra $A\subset B(H)$ containing the 1 dimensional scalars with the following property:
The algebra $A$ has trivial intersection with the set of commutator ...
- 356
1
vote
3
answers
229
views
Some non-trivial Baer *-rings
A Baer *-ring is an *-algebra whose lattice of projections is complete. I know two well-handed kinds of these structures:
1- W*-algebras (abstract case of von Neumann algebras).
2- The inverse ...
- 4,058
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?
- 356
1
vote
1
answer
164
views
A Baer *-ring which is not embedded into $B(H)$
Assume $A$ is a complex $*$-algebra which is also a Baer*-ring.
Q. Can we concluded that there exists a Hilbert space $H$ such that $A$ is embedded in $B(H)$ as a Baer*-ring? What about when $A$ ...
- 4,058