All Questions
6 questions
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 $...
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{...
0
votes
1
answer
185
views
A subset (or subgroup) associated to a group
Edit: According to comment conversations we revise the question.
Let $G$ be a group. We consider the following subset of $G$:
$$\{g\in G \mid e^{\lambda_g} \in \mathbb{C}\lambda (G)\},$$
where $\...
8
votes
1
answer
739
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^*_{\...
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 ...
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+...