Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a field and $G$ be a torsion-free group. Kaplansky's idempotent conjecture states that the group ring $K[G]$ does not contain any non-trivial idempotent, i.e. if $x^2=x$ then $x=0$ or $x=1$.

Is Kaplansky's idempotent conjecture known for Thompson's group $F$?

share|improve this question
4  
I think F is orderable and hence satisfies this conjecture. –  Mustafa Gokhan Benli Feb 24 at 17:16
    
Thanks Mustafa! –  Mahdi Teymuri Garakani Feb 25 at 15:02

2 Answers 2

up vote 11 down vote accepted

Thompson's group $F$ satisfies the idempotent conjecture, because it is torsion-free and orderable. For torsion-free groups it is known that the zero-divisor conjecture for group rings implies the idempotent conjecture. Malcev has proved in $1948$ that orderable groups satisfy the zero-divisor conjecture. Hence the claim follows for Thopson's group. For more details see What is the current status of the Kaplansky zero-divisor conjecture for group rings?.
A reference for properties of Thompson's group $F$ can be found here.

share|improve this answer
    
Thanks for the answer! –  Mahdi Teymuri Garakani Feb 25 at 15:01
    
You are welcome! –  Dietrich Burde Feb 27 at 13:15
1  
@DietrichBurde What about Kaplanski-Kadison conjecture for such groups? A more general question:Is there a free group $G$ which satisfies Kaplanski conjecture but does not satisfies the Kadison conjecture? –  Ali Taghavi Aug 18 at 6:10

They satisfy Baum-Connes conjecture, so by surjevtivity of the assembly map, the Thompson's group $F$ satisfies even the stronger conjecture of Kaplansky and Kadison: the reduced group $C^*$-algebra has no idempotents or projections.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.