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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 $F$?

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I think F is orderable and hence satisfies this conjecture. – Mustafa Gokhan Benli Feb 24 '14 at 17:16
Thanks Mustafa! – Mahdi Teymuri Garakani Feb 25 '14 at 15:02
up vote 12 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.

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Thanks for the answer! – Mahdi Teymuri Garakani Feb 25 '14 at 15:01
You are welcome! – Dietrich Burde Feb 27 '14 at 13:15
@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 '14 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.

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