Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$

Details Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$ Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$

2006-09-04

arxiv.org/abs/math/0609102v2

Mathematics

Operator Algebras

30 pages; typos corrected

Scientific paper

10.1007/s00222-007-0050-5

For $n\geq 3,$ let $\Gamma=SL_n(\mathbb Z).$ We prove the following superridigity result for $\Gamma$ in the context of operator algebras. Let $L(\Gamma)$ be the von Neumann algebra generated by the left regular representation of $\Gamma.$ Let $M$ be a finite factor and let $U(M)$ be its unitary group. Let $\pi: \Gamma\to U(M)$ be a group homomorphism such that $\pi(\Gamma)''=M.$ Then \begin{itemize} \item[(i)] either $M$ is finite dimensional, or \item [(ii)] there exists a subgroup of finite index $\Lambda$ of $\Gamma$ such that $\pi|_\Lambda$ extends to a homomorphism $U(L(\Lambda))\to U(M).$ \end{itemize} The result is deduced from a complete description of the tracial states on the full $C^*$--algebra of $\Gamma.$ As another application, we show that the full $C^*$--algebra of $\Gamma$ has no faithful tracial state.

Affiliated with

Also associated with

No associations

Rating

Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$ does not yet have a rating. At this time, there are no reviews or comments for this scientific paper.

If you have personal experience with Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$, we encourage you to share that experience with our LandOfFree.com community. Your opinion is very important and Operator-algebraic superrigidity for $SL_n(\mathbb Z),n\geq 3$ will most certainly appreciate the feedback.

Rate now

Comments { 0 }

Profile ID: LFWR-SCP-O-727334