# Stable isomorphism of group C$^*$-algebras

For a discrete group $$G$$, let $$C^*_r(G)$$ be its reduced group C$$^*$$-algebra.

Question: Do there exist discrete, torsion-free non-isomorphic groups $$G,H$$ such that $$C^*_r(G)$$ and $$C^*_r(H)$$ are stably isomorphic? Stable isomorphism means $$C^*_r(G)\otimes K\cong C^*_r(H)\otimes K$$ where $$K$$ are the compact operators on a separable, infinite-dimensional Hilbert space.

Motivation: It is a well known(?) open question whether or not there exist non-isomorphic, discrete torsion-free groups $$G,H$$ such that $$C^*_r(G)\cong C^*_r(H).$$ If you are like me and your intuition suggests there should exist such a pair of non-isomorphic groups, then you are naturally led to this "easier" version of the rigidity question. (There are a few classes of groups (e.g. abelian, finitely generated 2-step nilpotent) for which it is known that $$G\cong H$$ implies $$C^*_r(G)\cong C^*_r(H).$$)

For years I assumed one could follow this procedure to positively answer my question:

Write down two (smartly chosen) groups arising as (non-split) extensions $$0\to \mathbb{Z}^d\to G_i\to F\to 0$$ where $$F$$ is finite and the actions of $$F$$ on $$\mathbb{Z}^d$$ are the same but the groups are non-isomorphic. Then use a Packer-Raeburn type trick to untwist the actions and get matrix algebras over $$C^*_r(G_1)$$ and $$C^*_r(G_2)$$ isomorphic to each other. I recently decided to actually sit down and do this, and it doesn't work (at least in my examples...). So here I am.

Notice that for abelian C$$^*$$-algebras stably isomorphic implies isomorphic, so stably isomorphic group C$$^*$$-algebras of torsion-free abelian groups implies isomorphic groups.

• If we assume Kadison-Kaplansky, is it known that $M_2(C_r^*(G_1))\cong M_2(C_r^*(G_2))$ implies $C_r^*(G_1)\cong C_r^*(G_2)$? (I was thinking very half-heartedly about pushing matrix units from one algebra over to the other and then trying to play games with traces, but I don't pretend to have thought this through.) Nov 12 at 22:09
• Perhaps one could do something clever with Takai duality: $C^*_r(G) \otimes \mathcal K \simeq (C^*_r(G) \times_\alpha H) \times_{\hat\alpha} \hat H$. By clever, I mean realizing the right-hand-side above as another $C^*_r(G_2)$ where $G$ and $G_2$ are non-isomorphic. Nov 13 at 20:00