Discrete groups G whose full C*-algebra C*(G) is not quasidiagonal?

Question

Is there a known example of a countable discrete group G whose full group C*-algebra C*(G) is not quasidiagonal?

Let us recall that a separable C*-algebra A is quasidiagonal if it admits a faithful *-representation $\pi:A \to L(H)$ on a separable Hilbert space with the property that there is an increasing sequence $(p_n)$ of finite rank projections in $L(H)$ which converges strongly to $1_H$ and such that for each $a\in A$, $\lim_{n\to \infty}\|[\pi(a)p_n-p_n\pi(a)]\|=0$.

$G=SL_3(\mathbb{Z})$ is believed to be a candidate. Bekka has shown that $C^*(SL_3(\mathbb{Z}))$ is not residually finite dimensional.

Answer 1

I don't know much about the subject, but isn't this an example? There is a discrete property (T) group $G$ without nontrivial finite dimensional representations [2, Remark in the last page]. Then, its max C*-algebra contains a non-QD algebra as its direct summand, namely $C^*(G)(1-z)$ for the Kazhdan projection $z$ (corresponding to the trivial rep) [1, Corollary 17.2.6].

[1] N. P. Brown and N. Ozawa. -algebras and finite-dimensional approximations, volume 88 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2008.

[2] P. de la Harpe, A. G. Robertson, and A. Valette. On the spectrum of the sum of generators for a finitely generated group. Israel J. Math., 81(1-2):65–96, 1993.

Answer 2

Over the last week, I was discussing this question with Narutaka Ozawa. I could come up with a proof that quasi-diagonal, torsion generated Kazhdan groups must have a finite quotient. Here, a group is called quasi-diagonal, if it admits a faithful quasi-diagonal unitary representation. In particular, some Tarski monsters (as constructed by Ol'shanskii) cannot have a quasi-diagonal maximal group $C^{\ast}$-algebra.

Meanwhile, we could generalize this result and prove:

Theorem: Every infinite quasi-diagonal Kazhdan group has an infinite residually finite quotient.

The proof is a bit involved, but it is easy to prove a weaker result in a special case. Assume that $G$ has a quasi-diagonal unitary representation $\pi \colon G \to U(H)$ without fixed vectors, and show that there is at least one non-trivial finite quotient. Let $p_n$ be a sequence of finite rank projections, such that $\|[\pi(g),p_n]\|\to 0$.

Now, for a finite rank operator $T$, $$\|T\|_{HS} \leq {\rm rk}(T)^{1/2} \cdot \|T\|,$$ where $\|.\|_{HS}$ denotes the Hilbert-Schmidt norm. This easily implies that $p_n \cdot {\rm rk}(p_n)^{-1/2}$ is a sequence of unit vectors in $HS(H)$ (the Hilbert space of Hilbert-Schmidt operators) which is more and more invariant under the conjugation action. Hence, since $G$ is a Kazhdan group, there must exist a conjugation invariant Hilbert-Schmidt operator $T$. Since the eigenspaces of $T^*T$ are finite-dimensional and $G$-invariant, $G$ admits a non-trivial finite-dimensional representation. By Malcev's theorem, every finitely generated subgroup of $U(n)$ is residually finite, so that $G$ also admits at least one non-trivial finite quotient.

The remaining work involves showing that one can assume that $\pi$ has no fixed vectors and that there exist many finite quotients. We will upload the complete argument to the arXiv within the next weeks.