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'shanksii) 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.

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.