Type I group von Neumann algebra

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Let $G$ be a locally compact group. One says that $G$ is type I if its full group $C^*$ algebra is type I. Since being type I passes to quotients and generated von Neumann algebras, if $G$ is type I then reduced group $C^*$-algebra $C^*_r(G)$ and group von Neumann algebra $L(G)$ are also type I.

I have two "converse" questions: let $G$ be a general locally compact group.

  1. Assume that $C^*_r(G)$ is type I. Does it follow that $C^*(G)$ is type I?

  2. Assume that $L(G)$ is type I. Does it follow that $C^*_r(G)$ is type I?

I vaguely remember that the answer to 2. is negative, but I cannot find a source.

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Considering the visitors to this page, perhaps it is useful to list some preliminaries & definitions first.

A $C^*$-algebra $A$ is Type I iff it is GCR/postliminal iff $A^{**}$ is a Type I von Neumann algebra.

A von Neumann algebra is said to be atomic if it is generated by its minimal projections, equivalently if it is isomorphic to $(\bigoplus_{i\in J} B(H_i))_{\ell^{\infty}}$ for some set of Hilbert spaces $\{H_i:i\in J\}$. Clearly an atomic von Neumann algebra is Type I.

A locally compact group $G$ is said to be of class $[AR]$ if its regular representation is completely reducible a.k.a. its regular representation is atomic. $G$ is of $[AR]$ iff $VN(G)$ (another notation for $L(G)$) is an atomic von Neumann algebra, see Theorems 3.5 & 4.1 in [Taylor1983].

$C_r^*(G)=C^*(G)$ iff $G$ is amenable. If $G$ is not amenable, there exists a surjective homomorphism $\phi:C^*(G)\to C_r^*(G)$ that fixes each $f\in L^1(G)$.


For the 2nd question, there exist totally disconnected LCGs of $[AR]$ that aren't Type I, please see Theorem 4.1 and the remark after in [Blackadar1977]. Also, Baggett and Taylor had provided connected LCGs of $[AR]$ in [BaggettTaylor1978], one of which is amenable & not Type I. For these groups, $VN(G)$ is atomic so Type I, whereas $(C_r^*(G))^{**} = (C^*(G))^{**} = W^*(G)$ (the universal enveloping von Neumann algebra) is not Type I.

I don't have an answer for the 1st question, although if there exists a counterexample $G$, it must be non-amenable and non-Type I.