For a locally compact group $G$, I will denote by $L(G)$ its group von Neumann algebra, which is the von Neumann algebra acting on $L^2(G)$, generated by the image of left regular representation $\lambda(x)\xi(y)=\xi(x^{-1}y)$.
It is well known (by Plancherel theorem) that if $G$ is compact, then $L(G)$ is a direct product of matrix algebras $L(G)\simeq \prod_i M_{n_i}(\mathbb{C})$, where each summand $i$ corresponds to a (class of) irreducible representation of $G$, in particular $n_i<+\infty$. One of the summands corresponds to the trivial representation, hence we obtain $L(G)\simeq \mathbb{C}\oplus N$ for a von Neumann algebra $N$. In particular, there is a normal character $L(G)\rightarrow \mathbb{C}$ given by $(z,n)\mapsto z$.
My question is as follows: assume that $G$ is an arbitrary locally compact group and $L(G)$ admits a normal character $L(G)\rightarrow \mathbb{C}$ (equivalently $L(G)\simeq \mathbb{C}\oplus N$ for some $N$). Does it follow that $G$ is compact?
Clearly it is an important assumption that character $L(G)\rightarrow\mathbb{C}$ is normal, as e.g. $L(\mathbb{R})\simeq L^{\infty}(\mathbb{R})$ admits plenty of non-normal characters.
Best Answer
I realized that the proof for the discrete case carries over to the general case. The following is a proof:
Assume that $G$ is not compact. Let $\Omega$ be the net of compact subsets of $G$, ordered by inclusion. Since $G$ is not compact, we may, for each $\omega \in \Omega$, choose $g_\omega \in G \setminus \omega$.
I claim that $\lim_\Omega \lambda(g_\omega) = 0$ in the weak operator topology. Indeed, for any $\xi, \eta \in L^2(G)$ and any $\epsilon > 0$, choose $\epsilon’ > 0$ small enough s.t. $\|\xi\|\epsilon’ + (\|\eta\| + \epsilon’)\epsilon’ < \epsilon$. Now choose $\xi’, \eta’ \in C_c(G)$ s.t. $\|\xi - \xi’\| < \epsilon’$ and $\|\eta - \eta’\| < \epsilon’$. By the choice of $\epsilon’$, we have $|\langle \xi, \lambda(g)\eta \rangle - \langle \xi’, \lambda(g)\eta’ \rangle| < \epsilon$ for all $g \in G$. Since $\xi’, \eta’$ are both compactly supported, there is a compact set $K$ s.t. whenever $g \notin K$, we have $\langle \xi’, \lambda(g)\eta’ \rangle = 0$. But this means, whenever $\omega \supset K$,
$$|\langle \xi, \lambda(g_\omega)\eta \rangle| < |\langle \xi’, \lambda(g_\omega)\eta’ \rangle| + \epsilon = \epsilon$$
This implies $L(G)$ cannot admit a normal character. Indeed, assume otherwise that $\pi: L(G) \to \mathbb{C}$ is a normal character, then $\lim_\Omega \pi(\lambda(g_\omega)) = 0$. But $\lambda(g_\omega)$ is a unitary, so $|\pi(\lambda(g_\omega))| = 1$, a contradiction.
(More generally, this shows $L(G)$ cannot even admit a finite-dimensional summand.)