Normal character on a group von Neumann algebra

Question

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.

Answer 1

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.)

Answer 2

David Gao's answer is perfectly good, thank you!

Now I have a solution in greater generality of locally compact quantum groups; I'll write it in case someone is interested.

To be precise, the claim is: let $\mathbb{G}$ be a locally compact quantum group which admits a character $\omega\in L^1(\mathbb{G}):=L^{\infty}(\mathbb{G})_*$. Then $\mathbb{G}$ is discrete.

This claim generalised David Gao's answer, as we can take $\mathbb{G}=\widehat{H}$ for a locally compact group $H$, then $L^{\infty}(\mathbb{G})=L(H)$.

Proof: First, as J. De Ro observed in a comment, $\omega$ restricts to a non-zero (by normality) character on $C_0(\mathbb{G})$, hence by Theorem 3.1 in [Amenability and co-amenability for locally compact quantum groups; Bedos, Tuset], $\mathbb{G}$ is coamenable. In particular, $\mathbb{G}$ admits a continuous counit $\epsilon\in C_0(\mathbb{G})^*$.

Next, observe that $\omega$ is invariant under scaling group. I suspect this (or something very close) is written in literature, but I cannot find a good reference. One proof goes as follows: for $t\in \mathbb{R}$ consider $\hat{x}_t:=(\omega\tau_t \otimes id)W=\hat{\tau}_{-t}( (\omega\otimes id)W)\in L^{\infty}(\widehat{\mathbb{G}})$, where $W=L^{\infty}(\mathbb{G})\bar\otimes L^{\infty}(\widehat{\mathbb{G}})$ is the left Kac-Takesaki operator of $\mathbb{G}$. Since $\omega\tau_t\neq 0$, also $\widehat{x}_t\neq 0$. Now, $\hat{\Delta}(\hat{x}_t)=(\hat{\tau}_{-t}\otimes \hat{\tau}_{-t})((\omega\otimes id\otimes id)(W_{13}W_{12}))=(\hat{\tau}_{-t}\otimes \hat{\tau}_{-t})((\omega\otimes id)(W)\otimes (\omega\otimes id)(W))=\hat{x}_t\otimes \hat{x}_t$, using multiplicativity of $\omega$. By Theorem 3.9 (and its proof) in [From quantum groups to groups; Kalantar, Neufang] we have that $\hat{x}_t$ belongs to the intrinsic group $Gr(\widehat{\mathbb{G}})$, i.e. $\hat{x}_t$ is invertible. As discussed on page 6 of this paper, elements of $Gr(\widehat{\mathbb{G}})$ are invariant under scaling group. Thus $\hat{x}_t=(\omega\tau_t\otimes id)W=\hat{\tau}_{-t}((\omega\otimes id)W)=(\omega\otimes id)W$ and we get $\omega\tau_t=\omega$ as claimed.

Let $S,R$ be the antipode and unitary antipode of $\mathbb{G}$. Next we claim that $\omega\star\omega R$, restricted to $C_0(\mathbb{G})$ is equal to the counit. Take $\rho\in L^1(\widehat{\mathbb{G}})$. We have $$(\omega\star\omega R)((id\otimes \rho)W)=(\omega\otimes \omega R\otimes \rho)(W_{13}W_{23})=\omega S( (id\otimes \rho(\omega\otimes id)(W))W)=\omega ( (id\otimes \rho(\omega\otimes id)(W))W^*)=(\omega\otimes \omega\otimes \rho)(W_{13}W_{23}^*)=(\omega\otimes \rho)(WW^*)=\omega(1_{L^{\infty}(\mathbb{G})})\rho(1_{L^{\infty}(\widehat{\mathbb{G}})})=\rho(1_{L^{\infty}(\widehat{\mathbb{G}})})=\epsilon ((id\otimes \rho)(W))$$ In this calculation we have used $\omega R=\omega S$ (on domain of $S$), which follows from $\omega\tau_t=\omega$. Finally we obtain $M(C_0(\widehat{\mathbb{G}}))\ni 1_{L^{\infty}(\widehat{\mathbb{G}})}=(\epsilon\otimes id)W=(\omega\star\omega R\otimes id)W$. But $\omega\star\omega R\in L^1(\mathbb{G})$, so in fact $1_{L^{\infty}(\widehat{\mathbb{G}})}\in C_0(\widehat{\mathbb{G}})$, i.e. $\widehat{\mathbb{G}}$ is compact and $\mathbb{G}$ is discrete.