# Normal character on a group von Neumann algebra

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.

• If such a normal character exists, then $G$ must be at least be amenable. Commented Apr 26 at 12:12
• Yes, as its restriction to the reduced group $C^*$-algebra will give non-zero (by normality) character. Commented Apr 26 at 12:14
• A partial answer: in the discrete case, an infinite discrete group generates a diffuse vNa, so its group vNa admits no normal character. So this is at least true for discrete groups. Commented Apr 27 at 22:16

## 2 Answers

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.)

• Nice answer. Now I wonder if the same claim is true for locally compact quantum groups, i.e. if $G$ is a locally compact quantum group then $L^\infty(\hat{G})$ has a normal character if and only if $G$ is compact. Again, one implication is immediate. Commented Apr 28 at 7:08
• @J.DeRo I wouldn’t know. I’m not familiar with the theory of quantum groups. But I wouldn’t be surprised given how much parallels there are between quantum groups and classical groups. Commented Apr 28 at 8:25

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.