# is the conditional expectation faithful?

Let $G$ be locally compact group and let $H$ be a open subgroup in $G$. Then the full group $C^*$-algebra of $H$, $C^*(H)$, is a subalgebra of $C^*(G)$ and there is a conditional expectation $$E\colon C^*(G)\to C^*(H),$$ which is induced by restriction $f\in L^1(G) \mapsto f_{|H}\in L^1(H)$ of functions which are integrable w.r.t. the left Haar measure on $G$, see Rieffel, induced representations of $C^*$-algebras, Proposition 1.2.

Note that $E$ is'n faithful in general, there is an example in this paper in the section above definition 3.: Consider $G$ a nonamenable discrete group and $H$ an open subgroup consisting of the identity element of $G$. Then there are nonzero elements $c$ in the kernel of the left regular representation of $G$ (since $G$ is not amenable) and they satisfy $E(c^*c)=0$.

My Question: Now, let $G$ be a locally compact amenable group and $H$ be an open compact (amenable) subgroup.

Is then $E$ faithful?

I think yes (I have considered some examples), but I am stuck with a proof.

If I additionally assume $G$ (and $H$) to be discrete I can prove it considering $E$ as a conditional expectation $C_r^*(G)\to C_r^*(H)$ and then it is $\tau_G=\tau_H\circ E$, where $\tau_G$ and $\tau_H$ are the canonical faithful tracial states on $C_r^*(G)$, $C_r^*(H)$ respectively. It follows that $E$ must be faithful.

For the more general case, I thought about trying a similar strategy using the fact that $C_r^*(G_1)$ of a locally compact group $G_1$ which contains a non-trivial amenable open subgroup $H_1$ has a tracial state $\tau^{G_1}$ satisfying $$\tau^{G_1}( \lambda_{G_1}(f))=\int_{H_1}f(s)d\mu(s),$$ see corollary 4.1 in 'embedding theorems in group $C^*$-algebra' by Lee for this fact. If one can check that the tracial state $\tau^{H_1}$ is faithful, then this together with $\tau^{G_1}=\tau^{H_1}\circ E$ implies that $E$ is faithful. But I am stuck with proving faithfulness of the trace. Probably I am on the wrong track..Other strategies regarding my question are welcome.

## 1 Answer

As $G$ and $H$ are amenable, we have that $C^*(G) = C^*_r(G) \subseteq VN(G)$ the group von Neumann algebra. So let's prove the stronger result that $E:VN(G)\rightarrow VN(H)$ is faithful.

As $H$ is open, it is also closed (write $H$ as the complement of the union of cosets of $H$). It follows that $G/H$ is discrete topologically, and hence also in measure. So $L^2(G) = \ell^2(G/H) \otimes L^2(H)$ (or, $L^2(G)$ is the $\ell^2$ direct sum of $L^2(H)$, one for each coset).

Let $U:L^2(H)\rightarrow L^2(G)$ be the inclusion, an isometry. $U^*:L^2(G) \rightarrow L^2(H)$ is just restriction. Then $E(x) = U^* x U$. To see this, compute $(f*\xi|\eta)$ for $f\in L^1(G)$ and $\xi,\eta\in L^2(H)\subseteq L^2(G)$.

Let $x\in VN(G)$ with $0 = E(x^*x) = U^*x^*xU$, so $xU=0$. We want to show that $x^*x=0$, that is, $x=0$. As $x\in VN(G)$ it commutes with all right translation operators $\rho(s)$ for $s\in G$. Thus $0 = \rho(s)xU = x \rho(s)U$ and so $x\rho(s)\xi=0$ for all $s\in G, \xi\in L^2(H)$. The result now follows from the observation that the linear span of such vectors $\rho(s)\xi$ is dense in $L^2(G)$ (compute the orthogonal complement).

• I see it's time for me to make myself more familiar with von Neumann algebras. However, I understand, thank you very much. – Sabrina Gemsa Jun 10 '18 at 12:40
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