# Is the conditional expectation faithful?

Let $$G$$ be a locally compact group and let $$H$$ be an 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 $$E$$ isn’t faithful in general, there is an example in the paper right before Definition 1.3.: 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.

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).