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](https://doi.org/10.1016/0001-8708(74)90068-1), 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.



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> Is then $E$ faithful?
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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.



Is the conditional expectation faithful?
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).


