On the Relative Weak Asymptotic Homomorphism Property for Triples of Group von Neumann Algebras

Let $1\in B\subset N\subset M$ be a triple of finite von Neumann algebras gifted with a fixed, normal, finite, faithful and normalized trace $\tau$. Then $\mathbb{E}_{N}$ (resp. $\mathbb{E}_{B}$) denotes the $\tau$-preserving conditional expectation from $M$ onto $N$ (resp. $B$); we also set $M\ominus N=\{x\in M:\mathbb{E}_{N}(x)=0\}$.

Following [3], we say that the triple $B\subset N\subset M$ has the relative weak asymptotic homomorphism property if there exists a net of unitaries $(u_{i})_{i\in I}\subset U(B)$ such that, for all $x,y\in M\ominus N$,

$$\lim_{i\in I}\|\mathbb{E}_{B}(xu_{i}y)\|_{2}=0.$$

The one sided quasi-normalizer of $B$ in $M$ is the set of elements $x\in M$ for which there exist finitely many elements $x_{1},\ldots,x_{n}\in M$ such that $Bx\subset\sum_{i=1}^{n}x_{i}B$. It is denoted by $q\mathcal{N}_{M}^{(1)}(B)$.

Inspired by [2], the authors of [3] prove in Theorem 3.1 that the triple $B\subset N\subset M$ has the relative weak asymptotic homomorphism property if and only if $q\mathcal{N}_{M}^{(1)}(B)\subset N$. Furthermore, they also study the case of group algebras that we recall now.

Let $G$ be a discrete group and let $H$ be a subgroup of $G$. Then there is a natural analogue of the one sided quasi-normalizer for such a pair of groups: we denote by $q\mathcal{N}_{G}^{(1)}(H)$ the set of elements $g\in G$ for which there exist finitely many elements $g_{1},\ldots,g_{n}\in G$ such that $Hg\subset\cup_{i=1}^{n}g_{i}H$.

Thus, if $H<K<G$ is a triple of groups, if $B=L(H)\subset N=L(K)\subset M=L(G)$ denotes the triple of von Neumann algebras associated to $H<K<G$, it is reasonable to ask whether $B\subset N\subset M$ has the relative weak asymptotic homomorphism property if and only if $q\mathcal{N}_{G}^{(1)}(H)\subset K$. Corollary 5.4 in [3] states that this is indeed true when $K=H$, but the proof presented there relies heavily on the main theorem of the article. It is thus natural to look for a more direct and elementary proof of the above mentionned result, and the aim of the present note is to provide such a proof and to add three more equivalent conditions.

Before stating our result, let us fix some additional notations. For each element $g\in G$ we denote by $\lambda_{g}$ the unitary operator acting by left translation on $\ell^{2}(G)$, i.e. $(\lambda_{g}\xi)(g^{\prime})=\xi(g^{-1}g^{\prime})$ for every $\xi\in\ell^{2}(G)$ and every $g^{\prime}\in G$. We denote also by $L_{f}(G)$ the subalgebra of all elements of $L(G)$ with finite support, i.e. $L_{f}(G)$ is the linear span of $\lambda(G)$ in $B(\ell^{2}(G))$.

We fix a triple of groups $H<K<G$ for the rest of the article.

Let $\pi$ denote the quasi-regular representation of $G$ on $\ell^{2}(G/H)$; we denote by $[g]$ the equivalence class $[g]=gH$, so that $\pi(g)\xi([g^{\prime}])=\xi([g^{-1}g^{\prime}])$ for all $g,g^{\prime}\in G$ and $\xi\in\ell^{2}(G/H)$. Following [1], we say that a vector $\xi\in\ell^{2}(G/H)$ is $H$-compact if the norm closure of its $H$-orbit $\{\pi(h)\xi:h\in H\}$ is a compact subset of $\ell^{2}(G/H)$. The set of all $H$-compact vectors is a closed subspace of $\ell^{2}(G/H)$ that we denote by $\ell^{2}(G/H)_{c,H}$. We also set

$$\ell^{2}(G/H)^{H}=\{\xi\in\ell^{2}(G/H):\pi(h)\xi=\xi\ \forall h\in H\},$$

which is the subspace of all $H$-invariant vectors of $\ell^{2}(G/H)$. We observe that it is contained in $\ell^{2}(G/H)_{c,H}$.

Proof. (1) $\Rightarrow$ (2) is obvious.

For every $g\in G\setminus K$, and for every non empty finite set $F\subset G\setminus K$, there exists $h\in H$ such that $Fhg\cap H=\emptyset$.

Thus, let us assume that condition (3) does not hold. There exists $g\in G\setminus K$ and a non empty finite set $F\subset G\setminus K$ such that $Fhg\cap H\not=\emptyset$ for every $h\in H$. Then let $u\in U(B)$. One has:

	$\displaystyle\sum_{g^{\prime}\in F}\|\mathbb{E}_{B}(\lambda_{g^{\prime}}u\lambda_{g})\|_{2}^{2}$	$\displaystyle=$	$\displaystyle\sum_{g^{\prime}\in F}\left(\sum_{h\in H,g^{\prime}hg\in H}|u(h)|^{2}\right)$
		$\displaystyle=$	$\displaystyle\sum_{h\in H}\left(\sum_{g^{\prime}\in F,g^{\prime}hg\in H}|u(h)|^{2}\right)$
		$\displaystyle\geq$	$\displaystyle\sum_{h\in H}|u(h)|^{2}=\|u\|_{2}^{2}=1$

since, for every $h\in H$, one can find $g^{\prime}(h)\in F$ such that $g^{\prime}(h)hg\in H$. Hence there cannot exist a net $(u_{i})_{i\in I}\subset U(B)$ as above, and the triple $B\subset N\subset M$ does not have the relative weak asymptotic homomorphism property.

Let $s\in T$ be such that $\epsilon:=|\xi([s])|>0$. There exists then finitely many vectors $\xi_{1},\ldots,\xi_{n}\in\ell^{2}(G/H)$ such that, for every $h\in H$, there exists $1\leq j\leq n$ such that $\|\pi(h)\xi-\xi_{j}\|\leq\epsilon/2$. Set

$$F=\bigcup_{j=1}^{n}\{t\in T:|\xi_{j}([t])|\geq\epsilon/2\},$$

which is a finite set. Then we claim that $Hs\subset FH$. Indeed, if $h\in H$, let $t\in T$ be such that $[hs]=[t]$, and let $j$ be such that $\|\pi(h)\xi-\xi_{j}\|\leq\epsilon/2$. Then

$$\epsilon-|\xi_{j}([t])|=|\xi([s])|-|\xi_{j}([t])|\leq|\xi([s])-\xi_{j}([hs])|\leq\|\pi(h)\xi-\xi_{j}\|\leq\epsilon/2$$

hence $\epsilon/2\leq|\xi_{j}([t])|$ and $t\in F$. Thus $Hs\subset FH$, and condition (3) implies that $s\in K$. This proves that $\xi\in\ell^{2}(K/H)$.