# The (Hecke) double coset von Neumann algebra

It it well-known in the von Neumann algebra theory that for $\Gamma$ a non-trivial countable group, the von Neumann algebra $L(\Gamma)$ generating by $\Gamma$ acting by left multiplication on $l^2(\Gamma)$, is a ${\rm II}_1$ factor iff $\Gamma$ is an ICC group.

Now let $(G \subset \Gamma)$ be an inclusion of a finite group $G$ in a countable group $\Gamma$, Let $\mathbb{C}(G \backslash \Gamma / G)$ be the (Hecke) double coset algebra: the subalgebra of $\mathbb{C}G$, generated by the elements $a_{\gamma} = \sum_{\alpha \in G \gamma G} \alpha$ (well-defined because $G$ finite) with $\gamma \in \Gamma$. Let $L(\Gamma,G)$ be the von Neumann algebra generated by $\mathbb{C}(G \backslash \Gamma / G)$ acting by left multiplication on $l^2(G \backslash \Gamma)$.

Question: What's the necessary and sufficient condition on $(G \subset \Gamma)$ for $L(\Gamma,G)$ to be a ${\rm II}_1$ factor?

In the case that there are inclusions $(G \subset \Gamma)$ with $\Gamma$ ICC, $G \neq \{ e \}$ and $L(\Gamma,G)$ ${\rm II}_1$ factor:
Optional question 1: Is it true that $L(\Gamma,G) \simeq L(\Gamma)$?

Optional question 2: How to generalize the construction above for $G$ infinite?

• Are the tags 'hypergroups' and 'harmonic-analysis' really relevant to this question? – Ricardo Andrade Jun 8 '15 at 9:43
• @RicardoAndrade: The double coset algebra $\mathbb{C}(G \backslash \Gamma / G)$ is an example of hypergroup. All the objects of the post are close to noncommutative harmonic analysis, so an harmonic analyst could answer such questions or at least being interested in. – Sebastien Palcoux Jun 8 '15 at 10:06

$L(\Gamma,G)$ is the algebra of endomorphisms of the representation $l^2( \Gamma/G)$ (with $\Gamma$ acting by left multiplication). This answer your second optional question. Also, by classical results on $W^*$-categories, the category of normal representations of $L(\Gamma,G)$ will be equivalent to the category of unitary representations of $G$ that are retract of sums of copies of $l^2(\Gamma /G)$, which generally allow to determine the type in concrete situations but I don't know what a general criterion would be. In fact, in the general case this algebra can also be of type $III$ which make me think there is no simple criterion.
In the special case where $G$ is finite, $l^2(\Gamma /G)$ is a retract of $l^2(\Gamma)$ hence $L(\Gamma,G)$ is a corner of $L(\Gamma)$. If in addition $\Gamma$ is ICC, then $L(\Gamma)$ is a factor and hence any non trivial corner will be Morita equivalent to $L(\Gamma)$. So $L(\Gamma,G)$ will be Morita equivalent to $L(\Gamma)$ and hence of the same type.
Also note that in this situation $L(\Gamma)$ and $L(\Gamma,G)$ will often be isomorphic, but by a "non natural" isomorphism which will not going to be compatible with the natural Morita equivalence...
• By the same type, I meant type $II$, but it is also easy to check that it is type $II_1$ either because $l^2(\Gamma/G)$ is finitely generated as a $L^2(G)$ module or by constructing a trace explicitely. – Simon Henry Jun 8 '15 at 14:43
• Thank you! I note the typo that sometimes you have written $L^2(\Gamma)$ instead of $L(\Gamma)$. – Sebastien Palcoux Jun 8 '15 at 15:40
• What's your type ${\rm III}$ examples? – Sebastien Palcoux Jun 8 '15 at 15:59