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Let $(N \subset M)$ be a finite index unital inclusion of ${\rm II}_1$ factors, and $N \subset M \subset M_1$ the basic construction.
The subfactor $(N \subset M)$ is called self-dual if it is isomorphic to its dual $(M \subset M_1)$.

Let $R$ be the hyperfinite ${\rm II}_1$ factor. We will use the outer action of any finite group $G$ on $R$, and the fixed point subfactor $R^G \subset R$.

Remark: Let $G$ be a finite group. Then, $(R ^ {G} \subset R)$ is self-dual iff $G$ is self-dual, iff $G$ is abelian.

Consider the group-subgroup subfactors of the form $(R ^ {G \rtimes H} \subset R^H)$, with $G$ a finite abelian group and $H$ a subgroup of $Aut(G)$.

Question: Is a self-dual group-subgroup subfactor necessarily of this form? Is it also sufficient?

Remark: It is true at index $3$ and $5$ (see here p37). I've to check index $4$.

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    $\begingroup$ It is necessary, that the double coset algebra $H\backslash (G\rtimes H) /H$ is commutative, i.e. $(G\rtimes H,H)$ has to be a Gelfand pair. Namely, one two box space is $\mathbb{C}^N$ where $N$ is the number of double cosets, while the other is the double coset algebra itself. $\endgroup$ Commented Jun 13, 2016 at 18:04
  • $\begingroup$ @MarcelBischoff: It is a Gelfand pair in this case because $\forall h_1, h_2, h_3 \in H \subset G\rtimes H$ and $\forall g,g' \in G \subset G\rtimes H$, we have: $$ h_1gh_2g'h_3 = h_1 g \sigma_{h_2}(g') h_2 h_3 = h_1 \sigma_{h_2}(g') g h_2 h_3 = h_1 h_2 g' h_2^{-1} g h_2h_3 $$ which means that $HgHg'H = Hg'HgH$, i.e. the double coset algebra is commutative. $\endgroup$ Commented Jun 13, 2016 at 19:11
  • $\begingroup$ Ok, I see, I was already expecting that this is automatic. Is $G$ necessarily abelian? $\endgroup$ Commented Jun 13, 2016 at 19:14
  • $\begingroup$ @MarcelBischoff: No, see the answer here: mathoverflow.net/q/242271/34538 $\endgroup$ Commented Jun 15, 2016 at 16:59

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