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Let $(N \subset M)$ be a finite index 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 $4$.

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 $4$.

Let $(N \subset M)$ be a finite index 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.

Let $G$ be a finite abelian group and $H$ a subgroup of $Aut(G)$.

Question: Is every self-dual group-subgroup subfactor of the form $(R ^ {G \rtimes H} \subset R^H)$?

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

Remark (after an exchange with Marcel Bischoff): The question above asks if this form is necessary. We know that this form is not sufficient, because $S_4 = A_4 \rtimes \langle (1,2) \rangle$, and the double coset algebra for the inclusion $(\langle (1,2) \rangle \subset S_4)$ is not commutative (see here p45). Now the subfactor planar algebra $P=P(N \subset M)$ is selfdual only if $N' \cap M_1 = P _{2,+} \simeq P_{2,-} = M' \cap M_2$, but for this example, one is commutative and the other is not, so this $(N \subset M)$ is not self-dual.

Let $(N \subset M)$ be a finite index 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 $4$.

Let $(N \subset M)$ be a finite index 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.

Let $G$ be a finite abelian group and $H$ a subgroup of $Aut(G)$.

Questions: Is the subfactor $(R ^ {G \rtimes H} \subset R^H)$ self-dual?
Is every self-dual group-subgroup subfactor of this form?

Let $(N \subset M)$ be a finite index 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.

Let $G$ be a finite abelian group and $H$ a subgroup of $Aut(G)$.

Question: Is every self-dual group-subgroup subfactor of the form $(R ^ {G \rtimes H} \subset R^H)$?

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

Remark (after an exchange with Marcel Bischoff): The question above asks if this form is necessary. We know that this form is not sufficient, because $S_4 = A_4 \rtimes \langle (1,2) \rangle$, and the double coset algebra for the inclusion $(\langle (1,2) \rangle \subset S_4)$ is not commutative (see here p45). Now the subfactor planar algebra $P=P(N \subset M)$ is selfdual only if $N' \cap M_1 = P _{2,+} \simeq P_{2,-} = M' \cap M_2$, but for this example, one is commutative and the other is not, so this $(N \subset M)$ is not self-dual.