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Let amenable groups $\Gamma$ and $\Gamma'$. They act outerly of only one manner on the hyperfinite ${\rm II}_1$-factor $\mathcal{R}$.

Question: $(\mathcal{R} \subset \mathcal{R} \rtimes \Gamma) \simeq (\mathcal{R} \subset \mathcal{R} \rtimes \Gamma') \, \Rightarrow \, \Gamma \simeq \Gamma'$ ?

If $\Gamma$ is infinite, then it exists an outer action on $\mathcal{R}$ such that $\mathcal{R}^{\Gamma} = \mathbb{C}$ (see comments of this answer), so $(\mathcal{R}^{\Gamma} \subset \mathcal{R})$ can't be the dual of $(\mathcal{R} \subset \mathcal{R} \rtimes \Gamma)$.
Let the infinite amenable group $\Gamma = \bigcup_iG_i$ with $(G_i)_i$ an increasing sequence of finite groups.
Now, $(\mathcal{R}^{G_i} \subset \mathcal{R})$ is the dual of $(\mathcal{R} \subset \mathcal{R} \rtimes G_i)$ and $(\mathcal{R} \subset \overline{\bigcup_i\mathcal{R} \rtimes G_i}) \simeq (\mathcal{R} \subset \mathcal{R} \rtimes \Gamma)$.

Question: How to characterize the subfactor $(\mathcal{Q} \subset \mathcal{R})$ for being the dual of $(\mathcal{R} \subset \mathcal{R} \rtimes \Gamma)$?
(If I'm not mistaken $\bigcap_i \mathcal{R}^{G_i} = \mathcal{R}^{\Gamma}$, so we can't take $\mathcal{Q} = \bigcap_i \mathcal{R}^{G_i}$)

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  • $\begingroup$ You should be more precise about what you mean by "only one manner". Why do you conclude that $\mathcal R^\Gamma = \mathbb C$? $\endgroup$ Aug 6, 2014 at 19:29
  • $\begingroup$ @JessePeterson: An infinite amenable group $\Gamma$ acts outerly on $\mathcal{R}$, "only one manner" means "unique up to outer conjugacy". Now $\mathcal{R} \simeq \bigotimes_{\gamma \in \Gamma} \mathbb{M}_2(\mathbb{C})$, but $(\bigotimes_{\gamma \in \Gamma} \mathbb{M}_2(\mathbb{C}))^{\Gamma} = \mathbb{C}$ (by your comment), so by uniqueness (up to out. conj.), $\mathcal{R}^{\Gamma} = \mathbb{C}$. Is it correct? $\endgroup$ Aug 6, 2014 at 22:19
  • $\begingroup$ I don't believe it is correct. Why should uniqueness up to outer conjugacy imply $\mathcal R^\Gamma = \mathbb C$? $\endgroup$ Aug 6, 2014 at 23:13
  • $\begingroup$ How about $R \overline \otimes (\otimes_{\gamma \in \Gamma} \mathbb M_2(\mathbb C) )$, where $\Gamma$ acts trivially on the first copy of $R$? $\endgroup$ Aug 7, 2014 at 16:33
  • $\begingroup$ @JessePeterson: Yes sure. But let $\alpha$, $\beta$ be two outer action of $\Gamma$ on $\mathcal{R}$ such that $\mathcal{R}^{\Gamma} = \mathbb{C}$ for the first action. By outer conjugacy, there is an isomorphism $\Delta$ such that $\Delta \alpha_g \Delta^{-1} = \beta_g$. Now if $\forall g \, \, \beta_g(x) = x$ then $\alpha_g(y)=y$ with $y=\Delta^{-1}(x)$. But $y \in \mathbb{C}$ (by assumption) so $x \in \mathbb{C}$ too, and $\mathcal{R}^{\Gamma} = \mathbb{C}$ for the second action. Once again, there is something I don't understand. $\endgroup$ Aug 7, 2014 at 16:48

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The answer to the first question is yes by a result of Herman and Ocneanu (MR1055223) or of Enock and Nest (MR1387518).

We call an inclusion of II$_1$-factors depth 2 if $M_0'\cap M_2$ is abelian and $M_0'\cap M_3$ is a factor. (In fact, we only need to say $M_0'\cap M_3$ is a factor, and this works for arbitrary inclusions of factors, but I'll stick to the II$_1$ case.) Now for a depth 2 inclusion, Herman and Ocneanu say that the group $\Gamma$ can be recovered from unitaries which normalize $M_0$, using an extension of Proposition 1.7 of Pimsner-Popa's paper (MR0860811). But you should also check out Enock and Nest's article -- it's really very nice.

Now for the second part. There is not yet a completely satisfactory notion of the dual subfactor of an infinite index subfactor. Given an infinite index inclusion of II$_1$-factors $M_0\subset M_1$, the basic construction $M_2$ is a II$_\infty$-factor. So in this sense, it doesn't make sense to call $M_1\subset M_2$ the dual subfactor. This implies, as you noted above, that you can't get a II$_1$-factor from a downward basic construction either.

(However, Herman and Ocneanu have a nice notion of discrete and compact inclusions which are dual to each other. I'd recommend looking in their article.)

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  • $\begingroup$ The assumption "$M_0' \cap M_2$ is abelian" is checked for a group subfactor, but not for all the Kac algebra subfactors (so not checked for all the depth $2$ inclusion). $\endgroup$ Aug 6, 2014 at 3:58
  • $\begingroup$ For the first question, the converse ($\Leftarrow$) is true for an amenable group (because it acts outerly of only one manner), but not for non-amenable groups: inequivalent outer actions $\alpha$, $\alpha'$ imply inequivalent subfactors $(\mathcal{R} \subset \mathcal{R} \rtimes_{\alpha} \Gamma)$ and $(\mathcal{R} \subset \mathcal{R} \rtimes_{\alpha'} \Gamma)$, isn't it? $\endgroup$ Aug 6, 2014 at 4:18
  • $\begingroup$ "There is not yet a completely satisfactory notion of the dual subfactor of an infinite index subfactor." Is it possible to have a completely satisfactory notion of dual in the framework of "planar algebras" or "tensor categories"? $\endgroup$ Aug 6, 2014 at 13:15

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