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This post was inspired by this answer of Dave Penneys.

In the category of (irreducible hyperfinite II$_1$) subfactors, the morphisms of $(N \subset M)$ to $(N' \subset M')$ are usually defined as the $W^*$-morphisms $\phi: M \to M'$ with $\phi (N) \subset N'$.

Unfortunately, through this definition, the category of finite group is $\underline{not}$ a (natural) subcategory of the category of subfactors.
In fact, let $G$ and $G'$ be finite groups and $f: G \to G'$ be a surjective group-morphism, then in general, $f$ does $\underline{not}$ extend into a (usual) subfactor-morphism of $(R \subset R \rtimes G)$ to $(R \subset R \rtimes G')$.

Here is the explanation in the answer of Dave:
A II$_1$-factor is algebraically simple, so each morphism of II$_1$-factors is either injective or zero.
Thus every non-zero morphism is an isomorphism onto its image.
I don't think the canonical surjection $G\to G'=G/\ker(f)$ actually gives you a map of factors $R\rtimes G\to R\rtimes G'$. In particular, if we denote the implementing unitaries as $u_g$ for $g\in G$, the map $u_g\mapsto u_{g\ker(f)}$ does not extend to a non-zero map of II$_1$-factors if $\ker(f)$ is non-trivial. The element $u_g-u_{g'}$ would map to zero if $g,g'\in \ker(f)$, and a non-trivial map of II$_1$-factors must be injective.

Question: Is there an $\underline{other}$ (natural) definition of subfactor-morphisms such that the category of finite groups is a (natural) subcategory of this "new" category of subfactors ?

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The category of finite groups is also not a (natural) subcategory of the category of planar algebras because the subfactor planar algebras are simple (see here), so the subfactor planar algebra morphisms are also either injective or $0$... –  Sébastien Palcoux Jan 29 at 21:16
    
How are you having $G$ and $G'$ act on $R$? –  Jesse Peterson Jan 31 at 3:47
    
@JessePeterson : the finite groups $G$ and $G'$ act as outer automorphisms of the hyperfinite II$_1$ factor $R$. –  Sébastien Palcoux Jan 31 at 10:57
    
Are you taking specific actions or are you looking for something which holds for arbitrary actions? –  Jesse Peterson Jan 31 at 18:08
    
@JessePeterson : because the isomorphic class of $R \subset R \rtimes G$ does not depend on the choice of the action (as above), I could say "something which holds for arbitrary actions", but because I'm looking for "other" subfactor-morphisms, the choice of specific action is perhaps relevant. –  Sébastien Palcoux Jan 31 at 18:22

2 Answers 2

up vote 2 down vote accepted

Take $C$ to be the category of dualizable $N$-$N$-bimoduls, $N$ a factor. A subfactor $N\subset M$ (or $N_0\subset N$) with finite index and finite depth gives an algebra object $A$ in $C$, namely $A={}_NM_N$ (or ${}_NL^2M_N$ if you prefer) and conversely an algebra object (more precisely a Q-system) gives a subfactor $N\subset M$. Instead of building artificially a category of subfactors, you take the category of (simple) algebra objects (Q-systems) in $C$. Each finite group gives an object $A_G=\bigoplus_{g\in G} {}_NN^{\circ\alpha_g}_N$ with $\alpha_g$ automorphisms on $N$ such that $\alpha_g\alpha_h=\alpha_{gh}$ for $g,h\in G$ and ${}_NN^{\circ\alpha_g}_N$ is ${}_NN_N$ seen as a $N$-$N$ bimodule, where the right action is composed with $\alpha$. A morphism $H\to G$ gives a morphism $A_H\to A_G$ between algebra object. This category contains also finite groups, their duals, Kac-algebras and weak-C${}^\ast$ Hopf algebras.

If you want irreducible subfactors, you ask $A$ to be haploid, then you lose weak-C${}^\ast$ Hopf algebras.

This also tells you how a "category of subfactors" should work...

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Very interesting! Before accepting your answer, I need to check that this definition is coherent with the definition of normal intermediate subfactor of T. Teruya (see a simplified definition here) as explained in the remark of my answer, and also I need to check by myself that $G \to A_G$ is really the natural expected functor from the category of finite groups to the category of (simple) algebra objects in $C$ (and extends to the category Kac algebras). –  Sébastien Palcoux Oct 29 at 12:22
    
What's your reference for haploid? –  Sébastien Palcoux Oct 29 at 12:27
    
In general, the $W^∗$-isomorphism of subfactors implies the isomorphism of planar algebras but the converse is false. I would like to see how the isomorphism of algebra objects is positioned relative to these two last isomorphisms. –  Sébastien Palcoux Oct 29 at 12:51
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Haploid means that ${}_NN_N \prec {}_NM_N$ has multiplicity one, which is equivalent with irreducibility. –  Marcel Bischoff Oct 30 at 4:49
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An algebra subobject is an intermediate subfactor. The condition for sub algebra objects etc., we reviewed here in the type III case. The projections should be the same as in the type II case here: arxiv.org/abs/1407.4793 I never really thought about morphisms besides injections.... –  Marcel Bischoff Nov 1 at 2:31

This is an artificial answer, I'm looking for something more natural.

In this paper, T. Teruya introduced the notion of normal intermediate subfactors, generalizing exactly the notion of normal subgroups (see the post Jordan-Hölder theorem for subfactors for more details).

So we can generalize the group-morphisms to the subfactors as follows :
A (group-like) morphism for $(A \subset B)$ to $(C \subset D)$ is the data of:

  • a normal intermediate subfactor $(A \subset P \subset B)$
  • an intermediate subfactor $(C \subset Q \subset D)$
  • a $W^*$-isomorphism $\phi_l : (A \subset P) \to (Q \subset D)$ or $\phi_r : (P \subset B) \to (C \subset Q)$

Remark: This notion generalizes by construction the group-morphisms, unfortunately, it's a bit artificial, I would prefer a more natural definition of morphisms, without using 'ad hoc' the notion of normal intermediate subfactors, but such that the kernel of these natural morphisms are exactly the normal intermediate subfactors.

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