# What's the natural equivalence of subfactors in general?

Let $A$ be a factor and $\mathcal{C}_{A}$ be the category of all the subfactors $(M \subset N)$ such that $M$ and $N$ are isomorphic to $A$. The most famous of them is perhaps $\mathcal{C}_{R}$ with $R$ the hyperfinite $II_{1}$ factor.
One can define an equivalence relation called $\sim_{1}$, available on each such category, as follows:

Definition : Let $A$ be a factor, $(P \subset Q)$ and $(R \subset S)$ be two subfactors of $\mathcal{C}_{A}$, then:
$(P \subset Q) \sim_{1} (R \subset S)$ if it exists an isomorphism $\phi : Q \to S$ such that $\phi(P) = R$.
Temporary remark : I hope this is the most common definition (also called isomorphism of subfactors), because I don't find it written explicitly in the literature (certainly because it's obviously this one).

The purpose of this issue is to ask how naturally generalize $\sim_{1}$ into an equivalence relation $\sim$ available on the category $\mathcal{C}$ of all the subfactors, in order to verify the following specifications (even if $M \not\simeq N$) :

1. $(M \subset M) \sim (N \subset N)$
2. $(P \bar\otimes M \subset Q \bar\otimes M) \sim (P \bar\otimes N \subset Q \bar\otimes N)$
3. $[(P \subset Q) \sim (R \subset S) ]$ $\Leftrightarrow$ $[(P \bar\otimes M \subset Q \bar\otimes M) \sim (R \bar\otimes N \subset S \bar\otimes N) ]$
4. $(P^{G} \subset P) \sim (Q^{G} \subset Q)$ such that $G$ embeds into $Out(P)$ and $Out(Q)$.

Is it coherent ? (of course $3 \Rightarrow 2 \Rightarrow 1$)

Motivation (Jones' philosophy) : The purpose of this equivalence relation is that any equivalence class $[M \subset N]_{\sim}$ captures only the information (or symmetry) contained in the relative position of $M$ inside $N$ (forgetting the factors themselves), in order to obtain a kind of strictly group-like object.

Ambiguities : It appears that such a relation $\sim$ retricted to $\mathcal{C}_{R}$ would be coaser than $\sim_{1}$ :

• After Bisch-Nicoara-Popa and Owen Sizemore's comments (see below), one can have $(P \subset Q) \not\sim_{1} (P \subset Q)^{t}$ while $(P \subset Q) \sim (P \subset Q)^{t}$.
Examples: uncountably many non-isomorphic subfactors at index $6$ would be equivalent !
Is there a relevant difference between the relative position of $P$ inside $Q$, and $P^{t}$ inside $Q^{t}$ ?
• After Ocneanu and Jones' works, an amenable group $G$ acts outerly on $R$ by only one manner, but a non-amenable one, by at least two manners (see here).

Conclusion : the relation $\sim$ would be coaser, but maybe more natural regarding to the motivation.

The existence of what I have called ambiguities reinforces the purpose of the following question :

What's the natural equivalence of subfactors in general ?

• Once again, a seemingly non-justified downvote, without comment... – Sebastien Palcoux Jul 11 '13 at 19:47
• Sebastian, I didn't vote your question down, but it could use some more explanation. How exactly is your class of subfactors defined? How do you "easily" define your equivalence $\tilde_1$? What properties do you want from your equivalence? Do you have examples? I feel that this question is too open-ended to get a good answer, as phrased. – MTS Jul 11 '13 at 20:49
• Sebasitien: are you looking for a different equivalence relation than "have the same planar algebra"? – André Henriques Jul 11 '13 at 20:57
• Ok @MTS I will post some precisions. – Sebastien Palcoux Jul 12 '13 at 7:55
• What invariant does one use to distinguish those uncountably non-isomorphic index 6 subfactors? Using that invariant, you'll be able to sharpen you question and ask for an equivalence relation on subfactors that preserves that invariant. – André Henriques Jul 12 '13 at 8:49

Some attempts :

1. Common multiple : Let $(P \subset Q)$ and $(R \subset S)$ be two subfactors, then :
$(P \subset Q) \sim (R \subset S)$ if exists $\phi \in Aut(Q \bar\otimes S)$ such that $\phi(P\bar\otimes S) = Q\bar\otimes R$.

2. Absorption : We say that $M$ absorbs $P$ if $M \bar\otimes P \simeq M$.

• Definition : $(P \subset Q) \sim_{M} (R \subset S)$ if the factor $M$ absorbs $P,Q,R,S$ and $(M \bar\otimes P \subset M \bar\otimes Q) \sim_{1} (M \bar\otimes R \subset M \bar\otimes S)$.

• Existence : For all subfactors $(P \subset Q)$ and $(R \subset S)$ :
Does there always exist a factor $M$ absorbing $P,Q,R,S$ ? (see Owen's comments below)

• Invariance (?) : If two factors $M$ and $N$ absorb $P,Q,R,S$, then:
Is $[(P \subset Q) \sim_{M} (R \subset S)] \Leftrightarrow [(P \subset Q) \sim_{N} (R \subset S)]$ ?

3. Planar algebra (after André's comment above) :
This relation could be equivalent to "have the same planar algebra".

• About existence, yes given any finite collection of $II_1$ factors $P_1, ..., P_k$. Consider the the infinite tensor product factor $\bigotimes (P_1\otimes\cdots\otimes P_k)$. This will be absorbing for all the $P_i$. – Owen Sizemore Jul 11 '13 at 22:12
• However, this notion kills any information from the fundamental group. Specifically, consider the BNP examples $P\subset Q$. Then $P$ and $Q$ are the hyperfinite $II_1$, which we call $\mathcal{R}$, which is absorbing for itself. Then we have $P^t\otimes\mathcal{R} \subset Q^t\otimes\mathcal{R}\simeq P\otimes\mathcal{R}^t\subset Q\otimes\mathcal{R}^t\simeq P\otimes\mathcal{R}\subset Q\otimes \mathcal{R}$. This will always happen if the absorbing factor ($M$ above) has full fundamental group, and anything that absorbs $\mathcal{R}$ is a McDuff factor and has full fundamental group. – Owen Sizemore Jul 12 '13 at 13:07
• I'm not sure to well understand the first isomorphism, but anyway, it seems imply that the uncountably many non-isomorphic subfactors in BNP are in fact equivalent in this sense. This is good news ! – Sebastien Palcoux Jul 12 '13 at 13:43
• $P^t\subset Q^t=pM_n(P)p\subset pM_n(Q)p$ for an appropriate projection $p\in P$. Similarly for $P^t\otimes\mathcal{R}\subset Q^t\otimes\mathcal{R}$, but $p\in P\otimes\mathcal{R}$ in the last case. The isomorphism type only depends on the trace of $p$ and since $P\otimes\mathcal{R}$ is a factor we get a projection with correct trace in $\mathcal{R}$. So we can assume $p\in 1\otimes\mathcal{R}$. Then we just view $M_n(P)\otimes\mathcal{R}$ as $M_n(\mathbb{C})\otimes P \otimes\mathcal{R}= P\otimes M_n(\mathcal{R})$. Since $p\in P'$ we get the amplification is happening on $\mathcal{R}$. – Owen Sizemore Jul 12 '13 at 13:52