17

In a nonabelian setting the correct notion of kernel is given by the kernel pair, and the correct notion of cokernel is given by the cokernel pair. For example, in any category, a morphism $f : a \to b$ is a monomorphism iff its kernel pair exists and is trivial, and dually $f$ is an epimorphism iff its cokernel pair exists and is trivial. By comparison, the ...

answered Nov 2 '14 at 22:18

Qiaochu Yuan

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16

Overview of an explanation :
Jones-Wassermann subfactors for the loop algebra :
Let $\mathfrak{g} = \mathfrak{sl}_{2}$ be the Lie algebra, $L\mathfrak{g}$ its loop algebra and $\mathcal{L}\mathfrak{g} = L\mathfrak{g} \oplus \mathbb{C}\mathcal{L}$ the central extension :
$$[X^{a}_{n},X^{b}_{m}] = [X^{a},X^{b}]_{m+n} + m\delta_{ab}\delta_{m+n}\mathcal{L}$$ ...

answered Oct 11 '13 at 13:18

Sebastien Palcoux

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11

At the very least there's one Bisch-Haagerup subfactor standard invariant for every quotient group of $\mathbb{Z}/2 \ast \mathbb{Z}/3 \cong \mathrm{PSL}_2(\mathbb{Z})$. That's already way too much to try to classify. Even if you restrict to finite depth, there's way too many finite quotients of the modular group.
On the other hand, if you're very ...

10

For finite groups, the answer was given by Izumi in his paper "Characterization of isomorphic group-subgroup subfactors" (MR1920326). There he looks at the crossed product subfactor, but you can always take duals.
Edit after @Andre's comment:
The actual condition between the two pairs of subgroups is quite technical, and it would basically require ...

10

Given any two tracial von Neumann algebras $(N_1, \tau_1)$ and $(N_2, \tau_2)$ the $L^2$ space of the free product $(N_1 * N_2, \tau)$ canonically decomposes as
$$
L^2(N_1 * N_2, \tau) = \mathbb C \oplus_{n \in \mathbb N} \bigoplus_{i_1 \not= i_2, i_2 \not= i_3, \ldots, i_{n - 1} \not= i_n } \overline {\otimes}_{k = 1}^n L^2_0(N_{i_k}, \tau_{i_k}),
$$
where ...

10

I think this paper of Wakui says that the answer is "No". The Turaev-Viro invariants associated to the generalized $E6$ subfactors for $\mathbb{Z}/7$ don't seem to distinguish $L(7,1)$ and $L(7,2)$.

10

The answer to (1) (for second countable groups) is yes: every locally compact second countable group $G$ admits a countinuous, faithful outer action on the hyperfinite $II_1$ factor. This is attributed to Blattner, and is stated explicitly in Proposition 1 of the following article:
R. J. Plymen. "Automorphic group representations: The hyperfinite $II_1$ ...

9

If you also assume finite depth, then there's a hope (it's too vague to call it a conjecture) that all integer index subfactors can be classified "using only finite group theory." That is, if you had a black box which could answer all questions about finite groups and their cohomology you'd be able to understand all finite depth integer index subfactors. The ...

8

It is relatively easy to give an explicit sequence of inner automorphisms that converges to the flip automorphism $\sigma$. First of all, realize $R$ as an infinite tensor product of matrix algebras $(R,\varphi)=\bigotimes_n (M_{k_n}(\mathbb{C}),\varphi_n)$. For every $n\in \mathbb{N}$, we find a unitary $u_n\in M_{k_n}(\mathbb{C})\otimes M_{k_n}(\mathbb{C})$...

8

This isn't a full answer.
First letâ€™s translate this into purely group theoretic language.
The vertex at depth $0$ is the trivial $G$-rep, the vertex at depth $1$ is the trivial $H$-rep, the vertices at depth $0$ or $2$ are the $G$-irreps in $\mathrm{Ind}_H^G 1$, the vertices at depth $1$ or $3$ are the $H$-irreps in $\mathrm{Res}_H\mathrm{Ind}_H^G 1$, ...

7

You can construct a subfactor (under very mild assumptions) from an object $X$ in a rigid C*-tensor category, by taking the limit of inclusions
$$
{\rm End}(X^{\otimes n}) \simeq \{ \iota_X \otimes T \mid T \in {\rm End}(X^{\otimes n})\} \subset {\rm End}(X^{\otimes n+1})
$$
as $n \to \infty$. One good entry point is the Longo-Roberts paper [LR97], I guess. ...

6

The infinite depth subfactor coming from SU(3) at any index above 9 gives an example. Here the Q-system is $V_{(1,0)} \otimes V_{(0,1)} \cong V_{(1,1)} \oplus V_{(0,0)}$ so the only possible sub-objects are the whole thing or the trivial, so it's certainly maximal.

6

(Unitary) fusion categories are interesting in physics because they classify gapped phases on the boundary of 2+1D quantum states of matter. Similarly, unitary modular tensor categories are interesting in physics because they classify gapped phases of 2+1D quantum states of matter. I have two papers to explain the connections
arXiv:1405.5858 and arXiv:...

6

No. For an example, consider $M=L^\infty([0,1])$ with the Lebesgue measure, take $A$ to be the functions that are piecewise constant on dyadic intervals and $\alpha_n(f)=f\circ \phi_n^{-1}$ where $\phi_n(t)=k/2^n + (2^n t-k)^2/2^n$ if $t \in [k/2^n,(k+1)/2^n[$. In words, $\phi_n$ acts as some fixed transformation (here $t \mapsto t^2$, but it could be ...

answered Jul 11 '12 at 23:20

Mikael de la Salle

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6

1, and indeed its generalization to the amenable case, is in "Amenable tensor categories and their realizations as AFD bimodules" by Hayashi and Yamagami, see Section 7. I don't think 2 has appeared in the literature, though I'd expect that it's true.
In the fusion case, if the universal grading group is trivial, I think you can just look at the algebra ...

6

I'm not an expert in algebraic geometry, but I can say something about methods for solving pentagon equations that will hopefully be of use.
The primary way to determine whether or not there is a solution to the pentagon equations is to use Groebner basis methods. However, these begin to break down very quickly for algebraic varieties of the size that we ...

6

The answer is no. If $N_1$ and $N_2$ are both finite index subfactors of a nonamenable ${\rm II}_1$ factor $M \subset \mathcal B(L^2M)$ such that $N_1 \cap N_2 = \mathbb C$, then $N_1'$ and $N_2'$ are both finite and $N_1' \cap N_2'$ is nonamenable since it contains $M'$.
For an example of such a situation consider non-trivial finite groups $G$ and $H$, set ...

5

I'm absolutely ignorant about subfactors, but if I correctly understand what you said there (the theorem in the question), the answer to all three questions is yes, even with non maximal subgroups.
Namely, let $H\subset G$ be of finite index $n$. Since the intersection $K$ of all conjugates of $H$ in $G$ has index dividing $n!$ in $G$ (it is the kernel of ...

5

This is somewhere between a comment and an answer. But it is too long for a comment, so I put it here.
To me the natural thing to look at is the action $G\curvearrowright G/H$. $|G/H|$ captures the index, which you surely want to do, and the action should in some sense capture the position of $H$ inside $G$. The issue then would be to try to come up with an ...

5

In the formula, $tr(b)$ should be replaced by $tr(|b|)$, i.e., $\|b\|_1$.
$$b_\alpha*b_\beta=\sum_\gamma\frac{\|b_\alpha\|_1\|b_\beta\|_1\bar{c}_{\alpha\beta}^\gamma}{\sqrt{n}\|b_\gamma\|_1}b_\gamma.$$
Unfortunately the proof needs the irreducibility. I do not know how to generalize it to weak Kac algebras.
As Dave mentioned, one direction is clear. It ...

5

The answer of the main question is yes. Let $G$ be a finite group and $H$ a subgroup.
Definition: The group $G$ is called $H$-cyclic if $\exists g \in G$ such that $\langle H,g \rangle = G$.
Note that: $\langle H,g \rangle = G \Leftrightarrow \langle Hg \rangle = G$.
Ore's theorem for intervals (1938): If the interval $[H,G]$ is distributive, then $G$ is $...

answered Jan 31 '15 at 12:54

Sebastien Palcoux

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5

As Noah points out, you are looking for some (core-free) maximal subgroup $H<G$ such that $1_H^G$ has nonzero inner product with every irreducible character.
Say $G=L_2(p)$ with $p$ prime and $p \equiv 1 \bmod 8$. Then $G$ has a maximal subgroup $H \cong S_4$. Every element $h \in H$ has diagonalizable preimage in $SL_2(p)$, and the conjugacy class of $...

5

A positive answer to my question (2) is provided in the last section of Popa Takesaki "Topological structure of unitary and automorphism groups".
They show that for any locally compact group $G$, the action of $G$ on the $II_1$ factor $\{\mathit{CAR}(L^2G)\}''$ is appropriately continuous. Here, the double commutant is taken on $L^2(\mathit{CAR}(L^2G),tr)$ ...

answered Mar 12 '19 at 0:41

André Henriques

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5

I think that a noncommutative fusion ring of rank 5 does not exist. Namely, let $a$ and $b$ be the formal codegrees (see https://arxiv.org/pdf/0810.3242.pdf) of such ring. Then $a$ and $b$ are positive (EDIT: and rational, see the explanation by Noah) integers satisfying $\frac1a+\frac2b=1$ (see Proposition 2.10 in https://arxiv.org/pdf/1309.4822.pdf). It is ...

4

If a subfactor $N \subset M$ at index 6 is composed, i.e., it admits a non-trivial intermediate subfactor $N \subset P \subset M$, then the two components are necessarily of indices $2$ and $3$. There exists one subfactor at index $2$ given by the graph $A_{3}$, and two at index $3$ given by $A_{5}$ and $D_{4}$.
Then, the standard invariant of a composed ...

answered Jul 7 '13 at 11:05

Sebastien Palcoux

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4

Your answer will work, except if the $b_i$'s have trace zero as Theo points out. If the $b_i\in \mathcal{P}_{n_i}$ has trace zero, just use $1_{n_i}+b_i$ instead of $b_i$, where $1_{n_i}$ is $n_i$ parallel strands. Then you can cap this off to get a scalar as before. By what you remarked above, you'll be able to recover $1_{n_i}+b_i$, from which you can ...

4

This is not a proper answer, but is a bit too long for a comment. You should look at the following paper:
Groups and Lattices by P.P. Palfy
First, he states a stronger version of Ore's result:
Theorem: Let $G$ be any group. The subgroup lattice $\mathcal{L}(G)$
is distributive if
and only if the group
$G$
is locally cyclic. (i.e. every ...

3

The answer is no. In fact, the group planar algebra has no non-trivial planar ideals.
If $P_\bullet$ is a spherical planar algebra with non-zero modulus, and if $P_{0,\pm}$ are one dimensional, then every planar ideal is contained in the planar ideal $N_\bullet$ of negligible elements, i.e., those $x\in P_{n,\pm}$ for which tr$(xy)=0$ for all $y\in P_{n,\pm}...

3

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. So $\phi: M \to \phi(M)$ is an isomorphism that takes $\phi^{-1}(N')$ to $N'$.
I don't think the canonical surjection $G\to G'=G/\ker(f)$ actually gives you a map of factors $M\rtimes G\to M\...

3

Let $_RM_R:={}_R(L^2R\otimes_{\mathbb C} L^2R)_R$, where the first $R$ acts on the first $L^2R$ and the second $R$ acts on the second $L^2R$.
Its algebra of $R$-$R$-bimodule endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.
Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\...

answered Nov 25 '13 at 10:43

André Henriques

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