arXiv:1701.00097v1 [math.OA] 31 Dec 2016

TUBE ALGEBRA OF GROUP-TYPE SUBFACTORS
Dietmar Bisch, Paramita Das, Shamindra Kumar Ghosh and Narayan Rakshit
Abstract. We describe the tube algebra and its representations in the cases of diagonal and BischHaagerup subfactors possibly with a scalar 3-cocycle obstruction. We show that these categories are additively equivalent to the direct product over conjugacy classes of representation category of a centralizer subgroup (corresponding to the conjugacy class) twisted by a scalar 2-cocycle obtained from the 3-cocycle obstruction.
1. Introduction
Annular representations of planar algebras were introduced by Vaughan Jones in [Jon] to construct subfactors with principal graphs E6 and E8. In the same paper, he explicitly worked out the Temperley-Lieb example. These calculations helped in construction of new examples such as [Pet]. Recently, annular representations of subfactors and semisimple rigid C-tensor categories have become a very interesting area of research. The annular representation category turns out to be a nice braided tensor category - not necessarily semisimple - which is equivalent to the center of the original bimodule / C-tensor category in the case of finite depth / fusion categories (see [DGG2], [DGG3], [GJ]). For general depth, this category becomes equivalent to the center of a certain induced category (which is basically an extension where infinite direct sums are allowed) - see [NY], [PV]. There is also an analytic aspect of the annular representation category. Analytic properties, such as, amenabilty, Haagerup property, property (T) of the subfactor / C-tensor category can be reinterpreted in terms of annular representations.
In this paper, we deal with two group-type subfactors - the so-called diagonal and the BischHaagerup ones - possibly with 3-cocyle obstruction. The approximation properties of these two examples are well-known and depend on the associated group (see [Pop1], [Pop2] for diagonal and [BH] for Bisch-Haagerup). We determine the annular representation category. For this, we borrow techniques from [GJ], namely, we fix a `full' weight set in the object space and find the annular category over the weight set. It was shown in [GJ] that the annular representation categories over any two full weight sets are equivalent. Moreover, annular representations (in the sense of Vaughan Jones) of a subfactor N  M are the same as the annular representations of the N -N -bimodule category CNN generated by N L2(M )N .
For the diagonal subfactor, CNN is a pointed category, that is, the category of group graded vector spaces with a possibly nontrivial associator; the group is the one generated by the automorphisms used to build the diagonal subfactor and the associator is given by the 3-cocycle obstruction. When the cocycle is trivial, the annular representations were discussed in [GJ]. We consider the annular algebra over the irreducible bimodules, that is, Ocneanu's tube algebra. We show that the tube algebra is a direct sum over conjugacy classes of -algebras consisting of a matrix algebra tensored with the group algebra of the centralizer subgroup twisted by a 2-cocycle. We give the explicit dependence of the 2-cocycle on the 3-cocycle obstruction. As a result, the annular representation decomposes as (possibly infinite) direct sum of projective representations of the centralizer subgroups corresponding to the conjugacy classes.
Key words and phrases. Planar algebras, subfactors, group-type subfactors, fusion algebras, affine representations. 1

In the Bisch-Haagerup case, we consider the intermediate subfactor N  Q  M where N = QH

and M = Q  K with H, K being finite groups acting outerly on the II1 factor Q. The category of Q-Q-bimodules CQQ generated by QL2(Q)  L2(Q)Q and QL2(M )Q, is again a pointed category
N
equivalent to the category of G-graded vector spaces where G is the group generated by H and K

in Out(Q) with the associator given by the 3-cocycle obstruction. This category has special algebra

objects, namely, A = QL2(Qh)Q and B = QL2(Qk)Q. Now, the 2-category of bimodules

hH

kK

over N and M can be made equivalent to the 2-category of bimodules in CQQ over A and B. This

was a method suggested to us by Scott Morrison. However, we obtain the annular representations

straight from the actual bifinite N -N -bimodules using techniques in [GJ]. CNN unfortunately is

not pointed anymore. The set of isomorphism classes of irreducibles turns out to be complicated.

So, we consider a different weight set, namely N L2(Qg)N for g  G and obtain the annular algebra over it. As a -algebra, it turns out to be same as before, namely direct sum over conjugacy

classes of matrix algebras tensored with the group algebra of the centralizer subgroup twisted by a

2-cocycle. Thereby, the representations are also graded by the conjugacy classes and in each grade

the representations are the same as that of the corresponding centralizer subgroup twisted by the

2-cocyle.

Acknowledgement. The authors would like to thank Scott Morrison and Corey Jones for several useful discussions. A part of this work was completed during the trimester program on von Neumann algebras at the Hausdorff Institute of Mathematics and the authors would like to thank HIM for the opportunity. The first named author was supported by US NSF grants DMS-0653717 and DMS-1001560, and the Simons Foundation Collaboration Grant no. 359625.

2. Some basics on group cocycles

Let G be a group with identity element e and   Z3(G, S1) be a 3-cocycle of G, that is,  satisfies the following:
(2.1) (g1, g2, g3)(g1, g2g3, g4)(g2, g3, g4) = (g1g2, g3, g4)(g1, g2, g3g4) for all g1, g2, g3, g4  G
We will use Equation 2.1 at various instances in the article by denoting the particular elements of G which will correspond to g1, g2, g3, g4 simply by 1, 2, 3, 4 respectively. Up to 3-coboundary equivalence, we may consider  to be a normalized cocycle, i.e., (g1, g2, g3) = 1 whenever either g1, g2 or g3 is e; namely, if (g1, g2) = (g1, e, e)(e, e, g2) for all g1, g1  G, then () is normalized.
For a  G, let Ga denote the centralizer subgroup of a. The following result may be well-known to specialists but we include the statement for the sake of completeness.
Lemma 2.1. Ga × Ga  (g, h) -a (a, g, h)(g, a, h)(g, h, a) is a 2-cocycle of Ga.

Proof. Note that a(h2, h3)a(h1h2, h3)a(h1, h2h3)a(h1, h2) contain twelve terms involving . The product of the four -terms with a in the first place is

(a,
1

h1,
2

h32 )(a1 ,

h1h2,
23

h43 )(a1 ,

h1,
2

h2h3)(a,
34

h2,

h3)

=

(ah1,

h2,

h3)(h1,

h2,

h3)(a,

h2,

h3),

the product of the four -terms with a in the third place is

(h2,
2

h3,
3

a4 ) (h11

h2
2

,

h3,
3

a4 ) (h11 ,

h2 h3 ,
23

a4 ) (h1 ,

h2,

a)

=

(h1,

h2,

h3a)(h1,

h2,

h3)(h1,

h2,

a),

and the remaining product is

(h2,
2

a,
3

h3)(h1

4

1

h2
2

,

a,
3

h3)
4

(h1,
1

a,
2

h2h3
34

)(h1
1

,

a,
2

h2)
3

= (h1, h2, a)(h1, h2a, h3)(h1, h2, ah3)(h1, ah2, h3)(a, h2, h3)(h1a, h2, h3).

2

Now, since h1, h2, h3 commute with a, all terms in the grand product cancel amongst each other.

Instead of a, if we take xax-1 for any x  G, then it is natural to ask whether xax-1  (Adx × Adx) : Ga × Ga  S1 is coboundarily equivalent to a. The answer is yes; however, we will prove not just this, but a slightly general formula which will be useful later.

Proposition 2.2. For all a, x, y  G, there exists a,x,y : Ga  S1 such that (xax-1, xgy-1, yhz-1)(xgy-1, yay-1, yhz-1)(xgy-1, yhz-1, zaz-1)

= a,x,y(g)a,y,z (h)a,x,z(gh) a(g, h)
for all g, h  Ga. Thus, a,x,x is a scalar 1-cochain of Ga which implements the coboundary equivalence between xax-1  (Adx × Adx) and a.

Proof. We write out the three terms in the L.H.S. of the equation in the statement one by one

and expand them using Equation 2.1. In the successive steps, we just repeat the process, each

time expanding the last term coming from the previous step. In the final step, some of the terms

are decorated with numbers and strike-throughs, or underlines and alphabets, the explanation for

which is given below. These are just elementary cocycle calculations which has been exhibited,

down to the last detail. The first term is:

(xax-1, xgy-1, yhz-1)

= (x, ax-1, xghz-1)(x, agy-1, yhz-1)(x, ax-1, xgy-1)(ax-1, xgy-1, yhz-1)

= (x, ax-1, xghz-1)(x, agy-1, yhz-1)(x, ax-1, xgy-1)(a, x-1, xghz-1)(a, x-1, xgy-1)

(x-1, xgy-1, yhz-1)(a, gy-1, yhz-1)

= (x, ax-1, xghz-1)(x, agy-1, yhz-1)(x, ax-1, xgy-1)(a, x-1, xghz-1)(a, x-1, xgy-1)

(x-1, xgy-1, yhz-1)(ag, y-1, yhz-1)(a, g, y-1)(g, y-1, yhz-1)(a, g, hz-1)

= (x, ax-1, xghz-1)(x,agy-1, yhz-1 )1(x, ax-1, xgy-1)(a, x-1, xghz-1)(a, x-1, xgy-1)

A

A

B

B

(x-1, xgy-1, yhz-1)(ag, y-1,yhz-1 )2(a, g, y-1)(g,y-1, yhz-1 )7(ag, h, z-1 ) 3

C

(g,h,z-1)6(a, gh, z-1) (a, g, h)

C

The second term is:

(xgy-1, yay-1, yhz-1)

= (x, gy-1, yay-1)(x, gay-1, yhz-1)(x, gy-1, yahz-1)(gy-1, yay-1, yhz-1)

= (x, gy-1, yay-1)(x, gay-1, yhz-1)(x, gy-1, yahz-1)(g, y-1, yay-1)(y-1, yay-1, yhz-1)

(g, y-1, yahz-1)(g, ay-1, yhz-1)

= (x, gy-1, yay-1)(x, gay-1, yhz-1)(x, gy-1, yahz-1)(g, y-1, yay-1)(y-1, yay-1, yhz-1)

(g, y-1, yahz-1)(g, a, y-1)(a, y-1, yhz-1)(ga, y-1, yhz-1)(g, a, hz-1)

= (x, gy-1, yay-1)(x,gay-1, yhz-1)1(x,gy-1,yahz-1 )8(g, y-1, yay-1)(y-1, yay-1, yhz-1)

F

E

 5 (g, y-1, yahz-1)(g, 

a,

y

-1)(a,

y

-1,

yhz-1

  2 )(ga, y-1,yhz-1)(a, 

h,

z-1

)(g,ah, z-1 )

4

D

B

C

(ga, h, z-1) 3(g, a, h) 

The third term is:

(xgy-1, yhz-1, zaz-1)

= (x, gy-1, yhaz-1)(x, gy-1, yhz-1)(x, ghz-1, zaz-1)(gy-1, yhz-1, zaz-1)

3

= (x, gy-1, yhaz-1)(x, gy-1, yhz-1)(x, ghz-1, zaz-1)(g, y-1, yhaz-1)(g, y-1, yhz-1)

(y-1, yhz-1, zaz-1)(g, hz-1, zaz-1)

= (x, gy-1, yhaz-1)(x, gy-1, yhz-1)(x, ghz-1, zaz-1)(g, y-1, yhaz-1)(g, y-1, yhz-1)

(y-1, yhz-1, zaz-1)(g, h, z-1)(h, z-1, zaz-1)(gh, z-1, zaz-1)(g, h, az-1)

=

(x, gy-1,yhaz-1)8(x, 

gy-1,

yhz-1)(x,

ghz

-1 ,

zaz-1)(g,y-1, yhaz-1 )5(g,y-1, yhz-1 )

7

F
(y-1, yhz-1, zaz-1)(g,h,z-1)6(h, z-1, zaz-1)(gh, z-1, zaz-1)(gh, a, z-1)

E
(h, a, z-1)(g,ha, z-1) 4(g, h, a)

E

D

D
Thus each -term has been expressed as a product of 13 -terms. After combining these 39 terms

and noting that

(i) 8 pairs of terms cancel since g, h  Ga (the cancellations have been marked with numbers for the reader's convenience),

(ii) the last (boxed) terms on the R.H.S of the three expressions above can be combined to yield

a(g, h), we are left with the following 20 -terms which have been grouped under A,B, C,D, E, F for reasons

that will become apparent as we go along (namely, contribution towards defining the function  in

the statement of the Proposition.): A terms: = (x, ax-1, xgy-1)(x, ax-1, xghz-1) B terms: (a, x-1, xgy-1)(a, y-1, yhz-1)(a, x-1, xghz-1) C terms: (a, g, y-1)(a, h, z-1)(a, gh, z-1) D terms: (g, a, y-1)(h, a, z-1)(gh, a, z-1) E terms: (g, y-1, yay-1)(h, z-1, zaz-1)(gh, z-1, zaz-1) F terms: (x, gy-1, yay-1)(x, ghz-1, zaz-1)

The remaining 4 terms are:

(2.2)

(x-1, xgy-1, yhz-1)(y-1, yay-1, yhz-1)(x, gy-1, yhz-1)(y-1, yhz-1, zaz-1)

The second and fourth terms in expression 2.2 are again broken up using Equation 2.1 as follows:

(y-1, yay-1, yhz-1) = (y-1, y, ay-1)(y, ay-1, yhz-1)(y-1, y, ahz-1 ) 9

A

(y-1, yhz-1, zaz-1) = (y-1, y, hz-1)(y, hz-1, zaz-1)(y-1, y, haz-1 ) 9

G

F

and the first and the third terms in 2.2 taken together, is:

(x-1, xgy-1, yhz-1)(x, gy-1, yhz-1) = (x-1, x, gy-1)(x-1, x, ghz-1)

G

G

We now expand each of the terms in E, using Equation 2.1 again:

(g, y-1, yay-1) (h, z-1, zaz-1) (gh, z-1, zaz-1)

= (gy-1, y, ay-1) (g, y-1, y)

(y-1, y, ay-1)

(hz-1, z, az-1) (h, z-1, z)
(ghz-1, z, az-1) (gh, z-1, z) Call the terms in the first column as

E1

an((dzz--th11,,ezzt,,eaarzzm--s11  ))in1100the second

column as

E2.

The

new A

and F terms that popped up from breaking down 2.2, and the E1 and E2 terms are added to the

4

existing list: A terms: = (x, ax-1, xgy-1)(y, ay-1, yhz-1)(x, ax-1, xghz-1) B terms: (a, x-1, xgy-1)(a, y-1, yhz-1)(a, x-1, xghz-1) C terms: (a, g, y-1)(a, h, z-1)(a, gh, z-1) D terms: (g, a, y-1)(h, a, z-1)(gh, a, z-1) E1 terms: (gy-1, y, ay-1)(hz-1, z, az-1)(ghz-1, z, az-1) E2 terms: (g, y-1, y)(h, z-1, z)(gh, z-1, z) F terms: (x, gy-1, yay-1)(y, hz-1, zaz-1)(x, ghz-1, zaz-1) G terms: (x-1, x, gy-1)(y-1, y, hz-1)(x-1, x, ghz-1)

Thus we define a,x,y(g) = (x, ax-1, xgy-1)(a, x-1, xgy-1)(a, g, y-1)(g, a, y-1) (gy-1, y, ay-1)(g, y-1, y)(x, gy-1, yay-1)(x-1, x, gy-1). This fits the bill.

3. Diagonal Subfactors

In this section, we will describe the affine module category of the planar algebra of a diagonal
subfactor associated associated to a `G-kernel' where G is a finitely generated discrete group. Recall that G-kernel is simply an injective homomorphism  : G  Out(N ) where N is a II1 factor. If  : G  Aut(N ) is a lift of  (that is, (g) = g Inn(N ) for all g  G) and I is a set of generators of G, then the associated diagonal subfactor is given by

N  x  diag(i(x))iI  MI (N ) =: M.
Further for g1, g2  G, we may choose u(g1, g2)  U (N ) such that g1 g2 = Adu(g1,g2)g1g2 for all g1, g2  G. Associativity of multiplication in G gives us a 3-cocycle  : G×3  S1 such that

(3.1)

u(g1, g2) u(g1g2, g3) = (g1, g2, g3) g1(u(g2, g3)) u(g1, g2g3)

for g1, g2, g3  G. One may easily check that the coboundary class of the 3-cocyle  in H3(G, S1) does not depend on the choice of the lift  and the unitaries u(·, ·); this class is referred as the

obstruction of the G-kernel . It is well-known (see [Pop1]) that the standard invariant of the above subfactor N  M depends only on the group G, its generators and the 3-cocycle obstruction.

We will find the tube algebra of the category CNN of N -N bifinite bimodules coming from this

subfactor and then find the tube representations. Note that this will suffice since, by [GJ], the rep-

resentation category of the tube algebra of CNN is (tensor) equivalent to the category of annular rep-

resentations with respect to any full weight set in ob(CNN ) (in particular,

N L2(M )N

k
N :kN

which gives the affine modules of Jones). In fact, if   1, the affine modules were obtained in

[GJ].

All

simple

objects

in

CN,N

are

invertible.

This

is

clear

because

L2(M )

=
N -N

L2(Ni ).
iI

Here

the

notation N L2(N)N (for   Aut(N )) denotes the bimodule obtained from the Hilbert space L2(N )

with left N -action being the usual left multiplication whereas the right one is twisted by . This

bimodule depends only on the class defined by  in Out(N ) up to isomorphism, and the tensor

 and the contragradient of such bimodules correspond to multiplication and inverse in Out(N ).

N

Since idN is in the set {i : i  I}, we get all such index one bimodules corresponding to any g  G,

appearing as sub-bimodules of

N L2(M )N

k
N

as we vary k.

Moreover,

up to isomorphism these

are the only irreducible bimodules of CNN . Thus, the fusion algebra of CNN is just given by G. It

is then easy to verify that CNN is tensor equivalent to the category Vec(G, ) of G-graded vector

spaces with associativity constraint given by the 3-cocycle obstruction . So, our job boils down to

5

finding out the tube representations of Vec(G, ). However, we will work with bimodules in CNN instead, as the framework will be useful in the next section.
Since the standard invariant (and thereby the category CNN ) is independent of the lift , without loss of generality we assume e = idN . Further, we may set u(g1, e) = 1N = u(e, g2) for all g1, g2  G. These assumptions make the 3-cocycle  : G×3  S1 normalized.
For g  G, let Xg :=N L2(Ng )N . The morphism space in CNN from object U to object V , will be denoted by CNN (U, V ). The tube morphism from Xg1 to Xg2 is then given by Tg1,g2 := Tgs1,g2
sG
where Tgs1,g2 = CNN (Xg1 Xs, Xs Xg2). Clearly, Tg1,g2 = {0} if and only if g1 and g2 are conjugates of each other. Further, if g1 = sg2s-1, then Tgs1,g2 is one-dimensional; we will fix a distinguished element in this space, namely, a(g1, s, g2) defined by

Xg1  Xs  [1]g1  [1]s a(g-1,s,g2) [u(g1, s)u(s, g2)]s  [1]g2  Xs  Xg2.

N

N

N

N

It is an easy exercise to check that the above map is indeed an N -N -linear unitary.

Before we multiply two nonzero tube morphisms a(g1, s, g2) and a(g2, t, g3), we need to know the

one dimensional spaces CNN (XsXt, Xst) = C [1]s  [1]t -s,t [u(s, t)]st and CNN (Xst, XsXt) =

N

N

N

C [1]st -s,t [u(s, t)]s  [1]t} . Following the multiplication defined in [GJ, Section 3], we have
N

a(g2,

t,

g3)

·

a(g1,

s,

g2)

=

s,t


N

idXg3



idXs


N

a(g2,

t,

g3)



a(g1,

s,

g2)


N

idXt



idXg1


N

s,t.

Right from the definitions, one can easily see that a(g2, t, g3) · a(g1, s, g2) sends [1]g1  [1]st to
N

[g1(u(s, t))

u(g1, s)

u(s, g2)

s (u(g2, t)u(t, g3))

u(s,

t)]st


N

[1]g3

Now,

g1(u(s, t)) u(g1, s) u(s, g2) s (u(g2, t)u(t, g3)) u(s, t) = (g1, s, t) u(g1, st) u( g1s , t) u(s, g2) s(u(g2, t)) s(u(t, g3)) u(s, t)
=sg2
(using Equation 3.1 on the first two terms) = (g1, s, t) u(g1, st) (s, g2, t) u(s, g2t ) s(u(t, g3)) u(s, t)
=tg3
(using Equation 3.1 on the third, fourth and fifth terms) = (g1, s, t) u(g1, st) (s, g2, t) (s, t, g3) u(st, g3)
(using Equation 3.1 on the last three terms) = [(g1, s, t) (s, g2, t) (s, t, g3)] u(g1, st) u(st, g3)

Thus, multiplication is given by

a(g2, t, g3) · a(g1, s, g2) = [(g1, s, t) (s, g2, t) (s, t, g3)] a(g1, st, g3).

Next we will obtain the -structure on the tube algebra which we denote by # following the

notation in [GJ]. For this, we need a standard solution to the conjugate equations for the pair

(Xs, Xs-1 ).

We

set Rs

:=

s-1,s

:

Xe



Xs-1  Xs
N

and

Rs

:=

(s, s-1, s)s,s-1

: Xe  Xs Xs-1 .
N

It is completely routine to check that (Rs, Rs) satisfies the conjugate equation and is standard. Now

6

by [GJ],

(a(g1, s, g2))# =

idXs-1


N

idXg1

 (Rs)
N



idXs-1


N

(a(g1, s, g2))


N

idXs-1



Rs


N

idXg2


N

idXs-1

.

The map (a(g1, s, g2)) sends [1]s  [1]g2 to [u(s, g2)u(g1, s)]g1  [1]s. Using all the three maps

N

N

(a(g1, s, g2)), Rs and Rs, we can express the image of [1]g2  [1]s-1 under (a(g1, s, g2))# as

N

(s, s-1, s) u(s-1, s) s-1 (u(s, g2)u(g1, s)) s-1(g1 (u(s, s-1)))

 [1]g1 .

=(s-1 ,s,s-1 )

N s-1

We will simplify the first tensor component in the following way:

(s-1, s, s-1) u(s-1, s) s-1 (u(s, g2)) s-1 u(g1, s)g1 (u(s, s-1)) = (s-1, s, s-1) (s-1, s, g2) u(s-1, sg2) (g1, s, s-1) s-1(u( g1s , s-1))
=sg2

(using Equation 3.1 on the second and third, and fourth and fifth terms separately)

= (s-1, s, s-1) (s-1, s, g2) (g1, s, s-1) (s-1, sg2, s-1) u(g2, s-1) u(s-1, g1)

(using Equation 3.1 on the third and fifth terms)

= (s-1, s, s-1) (g1, s, s-1) (s, g2, s-1) (s-1, s, g2s-1 ) u(g2, s-1) u(s-1, g1)
=s-1g1

(using Equation 2.1 on the second and fourth terms)

= (g1, s, s-1) (s, g2, s-1) (s, s-1, g1) u(g2, s-1) u(s-1, g1)

(using Equation 2.1 on the first and fourth terms)

Hence, # is given by the formula: (a(g1, s, g2))# = (g1, s, s-1) (s, g2, s-1) (s, s-1, g1) a(g2, s-1, g1).

The canonical (faithful) trace  on the tube algebra (as defined in [GJ]) is given by (a(g1, s, g2)) = g1=g2s=e. Thus, the set {a(g1, s, g2) : g1, g2, s  G satisfying g1s = sg2} becomes an orthonormal basis with respect to the inner product arising from  and #.
To have a better understanding of the -algebra structure of the tube algebra, we will now set up some notations. Let C denote the set of conjugacy classes of G. For each C  C , we pick a representative gC  C and for each g  C, we fix wg  G such that g = wg gC wg-1 and wgC = e. Also for C  C , we will denote the centralizer subgroup of gC by GC := {s  G : gC = s gC s-1}, and C will denote the 2-cocycle on GC given by C (s, t) := gC (t-1, s-1) (recall the definition of gC in Lemma 2.1).
With the above notation, we give an alternate description of -algebra structure of the tube algebra in the following proposition which will be handy in classifying the representations.

Theorem 3.1.

(i) The tube algebra T

= ((Tg1,g2 ))fin. supp.

is isomorphic

to
C C

MC  [CGC ]C

as

a -algebra where [CGC]C is the 2-cocycle twisted group algebra and MC denotes the -algebra of

finitely supported matrices whose rows and columns are indexed by elements of C.

(ii) Every Hilbert space representation  : T  L(V ) decomposes over C  C uniquely (up to

isomorphism) as an orthogonal direct sum of submodules generated by the range of the projection (a(gC , e, gC )) (which is the gC th-space of V ). (We will call a representation of T `supported on C  C ' if it is generated by its vectors in the gCth-space.) The category of C-supported representations of T is additively equivalent to representation category of [CGC ]C .

7

Proof. (i) We will send the orthonormal basis of T (discussed above) via a map  to a canonical basis of CC MC  [CGC ]C in the follwoing way: for g1, g2  C and s  G such that g1s = sg2 (implying wg-11swg2  GC )
a(g1, s, g2) - gC ,wg1 ,wg2 (wg-11swg2 ) Eg2,g1  [wg-21s-1wg1 ] where we use the family of functions a,x,y : Ga  S1 a,x,yG appearing in Proposition 2.2.
To show  preserves multiplication, consider
(a(g2, t, g3)) (a(g1, t, g2))
= gC ,wg2 ,wg3 (wg-21twg3 ) Eg3,g2  [wg-31t-1wg2 ] gC ,wg1 ,wg2 (wg-11swg2 ) Eg2,g1  [wg-21s-1wg1 ] = gC,wg1 ,wg2 (wg-11swg2 ) gC,wg2 ,wg3 (wg-21twg3 ) C (wg-31t-1wg2 , wg-21s-1wg1 ) Eg3,g1  [wg-31(st)-1wg1 ] = gC ,wg1 ,wg2 (wg-11swg2 ) gC ,wg2 ,wg3 (wg-21twg3 ) gC ,wg1 ,wg3 (wg-11stwg3 ) gC (wg-11swg2 , wg-21twg3 )
(a(g1, st, g3)) = (g1, s, t)(s, g2, t)(s, t, g3) (a(g1, st, g3)) (using Proposition 2.2) =  (a(g2, t, g3) a(g1, s, g2)) . The map  is  preserving because [(a(g1, s, g2))]
= gC,wg1 ,wg2 (wg-11swg2 ) C (wg-11swg2 , wg-21s-1wg1 ) Eg1,g2  [wg-11swg2 ] = gC,wg1 ,wg2 (wg-11swg2 ) gC (wg-11swg2 , wg-21s-1wg1 ) gC,wg2 ,wg1 (wg-21s-1wg1 ) (a(g2, s-1, g1)) = gC,wg1 ,wg1 (e) (g1, s, s-1) (s, g2, s-1) (s, s-1, g1) (a(g2, s-1, g1)) (using Proposition 2.2) =  [a(g1, s, g2)]#
where we use gC,wg1,wg1 (e) = 1 at the very last step which follows directly from the definition of a,x,y in the proof of Proposition 2.2.
(ii) The decomposition follows easily from the -algebra structure described in part (i). Fix C  C . If  : T  L(W ) is C-supported, then we can define the representation  : [CGC ]C  L(WgC ) defined by (s) =  -1(EgC,gC  [s]) . Conversely, if  : [CGC ]C  L(U ) is a representation, then one can consider the unique extension
 : T  L(l2(C)  U ) defined by  -1(Eg1,g2  [s]) := g1C Eg1,g2  (s).
Remark 3.2. Note that the canonical trace  on T corresponds to the direct sum of the canonical traces on MC  [CGC]C . Also, the -algebra Te,e (by definition) is isomorphic to the fusion algebra which is basically the group algebra CG without any nontrivial 2-cocycle twist (since e is the constant function 1 which follows from its definition in Lemma 2.1). Thus, the analytic properties (such as, amenability, Haagerup, property (T)) of the bimodule category corresponding to the subfactor N  M corresponds exactly to that of the group G; this fact was obtained by Sorin Popa long time back in [Pop1] and [Pop2]. However, the analytic properties in the higher weight spaces (as defined in [GJ]) depend on the corresponding centralizer subgroup.
4. Bisch-Haagerup Subfactors
In this section, we intend to find the tube algebra of the Bisch-Haagerup subfactor N := QH  Q  K =: M where H and K act outerly on the II1-factor Q. It is well known that the planar
8

algebra of N  M depends on the group G generated by H and K in Out(Q) and the scalar 3-cocycle obstruction (up to 2-coboundary) (see [BH, BDG]).

We first lay down the strategy to achieve our goal. Instead of computing the tube algebra of CNN directly (unlike the case of diagonal subfactors because the irreducible bimodules of CNN for Bisch-Haagerup subfactors, are not so easy to work with), we will consider the affine annular algebra with respect to a particular full weight set (in the sense of [GJ, Definition 3.4]) in ob(CNN ), and then cut it down by the Irr CNN .
We need to set up some notations for this. Pick a representative map Out(Q)  G  g  g  Aut(Q) such that g = gInn(Q), and |H : H  Aut(Q), |K : K  Aut(Q) are homomorphisms. Now, if X =N L2(M )M , Y =N L2(Q)Q and Z =Q L2(M )M , then

m
X X N = Y ( Z Z )( Y Y )( Z Z )···( Z Z )Y.

Q

Q

Q

QQ

Q

We know that ( Y

Y

) =
Q-Q

hH

QL2(Qh )Q

and

(

Z

Z

)

=
Q-Q

QL2(Qk )Q.
kK

So,

XX

k
N

=

N -N

N L2(Qk1 h1 k2 h2 ···km )N

=
N -N

k1 ,k2 ,...K

h1,h2,...H

N L2(Qk1h1k2h2···km )N
k1,k2,...K h1,h2,...H

Since the subgroups H and K generate G, therefore the set  := Xg := N L2(Qg )N g  G forms

a full since

weight set Xg = Xgh

in for

CNN . all g

It is possible to reduce the indexing  G, h  H. However, we will not do

set G of the weight set that since by reducing

 further, the weight

set, one needs to work with coset representative which makes the calculations more cumbersome.

4.1. Morphism spaces in CNN .
For the affine annular algebra over G (indexing the above set), we do not need all morphism spaces of CNN . We will instead concentrate on morphisms between elements of  and their tensor products. Before that, we need more notations. Choose a map u : G × G  U (Q) such that g1 g2 = Adu(g1,g2)g1g2 and

(4.1)

u (H × H  K × K  G × {e}  {e} × G) = {1Q}.

Again, associativity of multiplication in G and condition 4.1 will give us a 3-cocycle  satisfying Equation 3.1 and

(4.2)

|H×H×H  1  |K×K×K .

This along with Equation 2.1, implies (g, l, l-1) = (gl, l-1, l) and (l-1, l, g) = (l, l-1, lg) for

all g  G, l  H  K. We will now prove a lemma on scalar cocycles which lets us choose the map

u in such a way that the 3-cocycle  gets simplified making our calculations easy.

Lemma 4.1. Any scalar 3-cocycle  of a group G generated by subgroups H and K, is coboundarily equivalent to  which satisfies the relation 4.2 as well as

(4.3)

(g, l, l-1) = 1 = (l-1, l, g) for all g  G, l  H  K.

Proof. Consider the subsets AH = (H × H×), AK = (K × K×), VH = (H× × H), and VK = (K× × K) of G × G, and the order 2 bijections G × G  (g1, g2)^ (g1, g2)^= (g1g2, g2-1)  G × G and G × G  (g1, g2) (g1, g2)= (g1-1, g1g2)  G × G (where H× = H \ {e} and K× = K \ {e}). Note that AH and AK (resp. VH and VK ) are separately closed under^(resp. ) and have no fixed points. Now, AH  VK = (K \ H) × (H \ K) (resp. AK  VH = (H \ K) × (K \ H)) is mapped into AH \ VK (resp. AK \ VH ) under^and into VK \ AH (resp. VH \ AK ) under. We choose
9

(i) a representative in each orbit of ^ inside AH  AK such that the representative of the orbit containing (k, h)  AH  VK is chosen as (kh, h-1) and the representative of the one containing (h, k)  AK  VH is chosen as (hk, k-1),
(ii) a representative in each orbit of  inside VH  VK such that the representative of the orbit containing (k, h)  AH  VK is chosen as (k-1, kh) and the representative of the one containing (h, k)  AK  VH is chosen as (h-1, hk).
Let A (resp. V ) be the set of representatives in AH  AK (resp. VH  VK ). From our choice, it can be verified that A  V = . Define  : G × G  T by:
(a) |G×G\(AV ) = 1, (b) (g, l) = (g, l, l-1) = (gl, l-1, l) for (g, l)  A, (c) (l, g) = (l-1, l, g) = (l, l-1, lg) for (l, g)  V . It follows that 2() is normalized since  is also so, and 2() satisfies the relation 4.2 since (H × H  K × K)  (A  V ) =  (where 2 denotes the 2-cochain map). Thus, the 3-cocycle  = 2() ·  is normalized and satisfies relation 4.2.
For relation 4.3, we consider g  G and l  H  K. Without loss of generality, we assume g = e = l. So, (g, l), (gl, l-1) ( resp. (l, g), (l-1, lg) ) is an orbit of ^(resp. ) in AH  AK (resp. VH  VK ), and  takes the value (g, l, l-1) = (gl, l-1, l) (resp. (l-1, l, g) = (l, l-1, lg)) on the representative of the orbit and 1 on the other. This implies
(g, l, l-1) = (gl, l-1) (g, l) (g, l, l-1) = 1
since (g, e) = 1 = (l, l-1), and similarly (l-1, l, g) = 1.

By the above lemma, without loss of generality, we may assume  satisfies:

(4.4)

(i) (g1, l, l-1) = 1 = (g1, l, l-1)

(ii) (g1, g2, l)

=

(g1, g2l, l-1)

(iii) (g1, l, g2)

=

(g1l, l-1, lg2)

(iv) (l, g1, g2)

=

(l-1, lg1, g2)

(v)

u(g1, l)

=

u(g1l, l-1)

(vi)

u(l, g2)

=

l u(l-1, lg2)

for all g1, g2  G, l  H  K (where (ii), (iii) and (iv) are immediate implication of (i) and 2.1). We will need the relation 4.4 only when l  H; however, we gave the general version, in case any
reader is interested to see the actual 2-category of N  M instead of just CNN .

Proposition 4.2. The morphism space CNN (Xg1, Xg2) is zero unless g1 and g2 give the same H-H double coset, and if they do, the space has a basis given by





   

Xg1

 [x]g1 - [h1 (x)u(h1, g1)u(g2, h2)]g2  Xg2

Bg1,g2 :=

g2

  

denoted

by

the

symbol

h1

h2

 

g1





h1, h2  H

   

such that

.

h1g1 = g2h2

  





Proof. By Frobenius reciprocity, dimC (CNN (Xg1 , Xg2 )) = dimC CNN N L2(N )N , Xg1  Xg2 .
N

Again Xg1  Xg2 =

N

Q-Q

g1 [Q  H] g2 where the left and right actions of Q on Q  H is twisted

by g1 and g2 respectively. Any element of CNN N L2(N )N , g1 [Q  H] g2 corresponds to an

10

element of Q  H (the image of ^1), say y = yh h. By N -N linearity, we will have g1(n) yh =
hH
yh h(g2(n)) for all n  N, h  H, equivalently
n -g11(yh) = -g11(yh) -g11(h(g2(n))) for all n  N, h  H.

The following is a well-known fact for the fixed-point subfactor N  Q of an outer action of H. For y  Q and   Aut(Q), the following are equivalent:

(i) y = 0 and ny = y(n) for all n  N = QH ,

(ii) y0 :=

y y

 U (Q) and Ady0    {h : h  H}.

By the above fact, y = 0 only when there exists h1, h2  H such that -g11h1g2h2  Inn(Q),

equivalently g1 and g2 generate the same H-H double coset. In particular, yh = 0 unless h belongs to

H y0

 g1Hg2-1.

=

yh yh

.

And This

for h  implies

H  g1Hg2-1, Ady0 h g2

for yh = 0, we have Ad-g11(y0) = g1 g1-1hg2, equivalently,

-g11 h g2 Ady0u(h,g2)

= =

g1-1hg2 where Adu(g1,g1-1hg2).

Hence, yh  C{u(g1, g1-1hg2)u(h, g2)}. Thus, the set

u(g1, h-2 1)u(h-1 1, g2) h-1 1 : h1, h2  H such that h1g1 = g2h2

forms a basis of the vector space V := {y  Q  H : g1(n)y = yg2(n) for all n  N }. To show that the set Bg1,g2 forms a basis for Hg1H = Hg2H, we need the following explicit isomorphism:

V  y - (y) := J yJ  CNN (Xg1 , Xg2 )

where J is the canonical anti-unitary of L2(Q). Set yh1,h2 := u(g1, h-2 1)u(h-1 1, g2) h-1 1 for h1g1 = g2h2. Then,

(yh1,h2)[x]g1 = h1 xu(g1, h-2 1)u(h-1 1, g2) g2 = h1 (x) h1 u(g1, h-2 1)u(h-1 1, g2) g2 .

We simplify h1 u(g1, h-2 1)u(h-1 1, g2) using Equations 3.1, 4.1 and 4.4 to get

(h1, g1, h-2 1)u(h1, g1)u(h1g1, h-2 1)u(h1, g1h-2 1) u(h1, h-1 1g2) = (h1, g1, h-2 1) u(h1, g1)u(g2h2, h-2 1) = (h1, g1, h-2 1) {u(h1, g1)u(g2, h2)} .

g2 Hence, (yh1,h2) is a unit scalar multiple of h1 h2 corresponding to (h1, h2).
g1

Remark 4.3. The maps

N L2(N )N  ^1 -Rg

i

[u(g-1, g)g-1 (bi)]g-1


N

[bi ]g



Xg-1


N

Xg

N L2(N )N  ^1 -Rg (g, g-1, g)

i

[u(g, g-1)g(bi)]g


N

[bi ]g-1



Xg


N

Xg-1

are standard solutions to conjugate equations for duality of Xg where {bi}i is a basis for the subfactor N  Q. We will also need the  of these maps, namely

Xg-1  Xg  [x]g-1  [y]g -Rg EN xg-1 (y)u(g-1, g)  N L2(N )N

N

N

Xg  Xg-1  [x]g  [y]g-1 -Rg (g, g-1, g) EN xg(y)u(g, g-1)  N L2(N )N

N

N

11

Proposition 4.4.

g3

h3 h4

g3

g2

g3

(i)

g2 := h3 h4  h1 h2 = (h3, h1, g1) (h3, g2, h2) (g3, h4, h2) h3h1 h4h2

h1 h2

g2

g1

g1

g1

 g2 

g1

(ii)  h1 h2  g1

= (h1, g1, h-2 1) h-1 1 gh2 -2 1

Proof. (i) The left side is given by [x]g1  [h3h1 (x) h3(u(h1, g1)u(g2, h2)) u(h3, g2)u(g3, h4)]g3 . Observe that
h3(u(h1, g1)u(g2, h2)) u(h3, g2)u(g3, h4) = (h3, h1, g1) u(h3h1, g1) u(h3, h1g1 ) h3 (u(g2, h2)) u(h3, g2) u(g3, h4)
=g2h2
(applying 3.1 and 4.1 on the first term) = (h3, h1, g1) (h3, g2, h2) u(h3h1, g1) u( h3g2 , h2) u(g3, h4)
=g3h4
(applying 3.1 on the second, third and fourth terms) = (h3, h1, g1) (h3, g2, h2) (g3, h4, h2) u(h3h1, g1) u(g3, h4h2)
(applying 3.1 and 4.1 on the last two terms)

which gives the required result.

g2

(ii) Note that h1 h2 is a unitary which follows right from its definition. Using part (i), one can g1

g1

g2

easily show that h-1 1

gh2 -2 1

is indeed the inverse of h1

h2 g1

where one uses the relations in 4.4.

Next, we will prove some facts about tensor product of two elements from . For g1, g2  G and g1hg2

h  H, we define h

: Xg1  Xg2  Xg1hg2 in the following way

g1 g2

N

Xg1


N

Xg2



[x]g1


N

[y]g2

-

|H

|-

1 2

[x

g1 (h (y))

u(g1, h)

u(g1h, g2)]g1hg2



Xg1hg2 .

Remark 4.5. With standard inner product computation, one can show that



g1hg2

h 

 

:

[z]g1hg2

-

|H

|-

1 2

g1 g2

i

[zu(g1h,

g2)u(g1,

h)g1 (bi)]g1


N

[h-1 (bi )]g2

=

|H

|-

1 2

i

[g1 (bi)]g1  [h-1
N

bi -g11(zu(g1h, g2)u(g1, h)) ]g2

where {bi}i is any basis of Q over N .
12

To see this, consider

g1hg2

h

[x]g1  [y]g2

g1 g2

N

, [z]g1hg2

=

|H

|-

1 2

tr

(x

g1 (h (y))

u(g1,

h)

u(g1h,

g2)

z)

=

|H |-

1 2

tr (x g1 (EN (yh-1 (bi))bi ) u(g1, h) u(g1h, g2) z)

i

=

|H |-

1 2

tr (x g1 (EN (yh-1 (bi))) (z u(g1h, g2) u(g1, h) g1(bi)))

i

=

|H |-

1 2

i

[x]g1 N [y]g2 , [h-1 (bi )]g2 , [z u(g1h, g2) u(g1, h) g1(bi)]g1



g1hg2



g1hg2

g1 g2

=

[x]g1

 [y]g2
N

,


 g1

h

 

[z]g1hg2

.

We will denote  

h

g2

g1

 by  g2

h . It is g1hg2

g1 g2

straightforward to check h

preserves inner product and thereby is an isometry. So, the

g1 g2 h

g1hg2  g1 g2

g1hg2

element

g1hg2

:=  

h

g1 g2 h  H.

h

h

g1hg2 g1

g2

 is a projection 

in End (Xg1  Xg2 ) for every
N

 

g1hg2

 





Proposition 4.6. The set

h

: h  H gives a resolution of the identity in End (Xg1 

 

g1

g2

 

N

Xg2 ).

g1 g2 h

Proof. It is enough to check
hH

g1hg2 h

=

idXg1 Xg2 . N

The left side acting on [x]g1  [y]g2
N

gives

g1 g2

= |H|-1 [g1 (bi)]g1  [h-1

i,h

N

bi -g11(x g1 (h(y))) ]g2

= |H|-1 [g1 (bi)]g1  [h-1

i,h

N

bi -g11(x)

y]g2 =

[g1 (bi)]g1  [EN

i

N

bi -g11(x)

y]g2 = [x]g1  [y]g2 .
N

13

Remark PSfrag 4.7. From Propositions 4.2 and 4.6, we may conclude that CNN (Xg1  Xg2, Xg3  Xg4) is

N

N

linearly spanned by the (linearly independent) set

g3 g4

 





   

h4

   

    

g3 g4

g3h4g4

g1h3g2

    

h1 h2 := h4

 h1 h2  h3

h1, h2, h3, h4  H .

   

g3h4g4

g1h3g2 g1

g2

   

   

h3

   





g1 g2

 

We will now prove two lemmas which will be very useful in finding the structure the annular algebra. As for notations, we will use the standard graphical representations of morphism where composition will be represented by stacking the morphisms vertically with the left most being in the top.

Lemma 4.8. (i)
(ii)

g1hs

g1hs

h

e h2

s = [(g1h, s, h2) (g1h, h1, t) (g1, h, h1)] g1hh1t

g1 h1 h2 t

hh1 g1 t

shg2

h s
h1 h2 t

g2 = (h1, th-2 1h, g2) (s, h2, h-2 1h) (h1, t, h-2 1h)

shg2
h1 e th-2 1hg2
h-2 1h t g2

Proof. (i) The left side acts on [x]g1  [y]t, gives
N

|H

|-

1 2

[xg1

(h

(h1 (y)u(h1,

t)u(s,

h2)))

u(g1, h)u(g1h,

s)]g1hs

whereas the right side yields

|H

|-

1 2

[xg1

(hh1 (y))

u(g1,

hh1)u(g1hh1,

t)u(g1hs,

h2)]g1hs

.

After striking out the similar terms, we will be left with

g1 (h (u(h1, t)u(s, h2))) u(g1, h)u(g1h, s) = u(g1, h)g1h (u(h1, t)u(s, h2)) u(g1h, s) = u(g1, h)g1h (u(h1, t)) (g1h, s, h2)u(g1h, sh2 )u(g1hs, h2)
=h1t
= (g1h, s, h2)u(g1, h)(g1h, h1, t)u(g1h, h1)u(g1hh1, t)u(g1hs, h2) = [(g1h, s, h2)(g1h, h1, t)(g1, h, h1)] (u(g1, hh1)u(g1hh1, t)u(g1hs, h2))

(ii) The action of left side on [x]t  [y]g2 is
N

|H

|-

1 2

[h1 (x)u(h1, t)u(s, h2)

s (h(y))

u(s, h)u(sh, g2)]shg2

=

|H

|-

1 2

h1

x t(h-2 1h(y)

u(h1, t)u(s, h2) u(s, h)u(sh, g2)
shg2

14

and the right side on the same is

|H

|-

1 2

h1

=

|H

|-

1 2

h1

x t(h-2 1h(y) u(t, h-2 1h)u(th-2 1h, g2) u(h1, th-2 1hg2) shg2 x t(h-2 1h(y) h1 (u(t, h-2 1h)) (h1, th-2 1h, g2)u(h1, th-2 1h)u(h1th-2 1h, g2) shg2

=

(h1,

th-2 1h,

g2)

|H |-

1 2

h1

x t(h-2 1h(y)

=

(h1,

th-2 1h,

g2)(h1,

t,

h-2 1h)

|H |-

1 2

(h1, t, h-2 1h)u(h1, t)u( h1t , h-2 1h) u(sh, g2)

=sh2

shg2

h1

x t(h-2 1h(y)

u(h1, t) (s, h2, h-2 1h)u(s, h2) u(s, h) u(sh, g2)

.
shg2

Lemma 4.9. g2h2g3

g1 g2h2g3

g1

h2

h1

(i)

g2 = [(g1h1, g2, h2) (g1h1, g2h2, g3)]

h1

g3

g1h1g2h2g3 h2

g1h1g2

g1h1g2 g3

g1h1g2

g1h1g2 g3

h1

g3

h2

(ii)

g2 = [(g1h1, g2, h2) (g1h1, g2h2, g3)]

g1h1g2h2g3

g1

h2

g2h2g3

h1 g1 g2h2g3

Proof. The left side acting on [x]g1h1g2  [y]g3 gives
N

|H |-1

i

[xu(g1h1,

g2)u(g1,

h1)g1 (bi)]g1


N

[-h11(bi )g2 (h2(y))

u(g2,

h2)u(g2h2,

g3 )]g2 h2 g3

= |H|-1 [xu(g1h1, g2)u(g1, h1)g1 (bi)]g1

i,j


N

[EN

(bi h1

(g2(h2 (y))

u(g2,

h2)u(g2h2,

g3))

bj )

-h11(bj )]g2h2g3

= |H|-1

j

[xu(g1h1,

g2)u(g1,

h1)g1

(h1

(g2(h2 (y))

u(g2,

h2)u(g2h2,

g3))

bj )]g1


N

[-h11(bj )]g2h2g3

= |H|-1 [x g1h1g2 (h2 (y)) u(g1h1, g2)u(g1, h1) g1 (h1 (u(g2, h2)u(g2h2, g3))) g1(bj )]g1

j


N

[-h11(bj )]g2h2g3 .

Simplifying the underlined expression, we get

u(g1h1, g2) g1h1 (u(g2, h2)u(g2h2, g3)) u(g1, h1) = (g1h1, g2, h2) u(g1h1g2, h2)u(g1h1, g2h2) g1h1 (u(g2h2, g3)) u(g1, h1) = (g1h1, g2, h2) u(g1h1g2, h2) (g1h1, g2h2, g3)u(g1h1g2h2, g3)u(g1h1, g2h2g3) u(g1, h1).

This is exactly what we wanted from the right side acting on [x]g1h1g2  [y]g3.
N
(ii) This follows from taking  on both sides.
15

4.2. The affine annular algebra over the weight set  indexed by G.

Let A denote the affine annular algebra of CNN with respect to G which indexes the weight set . In our set up, the indexing set G is more important rather than the set ; for instance, Xh and Xe are identical in   ob(CNN ) when h  H.

We will recall the definition of A here. For g1, g2  G, we have a vector space Ag1,g2 which is the

quotient of the vector space

CNN Xg1  W , W  Xg2 over the subspace generated by

W ob(CNN )

N

N

elements of the form

a



(idXg1


N

f)

-

(f


N

idXg2 )



a

for a  CNN

Xg1  Z , W  Xg2

N

N

and

f  CNN (W, Z). We denote the quotient map by g1,g2. We will also use the notation gW1,g2 (resp.,

gs1,g2) for the restriction map g1,g2

(resp., g1,g2

for

CNN Xg1 W , W Xg2

N

N

CNN Xg1 Xs , XsXg2

N

N

s  G). Further, AW g1,g2 and Asg1,g2 will denote the range of the maps gW1,g2 and gs1,g2 respectively.

Notation. For any two vectors v1 and v2 in any vector space, we will write v1  v2 when span v1 = span v2.

 

 s g2 

 





   

 

h2

 

   





   

 

sh2g2

   

Proposition 4.10. Ag1,g2 is linearly spanned by the set

gs1,g2

 

h1

e

 

h1, h2

 H, s  G

.





   

 

g1s

 

   





  

 

e

 

  











g1 s



We denote the above element by a(h1, g1, s, h2, g2). Note that h1g1 = s h2g2 s-1.

Proof. Since the weight set  = {Xs : s  G} is full, we may use the relation satisfied by the

quotient map g1,g2 to say Ag1,g2 = span sGAsg1,g2 . So, by Remark 4.7, Ag1,g2 can be linearly

 s g2 

 

h4

 

 

sh4g2

generated

by

elements

of

the

form

gs1 ,g2

 

h1

h2

 

for

h1,

h2,

h3,

h4



H,

s



G.





 g1h3s





 

h3

 

g1 s

16

g

h-1 1 h2

Using Proposition 4.4, we may write

h1gh2 

h1 h-2 1

g

g
ee g

satisfied by the quotient map and setting t := h3sh-2 1, we get

= idXg . Again, using the relation

 s g2 



 

h4

 

 

 

sh4g2

 

t

gs1 ,g2

 

h1

h2

 



gt 1 ,g2

 

h3

h2





 g1h3s

 

s







 

h3

 

 

g1 s

s g2



h4

 



sh4g2 

s

 


N

idXg2





h1

h2 g1h3s



idXg1


N

h-3 1

th-2 1 .



h3

 

g1 s

We then apply Lemma 4.8 (i) and (ii) to get

 s g2 



 

h4

 



 

sh4g2

 

gs1 ,g2

 

h1

h2

 



gt 1 ,g2

 







 g1h3s









 

h3

 



g1 s

t g2 h2h4 
th2h4g2

th2h4g2



h3

e h-3 1th2h4g2



g1t

 

h1 h2

e

 

.

g1th2

g1

t

 

e gh1t-2 1

 

where the three vertically stacked discs correspond to their composition. Once we apply the multiplication of these discs as stated in Proposition 4.4 (i), it becomes clear that the resultant (up to a unit scalar) is indeed of the form mentioned in the statement of this proposition.

We will next unravel the multiplication in A.

Remark 4.11. Multiplication of affine annular morphisms is given by

gt2,g3 (c)  gs1,g2 (d)

=

Xs Xt
g1 ,gN3

idXs


N

c



d


N

idXt

for c  CNN (Xg2  Xt , Xt  Xg3) and d  CNN (Xg1  Xs , Xs  Xg2). Using Proposition 4.6, we can rewrite the above as

gt2,g3 (c)  gs1,g2 (d)

 sht

s t

=

gs1h,tg3

 

h

hH

s

t


N

idXg3



idXs


N

c



d


N

idXt



idXg1


N

h . 
sht

Proposition 4.12.
(h2, g2, t)(t, h3, g3) a(h2, g2, t, h3, g3)  (h1, g1, s)(s, h2, g2) a(h1, g1, s, h2, g2) = h2=h2 (s, t, h3g3)(s, h2g2, t)(h1g1, s, t) [(h1, g1, st) (st, h3, g3) a(h1, g1, st, h3, g3)]
17

Proof. The above remark lets us express the element [a(h2, g2, t, h3, g3)  a(h1, g1, s, h2, g2)] as a sum over h  H of













 

sht

gs1h,tg3

  

h

s t
























N

idXg3



idXs 


N









t g3  s g2



h3

 

 

h2

 

th3g3

 

sh2g2

 

h2

e

 



 

h1

e



g2t

 

 

g1s


N

idXt

  



idXg1


N





e

 

 

e

 

g2 t

g1 s









s

t

 

 h .


sht

 











=: bh say

We could use Lemma 4.9 at three instances in the above expression of bh, and thereby we may rewrite bh up to a unit scalar as

(4.5)

sht g3 h3
shth3g3 h





idXs


N

h2

s

th3g3

h2

e   sh2g2t

g2t

e

g2t 
  h1

g1s t

sh2g2



e g1s


N

idXt





h g1sht .
e

s th3g3

sh2g2 t

g1 sht

In the above expression 4.5, using Lemma 4.8 (ii), we could make the disc in the fourth term pass through the bottom box in the third term to its top. As a result, expression 4.5 turns out to be a scalar multiple of

(4.6)

sht g3 h3
shth3g3 h



th3g3



idXs


N

h2

e g2t



s g2t

h2 sh2g2t
h1 e

g1s t h
 g1sht .

g1st

e

s th3g3

e

g1 sht

g1s t

Observe that the composition of the bottom box of the third term and the top box in the fourth

term (in expression 4.6) is idXg1st if h = e and zero otherwise; this follows from Proposition 4.6. Similarly, Lemma 4.8 (i) allows us to move the disc in the second term up through the bottom box

of the first term, and thereby the expression 4.6 becomes a scalar multiple of

(4.7)

sht g3

h3

(shh2g2t =)shth3g3



hh2

s g2t

s g2t

h2 sh2g2t
h1 e

g1s t h
 g1sht .

g1st

e

e

g1 sht

g1s t

Again, by Proposition 4.6, the composition of the bottom box of the first term and the top of the second term in expression 4.7 is idXshth3g3 if hh2 = h2 and zero otherwise.
18

We now consider the case h = e and h2 = h2 (= h2 say). The above discussion implies that in this case, [a(h2, g2, t, h3, g3)  a(h1, g1, s, h2, g2)] is indeed a scalar multiple of a(h1, g1, st, h3, g3). So, we need to gather the 3-cocycle arising at various steps. To obtain step 4.5, Lemma 4.9 will give the following six scalars
[(s, t, h3)(s, th3, g3)] [(sh2,g2, e)(sh2, g2, t)] [(g1, s, e)(g1, s, t)] .
Application of Lemma 4.8 (ii) (resp., (i)) while obtaining step 4.6 (resp., 4.7) from step 4.5 (resp., 4.6), yield
[(h1, g1s, t)(sh2g2,e,e)(h1, g1s, e)] (resp., (s,th3g3,e)(s, h2, g2t)(s,e,h2) ).
Thus, we obtained the equation
a(h2, g2, t, h3, g3)  a(h1, g1, s, h2, g2) = [(s, t, h3)(s, th3, g3)(sh2, g2, t)(g1, s, t)(h1, g1s, t)(s, h2, g2t)] a(h1, g1, st, h3, g3).
We will be done with the proof once we match the scalars. Applying the 3-cocycle relation 2.1 on first and second, third and sixth, fourth and fifth terms separately, we get
[(st, h3, g3)(s, t, h3g3)(t, h3, g3)] [(s, h2, g2)(s, h2g2, t)(h2, g2, t)] [(h1g1, s, t)(h1, g1, st)(h1, g1, s)]

Notation. We see that [(h1, g1, s)(s, h2, g2) a(h1, g1, s, h2, g2)] is better behaved with respect to multiplication than a(h1, g1, s, h2, g2). So, we set A(h1, g1, s, h2, g2) := [(h1, g1, s)(s, h2, g2) a(h1, g1, s, h2, g2)] and the above proposition translates as: A(h2, g2, t, h3, g3)  A(h1, g1, s, h2, g2) = h2=h2 [(s, t, h3g3)(s, h2g2, t)(h1g1, s, t)] A(h1, g1, st, h3, g3).

Next we will compute the canonical trace  on Ag,g for g  G. For this, we need orthonormal basis of CNN (N L2(N )N , Xs) for s  G with respect to the inner product given by

CNN (N L2(N )N , Xs) × CNN (N L2(N )N , Xs)  (c, d) - d  c  C.

By Proposition 4.2, CNN (N L2(N )N , Xs) is zero unless s  H. Now, Xe = Xh for all h  H. Since

NtheinQcluissiirornedmuacpib^1le,-thhe[s1p]ha.cehCNisNs(imN Lpl2y(Nth)eNc,oXnhd)itiisoonnael -edximpeecntsaitoinonal

and EN .

spanned

by

the

element

The definition of  then turns out to be (following [GJ])

Ag,g  gs,g(c) -

Rg 

sH

idXg-1


N

s


N

idXg



idXg-1


N

c



idXg-1

Xg
N

 s
N

 Rg  C.

Proposition 4.13.  (A(h1, g, s, h2, g)) = h1=h2 s=e.
19

Proof. For h  H, we need to compute the scalar

Rg 

idXg-1


N

h


N

idXg













idXg-1


N









h g

h2

 

hh2g

h1

e

 





gh 



e

 

gh

idXg-1

Xg
N

 h
N

 Rg (^1)

=

Rg 
i

idXg-1


N

h


N

idXg













idXg-1


N









h g

h2

 

hh2g

h1

e gh

   

[u (g-1 ,

g)g-1 (bi)]g-1


N

[bi ]g


N

[1]h



e

 

gh

= |H|-1 Rg 
i,j

idXg-1


N

h


N

idXg

[u(g-1, g)g-1 (bi)]g-1


N

[h1

(bi u(g,

h))

u(h1, gh)

u(hh2, g) h(bj)]h


N

[h-2 1

(bj )]g

= |H|-1 Rg
j

u(g-1, g)g-1 u(g, h) h-1 1 (u(h1, gh) u(hh2, g)h(bj ))

g-1


N

[h-2 1 (bj )]g

= |H|-1 u(g-1, g)g-1 u(g, h) h-1 1 (u(h1, gh) u(hh2, g)h(bj )) h-2 1(bj ) u(g-1, g)
j

= |H|-1 u(g-1, g)g-1 u(g, h) h-1 1 (u(h1, gh) u(hh2, g)) g-1 h-1 1h(bj )h-2 1 (bj ) u(g-1, g).
j

Pulling the sum over the last term, we get g-1 h-2 1 j h2h-1 1h(bj )bj

= h=h1h-2 1 |H| (which

is a standard fact in fixed-point subfactor of an outer action of finite group). Let us assume

h = h1h-2 1. But then, h1gh = hh2g will imply h1 = h2 and thereby h = e.

Under the assumption h = e and h1 = h2, in the above expression, the term in between u(g-1, g) and u(g-1, g), becomes 1. This gives the required result.

Corollary 4.14. The set {A(h1, g1, s, h2, g2) : h1, h2  H, s  G such that h1g1s = sh2g2} is a basis for Ag1,g2 .

Proof. This easily follows from that  is non-degenerate on A (which is a consequence of  being positive (see [GJ])).
20

We will now describe the -structure on A which we denote by #. From [GJ], the definition of (gs1,g2(c))# is the following:

gs2-,1g1

idXs-1


N

idXg1


N

Rs



idXs-1


N

c


N

idXs-1



Rs


N

idXg2


N

idXs-1

 Ag2,g1 .

Proposition 4.15. (A(h1, g1, s, h2, g2))#

= (h1g1, s, s-1) (s, h2g2, s-1) (s, s-1, h1g1) A(h2, g2, s-1, h1, g1).

Proof. Set A(h1, g1, s, h2, g2) := (h1g1, s, s-1) (s, h2g2, s-1) (s, s-1, h1g1) A(h2, g2, s-1, h1, g1). Now, we get an inner product ·, ·  defined as

A(h1, g1, s, h2, g2) , A(h3, g3, t, h4, g4)  :=  A(h3, g3, t, h4, g4) A(h1, g1, s, h2, g2)

and extended linearly in the first and conjugate-linearly in the second variable. In fact, the basis elements are orthonormal with respect to ·, · . Since   # =  (by positivity of  ([GJ])), it
will be enough to prove (A(h1, g1, s, h2, g2))#  A(h1, g1, s, h2, g2). This is equivalent to proving (a(h1, g1, s, h2, g2))#  a(h2, g2, s-1, h1, g1). This will follow from

g1 s e

idXs-1


N

idXg1


N

Rs

A

g1s



idXs-1


N

h-1 1

e sh2g2


N

idXs-1

h2

s g2



Rs


N

idXg2


N

idXs-1

C

B

s-1

g1

h1
s-1h1g1  h2 e
g2s-1

e

g2

s-1

The right side acting on [x]g2  [y]s-1 gives (up to a nonzero scalar)
N

(4.8)

i

h2 xg2(y) u(g2, s-1)

u(h2, g2s-1)u(s-1h1, g1)u(s-1, h1)s-1 (bi)

s-1


N

[h-1 1

(bi

)]g1

21

Next we compute the left side acting on [x]g2  [y]s-1 (up to a nonzero scalar) in the following way
N

-C

i

[u(s-1,

s)s-1 (bi)]s-1


N

[bi ]s


N

[x]g2


N

[y]s-1

-B [u(s-1, s)s-1 (bi)]s-1

i,j


N

h-1 1 (bi s(h2 (x)) u(s, h2)u(sh2, g2) ) u(h-1 1, sh2g2)u(g1, s)g1(bj )

g1


N

[bj ]s


N

[y]s-1

-A [u(s-1, s)s-1 (bi)]s-1 

i,j

N

h-1 1 (bi s(h2 (x)) u(s, h2)u(sh2, g2) ) u(h-1 1, sh2g2)u(g1, s)g1(bj )

EN
g1

bj s(y)u(s, s-1)



[u(s-1, s)s-1 (bi)]s-1

i


N

h-1 1 (bi s(h2 (x)) u(s, h2)u(sh2, g2) ) u(h-1 1, sh2g2)u(g1, s)g1

s(y)u(s, s-1)

g1

=

[u(s-1, s)s-1 (bi)]s-1 

i,k

N

EN bi s(h2 (x)) u(s, h2)u(sh2, g2)h1 u(h-1 1, sh2g2)u(g1, s)g1 s(y)u(s, s-1) bk h-1 1 (bk) g1

=

u(s-1, s)s-1 s(h2 (x)) u(s, h2)u(sh2, g2)h1 u(h-1 1, sh2g2)u(g1, s)g1 s(y)u(s, s-1) bk s-1

k


N

h-1 1 (bk) g1

Since the second tensor component matches with that of the expression in 4.8, we will now work with the first term.
u(s-1, s)s-1 s(h2 (x)) u(s, h2)u(sh2, g2)h1 u(h-1 1, sh2g2)u(g1, s)g1 s(y)u(s, s-1) bk = h2 (x) u(s-1, s) s-1 u(s, h2)u(sh2, g2)h1 u(h-1 1, sh2g2)u(g1, s)g1 s(y)u(s, s-1) s-1 (bk)

In the last expression, we pick y and using the intertwining relation between u and , we push it leftwards all the way to the right side of the term h2(x) and it becomes h2(g2(y)). This matches the first two and the last terms with that of the first tensor component of the expression 4.8. We are left with showing the u-terms in the middle, namely

(4.9)

u(s-1, s) s-1 u(s, h2)u(sh2, g2)h1 u(h-1 1, sh2g2)u(g1, s)g1 u(s, s-1)

is a nonzero multiple of the u-terms in 4.8, that is,

(4.10)

h2 u(g2, s-1) u(h2, g2s-1)u(s-1h1, g1)u(s-1, h1)

Taking the adjoint of 4.9 and 4.10 separately, we get the same automorphism h2g2 s-1-g11-h11-s-11. Hence, we are done.
22

In order to describe the representations of A, we need a few more notations. As in Section 3, C will denote the set of conjugacy classes, gC will be a representative of C  C and for g  C, we pick wg such that g = wggC wg-1. Also, C will be the 2-cocycle gC of GC . For C  C , set SC := {(h, g)  H × G : hg  C}.
Theorem 4.16. (i) The affine annular algebra A = ((Ag1,g2))fin. supp.is isomorphic as a -algebra to MSC  [CGC ]C where MSC denotes the -algebra of finitely supported matrices with rows
C C
and columns indexed by elements of SC. (ii) Every Hilbert space representation  : A  L(V ) decomposes uniquely (up to isomorphism) as an orthogonal direct sum of submodules V C := Range  (a(e, gC , e, e, gC )) for C  C . (We will call a representation of A `supported on C  C ' if it is generated by the range of the action of the projection a(e, gC , e, e, gC ).) The category of C-supported representations of C is additively equivalent to representation category of [CGC]C .

Proof. (i) Define the map  : A - MSC  [CGC ]C by
C C

a(h1, g1, s, h2, g2) - gC,wh1g1 ,wh2g2 E(h2,g2),(h1,g1)  [wh-21g2 s-1 wh1g1 ]

extended linearly where h1g1, h2g2  C. Using the formula for multiplication and # in Propositions 4.12 and 4.15 and the cocyle relation in Proposition 2.1, one can imitate the proof of Proposition
3.1 to show that the map  serves as the required isomorphism.

(ii) Let  : A  L(V ) be a Hilbert space representation. For C1, C2  C such that C1 = C2, we need to show V C1 and V C2 are orthogonal. Taking inner product of the generating vectors, we get (a(e, gC1 , s1, h1, g)) , (a(e, gC2 , s2, h2, g)) = (a(e, gC2 , s2, h2, g))# · a(e, gC1 , s1, h1, g)) , 
which is zero unless h1 = h2 but in that case C1 and C2 have to be the same; so, the inner product
is zero.

For the decomposition, it remains to show that V  V C. Let   Vg. Note that the identity

C C

a(h, g, e, h, g) of Ag,g is a sum of orthogonal projections. So,  = (a(h, g, e, h, g)). For h 

hH

hH

H, we have (a(h, g, e, h, g)) = (a(e, gC , e, h, g) where hg  C  C and  = (a(h, g, e, e, gC )) 

V C.

The proof of equivalence of C-supported representations with representations of [CG]C is exactly the same as the proof of Theorem 3.1.

Remark 4.17. To find the tube algebra T of CNN , we need to first choose a set of representatives in the isomorphism classes of simple objects in CNN . By Propositions 4.2 and 4.4, Xg1 and Xg2 are isomorphic if and only if g1 and g2 are in the same H-H double coset where the isomorphism
g2 is implemented by h1 h2 for any h1, h2  H satisfying h1g1 = g2h2. Now for g  G, the
g1 endomorphism space End(Xg) is isomorphic to the group algebra Hg := H  g-1Hg twisted by the scalar 2-cocycle Hg × Hg  (h1, h2)  (gh1g-1, gh2g-1, g) (gh1g-1, g, h2) (g, h1, h2)  S1 via
g CHg  Hg  h - ghg-1 h  End(Xg).
g

For g  G, fix a maximal set g of mutually orthogonal minimal projections in End(Xg). Let H\G/H be a set of representatives from all the H-H double cosets in G. Then, it follows
23

that

{Range(p) : p  g} is a set of representatives in the isomorphism classes of simple

gH \G/H

objects in CNN . Hence, the tube algebra T is isomorphic (as a -algebra) to

g½2,g2 (p2)  Ag1,g2  g½1,g1 (p1) .
g1,g2H\G/H p1g1 ,p2g2

Remark 4.18. By [GJ, Theorem 4.2], we know that the representation categories of the affine algebra A and tube algebra T are equivalent although as -algebras they are non-isomorphic. There is one
thing to notice that this representation category (appearing in Theorem 4.16) is also equivalent to
the category of tube representations of the diagonal subfactor (as in Theorem 3.1) corresponding
to the automorphisms g, g  H  K of the II1-factor Q. This equivalence can be seen in an alternative way:
Let AHB be an extremal bifinite bimodule and P be its corresponding subfactor planar algebra (namely the `unimodular bimodule planar algebra', in the sense of [DGG1]). By Theorem 4.2 of
[GJ], the category of (Jones) affine P -modules is equivalent to the representation category of the
tube algebra of CAA := the category of bifinite A-A-bimodules generated by AHB. By [DGG2, Remark 2.16], the affine module categories corresponding to P and its dual P are equivalent. On
the other hand, the dual planar algebra is isomorphic to the subfactor planar algebra associated to
the contragradient bimodule BHA. Thus the representation category of the tube algebra of CAA is equivalent to that of CBB := the category of bifinite B-B-bimodules generated by AHB.
Next, consider an intermediate extremal finite index subfactor N  Q  M . Let  denote the bifinite bimodule N L2(M )Q. It is easy to check that the category CNN of bifinite N -N -bimodules generated by  is the same as those which come from the subfactor N  M . Let CQQ denote the smallest C-tensor category of bifinite Q-Q-bimodules coming from the subfactor N  Q as
well as Q  M . One can verify that CQQ is the same as the category of bifinite Q-Q-bimodules generated by . Hence, from the previous paragraph, the category of tube representations of CNN is equivalent to that of CQQ.
Coming back to our context of Bisch-Haagerup subfactor N = QH  Q  K as set up in the beginning of this section, it remains to show that CQQ is the C-tensor category generated by the bimodules QL2(Qg )Q for g  H  K; this is an easy computation.

References

[BH]
[BDG]
[DGG1]
[DGG2]
[DGG3] [GJ]
[CJon] [Jon] [NY]
[Pet]
[Pop1]
[Pop2]

D Bisch and U. Haagerup, Composition of subfactors: New examples of infinite depth subfactors, Ann. Sci. Ecole Norm. Sup., 29, 329-383, 1996. D Bisch, P Das and S K Ghosh, The planar algebra of group-type subfactors, J. Funct. Anal., 257, 20-46, 2009. P Das, S K Ghosh and V P Gupta, Perturbations of Planar Algebras, Math. Scand. Vol 114, No. 1 (2014), arXiv:1009.0186. P Das, S K Ghosh and V P Gupta, Affine modules and the Drinfeld Center, Math. Scand. Vol 118, No. 1 (2016), arXiv:1010.0460. P Das, S K Ghosh and V P Gupta, Drinfeld Center of planar algebra, Internat. J. Math., 25(8), 2014. S. Ghosh, C. Jones, Annular representation theory for rigid C-tensor categories J. Funct. Anal., 270, 4, 1537-1584, 2016. C. Jones, Quantum G2 categories have property (T), submitted to Int. J. Math., arXiv:1504.08338v5. V F R Jones, The annular structure of subfactors, L'Enseignement Math., 38, 2001. S. Neshveyev, M. Yamashita, Drinfeld center and representation theory for monoidal categories, arXiv:1501.07390 to appear in Comm. Math. Phys. E. Peters, A planar algebra construction of the Haagerup subfactor Int. J. Math., Vol. 21, No.8 ,(2010) 987-1045. S. Popa, Sousfacteurs, actions des groupes et cohomologie, Serie I, Comptes Rend. Acad. Sci. Paris, 309, 771-776, 1989. S. Popa, Classification of amenable subfactors of type II, Acta Math., 172, 163-255 (1994).
24

[PV]

S. Popa, S. Vaes, Representation theory for subfactors, -lattices and C*-tensor categories, Comm. Math. Phys. 340 (2015), 1239-1280.

Department of Mathematics, Vanderbilt University, Nashville, USA E-mail address: dietmar.bisch@vanderbilt.edu
Stat-Math Unit, Indian Statistical Institute, Kolkata, INDIA E-mail address: paramita.das@isical.ac.in, shami@isical.ac.in, narayan753@gmail.com

25

