14c. Fixed points

[math] \newcommand{\mathds}{\mathbb}[/math]

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We discuss now the connection of all the above with the main examples of subfactors. We recall from chapter 13 that the main examples of subfactors are all of integer index, and appear as fixed point subfactors, according to the following result:

Theorem

Let [math]G[/math] be a compact quantum group, and [math]G\to Aut(P)[/math] be a minimal action on a [math]{\rm II}_1[/math] factor. Consider a Markov inclusion of finite dimensional algebras

[[math]] B_0\subset B_1 [[/math]]

and let [math]G\to Aut(B_1)[/math] be an action which leaves invariant [math]B_0[/math] and which is such that its restrictions to the centers of [math]B_0[/math] and [math]B_1[/math] are ergodic. We have then a subfactor

[[math]] (B_0\otimes P)^G\subset (B_1\otimes P)^G [[/math]]

of index [math]N=[B_1:B_0][/math], called generalized Wassermann subfactor, whose Jones tower is

[[math]] (B_1\otimes P)^G\subset(B_2\otimes P)^G\subset(B_3\otimes P)^G\subset\ldots [[/math]]

where [math]\{ B_i\}_{i\geq 1}[/math] are the algebras in the Jones tower for [math]B_0\subset B_1[/math], with the canonical actions of [math]G[/math] coming from the action [math]G\to Aut(B_1)[/math], and whose planar algebra is given by:

[[math]] P_k=(B_0'\cap B_k)^G [[/math]]

These subfactors generalize the Jones, Ocneanu, Wassermann and Popa subfactors.

Show Proof

This is something that we know well from chapter 13, whose proof basically comes by generalizing, several times, the results of Wassermann in .

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In view of the above result, what we have to do in relation with such subfactors is to further interpret the last formula there, that of the planar algebra, namely:

[[math]] P_k=(B_0'\cap B_k)^G [[/math]]

To be more precise, we will show here that, under suitable assumptions on the original inclusion [math]B_0\subset B_1[/math], we can associate a certain combinatorial planar algebra [math]P(B_0\subset B_1)[/math] to this inclusion, and then the planar algebra associated to the fixed point subfactor itself appears as a fixed point subalgebra of this planar algebra, as follows:

[[math]] P=P(B_0\subset B_1)^G [[/math]]

This is something quite technical, and we will do this in two steps. First we will explain, following Jones' paper , how to associate a planar algebra [math]P(B_0\subset B_1)[/math] to an inclusion of algebras [math]B_0\subset B_1[/math]. And then we will explain, following and subsequent papers, and notably , how to prove the above formula [math]P=P(B_0\subset B_1)^G[/math].

Getting started now, the idea will be that [math]P(B_0\subset B_1)[/math] appears as a joint generalization of the spin and tensor planar algebras, discussed above, which appear as follows:

[[math]] \mathcal S_N=P(\mathbb C\subset\mathbb C^N) [[/math]]

[[math]] \mathcal T_N=P(\mathbb C\subset M_N(\mathbb C)) [[/math]]

Thus, our first task will be that of getting back to the Markov inclusions [math]B_0\subset B_1[/math], from chapter 13, and further discuss the combinatorics of their basic construction, with planar algebra ideas in mind. As in chapter 13, it is most convenient to denote such inclusions by [math]A\subset B[/math], at least at a first stage of their study. Following the book of Goodman, de la Harpe and Jones , which is the standard reference for such things, we first have:

Definition

Associated to an inclusion [math]A\subset B[/math] of finite dimensional algebras are the following objects:

• The column vector [math](a_i)\in\mathbb N^s[/math] given by [math]A=\oplus_{i=1}^sM_{a_i}(\mathbb C)[/math].
• The column vector [math](b_j)\in\mathbb N^t[/math] given by [math]B=\oplus_{j=1}^tM_{b_j}(\mathbb C)[/math].
• The inclusion matrix [math](m_{ij})\in M_{s\times t}(\mathbb N)[/math], satisfying [math]m^ta=b[/math].

To be more precise here, in what regards the inclusion matrix, each minimal idempotent in [math]M_{a_i}(\mathbb C)\subset A[/math] splits, when regarded as an element of [math]B[/math], as a sum of minimal idempotents of [math]B[/math], and [math]m_{ij}\in\mathbb N[/math] is the number of such idempotents from [math]M_{b_j}(\mathbb C)[/math]. We have the following result, bringing traces into picture:

Proposition

For an inclusion [math]A\subset B[/math], the following are equivalent:

• [math]A\subset B[/math] commutes with the canonical traces.
• We have [math]mb=ra[/math], where [math]r=||b||^2/||a||^2[/math].

Show Proof

The weight vectors of the canonical traces of [math]A,B[/math] are given by:

[[math]] \tau_i=\frac{a_i^2}{||a||^2}\quad,\quad \tau_j=\frac{b_j^2}{||b||^2} [[/math]]

We can plug these values into the following standard compatibility formula:

[[math]] \frac{\tau_i}{a_i}=\sum_jm_{ij}\cdot\frac{\tau_j}{b_j} [[/math]]

We obtain in this way the condition in the statement.

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We will need as well the following basic facts, also from :

Definition

Associated to an inclusion of finite dimensional algebras [math]A\subset B[/math], with inclusion matrix [math]m\in M_{s\times t}(\mathbb N)[/math], are:

• The Bratteli diagram: this is the bipartite graph [math]\Gamma[/math] having as vertices the sets [math]\{1,\ldots,s\}[/math] and [math]\{1,\ldots,t\}[/math], the number of edges between [math]i,j[/math] being [math]m_{ij}[/math].
• The basic construction: this is the inclusion [math]B\subset A_1[/math] obtained from [math]A\subset B[/math] by reflecting the Bratteli diagram.
• The Jones tower: this is the tower of algebras [math]A\subset B\subset A_1\subset B_1\subset\ldots[/math] obtained by iterating the basic construction.

We know that for a Markov inclusion [math]A\subset B[/math] we have [math]m^ta=b[/math] and [math]mb=ra[/math], and so [math]mm^ta=ra[/math], which gives an eigenvector for the square matrix [math]mm^t\in M_s(\mathbb N)[/math]. When this latter matrix has positive entries, by Perron-Frobenius we obtain:

[[math]] ||mm^t||=r [[/math]]

This equality holds in fact without assumptions on [math]m[/math], and we have:

Theorem

Let [math]A\subset B[/math] be Markov, with inclusion matrix [math]m\in M_{s\times t}(\mathbb N)[/math].

• [math]r=\dim(B)/\dim(A)[/math] is an integer.
• [math]||m||=||m^t||=\sqrt{r}[/math].
• [math]||\ldots mm^tmm^t\ldots||=r^{k/2}[/math], for any product of lenght [math]k[/math].

Show Proof

Consider the vectors [math]a,b[/math], as in Definition 14.16. We know from definitions and from Proposition 14.17 that we have:

[[math]] b=m^ta\quad,\quad mb=ra\quad,\quad r=||b||^2/||a||^2 [[/math]]

(1) If we construct as above the Jones tower for [math]A\subset B[/math], we have, for any [math]k[/math]:

[[math]] \frac{\dim B_k}{\dim A_k} =\frac{\dim A_k}{\dim B_{k-1}} =r [[/math]]

On the other hand, we have as well the following well-known formula:

[[math]] \lim_{k\to\infty}(\dim A_k)^{1/2k} =\lim_{k\to\infty}(\dim B_k)^{1/2k} =||mm^t|| [[/math]]

By combining these two formulae we obtain the following formula:

[[math]] ||mm^t||=r [[/math]]

But from [math]r\in\mathbb Q[/math] and [math](mm^t)^ka=r^ka[/math] for any [math]k\in\mathbb N[/math], we get [math]r\in\mathbb N[/math], and we are done.

(2) This follows from the above equality [math]||mm^t||=r[/math], and from the following standard equalities, for any real rectangular matrix [math]r[/math]:

[[math]] ||m||^2 =||m^t||^2 =||mm^t|| [[/math]]

(3) Let [math]n[/math] be the length [math]k[/math] word in the statement. First, by applying the norm and by using the formula [math]||m||=||m^t||=\sqrt{r}[/math], we obtain the following inequality:

[[math]] ||n||\leq r^{k/2} [[/math]]

For the converse inequality, assume first that [math]k[/math] is even. Then [math]n[/math] has either [math]a[/math] or [math]b[/math] as eigenvector, depending on whether [math]n[/math] begins with a [math]m[/math] or with a [math]m^t[/math], in both cases with eigenvalue [math]r^{k/2}[/math], and this gives the desired inequality, namely:

[[math]] ||n||\geq r^{k/2} [[/math]]

Assume now that [math]k[/math] is odd, and let [math]\circ\in\{1,t\}[/math] be such that [math]n'=m^\circ n[/math] is alternating. Since [math]n'[/math] has even length, we already know that we have:

[[math]] ||n'||=r^{(k+1)/2} [[/math]]

On the other hand, we have as well the following estimate:

[[math]] ||n'|| \leq||m^\circ||\cdot ||n|| =\sqrt{r}||n|| [[/math]]

But this gives the reverse inequality [math]||n||\geq r^{k/2}[/math], as desired.

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The point now is that for a Markov inclusion, the basic construction and the Jones tower have a particularly simple form. Let us first work out the basic construction:

Proposition

The basic construction for a Markov inclusion [math]i:A\subset B[/math] of index [math]r\in\mathbb N[/math] is the inclusion [math]j:B\subset A_1[/math] obtained as follows:

• [math]A_1=M_r(\mathbb C)\otimes A[/math], as an algebra.
• [math]j:B\subset A_1[/math] is given by [math]mb=ra[/math].
• [math]ji:A\subset A_1[/math] is given by [math](mm^t)a=ra[/math].

Show Proof

With notations from the above, the weight vector of the algebra [math]A_1[/math] appearing from the basic construction is [math]mb=ra[/math], and this gives the result.

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We fix a Markov inclusion [math]i:A\subset B[/math]. We have the following result:

Proposition

The Jones tower associated to a Markov inclusion [math]i:A\subset B[/math], denoted as follows, with alternating letters,

[[math]] A\subset B\subset A_1\subset B_1\subset\ldots [[/math]]

is given by the following formulae:

• [math]A_k=M_r(\mathbb C)^{\otimes k}\otimes A[/math].
• [math]B_k=M_r(\mathbb C)^{\otimes k}\otimes B[/math].
• [math]A_k\subset B_k[/math] is [math]id_k\otimes i[/math].
• [math]B_k\subset A_{k+1}[/math] is [math]id_k\otimes j[/math].

Show Proof

This follows from Proposition 14.20, with the remark that if [math]i:A\subset B[/math] is Markov, then so is its basic construction [math]j:B\subset A_1[/math].

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Regarding now the relative commutants for this tower, we have here:

Proposition

The relative commutants for the Jones tower

[[math]] A\subset B\subset A_1\subset B_1\subset\ldots [[/math]]

associated to a Markov inclusion [math]A\subset B[/math] are given by:

• [math]A_s'\cap A_{s+k}=M_r(\mathbb C)^{\otimes k}\otimes (A'\cap A)[/math].
• [math]A_s'\cap B_{s+k}=M_r(\mathbb C)^{\otimes k}\otimes (A'\cap B)[/math].
• [math]B_s'\cap A_{s+k}=M_r(\mathbb C)^{\otimes k}\otimes (B'\cap A)[/math].
• [math]B_s'\cap B_{s+k}=M_r(\mathbb C)^{\otimes k}\otimes (B'\cap B)[/math].

Show Proof

The above assertions are all elementary, as follows:

(1,2) These assertions both follow from Proposition 14.21.

(3) In order to prove the formula in the statement, observe first that we have:

[[math]] \begin{eqnarray*} B'\cap A_1 &=&(B'\cap B_1)\cap A_1\\ &=&(M_r(\mathbb C)\otimes Z(B))\cap (M_r(\mathbb C)\otimes A)\\ &=&M_r(\mathbb C)\otimes (B'\cap A) \end{eqnarray*} [[/math]]

But this proves the assertion at [math]s=0,k=1[/math], and the general case follows from it.

(4) This is again clear, once again coming from Proposition 14.21.

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In order to further refine all this, let us formulate the following key definition:

Definition

We say that a Markov inclusion [math]A\subset B[/math] is abelian if [math][A,B]=0[/math], with the commutant being computed inside [math]B[/math].

In other words, we are asking for the commutation relation [math]ab=ba[/math], for any [math]a\in A[/math] and [math]b\in B[/math]. Note that this is the same as asking that [math]B[/math] is an [math]A[/math]-algebra, [math]A\subset Z(B)[/math]. As basic examples, observe that all inclusions with [math]A=\mathbb C[/math] or with [math]B=\mathbb C^n[/math] are abelian. The point with this notion is that it leads to the following simple statement:

Proposition

With [math]\tilde{B}_k=M_r(\mathbb C)^{\otimes k}\otimes Z(B)[/math], the relative commutants for the Jones tower [math]A\subset B\subset A_1\subset B_1\subset\ldots[/math] of an abelian inclusion are given by:

• [math]A_s'\cap A_{s+k}=A_k[/math].
• [math]A_s'\cap B_{s+k}=B_k[/math].
• [math]B_s'\cap A_{s+k}=A_k[/math].
• [math]B_s'\cap B_{s+k}=\tilde{B}_k[/math].

Show Proof

This follows from the fact that for an abelian inclusion we have:

[[math]] Z(A)=A\quad,\quad A'\cap B=B\quad,\quad B'\cap A=A [[/math]]

Thus, we are led to the conclusion in the statement.

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Getting back now to the fixed point subfactors, from Theorem 14.15, we can improve the planar algebra computation there, in the abelian case, as follows:

Theorem

The commutants for the tower [math]N\subset M\subset N_1\subset M_1\subset\ldots[/math] associated to an abelian fixed point subfactor [math](A\otimes P)^G\subset(B\otimes P)^G[/math] are:

• [math]N_s'\cap N_{s+k}=A_k^G[/math].
• [math]N_s'\cap M_{s+k}=B_k^G[/math].
• [math]M_s'\cap N_{s+k}=A_k^G[/math].
• [math]M_s'\cap M_{s+k}=\tilde{B}_k^G[/math].

Show Proof

This follows indeed by combining the planar algebra computation from Theorem 14.15 with the result about abelian inclusions from Proposition 14.24.

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In order to further advance now, the idea will be that of associating to the original inclusion [math]B_0\subset B_1[/math] a certain combinatorial planar algebra [math]P(B_0\subset B_1)[/math], as for the planar algebra associated to the fixed point subfactor itself to appear as follows:

[[math]] P=P(B_0\subset B_1)^G [[/math]]

As already mentioned, the idea will be that [math]P(B_0\subset B_1)[/math] appears as a joint generalization of the spin and tensor planar algebras, which appear as follows:

[[math]] \mathcal S_N=P(\mathbb C\subset\mathbb C^N) [[/math]]

[[math]] \mathcal T_N=P(\mathbb C\subset M_N(\mathbb C)) [[/math]]

In practice now, we will need for all this the notion of planar algebra of a bipartite graph, generalizing the algebras [math]\mathcal S_N,\mathcal T_N[/math], constructed by Jones in . So, let [math]\Gamma[/math] be a bipartite graph, with vertex set [math]\Gamma_a\cup\Gamma_b[/math]. It is useful to think of [math]\Gamma[/math] as being the Bratteli diagram of an inclusion [math]A\subset B[/math], in the sense of Definition 14.16. Our first task is to define the graded vector space [math]P[/math]. Since the elements of [math]P[/math] will be subject to a planar calculus, it is convenient to introduce them “in boxes”, as follows:

Definition

Associated to [math]\Gamma[/math] is the abstract vector space [math]P_k[/math] spanned by the [math]2k[/math]-loops based at points of [math]\Gamma_a[/math]. The basis elements of [math]P_k[/math] will be denoted

[[math]] x=\begin{pmatrix} e_1&e_2&\ldots&e_k\\ e_{2k}&e_{2k-1}&\ldots &e_{k+1} \end{pmatrix} [[/math]]

where [math]e_1,e_2,\ldots,e_{2k}[/math] is the sequence of edges of the corresponding [math]2k[/math]-loop.

Consider now the adjacency matrix of [math]\Gamma[/math], which is of the following type:

[[math]] M=\begin{pmatrix}0&m\\ m^t&0\end{pmatrix} [[/math]]

We pick an [math]M[/math]-eigenvalue [math]\gamma\neq 0[/math], and then a [math]\gamma[/math]-eigenvector, as follows:

[[math]] \eta:\Gamma_a\cup\Gamma_b\to\mathbb C-\{0\} [[/math]]

With this data in hand, we have the following construction, due to Jones :

Definition

Associated to any tangle is the multilinear map

[[math]] T(x_1\otimes\ldots\otimes x_r)=\gamma^c\sum_x\delta(x_1,\ldots ,x_r,x)\prod_m\mu(e_m)^{\pm 1}x [[/math]]

where the objects on the right are as follows:

• The sum is over the basis of [math]P_k[/math], and [math]c[/math] is the number of circles of [math]T[/math].
• [math]\delta=1[/math] if all strings of [math]T[/math] join pairs of identical edges, and [math]\delta=0[/math] if not.
• The product is over all local maxima and minima of the strings of [math]T[/math].
• [math]e_m[/math] is the edge of [math]\Gamma[/math] labelling the string passing through [math]m[/math] (when [math]\delta=1[/math]).
• [math]\mu(e)=\sqrt{\eta(e_f)/\eta(e_i)}[/math], where [math]e_i,e_f[/math] are the initial and final vertex of [math]e[/math].
• The [math]\pm[/math] sign is [math]+[/math] for a local maximum, and [math]-[/math] for a local minimum.

This looks quite similar to the calculus for the tensor and spin planar algebras. Let us work out the precise formula of the action, for 6 carefully chosen tangles:

(1) Let us look first at the identity [math]1_k[/math]. This tangle acts by the identity:

[[math]] 1_k\begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix}= \begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix} [[/math]]

(2) The multiplication tangle [math]M_k[/math] acts as follows:

[[math]] M_k\left( \begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix} \otimes\begin{pmatrix}h_1&\ldots&h_k\\ g_1&\ldots&g_k\end{pmatrix} \right)= \delta_{f_1g_1}\ldots\delta_{f_kg_k} \begin{pmatrix}h_1&\ldots&h_k\\ e_1&\ldots& e_k\end{pmatrix} [[/math]]

(3) Regarding now the inclusion [math]I_k[/math], the formula here is:

[[math]] I_k\begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix}= \sum_g\begin{pmatrix}f_1&\ldots&f_k&g\\ e_1&\ldots&e_k&g\end{pmatrix} [[/math]]

(4) The expectation tangle [math]U_k[/math] acts with a spin factor, as follows:

[[math]] U_k\begin{pmatrix}f_1&\ldots&f_k&h\\ e_1&\ldots&e_k&g\end{pmatrix}= \delta_{gh}\mu(g)^2\begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix} [[/math]]

(5) For the Jones projection [math]E_k[/math], the formula is as follows:

[[math]] E_k(1)=\sum_{egh}\mu(g)\mu(h)\begin{pmatrix}e_1&\ldots &e_k&h&h\\ e_1&\ldots&e_k&g&g\end{pmatrix} [[/math]]

(6) As for the shift [math]J_k[/math], its action is given by:

[[math]] J_k\begin{pmatrix}f_1&\ldots&f_k\\ e_1&\ldots&e_k\end{pmatrix}= \sum_{gh}\begin{pmatrix}g&h&f_1&\ldots&f_k\\ g&h&e_1&\ldots&e_k\end{pmatrix} [[/math]]

Summarizing, we have here formulae which are quite similar to those for the tensor and spin planar algebras. We have the following result, from Jones' paper :

Theorem

The graded linear space [math]P=(P_k)[/math], together with the action of the planar tangles given above, is a planar algebra.

Show Proof

This is something which is quite routine, starting from the above study of the main planar algebra tangles, which can be proved by using Theorem 14.11. Also, let us mention that all this generalizes the previous constructions of the spin and tensor planar algebras [math]\mathcal S_N,\mathcal T_N[/math], which appear respectively from the Bratteli diagrams of the inclusions [math]\mathbb C\subset\mathbb C^N[/math] and [math]\mathbb C\subset M_N(\mathbb C)[/math]. For full details on all this, we refer to Jones .

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Let us go back now to the Markov inclusions [math]A\subset B[/math], as before. We have here the following result, regarding such inclusions, also from Jones' paper :

Theorem

The planar algebra associated to the graph of [math]A\subset B[/math], with eigenvalue [math]\gamma=\sqrt{r}[/math] and eigenvector [math]\eta(i)=a_i/\sqrt{\dim A}[/math], [math]\eta(j)=b_j/\sqrt{\dim B}[/math], is as follows:

• The graded algebra structure is given by [math]P_{2k}=A'\cap A_k[/math], [math]P_{2k+1}=A'\cap B_k[/math].
• The elements [math]e_k[/math] are the Jones projections for [math]A\subset B\subset A_1\subset B_1\subset\ldots[/math]
• The expectation and shift are given by the above formulae.

Show Proof

As a first observation, [math]\eta[/math] is indeed a [math]\gamma[/math]-eigenvector for the adjacency matrix of the graph. Indeed, we have the following formulae:

[[math]] m^ta=b\quad,\quad mb=ra\quad,\quad \sqrt{r}=||b||/||a|| [[/math]]

By using these formulae, we have the following computation:

[[math]] \begin{eqnarray*} \begin{pmatrix}0&m\\ m^t&0\end{pmatrix}\begin{pmatrix}a/||a||\\ b/||b||\end{pmatrix} &=&\begin{pmatrix}\gamma^2a/||b||\\ b/||a||\end{pmatrix}\\ &=&\gamma\begin{pmatrix}\gamma a/||b||\\ b/\gamma||a||\end{pmatrix}\\ &=&\gamma\begin{pmatrix}a/||a||\\ b/||b||\end{pmatrix} \end{eqnarray*} [[/math]]

Since the algebra [math]A[/math] was supposed abelian, the Jones tower algebras [math]A_k,B_k[/math] are simply the span of the [math]4k[/math]-paths, respectively [math]4k+2[/math]-paths on [math]\Gamma[/math], starting at points of [math]\Gamma_a[/math]. With this description in hand, when taking commutants with [math]A[/math] we have to just have to restrict attention from paths to loops, and we obtain the above spaces [math]P_{2k},P_{2k+1}[/math]. See .

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In the particular case of the inclusions satisfying [math][A,B]=0[/math], we have:

Proposition

The “bipartite graph” planar algebra [math]P(A\subset B)[/math] associated to an abelian inclusion [math]A\subset B[/math] can be described as follows:

• As a graded algebra, this is the Jones tower [math]A\subset B\subset A_1\subset B_1\subset\ldots[/math]
• The Jones projections and expectations are the usual ones for this tower.
• The shifts correspond to the canonical identifications [math]A_1'\cap P_{k+2}=P_k[/math].

Show Proof

The first assertion is a reformulation of Theorem 14.28 in the abelian case, by using the identifications [math]A'\cap A_k=A_k[/math] and [math]A'\cap B_k=B_k[/math] from Proposition 14.24. The assertion on Jones projections follows as well from Theorem 14.28, and the assertion on expectations follows from the fact that their composition is the usual trace. Regarding now the third assertion, let us recall first from Proposition 14.24 that we have indeed identifications [math]A_1'\cap A_{k+1}=A_k[/math] and [math]A_1'\cap B_{k+1}=B_k[/math]. By using the path model for these algebras, as in the proof of Theorem 14.28, we obtain the result.

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In order to formulate now our main result, regarding the subfactors associated to the compact quantum groups [math]G[/math], we will need a few abstract notions. Let us start with:

Definition

Let [math]P_1,P_2[/math] be two finite dimensional algebras, coming with coactions [math]\alpha_i:P_i\to P_i\otimes L^\infty(G)[/math], and let [math]T:P_1\to P_2[/math] be a linear map.

• We say that [math]T[/math] is [math]G[/math]-equivariant if [math](T\otimes id)\alpha_1=\alpha_2T[/math].
• We say that [math]T[/math] is weakly [math]G[/math]-equivariant if [math]T(P_1^G)\subset P_2^G[/math].

Consider now a planar algebra [math]P=(P_k)[/math]. The annular category over [math]P[/math] is the collection of maps [math]T:P_k\to P_l[/math] coming from the “annular” tangles, having at most one input box. These maps form sets [math]Hom(k,l)[/math], and these sets form a category. We have:

Definition

A coaction of [math]L^\infty(G)[/math] on [math]P[/math] is a graded algebra coaction

[[math]] \alpha:P\to P\otimes L^\infty(G) [[/math]]

such that the annular tangles are weakly [math]G[/math]-equivariant.

This is something a bit technical, coming out of the known examples that we have. In fact, as we will show below, the examples are basically those coming from actions of quantum groups on Markov inclusions [math]A\subset B[/math], under the assumption [math][A,B]=0[/math]. For the moment, at the generality level of Definition 14.31, we have:

Proposition

If [math]G[/math] acts on a planar algebra [math]P[/math], then [math]P^G[/math] is a planar algebra.

Show Proof

The weak equivariance condition tells us that the annular category is contained in the suboperad [math]\mathcal P'\subset\mathcal P[/math] consisting of tangles which leave invariant [math]P^G[/math]. On the other hand the multiplicativity of [math]\alpha[/math] gives [math]M_k\in\mathcal P'[/math], for any [math]k[/math]. Now since [math]\mathcal P[/math] is generated by multiplications and annular tangles, we get [math]\mathcal P'=\mathcal P[/math], and we are done.

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Let us go back now to the abelian inclusions. We have the following key result:

Proposition

If [math]G[/math] acts on an abelian inclusion [math]A\subset B[/math], the canonical extension of this coaction to the Jones tower is a coaction of [math]G[/math] on the planar algebra [math]P(A\subset B)[/math].

Show Proof

We know from the above that, as a graded algebra, [math]P=P(A\subset B)[/math] coincides with the Jones tower for our inclusion, denoted as follows:

[[math]] A\subset B\subset A_1\subset B_1\subset\ldots [[/math]]

Thus the coaction in the statement can be regarded as a graded coaction, as follows:

[[math]] \alpha:P\to P\otimes L^\infty(G) [[/math]]

In order to finish, we have to prove that the annular tangles are weakly equivariant, as in Definition 14.31, and this can be done as follows:

(1) First, since the annular category is generated by [math]I_k,E_k,U_k,J_k[/math], we just have to prove that these 4 particular tangles are weakly equivariant. Now since [math]I_k,E_k,U_k[/math] are plainly equivariant, by construction of the coaction of [math]G[/math] on the Jones tower, it remains to prove that the shift [math]J_k[/math] is weakly equivariant.

(2) We know that the image of the fixed point subfactor shift [math]J_k'[/math] is formed by the [math]G[/math]-invariant elements of the relative commutant [math]A_1'\cap P_{k+2}=P_k[/math]. Now since this commutant is the image of the planar shift [math]J_k[/math], we have [math]Im(J_k)=Im(J_k')[/math], and this gives the result.

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With the above result in hand, we can now prove:

Proposition

Assume that [math]G[/math] acts on an abelian inclusion [math]A\subset B[/math]. Then the graded vector space of fixed points [math]P(A\subset B)^G[/math] is a planar subalgebra of [math]P(A\subset B)[/math].

Show Proof

This follows indeed from Proposition 14.33 and Proposition 14.34.

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We are now in position of stating and proving a main result, from :

Theorem

In the abelian case, the planar algebra of the fixed point subfactor

[[math]] (A\otimes P)^G\subset (B\otimes P)^G [[/math]]

is the fixed point algebra [math]P(A\subset B)^G[/math] of the bipartite graph algebra [math]P(A\subset B)[/math].

Show Proof

This basically follows from what we have, as follows:

(1) Let [math]P=P(A\subset B)[/math], and let [math]Q[/math] be the planar algebra of the fixed point subfactor. We know that we have an equality of graded algebras [math]Q=P^G[/math]. Thus, it remains to prove that the planar algebra structure on [math]Q[/math] coming from the fixed point subfactor agrees with the planar algebra structure of [math]P[/math], coming from Proposition 14.30.

(2) Since [math]\mathcal P[/math] is generated by the annular category [math]\mathcal A[/math] and by the multiplication tangles [math]M_k[/math], we just have to check that the annular tangles agree on [math]P,Q[/math]. Moreover, since [math]\mathcal A[/math] is generated by [math]I_k,E_k,U_k,J_k[/math], we just have to check that these tangles agree on [math]P,Q[/math].

(3) We know that [math]Q\subset P[/math] is an inclusion of graded algebras, that all the Jones projections for [math]P[/math] are contained in [math]Q[/math], and that the conditional expectations agree. Thus the tangles [math]I_k,E_k,U_k[/math] agree on [math]P,Q[/math], and the only verification left is that for the shift [math]J_k[/math].

(4) Now by using either the axioms of Popa in , or the construction of Jones in , it is enough to show that the image of the subfactor shift [math]J_k'[/math] coincides with that of the planar shift [math]J_k[/math]. But this follows as in the proof of Proposition 14.34.

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General references

Banica, Teo (2024). "Principles of operator algebras". arXiv:2208.03600 [math.OA].