Toral subgroups

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13a. Diagonal tori

In this final part of the present book we discuss a number of more advanced questions. There are plenty of topics here, on which serious work has gone into, and where there are interesting things to be said, to choose from. We have opted here to talk about:

(1) Toral subgroups. This is a fundamental question, coming from the fact that, while Lie theory is definitely not available for the arbitrary closed subgroups [math]G\subset U_N[/math], a notion of “maximal torus” is available. In addition, while all this is very interesting, things are quie recent, and there are just a handful of things known here. Which makes this topic an ideal one to start with, providing a glimpse at beautiful lands, not explored yet.

(2) Amenability, growth. This is something far more classical, and we definitely owe you some more explanations here, besides what was quickly said on these subjects in chapter 3 above. Passed the basics, that we will explain in detail, there are many things that have been done on these classical topics, to choose from. And we will choose here to talk about amenability and growth in relation with maximal tori, a hot topic.

(3) Homogeneous spaces. This is yet another thing that we have to talk about, imperatively, as a continuation of the various noncommutative geometry considerations scattered all across the present book, starting from chapter 1. Again, wide subject, a choice to be made here, and we will discuss, as a main topic, the construction and main properties of the “simplest” possible homogeneous spaces, over the easy quantum groups.

(4) Matrix models. This is perhaps the most beautiful of all “advanced topics” that can be discussed, the idea here, which is extremely simple, being that of looking for matrix models [math]U_{ij}\in M_K(C(T))[/math] for the standard coodinates [math]u_{ij}\in C(G)[/math] of a given closed subgroup [math]G\subset U_N[/math]. And not only this is mathematically natural, but physically speaking, this is expected to be of great use, in connection with statistical mechanics.

So, this will be our plan, for the present chapter and for the next 3 ones, introduction to (1-4). Regarding other topics, unfortunately left aside, it was particularly heartbreaking not to talk more about quantum permutation groups, although we will still meet such quantum groups on numerous occasions, when discussing (1-4). For even more about quantum permutations, I recommend my advanced book .

Getting started now, in relation with tori, we have seen on various occasions that the group duals [math]G=\widehat{\Gamma}[/math] can be thought of as being “tori”, in the compact quantum group framework. Also, given a closed subgroup [math]G\subset U_N^+[/math], the group dual subgroups [math]\widehat{\Lambda}\subset G[/math] can be thought of as being tori of [math]G[/math], and play a potentially important role.

Our purpose here will be that of understanding how the structure of a closed subgroup [math]G\subset U_N^+[/math] can be recovered from the knowledge of these tori [math]\widehat{\Lambda}\subset G[/math]. Let us start with a basic statement, regarding the classical and group dual cases:

Proposition

Let [math]G\subset U_N^+[/math] be a compact quantum group, and consider the group dual subgroups [math]\widehat{\Lambda}\subset G[/math], also called toral subgroups, or simply “tori”.

• In the classical case, where [math]G\subset U_N[/math] is a compact Lie group, these are the usual tori, that is, the closed abelian subgroups of [math]G[/math].
• In the group dual case, [math]G=\widehat{\Gamma}[/math] with [math]\Gamma= \lt g_1,\ldots,g_N \gt [/math] being a discrete group, these are the duals of the various quotients [math]\Gamma\to\Lambda[/math].

Show Proof

Both these assertions are clear, as follows:

(1) This follows indeed from the fact that a closed subgroup [math]H\subset U_N^+[/math] is at the same time classical, and a group dual, precisely when it is classical and abelian.

(2) This follows from the general propreties of the Pontrjagin duality, and more precisely from the fact that the subgroups [math]\widehat{\Lambda}\subset\widehat{\Gamma}[/math] correspond to the quotients [math]\Gamma\to\Lambda[/math].

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Based on the above simple facts, regarding the simplest compact quantum groups that we know, namely the compact groups and the discrete group duals, we can see that there are two potential motivations for the study of toral subgroups [math]\widehat{\Lambda}\subset G[/math], as follows:

(1) First, it is well-known that the fine structure of a compact Lie group [math]G\subset U_N[/math] is partly encoded by its maximal torus. Thus, in view of Proposition 13.1, the various tori [math]\widehat{\Lambda}\subset G[/math] encode interesting information about a quantum group [math]G\subset U_N^+[/math], both in the classical and the group dual case. We can expect this to hold in general.

(2) Also, any action [math]G\curvearrowright X[/math] on some geometric object, such as a manifold, will produce actions of its tori on the same object, [math]\widehat{\Lambda}\curvearrowright X[/math]. And, due to the fact that [math]\Lambda[/math] are familiar objects, namely discrete groups, these latter actions are easier to study, and this can ultimately lead to results about the action [math]G\curvearrowright X[/math] itself.

At a more concrete level now, most of the tori that we met appear as diagonal tori, in the sense of chapter 2 above. Let us first review this material. We first have:

Theorem

Given a closed subgroup [math]G\subset U_N^+[/math], consider its “diagonal torus”, which is the closed subgroup [math]T\subset G[/math] constructed as follows:

[[math]] C(T)=C(G)\Big/\left \lt u_{ij}=0\Big|\forall i\neq j\right \gt [[/math]]

This torus is then a group dual, [math]T=\widehat{\Lambda}[/math], where [math]\Lambda= \lt g_1,\ldots,g_N \gt [/math] is the discrete group generated by the elements [math]g_i=u_{ii}[/math], which are unitaries inside [math]C(T)[/math].

Show Proof

This is something going back to , that we know from chapter 2. The idea indeed is that since [math]u[/math] is unitary, its diagonal entries [math]g_i=u_{ii}[/math] are unitaries inside [math]C(T)[/math]. Moreover, from [math]\Delta(u_{ij})=\sum_ku_{ik}\otimes u_{kj}[/math] we obtain, when passing inside the quotient:

[[math]] \Delta(g_i)=g_i\otimes g_i [[/math]]

It follows that we have [math]C(T)=C^*(\Lambda)[/math], modulo identifying as usual the [math]C^*[/math]-completions of the various group algebras, and so that we have [math]T=\widehat{\Lambda}[/math], as claimed.

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Alternatively, we have the following construction for the diagonal torus:

Proposition

The diagonal torus [math]T\subset G[/math] can be defined as well by

[[math]] T=G\cap\mathbb T_N^+ [[/math]]

where [math]\mathbb T_N^+\subset U_N^+[/math] is the free complex torus, appearing as

[[math]] \mathbb T_N^+=\widehat{F_N} [[/math]]

with [math]F_N= \lt g_1,\ldots,g_N \gt [/math] being the free group on [math]N[/math] generators.

Show Proof

As a main particular case of Theorem 13.2, that we know as well from chapter 2, the biggest quantum group produces the biggest torus, and so we have:

[[math]] C(\mathbb T_N^+)=C(U_N^+)\Big/\left \lt u_{ij}=0\Big|\forall i\neq j\right \gt [[/math]]

Thus, by intersecting with [math]G[/math] we obtain the diagonal torus of [math]G[/math].

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Most of our computations so far of diagonal tori, that we will recall in a moment, concern various classes of easy quantum groups. In the general easy case, we have:

Proposition

For an easy quantum group [math]G\subset U_N^+[/math], coming from a category of partitions [math]D\subset P[/math], the associated diagonal torus is [math]T=\widehat{\Gamma}[/math], with:

[[math]] \Gamma=F_N\Big/\left \lt g_{i_1}\ldots g_{i_k}=g_{j_1}\ldots g_{j_l}\Big|\forall i,j,k,l,\exists\pi\in D(k,l),\delta_\pi\begin{pmatrix}i\\ j\end{pmatrix}\neq0\right \gt [[/math]]

Moreover, we can just use partitions [math]\pi[/math] which generate the category [math]D[/math].

Show Proof

Let [math]g_i=u_{ii}[/math] be the standard coordinates on the diagonal torus [math]T[/math], and set [math]g=diag(g_1,\ldots,g_N)[/math]. We have then the following computation:

[[math]] \begin{eqnarray*} C(T) &=&\left[C(U_N^+)\Big/\left \lt T_\pi\in Hom(u^{\otimes k},u^{\otimes l})\Big|\forall\pi\in D\right \gt \right]\Big/\left \lt u_{ij}=0\Big|\forall i\neq j\right \gt \\ &=&\left[C(U_N^+)\Big/\left \lt u_{ij}=0\Big|\forall i\neq j\right \gt \right]\Big/\left \lt T_\pi\in Hom(u^{\otimes k},u^{\otimes l})\Big|\forall\pi\in D\right \gt \\ &=&C^*(F_N)\Big/\left \lt T_\pi\in Hom(g^{\otimes k},g^{\otimes l})\Big|\forall\pi\in D\right \gt \end{eqnarray*} [[/math]]

The associated discrete group, [math]\Gamma=\widehat{T}[/math], is therefore given by:

[[math]] \Gamma=F_N\Big/\left \lt T_\pi\in Hom(g^{\otimes k},g^{\otimes l})\Big|\forall\pi\in D\right \gt [[/math]]

Now observe that, with [math]g=diag(g_1,\ldots,g_N)[/math] as above, we have:

[[math]] T_\pi g^{\otimes k}(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j_1\ldots j_l}\delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l}\cdot g_{i_1}\ldots g_{i_k} [[/math]]

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

[[math]] g^{\otimes l}T_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j_1\ldots j_l}\delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l}\cdot g_{j_1}\ldots g_{j_l} [[/math]]

We conclude that the relation [math]T_\pi\in Hom(g^{\otimes k},g^{\otimes l})[/math] reformulates as follows:

[[math]] \begin{eqnarray*} &&\sum_{j_1\ldots j_l}\delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l}\cdot g_{i_1}\ldots g_{i_k}\\ &=&\sum_{j_1\ldots j_l}\delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}e_{j_1}\otimes\ldots\otimes e_{j_l}\cdot g_{j_1}\ldots g_{j_l} \end{eqnarray*} [[/math]]

Thus, the following condition must be satisfied:

[[math]] \delta_\pi\begin{pmatrix}i_1&\ldots&i_k\\ j_1&\ldots&j_l\end{pmatrix}\neq0\implies g_{i_1}\ldots g_{i_k}=g_{j_1}\ldots g_{j_l} [[/math]]

Thus, we obtain the formula in the statement. Finally, the last assertion follows from Tannakian duality, because we can replace everywhere [math]D[/math] by a generating subset.

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In practice now, in the continuous case we have the following result:

Theorem

The diagonal tori of the basic unitary quantum groups, namely

[[math]] \xymatrix@R=15mm@C=17mm{ U_N\ar[r]&U_N^*\ar[r]&U_N^+\\ O_N\ar[r]\ar[u]&O_N^*\ar[r]\ar[u]&O_N^+\ar[u]} [[/math]]

and of their [math]q=-1[/math] twists as well, are the standard cube and torus, namely [math]T_N=\mathbb Z_2^N[/math] and [math]\mathbb T_N=\mathbb T^N[/math] in the classical case, and their liberations in general, which are as follows:

[[math]] \xymatrix@R=15mm@C=17mm{ \mathbb T_N\ar[r]&\mathbb T_N^*\ar[r]&\mathbb T_N^+\\ T_N\ar[r]\ar[u]&T_N^*\ar[r]\ar[u]&T_N^+\ar[u]} [[/math]]

Also, for the quantum groups [math]B_N,B_N^+,C_N,C_N^+[/math], the diagonal torus collapses to [math]\{1\}[/math].

Show Proof

We have several assertions here, the idea being as follows:

(1) The main assertion, regarding the basic unitary quantum groups, is something that we already know, from chapter 2 above, with the various liberations [math]T_N^\times,\mathbb T_N^\times[/math] of the basic tori [math]T_N,\mathbb T_N[/math] in the statement being by definition those appearing there.

(2) Regarding the invariance under twisting, this is best seen by using Proposition 13.4. Indeed, the computation in the proof there applies in the same way to the general quizzy case, and shows that the diagonal torus is invariant under twisting.

(3) In the bistochastic case the fundamental corepresentation [math]g=diag(g_1,\ldots,g_N)[/math] of the diagonal torus must be bistochastic, and so [math]g_1=\ldots=g_N=1[/math], as desired.

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Regarding now the discrete case, the result here is as follows:

Theorem

The diagonal tori of the basic quantum reflection groups, namely

[[math]] \xymatrix@R=15mm@C=17mm{ K_N\ar[r]&K_N^*\ar[r]&K_N^+\\ H_N\ar[r]\ar[u]&H_N^*\ar[r]\ar[u]&H_N^+\ar[u]} [[/math]]

are the same as those for [math]O_N^\times,U_N^\times[/math] described above. Also, for [math]S_N,S_N^+[/math] we have [math]T=\{1\}[/math].

Show Proof

The first assertion follows from the general fact that the diagonal torus of [math]G_N\subset U_N^+[/math] equals the diagonal torus of the discrete version, namely:

[[math]] G_N^d=G_N\cap K_N^+ [[/math]]

Indeed, this fact follows from definitions, for instance via Proposition 13.3. As for the second assertion, this follows from the following inclusions:

[[math]] S_N\subset B_N\quad,\quad S_N^+\subset B_N^+ [[/math]]

Indeed, by using the last assertion in Theorem 13.5, we obtain the result.

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As a conclusion, the diagonal torus [math]T\subset G[/math] is usually a quite interesting object, but for certain quantum groups like the bistochastic ones, or the quantum permutation ones, this torus collapses to [math]\{1\}[/math], and so it cannot be of use in the study of [math]G[/math].

13b. The skeleton

In order to deal with the above issue, regarding the diagonal torus, the idea, from , , will be that of using the following generalization of Theorem 13.2:

Theorem

Given a closed subgroup [math]G\subset U_N^+[/math] and a matrix [math]Q\in U_N[/math], we let [math]T_Q\subset G[/math] be the diagonal torus of [math]G[/math], with fundamental representation spinned by [math]Q[/math]:

[[math]] C(T_Q)=C(G)\Big/\left \lt (QuQ^*)_{ij}=0\Big|\forall i\neq j\right \gt [[/math]]

This torus is then a group dual, given by [math]T_Q=\widehat{\Lambda}_Q[/math], where [math]\Lambda_Q= \lt g_1,\ldots,g_N \gt [/math] is the discrete group generated by the elements

[[math]] g_i=(QuQ^*)_{ii} [[/math]]

which are unitaries inside the quotient algebra [math]C(T_Q)[/math].

Show Proof

This follows from Theorem 13.2, because, as said in the statement, [math]T_Q[/math] is by definition a diagonal torus. Equivalently, since [math]v=QuQ^*[/math] is a unitary corepresentation, its diagonal entries [math]g_i=v_{ii}[/math], when regarded inside [math]C(T_Q)[/math], are unitaries, and satisfy:

[[math]] \Delta(g_i)=g_i\otimes g_i [[/math]]

Thus [math]C(T_Q)[/math] is a group algebra, and more specifically we have [math]C(T_Q)=C^*(\Lambda_Q)[/math], where [math]\Lambda_Q= \lt g_1,\ldots,g_N \gt [/math] is the group in the statement, and this gives the result.

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Summarizing, associated to any closed subgroup [math]G\subset U_N^+[/math] is a whole family of tori, indexed by the unitaries [math]U\in U_N[/math]. We use the following terminology:

Definition

Let [math]G\subset U_N^+[/math] be a closed subgroup.

• The tori [math]T_Q\subset G[/math] constructed above are called standard tori of [math]G[/math].
• The collection of tori [math]T=\left\{T_Q\subset G\big|Q\in U_N\right\}[/math] is called skeleton of [math]G[/math].

This might seem a bit awkward, but in view of various results, examples and counterexamples, to be presented below, this is perhaps the best terminology. As a first general result now regarding these tori, coming from Woronowicz , we have:

Theorem

Any torus [math]T\subset G[/math] appears as follows, for a certain [math]Q\in U_N[/math]:

[[math]] T\subset T_Q\subset G [[/math]]

In other words, any torus appears inside a standard torus.

Show Proof

Given a torus [math]T\subset G[/math], we have an inclusion as follows:

[[math]] T\subset G\subset U_N^+ [[/math]]

On the other hand, we know from chapter 3 above that each torus [math]T\subset U_N^+[/math] has a fundamental corepresentation as follows, with [math]Q\in U_N[/math]:

[[math]] u=Q \begin{pmatrix}g_1\\&\ddots\\&&g_N\end{pmatrix} Q^* [[/math]]

But this shows that we have [math]T\subset T_Q[/math], and this gives the result.

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Let us do now some computations, following , where the standard tori were introduced. In the classical case, the result is as follows:

Proposition

For a closed subgroup [math]G\subset U_N[/math] we have

[[math]] T_Q=G\cap(Q^*\mathbb T^NQ) [[/math]]

where [math]\mathbb T^N\subset U_N[/math] is the group of diagonal unitary matrices.

Show Proof

This is indeed clear at [math]Q=1[/math], where [math]\Gamma_1[/math] appears by definition as the dual of the compact abelian group [math]G\cap\mathbb T^N[/math]. In general, this follows by conjugating by [math]Q[/math].

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In the group dual case now, still following , we have the following result:

Proposition

Given a finitely generated discrete group

[[math]] \Gamma= \lt g_1,\ldots,g_N \gt [[/math]]

consider its dual compact quantum group [math]G=\widehat{\Gamma}[/math], diagonally embedded into [math]U_N^+[/math]. We have

[[math]] \Lambda_Q=\Gamma/\left \lt g_i=g_j\Big|\exists k,Q_{ki}\neq0,Q_{kj}\neq0\right \gt [[/math]]

with the embedding [math]T_Q\subset G=\widehat{\Gamma}[/math] coming from the quotient map [math]\Gamma\to\Lambda_Q[/math].

Show Proof

Assume indeed that [math]\Gamma= \lt g_1,\ldots,g_N \gt [/math] is a discrete group, with dual [math]\widehat{\Gamma}\subset U_N^+[/math] coming via [math]u=diag(g_1,\ldots,g_N)[/math]. With [math]v=QuQ^*[/math], we have the following computation:

[[math]] \begin{eqnarray*} \sum_s\bar{Q}_{si}v_{sk} &=&\sum_{st}\bar{Q}_{si}Q_{st}\bar{Q}_{kt}g_t\\ &=&\sum_t\delta_{it}\bar{Q}_{kt}g_t\\ &=&\bar{Q}_{ki}g_i \end{eqnarray*} [[/math]]

Thus the condition [math]v_{ij}=0[/math] for [math]i\neq j[/math] gives [math]\bar{Q}_{ki}v_{kk}=\bar{Q}_{ki}g_i[/math], which tells us that:

[[math]] Q_{ki}\neq0\implies g_i=v_{kk} [[/math]]

Now observe that this latter equality reads:

[[math]] g_i=\sum_j|Q_{kj}|^2g_j [[/math]]

We conclude from this that we have, as desired:

[[math]] Q_{ki}\neq0,Q_{kj}\neq0\implies g_i=g_j [[/math]]

As for the converse, this is elementary to establish as well. See .

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According to the above results, we can expect the skeleton [math]T[/math] to encode various algebraic and analytic properties of [math]G[/math]. We will discuss this in what follows, with a number of results and conjectures, following . We first have the following result:

Theorem

The following hold, both over the category of compact Lie groups, and over the category of duals of finitely generated discrete groups:

• Injectivity: the construction [math]G\to T[/math] is injective, in the sense that [math]G\neq H[/math] implies, for some [math]Q\in U_N[/math]:
  [[math]] T_Q(G)\neq T_Q(H) [[/math]]
• Monotony: the construction [math]G\to T[/math] is increasing, in the sense that passing to a subgroup [math]H\subset G[/math] decreases at least one of the tori [math]T_Q[/math]:
  [[math]] T_Q(H)\neq T_Q(G) [[/math]]
• Generation: any closed quantum subgroup [math]G\subset U_N^+[/math] is generated by its tori, or, equivalently, has the following generation property:
  [[math]] G= \lt T_Q|Q\in U_N \gt [[/math]]

Show Proof

We have two cases to be investigated, as follows:

(1) Assume first that we are in the classical case, [math]G\subset U_N[/math]. In order to prove the generation property we use the following formula, established above:

[[math]] T_Q=G\cap Q^*\mathbb T^NQ [[/math]]

Now since any group element [math]U\in G[/math] is unitary, and so diagonalizable by basic linear algebra, we can write, for certain matrices [math]Q\in U_N[/math] and [math]D\in\mathbb T^N[/math]:

[[math]] U=Q^*DQ [[/math]]

But this shows that we have [math]U\in T_Q[/math], for this precise value of the spinning matrix [math]Q\in U_N[/math], used in the construction of the standard torus [math]T_Q[/math]. Thus we have proved the generation property, and the injectivity and monotony properties follow from this.

(2) Regarding now the group duals, here everything is trivial. Indeed, when the group duals are diagonally embedded we can take [math]Q=1[/math], and when the group duals are embedded by using a spinning matrix [math]Q\in U_N[/math], we can use precisely this matrix [math]Q[/math].

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As explained in , it is possible to go beyond the above verifications, notably with some results regarding the half-classical and the free cases. We will be back to this in chapter 14 below, with a number of more specialized statements, also from , and which are for the moment conjectural as well, on the question of recovering the fine analytic properties of [math]G[/math] out of the fine analytic properties of its tori.

13c. Generation questions

Let us focus now on the generation property, from Theorem 13.12 (3), which is perhaps the most important, in view of the various potential applications. In order to discuss the general case, we will need some abstract theory. Let us start with:

Proposition

Given a closed subgroup [math]G\subset U_N^+[/math] and a matrix [math]Q\in U_N[/math], the corresponding standard torus and its Tannakian category are given by

[[math]] T_Q=G\cap\mathbb T_Q\quad,\quad C_{T_Q}= \lt C_G,C_{\mathbb T_Q} \gt [[/math]]

where [math]\mathbb T_Q\subset U_N^+[/math] is the dual of the free group [math]F_N= \lt g_1,\ldots,g_N \gt [/math], with the fundamental corepresentation of [math]C(\mathbb T_Q)[/math] being the matrix [math]u=Qdiag(g_1,\ldots,g_N)Q^*[/math].

Show Proof

The first assertion comes from the fact, that we know from chapter 7, that given two closed subgroups [math]G,H\subset U_N^+[/math], the corresponding quotient algebra [math]C(U_N^+)\to C(G\cap H)[/math] appears by dividing by the kernels of the following quotient maps:

[[math]] C(U_N^+)\to C(G)\quad,\quad C(U_N^+)\to C(H) [[/math]]

Indeed, the construction of [math]T_Q[/math] from Theorem 13.7 amounts precisely in performing this operation, with [math]H=\mathbb T_Q[/math], and so we obtain, as claimed:

[[math]] T_Q=G\cap\mathbb T_Q [[/math]]

As for the Tannakian category formula, this follows from this, and from the following general Tannakian duality formula, that we know as well from chapter 7:

[[math]] C_{G\cap H}= \lt C_G,C_H \gt [[/math]]

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

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We have the following Tannakian reformulation of the toral generation property:

Theorem

Given a closed subgroup [math]G\subset U_N^+[/math], the subgroup

[[math]] G'= \lt T_Q|Q\in U_N \gt [[/math]]

generated by its standard tori has the following Tannakian category:

[[math]] C_{G'}=\bigcap_{Q\in U_N} \lt C_G,C_{\mathbb T_Q} \gt [[/math]]

In particular we have [math]G=G'[/math] when this intersection reduces to [math]C_G[/math].

Show Proof

Consider the subgroup [math]G'\subset G[/math] constructed in the statement. We have:

[[math]] C_{G'}=\bigcap_{Q\in U_N}C_{T_Q} [[/math]]

Together with the formula in Proposition 13.13, this gives the result.

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Let us further discuss now the toral generation property, with some modest results, regarding its behaviour with respect to product operations. We first have:

Proposition

Given two closed subgroups [math]G,H\subset U_N^+[/math], and [math]Q\in U_N[/math], we have:

[[math]] \lt T_Q(G),T_Q(H) \gt \subset T_Q( \lt G,H \gt ) [[/math]]

Also, the toral generation property is stable under the operation [math] \lt \,, \gt [/math].

Show Proof

The first assertion can be proved either by using Theorem 13.14, or directly. For the direct proof, which is perhaps the simplest, we have:

[[math]] \begin{eqnarray*} T_Q(G) &=&G\cap\mathbb T_Q\subset \lt G,H \gt \cap\mathbb T_Q\\ &=&T_Q( \lt G,H \gt ) \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} T_Q(H) &=&H\cap\mathbb T_Q\subset \lt G,H \gt \cap\mathbb T_Q\\ &=&T_Q( \lt G,H \gt ) \end{eqnarray*} [[/math]]

Now since [math]A,B\subset C[/math] implies [math] \lt A,B \gt \subset C[/math], this gives the result. Regarding now the second assertion, we have the following computation:

[[math]] \begin{eqnarray*} \lt G,H \gt &=& \lt \lt T_Q(G)|Q\in U_N \gt , \lt T_Q(H)|Q\in U_N \gt \gt \\ &=& \lt T_Q(G),T_Q(H)|Q\in U_N \gt \\ &=& \lt \lt T_Q(G),T_Q(H) \gt |Q\in U_N \gt \\ &\subset& \lt T_Q( \lt G,H \gt )|Q\in U_N \gt \end{eqnarray*} [[/math]]

Thus the quantum group [math] \lt G,H \gt [/math] is generated by its tori, as claimed.

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Along the same lines, we have as well the following result:

Proposition

We have the following formula, for any [math]G,H[/math] and [math]R,S[/math]:

[[math]] T_{R\otimes S}(G\times H)=T_R(G)\times T_S(H) [[/math]]

Also, the toral generation property is stable under usual products [math]\times[/math].

Show Proof

The product formula in the statement is clear from definitions. Regarding now the second assertion, we have the following computation:

[[math]] \begin{eqnarray*} \lt T_Q(G\times H)|Q\in U_{MN} \gt &\supset& \lt T_{R\otimes S}(G\times H)|R\in U_M,S\in U_N \gt \\ &=& \lt T_R(G)\times T_S(H)|R\in U_M,S\in U_N \gt \\ &=& \lt T_R(G)\times\{1\},\{1\}\times T_S(H)|R\in U_M,S\in U_N \gt \\ &=& \lt T_R(G)|R\in U_M \gt \times \lt T_G(H)|H\in U_N \gt \\ &=&G\times H \end{eqnarray*} [[/math]]

Thus the quantum group [math]G\times H[/math] is generated by its tori, as claimed.

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In order to get beyond these results, let us discuss now some weaker versions of the generation property, related to the classification program for the compact quantum groups, explained in chapters 11-12. We have here the following technical definition:

Definition

A closed subgroup [math]G_N\subset U_N^+[/math], with classical version [math]G_N^c=G_N\cap U_N[/math], is said to be weakly generated by its tori when:

[[math]] G_N= \lt G_N^c,(T_Q)_{Q\in U_N} \gt [[/math]]

When the following even weaker condition, involving [math]T_1[/math] only, is satisfied,

[[math]] G_N= \lt G_N^c,T_1 \gt [[/math]]

we say that [math]G_N[/math] appears as a diagonal liberation of its classical version [math]G_N^c[/math].

According to our results above, the first property is satisfied for the compact groups, for the discrete group duals, and is stable under generation, and direct products. Regarding the second property, this is something quite interesting, related to many things. The idea here, from Chirvasitu and subsequent papers, is that such formulae can be proved by recurrence on [math]N\in\mathbb N[/math]. In order to discuss this, let us start with:

Proposition

Assume that [math]G=(G_N)[/math] is weakly uniform, let [math]n\in\{2,3,\ldots,\infty\}[/math] be minimal such that [math]G_n[/math] is not classical, and consider the following conditions:

• Strong generation: [math]G_N= \lt G_N^c,G_n \gt [/math], for any [math]N \gt n[/math].
• Usual generation: [math]G_N= \lt G_N^c,G_{N-1} \gt [/math], for any [math]N \gt n[/math].
• Initial step generation: [math]G_{n+1}= \lt G_{n+1}^c,G_n \gt [/math].

We have then [math](1)\iff(2)\implies(3)[/math], and [math](3)[/math] is in general strictly weaker.

Show Proof

All the implications and non-implications are elementary, as follows:

[math](1)\implies(2)[/math] This follows from [math]G_n\subset G_{N-1}[/math] for [math]N \gt n[/math], coming from uniformity.

[math](2)\implies(1)[/math] By using twice the usual generation, and then the uniformity, we have:

[[math]] \begin{eqnarray*} G_N &=& \lt G_N^c,G_{N-1} \gt \\ &=& \lt G_N^c,G_{N-1}^c,G_{N-2} \gt \\ &=& \lt G_N^c,G_{N-2} \gt \end{eqnarray*} [[/math]]

Thus we have a descent method, and we end up with the strong generation condition.

[math](2)\implies(3)[/math] This is clear, because (2) at [math]N=n+1[/math] is precisely (3).

[math](3)\hskip2.3mm\not\hskip-2.3mm\implies(2)[/math] In order to construct counterexamples here, the simplest is to use group duals. Indeed, with [math]G_N=\widehat{\Gamma_N}[/math] and [math]\Gamma_N= \lt g_1,\ldots,g_N \gt [/math], the uniformity condition tells us that we must be in a projective limit situation, as follows:

[[math]] \Gamma_1\leftarrow\Gamma_2\leftarrow\Gamma_3\leftarrow\Gamma_4\leftarrow\ldots [[/math]]

[[math]] \Gamma_{N-1}=\Gamma_N/ \lt g_N=1 \gt [[/math]]

Now by assuming for instance that [math]\Gamma_2[/math] is given and not abelian, there are many ways of completing the sequence, and so the uniqueness coming from (2) can only fail.

■

Let us introduce now a few more notions, as follows:

Proposition

Assume that [math]G=(G_N)[/math] is weakly uniform, let [math]n\in\{2,3,\ldots,\infty\}[/math] be as above, and consider the following conditions, where [math]I_N\subset G_N[/math] is the diagonal torus:

• Strong diagonal liberation: [math]G_N= \lt G_N^c,I_n \gt [/math], for any [math]N\geq n[/math].
• Technical condition: [math]G_N= \lt G_N^c,I_{N-1} \gt [/math] for any [math]N \gt n[/math], and [math]G_n= \lt G_n^c,I_n \gt [/math].
• Diagonal liberation: [math]G_N= \lt G_N^c,I_N \gt [/math], for any [math]N[/math].
• Initial step diagonal liberation: [math]G_n= \lt G_n^c,I_n \gt [/math].

We have then [math](1)\implies(2)\implies(3)\implies(4)[/math].

Show Proof

Our claim is that when assuming that [math]G=(G_N)[/math] is weakly uniform, so is the family of diagonal tori [math]I=(I_N)[/math]. Indeed, we have the following computation:

[[math]] \begin{eqnarray*} I_N\cap U_{N-1}^+ &=&(G_N\cap\mathbb T_N^+)\cap U_{N-1}^+\\ &=&(G_N\cap U_{N-1}^+)\cap(\mathbb T_N^+\cap U_{N-1}^+)\\ &=&G_{N-1}\cap\mathbb T_{N-1}^+\\ &=&I_{N-1} \end{eqnarray*} [[/math]]

Thus our claim is proved, and this gives the various implications in the statement.

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Based on the above technical results, we can now formulate a key theoretical observation, in relation with the various generation properties that we have, as follows:

Theorem

If [math]G=(G_N)[/math] is weakly uniform, and with [math]n\in\{2,3,\ldots,\infty\}[/math] being as above, the following conditions are equivalent, modulo their initial steps:

• Generation: [math]G_N= \lt G_N^c,G_{N-1} \gt [/math], for any [math]N \gt n[/math].
• Strong generation: [math]G_N= \lt G_N^c,G_n \gt [/math], for any [math]N \gt n[/math].
• Diagonal liberation: [math]G_N= \lt G_N^c,I_N \gt [/math], for any [math]N\geq n[/math].
• Strong diagonal liberation: [math]G_N= \lt G_N^c,I_n \gt [/math], for any [math]N\geq n[/math].

Show Proof

Our first claim is that generation plus initial step diagonal liberation imply the technical diagonal liberation condition. Indeed, the recurrence step goes as follows:

[[math]] \begin{eqnarray*} G_N &=& \lt G_N^c,G_{N-1} \gt \\ &=& \lt G_N^c,G_{N-1}^c,I_{N-1} \gt \\ &=& \lt G_N^c,I_{N-1} \gt \end{eqnarray*} [[/math]]

In order to pass now from the technical diagonal liberation condition to the strong diagonal liberation condition itself, observe that we have:

[[math]] \begin{eqnarray*} G_N &=& \lt G_N^c,G_{N-1} \gt \\ &=& \lt G_N^c,G_{N-1}^c,I_{N-1} \gt \\ &=& \lt G_N^c,I_{N-1} \gt \end{eqnarray*} [[/math]]

With this condition in hand, we have then as well:

[[math]] \begin{eqnarray*} G_N &=& \lt G_N^c,G_{N-1} \gt \\ &=& \lt G_N^c,G_{N-1}^c,I_{N-2} \gt \\ &=& \lt G_N^c,I_{N-2} \gt \end{eqnarray*} [[/math]]

This procedure can be of course be continued. Thus we have a descent method, and we end up with the strong diagonal liberation condition, as desired. In the other sense now, we want to prove that we have the following formula, at any [math]N\geq n[/math]:

[[math]] G_N= \lt G_N^c,G_{N-1} \gt [[/math]]

At [math]N=n+1[/math] this is something that we already know. At [math]N=n+2[/math] now, we have:

[[math]] \begin{eqnarray*} G_{n+2} &=& \lt G_{n+2}^c,I_n \gt \\ &=& \lt G_{n+2}^c,G_{n+1}^c,I_n \gt \\ &=& \lt G_{n+2}^c,G_{n+1} \gt \end{eqnarray*} [[/math]]

This procedure can be of course be continued. Thus, we have a descent method, and we end up with the strong generation condition, as desired.

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It is possible to prove that many interesting quantum groups have the above properties, and hence appear as diagonal liberations, but the whole subject is quite technical. Here is however a statement, collecting most of the known results on the subject:

Theorem

The basic quantum unitary and reflection groups are as follows:

• [math]O_N^*,U_N^*[/math] appear via diagonal liberation.
• [math]O_N^+,U_N^+[/math] appear via diagonal liberation.
• [math]H_N^*,K_N^*[/math] appear via diagonal liberation.
• [math]H_N^+,K_N^+[/math] do not appear via diagonal liberation.

In addition, [math]B_N^+,C_N^+,S_N^+[/math] do not appear either via diagonal liberation.

Show Proof

All this is quite technical, the idea being as follows:

(1) The half-classical quantum groups [math]O_N^*,U_N^*[/math] are not uniform, and so cannot be investigated with the above techniques. However, these quantum groups can be studied by using the matrix model technology in , , which will be briefly discussed in chapter 16 below, and this leads to the following generation formulae:

[[math]] O_N^*= \lt O_N,T_N^* \gt [[/math]]

[[math]] U_N^*= \lt U_N,T_N^* \gt [[/math]]

But these two formulae imply the following generation formula, as desired:

[[math]] U_N^*= \lt U_N,\mathbb T_N^* \gt [[/math]]

(2) The quantum groups [math]O_N^+,U_N^+[/math] are uniform, and a quite technical computation, from Chirvasitu et al. , , shows that the generation conditions from Theorem 13.20 are satisfied for [math]O_N^+[/math]. Thus we obtain the following generation formula:

[[math]] O_N^+= \lt O_N,T_N^+ \gt [[/math]]

From this we can deduce via the maximality results in that we have:

[[math]] U_N^+= \lt U_N,T_N^+ \gt [[/math]]

But this implies the following generation formula, as desired:

[[math]] U_N^+= \lt U_N,\mathbb T_N^+ \gt [[/math]]

(3) The situation for [math]H_N^*,K_N^*[/math] is quite similar to the one for [math]O_N^*,U_N^*[/math], explained above. Indeed, the technology in , applies, and this leads to:

[[math]] H_N^*= \lt H_N,T_N^* \gt [[/math]]

[[math]] K_N^*= \lt K_N,T_N^* \gt [[/math]]

Thus, we have as well the following formula, as desired:

[[math]] K_N^*= \lt K_N,\mathbb T_N^* \gt [[/math]]

As a comment here, in fact these results are stronger than the above-mentioned ones for the quantum groups [math]O_N^*,U_N^*[/math], via some standard generation formulae.

(4) This is something subtle as well, coming from the quantum groups [math]H_N^{[\infty]},K_N^{[\infty]}[/math] from Raum-Weber , discussed before. The idea here is that the following relations, related to the defining relations for [math]H_N^{[\infty]},K_N^{[\infty]}[/math], are trivially satisfied for real reflections:

[[math]] g_ig_ig_j=g_jg_ig_i [[/math]]

Thus, the diagonal tori of these quantum groups coincide with those for [math]H_N^+,K_N^+[/math], and so the diagonal liberation procedure “stops” at [math]H_N^{[\infty]},K_N^{[\infty]}[/math].

Finally, regarding the last assertion, here [math]B_N^+,C_N^+,S_N^+[/math] do not appear indeed via diagonal liberation, and this because of a trivial reason, namely [math]T=\{1\}[/math].

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13d. Fourier liberation

All the above was quite technical, but regardless of the difficulties involved there, and also of the various positive results on the subject, the notion of diagonal liberation is obviously not the good one, in general. As a conjectural solution to these difficulties, we have the notion of Fourier liberation, that we will discuss now.

Let us start with the following basic fact, which generalizes the construction of the embedding [math]\widehat{D}_\infty\subset S_4^+[/math], that we met in chapter 9, when proving that we have [math]S_4^+\neq S_4[/math]:

Proposition

Consider a discrete group generated by elements of finite order, written as a quotient group, as follows:

[[math]] \mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}\to\Gamma [[/math]]

We have then an embedding of quantum groups [math]\widehat{\Gamma}\subset S_N^+[/math], where [math]N=N_1+\ldots+N_k[/math].

Show Proof

We have a sequence of embeddings and isomorphisms as follows:

[[math]] \begin{eqnarray*} \widehat{\Gamma} &\subset&\widehat{\mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}}\\ &=&\widehat{\mathbb Z_{N_1}}\,\hat{*}\,\ldots\,\hat{*}\,\widehat{\mathbb Z_{N_k}}\\ &\simeq&\mathbb Z_{N_1}\,\hat{*}\,\ldots\,\hat{*}\,\mathbb Z_{N_k}\\ &\subset&S_{N_1}\,\hat{*}\,\ldots\,\hat{*}\,S_{N_k}\\ &\subset&S_{N_1}^+\,\hat{*}\,\ldots\,\hat{*}\,S_{N_k}^+\\ &\subset&S_N^+ \end{eqnarray*} [[/math]]

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

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The above result is quite abstract, and it is worth working out the details, with an explicit formula for the associated magic matrix. Let us start with a study of the simplest situation, where [math]k=1[/math], and where [math]\Gamma=\mathbb Z_{N_1}[/math]. The result here is as follows:

Proposition

The magic matrix for the quantum permutation group

[[math]] \widehat{\mathbb Z}_N\simeq\mathbb Z_N\subset S_N\subset S_N^+ [[/math]]

with standard Fourier isomorphism on the left, is given by the formula

[[math]] u=FIF^* [[/math]]

where [math]F=\frac{1}{\sqrt{N}}(w^{ij})[/math] with [math]w=e^{2\pi i/N}[/math] is the Fourier matrix, and where

[[math]] I=\begin{pmatrix} 1\\ &g\\ &&\ddots\\ &&&g^{N-1} \end{pmatrix} [[/math]]

is the diagonal matrix formed by the elements of [math]\mathbb Z_N[/math], regarded as elements of [math]C^*(\mathbb Z_N)[/math].

Show Proof

The magic matrix for the quantum group [math]\mathbb Z_N\subset S_N\subset S_N^+[/math] is given by:

[[math]] v_{ij} =\chi\left(\sigma\in\mathbb Z_N\Big|\sigma(j)=i\right) =\delta_{i-j} [[/math]]

Let us apply now the Fourier transform. According to our Pontrjagin duality conventions from chapter 1 above, in one sense this is given by the following formula:

[[math]] \Phi:C(\mathbb Z_N)\to C^*(\mathbb Z_N)\quad,\quad \delta_i\to\frac{1}{N}\sum_kw^{ik}g^k [[/math]]

As for the inverse isomorphism, this is given by the following formula:

[[math]] \Psi:C^*(\mathbb Z_N)\to C(\mathbb Z_N)\quad,\quad g^i\to\sum_kw^{-ik}\delta_k [[/math]]

Here [math]w=e^{2\pi i/N}[/math], and we use the standard Fourier analysis convention that the indices are [math]0,1,\ldots,N-1[/math]. With [math]F=\frac{1}{\sqrt{N}}(w^{ij})[/math] and [math]I=diag(g^i)[/math] as above, we have:

[[math]] \begin{eqnarray*} u_{ij} &=&\Phi(v_{ij})\\ &=&\frac{1}{N}\sum_kw^{(i-j)k}g^k\\ &=&\frac{1}{N}\sum_kw^{ik}g^kw^{-jk}\\ &=&\sum_kF_{ik}I_{kk}(F^*)_{kj}\\ &=&(FIF^*)_{ij} \end{eqnarray*} [[/math]]

Thus, the magic matrix that we are looking for is [math]u=FIF^*[/math], as claimed.

■

With the above result in hand, we can complement Proposition 13.22 with:

Proposition

Given a quotient group [math]\mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}\to\Gamma[/math], the magic matrix for the subgroup [math]\widehat{\Gamma}\subset S_N^+[/math] found in Proposition 13.22, with [math]N=N_1+\ldots+N_k[/math], is given by

[[math]] u=\begin{pmatrix} F_{N_1}I_1F_{N_1}^*\\ &\ddots\\ &&F_{N_k}I_kF_{N_k}^* \end{pmatrix} [[/math]]

where [math]F_N=\frac{1}{\sqrt{N}}(w_N^{ij})[/math] with [math]w_N=e^{2\pi i/N}[/math] are Fourier matrices, and where

[[math]] I_r=\begin{pmatrix} 1\\ &g_r\\ &&\ddots\\ &&&g_r^{N_r-1} \end{pmatrix} [[/math]]

with [math]g_1,\ldots,g_k[/math] being the standard generators of [math]\Gamma[/math].

Show Proof

This follows indeed from Proposition 13.22 and Proposition 13.23.

■

Following Bichon , let us prove now that this construction provides us with all the group duals [math]\widehat{\Gamma}\subset S_N^+[/math]. The idea will be that of using orbit theory, which is as follows:

Theorem

Given a closed subgroup [math]G\subset S_N^+[/math], with standard coordinates denoted [math]u_{ij}\in C(G)[/math], the following defines an equivalence relation on [math]\{1,\ldots,N\}[/math],

[[math]] i\sim j\iff u_{ij}\neq0 [[/math]]

that we call orbit decomposition associated to the action [math]G\curvearrowright\{1,\ldots,N\}[/math]. In the classical case, [math]G\subset S_N[/math], this is the usual orbit equivalence coming from the action of [math]G[/math].

Show Proof

We first check the fact that we have indeed an equivalence relation:

(1) The condition [math]i\sim i[/math] follows indeed from [math]\varepsilon(u_{ij})=\delta_{ij}[/math], which gives:

[[math]] \varepsilon(u_{ii})=1 [[/math]]

(2) The condition [math]i\sim j\implies j\sim i[/math] follows from [math]S(u_{ij})=u_{ji}[/math], which gives:

[[math]] u_{ij}\neq0\implies u_{ji}\neq0 [[/math]]

(3) As for the condition [math]i\sim j,j\sim k\implies i\sim k[/math], this follows from:

[[math]] \Delta(u_{ik})=\sum_ju_{ij}\otimes u_{jk} [[/math]]

Indeed, in this formula, the right-hand side is by definition a sum of projections, so assuming that we have [math]u_{ij}\neq0,u_{jk}\neq0[/math] for a certain index [math]j[/math], we obtain:

[[math]] u_{ij}\otimes u_{jk} \gt 0 [[/math]]

Thus we have [math]\Delta(u_{ik}) \gt 0[/math], which gives [math]u_{ik}\neq0[/math], as desired. Finally, in the classical case, [math]G\subset S_N[/math], the standard coordinates are the following characteristic functions:

[[math]] u_{ij}=\chi\left(\sigma\in G\Big|\sigma(j)=i\right) [[/math]]

Thus [math]u_{ij}\neq0[/math] is equivalent to the existence of an element [math]\sigma\in G[/math] such that [math]\sigma(j)=i[/math]. But this means precisely that [math]i,j[/math] must be in the same orbit of [math]G[/math], as claimed.

■

Generally speaking, the theory from the classical case extends well to the quantum group setting, and we have in particular the following result, also from Bichon :

Theorem

Given a closed subgroup [math]G\subset S_N^+[/math], with magic matrix [math]u=(u_{ij})[/math], consider the associated coaction map, on the space [math]X=\{1,\ldots,N\}[/math]:

[[math]] \Phi:C(X)\to C(X)\otimes C(G)\quad,\quad e_i\to\sum_je_j\otimes u_{ji} [[/math]]

The following three subalgebras of [math]C(X)[/math] are then equal,

[[math]] Fix(u)=\left\{\xi\in C(X)\Big|u\xi=\xi\right\} [[/math]]

[[math]] Fix(\Phi)=\left\{\xi\in C(X)\Big|\Phi(\xi)=\xi\otimes1\right\} [[/math]]

[[math]] F=\left\{\xi\in C(X)\Big|i\sim j\implies \xi(i)=\xi(j)\right\} [[/math]]

where [math]\sim[/math] is the orbit equivalence relation constructed in Theorem 13.25.

Show Proof

There are several assertions here, the idea being as follows:

(1) The fact that we have the equality [math]Fix(u)=Fix(\Phi)[/math] is standard, with this being valid for any corepresentation of a compact quantum group [math]u=(u_{ij})[/math].

(2) Regarding now the equality with the algebra [math]F[/math], we know from Theorem 13.25 that the magic unitary [math]u=(u_{ij})[/math] is block-diagonal, with respect to the orbit decomposition there. But this shows that the algebra [math]Fix(u)=Fix(\Phi)[/math] decomposes as well with respect to the orbit decomposition, and so in order to prove the result, we are left with a study in the transitive case, where the result is clear. For details here, see .

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We have as well the following result, of analytic flavor:

Proposition

For a closed subgroup [math]G\subset S_N^+[/math], the following are equivalent:

• [math]G[/math] is transitive.
• [math]Fix(u)=\mathbb C\xi[/math], where [math]\xi[/math] is the all-one vector.
• [math]\int_Gu_{ij}=\frac{1}{N}[/math], for any [math]i,j[/math].

Show Proof

This is well-known in the classical case. In general, the proof is as follows:

[math](1)\iff(2)[/math] This follows from the identifications in Theorem 13.26.

[math](2)\iff(3)[/math] This is clear from the general properties of the Haar integration.

■

As a final comment here, the theory of quantum group orbits and transitivity, originally developed by Bichon in , has an interesting extension into a theory of quantum group orbitals and 2-transitivity, developed by Lupini-Man\v cinska-Roberson in .

Now back to the tori, we have the following key result, from :

Theorem

Consider a quotient group as follows, with [math]N=N_1+\ldots+N_k[/math]:

[[math]] \mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}\to\Gamma [[/math]]

We have then [math]\widehat{\Gamma}\subset S_N^+[/math], and any group dual subgroup of [math]S_N^+[/math] appears in this way.

Show Proof

This result, from , can be proved in two steps, as follows:

(1) The fact that we have a subgroup as in the statement is something that we already know. Conversely, assume that we have a group dual subgroup [math]\widehat{\Gamma}\subset S_N^+[/math]. The corresponding magic unitary must be of the following form, with [math]U\in U_N[/math]:

[[math]] u=U \begin{pmatrix}g_1\\&\ddots\\&&g_N\end{pmatrix} U^* [[/math]]

Consider now the orbit decomposition for [math]\widehat{\Gamma}\subset S_N^+[/math], coming from Theorem 13.25:

[[math]] N=N_1+\ldots+N_k [[/math]]

We conclude that [math]u[/math] has a [math]N=N_1+\ldots+N_k[/math] block-diagonal pattern, and so that [math]U[/math] has as well this [math]N=N_1+\ldots+N_k[/math] block-diagonal pattern.

(2) But this discussion reduces our problem to its [math]k=1[/math] particular case, with the statement here being that the cyclic group [math]\mathbb Z_N[/math] is the only transitive group dual [math]\widehat{\Gamma}\subset S_N^+[/math]. The proof of this latter fact being elementary, we obtain the result. See .

■

Here is a related result, from , which is useful for our purposes:

Theorem

For the quantum permutation group [math]S_N^+[/math], we have:

• Given [math]Q\in U_N[/math], the quotient [math]F_N\to\Lambda_Q[/math] comes from the following relations:
  [[math]] \begin{cases} g_i=1&{\rm if}\ \sum_lQ_{il}\neq 0\\ g_ig_j=1&{\rm if}\ \sum_lQ_{il}Q_{jl}\neq 0\\ g_ig_jg_k=1&{\rm if}\ \sum_lQ_{il}Q_{jl}Q_{kl}\neq 0 \end{cases} [[/math]]
• Given a decomposition [math]N=N_1+\ldots+N_k[/math], for the matrix [math]Q=diag(F_{N_1},\ldots,F_{N_k})[/math], where [math]F_N=\frac{1}{\sqrt{N}}(\xi^{ij})_{ij}[/math] with [math]\xi=e^{2\pi i/N}[/math] is the Fourier matrix, we obtain:
  [[math]] \Lambda_Q=\mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k} [[/math]]
• Given a matrix [math]Q\in U_N[/math], there exists a decomposition [math]N=N_1+\ldots+N_k[/math], such that [math]\Lambda_Q[/math] appears as quotient of [math]\mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}[/math].

Show Proof

This is more or less equivalent to Theorem 13.28, and the proof can be deduced either from Theorem 13.28, or from some direct computations, as follows:

(1) Fix a unitary matrix [math]Q\in U_N[/math], and consider the following quantities:

[[math]] \begin{cases} c_i=\sum_lQ_{il}\\ c_{ij}=\sum_lQ_{il}Q_{jl}\\ d_{ijk}=\sum_l\bar{Q}_{il}\bar{Q}_{jl}Q_{kl} \end{cases} [[/math]]

We write [math]w=QvQ^*[/math], where [math]v[/math] is the fundamental corepresentation of [math]C(S_N^+)[/math]. Assume [math]X\simeq\{1,\ldots,N\}[/math], and let [math]\alpha[/math] be the coaction of [math]C(S_N^+)[/math] on [math]C(X)[/math]. Let us set:

[[math]] \varphi_i=\sum_l\bar{Q}_{il}\delta_l\in C(X) [[/math]]

Also, let [math]g_i=(QvQ^*)_{ii}\in C^*(\Lambda_Q)[/math]. If [math]\beta[/math] is the restriction of [math]\alpha[/math] to [math]C^*(\Lambda_Q)[/math], then:

[[math]] \beta(\varphi_i)=\varphi_i\otimes g_i [[/math]]

Now recall that [math]C(X)[/math] is the universal [math]C^*[/math]-algebra generated by elements [math]\delta_1,\ldots,\delta_N[/math] which are pairwise orthogonal projections. Writing these conditions in terms of the linearly independent elements [math]\varphi_i[/math] by means of the formulae [math]\delta_i=\sum_lQ_{il}\varphi_l[/math], we find that the universal relations for [math]C(X)[/math] in terms of the elements [math]\varphi_i[/math] are as follows:

[[math]] \begin{cases} \sum_ic_i\varphi_i=1\\ \varphi_i^*=\sum_jc_{ij}\varphi_j\\ \varphi_i\varphi_j=\sum_kd_{ijk}\varphi_k \end{cases} [[/math]]

Let [math]\tilde{\Lambda}_Q[/math] be the group in the statement. Since [math]\beta[/math] preserves these relations, we get:

[[math]] \begin{cases} c_i(g_i-1)=0\\ c_{ij}(g_ig_j-1)=0\\ d_{ijk}(g_ig_j-g_k)=0 \end{cases} [[/math]]

We conclude from this that [math]\Lambda_Q[/math] is a quotient of [math]\tilde{\Lambda}_Q[/math]. On the other hand, it is immediate that we have a coaction map as follows:

[[math]] C(X)\to C(X)\otimes C^*(\tilde{\Lambda}_Q) [[/math]]

Thus [math]C(\tilde{\Lambda}_Q)[/math] is a quotient of [math]C(S_N^+)[/math]. Since [math]w[/math] is the fundamental corepresentation of [math]S_N^+[/math] with respect to the basis [math]\{\varphi_i\}[/math], it follows that the generator [math]w_{ii}[/math] is sent to [math]\tilde{g}_i\in\tilde{\Lambda}_Q[/math], while [math]w_{ij}[/math] is sent to zero. We conclude that [math]\tilde{\Lambda}_Q[/math] is a quotient of [math]\Lambda_Q[/math]. Since the above quotient maps send generators on generators, we conclude that [math]\Lambda_Q=\tilde{\Lambda}_Q[/math], as desired.

(2) We apply the result found in (1), with the [math]N[/math]-element set [math]X[/math] used in the proof there chosen to be the following set:

[[math]] X=\mathbb Z_{N_1}\sqcup\ldots\sqcup\mathbb Z_{N_k} [[/math]]

With this choice, we have [math]c_i=\delta_{i0}[/math] for any [math]i[/math]. Also, we have [math]c_{ij}=0[/math], unless [math]i,j,k[/math] belong to the same block to [math]Q[/math], in which case [math]c_{ij}=\delta_{i+j,0}[/math], and also [math]d_{ijk} =0[/math], unless [math]i,j,k[/math] belong to the same block of [math]Q[/math], in which case [math]d_{ijk}=\delta_{i+j,k}[/math]. We conclude from this that [math]\Lambda_Q[/math] is the free product of [math]k[/math] groups which have generating relations as follows:

[[math]] g_ig_j=g_{i+j}\quad,\quad g_i^{-1}=g_{-i} [[/math]]

But this shows that our group is [math]\Lambda_Q=\mathbb Z_{N_1}*\ldots*\mathbb Z_{N_k}[/math], as stated.

(3) This follows indeed from (2). See .

■

Summarizing, for quantum permutation groups, the standard tori parametrized by Fourier matrices play a special role. This suggests formulating the following definition:

Definition

Consider a closed subgroup [math]G\subset U_N^+[/math].

• Its standard tori [math]T_F[/math], with [math]F=F_{N_1}\otimes\ldots\otimes F_{N_k}[/math], and [math]N=N_1+\ldots+N_k[/math] being regarded as a partition, are called Fourier tori.
• In the case where we have [math]G_N= \lt G_N^c,(T_F)_F \gt [/math], we say that [math]G_N[/math] appears as a Fourier liberation of its classical version [math]G_N^c[/math].

We believe that the easy quantum groups should appear as Fourier liberations. With respect to Theorem 13.21 above, the situation in the free case is as follows:

(1) [math]O_N^+,U_N^+[/math] are diagonal liberations, so they are Fourier liberations as well.

(2) [math]B_N^+,C_N^+[/math] are Fourier liberations too, by using the results in chapter 7.

(3) [math]S_N^+[/math] is a Fourier liberation too, being generated by its tori .

(4) [math]H_N^+,K_N^+[/math] remain to be investigated, by using the general theory in .

Finally, as a word of warning here, observe that an arbitrary classical group [math]G_N\subset U_N[/math] is not necessarily generated by its Fourier tori, and nor is an arbitrary discrete group dual, with spinned embedding. Thus, the Fourier tori, and the related notion of Fourier liberation, remain something quite technical, in connection with the easy case.

General references

Banica, Teo (2024). "Introduction to quantum groups". arXiv:1909.08152 [math.CO].