Quantum permutations

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

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

9a. Magic matrices

The quantum groups that we considered so far, namely [math]O_N,U_N[/math] and their liberations and twists, are of “continuous” nature. In this third part of the present book we discuss the “discrete” examples, such as the symmetric group [math]S_N[/math], the hyperoctahedral group [math]H_N[/math], and more complicated reflection groups, and their liberations. We will see that most of these groups are easy, have indeed liberations, and equal their own twists.

In this chapter, to start with, we discuss the symmetric group [math]S_N[/math], and its free version [math]S_N^+[/math]. All this is very standard, following the paper of Wang and the subsequent paper , both from the late 90s, and the paper , from the mid 00s. At the level of more advanced results, we will have as well a look into , from the late 00s, dealing more generally with quantum permutation groups [math]S_F^+[/math] of finite quantum spaces [math]F[/math], which are subject to some interesting twisting results, bringing us back to [math]O_N,U_N[/math].

In order to get started, we need to have a matrix group look at the symmetric group [math]S_N[/math]. But this is something obvious, obtained by looking at [math]S_N[/math] as being the permutation group of the [math]N[/math] coordinate axes of [math]\mathbb R^N[/math]. Indeed, we obtain in this way an embedding [math]S_N\subset O_N[/math], which is given by the standard permutation matrices, [math]\sigma(e_j)=e_{\sigma(j)}[/math].

Based on this, we have the following functional analytic description of [math]S_N[/math]:

Proposition

Consider the symmetric group [math]S_N[/math].

• The standard coordinates [math]v_{ij}\in C(S_N)[/math], coming from the embedding [math]S_N\subset O_N[/math] given by the permutation matrices, are given by:
  [[math]] v_{ij}=\chi\left(\sigma\Big|\sigma(j)=i\right) [[/math]]
• The matrix [math]v=(v_{ij})[/math] is magic, in the sense that its entries are orthogonal projections, summing up to [math]1[/math] on each row and each column.
• The algebra [math]C(S_N)[/math] is isomorphic to the universal commutative [math]C^*[/math]-algebra generated by the entries of a [math]N\times N[/math] magic matrix.

Show Proof

These results are all elementary, as follows:

(1) We recall that the canonical embedding [math]S_N\subset O_N[/math], coming from the standard permutation matrices, is given by [math]\sigma(e_j)=e_{\sigma(j)}[/math]. Thus, we have [math]\sigma=\sum_je_{\sigma(j)j}[/math], and it follows that the standard coordinates on [math]S_N\subset O_N[/math] are given by:

[[math]] v_{ij}(\sigma)=\delta_{i,\sigma(j)} [[/math]]

(2) Any characteristic function [math]\chi\in\{0,1\}[/math] being a projection in the operator algebra sense ([math]\chi^2=\chi^*=\chi[/math]), we have indeed a matrix of projections. As for the sum 1 condition on rows and columns, this is clear from the formula of the elements [math]v_{ij}[/math].

(3) Consider the universal algebra in the statement, namely:

[[math]] A=C^*_{comm}\left((w_{ij})_{i,j=1,\ldots,N}\Big|w={\rm magic}\right) [[/math]]

We have a quotient map [math]A\to C(S_N)[/math], given by [math]w_{ij}\to v_{ij}[/math]. On the other hand, by using the Gelfand theorem we can write [math]A=C(X)[/math], with [math]X[/math] being a compact space, and by using the coordinates [math]w_{ij}[/math] we have [math]X\subset O_N[/math], and then [math]X\subset S_N[/math]. Thus we have as well a quotient map [math]C(S_N)\to A[/math] given by [math]v_{ij}\to w_{ij}[/math], and this gives (3).

■

With the above result in hand, we can now formulate, following Wang :

Theorem

The following is a Woronowicz algebra, with magic meaning formed of projections, which sum up to [math]1[/math] on each row and each column,

[[math]] C(S_N^+)=C^*\left((u_{ij})_{i,j=1,\ldots,N}\Big|u={\rm magic}\right) [[/math]]

and the underlying compact quantum group [math]S_N^+[/math] is called quantum permutation group.

Show Proof

The algebra [math]C(S_N^+)[/math] is indeed well-defined, because the magic condition forces [math]||u_{ij}||\leq1[/math], for any [math]C^*[/math]-norm. Our claim now is that, by using the universal property of this algebra, we can define maps [math]\Delta,\varepsilon,S[/math]. Consider indeed the following matrix:

[[math]] U_{ij}=\sum_ku_{ik}\otimes u_{kj} [[/math]]

As a first observation, we have [math]U_{ij}=U_{ij}^*[/math]. In fact the entries [math]U_{ij}[/math] are orthogonal projections, because we have as well:

[[math]] \begin{eqnarray*} U_{ij}^2 &=&\sum_{kl}u_{ik}u_{il}\otimes u_{kj}u_{lj}\\ &=&\sum_ku_{ik}\otimes u_{kj}\\ &=&U_{ij} \end{eqnarray*} [[/math]]

In order to prove now that the matrix [math]U=(U_{ij})[/math] is magic, it remains to verify that the sums on the rows and columns are 1. For the rows, this can be checked as follows:

[[math]] \begin{eqnarray*} \sum_jU_{ij} &=&\sum_{jk}u_{ik}\otimes u_{kj}\\ &=&\sum_ku_{ik}\otimes1\\ &=&1\otimes1 \end{eqnarray*} [[/math]]

For the columns the computation is similar, as follows:

[[math]] \begin{eqnarray*} \sum_iU_{ij} &=&\sum_{ik}u_{ik}\otimes u_{kj}\\ &=&\sum_k1\otimes u_{kj}\\ &=&1\otimes1 \end{eqnarray*} [[/math]]

Thus the matrix [math]U=(U_{ij})[/math] is magic indeed, and so we can define a comultiplication map by setting [math]\Delta(u_{ij})=U_{ij}[/math]. By using a similar reasoning, we can define as well a counit map by [math]\varepsilon(u_{ij})=\delta_{ij}[/math], and an antipode map by [math]S(u_{ij})=u_{ji}[/math]. Thus the Woronowicz algebra axioms from chapter 2 are satisfied, and this finishes the proof.

■

The terminology in the above result comes from the comparison with Proposition 9.1 (3), which tells us that we have an inclusion [math]S_N\subset S_N^+[/math], and that this inclusion is a liberation, in the sense that the classical version of [math]S_N^+[/math], obtained at the algebra level by dividing by the commutator ideal, is the usual symmetric group [math]S_N[/math]. The terminology is further motivated by the following result, also from Wang's paper :

Proposition

The quantum permutation group [math]S_N^+[/math] acts on [math]X=\{1,\ldots,N\}[/math], the corresponding coaction map [math]\Phi:C(X)\to C(X)\otimes C(S_N^+)[/math] being given by:

[[math]] \Phi(\delta_i)=\sum_j\delta_j\otimes u_{ji} [[/math]]

In fact, [math]S_N^+[/math] is the biggest compact quantum group acting on [math]X[/math], by leaving the counting measure invariant, in the sense that

[[math]] (tr\otimes id)\Phi=tr(.)1 [[/math]]

where [math]tr[/math] is the standard trace, given by [math]tr(\delta_i)=\frac{1}{N},\forall i[/math].

Show Proof

Our claim is that given a compact matrix quantum group [math]G[/math], the formula [math]\Phi(\delta_i)=\sum_j\delta_j\otimes u_{ji}[/math] defines a morphism of algebras, which is a coaction map, leaving the trace invariant, precisely when the matrix [math]u=(u_{ij})[/math] is a magic corepresentation of [math]C(G)[/math]. Indeed, let us first determine when [math]\Phi[/math] is multiplicative. We have:

[[math]] \Phi(\delta_i)\Phi(\delta_k) =\sum_{jl}\delta_j\delta_l\otimes u_{ji}u_{lk} =\sum_j\delta_j\otimes u_{ji}u_{jk} [[/math]]

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

[[math]] \Phi(\delta_i\delta_k) =\delta_{ik}\Phi(\delta_i) =\delta_{ik}\sum_j\delta_j\otimes u_{ji} [[/math]]

Thus, the multiplicativity of [math]\Phi[/math] is equivalent to the following conditions:

[[math]] u_{ji}u_{jk}=\delta_{ik}u_{ji}\quad,\quad\forall i,j,k [[/math]]

Regarding now the unitality of [math]\Phi[/math], we have the following formula:

[[math]] \begin{eqnarray*} \Phi(1) &=&\sum_i\Phi(\delta_i)\\ &=&\sum_{ij}\delta_j\otimes u_{ji}\\ &=&\sum_j\delta_j\otimes\left(\sum_iu_{ji}\right) \end{eqnarray*} [[/math]]

Thus [math]\Phi[/math] is unital when the following conditions are satisfied:

[[math]] \sum_iu_{ji}=1\quad,\quad\forall i [[/math]]

Finally, the fact that [math]\Phi[/math] is a [math]*[/math]-morphism translates into:

[[math]] u_{ij}=u_{ij}^*\quad,\quad\forall i,j [[/math]]

Summing up, in order for [math]\Phi(\delta_i)=\sum_j\delta_j\otimes u_{ji}[/math] to be a morphism of [math]C^*[/math]-algebras, the elements [math]u_{ij}[/math] must be projections, summing up to 1 on each row of [math]u[/math]. Regarding now the preservation of the trace condition, observe that we have:

[[math]] (tr\otimes id)\Phi(\delta_i)=\frac{1}{N}\sum_ju_{ji} [[/math]]

Thus the trace is preserved precisely when the elements [math]u_{ij}[/math] sum up to 1 on each of the columns of [math]u[/math]. We conclude from this that [math]\Phi(\delta_i)=\sum_j\delta_j\otimes u_{ji}[/math] is a morphism of [math]C^*[/math]-algebras preserving the trace precisely when [math]u[/math] is magic, and since the coaction conditions on [math]\Phi[/math] are equivalent to the fact that [math]u[/math] must be a corepresentation, this finishes the proof of our claim. But this claim proves all the assertions in the statement.

■

As a perhaps quite surprising result now, also from Wang , we have:

Theorem

We have an embedding of compact quantum groups

[[math]] S_N\subset S_N^+ [[/math]]

given at the algebra level, [math]C(S_N^+)\to C(S_N)[/math], by the formula

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

and this embedding is an isomorphism at [math]N\leq3[/math], but not at [math]N\geq4[/math], where [math]S_N^+[/math] is non-classical, infinite compact quantum group.

Show Proof

The fact that we have indeed an embedding as above is clear from Proposition 9.1 and Theorem 9.2. Note that this follows as well from Proposition 9.3. Regarding now the second assertion, we can prove this in four steps, as follows:

\underline{Case [math]N=2[/math]}. The result here is trivial, the [math]2\times2[/math] magic matrices being by definition as follows, with [math]p[/math] being a projection:

[[math]] U=\begin{pmatrix}p&1-p\\1-p&p\end{pmatrix} [[/math]]

Indeed, this shows that the entries of a [math]2\times2[/math] magic matrix must pairwise commute, and so the algebra [math]C(S_2^+)[/math] follows to be commutative, which gives the result.

\underline{Case [math]N=3[/math]}. This is more tricky, and we present here a simple, recent proof, from . By using the same abstract argument as in the [math]N=2[/math] case, and by permuting rows and columns, it is enough to check that [math]u_{11},u_{22}[/math] commute. But this follows from:

[[math]] \begin{eqnarray*} u_{11}u_{22} &=&u_{11}u_{22}(u_{11}+u_{12}+u_{13})\\ &=&u_{11}u_{22}u_{11}+u_{11}u_{22}u_{13}\\ &=&u_{11}u_{22}u_{11}+u_{11}(1-u_{21}-u_{23})u_{13}\\ &=&u_{11}u_{22}u_{11} \end{eqnarray*} [[/math]]

Indeed, by applying the involution to this formula, we obtain from this that we have as well [math]u_{22}u_{11}=u_{11}u_{22}u_{11}[/math]. Thus we get [math]u_{11}u_{22}=u_{22}u_{11}[/math], as desired.

\underline{Case [math]N=4[/math]}. In order to prove our various claims about [math]S_4^+[/math], consider the following matrix, with [math]p,q[/math] being projections, on some infinite dimensional Hilbert space:

[[math]] U=\begin{pmatrix} p&1-p&0&0\\ 1-p&p&0&0\\ 0&0&q&1-q\\ 0&0&1-q&q \end{pmatrix} [[/math]]

This matrix is magic, and if we choose [math]p,q[/math] as for the algebra [math] \lt p,q \gt [/math] to be not commutative, and infinite dimensional, we conclude that [math]C(S_4^+)[/math] is not commutative and infinite dimensional as well, and in particular is not isomorphic to [math]C(S_4)[/math].

\underline{Case [math]N\geq5[/math]}. Here we can use the standard embedding [math]S_4^+\subset S_N^+[/math], obtained at the level of the corresponding magic matrices in the following way:

[[math]] u\to\begin{pmatrix}u&0\\ 0&1_{N-4}\end{pmatrix} [[/math]]

Indeed, with this embedding in hand, the fact that [math]S_4^+[/math] is a non-classical, infinite compact quantum group implies that [math]S_N^+[/math] with [math]N\geq5[/math] has these two properties as well.

■

9b. Representations

In order to study now [math]S_N^+[/math], we can use our various methods developed in chapters 1-4 above. Let us begin with some basic algebraic results, as follows:

Proposition

The quantum groups [math]S_N^+[/math] have the following properties:

• We have [math]S_N^+\,\hat{*}\,S_M^+\subset S_{N+M}^+[/math], for any [math]N,M[/math].
• In particular, we have an embedding [math]\widehat{D_\infty}\subset S_4^+[/math].
• [math]S_4\subset S_4^+[/math] are distinguished by their spinned diagonal tori.
• The half-classical version [math]S_N^*=S_N^+\cap O_N^*[/math] collapses to [math]S_N[/math].

Show Proof

These results are all elementary, the proofs being as follows:

(1) If we denote by [math]u,v[/math] the fundamental corepresentations of [math]C(S_N^+),C(S_M^+)[/math], the fundamental corepresentation of [math]C(S_N^+\,\hat{*}\,S_M^+)[/math] is by definition:

[[math]] w=\begin{pmatrix}u&0\\0&v\end{pmatrix} [[/math]]

But this matrix is magic, because both [math]u,v[/math] are magic. Thus by universality of [math]C(S_{N+M}^+)[/math] we obtain a quotient map as follows, as desired:

[[math]] C(S_{N+M}^+)\to C(S_N^+\,\hat{*}\,S_M^+) [[/math]]

(2) This result, which refines our [math]N=4[/math] trick from the proof of Theorem 9.4, follows from (1) with [math]N=M=2[/math]. Indeed, we have the following computation:

[[math]] \begin{eqnarray*} S_2^+\,\hat{*}\,S_2^+ &=&S_2\,\hat{*}\, S_2\\ &=&\mathbb Z_2\,\hat{*}\, \mathbb Z_2\\ &\simeq&\widehat{\mathbb Z_2}\,\hat{*}\, \widehat{\mathbb Z_2}\\ &=&\widehat{\mathbb Z_2*\mathbb Z_2}\\ &=&\widehat{D_\infty} \end{eqnarray*} [[/math]]

(3) As a first observation here, the quantum groups [math]S_4\subset S_4^+[/math] are not distinguished by their diagonal torus, which is [math]\{1\}[/math] for both of them. However, according to the general results of Woronowicz in , the group dual [math]\widehat{D_\infty}\subset S_4^+[/math] that we found in (2) must be a subgroup of the diagonal torus of the following compact quantum group, with the standard unitary representations being spinned by a certain unitary [math]F\in U_4[/math]:

[[math]] (S_4^+,FuF^*) [[/math]]

Now since this group dual [math]\widehat{D_\infty}[/math] is not classical, it cannot be a subgroup of the diagonal torus of [math](S_4,FuF^*)[/math]. Thus, the diagonal torus spinned by [math]F[/math] distinguishes [math]S_4\subset S_4^+[/math].

(4) Consider the following compact quantum group, with the intersection operation being taken inside [math]U_N^+[/math], whose coordinates satisfy [math]abc=cba[/math]:

[[math]] S_N^*=S_N^+\cap O_N^* [[/math]]

In order to prove that we have [math]S_N^*=S_N[/math], it is enough to prove that [math]S_N^*[/math] is classical. And here, we can use the fact that for a magic matrix, the entries in each row sum up to 1. Indeed, by making [math]c[/math] vary over a full row of [math]u[/math], we obtain [math]abc=cba\implies ab=ba[/math].

■

Summarizing, we have some advances on the quantum permutations, including a more conceptual explanation for our main observation so far, namely [math]S_4^+\neq S_4[/math]. At the representation theory level now, we have the following result, from :

Theorem

For the quantum groups [math]S_N,S_N^+[/math], the intertwining spaces for the tensor powers of the fundamental corepresentation [math]u=(u_{ij})[/math] are given by

[[math]] Hom(u^{\otimes k},u^{\otimes l})=span\left(T_\pi\Big|\pi\in D(k,l)\right) [[/math]]

with [math]D=P,NC[/math]. In other words, [math]S_N,S_N^+[/math] are easy, coming from the categories [math]P,NC[/math].

Show Proof

We use the Tannakian duality results from chapter 4 above:

(1) [math]S_N^+[/math]. According to Theorem 9.2, the algebra [math]C(S_N^+)[/math] appears as follows:

[[math]] C(S_N^+)=C(O_N^+)\Big\slash\Big \lt u={\rm magic}\Big \gt [[/math]]

Consider the one-block partition [math]\mu\in P(2,1)[/math]. The linear map associated to it is:

[[math]] T_\mu(e_i\otimes e_j)=\delta_{ij}e_i [[/math]]

We have [math]T_\mu=(\delta_{ijk})_{i,jk}[/math], and we obtain the following formula:

[[math]] (T_\mu u^{\otimes 2})_{i,jk} =\sum_{lm}(T_\mu)_{i,lm}(u^{\otimes 2})_{lm,jk} =u_{ij}u_{ik} [[/math]]

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

[[math]] (uT_\mu)_{i,jk} =\sum_lu_{il}(T_\mu)_{l,jk} =\delta_{jk}u_{ij} [[/math]]

Thus, the relation defining [math]S_N^+\subset O_N^+[/math] reformulates as follows:

[[math]] T_\mu\in Hom(u^{\otimes 2},u)\iff u_{ij}u_{ik}=\delta_{jk}u_{ij},\forall i,j,k [[/math]]

The condition on the right being equivalent to the magic condition, we obtain:

[[math]] C(S_N^+)=C(O_N^+)\Big\slash\Big \lt T_\mu\in Hom(u^{\otimes 2},u)\Big \gt [[/math]]

By using now the general easiness theory from chapter 7, we conclude that the quantum group [math]S_N^+[/math] is indeed easy, with the corresponding category of partitions being:

[[math]] D= \lt \mu \gt [[/math]]

But this latter category is [math]NC[/math], as one can see by “chopping” the noncrossing partitions into [math]\mu[/math]-shaped components. Thus, we are led to the conclusion in the statement.

(2) [math]S_N[/math]. Here the first part of the proof is similar, leading to the following formula:

[[math]] C(S_N)=C(O_N)\Big\slash\Big \lt T_\mu\in Hom(u^{\otimes 2},u)\Big \gt [[/math]]

But this shows that [math]S_N[/math] is easy, the corresponding category of partitions being:

[[math]] D = \lt \mu,P_2 \gt = \lt NC,P_2 \gt =P [[/math]]

Alternatively, this latter formula follows directly for the result for [math]S_N^+[/math] proved above, via [math]S_N=S_N^+\cap O_N[/math], and the functoriality results explained in chapter 7.

■

In order to discuss the representations of [math]S_N^+[/math], we will need linear independence results for the vectors [math]\xi_\pi[/math] associated to the partitions [math]\pi\in NC[/math]. This is something which is more technical than the previous results for pairings. Let us start with:

Proposition

We have a bijection [math]NC(k)\simeq NC_2(2k)[/math], constructed as follows:

• The application [math]NC(k)\to NC_2(2k)[/math] is the “fattening” one, obtained by doubling all the legs, and doubling all the strings as well.
• Its inverse [math]NC_2(2k)\to NC(k)[/math] is the “shrinking” application, obtained by collapsing pairs of consecutive neighbors.

Show Proof

This is something elementary, coming from the fact that both compositions of the operations in the statement are the identity, that we know from chapter 8.

■

The point now is that we have the following result, due to Jones :

Theorem

Consider the Temperley-Lieb algebra of index [math]N\geq4[/math], defined as

[[math]] TL_N(k)=span(NC_2(k,k)) [[/math]]

with product given by the rule [math]\bigcirc=N[/math], when concatenating.

• We have a representation [math]i:TL_N(k)\to B((\mathbb C^N)^{\otimes k})[/math], given by [math]\pi\to T_\pi[/math].
• [math]Tr(T_\pi)=N^{loops( \lt \pi \gt )}[/math], where [math]\pi\to \lt \pi \gt [/math] is the closing operation.
• The linear form [math]\tau=Tr\circ i:TL_N(k)\to\mathbb C[/math] is a faithful positive trace.
• The representation [math]i:TL_N(k)\to B((\mathbb C^N)^{\otimes k})[/math] is faithful.

In particular, the vectors [math]\left\{\xi_\pi|\pi\in NC(k)\right\}\subset(\mathbb C^N)^{\otimes k}[/math] are linearly independent.

Show Proof

All this is quite standard, but advanced, the idea being as follows:

(1) This is clear from the categorical properties of [math]\pi\to T_\pi[/math].

(2) This follows indeed from the following computation:

[[math]] \begin{eqnarray*} Tr(T_\pi) &=&\sum_{i_1\ldots i_k}\delta_\pi\binom{i_1\ldots i_k}{i_1\ldots i_k}\\ &=&\#\left\{i_1,\ldots,i_k\in\{1,\ldots,N\}\Big|\ker\binom{i_1\ldots i_k}{i_1\ldots i_k}\geq\pi\right\}\\ &=&N^{loops( \lt \pi \gt )} \end{eqnarray*} [[/math]]

(3) The traciality of [math]\tau[/math] is clear. Regarding now the faithfulness, this is something well-known, and we refer here to Jones' paper .

(4) This follows from (3) above, via a standard positivity argument. As for the last assertion, this follows from (4), by fattening the partitions.

■

Alternatively, the linear independence of the vectors [math]\xi_\pi[/math] with [math]\pi\in NC(k)[/math], that we will need in what follows, comes the following result of Di Francesco :

Theorem

The determinant of the Gram matrix for [math]S_N^+[/math] is given by

[[math]] \det(G_{kN})=(\sqrt{N})^{a_k}\prod_{r=1}^kP_r(\sqrt{N})^{d_{kr}} [[/math]]

where [math]P_r[/math] are the Chebycheff polynomials, given by

[[math]] P_0=1\quad,\quad P_1=X\quad,\quad P_{r+1}=XP_r-P_{r-1} [[/math]]

and [math]d_{kr}=f_{kr}-f_{k,r+1}[/math], with [math]f_{kr}[/math] being the following numbers, depending on [math]k,r\in\mathbb Z[/math],

[[math]] f_{kr}=\binom{2k}{k-r}-\binom{2k}{k-r-1} [[/math]]

with the convention [math]f_{kr}=0[/math] for [math]k\notin\mathbb Z[/math], and where [math]a_k[/math] are the following numbers:

[[math]] a_k=\sum_{\pi\in P(k)}(2|\pi|-k) [[/math]]

In particular, [math]\left\{\xi_\pi|\pi\in NC(k)\right\}\subset(\mathbb C^N)^{\otimes k}[/math] are linearly independent at [math]N\geq4[/math].

Show Proof

This looks very similar to the result for [math]O_N^+[/math] from chapter 8, also due to Di Francesco, and can be in fact deduced from that result, by shrinking the partitions, according to the shrinking formulae explained in chapter 8. See .

■

Long story short, we need to know that the vectors [math]\xi_\pi[/math] with [math]\pi\in NC(k)[/math] are linearly independent at [math]N\geq4[/math], and this is something non-trivial, but proofs of this fact do exist, coming from the work of Jones and Di Francesco , so the result holds. There are some other proofs as well, but none is any simpler. Be said in passing, I can feel that you might say at this point that you would like a full, simple proof, we're not mathematical physicists, right. To which I would answer, welcome to mathematical physics.

We can work out now the representation theory of [math]S_N^+[/math], as follows:

Theorem

The quantum groups [math]S_N^+[/math] with [math]N\geq4[/math] have the following properties:

• The moments of the main character are the Catalan numbers:
  [[math]] \int_{S_N^+}\chi^k=C_k [[/math]]
• The fusion rules for representations are as follows, exactly as for [math]SO_3[/math]:
  [[math]] r_k\otimes r_l=r_{|k-l|}+r_{|k-l|+1}+\ldots+r_{k+l} [[/math]]
• The dimensions of the irreducible representations are given by
  [[math]] \dim(r_k)=\frac{q^{k+1}-q^{-k}}{q-1} [[/math]]
  where [math]q,q^{-1}[/math] are the roots of [math]X^2-(N-2)X+1=0[/math].

Show Proof

The proof, from , based on Theorems 9.8 or 9.9, goes as follows:

(1) We have indeed the following computation, coming from the [math]SU_2[/math] computations from chapter 5, and from Theorem 9.6, Proposition 9.7, and Theorems 9.8 or 9.9:

[[math]] \begin{eqnarray*} \int_{S_N^+}\chi^k &=&\dim(Fix(u^{\otimes k}))\\ &=&|NC(k)|\\ &=&|NC_2(2k)|\\ &=&C_k \end{eqnarray*} [[/math]]

(2) This is standard, by using the formula in (1), and the known theory of [math]SO_3[/math]. Let [math]A=span(\chi_k|k\in\mathbb N)[/math] be the algebra of characters of [math]SO_3[/math]. We can define a morphism as follows, where [math]f[/math] is the character of the fundamental representation of [math]S_N^+[/math]:

[[math]] \Psi:A\to C(S_N^+)\quad,\quad \chi_1\to f-1 [[/math]]

The elements [math]f_k=\Psi(\chi_k)[/math] verify then the following formulae:

[[math]] f_kf_l=f_{|k-l|}+f_{|k-l|+1}+\ldots+f_{k+l} [[/math]]

We prove now by recurrence that each [math]f_k[/math] is the character of an irreducible corepresentation [math]r_k[/math] of [math]C(S_N^+)[/math], non-equivalent to [math]r_0,\ldots,r_{k-1}[/math]. At [math]k=0,1[/math] this is clear, so assume that the result holds at [math]k-1[/math]. By integrating characters we have, exactly as for [math]SO_3[/math]:

[[math]] r_{k-2},r_{k-1}\subset r_{k-1}\otimes r_1 [[/math]]

Thus there exists a certain corepresentation [math]r_k[/math] such that:

[[math]] r_{k-1}\otimes r_1=r_{k-2}+r_{k-1}+r_k [[/math]]

Once again by integrating characters, we conclude that [math]r_k[/math] is irreducible, and non-equivalent to [math]r_1,\ldots,r_{k-1}[/math], as for [math]SO_3[/math], which proves our claim. Finally, since any irreducible representation of [math]S_N^+[/math] must appear in some tensor power of [math]u[/math], and we have a formula for decomposing each [math]u^{\otimes k}[/math] into sums of representations [math]r_l[/math], we conclude that these representations [math]r_l[/math] are all the irreducible representations of [math]S_N^+[/math].

(3) From the Clebsch-Gordan rules we have, in particular:

[[math]] r_kr_1=r_{k-1}+r_k+r_{k+1} [[/math]]

We are therefore led to a recurrence, and the initial data being [math]\dim(r_0)=1[/math] and [math]\dim(r_1)=N-1=q+1+q^{-1}[/math], we are led to the following formula:

[[math]] \dim(r_k)=q^k+q^{k-1}+\ldots+q^{1-k}+q^{-k} [[/math]]

In more compact form, this gives the formula in the statement.

■

The above result is quite surprising, and raises a massive number of questions. We would like to better understand the relation with [math]SO_3[/math], and more generally see what happens at values [math]N=n^2[/math] with [math]n\geq2[/math], and also compute the law of [math]\chi[/math], and so on.

9c. Twisted extension

As a first topic to be discussed, one way of understanding the relation with [math]SO_3[/math] comes from noncommutative geometry considerations. We recall that, according to the general theory from chapter 1, each finite dimensional [math]C^*[/math]-algebra [math]A[/math] can be written as [math]A=C(F)[/math], with [math]F[/math] being a “finite quantum space”. To be more precise, we have:

Definition

A finite quantum space [math]F[/math] is the abstract dual of a finite dimensional [math]C^*[/math]-algebra [math]A[/math], according to the following formula:

[[math]] C(F)=A [[/math]]

The number of elements of such a space is by definition the number [math]|F|=\dim A[/math]. By decomposing the algebra [math]A[/math], we have a formula of the following type:

[[math]] C(F)=M_{n_1}(\mathbb C)\oplus\ldots\oplus M_{n_k}(\mathbb C) [[/math]]

With [math]n_1=\ldots=n_k=1[/math] we obtain in this way the space [math]F=\{1,\ldots,k\}[/math]. Also, when [math]k=1[/math] the equation is [math]C(F)=M_n(\mathbb C)[/math], and the solution will be denoted [math]F=M_n[/math].

In order to talk about the quantum symmetry group [math]S_F^+[/math], we must use universal coactions. As in Proposition 9.3, we must endow our space [math]F[/math] with its counting measure:

Definition

We endow each finite quantum space [math]F[/math] with its counting measure, corresponding as the algebraic level to the integration functional

[[math]] tr:C(F)\to B(l^2(F))\to\mathbb C [[/math]]

obtained by applying the regular representation, and then the normalized matrix trace.

To be more precise, consider the algebra [math]A=C(F)[/math], which is by definition finite dimensional. We can make act [math]A[/math] on itself, by left multiplication:

[[math]] \pi:A\to\mathcal L(A)\quad,\quad a\to(b\to ab) [[/math]]

The target of [math]\pi[/math] being a matrix algebra, [math]\mathcal L(A)\simeq M_N(\mathbb C)[/math] with [math]N=\dim A[/math], we can further compose with the normalized matrix trace, and we obtain [math]tr[/math]:

[[math]] tr=\frac{1}{N}\,Tr\circ\pi [[/math]]

Let us study the quantum group actions [math]G\curvearrowright F[/math]. We denote by [math]\mu,\eta[/math] the multiplication and unit map of the algebra [math]C(F)[/math]. Following , , we first have:

Proposition

Consider a linear map [math]\Phi:C(F)\to C(F)\otimes C(G)[/math], written as

[[math]] \Phi(e_i)=\sum_je_j\otimes u_{ji} [[/math]]

with [math]\{e_i\}[/math] being a linear space basis of [math]C(F)[/math], orthonormal with respect to [math]tr[/math].

• [math]\Phi[/math] is a linear space coaction [math]\iff[/math] [math]u[/math] is a corepresentation.
• [math]\Phi[/math] is multiplicative [math]\iff[/math] [math]\mu\in Hom(u^{\otimes 2},u)[/math].
• [math]\Phi[/math] is unital [math]\iff[/math] [math]\eta\in Hom(1,u)[/math].
• [math]\Phi[/math] leaves invariant [math]tr[/math] [math]\iff[/math] [math]\eta\in Hom(1,u^*)[/math].
• If these conditions hold, [math]\Phi[/math] is involutive [math]\iff[/math] [math]u[/math] is unitary.

Show Proof

This is a bit similar to the proof of Proposition 9.3, as follows:

(1) There are two axioms to be processed here. First, we have:

[[math]] \begin{eqnarray*} (id\otimes\Delta)\Phi=(\Phi\otimes id)\Phi &\iff&\sum_je_j\otimes\Delta(u_{ji})=\sum_k\Phi(e_k)\otimes u_{ki}\\ &\iff&\sum_je_j\otimes\Delta(u_{ji})=\sum_{jk}e_j\otimes u_{jk}\otimes u_{ki}\\ &\iff&\Delta(u_{ji})=\sum_ku_{jk}\otimes u_{ki} \end{eqnarray*} [[/math]]

As for the axiom involving the counit, here we have as well, as desired:

[[math]] \begin{eqnarray*} (id\otimes\varepsilon)\Phi=id &\iff&\sum_j\varepsilon(u_{ji})e_j=e_i\\ &\iff&\varepsilon(u_{ji})=\delta_{ji} \end{eqnarray*} [[/math]]

(2) We have the following formula:

[[math]] \begin{eqnarray*} \Phi(e_i) &=&\sum_je_j\otimes u_{ji}\\ &=&\left(\sum_{ij}e_{ji}\otimes u_{ji}\right)(e_i\otimes 1)\\ &=&u(e_i\otimes 1) \end{eqnarray*} [[/math]]

By using this formula, we obtain the following identity:

[[math]] \begin{eqnarray*} \Phi(e_ie_k) &=&u(e_ie_k\otimes 1)\\ &=&u(\mu\otimes id)(e_i\otimes e_k\otimes 1) \end{eqnarray*} [[/math]]

On the other hand, we have as well the following identity, as desired:

[[math]] \begin{eqnarray*} \Phi(e_i)\Phi(e_k) &=&\sum_{jl}e_je_l\otimes u_{ji}u_{lk}\\ &=&(\mu\otimes id)\sum_{jl}e_j\otimes e_l\otimes u_{ji}u_{lk}\\\ &=&(\mu\otimes id)\left(\sum_{ijkl}e_{ji}\otimes e_{lk}\otimes u_{ji}u_{lk}\right)(e_i\otimes e_k\otimes 1)\\ &=&(\mu\otimes id)u^{\otimes 2}(e_i\otimes e_k\otimes 1) \end{eqnarray*} [[/math]]

(3) The formula [math]\Phi(e_i)=u(e_i\otimes1)[/math] found above gives by linearity [math]\Phi(1)=u(1\otimes1)[/math], which shows that [math]\Phi[/math] is unital precisely when [math]u(1\otimes1)=1\otimes1[/math], as desired.

(4) This follows from the following computation, by applying the involution:

[[math]] \begin{eqnarray*} (tr\otimes id)\Phi(e_i)=tr(e_i)1 &\iff&\sum_jtr(e_j)u_{ji}=tr(e_i)1\\ &\iff&\sum_ju_{ji}^*1_j=1_i\\ &\iff&(u^*1)_i=1_i\\ &\iff&u^*1=1 \end{eqnarray*} [[/math]]

(5) Assuming that (1-4) are satisfied, and that [math]\Phi[/math] is involutive, we have:

[[math]] \begin{eqnarray*} (u^*u)_{ik} &=&\sum_lu_{li}^*u_{lk}\\ &=&\sum_{jl}tr(e_j^*e_l)u_{ji}^*u_{lk}\\ &=&(tr\otimes id)\sum_{jl}e_j^*e_l\otimes u_{ji}^*u_{lk}\\ &=&(tr\otimes id)(\Phi(e_i)^*\Phi(e_k))\\ &=&(tr\otimes id)\Phi(e_i^*e_k)\\ &=&tr(e_i^*e_k)1\\ &=&\delta_{ik} \end{eqnarray*} [[/math]]

Thus [math]u^*u=1[/math], and since we know from (1) that [math]u[/math] is a corepresentation, it follows that [math]u[/math] is unitary. The proof of the converse is standard too, by using similar tricks.

■

Still following , , we have the following result, extending the basic theory of [math]S_N^+[/math] to the present finite noncommutative space setting:

Theorem

Given a finite quantum space [math]F[/math], there is a universal compact quantum group [math]S_F^+[/math] acting on [math]F[/math], leaving the counting measure invariant. We have

[[math]] C(S_F^+)=C(U_N^+)\Big/\Big \lt \mu\in Hom(u^{\otimes2},u),\eta\in Fix(u)\Big \gt [[/math]]

where [math]N=|F|[/math] and where [math]\mu,\eta[/math] are the multiplication and unit maps of [math]C(F)[/math]. For [math]F=\{1,\ldots,N\}[/math] we have [math]S_F^+=S_N^+[/math]. Also, for the space [math]F=M_2[/math] we have [math]S_F^+=SO_3[/math].

Show Proof

This result is from , the idea being as follows:

(1) This follows from Proposition 9.13 above, by using the standard fact that the complex conjugate of a corepresentation is a corepresentation too.

(2) Regarding now the main example, for [math]F=\{1,\ldots,N\}[/math] we obtain indeed the quantum permutation group [math]S_N^+[/math], due to the abstract result in Proposition 9.3 above.

(3) In order to do now the computation for [math]F=M_2[/math], we use some standard facts about [math]SU_2,SO_3[/math]. We have an action by conjugation [math]SU_2\curvearrowright M_2(\mathbb C)[/math], and this action produces, via the canonical quotient map [math]SU_2\to SO_3[/math], an action [math]SO_3\curvearrowright M_2(\mathbb C)[/math]. On the other hand, it is routine to check, by using arguments like those in the proof of Theorem 9.4 at [math]N=2,3[/math], that any action [math]G\curvearrowright M_2(\mathbb C)[/math] must come from a classical group. We conclude that the action [math]SO_3\curvearrowright M_2(\mathbb C)[/math] is universal, as claimed.

■

Regarding now the representation theory of these generalized quantum permutation groups [math]S_F^+[/math], the result here, from , is very similar to the one for [math]S_N^+[/math], as follows:

Theorem

The quantum groups [math]S_F^+[/math] have the following properties:

• The associated Tannakian categories are [math]TL_N[/math], with [math]N=|F|[/math].
• The main character follows the Marchenko-Pastur law [math]\pi_1[/math], when [math]N\geq4[/math].
• The fusion rules for [math]S_F^+[/math] with [math]|F|\geq4[/math] are the same as for [math]SO_3[/math].

Show Proof

Once again this result is from , the idea being as follows:

(1) Our first claim is that the fundamental representation is equivalent to its adjoint, [math]u\sim\bar{u}[/math]. Indeed, let us go back to the coaction formula from Proposition 9.13:

[[math]] \Phi(e_i)=\sum_je_j\otimes u_{ji} [[/math]]

We can pick our orthogonal basis [math]\{e_i\}[/math] to be the stadard multimatrix basis of [math]C(F)[/math], so that we have [math]e_i^*=e_{i^*}[/math], for a certain involution [math]i\to i^*[/math] on the index set. With this convention made, by conjugating the above formula of [math]\Phi(e_i)[/math], we obtain:

[[math]] \Phi(e_{i^*})=\sum_je_{j^*}\otimes u_{ji}^* [[/math]]

Now by interchanging [math]i\leftrightarrow i^*[/math] and [math]j\leftrightarrow j^*[/math], this latter formula reads:

[[math]] \Phi(e_i)=\sum_je_j\otimes u_{j^*i^*}^* [[/math]]

We therefore conclude, by comparing with the original formula, that we have:

[[math]] u_{ji}^*=u_{j^*i^*} [[/math]]

But this shows that we have an equivalence [math]u\sim\bar{u}[/math], as claimed. Now with this result in hand, the proof goes as for the proof for [math]S_N^+[/math]. To be more precise, the result follows from the fact that the multiplication and unit of any complex algebra, and in particular of [math]C(F)[/math], can be modelled by the following two diagrams:

[[math]] m=|\cup|\quad,\quad u=\cap [[/math]]

Indeed, this is certainly true algebrically, and this is something well-known. As in what regards the [math]*[/math]-structure, things here are fine too, because our choice for the trace leads to the following formula, which must be satisfied as well:

[[math]] \mu\mu^*=N\cdot id [[/math]]

But the above diagrams [math]m,u[/math] generate the Temperley-Lieb algebra [math]TL_N[/math], as stated.

(2) The proof here is exactly as for [math]S_N^+[/math], by using moments. To be more precise, according to (1) these moments are the Catalan numbers, which are the moments of [math]\pi_1[/math].

(3) Once again same proof as for [math]S_N^+[/math], by using the fact that the moments of [math]\chi[/math] are the Catalan numbers, which naturally leads to the Clebsch-Gordan rules.

■

It is quite clear now that our present formalism, and the above results, provide altogether a good and conceptual explanation for our [math]SO_3[/math] result regarding [math]S_N^+[/math]. To be more precise, we can merge and reformulate the above results in the following way:

Theorem

The quantun groups [math]S_F^+[/math] have the following properties:

• For [math]F=\{1,\ldots,N\}[/math] we have [math]S_F^+=S_N^+[/math].
• For the space [math]F=M_N[/math] we have [math]S_F^+=PO_N^+=PU_N^+[/math].
• In particular, for the space [math]F=M_2[/math] we have [math]S_F^+=SO_3[/math].
• The fusion rules for [math]S_F^+[/math] with [math]|F|\geq4[/math] are independent of [math]F[/math].
• Thus, the fusion rules for [math]S_F^+[/math] with [math]|F|\geq4[/math] are the same as for [math]SO_3[/math].

Show Proof

This is basically a compact form of what has been said above, with a new result added, and with some technicalities left aside:

(1) This is something that we know from Theorem 9.14.

(2) We know from chapter 4 that the inclusion [math]PO_N^+\subset PU_N^+[/math] is an isomorphism, with this coming from the formula [math]\widetilde{O}_N^+=U_N^+[/math], but we will actually reprove this result. Consider indeed the standard vector space action [math]U_N^+\curvearrowright\mathbb C^N[/math], and then its adjoint action [math]PU_N^+\curvearrowright M_N(\mathbb C)[/math]. By universality of [math]S_{M_N}^+[/math], we have inclusions as follows:

[[math]] PO_N^+\subset PU_N^+\subset S_{M_N}^+ [[/math]]

On the other hand, the main character of [math]O_N^+[/math] with [math]N\geq2[/math] being semicircular, the main character of [math]PO_N^+[/math] must be Marchenko-Pastur. Thus the inclusion [math]PO_N^+\subset S_{M_N}^+[/math] has the property that it keeps fixed the law of main character, and by Peter-Weyl theory we conclude that this inclusion must be an isomorphism, as desired.

(3) This is something that we know from Theorem 9.14, and that can be deduced as well from (2), by using the formula [math]PO_2^+=SO_3[/math], which is something elementary.

(4) This is something that we know from Theorem 9.15.

(5) This follows from (3,4), as already pointed out in Theorem 9.15.

■

All this is certainly conceptual, but perhaps a bit too abstract. At [math]N=4[/math] we can formulate a more concrete result on the subject, by using the following construction:

Definition

[math]C(SO_3^{-1})[/math] is the universal [math]C^*[/math]-algebra generated by the entries of a [math]3\times 3[/math] orthogonal matrix [math]a=(a_{ij})[/math], with the following relations:

• Skew-commutation: [math]a_{ij}a_{kl}=\pm a_{kl}a_{ij}[/math], with sign [math]+[/math] if [math]i\neq k,j\neq l[/math], and [math]-[/math] otherwise.
• Twisted determinant condition: [math]\Sigma_{\sigma\in S_3}a_{1\sigma(1)}a_{2\sigma(2)}a_{3\sigma(3)}=1[/math].

Observe the similarity with the twisting constructions from chapter 7. However, [math]SO_3[/math] being not easy, we are not exactly in the Schur-Weyl twisting framework from there.

Our first task would be to prove that [math]C(SO_3^{-1})[/math] is a Woronowicz algebra. This is of course possible, by doing some computations, but we will not need to do these computations, because the result follows from the following theorem, from :

Theorem

We have an isomorphism of compact quantum groups

[[math]] S_4^+=SO_3^{-1} [[/math]]

given by the Fourier transform over the Klein group [math]K=\mathbb Z_2\times\mathbb Z_2[/math].

Show Proof

Consider indeed the matrix [math]a^+=diag(1,a)[/math], corresponding to the action of [math]SO_3^{-1}[/math] on [math]\mathbb C^4[/math], and apply to it the Fourier transform over the Klein group [math]K=\mathbb Z_2\times\mathbb Z_2[/math]:

[[math]] u= \frac{1}{4} \begin{pmatrix} 1&1&1&1\\ 1&-1&-1&1\\ 1&-1&1&-1\\ 1&1&-1&-1 \end{pmatrix} \begin{pmatrix} 1&0&0&0\\ 0&a_{11}&a_{12}&a_{13}\\ 0&a_{21}&a_{22}&a_{23}\\ 0&a_{31}&a_{32}&a_{33} \end{pmatrix} \begin{pmatrix} 1&1&1&1\\ 1&-1&-1&1\\ 1&-1&1&-1\\ 1&1&-1&-1 \end{pmatrix} [[/math]]

It is routine to check that this matrix is magic, and vice versa, i.e. that the Fourier transform over [math]K[/math] converts the relations in Definition 9.17 into the magic relations in Definition 9.1. Thus, we obtain the identification from the statement.

■

Yet another extension of Theorem 9.10, which is however quite technical, comes by looking at the general case [math]N=n^2[/math], with [math]n\geq2[/math]. It is possible indeed to complement Theorem 9.16 above with a general twisting result of the following type:

[[math]] G^+(\widehat{F}_\sigma)=G^+(\widehat{F})^\sigma [[/math]]

To be more precise, this formula is valid indeed, for any finite group [math]F[/math] and any [math]2[/math]-cocycle [math]\sigma[/math] on it. In the case [math]F=\mathbb Z_n^2[/math] with Fourier cocycle on it, this leads to the conclusion that [math]PO_n^+[/math] appears as a cocycle twist of [math]S_{n^2}^+[/math]. For details here, we refer to .

We have the following interesting probabilistic fact, from as well:

Theorem

The following families of variables have the same joint law,

• [math]\{u_{ij}^2\}\in C(O_n^+)[/math],
• [math]\{X_{ij}=\frac{1}{n}\sum_{ab}p_{ia,jb}\}\in C(S_{n^2}^+)[/math],

where [math]u=(u_{ij})[/math] and [math]p=(p_{ia,jb})[/math] are the corresponding fundamental corepresentations.

Show Proof

As explained in , this result can be obtained via the above twisting methods. An alternative approach, also from , is by using the Weingarten formula for our two quantum groups, and the shrinking operation [math]\pi\to\pi'[/math]. Indeed, we have:

[[math]] \begin{eqnarray*} \int_{O_n^+}u_{ij}^{2k}&=&\sum_{\pi,\sigma\in NC_2(2k)}W_{2k,n}(\pi,\sigma)\\ \int_{S_{n^2}^+}X_{ij}^k&=&\sum_{\pi,\sigma\in NC_2(2k)}n^{|\pi'|+|\sigma'|-k}W_{k,n^2}(\pi',\sigma') \end{eqnarray*} [[/math]]

The point now is that, via the shrinking operation [math]\pi\to\pi'[/math], the Gram matrices of [math]NC_2(2k),NC(k)[/math] are related by the following formula, [math]\Delta_{kn}[/math] being the diagonal of [math]G_{kn}[/math]:

[[math]] G_{2k,n}(\pi,\sigma)=n^k(\Delta_{kn}^{-1}G_{k,n^2}\Delta_{kn}^{-1})(\pi',\sigma') [[/math]]

Thus in our moment formulae above the summands coincide, and so the moments are equal, as desired. The proof in general, dealing with joint moments, is similar.

■

The above result is quite interesting, because it makes a connection between free hyperspherical and free hypergeometric laws. We refer here to , .

9d. Poisson laws

Let us go back now to our main result so far, namely Theorem 9.10, and further build on that, with probabilistic results. We have the following result:

Theorem

The spectral measure of the main character of [math]S_N^+[/math] with [math]N\geq4[/math] is the Marchenko-Pastur law of parameter [math]1[/math], having the following density:

[[math]] \pi_1=\frac{1}{2\pi}\sqrt{4x^{-1}-1}dx [[/math]]

Also, [math]S_4^+[/math] is coamenable, and [math]S_N^+[/math] with [math]N\geq5[/math] is not coamenable.

Show Proof

Here the first assertion follows from the following formula, which can be established by doing some calculus, and more specifically by setting [math]x=4\sin^2t[/math]:

[[math]] \frac{1}{2\pi}\int_0^4\sqrt{1-4x^{-1}}x^kdx=C_k [[/math]]

As for the second assertion, this follows from this, which shows that the spectrum of the main character is [math][0,4][/math], and from the Kesten criterion from chapter 3.

■

Our next purpose will be that of understanding, probabilistically speaking, the liberation operation [math]S_N\to S_N^+[/math]. In what regards [math]S_N[/math], we have the following basic result:

Theorem

Consider the symmetric group [math]S_N[/math], regarded as a compact group of matrices, [math]S_N\subset O_N[/math], via the standard permutation matrices.

• The main character [math]\chi\in C(S_N)[/math], defined as usual as [math]\chi=\sum_iu_{ii}[/math], counts the number of fixed points, [math]\chi(\sigma)=\#\{i|\sigma(i)=i\}[/math].
• The probability for a permutation [math]\sigma\in S_N[/math] to be a derangement, meaning to have no fixed points at all, becomes, with [math]N\to\infty[/math], equal to [math]1/e[/math].
• The law of the main character [math]\chi\in C(S_N)[/math] becomes, with [math]N\to\infty[/math], a Poisson law of parameter [math]1[/math], with respect to the counting measure.

Show Proof

This is something very classical, and beautiful, as follows:

(1) We have indeed the following computation:

[[math]] \chi(\sigma) =\sum_iu_{ii}(\sigma) =\sum_i\delta_{\sigma(i)i} =\#\left\{i\big|\sigma(i)=i\right\} [[/math]]

(2) This is best viewed by using the inclusion-exclusion principle. Let us set:

[[math]] S_N^{i_1\ldots i_k}=\left\{\sigma\in S_N\Big|\sigma(i_1)=i_1,\ldots,\sigma(i_k)=i_k\right\} [[/math]]

By using the inclusion-exclusion principle, we have, as desired:

[[math]] \begin{eqnarray*} \mathbb P(\chi=0) &=&\frac{1}{N!}|(S_1\cup\ldots\cup S_N)^c|\\ &=&\frac{1}{N!}\left(|S_N|-\sum_i|S_N^i|+\sum_{i \lt j}|S_N^{ij}|-\ldots+(-1)^N\sum_{i_1 \lt \ldots \lt i_N}|S_N^{i_1\ldots i_N}|\right)\\ &=&\frac{1}{N!}\sum_{k=0}^N(-1)^k\binom{N}{k}(N-k)!\\ &=&1-\frac{1}{1!}+\frac{1}{2!}-\ldots+(-1)^{N-1}\frac{1}{(N-1)!}+(-1)^N\frac{1}{N!} \end{eqnarray*} [[/math]]

(3) This follows by generalizing the computation in (2). To be more precise, a similar application of the inclusion-exclusion principle gives the following formula:

[[math]] \lim_{N\to\infty}\mathbb P(\chi=k)=\frac{1}{k!e} [[/math]]

Thus, we obtain in the limit a Poisson law, as stated.

■

In order to talk about free analogues of this, we will need some theory:

Theorem

The following Poisson type limits converge, for any [math]t \gt 0[/math],

[[math]] p_t=\lim_{n\to\infty}\left(\left(1-\frac{t}{n}\right)\delta_0+\frac{t}{n}\delta_1\right)^{*n}\quad,\quad \pi_t=\lim_{n\to\infty}\left(\left(1-\frac{t}{n}\right)\delta_0+\frac{t}{n}\delta_1\right)^{\boxplus n} [[/math]]

the limiting measures being the Poisson law [math]p_t[/math], and the Marchenko-Pastur law [math]\pi_t[/math],

[[math]] p_t=\frac{1}{e^t}\sum_{k=0}^\infty\frac{t^k\delta_k}{k!}\quad,\quad \pi_t=\max(1-t,0)\delta_0+\frac{\sqrt{4t-(x-1-t)^2}}{2\pi x}\,dx [[/math]]

whose moments are given by [math]M_k(p_t)=\sum_{\pi\in D(k)}t^{|\pi|}[/math], with [math]D=P,NC[/math].

Show Proof

This is something standard, which follows by using either [math]\log F,R[/math] and calculus, or classical and free cumulants. The point indeed is that the limiting measures must be those having classical and free cumulants [math]t,t,t,\ldots[/math] But this gives all the assertions, the density computations being standard. We refer here to , , , , but we will be back to this in chapter 10 below, with full details, directly in a more general setting.

■

We can now formulate a conceptual result about [math]S_N\to S_N^+[/math], as follows:

Theorem

The law of the main character [math]\chi_u[/math] is as follows:

• For [math]S_N[/math] with [math]N\to\infty[/math] we obtain a Poisson law [math]p_1[/math].
• For [math]S_N^+[/math] with [math]N\geq4[/math] we obtain a Marchenko-Pastur, or free Poisson law [math]\pi_1[/math].

In addition, these laws are related by the Bercovici-Pata correspondence.

Show Proof

This follows indeed from the computations that we have, from Theorem 9.20 and Theorem 9.21, by using the various theoretical results from Theorem 9.22.

■

As in the continuous case, our purpose now will be that of extending this result to the truncated characters. In order to discuss the classical case, we first have:

Proposition

Consider the symmetric group [math]S_N[/math], together with its standard matrix coordinates [math]u_{ij}=\chi(\sigma\in S_N|\sigma(j)=i)[/math]. We have the formula

[[math]] \int_{S_N}u_{i_1j_1}\ldots u_{i_kj_k}=\begin{cases} \frac{(N-|\ker i|)!}{N!}&{\rm if}\ \ker i=\ker j\\ 0&{\rm otherwise} \end{cases} [[/math]]

where [math]\ker i[/math] denotes as usual the partition of [math]\{1,\ldots,k\}[/math] whose blocks collect the equal indices of [math]i[/math], and where [math]|.|[/math] denotes the number of blocks.

Show Proof

According to the definition of [math]u_{ij}[/math], the integrals in the statement are given by:

[[math]] \int_{S_N}u_{i_1j_1}\ldots u_{i_kj_k}=\frac{1}{N!}\#\left\{\sigma\in S_N\Big|\sigma(j_1)=i_1,\ldots,\sigma(j_k)=i_k\right\} [[/math]]

The existence of [math]\sigma\in S_N[/math] as above requires [math]i_m=i_n\iff j_m=j_n[/math]. Thus, the integral vanishes when [math]\ker i\neq\ker j[/math]. As for the case [math]\ker i=\ker j[/math], if we denote by [math]b\in\{1,\ldots,k\}[/math] the number of blocks of this partition, we have [math]N-b[/math] points to be sent bijectively to [math]N-b[/math] points, and so [math](N-b)![/math] solutions, and the integral is [math]\frac{(N-b)!}{N!}[/math], as claimed.

■

We can now compute the laws of truncated characters, and we obtain:

Proposition

For the symmetric group [math]S_N\subset O_N[/math], regarded as a compact group of matrices, [math]S_N\subset O_N[/math], via the standard permutation matrices, the truncated character

[[math]] \chi_t=\sum_{i=1}^{[tN]}u_{ii} [[/math]]

counts the number of fixed points among [math]\{1,\ldots,[tN]\}[/math], and its law with respect to the counting measure becomes, with [math]N\to\infty[/math], a Poisson law of parameter [math]t[/math].

Show Proof

With [math]S_{k,b}[/math] being the Stirling numbers, we have:

[[math]] \begin{eqnarray*} \int_{S_N}\chi_t^k &=&\sum_{i_1\ldots i_k=1}^{[tN]}\int_{S_N}u_{i_1i_1}\ldots u_{i_ki_k}\\ &=&\sum_{\pi\in P_k}\frac{[tN]!}{([tN]-|\pi|!)}\cdot\frac{(N-|\pi|!)}{N!}\\ &=&\sum_{b=1}^{[tN]}\frac{[tN]!}{([tN]-b)!}\cdot\frac{(N-b)!}{N!}\cdot S_{k,b} \end{eqnarray*} [[/math]]

In particular with [math]N\to\infty[/math] we obtain the following formula:

[[math]] \lim_{N\to\infty}\int_{S_N}\chi_t^k=\sum_{b=1}^kS_{k,b}t^b [[/math]]

But this is a Poisson([math]t[/math]) moment, and so we are done.

■

We can now finish our computations, and generalize Theorem 9.23, as follows:

Theorem

The laws of truncated characters [math]\chi_t=\sum_{i=1}^{[tN]}u_{ii}[/math] are as follows:

• For [math]S_N[/math] with [math]N\to\infty[/math] we obtain a Poisson law [math]p_t[/math].
• For [math]S_N^+[/math] with [math]N\to\infty[/math] we obtain a free Poisson law [math]\pi_t[/math].

In addition, these laws are related by the Bercovici-Pata correspondence.

Show Proof

This follows from the above results:

(1) This is something that we already know, from Proposition 9.25.

(2) This is something that we know so far only at [math]t=1[/math], from Theorem 9.23. In order to deal with the general [math]t\in(0,1][/math] case, we can use the same method as for the orthogonal and unitary quantum groups, from chapter 8, and we obtain the following moments:

[[math]] M_k=\sum_{\pi\in NC(k)}t^{|\pi|} [[/math]]

But these numbers being the moments of the free Poisson law of parameter [math]t[/math], as explained in Theorem 9.22 above, we obtain the result. See .

■

Summarizing, the liberation operation [math]S_N\to S_N^+[/math] has many common features with the liberation operations [math]O_N\to O_N^+[/math] and [math]U_N\to U_N^+[/math], studied in chapter 8 above.

General references

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