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Summarizing, in connection with our [math](S,T,U,K)[/math] program, we have so far triples of type [math](S,T,U)[/math], along with some correspondences between [math]S,T,U[/math]. In order to introduce now the reflection groups [math]K[/math], things are more tricky, involving quantum permutation groups. Following Wang , these quantum groups are introduced as follows:

Theorem

The following construction, where “magic” means formed of projections, which sum up to [math]1[/math] on each row and column,

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

produces a quantum group liberation of [math]S_N[/math]. Moreover, the inclusion

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

is an isomorphism at [math]N\leq3[/math], but not at [math]N\geq4[/math], where [math]S_N^+[/math] is not classical, nor finite.

Show Proof

We have several things to be proved, the idea being as follows:

(1) The quantum group assertion follows by using the same arguments as those in the proof of Theorem 2.12. Consider indeed the following matrix:

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

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

[[math]] U_{ij}^2 =\sum_{kl}u_{ik}u_{il}\otimes u_{kj}u_{lj} =\sum_ku_{ik}\otimes u_{kj} =U_{ij} [[/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]] \sum_jU_{ij} =\sum_{jk}u_{ik}\otimes u_{kj} =\sum_ku_{ik}\otimes1 =1\otimes1 [[/math]]

For the columns the computation is similar, as follows:

[[math]] \sum_iU_{ij} =\sum_{ik}u_{ik}\otimes u_{kj} =\sum_k1\otimes u_{kj} =1\otimes1 [[/math]]

Thus the [math]U=(U_{ij})[/math] is magic, and so we can define a comultiplication map by using the universality property of [math]C(S_N^+)[/math], 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 Definition 2.1 are satisfied, and this finishes the proof of the first assertion, stating that [math]S_N^+[/math] is indeed a compact quantum group.

(2) Observe now that we have an embedding of compact quantum groups [math]S_N\subset S_N^+[/math], obtained by using the standard coordinates of [math]S_N[/math], viewed as an algebraic group:

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

By using the Gelfand theorem and working out the details, as we did with the free spheres are free unitary groups, the embedding [math]S_N\subset S_N^+[/math] is indeed a liberation.

(3) Finally, regarding the last assertion, the study here is 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]}. By using the same argument as in the [math]N=2[/math] case, and 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, this gives [math]u_{22}u_{11}=u_{11}u_{22}u_{11}[/math], and then [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.

■

The above result came as a surprise at the time of , and there has been a lot of work since then, in order to understand what the quantum permutations really are, at [math]N\geq4[/math]. We will be back to this, with further details, on several occasions. For the moment, let us just record the following alternative approach to [math]S_N^+[/math], also from Wang , which shows that we are not wrong with our formalism:

Proposition

The quantum group [math]S_N^+[/math] acts on the set [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(e_i)=\sum_je_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(e_i)=\frac{1}{N},\forall i[/math].

Show Proof

Our claim is that given a compact matrix quantum group [math]G[/math], the following formula 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]:

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

Indeed, let us first determine when [math]\Phi[/math] is multiplicative. We have:

[[math]] \Phi(e_i)\Phi(e_k) =\sum_{jl}e_je_l\otimes u_{ji}u_{lk} =\sum_je_j\otimes u_{ji}u_{jk} [[/math]]

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

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

We conclude that 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]] \Phi(1) =\sum_i\Phi(e_i) =\sum_{ij}e_j\otimes u_{ji} =\sum_je_j\otimes\left(\sum_iu_{ji}\right) [[/math]]

Thus [math]\Phi[/math] is unital when [math]\sum_iu_{ji}=1[/math], [math]\forall j[/math]. Finally, the fact that [math]\Phi[/math] is a [math]*[/math]-morphism translates into [math]u_{ij}=u_{ij}^*[/math], [math]\forall i,j[/math]. Summing up, in order for [math]\Phi(e_i)=\sum_je_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(e_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(e_i)=\sum_je_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.

■

With the above results in hand, we can now introduce the quantum reflections:

Theorem

The following constructions produce compact quantum groups,

[[math]] \begin{eqnarray*} C(H_N^+)&=&C^*\left((u_{ij})_{i,j=1,\ldots,N}\Big|u_{ij}=u_{ij}^*,\,(u_{ij}^2)={\rm magic}\right)\\ C(K_N^+)&=&C^*\left((u_{ij})_{i,j=1,\ldots,N}\Big|[u_{ij},u_{ij}^*]=0,\,(u_{ij}u_{ij}^*)={\rm magic}\right) \end{eqnarray*} [[/math]]

which appear respectively as liberations of the reflection groups [math]H_N[/math] and [math]K_N[/math].

Show Proof

This can be proved in the usual way, with the first assertion coming from the fact that if [math]u[/math] satisfies the relations in the statement, then so do the matrices [math]u^\Delta,u^\varepsilon,u^S[/math], and with the second assertion coming as in the sphere case. See , .

■

Summarizing, we are done with our construction task for the quadruplets [math](S,T,U,K)[/math], in the free real and complex cases, and we can now formulate:

Proposition

We have a quadruplet as follows, called free real,

[[math]] \xymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb R,+}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&T_N^+\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ O_N^+\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&H_N^+\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

and a quadruplet as follows, called free complex:

[[math]] \xymatrix@R=50pt@C=50pt{ S^{N-1}_{\mathbb C,+}\ar@{-}[r]\ar@{-}[d]\ar@{-}[dr]&\mathbb T_N^+\ar@{-}[l]\ar@{-}[d]\ar@{-}[dl]\\ U_N^+\ar@{-}[u]\ar@{-}[ur]\ar@{-}[r]&K_N^+\ar@{-}[l]\ar@{-}[ul]\ar@{-}[u] } [[/math]]

Show Proof

This is more of an empty statement, coming from the various constructions above, from chapter 1, and from the present chapter.

■

Going ahead now, we must construct correspondences between our objects [math](S,T,U,K)[/math], completing the work for the pairs [math](S,T)[/math] started in chapter 1. This will take some time, and we will need some preliminaries. To start with, let us record the following result, which refines the various liberation statements formulated above:

Theorem

The quantum unitary and reflection groups are as follows,

[[math]] \xymatrix@R=20pt@C=20pt{ &K_N^+\ar[rr]&&U_N^+\\ H_N^+\ar[rr]\ar[ur]&&O_N^+\ar[ur]\\ &K_N\ar[rr]\ar[uu]&&U_N\ar[uu]\\ H_N\ar[uu]\ar[ur]\ar[rr]&&O_N\ar[uu]\ar[ur] } [[/math]]

and in this diagram, any face [math]P\subset Q,R\subset S[/math] has the property [math]P=Q\cap R[/math].

Show Proof

The fact that we have inclusions as in the statement follows from the definition of the various quantum groups involved. As for the various intersection claims, these follow as well from definitions. For some further details on all this, we refer to .

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General references

Banica, Teo (2024). "Affine noncommutative geometry". arXiv:2012.10973 [math.QA].