guide:4b9e1b1b2b

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

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Following Wang , let us discuss now the construction and basic properties of the quantum permutation group [math]S_N^+[/math]. Let us first look at [math]S_N[/math]. We have here:

Theorem

The algebra of functions on [math]S_N[/math] has the following presentation,

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

and the multiplication, unit and inversion map of [math]S_N[/math] appear from the maps

[[math]] \Delta(u_{ij})=\sum_ku_{ik}\otimes u_{kj}\quad,\quad \varepsilon(u_{ij})=\delta_{ij}\quad,\quad S(u_{ij})=u_{ji} [[/math]]

defined at the algebraic level, of functions on [math]S_N[/math], by transposing.

Show Proof

This is something that we know from chapter 11, coming from the Gelfand theorem, applied to the universal algebra in the statement. Indeed, that algebra follows to be of the form [math]A=C(X)[/math], with [math]X[/math] being a certain compact space. Now since we have coordinates [math]u_{ij}:X\to\mathbb R[/math], we have an embedding [math]X\subset M_N(\mathbb R)[/math]. Also, since we know that these coordinates form a magic matrix, the elements [math]g\in X[/math] must be 0-1 matrices, having exactly one 1 entry on each row and each column, and so [math]X=S_N[/math], as desired.

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Following now Wang , we can liberate [math]S_N[/math], as follows:

Theorem

The following universal [math]C^*[/math]-algebra, with magic meaning as usual formed by projections [math](p^2=p^*=p)[/math], summing 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]]

is a Woronowicz algebra, with comultiplication, counit and antipode given by:

[[math]] \Delta(u_{ij})=\sum_ku_{ik}\otimes u_{kj}\quad,\quad \varepsilon(u_{ij})=\delta_{ij}\quad,\quad S(u_{ij})=u_{ji} [[/math]]

Thus the space [math]S_N^+[/math] is a compact quantum group, called quantum permutation group.

Show Proof

As a first observation, the universal [math]C^*[/math]-algebra in the statement is indeed well-defined, because the conditions [math]p^2=p^*=p[/math] satisfied by the coordinates give:

[[math]] ||u_{ij}||\leq1 [[/math]]

In order to prove now that we have a Woronowicz algebra, we must construct maps [math]\Delta,\varepsilon,S[/math] given by the formulae in the statement. Consider the following matrices:

[[math]] u^\Delta_{ij}=\sum_ku_{ik}\otimes u_{kj}\quad,\quad u^\varepsilon_{ij}=\delta_{ij}\quad,\quad u^S_{ij}=u_{ji} [[/math]]

Our claim is that, since [math]u[/math] is magic, so are these three matrices. Indeed, regarding [math]u^\Delta[/math], its entries are idempotents, as shown by the following computation:

[[math]] (u_{ij}^\Delta)^2 =\sum_{kl}u_{ik}u_{il}\otimes u_{kj}u_{lj} =\sum_{kl}\delta_{kl}u_{ik}\otimes\delta_{kl}u_{kj} =u_{ij}^\Delta [[/math]]

These elements are self-adjoint as well, as shown by the following computation:

[[math]] (u_{ij}^\Delta)^* =\sum_ku_{ik}^*\otimes u_{kj}^* =\sum_ku_{ik}\otimes u_{kj} =u_{ij}^\Delta [[/math]]

The row and column sums for the matrix [math]u^\Delta[/math] can be computed as follows:

[[math]] \sum_ju_{ij}^\Delta =\sum_{jk}u_{ik}\otimes u_{kj} =\sum_ku_{ik}\otimes 1 =1 [[/math]]

[[math]] \sum_iu_{ij}^\Delta =\sum_{ik}u_{ik}\otimes u_{kj} =\sum_k1\otimes u_{kj} =1 [[/math]]

Thus, [math]u^\Delta[/math] is magic. Regarding now [math]u^\varepsilon,u^S[/math], these matrices are magic too, and this for obvious reasons. Thus, all our three matrices [math]u^\Delta,u^\varepsilon,u^S[/math] are magic, so we can define [math]\Delta,\varepsilon,S[/math] by the formulae in the statement, by using the universality property of [math]C(S_N^+)[/math].

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Our first task now is to make sure that Theorem 13.29 produces indeed a new quantum group, which does not collapse to [math]S_N[/math]. Following Wang , we have:

Theorem

We have an embedding [math]S_N\subset S_N^+[/math], given at the algebra level by:

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

This 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

The fact that we have indeed an embedding as above follows from Theorem 13.28. Observe that in fact more is true, because Theorems 13.28 and 13.29 give:

[[math]] C(S_N)=C(S_N^+)\Big/\Big \lt ab=ba\Big \gt [[/math]]

Thus, the inclusion [math]S_N\subset S_N^+[/math] is a “liberation”, in the sense that [math]S_N[/math] is the classical version of [math]S_N^+[/math]. We will often use this basic fact, in what follows. Regarding now the second assertion, we can prove this in four steps, as follows:

\underline{Case [math]N=2[/math]}. The fact that [math]S_2^+[/math] is indeed classical, and hence collapses to [math]S_2[/math], is trivial, because the [math]2\times2[/math] magic matrices are as follows, with [math]p[/math] being a projection:

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

Thus [math]C(S_2^+)[/math] is commutative, and equals its biggest commutative quotient, [math]C(S_2)[/math].

\underline{Case [math]N=3[/math]}. 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 conjugating, [math]u_{22}u_{11}=u_{11}u_{22}u_{11}[/math], so [math]u_{11}u_{22}=u_{22}u_{11}[/math], as desired.

\underline{Case [math]N=4[/math]}. Consider the following matrix, with [math]p,q[/math] being projections:

[[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 we can choose [math]p,q\in B(H)[/math] as for the algebra [math] \lt p,q \gt [/math] to be noncommutative and infinite dimensional. We conclude that [math]C(S_4^+)[/math] is noncommutative and infinite dimensional as well, and so [math]S_4^+[/math] is non-classical and infinite, as claimed.

\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 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.

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As a first observation, as a matter of doublechecking our findings, we are not wrong with our formalism, because as explained once again in , we have as well:

Theorem

The quantum permutation 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]]

[[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]]

Similarly, [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, 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 this gives the result.

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General references

Banica, Teo (2024). "Graphs and their symmetries". arXiv:2406.03664 [math.CO].