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Many interesting examples of quantum permutation groups appear as particular cases of the following general construction from , involving finite graphs:

Proposition

Given a finite graph [math]X[/math], with adjacency matrix [math]d\in M_N(0,1)[/math], the following construction produces a quantum permutation group,

[[math]] C(G^+(X))=C(S_N^+)\Big/\Big \lt du=ud\Big \gt [[/math]]

whose classical version [math]G(X)[/math] is the usual automorphism group of [math]X[/math].

Show Proof

The fact that we have a quantum group comes from the fact that [math]du=ud[/math] reformulates as [math]d\in End(u)[/math], which makes it clear that we are dividing by a Hopf ideal. Regarding the second assertion, we must establish here the following equality:

[[math]] C(G(X))=C(S_N)\Big/\Big \lt du=ud\Big \gt [[/math]]

For this purpose, recall that [math]u_{ij}(\sigma)=\delta_{\sigma(j)i}[/math]. By using this formula, we have:

[[math]] \begin{eqnarray*} (du)_{ij}(\sigma) &=&\sum_kd_{ik}u_{kj}(\sigma)\\ &=&\sum_kd_{ik}\delta_{\sigma(j)k}\\ &=&d_{i\sigma(j)} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} (ud)_{ij}(\sigma) &=&\sum_ku_{ik}(\sigma)d_{kj}\\ &=&\sum_k\delta_{\sigma(k)i}d_{kj}\\ &=&d_{\sigma^{-1}(i)j} \end{eqnarray*} [[/math]]

Thus the condition [math]du=ud[/math] reformulates as [math]d_{ij}=d_{\sigma(i)\sigma(j)}[/math], and we are led to the usual notion of an action of a permutation group on [math]X[/math], as claimed.

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Let us work out some basic examples. We have the following result:

Theorem

The construction [math]X\to G^+(X)[/math] has the following properties:

• For the [math]N[/math]-point graph, having no edges at all, we obtain [math]S_N^+[/math].
• For the [math]N[/math]-simplex, having edges everywhere, we obtain as well [math]S_N^+[/math].
• We have [math]G^+(X)=G^+(X^c)[/math], where [math]X^c[/math] is the complementary graph.
• For a disconnected union, we have [math]G^+(X)\,\hat{*}\,G^+(Y)\subset G^+(X\sqcup Y)[/math].
• For the square, we obtain a non-classical, proper subgroup of [math]S_4^+[/math].

Show Proof

All these results are elementary, the proofs being as follows:

(1) This follows from definitions, because here we have [math]d=0[/math].

(2) Here [math]d=\mathbb I[/math] is the all-one matrix, and the magic condition gives [math]u\mathbb I=\mathbb Iu=N\mathbb I[/math]. We conclude that [math]du=ud[/math] is automatic in this case, and so [math]G^+(X)=S_N^+[/math].

(3) The adjacency matrices of [math]X,X^c[/math] being related by the formula [math]d_X+d_{X^c}=\mathbb I[/math]. We can use here the above formula [math]u\mathbb I=\mathbb Iu=N\mathbb I[/math], and we conclude that [math]d_Xu=ud_X[/math] is equivalent to [math]d_{X^c}u=ud_{X^c}[/math]. Thus, we obtain, as claimed, [math]G^+(X)=G^+(X^c)[/math].

(4) The adjacency matrix of a disconnected union is given by [math]d_{X\sqcup Y}=diag(d_X,d_Y)[/math]. Now let [math]w=diag(u,v)[/math] be the fundamental corepresentation of [math]G^+(X)\,\hat{*}\,G^+(Y)[/math]. Then [math]d_Xu=ud_X[/math] and [math]d_Yv=vd_Y[/math] imply, as desired, [math]d_{X\sqcup Y}w=wd_{X\sqcup Y}[/math].

(5) We know from (3) that we have [math]G^+(\square)=G^+(|\ |)[/math]. We know as well from (4) that we have [math]\mathbb Z_2\,\hat{*}\,\mathbb Z_2\subset G^+(|\ |)[/math]. It follows that [math]G^+(\square)[/math] is non-classical. Finally, the inclusion [math]G^+(\square)\subset S_4^+[/math] is indeed proper, because [math]S_4\subset S_4^+[/math] does not act on the square.

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In order to further advance, and to explicitely compute various quantum automorphism groups, we can use the spectral decomposition of [math]d[/math], as follows:

Proposition

A closed subgroup [math]G\subset S_N^+[/math] acts on a graph [math]X[/math] precisely when

[[math]] P_\lambda u=uP_\lambda\quad,\quad\forall\lambda\in\mathbb R [[/math]]

where [math]d=\sum_\lambda\lambda\cdot P_\lambda[/math] is the spectral decomposition of the adjacency matrix of [math]X[/math].

Show Proof

Since [math]d\in M_N(0,1)[/math] is a symmetric matrix, we can consider indeed its spectral decomposition, [math]d=\sum_\lambda\lambda\cdot P_\lambda[/math]. We have then the following formula:

[[math]] \lt d \gt =span\left\{P_\lambda\Big|\lambda\in\mathbb R\right\} [[/math]]

But this shows that we have the following equivalence:

[[math]] d\in End(u)\iff P_\lambda\in End(u),\forall\lambda\in\mathbb R [[/math]]

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

■

In order to exploit this, we will often combine it with the following standard fact:

Proposition

Consider a closed subgroup [math]G\subset S_N^+[/math], with associated coaction map

[[math]] \Phi:\mathbb C^N\to \mathbb C^N\otimes C(G) [[/math]]

For a linear subspace [math]V\subset\mathbb C^N[/math], the following are equivalent:

• The magic matrix [math]u=(u_{ij})[/math] commutes with [math]P_V[/math].
• [math]V[/math] is invariant, in the sense that [math]\Phi(V)\subset V\otimes C(G)[/math].

Show Proof

Let [math]P=P_V[/math]. For any [math]i\in\{1,\ldots,N\}[/math] we have the following formula:

[[math]] \begin{eqnarray*} \Phi(P(e_i)) &=&\Phi\left(\sum_kP_{ki}e_k\right)\\ &=&\sum_{jk}P_{ki}e_j\otimes u_{jk}\\ &=&\sum_je_j\otimes (uP)_{ji} \end{eqnarray*} [[/math]]

On the other hand the linear map [math](P\otimes id)\Phi[/math] is given by a similar formula:

[[math]] \begin{eqnarray*} (P\otimes id)(\Phi(e_i)) &=&\sum_kP(e_k)\otimes u_{ki}\\ &=&\sum_{jk}P_{jk}e_j\otimes u_{ki}\\ &=&\sum_je_j\otimes (Pu)_{ji} \end{eqnarray*} [[/math]]

Thus [math]uP=Pu[/math] is equivalent to [math]\Phi P=(P\otimes id)\Phi[/math], and the conclusion follows.

■

We have as well the following useful complementary result, from :

Proposition

Let [math]p\in M_N(\mathbb C)[/math] be a matrix, and consider its “color” decomposition, obtained by setting [math](p_c)_{ij}=1[/math] if [math]p_{ij}=c[/math] and [math](p_c)_{ij}=0[/math] otherwise:

[[math]] p=\sum_{c\in\mathbb C}c\cdot p_c [[/math]]

Then [math]u=(u_{ij})[/math] commutes with [math]p[/math] if and only if it commutes with all matrices [math]p_c[/math].

Show Proof

Consider the multiplication and counit maps of the algebra [math]\mathbb C^N[/math]:

[[math]] M:e_i\otimes e_j\to e_ie_j [[/math]]

[[math]] C:e_i\to e_i\otimes e_i [[/math]]

Since [math]M,C[/math] intertwine [math]u,u^{\otimes 2}[/math], their iterations [math]M^{(k)},C^{(k)}[/math] intertwine [math]u,u^{\otimes k}[/math], and so:

[[math]] \begin{eqnarray*} p^{(k)} &=&M^{(k)}p^{\otimes k}C^{(k)}\\ &=&\sum_{c\in\mathbb C}c^kp_c\\ &\in&End(u) \end{eqnarray*} [[/math]]

Let [math]S=\{c\in\mathbb C|p_c\neq 0\}[/math], and [math]f(c)=c[/math]. By Stone-Weierstrass we have [math]S= \lt f \gt [/math], and so for any [math]e\in S[/math] the Dirac mass at [math]e[/math] is a linear combination of powers of [math]f[/math]:

[[math]] \begin{eqnarray*} \delta_e &=&\sum_{k}\lambda_kf^k\\ &=&\sum_{k}\lambda_k \left(\sum_{c\in S}c^k\delta_c\right)\\ &=&\sum_{c\in S}\left(\sum_{k}\lambda_kc^k\right)\delta_c \end{eqnarray*} [[/math]]

The corresponding linear combination of matrices [math]p^{(k)}[/math] is given by:

[[math]] \begin{eqnarray*} \sum_k\lambda_kp^{(k)} &=&\sum_k\lambda_k \left(\sum_{c\in S}c^kp_c\right)\\ &=&\sum_{c\in S}\left(\sum_{k}\lambda_kc^k\right)p_c \end{eqnarray*} [[/math]]

The Dirac masses being linearly independent, in the first formula all coefficients in the right term are 0, except for the coefficient of [math]\delta_e[/math], which is 1. Thus the right term in the second formula is [math]p_e[/math], and it follows that we have [math]p_e\in End(u)[/math], as claimed.

■

The above results can be combined, and we are led into a “color-spectral” decomposition method for [math]d[/math], which can lead to a number of nontrivial results. In fact, all this is best understood in terms of Jones' planar algebras . We refer here to .

As a basic application of this, we can further study [math]G^+(\square)[/math], as follows:

Proposition

The quantum automorphism group of the [math]N[/math]-cycle is as follows:

• At [math]N\neq 4[/math] we have [math]G^+(X)=D_N[/math].
• At [math]N=4[/math] we have [math]D_4\subset G^+(X)\subset S_4^+[/math], with proper inclusions.

Show Proof

We already know that the results hold at [math]N\leq4[/math], so let us assume [math]N\geq5[/math]. Given a [math]N[/math]-th root of unity, [math]w^N=1[/math], consider the following vector:

[[math]] \xi=(w^i) [[/math]]

This is an eigenvector of [math]d[/math], with eigenvalue [math]w+w^{N-1}[/math]. With [math]w=e^{2\pi i/N}[/math], it follows that [math]1,f,f^2,\ldots ,f^{N-1}[/math] are eigenvectors of [math]d[/math]. More precisely, the invariant subspaces of [math]d[/math] are as follows, with the last subspace having dimension 1 or 2, depending on [math]N[/math]:

[[math]] \mathbb C 1,\, \mathbb C f\oplus\mathbb C f^{N-1},\, \mathbb C f^2\oplus\mathbb C f^{N-2},\ldots [[/math]]

Consider now the associated coaction [math]\Phi:\mathbb C^N\to \mathbb C^N\otimes C(G)[/math], and write:

[[math]] \Phi(f)=f\otimes a+f^{N-1}\otimes b [[/math]]

By taking the square of this equality we obtain:

[[math]] \Phi(f^2)=f^2\otimes a^2+f^{N-2}\otimes b^2+1\otimes(ab+ba) [[/math]]

It follows that [math]ab=-ba[/math], and that [math]\Phi(f^2)[/math] is given by the following formula:

[[math]] \Phi(f^2)=f^2\otimes a^2+f^{N-2}\otimes b^2 [[/math]]

By multiplying this with [math]\Phi(f)[/math] we obtain:

[[math]] \Phi(f^3)=f^3\otimes a^3+f^{N-3}\otimes b^3+f^{N-1}\otimes ab^2+f\otimes ba^2 [[/math]]

Now since [math]N\geq 5[/math] implies that [math]1,N-1[/math] are different from [math]3,N-3[/math], we must have [math]ab^2=ba^2=0[/math]. By using this and [math]ab=-ba[/math], we obtain by recurrence on [math]k[/math] that:

[[math]] \Phi(f^k)=f^k\otimes a^k+f^{N-k}\otimes b^k [[/math]]

In particular at [math]k=N-1[/math] we obtain:

[[math]] \Phi(f^{N-1})=f^{N-1}\otimes a^{N-1}+f\otimes b^{N-1} [[/math]]

On the other hand we have [math]f^*=f^{N-1}[/math], so by applying [math]*[/math] to [math]\Phi(f)[/math] we get:

[[math]] \Phi(f^{N-1})=f^{N-1}\otimes a^*+f\otimes b^* [[/math]]

Thus [math]a^*=a^{N-1}[/math] and [math]b^*=b^{N-1}[/math]. Together with [math]ab^2=0[/math] this gives:

[[math]] \begin{eqnarray*} (ab)(ab)^* &=&abb^*a^*\\ &=&ab^Na^{N-1}\\ &=&(ab^2)b^{N-2}a^{N-1}\\ &=&0 \end{eqnarray*} [[/math]]

From positivity we get from this [math]ab=0[/math], and together with [math]ab=-ba[/math], this shows that [math]a,b[/math] commute. On the other hand [math]C(G)[/math] is generated by the coefficients of [math]\Phi[/math], which are powers of [math]a,b[/math], and so [math]C(G)[/math] must be commutative, and we obtain the result.

■

Summarizing, this was a bad attempt in understanding [math]G^+(\square)[/math], which appears to be “exceptional” among the quantum symmetry groups of the [math]N[/math]-cycles.

An alternative approach to [math]G^+(\square)[/math] comes by regarding the square as the [math]N=2[/math] particular case of the [math]N[/math]-hypercube [math]\square_N[/math]. Indeed, the usual symmetry group of [math]\square_N[/math] is the hyperoctahedral group [math]H_N[/math], so we should have a formula of the following type:

[[math]] G(\square)=H_2^+ [[/math]]

In order to clarify this, let us start with the following simple fact:

Proposition

We have an embedding as follows, [math]g_i[/math] being the generators of [math]\mathbb Z_2^N[/math],

[[math]] \widehat{\mathbb Z_2^N}\subset S^{N-1}_{\mathbb R,+}\quad,\quad x_i=\frac{g_i}{\sqrt{N}} [[/math]]

whose image is the geometric hypercube:

[[math]] \square_N=\left\{x\in\mathbb R^N\Big|x_i=\pm\frac{1}{\sqrt{N}},\forall i\right\} [[/math]]

Show Proof

This is something that we already know, from chapter 1 above. Consider indeed the following standard group algebra generators:

[[math]] g_i\in C^*(\mathbb Z_2^N)=C(\widehat{\mathbb Z_2^N}) [[/math]]

These generators satisfy satisfy then [math]g_i=g_i^*[/math], [math]g_i^2=1[/math], and when rescaling by [math]1/\sqrt{N}[/math], we obtain the relations defining [math]\square_N[/math].

■

We can now study the quantum groups [math]G^+(\square_N)[/math], and we are led to the quite surprising conclusion, from , that these are the twisted orthogonal groups [math]\bar{O}_N[/math]:

Theorem

With [math]\mathbb Z_2^N= \lt g_1,\ldots,g_N \gt [/math] we have a coaction map

[[math]] \Phi:C^*(\mathbb Z_2^N)\to C^*(\mathbb Z_2^N)\otimes C(\bar{O}_N)\quad,\quad g_i\to\sum_jg_j\otimes u_{ji} [[/math]]

which makes [math]\bar{O}_N[/math] the quantum isometry group of the hypercube [math]\square_N=\widehat{\mathbb Z_2^N}[/math], as follows:

• With [math]\square_N[/math] viewed as an algebraic manifold, [math]\square_N\subset S^{N-1}_\mathbb R\subset S^{N-1}_{\mathbb R,+}[/math].
• With [math]\square_N[/math] viewed as a graph, with [math]2^N[/math] vertices and [math]2^{N-1}N[/math] edges.
• With [math]\square_N[/math] viewed as a metric space, with metric coming from [math]\mathbb R^N[/math].

Show Proof

Observe first that [math]\square_N[/math] is indeed an algebraic manifold, so (1) as formulated above makes sense, in the general framework of chapter 2. The cube [math]\square_N[/math] is also a graph, as indicated, and so (2) makes sense as well, in the framework of Proposition 10.1. Finally, (3) makes sense as well, because we can define the quantum isometry group of a finite metric space exactly as for graphs, but with [math]d[/math] being this time the distance matrix.

(1) In order for [math]G\subset O_N^+[/math] to act affinely on [math]\square_N[/math], the variables [math]G_i=\sum_jg_j\otimes u_{ji}[/math] must satisfy the same relations as the generators [math]g_i\in \mathbb Z_2^N[/math]. The self-adjointness condition being automatic, the relations to be checked are therefore:

[[math]] G_i^2=1\quad,\quad G_iG_j=G_jG_i [[/math]]

We have the following computation:

[[math]] \begin{eqnarray*} G_i^2 &=&\sum_{kl}g_kg_l\otimes u_{ik}u_{il}\\ &=&1+\sum_{k \lt l}g_kg_l\otimes(u_{ik}u_{il}+u_{il}u_{ik}) \end{eqnarray*} [[/math]]

As for the commutators, these are given by:

[[math]] \begin{eqnarray*} \left[G_i,G_j\right] &=&\sum_{k \lt l}g_kg_l\otimes(u_{ik}u_{jl}-u_{jk}u_{il}+u_{il}u_{jk}-u_{jl}u_{ik}) \end{eqnarray*} [[/math]]

From the first relation we obtain [math]ab=0[/math] for [math]a\neq b[/math] on the same row of [math]u[/math], and by using the antipode, the same happens for the columns. From the second relation we obtain:

[[math]] [u_{ik},u_{jl}]=[u_{jk},u_{il}]\quad,\quad\forall k\neq l [[/math]]

We use the Bhowmick-Goswami trick . By applying the antipode we obtain:

[[math]] [u_{lj},u_{ki}]=[u_{li},u_{kj}] [[/math]]

By relabelling, this gives the following formula:

[[math]] [u_{ik},u_{jl}]=[u_{il},u_{jk}]\quad,\quad j\neq i [[/math]]

Thus for [math]i\neq j,k\neq l[/math] we must have:

[[math]] [u_{ik},u_{jl}]=[u_{jk},u_{il}]=0 [[/math]]

We are therefore led to [math]G\subset\bar{O}_N[/math], as claimed.

(2) We can use here the fact that the cube [math]\square_N[/math], when regarded as a graph, is the Cayley graph of the group [math]\mathbb Z_2^N[/math]. The eigenvectors and eigenvalues of [math]\square_N[/math] are as follows:

[[math]] \begin{eqnarray*} v_{i_1\ldots i_N}&=&\sum_{j_1\ldots j_N} (-1)^{i_1j_1 +\ldots+i_Nj_N}g_1^{j_1}\ldots g_N^{j_N}\\ \lambda_{i_1\ldots i_N}&=&(-1)^{i_1}+\ldots +(-1)^{i_N} \end{eqnarray*} [[/math]]

With this picture in hand, and by using Proposition 10.3 and Proposition 10.4 above, the result follows from the same computations as in the proof of (1). See .

(3) Our claim here is that we obtain the same symmetry group as in (2). Indeed, observe that distance matrix of the cube has a color decomposition as follows:

[[math]] d=d_1+\sqrt{2}d_2+\sqrt{3}d_3+\ldots+ \sqrt{N}d_N [[/math]]

Since the powers of [math]d_1[/math] can be computed by counting loops on the cube, we have formulae as follows, with [math]x_{ij}\in\mathbb N[/math] being certain positive integers:

[[math]] \begin{eqnarray*} d_1^2&=&x_{21}1_N+x_{22}d_2\\ d_1^3&=&x_{31}1_N+x_{32}d_2+ x_{33}d_3\\ &\ldots&\\ d_1^N&=&x_{N1}1_N+x_{N2}d_2+x_{N3}d_3+\ldots+x_{NN}d_N \end{eqnarray*} [[/math]]

But this shows that we have [math] \lt d \gt = \lt d_1 \gt [/math]. Now since [math]d_1[/math] is the adjacency matrix of [math]\square_N[/math], viewed as graph, this proves our claim, and we obtain the result via (2).

■

Now back to our questions regarding the square, we have [math]G^+(\square)=\bar{O}_2[/math], and this formula appears as the [math]N=2[/math] particular case of a general formula, namely [math]G^+(\square_N)=\bar{O}_N[/math]. This is quite conceptual, but still not ok. The problem is that we have [math]G(\square_N)=H_N[/math], and so for our theory to be complete, we would need a formula of type [math]H_N^+=\bar{O}_N[/math]. And this latter formula is obviously wrong, because for [math]\bar{O}_N[/math] the character computations lead to Gaussian laws, who cannot appear as liberations of the character laws for [math]H_N[/math], that we have not computed yet, but which can only be something Poisson-related.

General references

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