14c. Reflection groups

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

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We know that we have results involving free wreath products, which replace the usual wreath products from the classical case. In particular, the quantum symmetry group of the graph formed by [math]N[/math] segments is the hyperoctahedral quantum group [math]H_N^+=\mathbb Z_2\wr_*S_N^+[/math], appearing as a free analogue of the usual hyperoctahedral group [math]H_N=\mathbb Z_2\wr S_N[/math].

The free analogues of the reflection groups [math]H_N^s[/math] can be constructed as follows:

Definition

The algebra [math]C(H_N^{s+})[/math] is the universal [math]C^*[/math]-algebra generated by [math]N^2[/math] normal elements [math]u_{ij}[/math], subject to the following relations,

• [math]u=(u_{ij})[/math] is unitary,
• [math]u^t=(u_{ji})[/math] is unitary,
• [math]p_{ij}=u_{ij}u_{ij}^*[/math] is a projection,
• [math]u_{ij}^s=p_{ij}[/math],

with Woronowicz algebra maps [math]\Delta,\varepsilon,S[/math] constructed by universality.

Here we allow the value [math]s=\infty[/math], with the convention that the last axiom simply disappears in this case. Observe that at [math]s \lt \infty[/math] the normality condition is actually redundant, because a partial isometry [math]a[/math] subject to the relation [math]aa^*=a^s[/math] is normal.

Observe also that we have an inclusion of quantum groups [math]H_N^s\subset H_N^{s+}[/math] which is a liberation, in the sense that the classical version of [math]H_N^{s+}[/math], obtained by dividing by the commutator ideal, is the group [math]H_N^s[/math]. Indeed, this follows exactly as for [math]S_N\subset S_N^+[/math].

In analogy with the results from the real case, we have the following result:

Proposition

The algebras [math]C(H_N^{s+})[/math] with [math]s=1,2,\infty[/math], and their presentation relations in terms of the entries of the matrix [math]u=(u_{ij})[/math], are as follows:

• For [math]C(H_N^{1+})=C(S_N^+)[/math], the matrix [math]u[/math] is magic: all its entries are projections, summing up to [math]1[/math] on each row and column.
• For [math]C(H_N^{2+})=C(H_N^+)[/math] the matrix [math]u[/math] is cubic: it is orthogonal, and the products of pairs of distinct entries on the same row or the same column vanish.
• For [math]C(H_N^{\infty+})=C(K_N^+)[/math] the matrix [math]u[/math] is unitary, its transpose is unitary, and all its entries are normal partial isometries.

Show Proof

This is something elementary, the idea being as follows:

(1) This follows from definitions, and from some standard operator algebra tricks.

(2) This follows again from definitions, and standard operator algebra tricks.

(3) This is just a translation of the definition of [math]C(H_N^{s+})[/math], at [math]s=\infty[/math].

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Let us prove now that [math]H_N^{s+}[/math] with [math]s \lt \infty[/math] is a quantum permutation group. For this purpose, we must change the fundamental representation. Let us start with:

Definition

A [math](s,N)[/math]-sudoku matrix is a magic unitary of size [math]sN[/math], of the form

[[math]] m=\begin{pmatrix} a^0&a^1&\ldots&a^{s-1}\\ a^{s-1}&a^0&\ldots&a^{s-2}\\ \vdots&\vdots&&\vdots\\ a^1&a^2&\ldots&a^0 \end{pmatrix} [[/math]]

where [math]a^0,\ldots,a^{s-1}[/math] are [math]N\times N[/math] matrices.

The basic examples of such matrices come from the group [math]H_n^s[/math]. Indeed, with [math]w=e^{2\pi i/s}[/math], each of the [math]N^2[/math] matrix coordinates [math]u_{ij}:H_N^s\to\mathbb C[/math] decomposes as follows:

[[math]] u_{ij}=\sum_{r=0}^{s-1}w^ra^r_{ij} [[/math]]

Here each [math]a^r_{ij}[/math] is a function taking values in [math]\{0,1\}[/math], and so a projection in the [math]C^*[/math]-algebra sense, and it follows from definitions that these projections form a sudoku matrix. Now with this sudoku matrix notion in hand, we have the following result:

Theorem

The following happen:

• The algebra [math]C(H_N^s)[/math] is isomorphic to the universal commutative [math]C^*[/math]-algebra generated by the entries of a [math](s,N)[/math]-sudoku matrix.
• The algebra [math]C(H_N^{s+})[/math] is isomorphic to the universal [math]C^*[/math]-algebra generated by the entries of a [math](s,N)[/math]-sudoku matrix.

Show Proof

The first assertion follows from the second one. In order to prove now the second assertion, consider the universal algebra in the statement:

[[math]] A=C^*\left(a_{ij}^p\ \Big\vert \left(a^{q-p}_{ij}\right)_{pi,qj}=(s,N)-\mbox{sudoku }\right) [[/math]]

Consider also the algebra [math]C(H_N^{s+})[/math]. According to Definition 14.19, this algebra is presented by certain relations [math]R[/math], that we will call here level [math]s[/math] cubic conditions:

[[math]] C(H_N^{s+})=C^*\left(u_{ij}\ \Big\vert\ u=N\times N\mbox{ level $s$ cubic }\right) [[/math]]

We will construct now a pair of inverse morphisms between these algebras.

(1) Our first claim is that [math]U_{ij}=\sum_pw^{-p}a^p_{ij}[/math] is a level [math]s[/math] cubic unitary. Indeed, by using the sudoku condition, the verification of (1-4) in Definition 14.19 is routine.

(2) Our second claim is that the elements [math]A^p_{ij}=\frac{1}{s}\sum_rw^{rp}u^r_{ij}[/math], with the convention [math]u_{ij}^0=p_{ij}[/math], form a level [math]s[/math] sudoku unitary. Once again, the proof here is routine.

(3) According to the above, we can define a morphism [math]\Phi:C(H_N^{s+})\to A[/math] by the formula [math]\Phi(u_{ij})=U_{ij}[/math], and a morphism [math]\Psi:A\to C(H_N^{s+})[/math] by the formula [math]\Psi(a^p_{ij})=A^p_{ij}[/math].

(4) It is then routine to check that [math]\Phi,\Psi[/math] are inverse morphisms, by a direct computation of their compositions. Thus, we have an isomorphism [math]C(H_N^{s+})=A[/math], as claimed.

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In order to further advance, we will need the following simple fact:

Proposition

A [math]sN\times sN[/math] magic unitary commutes with the matrix

[[math]] \Sigma= \begin{pmatrix} 0&I_N&0&\ldots&0\\ 0&0&I_N&\ldots&0\\ \vdots&\vdots&&\ddots&\\ 0&0&0&\ldots&I_N\\ I_N&0&0&\ldots&0 \end{pmatrix} [[/math]]

precisely when it is a sudoku matrix in the sense of Definition 14.21.

Show Proof

This follows from the fact that commutation with [math]\Sigma[/math] means that the matrix is circulant. Thus, we obtain the sudoku relations from Definition 14.21.

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Now let [math]Z_s[/math] be the oriented cycle with [math]s[/math] vertices, and consider the graph [math]NZ_s[/math] consisting of [math]N[/math] disjoint copies of it. Observe that, with a suitable labeling of the vertices, the adjacency matrix of this graph is the above matrix [math]\Sigma[/math]. We obtain from this:

Theorem

We have the following results:

• [math]H_N^s[/math] is the symmetry group of [math]NZ_s[/math].
• [math]H_N^{s+}[/math] is the quantum symmetry group of [math]NZ_s[/math].

Show Proof

This is something elementary, the idea being as follows:

(1) This follows indeed from definitions.

(2) This follows from Theorem 14.18 and Proposition 14.23, because the algebra [math]C(H_N^{s+})[/math] is the quotient of the algebra [math]C(S_{sN}^+)[/math] by the relations making the fundamental corepresentation commute with the adjacency matrix of [math]NZ_s[/math].

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Next in line, we must talk about wreath products. We have here:

Theorem

We have isomorphisms as follows,

[[math]] H_N^s=\mathbb Z_s\wr S_N\quad,\quad H_N^{s+}=\mathbb Z_s\wr_*S_N^+ [[/math]]

with [math]\wr[/math] being a wreath product, and [math]\wr_*[/math] being a free wreath product.

Show Proof

This follows from the following formulae, valid for any connected graph [math]X[/math], and explained before in this chapter, applied to the graph [math]Z_s[/math]:

[[math]] G(NX)=G(X)\wr S_N\quad,\quad G^+(NX)=G^+(X)\wr_*S_N^+ [[/math]]

Alternatively, (1) follows from definitions, and (2) can be proved directly, by constructing a pair of inverse morphisms. For details here, we refer to the literature.

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Regarding now the easiness property of [math]H_N^s,H_N^{s+}[/math], we have here:

Theorem

The quantum groups [math]H_N^s,H_N^{s+}[/math] are easy, the corresponding categories

[[math]] P^s\subset P\quad,\quad NC^s\subset NC [[/math]]

consisting of partitions satisfying [math]\#\circ=\#\bullet(s)[/math], as a weighted sum, in each block.

Show Proof

This is something quite routine, the idea being as follows:

(1) We already know this for the reflection group [math]H_N^s[/math], from chapter 12, and the idea is that the computation there works for [math]H_N^{s+}[/math] too, with minimal changes. Indeed, at [math]s=1[/math], to start with, this is something that we already know, from chapter 13.

(2) At [math]s=2[/math] now, we know that [math]H_N^+\subset O_N^+[/math] appears via the cubic relations, namely:

[[math]] u_{ij}u_{ik}=u_{ji}u_{ki}=0\quad,\quad\forall j\neq k [[/math]]

We conclude, exactly as in chapter 12, that [math]H_N^+[/math] is indeed easy, coming from:

[[math]] D= \lt H \gt =NC_{even} [[/math]]

(3) Regarding now that case [math]s=\infty[/math], for the quantum group [math]K_N^+[/math], the proof here is similar, leading this time to the category [math]\mathcal{NC}_{even}[/math] of noncrossing matching partitions.

(4) Summarizing, we have the result at [math]s=1,2,\infty[/math]. But the passage to the general case [math]s\in\mathbb N[/math] is then routine, by using functoriality, and the result at [math]s=\infty[/math].

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All the above is very nice, and at the first glance, it looks like a complete theory of quantum reflection groups. However, there is a skeleton in the closet, coming from: \begin{fact} The symmetry groups of the hypercube [math]\square_N\subset\mathbb R^N[/math] are

[[math]] G(\square_N)=H_N\quad,\quad G^+(\square_N)\neq H_N^+ [[/math]]

with the problem coming from the fact that [math]H_N^+[/math] does not act on [math]\square_N[/math]. \end{fact} Excited about this? We are here at the heart of quantum algebra, with all sorts of new phenomena, having no classical counterpart, waiting to be explored. In answer, we will prove that we have an equality as follows, with [math]O_N^{-1}[/math] being a certain twist of [math]O_N[/math]:

[[math]] G^+(\square_N)=O_N^{-1} [[/math]]

In order to introduce this new quantum group [math]O_N^{-1}[/math], we will need:

Proposition

There is a signature map [math]\varepsilon:P_{even}\to\{-1,1\}[/math], given by

[[math]] \varepsilon(\tau)=(-1)^c [[/math]]

where [math]c[/math] is the number of switches needed to make [math]\tau[/math] noncrossing. In addition:

• For [math]\tau\in S_k[/math], this is the usual signature.
• For [math]\tau\in P_2[/math] we have [math](-1)^c[/math], where [math]c[/math] is the number of crossings.
• For [math]\tau\leq\pi\in NC_{even}[/math], the signature is [math]1[/math].

Show Proof

The fact that [math]\varepsilon[/math] is indeed well-defined comes from the fact that the number [math]c[/math] in the statement is well-defined modulo 2, which is standard combinatorics. In order to prove now the remaining assertion, observe that any partition [math]\tau\in P(k,l)[/math] can be put in “standard form”, by ordering its blocks according to the appearence of the first leg in each block, counting clockwise from top left, and then by performing the switches as for block 1 to be at left, then for block 2 to be at left, and so on. With this convention:

(1) For [math]\tau\in S_k[/math] the standard form is [math]\tau'=id[/math], and the passage [math]\tau\to id[/math] comes by composing with a number of transpositions, which gives the signature.

(2) For a general [math]\tau\in P_2[/math], the standard form is of type [math]\tau'=|\ldots|^{\cup\ldots\cup}_{\cap\ldots\cap}[/math], and the passage [math]\tau\to\tau'[/math] requires [math]c[/math] mod 2 switches, where [math]c[/math] is the number of crossings.

(3) Assuming that [math]\tau\in P_{even}[/math] comes from [math]\pi\in NC_{even}[/math] by merging a certain number of blocks, we can prove that the signature is 1 by proceeding by recurrence.

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

Definition

Associated to a partition [math]\pi\in P_{even}(k,l)[/math] is the linear map

[[math]] \bar{T}_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j_1\ldots j_l}\bar{\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} [[/math]]

where the signed Kronecker symbols

[[math]] \bar{\delta}_\pi\in\{-1,0,1\} [[/math]]

are given by [math]\bar{\delta}_\pi=\varepsilon(\tau)[/math] if [math]\tau\geq\pi[/math], and [math]\bar{\delta}_\pi=0[/math] otherwise, with [math]\tau=\ker(^i_j)[/math].

In other words, what we are doing here is to add signatures to the usual formula of [math]T_\pi[/math]. Indeed, observe that the usual formula for [math]T_\pi[/math] can be written as folllows:

[[math]] T_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{j:\ker(^i_j)\geq\pi}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

Now by inserting signs, coming from the signature map [math]\varepsilon:P_{even}\to\{\pm1\}[/math], we are led to the following formula, which coincides with the one above:

[[math]] \bar{T}_\pi(e_{i_1}\otimes\ldots\otimes e_{i_k})=\sum_{\tau\geq\pi}\varepsilon(\tau)\sum_{j:\ker(^i_j)=\tau}e_{j_1}\otimes\ldots\otimes e_{j_l} [[/math]]

Getting now to quantum groups, we have the following construction:

Theorem

Given a category of partitions [math]D\subset P_{even}[/math], the construction

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

produces via Tannakian duality a quantum group [math]G_N^{-1}[/math], for any [math]N\in\mathbb N[/math].

Show Proof

It is routine to check that the assignement [math]\pi\to\bar{T}_\pi[/math] is categorical, in the sense that we have the following formulae, where [math]c(\pi,\sigma)[/math] are certain positive integers:

[[math]] \bar{T}_\pi\otimes\bar{T}_\sigma=\bar{T}_{[\pi\sigma]}\quad,\quad \bar{T}_\pi \bar{T}_\sigma=N^{c(\pi,\sigma)}\bar{T}_{[^\sigma_\pi]}\quad,\quad \bar{T}_\pi^*=\bar{T}_{\pi^*} [[/math]]

But with this, the result follows from the Tannakian results from chapter 13.

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We can unify the easy quantum groups, or at least the examples coming from categories [math]D\subset P_{even}[/math], with the quantum groups constructed above, as follows:

Definition

A quantum group [math]G[/math] is called [math]q[/math]-easy, or quizzy, with deformation parameter [math]q=\pm1[/math], when its tensor category appears as

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

for a certain category of partitions [math]D\subset P_{even}[/math], where, for [math]q=-1,1[/math]:

[[math]] \dot{T}=\bar{T},T [[/math]]

The Schur-Weyl twist of [math]G[/math] is the quizzy quantum group [math]G^{-1}[/math] obtained via [math]q\to-q[/math].

As an illustration for all this, which might seem quite abstract, we can now twist the orthogonal group [math]O_N[/math], and the unitary group [math]U_N[/math] too. The result here is as follows:

Theorem

The twists of [math]O_N,U_N[/math] are obtained by replacing the commutation relations [math]ab=ba[/math] between the coordinates [math]u_{ij}[/math] and their adjoints [math]u_{ij}^*[/math] with the relations

[[math]] ab=\pm ba [[/math]]

with anticommutation on rows and columns, and commutation otherwise.

Show Proof

The basic crossing, [math]\ker\binom{ij}{ji}[/math] with [math]i\neq j[/math], comes from the transposition [math]\tau\in S_2[/math], so its signature is [math]-1[/math]. As for its degenerated version [math]\ker\binom{ii}{ii}[/math], this is noncrossing, so here the signature is [math]1[/math]. We conclude that the linear map associated to the basic crossing is:

[[math]] \bar{T}_{\slash\!\!\!\backslash}(e_i\otimes e_j) =\begin{cases} -e_j\otimes e_i&{\rm for}\ i\neq j\\ e_j\otimes e_i&{\rm otherwise} \end{cases} [[/math]]

We can proceed now as in the untwisted case, and since the intertwining relations coming from [math]\bar{T}_{\slash\!\!\!\backslash}[/math] correspond to the relations defining [math]O_N^{-1},U_N^{-1}[/math], we obtain the result.

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Getting back now to graphs, we have the following result, from :

Theorem

The quantum symmetry group of the [math]N[/math]-hypercube is

[[math]] G^+(\square_N)=O_N^{-1} [[/math]]

with the corresponding coaction map on the vertex set being given by

[[math]] \Phi:C^*(\mathbb Z_2^N)\to C^*(\mathbb Z_2^N)\otimes C(O_N^{-1})\quad,\quad g_i\to\sum_jg_j\otimes u_{ji} [[/math]]

via the standard identification [math]\square_N=\widehat{\mathbb Z_2^N}[/math]. In particular we have [math]G^+(\square)=O_2^{-1}[/math].

Show Proof

This follows from a standard algebraic study, done in , as follows:

(1) Our first claim is that [math]\square_N[/math] is the Cayley graph of [math]\mathbb Z_2^N= \lt \tau_1,\ldots ,\tau_N \gt [/math]. Indeed, the vertices of this latter Cayley graph are the products of the following form:

[[math]] g=\tau_1^{i_1}\ldots\tau_N^{i_N} [[/math]]

The sequence of exponents defining such an element determines a point of [math]\mathbb R^N[/math], which is a vertex of the cube. Thus the vertices of the Cayley graph are the vertices of the cube, and in what regards the edges, this is something that we know too, from chapter 4.

(2) Our second claim now, which is something routine, coming from an elementary computation, is that when identifying the vector space spanned by the vertices of [math]\square_N[/math] with the algebra [math]C^*(\mathbb Z_2^N)[/math], the eigenvectors and eigenvalues of [math]\square_N[/math] are given by:

[[math]] v_{i_1\ldots i_N}=\sum_{j_1\ldots j_N} (-1)^{i_1j_1 +\ldots+i_Nj_N}\tau_1^{j_1}\ldots\tau_N^{j_N} [[/math]]

[[math]] \lambda_{i_1\ldots i_N}=(-1)^{i_1}+\ldots +(-1)^{i_N} [[/math]]

(3) We prove now that the quantum group [math]O_N^{-1}[/math] acts on the cube [math]\square_N[/math]. For this purpose, observe first that we have a map as follows:

[[math]] \Phi :C^*(\mathbb Z_2^N)\to C^*(\mathbb Z_2^N)\otimes C(O_N^{-1}) \quad,\quad\tau_i\to\sum_j\tau_j\otimes u_{ji} [[/math]]

It is routine to check that for [math]i_1\neq i_2\neq\ldots\neq i_l[/math] we have:

[[math]] \Phi(\tau_{i_1}\ldots\tau_{i_l})=\sum_{j_1\neq\ldots\neq j_l}\tau_{j_1}\ldots\tau_{j_l} \otimes u_{j_1i_1}\ldots u_{j_li_l} [[/math]]

In terms of eigenspaces [math]E_s[/math] of the adjacency matrix, this gives, as desired:

[[math]] \Phi(E_s)\subset E_s\otimes C(O_N^{-1}) [[/math]]

(4) Conversely now, consider the universal coaction on the cube:

[[math]] \Psi:C^*(\mathbb Z_2^N)\to C^*(\mathbb Z_2^N)\otimes C(G)\quad,\quad\tau_i\to\sum_j\tau_j\otimes u_{ji} [[/math]]

By applying [math]\Psi[/math] to the relation [math]\tau_i\tau_j=\tau_j\tau_i[/math] we get [math]u^tu=1[/math], so the matrix [math]u=(u_{ij})[/math] is orthogonal. By applying [math]\Psi[/math] to the relation [math]\tau_i^2=1[/math] we get:

[[math]] 1\otimes\sum_ku_{ki}^2+\sum_{k \lt l}\tau_k\tau_l\otimes(u_{ki}u_{li}+u_{li}u_{ki})=1\otimes 1 [[/math]]

This gives [math]u_{ki}u_{li}=-u_{li}u_{ki}[/math] for [math]i\neq j[/math], [math]k\neq l[/math], and by using the antipode we get [math]u_{ik}u_{il}=-u_{il}u_{ik}[/math] for [math]k\neq l[/math]. Also, by applying [math]\Psi[/math] to [math]\tau_i\tau_j=\tau_j\tau_i[/math] with [math]i\neq j[/math] we get:

[[math]] \sum_{k \lt l}\tau_k\tau_l\otimes(u_{ki}u_{lj}+u_{li}u_{kj})=\sum_{k \lt l}\tau_k\tau_l\otimes (u_{kj}u_{li}+u_{lj}u_{ki}) [[/math]]

Identifying coefficients, it follows that for [math]i\neq j[/math] and [math]k\neq l[/math], we have:

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

In other words, we have [math][u_{ki},u_{lj}]=[u_{kj},u_{li}][/math]. By using the antipode we get:

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

Now by combining these relations we get:

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

Thus [math][u_{il},u_{jk}]=0[/math], so the elements [math]u_{ij}[/math] satisfy the relations for [math]C(O_N^{-1})[/math], as desired.

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Many other things can be said about twists, reflection groups, and their actions on graphs, and for an introduction to this, we recommend , followed by .

General references

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