16a. Deformations

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

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In this chapter we go back to the usual complex Hadamard matrices, [math]H\in M_N(\mathbb C)[/math]. We know that associated to any such matrix is a certain quantum permutation group [math]G\subset S_N^+[/math], which describes the symmetries of the matrix. The main example for this construction [math]H\to G[/math] is, as expected, [math]F_N\to\mathbb Z_N[/math], and more generally, [math]F_G\to G[/math], for any finite abelian group [math]G[/math]. There are of course many things that can be said about the correspondence [math]H\to G[/math], but the main question remains the explicit computation of [math]G[/math], in terms of [math]H[/math]. Here we discuss this question for the deformed Fourier matrices.

Contrary to many other things discussed in this book, this is something that has been intensively studied, and not that the known results are fully satisfactory, but at least they lie at the level of what the experts can do. The story of the subject is as follows:

(1) The origins of the question go back to some discussions, and even papers, written by Bichon, Nicoara, Schlenker and myself in the mid 00s, containing a few mistakes, which ruined the thing, initially. Be said in passing, regarding wrong papers, never ever do that, if possible, and for good reason. Not with respect to mathematics and the community, who are legendary slow anyway in digesting new things, but with respect to yourself, and your business. Believe me, with any wrong paper, you dig your own grave.

(2) Towards the end of the 00s, some computations by Nicoara and his students on one hand, and some computations of Burstein, a student of Jones, on the other , done in the commuting square and subfactor context, showed that the problem for the deformed Fourier matrices is very interesting, and far more complicated than previously thought. In the context of the correspondence [math]H\to G[/math], as above, the study was done short after, in a joint paper by Bichon and myself , that we will explain in what follows.

(3) Finally, and as a third piece of the story, the paper , which contains several exciting things, had several follow-ups, both by Bichon and by myself, which are extremely technical, and barely readable, and that you will certainly be able to find on the internet, if interested, just by following citations, as usual. These papers are, needless to say, correct, but really tough, and the problem for younger generations is that of going beyond that. In my opinion, and Bichon's too, this is certainly possible, and very interesting.

Getting to work now, following , we would like to discuss the computation of the quantum groups associated to the Di\c t\u a deformations of the tensor products of Fourier matrices. Let us begin by recalling the construction of the Fourier matrix models:

Definition

Associated to a finite abelian group [math]G[/math] is the matrix model

[[math]] \pi:C(G)\to M_G(\mathbb C) [[/math]]

coming from the following magic matrix,

[[math]] (U_{ij})_{kl}=\frac{1}{N}F_{i-j,k-l} [[/math]]

where [math]F=F_G[/math] is the Fourier matrix of [math]G[/math].

Let us recall as well the construction of the deformed Fourier models:

Definition

Given two finite abelian groups [math]G,H[/math], we consider the corresponding deformed Fourier matrix, given by the formula

[[math]] (F_G\otimes_Q F_H)_{ia,jb}=Q_{ib}(F_G)_{ij}(F_H)_{ab} [[/math]]

and we factorize the associated representation [math]\pi_Q[/math] of the algebra [math]C(S_{G\times H}^+)[/math],

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C(G_Q)\ar[ur]_\pi&} [[/math]]

with [math]C(G_Q)[/math] being the Hopf image of this representation [math]\pi_Q[/math].

Explicitely computing the above quantum permutation group [math]G_Q\subset S_{G\times H}^+[/math], as function of the parameter matrix [math]Q\in M_{G\times H}(\mathbb T)[/math], will be our main purpose, in what follows. In order to do so, we will need the following elementary result:

Proposition

If [math]G[/math] is a finite abelian group then

[[math]] C(G)=C(S_G^+)\Big/\left \lt u_{ij}=u_{kl}\Big|\forall i-j=k-l\right \gt [[/math]]

with all the indices taken inside [math]G[/math].

Show Proof

As a first observation, the quotient algebra in the statement is commutative, because we have the following relations:

[[math]] u_{ij}u_{kl}=u_{ij}u_{i,l-k+i}=\delta_{j,l-k+i}u_{ij} [[/math]]

[[math]] u_{kl}u_{ij}=u_{i,l-k+i}u_{ij}=\delta_{j,l-k+i}u_{ij} [[/math]]

Thus if we denote the algebra in the statement by [math]C(H)[/math], we have [math]H\subset S_G[/math]. Now since [math]u_{ij}(\sigma)=\delta_{i\sigma(j)}[/math] for any [math]\sigma\in H[/math], we obtain:

[[math]] i-j=k-l\implies(\sigma(j)=i\iff\sigma(l)=k) [[/math]]

But this condition tells us precisely that [math]\sigma(i)-i[/math] must be independent on [math]i[/math], and so, for some [math]g\in G[/math], we have [math]\sigma(i)=i+g[/math]. Thus we have [math]\sigma\in G[/math], as desired.

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In order to factorize the representation in Definition 16.2, we will need:

Definition

Gives two Hopf algebra quotients, as follows,

[[math]] C(S_M^+)\to A\quad,\quad C(S_N^+)\to B [[/math]]

with fundamental corepresentations denoted [math]u,v[/math], we let

[[math]] A*_wB=A^{*N}*B/ \lt [u_{ab}^{(i)},v_{ij}]=0 \gt [[/math]]

with the Hopf algebra structure making [math]w_{ia,jb}=u_{ab}^{(i)}v_{ij}[/math] a corepresentation.

The fact that we have indeed a Hopf algebra follows from the fact that [math]w[/math] is magic. In terms of quantum groups, let us write:

[[math]] A=C(G)\quad,\quad B=C(H) [[/math]]

We can write then the Hopf algebra constructed above as follows:

[[math]] A*_wB=C(G\wr_*H) [[/math]]

In other words, we make the following convention:

[[math]] C(G)*_wC(H)=C(G\wr_*H) [[/math]]

The [math]\wr_*[/math] operation is then the free analogue of [math]\wr[/math], the usual wreath product. For details regarding this construction, we refer to , or to the book . Now with this notion in hand, we can factorize representation [math]\pi_Q[/math] in Definition 16.2, as follows:

Theorem

We have a factorization as follows,

[[math]] \xymatrix@R=40pt@C=40pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C(H\wr_*G)\ar[ur]_\pi&} [[/math]]

given on the standard generators by the formulae

[[math]] U_{ab}^{(i)}=\sum_jW_{ia,jb}\quad,\quad V_{ij}=\sum_aW_{ia,jb} [[/math]]

independently of [math]b[/math], where [math]W[/math] is the magic matrix producing [math]\pi_Q[/math].

Show Proof

With [math]K=F_G,L=F_H[/math] and [math]M=|G|,N=|H|[/math], the formula of the magic matrix [math]W\in M_{G\times H}(M_{G\times H}(\mathbb C))[/math] associated to [math]H=K\otimes_QL[/math] is as follows:

[[math]] \begin{eqnarray*} (W_{ia,jb})_{kc,ld} &=&\frac{1}{MN}\cdot\frac{Q_{ic}Q_{jd}}{Q_{id}Q_{jc}}\cdot\frac{K_{ik}K_{jl}}{K_{il}K_{jk}}\cdot\frac{L_{ac}L_{bd}}{L_{ad}L_{bc}}\\ &=&\frac{1}{MN}\cdot\frac{Q_{ic}Q_{jd}}{Q_{id}Q_{jc}}\cdot K_{i-j,k-l}L_{a-b,c-d} \end{eqnarray*} [[/math]]

Our claim now is that the representation [math]\pi_Q[/math] constructed in Definition 16.2 can be factorized in three steps, up to the factorization in the statement, as follows:

[[math]] \xymatrix@R=70pt@C=60pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[d]&&M_{G\times H}(\mathbb C)\\ C(S_H^+\wr_*S_G^+)\ar[r]\ar@{. \gt }[rru]&C(S_H^+\wr_*G)\ar[r]\ar@{. \gt }[ur]&C(H\wr_*G)\ar@{. \gt }[u]} [[/math]]

Indeed, these factorizations can be constructed as follows:

(1) The construction of the map on the left is standard, by checking the relations for the free wreath product, and this produces the first factorization.

(2) Regarding the second factorization, the one in the middle, this comes from the fact that since the elements [math]V_{ij}[/math] depend on [math]i-j[/math], they satisfy the defining relations for the quotient algebra [math]C(S_G^+)\to C(G)[/math], coming from Proposition 16.3.

(3) Finally, regarding the third factorization, the one on the right, observe that the above matrix [math]W_{ia,jb}[/math] depends only on [math]i,j[/math] and on [math]a-b[/math]. By summing over [math]j[/math] we obtain that the elements [math]U_{ab}^{(i)}[/math] depend only on [math]a-b[/math], and we are done.

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Summarizing, we already have some advances on our problem, the quantum group that we want to compute appearing as a subgroup of a certain free wreath product. In order to further factorize the above representation, we use:

Definition

If [math]H\curvearrowright\Gamma[/math] is a finite group acting by automorphisms on a discrete group, the corresponding crossed coproduct Hopf algebra is

[[math]] C^*(\Gamma)\rtimes C(H)=C^*(\Gamma)\otimes C(H) [[/math]]

with comultiplication given by the following formula,

[[math]] \Delta(r\otimes\delta_k)=\sum_{h\in H}(r\otimes\delta_h)\otimes(h^{-1}\cdot r\otimes\delta_{h^{-1}k}) [[/math]]

for [math]r\in\Gamma[/math] and [math]k\in H[/math]. The corresponding quantum group is denoted [math]\widehat{\Gamma}\rtimes H[/math].

Observe that [math]C(H)[/math] is a subcoalgebra, and that [math]C^*(\Gamma)[/math] is not a subcoalgebra. Now back to the factorization in Theorem 16.5, the point is that we have:

Proposition

With [math]L=F_H,N=|H|[/math] we have an isomorphism

[[math]] C(H\wr_*G)\simeq C^*(H)^{*G}\rtimes C(G) [[/math]]

given by [math]v_{ij}\to1\otimes v_{ij}[/math] and by

[[math]] u_{ab}^{(i)}=\frac{1}{N}\sum_cL_{b-a,c}c^{(i)}\otimes 1 [[/math]]

on the standard generators.

Show Proof

We know that the algebra [math]C(H\wr_*G)[/math], constructed according to our above conventions, is the quotient of [math]C(H)^{*G}*C(G)[/math] by the following relations:

[[math]] [u_{ab}^{(i)},v_{ij}]=0 [[/math]]

Now since the variable [math]v_{ij}[/math] depends only on [math]j-i[/math], we obtain:

[[math]] [u_{ab}^{(i)},v_{kl}] =[u_{ab}^{(i)},v_{i,l-k+i}] =0 [[/math]]

Thus, we are in a usual tensor product situation, and we have:

[[math]] C(H\wr_*G)=C(H)^{*G}\otimes C(G) [[/math]]

Consider now the Fourier transform over [math]H[/math], which is a map as follows:

[[math]] \Phi:C(H)\to C^*(H) [[/math]]

We can compose the above identification with the following map:

[[math]] \Psi=\Phi^{*G}\otimes id [[/math]]

Thus, we obtain an isomorphism as in the statement. Now observe that we have:

[[math]] \Phi(u_{ab})=\frac{1}{N}\sum_cL_{b-a,c}c [[/math]]

Thus the formula for the image of [math]u_{ab}^{(i)}[/math] is indeed the one in the statement.

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Here is now our key result, which will lead to further factorizations:

Proposition

With [math]c^{(i)}=\sum_aL_{ac}u_{a0}^{(i)}[/math] and [math]\varepsilon_{ke}=\sum_iK_{ik}e_{ie}[/math] we have:

[[math]] \pi(c^{(i)})(\varepsilon_{ke})=\frac{Q_{i,e-c}Q_{i-k,e}}{Q_{ie}Q_{i-k,e-c}}\varepsilon_{k,e-c} [[/math]]

In particular if [math]c_1+\ldots+c_s=0[/math] then the matrix

[[math]] \pi(c_1^{(i_1)}\ldots c_s^{(i_s)}) [[/math]]

is diagonal, for any choice of the indices [math]i_1,\ldots,i_s[/math].

Show Proof

With [math]c^{(i)}[/math] as in the statement, we have the following formula:

[[math]] \begin{eqnarray*} \pi(c^{(i)}) &=&\sum_aL_{ac}\pi(u_{a0}^{(i)})\\ &=&\sum_{aj}L_{ac}W_{ia,j0} \end{eqnarray*} [[/math]]

On the other hand, in terms of the basis in the statement, we have:

[[math]] W_{ia,jb}(\varepsilon_{ke})=\frac{1}{N}\delta_{i-j,k}\sum_d\frac{Q_{id}Q_{je}}{Q_{ie}Q_{jd}}L_{a-b,d-e}\varepsilon_{kd} [[/math]]

We therefore obtain, as desired:

[[math]] \begin{eqnarray*} \pi(c^{(i)})(\varepsilon_{ke}) &=&\frac{1}{N}\sum_{ad}L_{ac}\frac{Q_{id}Q_{i-k,e}}{Q_{ie}Q_{i-k,d}}L_{a,d-e}\varepsilon_{kd}\\ &=&\frac{1}{N}\sum_d\frac{Q_{id}Q_{i-k,e}}{Q_{ie}Q_{i-k,d}}\varepsilon_{kd}\sum_aL_{a,d-e+c}\\ &=&\sum_d\frac{Q_{id}Q_{i-k,e}}{Q_{ie}Q_{i-k,d}}\varepsilon_{kd}\delta_{d,e-c}\\ &=&\frac{Q_{i,e-c}Q_{i-k,e}}{Q_{ie}Q_{i-k,e-c}}\varepsilon_{k,e-c} \end{eqnarray*} [[/math]]

Regarding now the last assertion, this follows from the fact that each matrix of type [math]\pi(c_r^{(i_r)})[/math] acts on the standard basis elements [math]\varepsilon_{ke}[/math] by preserving the left index [math]k[/math], and by rotating by [math]c_r[/math] the right index [math]e[/math]. Thus when we assume [math]c_1+\ldots+c_s=0[/math] all these rotations compose up to the identity, and we obtain indeed a diagonal matrix.

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We have now all needed ingredients for refining Theorem 16.5, as follows:

Theorem

We have a factorization as follows,

[[math]] \xymatrix@R=40pt@C=30pt {C(S_{G\times H}^+)\ar[rr]^{\pi_Q}\ar[rd]&&M_{G\times H}(\mathbb C)\\&C^*(\Gamma_{G,H})\rtimes C(G)\ar[ur]_\rho&} [[/math]]

where the group on the bottom is given by

[[math]] \Gamma_{G,H}=H^{*G}\Big/\left \lt [c_1^{(i_1)}\ldots c_s^{(i_s)},d_1^{(j_1)}\ldots d_s^{(j_s)}]=1\Big|\sum_rc_r=\sum_rd_r=0\right \gt [[/math]]

with the above conventions and notations.

Show Proof

Assume that we have a representation, as follows:

[[math]] \pi:C^*(\Gamma)\rtimes C(G)\to M_L(\mathbb C) [[/math]]

Let [math]\Lambda[/math] be a [math]G[/math]-stable normal subgroup of [math]\Gamma[/math], so that [math]G[/math] acts on [math]\Gamma/\Lambda[/math], and we can form the product [math]C^*(\Gamma/\Lambda)\rtimes C(G)[/math], and assume that [math]\pi[/math] is trivial on [math]\Lambda[/math]. Then [math]\pi[/math] factorizes as:

[[math]] \xymatrix@R=40pt@C=30pt {C^*(\Gamma)\rtimes C(G)\ar[rr]^\pi\ar[rd]&&M_L(\mathbb C)\\&C^*(\Gamma/\Lambda)\rtimes C(G)\ar[ur]_\rho} [[/math]]

With [math]\Gamma=H^{*G}[/math], and by using the above results, this gives the result.

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In what follows we will restrict attention to the case where the parameter matrix [math]Q[/math] is generic, and we prove that, in this case, the representation in Theorem 16.9 is the minimal one. Our starting point is the group [math]\Gamma_{G,H}[/math] found above. Let us formulate:

Definition

Associated to two finite abelian groups [math]G,H[/math] is the discrete group

[[math]] \Gamma_{G,H}=H^{*G}\Big/\left \lt [c_1^{(i_1)}\ldots c_s^{(i_s)},d_1^{(j_1)}\ldots d_s^{(j_s)}]=1\Big|\sum_rc_r=\sum_rd_r=0\right \gt [[/math]]

where the superscripts refer to the [math]G[/math] copies of [math]H[/math], inside the free product.

We will need a more convenient description of this group. The idea here is that the above commutation relations can be realized inside a suitable semidirect product. Given a group acting on another group, [math]H\curvearrowright G[/math], we denote as usual by [math]G\rtimes H[/math] the semidirect product of [math]G[/math] by [math]H[/math], which is the set [math]G\times H[/math], with multiplication as follows:

[[math]] (a,s)(b,t)=(as(b),st) [[/math]]

Now given a group [math]G[/math], and a finite abelian group [math]H[/math], we can make [math]H[/math] act on [math]G^H[/math], in the obvious way, and then form the following crossed product:

[[math]] K=G^H\rtimes H [[/math]]

Since the elements of type [math](g,\ldots,g)[/math] are invariant under the action of [math]H[/math], we can form as well the following crossed product:

[[math]] K'=(G^H/G)\rtimes H [[/math]]

We can identify [math]G^H/G\simeq G^{|H|-1}[/math] via the following map:

[[math]] (1,g_1,\ldots,g_{|H|-1})\to(g_1,\ldots,g_{|H|-1}) [[/math]]

Thus, we obtain a crossed product [math]G^{|H|-1}\rtimes H[/math]. With these notations, we have the following result, regarding the group from Definition 16.10:

Proposition

The group [math]\Gamma_{G,H}[/math] has the following properties:

• We have an isomorphism as follows:
  [[math]] \Gamma_{G,H}\simeq\mathbb Z^{(|G|-1)(|H|-1)}\rtimes H [[/math]]
• We have as well an isomorphism as follows,
  [[math]] \Gamma_{G,H}\subset\mathbb Z^{(|G|-1)|H|}\rtimes H [[/math]]
  given on the standard generators by the formulae
  [[math]] c^{(0)}\to(0,c)\quad,\quad c^{(i)}\to(b_{i0}-b_{ic},c) [[/math]]
  where [math]b_{ic}[/math] are the standard generators of [math]\mathbb Z^{(|G|-1)|H|}[/math].

Show Proof

We prove these assertions at the same time. We must prove that we have group morphisms, given by the formulae in the statement, as follows:

[[math]] \begin{eqnarray*} \Gamma_{G,H} &\simeq&\mathbb Z^{(|G|-1)(|H|-1)}\rtimes H\\ &\subset&\mathbb Z^{(|G|-1)|H|}\rtimes H \end{eqnarray*} [[/math]]

Our first claim is that the formula in (2) defines a morphism as follows:

[[math]] \Gamma_{G,H}\to\mathbb Z^{(|G|-1)|H|}\rtimes H [[/math]]

Indeed, we know that the elements [math](0,c)[/math] produce a copy of [math]H[/math]. Also, we have a group embedding as follows:

[[math]] H\subset\mathbb Z^{|H|}\rtimes H\quad,\quad c\to(b_0-b_c,c) [[/math]]

Thus the elements [math]C^{(i)}=(b_{i0}-b_{ic},c)[/math] produce a copy of [math]H[/math], for any [math]i\neq 0[/math]. In order to check now the commutation relations, observe that we have:

[[math]] C_1^{(i_1)}\ldots C_s^{(i_s)} =\left(b_{i_10}-b_{i_1c_1}+b_{i_2c_1}-b_{i_2,c_1+c_2}+\ldots+b_{i_s,c_1+\ldots+c_{s-1}}-b_{i_s,c_1+\ldots+c_s},\sum_rc_r\right) [[/math]]

Thus [math]\sum_rc_r=0[/math] implies the following condition:

[[math]] C_1^{(i_1)}\ldots C_s^{(i_s)}\in\mathbb Z^{(|G|-1)|H|} [[/math]]

Since we are now inside an abelian group, we have the commutation relations, and our claim is proved. By using the general crossed product considerations before the statement, it is routine to construct an embedding as follows:

[[math]] \mathbb Z^{(|G|-1)(|H|-1)}\rtimes H\subset \mathbb Z^{(|G|-1)|H|}\rtimes H [[/math]]

To be more precise, we would like this embedding to be such that we have group morphisms whose composition is the group morphism just constructed, as follows:

[[math]] \begin{eqnarray*} \Gamma_{G,H} &\to&\mathbb Z^{(|G|-1)(|H|-1)}\rtimes H\\ &\subset&\mathbb Z^{(|G|-1)|H|}\rtimes H \end{eqnarray*} [[/math]]

It remains to prove that the map on the left is injective. For this purpose, consider the following morphism:

[[math]] \Gamma_{G,H}\to H\quad,\quad c^{(i)}\to c [[/math]]

The kernel [math]T[/math] of this morphism is formed by the elements of type [math]c_1^{(i_1)} \ldots c_s^{(i_s)}[/math], with [math]\sum_rc_r=0[/math]. We therefore obtain an exact sequence, as follows:

[[math]] 1\to T\to\Gamma_{G,H}\to H\to1 [[/math]]

This sequence splits by [math]c\to c^{(0)}[/math], so we have:

[[math]] \Gamma_{G,H}\simeq T\rtimes H [[/math]]

Now by the definition of [math]\Gamma_{G,H}[/math], the subgroup [math]T[/math] constructed above is abelian, and is moreover generated by the following elements:

[[math]] (-c)^{(0)}c^{(i)}\quad,\quad c\neq0 [[/math]]

Finally, the fact that [math]T[/math] is freely generated by these elements follows from the computation in the proof of Proposition 16.13 below.

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General references

Banica, Teo (2024). "Invitation to Hadamard matrices". arXiv:1910.06911 [math.CO].