8b. Master matrices

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

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Let us discuss now some defect computations for an interesting class of Hadamard matrices, namely the “master” ones, introduced by Avan et al. in :

Definition

A master Hadamard matrix is an Hadamard matrix of the form

[[math]] H_{ij}=\lambda_i^{n_j} [[/math]]

with [math]\lambda_i\in\mathbb T,n_j\in\mathbb R[/math]. The associated “master function” is [math]f(z)=\sum_jz^{n_j}[/math].

Observe that with [math]\lambda_i=e^{im_i}[/math] we have [math]H_{ij}=e^{im_in_j}[/math]. The basic example of such a matrix is the Fourier matrix [math]F_N[/math], having master function as follows:

[[math]] f(z)=\frac{z^N-1}{z-1} [[/math]]

Observe that, in terms of [math]f[/math], the Hadamard condition on [math]H[/math] is simply:

[[math]] f\left(\frac{\lambda_i}{\lambda_j}\right)=N\delta_{ij} [[/math]]

These matrices were introduced in , the motivating remark there being the fact that the following operator defines a representation of the Temperley-Lieb algebra :

[[math]] R=\sum_{ij}e_{ij}\otimes\Lambda^{n_i-n_j} [[/math]]

At the level of examples, the first observation, from , is that the standard [math]4\times 4[/math] complex Hadamard matrices are, with 2 exceptions, master Hadamard matrices:

Proposition

The following complex Hadamard matrix, with [math]|q|=1[/math],

[[math]] F_{2,2}^q=\begin{pmatrix} 1&1&1&1\\ 1&-1&1&-1\\ 1&q&-1&-q\\ 1&-q&-1&q \end{pmatrix} [[/math]]

is a master Hadamard matrix, for any [math]q\neq\pm1[/math].

Show Proof

We use the exponentiation convention [math](e^{it})^r=e^{itr}[/math], for [math]t\in[0,2\pi)[/math] and [math]r\in\mathbb R[/math]. Since we have [math]q^2\neq1[/math], we can find [math]k\in\mathbb R[/math] such that:

[[math]] q^{2k}=-1 [[/math]]

In terms of this parameter [math]k\in\mathbb R[/math], our matrix becomes:

[[math]] F_{2,2}^q=\begin{pmatrix} 1^0&1^1&1^{2k}&1^{2k+1}\\ (-1)^0&(-1)^1&(-1)^{2k}&(-1)^{2k+1}\\ q^0&q^1&q^{2k}&q^{2k+1}\\ (-q)^0&(-q)^1&(-q)^{2k}&(-q)^{2k+1}\\ \end{pmatrix} [[/math]]

Now let us pick [math]\lambda\neq1[/math] and write, by using our exponentiation convention above:

[[math]] 1=\lambda^x\quad,\quad -1=\lambda^y [[/math]]

[[math]] q=\lambda^z\quad,\quad -q=\lambda^t [[/math]]

But this gives the formula in the statement.

■

Observe that the above result shows that any Hamadard matrix at [math]N\leq5[/math] is master Hadamard. We have the following generalization of it, once again from :

Theorem

The deformed Fourier matrices [math]F_M\otimes_QF_N[/math] are master Hadamard, for any parameter matrix [math]Q\in M_{M\times N}(\mathbb T)[/math] of the form

[[math]] Q_{ib}=q^{i(Np_b+b)} [[/math]]

where [math]q=e^{2\pi i/MNk}[/math] with [math]k\in\mathbb N[/math], and [math]p_0,\ldots,p_{N-1}\in\mathbb R[/math].

Show Proof

The main construction in , in connection with deformations, that we will follow here, is in terms of master functions as follows:

[[math]] f(z)=f_M(z^{Nk})f_N(z) [[/math]]

Here [math]k\in\mathbb N[/math], and the functions on the right are by definition as follows:

[[math]] f_M(z)=\sum_iz^{Mr_i+i}\quad,\quad f_N(z)=\sum_az^{Np_a+a} [[/math]]

We use the eigenvalues [math]\lambda_{ia}=q^iw^a[/math], where [math]w=e^{2\pi i/N}[/math], and where [math]q^{Nk}=\nu[/math], where [math]\nu^M=1[/math]. We have [math]f(z)=f_M(z^{Nk})f_N(z)[/math], so the exponents are:

[[math]] n_{jb}=Nk(Mr_j+j)+Np_b+b [[/math]]

Thus the associated master Hadamard matrix is given by:

[[math]] \begin{eqnarray*} H_{ia,jb} &=&(q^iw^a)^{Nk(Mr_j+j)+Np_b+b}\\ &=&\nu^{ij}q^{i(Np_b+b)}w^{a(Np_b+b)}\\ &=&\nu^{ij}w^{ab}q^{i(Np_b+b)} \end{eqnarray*} [[/math]]

Now let us recall that we have the following formula, for the tensor product:

[[math]] (F_M\otimes F_N)_{ia,jb}=\nu^{ij}w^{ab} [[/math]]

Thus we have as claimed [math]H=F_M\otimes_QF_N[/math], with [math]Q[/math] being as follows:

[[math]] Q_{ib}=q^{i(Np_b+b)} [[/math]]

Finally, observe that [math]Q[/math] itself is a “master matrix” in our sense, because the indices split. Thus, we are led to the conclusions in the statement.

■

In view of the above examples, and of the lack of other known examples of master Hadamard matrices, the following conjecture was made in :

\begin{conjecture}[Master Hadamard Conjecture] The master Hadamard matrices appear as Di\c t\u a deformations of [math]F_N[/math]. \end{conjecture} There is a relation here with the notions of defect and isolation, that we would like to discuss now. First, we have the following defect computation:

Theorem

The defect of a master Hadamard matrix is given by

[[math]] d(H)=\dim_\mathbb R\left\{B\in M_N(\mathbb C)\Big|\bar{B}=\frac{1}{N}BL, (BR)_{i,ij}=(BR)_{j,ij}\ \forall i,j\right\} [[/math]]

where the matrices on the right are given by

[[math]] L_{ij}=f\left(\frac{1}{\lambda_i\lambda_j}\right)\quad,\quad R_{i,jk}=f\left(\frac{\lambda_j}{\lambda_i\lambda_k}\right) [[/math]]

with [math]f[/math] being the master function.

Show Proof

The first order deformation equations from chapter 7 are as follows:

[[math]] \sum_kH_{ik}\bar{H}_{jk}(A_{ik}-A_{jk})=0 [[/math]]

In our case, with [math]H_{ij}=\lambda_i^{n_j}[/math] we have the following formula:

[[math]] H_{ij}\bar{H}_{jk}=\left(\frac{\lambda_i}{\lambda_j}\right)^{n_k} [[/math]]

Thus, the defect is given by the following formula:

[[math]] d(H)=\dim_\mathbb R\left\{A\in M_N(\mathbb R)\Big|\sum_kA_{ik}\left(\frac{\lambda_i}{\lambda_j}\right)^{n_k}=\sum_kA_{jk}\left(\frac{\lambda_i}{\lambda_j}\right)^{n_k}\ \forall i,j\right\} [[/math]]

Now, pick [math]A\in M_N(\mathbb C)[/math] and set [math]B=AH^t[/math]. We have the following formula:

[[math]] A=\frac{1}{N}B\bar{H} [[/math]]

By using this formula, we have the following computation:

[[math]] \begin{eqnarray*} A\in M_N(\mathbb R) &\iff&B\bar{H}=\bar{B}H\\ &\iff&\bar{B}=\frac{1}{N}B\bar{H}H^* \end{eqnarray*} [[/math]]

On the other hand, the matrix on the right is given by:

[[math]] (\bar{H}H^*)_{ij} =\sum_k\bar{H}_{ik}\bar{H}_{jk} =\sum_k(\lambda_i\lambda_j)^{-n_k} =L_{ij} [[/math]]

Thus [math]A\in M_N(\mathbb R)[/math] if and only the condition [math]\bar{B}=\frac{1}{N}BL[/math] in the statement is satisfied. Regarding now the second condition on [math]A[/math], observe that with [math]A=\frac{1}{N}B\bar{H}[/math] we have:

[[math]] \begin{eqnarray*} \sum_kA_{ik}\left(\frac{\lambda_i}{\lambda_j}\right)^{n_k} &=&\frac{1}{N}\sum_{ks}B_{is}\left(\frac{\lambda_i}{\lambda_j\lambda_s}\right)^{n_k}\\ &=&\frac{1}{N}\sum_sB_{is}R_{s,ij}\\ &=&\frac{1}{N}(BR)_{i,ij} \end{eqnarray*} [[/math]]

Thus the second condition on [math]A[/math] simply reads:

[[math]] (BR)_{i,ij}=(BR)_{j,ij} [[/math]]

But this leads to the conclusions in the statement.

■

We refer to and related papers for more on the master Hadamard matrices. In what follows we will not discuss them further, but we will be back to a related topic, namely Temperley-Lieb algebra representations coming from the complex Hadamard matrices, in chapters 13-16 below, when talking about quantum permutation groups.

General references

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