8a. Deformed products

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We have seen in the previous chapter that the defect theory of Tadej-\.Zyczkowski can be successfully applied to the real Hadamard matrices, and to the generalized Fourier matrices. Following Avan et al. , McNulty-Weigert , Tadej-\.Zyczkowski , , and and other papers, we discuss here a number of more specialized questions, once again in relation with deformations and the defect, regarding the following matrices:

-- The tensor products. The main problem here, which quite surprisingly is non-trivial, and even open, is that of computing the defect of the tensor products.

-- The Di\c t\u a deformations of such tensor products. Here the problem is more complicated than for the tensor products, but a few things, however, can be said.

-- The Butson and the regular matrices. Here we have already met, in chapter 6, a conjecture about regular matrices and deformation, so again, things to be done.

-- The master Hadamard matrices. These are some interesting complex Hadamard matrices, introduced by Avan et al. in , generalizing the Fourier matrices.

-- The McNulty-Weigert matrices. These are again interesting complex Hadamard matrices, introduced by McNulty-Weigert in , which are quite often isolated.

-- The partial Hadamard matrices. Here there are, again, many things to be done, following , inspired by the theory from the square matrix case.

Let us begin with the tensor products. As already mentioned, this is a very interesting topic, which is far from being trivial, and to start with, we have the following result, coming straight from the general defect equations, found in chapter 7:

Proposition

For a tensor product [math]L=H\otimes K[/math] we have

[[math]] d(L)\geq d(H)d(K) [[/math]]

coming from an inclusion of linear spaces, as follows:

[[math]] \widetilde{T}_HX_M\otimes\widetilde{T}_KX_N\subset\widetilde{T}_LX_{MN} [[/math]]

Moreover, the above inequality is not an equality, in general.

Show Proof

We have several things to be proved, the idea being as follows:

(1) Let us first prove that we have the inclusion of linear spaces in the statement. For this purpose, we use the defect equations found in chapter 7, namely:

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

For a tensor product [math]A=B\otimes C[/math], we have the following formula:

[[math]] \begin{eqnarray*} \sum_{kc}(H\otimes K)_{ia,kc}\overline{(H\otimes K)}_{jb,kc}A_{ia,kc} &=&\sum_{kc}H_{ik}K_{ac}\cdot\bar{H}_{jk}\bar{K}_{bc}\cdot B_{ik}C_{ac}\\ &=&\sum_kH_{ik}\bar{H}_{jk}B_{ik}\sum_cK_{ac}\bar{K}_{bc}C_{ac} \end{eqnarray*} [[/math]]

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

[[math]] \begin{eqnarray*} \sum_{kc}(H\otimes K)_{ia,kc}\overline{(H\otimes K)}_{jb,kc}A_{jb,kc} &=&\sum_{kc}H_{ik}K_{ac}\cdot\bar{H}_{jk}bar{K}_{bc}\cdot B_{jk}C_{bc}\\ &=&\sum_kH_{ik}\bar{H}_{jk}B_{jk}\sum_cK_{ac}\bar{K}_{bc}C_{bc} \end{eqnarray*} [[/math]]

Now by assuming [math]B\in\widetilde{T}_HX_M[/math] and [math]C\in\widetilde{T}_KX_N[/math], the two quantities on the right in the above formulae are equal. Thus we have indeed [math]A\in\widetilde{T}_LX_{MN}[/math], as desired.

(2) The defect inequality [math]d(L)\geq d(H)d(K)[/math] follows from (1).

(3) Regarding now the equality case, this does not happen, even in very simple cases. For instance if we consider two Fourier matrices [math]F_2[/math], we know from chapter 7 that:

[[math]] d(F_2\otimes F_2)=10 \gt 9=d(F_2)^2 [[/math]]

There are of course many other counterexamples that can be constructed.

■

Generally speaking, it is quite hard to go beyond the above result. In fact, besides the isotypic decomposition results from chapter 7, valid for the Fourier matrices, there does not seem to be anything conceptual on this subject. We will be back to this, however, in Theorem 8.3 below, with a slight advance on all this.

In what regards now the computation of the defect for the Di\c t\u a deformations, which generalize the usual tensor products, this is an even more difficult question. Our only result here will concern the case where the deformation matrix is generic:

Definition

A rectangular matrix [math]Q\in M_{M\times N}(\mathbb T)[/math] is called “dephased and elsewhere generic” if the entries on its first row and column are all equal to [math]1[/math], and the remaining [math](M-1)(N-1)[/math] entries are algebrically independent over [math]\mathbb Q[/math].

Here the last condition takes of course into account the fact that the entries of [math]Q[/math] themselves have modulus 1, the independence assumption being modulo this fact. With this convention made, we have the following result:

Theorem

Assume that [math]H\in X_M,K\in X_N[/math] are dephased, of Butson type, and that [math]Q\in M_{M\times N}(\mathbb T)[/math] is dephased and elsewhere generic. We have then

[[math]] A=(A_{ia,kc})\in \widetilde{T}_{H\otimes_QK}X_{MN} [[/math]]

when the following equations are satisfied,

[[math]] A_{ac}^{ij}=A_{bc}^{ij}\quad,\quad A_{ac}^{ij}=\overline{A_{ac}^{ji}}\quad,\quad (A_{xy}^{ii})_{xy}\in\widetilde{T}_KX_N [[/math]]

for any [math]a,b,c[/math] and [math]i\neq j[/math], where:

[[math]] A_{ac}^{ij}=\sum_kH_{ik}\bar{H}_{jk}A_{ia,kc} [[/math]]

Show Proof

Consider the standard system of equations for the enveloping tangent space in the statement, coming from the results in chapter 7, namely:

[[math]] \sum_{kc}(H\otimes_QK)_{ia,kc}\overline{(H\otimes_QK)}_{jb,kc}(A_{ia,kc}-A_{jb,kc})=0 [[/math]]

We have the following formula, for our matrix:

[[math]] (H\otimes_QK)_{ia,jb}=q_{ib}H_{ij}K_{ab} [[/math]]

Thus, our system of equations is as follows:

[[math]] \sum_cq_{ic}\bar{q}_{jc}K_{ac}\bar{K}_{bc}\sum_kH_{ik}\bar{H}_{jk}(A_{ia,kc}-A_{jb,kc})=0 [[/math]]

Consider now the variables in the statement, namely:

[[math]] A_{ac}^{ij}=\sum_kH_{ik}\bar{H}_{jk}A_{ia,kc} [[/math]]

The conjugates of these variables are given by:

[[math]] \overline{A_{ac}^{ij}} =\sum_k\bar{H}_{ik}H_{jk}A_{ia,kc} =\sum_kH_{jk}\bar{H}_{ik}A_{ia,kc} [[/math]]

Thus, in terms of these variables, our system becomes simply:

[[math]] \sum_cq_{ic}\bar{q}_{jc}K_{ac}\bar{K}_{bc}(A_{ac}^{ij}-\overline{A_{bc}^{ji}})=0 [[/math]]

More precisely, the above equations must hold for any [math]i,j,a,b[/math]. By distinguishing now two cases, depending on whether [math]i,j[/math] are equal or not, the situation is as follows:

(1) Case [math]i\neq j[/math]. In this case, let us look at the row vector of parameters, namely:

[[math]] (q_{ic}\bar{q}_{jc})_c=(1,q_{i1}\bar{q}_{j1},\ldots,q_{iM}\bar{q}_{jM}) [[/math]]

Since the matrix [math]Q[/math] was assumed to be dephased and elsewhere generic, and because of our assumption [math]i\neq j[/math], the entries of the above vector are linearly independent over [math]\bar{\mathbb Q}[/math]. But, since by linear algebra we can restrict the attention to the computation of the solutions over [math]\bar{\mathbb Q}[/math], the [math]i\neq j[/math] part of our system simply becomes:

[[math]] A_{ac}^{ij}=\overline{A_{bc}^{ji}}\quad,\quad\forall a,b,c,\forall i\neq j [[/math]]

Now by making now [math]a,b,c[/math] vary, we are led to the following equations:

[[math]] A_{ac}^{ij}=A_{bc}^{ij},\quad A_{ac}^{ij}=\overline{A_{ac}^{ji}},\quad\forall a,b,c,i\neq j [[/math]]

(2) Case [math]i=j[/math]. In this case the [math]q[/math] parameters cancel, and our equations become:

[[math]] \sum_cK_{ac}\bar{K}_{bc}(A_{ac}^{ii}-\overline{A_{bc}^{ii}})=0,\quad\forall a,b,c,i [[/math]]

Now observe that we have the following formula:

[[math]] A_{ac}^{ii}=\sum_kA_{ia,kc} [[/math]]

Thus, our equations simply become:

[[math]] \sum_cK_{ac}\bar{K}_{bc}(A_{ac}^{ii}-A_{bc}^{ii})=0,\quad\forall a,b,c,i [[/math]]

But these are precisely the equations for the space [math]\widetilde{T}_KX_N[/math], and we are done.

■

Let us go back now to usual tensor products, and look at the affine cones. In view of the inclusion from Proposition 8.1, the problem is that of finding the biggest subcone of [math]T_{H\otimes K}^\circ X_{MN}[/math], obtained by gluing [math]T_H^\circ X_M,T_K^\circ X_N[/math]. Our answer here, taking into account the two “semi-trivial” cones coming from left and right Di\c t\u a deformations, is as follows:

Theorem

The cones [math]T_H^\circ X_M=\{B\}[/math] and [math]T_K^\circ X_N=\{C\}[/math] glue via the formulae

[[math]] A_{ia,jb}=\lambda B_{ij}+\psi_jC_{ab}+X_{ia}+Y_{jb}+F_{aj} [[/math]]

[[math]] A_{ia,jb}=\phi_bB_{ij}+\mu C_{ab}+X_{ia}+Y_{jb}+E_{ib} [[/math]]

producing in this way two subcones of the affine cone [math]T_{H\otimes K}^\circ X_{MN}=\{A\}[/math].

Show Proof

The idea will be that [math]X_{ia},Y_{jb}[/math] are the trivial parameters, and that [math]E_{ib},F_{aj}[/math] are the Di\c t\u a parameters. Given a matrix [math]A=(A_{ia,jb})[/math], consider the following quantity:

[[math]] P=\sum_{kc}H_{ik}\bar{H}_{jk}K_{ac}\bar{K}_{bc}q^{A_{ia,kc}-A_{jb,kc}} [[/math]]

Let us prove now the first statement, namely that for any choice of matrices [math]B\in T_H^\circ X_M,C\in T_H^\circ X_N[/math] and of parameters [math]\lambda,\psi_j,X_{ia},Y_{jb},F_{aj}[/math], the first matrix [math]A=(A_{ia,jb})[/math] constructed in the statement belongs indeed to [math]T_{H\otimes K}^\circ X_{MN}[/math]. We have:

[[math]] A_{ia,kc}=\lambda B_{ik}+\psi_kC_{ac}+X_{ia}+Y_{kc}+F_{ak} [[/math]]

[[math]] A_{jb,kc}=\lambda B_{jk}+\psi_kC_{bc}+X_{jb}+Y_{kc}+F_{bk} [[/math]]

Now by substracting these equations, we obtain:

[[math]] A_{ia,kc}-A_{jb,kc} =\lambda(B_{ik}-B_{jk})+\psi_k(C_{ac}-C_{bc})+(X_{ia}-X_{jb})+(F_{ak}-F_{bk}) [[/math]]

It follows that the above quantity [math]P[/math] is given by:

[[math]] \begin{eqnarray*} P &=&\sum_{kc}H_{ik}\bar{H}_{jk}K_{ac}\bar{K}_{bc}q^{\lambda(B_{ik}-B_{jk})+\psi_k(C_{ac}-C_{bc})+(X_{ia}-X_{jb})+(F_{ak}-F_{bk})}\\ &=&q^{X_{ia}-X_{jb}}\sum_kH_{ik}\bar{H}_{jk}q^{F_{ak}-F_{bk}}q^{\lambda(B_{ik}-B_{jk})}\sum_cK_{ac}\bar{K}_{bc}(q^{\psi_k})^{C_{ac}-C_{bc}}\\ &=&\delta_{ab}q^{X_{ia}-X_{ja}}\sum_kH_{ik}\bar{H}_{jk}(q^\lambda)^{B_{ik}-B_{jk}}\\ &=&\delta_{ab}\delta_{ij} \end{eqnarray*} [[/math]]

We conclude that we have, as claimed:

[[math]] A\in T_{H\otimes K}^\circ X_{MN} [[/math]]

In the second case now, the proof is similar. First, we have:

[[math]] A_{ia,kc}=\phi_cB_{ik}+\mu C_{ac}+X_{ia}+Y_{kc}+E_{ic} [[/math]]

[[math]] A_{jb,kc}=\phi_cB_{jk}+\mu C_{bc}+X_{jb}+Y_{kc}+E_{jc} [[/math]]

Thus by substracting, we obtain:

[[math]] A_{ia,kc}-A_{jb,kc} =\phi_c(B_{ik}-B_{jk})+\mu(C_{ac}-C_{bc})+(X_{ia}-X_{jb})+(E_{ic}-E_{jc}) [[/math]]

It follows that the above quantity [math]P[/math] is given by:

[[math]] \begin{eqnarray*} P &=&\sum_{kc}H_{ik}\bar{H}_{jk}K_{ac}\bar{K}_{bc}q^{\phi_c(B_{ik}-B_{jk})+\mu(C_{ac}-C_{bc})+(X_{ia}-X_{jb})+(E_{ic}-E_{jc})}\\ &=&q^{X_{ia}-X_{jb}}\sum_cK_{ac}\bar{K}_{bc}q^{E_{ic}-E_{jc}}q^{\mu(C_{ac}-C_{bc})}\sum_kH_{ik}\bar{H}_{jk}(q^{\phi_c})^{B_{ik}-B_{jk}}\\ &=&\delta_{ij}q^{X_{ia}-X_{ib}}\sum_cK_{ac}\bar{K}_{bc}(q^\mu)^{C_{ac}-C_{bc}}\\ &=&\delta_{ij}\delta_{ab} \end{eqnarray*} [[/math]]

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

■

We believe Theorem 8.4 to be “optimal”, in the sense that nothing more can be said about the affine tangent spaces of type [math]T_{H\otimes K}^\circ X_{MN}[/math], in the general case, besides what has been said there. However, this is something rather conjectural. As a continuation now of all this, bringing us into some concrete, interesting mathematics, let us discuss some rationality questions, in relation with the following definition:

Definition

The rational defect of [math]H\in X_N[/math] is the following number:

[[math]] d_\mathbb Q(H)=\dim_\mathbb Q(\widetilde{T}_HC_N\cap M_N(\mathbb Q)) [[/math]]

The vector space on the right is called rational enveloping tangent space at [math]H[/math].

As a first observation, this notion can be extended to all the tangent cones at [math]H[/math], and by using an arbitrary field [math]\mathbb K\subset\mathbb C[/math] instead of [math]\mathbb Q[/math]. Indeed, we can set:

[[math]] T_H^*X_N(\mathbb K)=T_H^*X_N\cap M_N(\mathbb K) [[/math]]

However, in what follows we will be interested only in the objects constructed in Definition 8.5. It follows from definitions that [math]d_\mathbb Q(H)\leq d(H)[/math], and we have:

\begin{conjecture}[Rationality] For the Butson matrices we have:

[[math]] d_\mathbb Q(H)=d(H) [[/math]]

That is, for such matrices, the defect equals the rational defect. \end{conjecture} More generally, we believe that the above equality should hold in the regular matrix case. However, since the regular matrix case is not known to fully cover the Butson matrix case, as explained in chapter 6, we prefer to state our conjecture as above. As a first piece of evidence now, we have the following elementary result:

Theorem

The rationality conjecture holds for [math]H\in H_N(l)[/math] with [math]l=2,3,4,6[/math].

Show Proof

Let us recall that the equations for the enveloping tangent space are:

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

With these equations in hand, the proof goes as follows:

\underline{Case [math]l=2[/math]}. Here the above equations are all real, and have [math]\pm1[/math] coefficients, so in particular, have rational coefficients.

\underline{Case [math]l=3[/math]}. Here we can use the fact that, with [math]w=e^{2\pi i/3}[/math], the real solutions of [math]x+wy+w^2z=0[/math] are those satisfying [math]x=y=z[/math]. We conclude that once again our system, after some manipulations, is equivalent to a real system having rational coefficients.

\underline{Case [math]l=4[/math]}. Here the coefficients are [math]1,i,-1,-i[/math], so by taking the real and imaginary parts, we reach once again to a system with rational coefficients.

\underline{Case [math]l=6[/math]}. Here the study is similar to the study at [math]l=3[/math].

Thus, in all cases under investigation, [math]l=2,3,4,6[/math], we have a real system with rational coefficients, and the result follows from standard linear algebra.

■

Observe that the above method cannot work at [math]l=5[/math], where the equation [math]a+wb+w^2c+w^3d+w^4e=0[/math] with [math]w=e^{2\pi i/5}[/math] and [math]a,b,c,d,e\in\mathbb R[/math] can have exotic solutions. Let us prove now that Conjecture 8.6 is verified for the Fourier matrices. We say that a matrix [math]L^{rs}[/math] over the group [math]\mathbb Z_{p^r}\times\mathbb Z_{p^s}[/math] is dephased if its nonzero entries belong to:

[[math]] X_{rs}=(\mathbb Z_{p^r}-\mathbb Z_{p^{r-1}})\times(\mathbb Z_{p^s}-\mathbb Z_{p^{s-1}}) [[/math]]

Here, and in what follows, we use the convention [math]\mathbb Z_{p^{-1}}=\emptyset[/math]. We have:

Proposition

For [math]F=F_{p^a}[/math], the elements [math]A\in\widetilde{T}_FC_N[/math] are the solutions of

[[math]] A_{ij}=\sum_{r+s\leq a}L^{rs}_{p^{a-r}i,p^{a-s}j} [[/math]]

where the [math]L[/math] variables are free, and form dephased matrices [math]L^{rs}[/math].

Show Proof

The number of [math]L[/math] variables is given by:

[[math]] \begin{eqnarray*} d &=&\sum_{r+s\leq a}|\mathbb Z_{p^r}-\mathbb Z_{p^{r-1}}|\cdot|\mathbb Z_{p^s}-\mathbb Z_{p^{s-1}}|\\ &=&\sum_{r\leq a}p^{a-r}|\mathbb Z_{p^r}-\mathbb Z_{p^{r-1}}|\\ &=&p^a+\sum_{r=1}^ap^{a-r}(p^r-p^{r-1})\\ &=&p^a+a(p-1)p^{a-1}\\ &=&(p+ap-a)p^{a-1} \end{eqnarray*} [[/math]]

Thus the number of [math]L[/math] variables equals the defect [math]d(F)[/math], so it is indeed the good one. As for the proof now, in the general case, this is quite similar to the one at [math]a=1,2[/math]. More precisely, consider the map [math]L\to A[/math]. This map is linear, and in view of the above calculation, it is enough to prove that this map is injective, and has the correct target:

(1) For the injectivity part, recall that at [math]a=2[/math] the formula in the statement reads:

[[math]] A_{ij}=L^{00}_{00}+L^{01}_{0,pj}+L^{10}_{pi,0}+L^{02}_{0j}+L^{20}_{i0}+L^{11}_{pi,pj} [[/math]]

Now assume [math]A=0[/math]. Then with [math]i=j=0[/math] we get [math]L^{00}_{00}=0[/math]. Using this, with [math]i=0[/math] and [math]pj=0,j\neq 0[/math] we get [math]L^{00}_{00}+L^{02}_{0j}=0[/math], and so [math]L^{02}_{0j}=0[/math]. So, with [math]i=0[/math] and [math]pj\neq 0[/math] we therefore obtain [math]L^{00}_{00}+L^{02}_{0j}+L^{01}_{0,pj}=0[/math], and so [math]L^{01}_{0,pj}=0[/math]. Now the same method gives as well succesively [math]L^{20}_{i0}=0[/math] and [math]L^{10}_{pi,0}=0[/math], so we are left with [math]A_{ij}=L^{11}_{pi,pj}[/math], so we must have [math]L^{11}_{pi,pj}=0[/math] as well, and we are done. This method works of course for any [math]a\in\mathbb N[/math].

(2) Regarding now the “target” part, we must prove [math]A\in\widetilde{T}_FC_N[/math]. The equations are:

[[math]] \sum_kw^{(i-j)k}\left(\sum_{r+s\leq a}L^{rs}_{p^{a-r}i,p^{a-s}k}-L^{rs}_{p^{a-r}j,p^{a-s}k}\right)=0 [[/math]]

So, for any indices [math]i,j[/math] and any [math]r+s\leq a[/math], we must prove that we have:

[[math]] \sum_kw^{(i-j)k}\left(L^{rs}_{p^{a-r}i,p^{a-s}k}-L^{rs}_{p^{a-r}j,p^{a-s}k}\right)=0 [[/math]]

In order to do this, consider the following quantity:

[[math]] X_{il}=\frac{1}{p^a}\sum_kw^{lk}L^{rs}_{p^{a-r}i,p^{a-s}k} [[/math]]

We must prove [math]X_{i,i-j}=X_{j,i-j}[/math]. But, with [math]k=m+p^sn[/math], we have:

[[math]] \begin{eqnarray*} X_{il} &=&\frac{1}{p^a}\sum_nw^{lp^sn}\sum_mw^{lm}L^{rs}_{p^{a-r}i,p^{a-s}m}\\ &=&\delta_{l0}\sum_mw^{lm}L^{rs}_{p^{a-r}i,p^{a-s}m} \end{eqnarray*} [[/math]]

Thus we have [math]l\neq 0\implies X_{il}=0[/math], and so [math]X_{i,i-j}=X_{j,i-j}[/math] and we are done.

■

By using the above result, we obtain:

Proposition

For an isotypic Fourier matrix, [math]H=F_N[/math] with [math]N=p^a[/math], we have

[[math]] T_H^\circ C_N=T_HC_N=\widetilde{T}_HC_N=\left\{A\in M_N(\mathbb R)\Big|A_{ij}=\sum_{r+s\leq a}L^{rs}_{p^{a-r}i,p^{a-s}j}\right\} [[/math]]

where the [math]L[/math] variables are free, and form dephased matrices [math]L^{rs}[/math].

Show Proof

We just have to show that the defect of [math]F_N[/math] is exhausted by affine deformations. With [math]k=m+p^sn[/math], as in the proof of Proposition 8.8, we have:

[[math]] \begin{eqnarray*} \sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}} &=&\sum_kw^{(i-j)k}\prod_{r+s\leq a}q^{L^{rs}_{p^{a-r}i,p^{a-s}k}-L^{rs}_{p^{a-r}j,p^{a-s}k}}\\ &=&\sum_nw^{(i-j)p^sn}\sum_mw^{(i-j)m}\prod_{r+s\leq a}q^{L^{rs}_{p^{a-r}i,p^{a-s}m}-L^{rs}_{p^{a-r}j,p^{a-s}m}}\\ &=&\delta_{ij}p^a\sum_mw^{(i-j)m}\prod_{r+s\leq a}q^{L^{rs}_{p^{a-r}i,p^{a-s}m}-L^{rs}_{p^{a-r}j,p^{a-s}m}} \end{eqnarray*} [[/math]]

Now since this quantity vanishes for [math]i\neq j[/math], this gives the result.

■

Observe that the above result shows that Conjecture 8.6 holds for the isotypic Fourier matrices. We will see in what follows that the same happens for any Fourier matrix. In order now to discuss the general case, [math]H=F_N[/math], we will need:

Proposition

If [math]G=H\times K[/math] is such that [math](|H|,|K|)=1[/math], the canonical inclusion

[[math]] \widetilde{T}_{F_H}C_{|H|}\otimes\widetilde{T}_{F_K}C_{|K|}\subset\widetilde{T}_{F_G}C_{|G|} [[/math]]

constructed in Proposition 8.1 is an isomorphism.

Show Proof

We have [math]F_G=F_{H\times K}[/math], and the defect of this matrix is given by:

[[math]] \begin{eqnarray*} d(F_{H\times K}) &=&\sum_{(h,k)\in H\times K}\frac{|H\times K|}{ord(h,k)}\\ &=&\sum_{(h,k)\in H\times K}\frac{|H\times K|}{ord(h)ord(k)}\\ &=&d(F_H)d(F_K) \end{eqnarray*} [[/math]]

Thus the inclusion in the statement must be indeed an isomorphism.

■

With the above result in hand, the idea now will be simply to “glue” the various isotypic formulae coming from Proposition 8.9. Indeed, let us recall from there that in the isotypic case, [math]N=p^a[/math], the parameter set for the enveloping tangent space is:

[[math]] X(p^a)=\bigsqcup_{r+s\leq a}(\mathbb Z_{p^r}-\mathbb Z_{p^{r-1}})\times (\mathbb Z_{p^s}-\mathbb Z_{p^{s-1}}) [[/math]]

Now since the defect is multiplicative over isotypic components, the parameter set in the general case, [math]N=p_1^{a_1}\ldots p_k^{a_k}[/math], will be simply given by:

[[math]] X(p_1^{a_1}\ldots p_k^{a_k})=X(p_1^{a_1})\times\ldots\times X(p_k^{a_k}) [[/math]]

We can obtain from this an even simpler description of the parameter set, just by expanding the product, and gluing the group components. Indeed, let us start with:

Definition

Given a finite abelian group [math]G=\mathbb Z_{p_1^{r_1}}\times\ldots\times\mathbb Z_{p_k^{r_k}}[/math] we set:

[[math]] G^\circ=(\mathbb Z_{p_1^{r_1}}-\mathbb Z_{p_1^{r_1-1}})\times\ldots\times(\mathbb Z_{p_k^{r_k}}-\mathbb Z_{p_k^{r_k-1}}) [[/math]]

A matrix [math]L\in M_{G\times H}(\mathbb R)[/math] will be called dephased if [math]L_{ij}=0[/math] for any [math](i,j)\not\in G^\circ\times H^\circ[/math].

Observe now that, with the above notation [math]G^\circ[/math], the parameter set discussed above is given by the following simple formula:

[[math]] X(N)=\bigsqcup_{G\times H\subset\mathbb Z_N}G^\circ\times H^\circ [[/math]]

In addition, we can see that the collection of dephased matrices [math]L\in M_{G\times H}(\mathbb R)[/math] , over all possible configurations [math]G\times H\subset\mathbb Z_N[/math], takes its parameters precisely in [math]X(N)[/math]. In order now to formulate our main result, we will need one more definition, as follows:

Definition

Given [math]N=p_1^{a_1}\ldots p_k^{a_k}[/math] and a subgroup [math]G\subset \mathbb Z_N[/math], we set

[[math]] \varphi_G(i_1,\ldots,i_k)=(p_1^{a_1-r_1}i_1,\ldots p_k^{a_k-r_k}i_k) [[/math]]

where the exponents [math]r_i\leq a_i[/math] are given by [math]G=\mathbb Z_{p_1^{r_1}}\times\ldots\times\mathbb Z_{p_k^{r_k}}[/math].

Observe that in the case [math]k=1[/math] this function is precisely the one appearing in Proposition 8.9. In fact, we have the following generalization of Proposition 8.9:

Theorem

For [math]H=F_N[/math] the vectors [math]A\in\widetilde{T}_HC_N[/math] appear as plain sums of type

[[math]] A_{ij}=\sum_{G\times H\subset\mathbb Z_N}L^{GH}_{\varphi_G(i)\varphi_H(j)} [[/math]]

where the [math]L[/math] variables form dephased matrices [math]L^{GH}\in M_{G\times H}(\mathbb R)[/math].

Show Proof

According to the above discussion, we just have to glue the various isotypic formulae coming from Proposition 8.9. The gluing formula reads:

[[math]] \begin{eqnarray*} A_{i_1\ldots i_k,j_1\ldots j_k} &=&A_{i_1j_1}\ldots A_{i_kj_k}\\ &=&\left(\sum_{r_1+s_1\leq a_1}L^{r_1s_1p_1}_{p_1^{a_1-r_1}i_1,p_1^{a_1-s_1}j_1}\ldots\sum_{r_k+s_k\leq a_k}L^{r_ks_kp_k}_{p_k^{a_k-r_k}i_k,p_k^{a_k-s_k}j_k}\right)\\ &=&\sum_{r_1+s_1\leq a_1}\ldots \sum_{r_k+s_k\leq a_k}L^{r_1s_1p_1}_{p_1^{a_1-r_1}i_1,p_1^{a_1-s_1}j_1}\ldots L^{r_ks_kp_k}_{p_k^{a_k-r_k}i_k,p_k^{a_k-s_k}j_k} \end{eqnarray*} [[/math]]

Now, let us introduce the following variables:

[[math]] L^{r_1\ldots r_k,s_1\ldots s_k}_{i_1\ldots i_k,j_1\ldots j_k}=L^{r_1s_1}_{i_1j_1}\ldots L^{r_ks_k}_{i_kj_k} [[/math]]

In terms of these new variables, the gluing formula reads:

[[math]] A_{i_1\ldots i_k,j_1\ldots j_k}=\sum_{r_1+s_1\leq a_1}\ldots \sum_{r_k+s_k\leq a_k}L^{r_1\ldots r_k,s_1\ldots s_k}_{p_1^{a_1-r_1}i_1,\ldots p_k^{a_k-r_k}i_k,p_1^{a_1-r_1}j_1\ldots p_k^{a_k-r_k}j_k} [[/math]]

Together with the fact that the new [math]L[/math] variables form dephased matrices, in the sense of Definition 8.11, this gives the result.

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As a main consequence, we have the following result:

Theorem

The rationality conjecture holds for the Fourier matrices.

Show Proof

Indeed, the formula in Theorem 8.13 shows that for [math]H=F_N[/math] the rational defect, as constructed in Definition 8.5, counts the same variables as the usual defect.

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General references

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