Geometry, defect

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

This article was automatically generated from a tex file and may contain conversion errors. If permitted, you may login and edit this article to improve the conversion.

7a. Affine deformations

We have seen so far that some theory for the complex Hadamard matrices [math]H\in M_N(\mathbb T)[/math] can be developed with some inspiration for the real case, [math]H\in M_N(\pm1)[/math], by looking at the Butson matrix case, [math]H\in M_N(\mathbb Z_l)[/math] with [math]l \lt \infty[/math]. However, all this root of unity business ultimately leads into questions of HC flavor, and to put it squarely, wrong way. In this chapter we take a radically different approach to the study of the complex Hadamard matrices. Let us recall that the complex Hadamard manifold appears as:

[[math]] X_N=M_N(\mathbb T)\cap\sqrt{N}U_N [[/math]]

This intersection is far from being smooth, and given a point [math]H\in X_N[/math], the problem is that of understanding the structure of [math]X_N[/math] around [math]H[/math], which is often singular. And this is what we will do, real algebraic geometry, for studying [math]X_N[/math] and its singularities. For this purpose, let us begin with some notations. We denote by [math]X_p[/math] an unspecified neighborhood of a point in a manifold, [math]p\in X[/math]. Also, for [math]q\in\mathbb T_1[/math], meaning that [math]q\in\mathbb T[/math] is close to [math]1[/math], we define [math]q^r[/math] with [math]r\in\mathbb R[/math] by [math](e^{it})^r=e^{itr}[/math]. With these conventions, we have:

Proposition

For [math]H\in X_N[/math] and [math]A\in M_N(\mathbb R)[/math], the following are equivalent:

• The following is an Hadamard matrix, for any [math]q\in\mathbb T_1[/math]:
  [[math]] H_{ij}^q=H_{ij}q^{A_{ij}} [[/math]]
• The following equations hold, for any [math]i\neq j[/math] and any [math]q\in\mathbb T_1[/math]:
  [[math]] \sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}}=0 [[/math]]
• The following equations hold, for any [math]i\neq j[/math] and any [math]\varphi:\mathbb R\to\mathbb C[/math]:
  [[math]] \sum_kH_{ik}\bar{H}_{jk}\varphi(A_{ik}-A_{jk})=0 [[/math]]
• For any [math]i\neq j[/math] and any [math]r\in\mathbb R[/math], with [math]E_{ij}^r=\{k|A_{ik}-A_{jk}=r\}[/math], we have:
  [[math]] \sum_{k\in E_{ij}^r}H_{ik}\bar{H}_{jk}=0 [[/math]]

If these conditions are satisfied, we call the matrix [math]H^q[/math] an affine deformation of [math]H[/math].

Show Proof

These equivalences are all elementary, and can be proved as follows:

[math](1)\iff(2)[/math] Indeed, the scalar products between the rows of [math]H^q[/math] are:

[[math]] \begin{eqnarray*} \lt H^q_i,H^q_j \gt &=&\sum_kH_{ik}q^{A_{ik}}\bar{H}_{jk}\bar{q}^{A_{jk}}\\ &=&\sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}} \end{eqnarray*} [[/math]]

[math](2)\implies(4)[/math] This follows from the following formula, and from the fact that the power functions [math]\{q^r|r\in\mathbb R\}[/math] over the unit circle [math]\mathbb T[/math] are linearly independent:

[[math]] \sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}}=\sum_{r\in\mathbb R}q^r\sum_{k\in E_{ij}^r}H_{ik}\bar{H}_{jk} [[/math]]

[math](4)\implies(3)[/math] This follows from the following formula:

[[math]] \sum_kH_{ik}\bar{H}_{jk}\varphi(A_{ik}-A_{jk})=\sum_{r\in\mathbb R}\varphi(r)\sum_{k\in E_{ij}^r}H_{ik}\bar{H}_{jk} [[/math]]

[math](3)\implies(2)[/math] This simply follows by taking [math]\varphi(r)=q^r[/math].

■

In order to understand the above deformations, which are “affine” in a certain sense, as suggested at the end of the statement, it is convenient to enlarge the attention to all types of deformations. We keep using the neighborhood notation [math]X_p[/math] introduced above, and we consider functions of type [math]f:X_p\to Y_q[/math], which by definition satisfy [math]f(p)=q[/math]. We have the following definition, further clarifying the terminology in Proposition 7.1:

Definition

Let [math]H\in M_N(\mathbb C)[/math] be a complex Hadamard matrix.

• A deformation of [math]H[/math] is a smooth function [math]f:\mathbb T_1\to (X_N)_H[/math].
• The deformation is called “affine” if [math]f_{ij}(q)=H_{ij}q^{A_{ij}}[/math], with [math]A\in M_N(\mathbb R)[/math].
• We call “trivial” the deformations of type [math]f_{ij}(q)=H_{ij}q^{a_i+b_j}[/math], with [math]a,b\in\mathbb R^N[/math].

Here the adjective “affine”, which is used in the same context as in Proposition 7.1, comes from the formula [math]f_{ij}(e^{it})=H_{ij}e^{iA_{ij}t}[/math], because the function [math]t\to A_{ij}t[/math] which produces the exponent is indeed affine. As for the adjective “trivial”, this comes from the fact that the affine deformations of type [math]f(q)=(H_{ij}q^{a_i+b_j})_{ij}[/math] are obtained from [math]H[/math] by multiplying the rows and columns by certain numbers in [math]\mathbb T[/math], so are automatically Hadamard.

The basic example of an affine deformation comes from the Di\c t\u a deformations [math]H\otimes_QK[/math], by taking all parameters [math]q_{ij}\in\mathbb T[/math] to be powers of [math]q\in\mathbb T[/math]. As an example, here are the exponent matrices coming from the left and right Di\c t\u a deformations of [math]F_2\otimes F_2[/math]:

[[math]] A_l= \begin{pmatrix} a&a&b&b\\ c&c&d&d\\ a&a&b&b\\ c&c&d&d \end{pmatrix}\qquad,\qquad A_r= \begin{pmatrix} a&b&a&b\\ a&b&a&b\\ c&d&c&d\\ c&d&c&d \end{pmatrix} [[/math]]

There are of course many other examples, which are less trivial, as for instance the Haagerup matrix, that we met in chapters 5-6, which is as follows:

[[math]] H_6^q=\begin{pmatrix} 1&1&1&1&1&1\\ 1&-1&i&i&-i&-i\\ 1&i&-1&-i&q&-q\\ 1&i&-i&-1&-q&q\\ 1&-i&\bar{q}&-\bar{q}&i&-1\\ 1&-i&-\bar{q}&\bar{q}&-1&i \end{pmatrix} [[/math]]

Observe that this is indeed an affine deformation of [math]H_6=H_6^1[/math], in the sense of Definition 7.2 (2), the corresponding matrix of exponents being as follows:

[[math]] A=\begin{pmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&1&1\\ 0&0&0&0&1&1\\ 0&0&-1&-1&0&0\\ 0&0&-1&-1&0&0 \end{pmatrix} [[/math]]

We will see that there are many other interesting examples of affine deformations, and that some general theory for such deformations can be developed. In order to investigate now the above types of deformations, we will use the corresponding tangent vectors. So, let us recall that the complex Hadamard matrix manifold [math]X_N[/math] is given by:

[[math]] X_N=M_N(\mathbb T)\cap\sqrt{N}U_N [[/math]]

This observation leads to the following definition, where in the first part we denote by [math]T_pX[/math] the tangent space to a point in a smooth manifold, [math]p\in X[/math]:

Definition

Associated to a point [math]H\in X_N[/math] are the following objects:

• The enveloping tangent space: [math]\widetilde{T}_HX_N=T_HM_N(\mathbb T)\cap T_H\sqrt{N}U_N[/math].
• The tangent cone [math]T_HX_N[/math]: the set of tangent vectors to the deformations of [math]H[/math].
• The affine tangent cone [math]T_H^\circ X_N[/math]: same as above, using affine deformations only.
• The trivial tangent cone [math]T_H^\times X_N[/math]: as above, using trivial deformations only.

Observe that [math]\widetilde{T}_HX_N,T_H^\times X_N[/math] are real linear spaces, and that [math]T_HX_N,T_H^\circ X_N[/math] are two-sided cones, in the sense that they satisfy the following condition:

[[math]] \lambda\in\mathbb R,A\in T\implies\lambda A\in T [[/math]]

Observe also that we have inclusions of cones, as follows:

[[math]] T_H^\times X_N\subset T_H^\circ X_N\subset T_HX_N\subset\widetilde{T}_HX_N [[/math]]

In more algebraic terms now, these various tangent cones are best described by the corresponding matrices, and we have here the following result:

Theorem

The cones [math]T_H^\times X_N\subset T_H^\circ X_N\subset T_HX_N\subset\widetilde{T}_HX_N[/math] are as follows:

• [math]\widetilde{T}_HX_N[/math] can be identified with the linear space formed by the matrices [math]A\in M_N(\mathbb R)[/math] satisfying the following condition:
  [[math]] \sum_kH_{ik}\bar{H}_{jk}(A_{ik}-A_{jk})=0 [[/math]]
  .
• [math]T_HX_N[/math] consists of those matrices [math]A\in M_N(\mathbb R)[/math] appearing as [math]A_{ij}=g_{ij}'(0)[/math], where [math]g:M_N(\mathbb R)_0\to M_N(\mathbb R)_0[/math] satisfies:
  [[math]] \sum_kH_{ik}\bar{H}_{jk}e^{i(g_{ik}(t)-g_{jk}(t))}=0 [[/math]]
• [math]T^\circ_HX_N[/math] is formed by the matrices [math]A\in M_N(\mathbb R)[/math] satisfying the following condition, for any [math]i\neq j[/math] and any [math]q\in\mathbb T[/math]:
  [[math]] \sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}}=0 [[/math]]
• [math]T^\times_HX_N[/math] is formed by the matrices [math]A\in M_N(\mathbb R)[/math] which are of the form [math]A_{ij}=a_i+b_j[/math], for certain vectors [math]a,b\in\mathbb R^N[/math].

Show Proof

All these assertions can be deduced by using basic differential geometry:

(1) This result is well-known, the idea being as follows. First, [math]M_N(\mathbb T)[/math] is defined by the algebraic relations [math]|H_{ij}|^2=1[/math], and with [math]H_{ij}=X_{ij}+iY_{ij}[/math] we have:

[[math]] d|H_{ij}|^2 =d(X_{ij}^2+Y_{ij}^2) =2(X_{ij}\dot{X}_{ij}+Y_{ij}\dot{Y}_{ij}) [[/math]]

Consider now an arbitrary vector [math]\xi\in T_HM_N(\mathbb C)[/math], written as follows:

[[math]] \xi=\sum_{ij}\alpha_{ij}\dot{X}_{ij}+\beta_{ij}\dot{Y}_{ij} [[/math]]

This vector belongs then to the tangent space [math]T_HM_N(\mathbb T)[/math] if and only if we have:

[[math]] \lt \xi,d|H_{ij}|^2 \gt =0 [[/math]]

We therefore obtain the following formula, for the tangent cone:

[[math]] T_HM_N(\mathbb T)=\left\{\sum_{ij}A_{ij}(Y_{ij}\dot{X}_{ij}-X_{ij}\dot{Y}_{ij})\Big|A_{ij}\in\mathbb R\right\} [[/math]]

We also know that the rescaled unitary group [math]\sqrt{N}U_N[/math] is defined by the following algebraic relations, where [math]H_1,\ldots,H_N[/math] are the rows of [math]H[/math]:

[[math]] \lt H_i,H_j \gt =N\delta_{ij} [[/math]]

The relations [math] \lt H_i,H_i \gt =N[/math] being automatic for the matrices [math]H\in M_N(\mathbb T)[/math], if for [math]i\neq j[/math] we let [math]L_{ij}= \lt H_i,H_j \gt [/math], then we have:

[[math]] \widetilde{T}_HC_N=\left\{\xi\in T_HM_N(\mathbb T)\Big| \lt \xi,\dot{L}_{ij} \gt =0,\,\forall i\neq j\right\} [[/math]]

On the other hand, differentiating the formula of [math]L_{ij}[/math] gives:

[[math]] \dot{L}_{ij}=\sum_k(X_{ik}+iY_{ik})(\dot{X}_{jk}-i\dot{Y}_{jk})+(X_{jk}-iY_{jk})(\dot{X}_{ik}+i\dot{Y}_{ik}) [[/math]]

Now if we pick [math]\xi\in T_HM_N(\mathbb T)[/math], written as above in terms of [math]A\in M_N(\mathbb R)[/math], we obtain:

[[math]] \lt \xi,\dot{L}_{ij} \gt =i\sum_k\bar{H}_{ik}H_{jk}(A_{ik}-A_{jk}) [[/math]]

Thus we have reached to the description of [math]\widetilde{T}_HX_N[/math] in the statement.

(2) We pick an arbitrary deformation, written as [math]f_{ij}(e^{it})=H_{ij}e^{ig_{ij}(t)}[/math]. Observe first that the Hadamard condition corresponds to the equations in the statement, namely:

[[math]] \sum_kH_{ik}\bar{H}_{jk}e^{i(g_{ik}(t)-g_{jk}(t))}=0 [[/math]]

Observe also that by differentiating this formula at [math]t=0[/math], we obtain:

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

Thus the matrix [math]A_{ij}=g_{ij}'(0)[/math] belongs indeed to [math]\widetilde{T}_HX_N[/math], so we obtain in this way a certain map, as follows:

[[math]] T_HX_N\to\widetilde{T}_HX_N [[/math]]

In order to check that this map is indeed the correct one, we have to verify that, for any [math]i,j[/math], the tangent vector to our deformation is given by:

[[math]] \xi_{ij}=g_{ij}'(0)(Y_{ij}\dot{X}_{ij}-X_{ij}\dot{Y}_{ij}) [[/math]]

But this latter verification is just a one-variable problem. So, by dropping all [math]i,j[/math] indices, which is the same as assuming [math]N=1[/math], we have to check that for any point [math]H\in\mathbb T[/math], written [math]H=X+iY[/math], the tangent vector to the deformation [math]f(e^{it})=He^{ig(t)}[/math] is:

[[math]] \xi=g'(0)(Y\dot{X}-X\dot{Y}) [[/math]]

But this is clear, because the unit tangent vector at [math]H\in\mathbb T[/math] is [math]\eta=-i(Y\dot{X}-X\dot{Y})[/math], and its coefficient coming from the deformation is:

[[math]] (e^{ig(t)})'_{|t=0}=-ig'(0) [[/math]]

(3) Observe first that by taking the derivative at [math]q=1[/math] of the condition (2) in Proposition 7.1, of just by using the condition (3) there with the function [math]\varphi(r)=r[/math], we get:

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

Thus we have a map [math]T_H^\circ X_N\to\widetilde{T}_HX_N[/math], and the fact that is map is indeed the correct one comes for instance from the computation in (2), with [math]g_{ij}(t)=A_{ij}t[/math].

(4) Observe first that the Hadamard matrix condition is satisfied, because:

[[math]] \sum_kH_{ik}\bar{H}_{jk}q^{A_{ik}-A_{jk}} =q^{a_i-a_j}\sum_kH_{ik}\bar{H}_{jk} =\delta_{ij} [[/math]]

As for the fact that [math]T_H^\times X_N[/math] is indeed the space in the statement, this is clear.

■

Let [math]Z_N\subset X_N[/math] be the real algebraic manifold formed by all the dephased [math]N\times N[/math] complex Hadamard matrices. Observe that we have a quotient map [math]X_N\to Z_N[/math], obtained by dephasing. With this notation, we have the following refinement of (4) above:

Proposition

We have a direct sum decomposition of cones

[[math]] T_H^\circ X_N=T_H^\times X_N\oplus T_H^\circ Z_N [[/math]]

where at right we have the affine tangent cone to the dephased manifold [math]X_N\to Z_N[/math].

Show Proof

If we denote by [math]M_N^\circ(\mathbb R)[/math] the set of matrices having [math]0[/math] outside the first row and column, we have a direct sum decomposition, as follows:

[[math]] \widetilde{T}_H^\circ X_N=M_N^\circ(\mathbb R)\oplus\widetilde{T}_H^\circ Z_N [[/math]]

Now by looking at the affine cones, and using Theorem 7.4, this gives the result.

■

Summarizing, we have so far a number of theoretical results about the tangent cones [math]T_HX_N[/math] that we are interested in, and their versions coming from the trivial and affine deformations, and from the intersection formula [math]X_N=M_N(\mathbb T)\cap\sqrt{N}U_N[/math] as well. In practice now, passed a few special cases where all these cones collapse to the trivial cone [math]T_N^\times X_N[/math], which by Proposition 7.5 means that the image of [math]H\in X_N[/math] must be isolated in the dephased manifold [math]X_N\to Z_N[/math], things are quite difficult to compute. However, as a concrete numerical invariant arising from all this, which can be effectively computed in many cases of interest, we have, following Tadej-\.Zyczkowski :

Definition

The real dimension [math]d(H)[/math] of the enveloping tangent space

[[math]] \widetilde{T}_HX_N=T_HM_N(\mathbb T)\cap T_H\sqrt{N}U_N [[/math]]

is called undephased defect of a complex Hadamard matrix [math]H\in X_N[/math].

In view of Proposition 7.5, it is sometimes convenient to replace [math]d(H)[/math] by the following related quantity, also introduced in , and called dephased defect of [math]H[/math]:

[[math]] d'(H)=d(H)-2N+1 [[/math]]

In what follows we will rather use the quantity [math]d(H)[/math] defined before, which behaves better with respect to a number of operations, and simply call it “defect” of [math]H[/math]. We already know, from Theorem 7.4, what is the precise geometric meaning of the defect, and how to compute it. Let us record again these results, that we will use many times in what follows, in a slightly different form, closer to the spirit of :

Theorem

The defect [math]d(H)[/math] is the real dimension of the linear space

[[math]] \widetilde{T}_HX_N=\left\{A\in M_N(\mathbb R)\Big|\sum_kH_{ik}\bar{H}_{jk}(A_{ik}-A_{jk})=0,\forall i,j\right\} [[/math]]

and the elements of this space are those making [math]H^q_{ij}=H_{ij}q^{A_{ij}}[/math] Hadamard at order [math]1[/math].

Show Proof

Here the first assertion is something that we already know, from Theorem 7.4 (1), and the second assertion follows either from Theorem 7.4 and its proof, or directly from the definition of the enveloping tangent space [math]\widetilde{T}_HX_N[/math], as used in Definition 7.6.

■

Still following , here are a few basic properties of the defect:

Proposition

Let [math]H\in X_N[/math] be a complex Hadamard matrix.

• If [math]H\simeq\widetilde{H}[/math] then [math]d(H)=d(\widetilde{H})[/math].
• We have [math]2N-1\leq d(H)\leq N^2[/math].
• If [math]d(H)=2N-1[/math], the image of [math]H[/math] in the dephased manifold [math]X_N\to Z_N[/math] is isolated.

Show Proof

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

(1) If we let [math]K_{ij}=a_ib_jH_{ij}[/math] with [math]|a_i|=|b_j|=1[/math] be a trivial deformation of our matrix [math]H[/math], the equations for the enveloping tangent space for [math]K[/math] are:

[[math]] \sum_ka_ib_kH_{ik}\bar{a}_j\bar{b}_k\bar{H}_{jk}(A_{ik}-A_{jk})=0 [[/math]]

By simplifying we obtain the equations for [math]H[/math], so [math]d(H)[/math] is invariant under trivial deformations. Since [math]d(H)[/math] is invariant as well by permuting rows or columns, we are done.

(2) Consider the inclusions [math]T_H^\times X_N\subset T_HX_N\subset\widetilde{T}_HX_N[/math]. Since [math]\dim(T_H^\times X_N)=2N-1[/math], the inequality at left holds indeed. As for the inequality at right, this is clear.

(3) If [math]d(H)=2N-1[/math] then [math]T_HX_N=T_H^\times X_N[/math], so any deformation of [math]H[/math] is trivial. Thus the image of [math]H[/math] in the quotient manifold [math]X_N\to Z_N[/math] is indeed isolated, as stated.

■

7b. Defect computations

As an illustration for the above notions, let us discuss now the computation of the defect for the most basic examples of complex Hadamard matrices that we know, namely the real ones, and the Fourier ones. In order to deal with the real case, it is convenient to modify the general formula from Theorem 7.7, via a change of variables, as follows:

Proposition

We have a linear space isomorphism as follows,

[[math]] \widetilde{T}_HX_N\simeq\left\{E\in M_N(\mathbb C)\Big|E=E^*,(EH)_{ij}\bar{H}_{ij}\in\mathbb R,\forall i,j\right\} [[/math]]

the correspondences [math]A\to E[/math] and [math]E\to A[/math] being given by the formulae

[[math]] E_{ij}=\sum_kH_{ik}\bar{H}_{jk}A_{ik}\quad,\quad A_{ij}=(EH)_{ij}\bar{H}_{ij} [[/math]]

with [math]A\in\widetilde{T}_HX_N[/math] being the usual components, from Theorem 7.7.

Show Proof

Given a matrix [math]A\in M_N(\mathbb C)[/math], if we set [math]R_{ij}=A_{ij}H_{ij}[/math] and [math]E=RH^*[/math], the correspondence [math]A\to R\to E[/math] is then bijective onto [math]M_N(\mathbb C)[/math], and we have:

[[math]] E_{ij}=\sum_kH_{ik}\bar{H}_{jk}A_{ik} [[/math]]

In terms of these new variables, the equations in Theorem 7.7 become:

[[math]] E_{ij}=\bar{E}_{ji} [[/math]]

Thus, when taking into account these conditions, we are simply left with the conditions [math]A_{ij}\in\mathbb R[/math]. But these correspond to the conditions [math](EH)_{ij}\bar{H}_{ij}\in\mathbb R[/math], as claimed.

■

With the above result in hand, we can now compute the defect of the real Hadamard matrices. The result here, from Szöllősi , is as follows:

Theorem

For any real Hadamard matrix [math]H\in M_N(\pm1)[/math] we have

[[math]] \widetilde{T}_HX_N\simeq M_N(\mathbb R)^{symm} [[/math]]

and so the corresponding defect is [math]d(H)=N(N+1)/2[/math].

Show Proof

We use Proposition 7.9. Since [math]H[/math] is now real the condition [math](EH)_{ij}\bar{H}_{ij}\in\mathbb R[/math] there simply tells us that [math]E[/math] must be real, and this gives the result.

■

As another computation now, let us discuss the case [math]N=4[/math]. Here we know from chapter 5 that the only complex Hadamard matrices are, up to equivalence, the Di\c t\u a deformations of [math]F_4[/math]. To be more precise, we have the following result:

Proposition

The complex Hadamard matrices at [math]N=4[/math] are, up to equivalence, the following matrices, appearing as Di\c t\u a deformations of [math]F_4[/math]:

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

At [math]q\in\{1,i,-1,-i\}[/math] we obtain tensor products of Fourier matrices, as follows:

• At [math]q=1[/math] we have [math]F_{2,2}^q=F_2\otimes F_2[/math].
• At [math]q=-1[/math] we have [math]F_{2,2}^q\simeq F_2\otimes F_2[/math].
• At [math]q=\pm i[/math] we have [math]F_{2,2}^q\simeq F_4[/math].

Show Proof

The first assertion is something that we already know, from chapter 5. Regarding now the [math]q=1,i,-1,-i[/math] specializations, the situation here is as follows:

(1) This is clear from definitions.

(2) This follows from (1), by permuting the third and the fourth columns:

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

(3) This follows from the following computation:

[[math]] F_{2,2}^{\pm i} =\begin{pmatrix} 1&1&1&1\\ 1&-1&\pm i&\mp i\\ 1&1&-1&-1\\ 1&-1&\mp i&\pm i \end{pmatrix} \sim\begin{pmatrix} 1&1&1&1\\ 1&i&-1&-i\\ 1&-1&1&-1\\ 1&-i&-1&i \end{pmatrix} =F_4 [[/math]]

Here we have interchanged the second column with the third one in the case [math]q=i[/math], and we have used a cyclic permutation of the last 3 columns in the case [math]q=-i[/math].

■

Let us compute now the defect of the above matrices. We will work out everything in detail, as an illustration for how the equations in Theorem 7.7 work. The result is:

Theorem

The defect of the [math]4\times4[/math] complex Hadamard matrices is given by

[[math]] d(F_{2,2}^q)= \begin{cases} 10&(q=\pm1)\\ 8&(q\neq\pm1) \end{cases} [[/math]]

with [math]F_{2,2}^q[/math], depending on [math]q\in\mathbb T[/math], being the matrix in Proposition 7.11.

Show Proof

Our starting point are the equations in Theorem 7.7, namely:

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

Since the [math]i \gt j[/math] equations are equivalent to the [math]i \lt j[/math] ones, and the [math]i=j[/math] equations are trivial, we just have to write down the equations corresponding to indices [math]i \lt j[/math]. And, with [math]ij=01,02,03,12,13,23[/math], these equations are:

[[math]] \begin{eqnarray*} (A_{00}-A_{10})-(A_{01}-A_{11})+\bar{q}(A_{02}-A_{12})-\bar{q}(A_{03}-A_{13})&=&0\\ (A_{00}-A_{20})+(A_{01}-A_{21})-(A_{02}-A_{22})-(A_{03}-A_{23})&=&0\\ (A_{00}-A_{30})-(A_{01}-A_{31})-\bar{q}(A_{02}-A_{32})+\bar{q}(A_{03}-A_{33})&=&0\\ (A_{10}-A_{20})-(A_{11}-A_{21})-q(A_{12}-A_{22})+q(A_{13}-A_{23})&=&0\\ (A_{10}-A_{30})+(A_{11}-A_{31})-(A_{12}-A_{32})-(A_{13}-A_{33})&=&0\\ (A_{20}-A_{30})-(A_{21}-A_{31})+\bar{q}(A_{22}-A_{32})-\bar{q}(A_{23}-A_{33})&=&0 \end{eqnarray*} [[/math]]

Assume first [math]q\neq\pm 1[/math]. Then [math]q[/math] is not real, and appears in 4 of the above equations. But these 4 equations can be written in the following way:

[[math]] \begin{eqnarray*} (A_{00}-A_{01})-(A_{10}-A_{11})+\bar{q}((A_{02}-A_{03})-(A_{12}-A_{13}))&=&0\\ (A_{00}-A_{01})-(A_{30}-A_{31})-\bar{q}((A_{02}-A_{03})-(A_{32}-A_{33}))&=&0\\ (A_{10}-A_{11})-(A_{20}-A_{21})-q((A_{12}-A_{13})-(A_{22}-A_{23}))&=&0\\ (A_{20}-A_{21})-(A_{30}-A_{31})+\bar{q}((A_{22}-A_{23})-(A_{32}-A_{33}))&=&0 \end{eqnarray*} [[/math]]

Now since the unknowns are real, and [math]q[/math] is not, we conclude that the terms between braces in the left part must be all equal, and that the same must happen at right:

[[math]] \begin{eqnarray*} A_{00}-A_{01}&=&A_{10}-A_{11}=A_{20}-A_{21}=A_{30}-A_{31}\\ A_{02}-A_{03}&=&A_{12}-A_{13}=A_{22}-A_{23}=A_{32}-A_{33} \end{eqnarray*} [[/math]]

Thus, the equations involving [math]q[/math] tell us that [math]A[/math] must be of the following form:

[[math]] A=\begin{pmatrix} a&a+x&e+y&e\\ b&b+x&f+y&f\\ c&c+x&g+y&g\\ d&d+x&h+y&h\end{pmatrix} [[/math]]

Let us plug now these values in the remaining 2 equations. We obtain:

[[math]] \begin{eqnarray*} a-c+a+x-c-x-e-y+g+y-e+g&=&0\\ b-d+b+x-d-x-f-y+h+y-f+h&=&0 \end{eqnarray*} [[/math]]

Thus we must have [math]a+g=c+e[/math] and [math]b+h=d+f[/math], which are independent conditions. We conclude that the dimension of the space of solutions is [math]10-2=8[/math], as claimed. Assume now [math]q=\pm 1[/math]. For simplicity we set [math]q=1[/math], and we compute the dephased defect. The dephased equations, obtained by setting [math]A_{i0}=A_{0j}=0[/math] in our system, are:

[[math]] \begin{eqnarray*} A_{11}-A_{12}+A_{13}&=&0\\ -A_{21}+A_{22}+A_{23}&=&0\\ A_{31}+A_{32}-A_{33}&=&0\\ -A_{11}+A_{21}-A_{12}+A_{22}+A_{13}-A_{23}&=&0\\ A_{11}-A_{31}-A_{12}+A_{32}-A_{13}+A_{33}&=&0\\ -A_{21}+A_{31}+A_{22}-A_{32}-A_{23}+A_{33}&=&0 \end{eqnarray*} [[/math]]

The first three equations tell us that our matrix must be of the following form:

[[math]] A=\begin{pmatrix}a&a+b&b\\ c+d&c&d\\ e&f&e+f\end{pmatrix} [[/math]]

Now by plugging these values in the last three equations, these become:

[[math]] \begin{eqnarray*} -a+c+d-a-b+c+b-d&=&0\\ a-e-a-b+f-b+e+f&=&0\\ -c-d+e+c-f-d+e+f&=&0 \end{eqnarray*} [[/math]]

Thus we must have [math]a=c[/math], [math]b=f[/math], [math]d=e[/math], and since these conditions are independent, the dephased defect is 3, and so the undephased defect is [math]3+7=10[/math], as claimed.

■

In general, the defect computation for the Di\c t\u a deformations, of even for the usual tensor products, is a difficult question. We will be back to this in chapter 8 below.

7c. Fourier matrices

Let us discuss now a fundamental question, namely the computation of the defect of the Fourier matrix [math]F_G[/math]. The main idea here goes back to a 1989 preprint of Karabegov , with some supplementary contributions from Nicoara , in 2006, and then the main formula, in the cyclic group case, was obtained by Tadej-\.Zyczkowski in , and the corresponding deformations of [math]F_G[/math] were studied by Nicoara-White in . As a first result on this subject, we have, following Tadej-\.Zyczkowski :

Theorem

For a Fourier matrix [math]F=F_G[/math], the matrices [math]A\in\widetilde{T}_FX_N[/math] with [math]N=|G|[/math], are those of the form [math]A=PF^*[/math], with [math]P\in M_N(\mathbb C)[/math] satisfying

[[math]] P_{ij}=P_{i+j,j}=\bar{P}_{i,-j} [[/math]]

where the indices [math]i,j[/math] are by definition taken in the group [math]G[/math].

Show Proof

We use the system of equations in Theorem 7.7, namely:

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

By decomposing our finite abelian group as [math]G=\mathbb Z_{N_1}\times\ldots\times\mathbb Z_{N_r}[/math] we can assume:

[[math]] F=F_{N_1}\otimes\ldots\otimes F_{N_r} [[/math]]

Thus with [math]w_k=e^{2\pi i/k}[/math] we have the following formula:

[[math]] F_{i_1\ldots i_r,j_1\ldots j_r}=(w_{N_1})^{i_1j_1}\ldots (w_{N_r})^{i_rj_r} [[/math]]

With [math]N=N_1\ldots N_r[/math] and [math]w=e^{2\pi i/N}[/math], we obtain the following formula:

[[math]] F_{i_1\ldots i_r,j_1\ldots j_r}=w^{\left(\frac{i_1j_1}{N_1}+\ldots+\frac{i_rj_r}{N_r}\right)N} [[/math]]

Thus the matrix of our system of equations is given by:

[[math]] F_{i_1\ldots i_r,k_1\ldots k_r}\bar{F}_{j_1\ldots j_r,k_1\ldots k_r}=w^{\left(\frac{(i_1-j_1)k_1}{N_1}+\ldots+\frac{(i_r-j_r)k_r}{N_r}\right)N} [[/math]]

Now by plugging in a multi-indexed matrix [math]A[/math], our system becomes:

[[math]] \sum_{k_1\ldots k_r}w^{\left(\frac{(i_1-j_1)k_1}{N_1}+\ldots+\frac{(i_r-j_r)k_r}{N_r}\right)N}(A_{i_1\ldots i_r,k_1\ldots k_r}-A_{j_1\ldots j_r,k_1\ldots k_r})=0 [[/math]]

Now observe that in the above formula we have in fact two matrix multiplications, so our system can be simply written as:

[[math]] (AF)_{i_1\ldots i_r,i_1-j_1\ldots i_r-j_r}-(AF)_{j_1\ldots j_r,i_1-j_1\ldots i_r-j_r}=0 [[/math]]

Now recall that our indices have a “cyclic” meaning, so they belong in fact to the group [math]G[/math]. So, with [math]P=AF[/math], and by using multi-indices, our system is simply:

[[math]] P_{i,i-j}=P_{j,i-j} [[/math]]

With [math]i=I+J,j=I[/math] we obtain the condition [math]P_{I+J,J}=P_{IJ}[/math] in the statement. In addition, [math]A=PF^*[/math] must be a real matrix. But, if we set [math]\tilde{P}_{ij}=\bar{P}_{i,-j}[/math], we have:

[[math]] \begin{eqnarray*} \overline{(PF^*)}_{i_1\ldots i_r,j_1\ldots j_r} &=&\sum_{k_1\ldots k_r}\bar{P}_{i_1\ldots i_r,k_1\ldots k_r}F_{j_1\ldots j_r,k_1\ldots k_r}\\ &=&\sum_{k_1\ldots k_r}\tilde{P}_{i_1\ldots i_r,-k_1\ldots -k_r}(F^*)_{-k_1\ldots -k_r,j_1\ldots j_r}\\ &=&(\tilde{P}F^*)_{i_1\ldots i_r,j_1\ldots j_r} \end{eqnarray*} [[/math]]

Thus we have [math]\overline{PF^*}=\tilde{P}F^*[/math], so the fact that the matrix [math]PF^*[/math] is real, which means by definition that we have [math]\overline{PF^*}=PF^*[/math], can be reformulated as [math]\tilde{P}F^*=PF^*[/math], and hence as [math]\tilde{P}=P[/math]. So, we obtain the conditions [math]P_{ij}=\bar{P}_{i,-j}[/math] in the statement.

■

We can now compute the defect, and we are led to the following formula:

Theorem

The defect of a Fourier matrix [math]F_G[/math] is given by

[[math]] d(F_G)=\sum_{g\in G}\frac{|G|}{ord(g)} [[/math]]

and equals as well the number of [math]1[/math] entries of the matrix [math]F_G[/math].

Show Proof

According to the formula [math]A=PF^*[/math] from Theorem 7.13, the defect [math]d(F_G)[/math] is the dimension of the real vector space formed by the matrices [math]P\in M_N(\mathbb C)[/math] satisfying:

[[math]] P_{ij}=P_{i+j,j}=\bar{P}_{i,-j} [[/math]]

Here, and in what follows, the various indices [math]i,j,\ldots[/math] will be taken in [math]G[/math]. Now the point is that, in terms of the columns of our matrix [math]P[/math], the above conditions are:

(1) The entries of the [math]j[/math]-th column of [math]P[/math], say [math]C[/math], must satisfy [math]C_i=C_{i+j}[/math].

(2) The [math](-j)[/math]-th column of [math]P[/math] must be conjugate to the [math]j[/math]-th column of [math]P[/math].

Thus, in order to count the above matrices [math]P[/math], we can basically fill the columns one by one, by taking into account the above conditions. In order to do so, consider the subgroup [math]G_2=\{j\in G|2j=0\}[/math], and then write [math]G[/math] as a disjoint union, as follows:

[[math]] G=G_2\sqcup X\sqcup(-X) [[/math]]

With this notation, the algorithm is as follows. First, for any [math]j\in G_2[/math] we must fill the [math]j[/math]-th column of [math]P[/math] with real numbers, according to the periodicity rule:

[[math]] C_i=C_{i+j} [[/math]]

Then, for any [math]j\in X[/math] we must fill the [math]j[/math]-th column of [math]P[/math] with complex numbers, according to the same periodicity rule [math]C_i=C_{i+j}[/math]. And finally, once this is done, for any [math]j\in X[/math] we just have to set the [math](-j)[/math]-th column of [math]P[/math] to be the conjugate of the [math]j[/math]-th column.

So, let us compute the number of choices for filling these columns. Our claim is that, when uniformly distributing the choices for the [math]j[/math]-th and [math](-j)[/math]-th columns, for [math]j\notin G_2[/math], there are exactly [math][G: \lt j \gt ][/math] choices for the [math]j[/math]-th column, for any [math]j[/math]. Indeed:

(1) For the [math]j[/math]-th column with [math]j\in G_2[/math] we must simply pick [math]N[/math] real numbers subject to the condition [math]C_i=C_{i+j}[/math] for any [math]i[/math], so we have indeed [math][G: \lt j \gt ][/math] such choices.

(2) For filling the [math]j[/math]-th and [math](-j)[/math]-th column, with [math]j\notin G_2[/math], we must pick [math]N[/math] complex numbers subject to the condition [math]C_i=C_{i+j}[/math] for any [math]i[/math]. Now since there are [math][G: \lt j \gt ][/math] choices for these numbers, so a total of [math]2[G: \lt j \gt ][/math] choices for their real and imaginary parts, on average over [math]j,-j[/math] we have [math][G: \lt j \gt ][/math] choices, and we are done again.

Summarizing, the dimension of the vector space formed by the matrices [math]P[/math], which is equal to the number of choices for the real and imaginary parts of the entries of [math]P[/math], is:

[[math]] d(F_G)=\sum_{j\in G}[G: \lt j \gt ] [[/math]]

But this is exactly the number in the statement. Regarding now the second assertion, according to the definition of [math]F_G[/math], the number of [math]1[/math] entries of [math]F_G[/math] is given by:

[[math]] \begin{eqnarray*} \#(1\in F_G) &=&\#\left\{(g,\chi)\in G\times\widehat{G}\Big|\chi(g)=1\right\}\\ &=&\sum_{g\in G}\#\left\{\chi\in\widehat{G}\Big|\chi(g)=1\right\}\\ &=&\sum_{g\in G}\frac{|G|}{ord(g)} \end{eqnarray*} [[/math]]

Thus, the second assertion follows from the first one.

■

Let us finish now the work, and explicitely compute the defect of [math]F_G[/math]. It is convenient to consider the following quantity, which behaves better:

[[math]] \delta(G)=\sum_{g\in G}\frac{1}{ord(g)} [[/math]]

As a first example, consider a cyclic group [math]G=\mathbb Z_N[/math], with [math]N=p^a[/math] power of a prime. The count here is very simple, over sets of elements having a given order:

[[math]] \begin{eqnarray*} \delta(\mathbb Z_{p^a}) &=&1+(p-1)p^{-1}+(p^2-p)p^{-2}+\ldots+(p^a-p^{a-1})p^{-1}\\ &=&1+a-\frac{a}{p} \end{eqnarray*} [[/math]]

In order to extend this kind of count to the general abelian case, we use two ingredients. First is the following result, which splits the computation over isotypic components:

Proposition

For any finite groups [math]G,H[/math] we have:

[[math]] \delta(G\times H)\geq\delta(G)\delta(H) [[/math]]

In addition, if [math](|G|,|H|)=1[/math], we have equality.

Show Proof

Indeed, we have the following estimate, coming from definitions:

[[math]] \begin{eqnarray*} \delta(G\times H) &=&\sum_{gh}\frac{1}{ord(g,h)}\\ &=&\sum_{gh}\frac{1}{[ord(g),ord(h)]}\\ &\geq&\sum_{gh}\frac{1}{ord(g)\cdot ord(h)}\\ &=&\delta(G)\delta(H) \end{eqnarray*} [[/math]]

Regarding the last assertion, in the case [math](|G|,|H|)=1[/math], the least common multiple appearing on the right becomes a product:

[[math]] [ord(g),ord(h)]=ord(g)\cdot ord(h) [[/math]]

Thus, we have equality in this case, as desired.

■

We deduce from this that we have the following result:

Proposition

For a finite abelian group [math]G[/math] we have

[[math]] \delta(G)=\prod_p\delta(G_p) [[/math]]

where [math]G_p[/math] with [math]G=\times_pG_p[/math] are the isotypic components of [math]G[/math].

Show Proof

This is clear from Proposition 7.15, the order of [math]G_p[/math] being a power of [math]p[/math].

■

As an illustration for the above results, we can recover in this way the following key defect computation, from Tadej-\.Zyczkowski :

Theorem

The defect of a usual Fourier matrix [math]F_N[/math] is given by

[[math]] d(F_N)=N\prod_{i=1}^s\left(1+a_i-\frac{a_i}{p_i}\right) [[/math]]

where [math]N=p_1^{a_1}\ldots p_s^{a_s}[/math] is the decomposition of [math]N[/math] into prime factors.

Show Proof

The underlying group here is the cyclic group [math]G=\mathbb Z_N[/math], whose isotypic components are the following cyclic groups:

[[math]] G_{p_i}=\mathbb Z_{p_i^{a_i}} [[/math]]

By applying now Proposition 7.16, and by using the computation for cyclic [math]p[/math]-groups performed before Proposition 7.15, we obtain:

[[math]] d(F_N)=N\prod_{i=1}^s\left(1+p_i^{-1}(p_i-1)a_i\right) [[/math]]

But this is exactly the formula in the statement.

■

Now back to the general case, where we have an arbitrary Fourier matrix [math]F_G[/math], we will need, as a second ingredient for our computation, the following result:

Proposition

For the [math]p[/math]-groups, the quantities

[[math]] c_k=\#\left\{g\in G\Big|ord(g)\leq p^k\right\} [[/math]]

are multiplicative, in the sense that [math]c_k(G\times H)=c_k(G)c_k(H)[/math].

Show Proof

Indeed, for a product of [math]p[/math]-groups we have:

[[math]] \begin{eqnarray*} c_k(G\times H) &=&\#\left\{(g,h)\Big|ord(g,h)\leq p^k\right\}\\ &=&\#\left\{(g,h)\Big|ord(g)\leq p^k,ord(h)\leq p^k\right\}\\ &=&\#\left\{g\Big|ord(g)\leq p^k\right\}\#\left\{h\Big|ord(h)\leq p^k\right\} \end{eqnarray*} [[/math]]

We recognize at right [math]c_k(G)c_k(H)[/math], and we are done.

■

Let us compute now [math]\delta[/math] in the general isotypic case. We have here:

Proposition

For [math]G=\mathbb Z_{p^{a_1}}\times\ldots\times\mathbb Z_{p^{a_r}}[/math] with [math]a_1\leq a_2\leq\ldots\leq a_r[/math] we have

[[math]] \delta(G)=1+\sum_{k=1}^rp^{(r-k)a_{k-1}+(a_1+\ldots+a_{k-1})-1}(p^{r-k+1}-1)[a_k-a_{k-1}]_{p^{r-k}} [[/math]]

with the convention [math]a_0=0[/math], and with the notation [math][a]_q=1+q+q^2+\ldots+q^{a-1}[/math].

Show Proof

First, in terms of the numbers [math]c_k[/math], we have the following formula:

[[math]] \delta(G)=1+\sum_{k\geq 1}\frac{c_k-c_{k-1}}{p^k} [[/math]]

In the case of a cyclic group [math]G=\mathbb Z_{p^a}[/math] we have [math]c_k=p^{\min(k,a)}[/math]. Thus, in the general isotypic case [math]G=\mathbb Z_{p^{a_1}}\times\ldots\times\mathbb Z_{p^{a_r}}[/math] we have the following formula:

[[math]] \begin{eqnarray*} c_k &=&p^{\min(k,a_1)}\ldots p^{\min(k,a_r)}\\ &=&p^{\min(k,a_1)+\ldots+\min(k,a_r)} \end{eqnarray*} [[/math]]

Now observe that the exponent on the right is a piecewise linear function of [math]k[/math]. More precisely, by assuming [math]a_1\leq a_2\leq\ldots\leq a_r[/math] as in the statement, the exponent is linear on each of the intervals [math][0,a_1],[a_1,a_2],\ldots,[a_{r-1},a_r][/math]. So, the quantity [math]\delta(G)[/math] to be computed will be 1 plus the sum of [math]2r[/math] geometric progressions, 2 for each interval.

In practice now, the numbers [math]c_k[/math] are as follows:

[[math]] c_0=1,c_1=p^r,c_2=p^{2r},\ldots,c_{a_1}=p^{ra_1}, [[/math]]

[[math]] c_{a_1+1}=p^{a_1+(r-1)(a_1+1)},c_{a_1+2}=p^{a_1+(r-1)(a_1+2)},\ldots,c_{a_2}=p^{a_1+(r-1)a_2}, [[/math]]

[[math]] c_{a_2+1}=p^{a_1+a_2+(r-2)(a_2+1)},c_{a_2+2}=p^{a_1+a_2+(r-2)(a_2+2)},\ldots,c_{a_3}=p^{a_1+a_2+(r-2)a_3}, [[/math]]

[[math]] \vdots [[/math]]

[[math]] c_{a_{r-1}+1}=p^{a_1+\ldots+a_{r-1}+(a_{r-1}+1)},c_{a_{r-1}+2}=p^{a_1+\ldots+a_{r-1}+(a_{r-1}+2)},\ldots,c_{a_r}=p^{a_1+\ldots+a_r} [[/math]]

Now by separating the positive and negative terms in the above formula of [math]\delta(G)[/math], we have indeed [math]2r[/math] geometric progressions to be summed, as follows:

[[math]] \begin{eqnarray*} \delta(G) &=&1+(p^{r-1}+p^{2r-2}+p^{3r-3}+\ldots+p^{a_1r-a_1})\\ &&-(p^{-1}+p^{r-2}+p^{2r-3}+\ldots+p^{(a_1-1)r-a_1})\\ &&+(p^{(r-1)(a_1+1)-1}+p^{(r-1)(a_1+2)-2}+\ldots+p^{a_1+(r-2)a_2})\\ &&-(p^{a_1r-a_1-1}+p^{(r-1)(a_1+1)-2}+\ldots+p^{a_1+(r-1)(a_2-1)-a_2})\\ &&\vdots\\ &&+(p^{a_1+\ldots+a_{r-1}}+p^{a_1+\ldots+a_{r-1}}+\ldots+p^{a_1+\ldots+a_{r-1}})\\ &&-(p^{a_1+\ldots+a_{r-1}-1}+p^{a_1+\ldots+a_{r-1}-1}+\ldots+p^{a_1+\ldots+a_{r-1}-1}) \end{eqnarray*} [[/math]]

Now by performing all the sums, we obtain the following formula:

[[math]] \begin{eqnarray*} \delta(G) &=&1+p^{-1}(p^r-1)\frac{p^{(r-1)a_1}-1}{p^{r-1}-1}\\ &&+p^{(r-2)a_1+(a_1-1)}(p^{r-1}-1)\frac{p^{(r-2)(a_2-a_1)}-1}{p^{r-2}-1}\\ &&+p^{(r-3)a_2+(a_1+a_2-1)}(p^{r-2}-1)\frac{p^{(r-3)(a_3-a_2)}-1}{p^{r-3}-1}\\ &&\vdots\\ &&+p^{a_1+\ldots+a_{r-1}-1}(p-1)(a_r-a_{r-1}) \end{eqnarray*} [[/math]]

By looking now at the general term, we get the formula in the statement.

■

Let us go back now to the general defect formula in Theorem 7.14. By putting it together with the various results above, we obtain:

Theorem

For a finite abelian group [math]G[/math], decomposed as [math]G=\times_pG_p[/math], we have

[[math]] d(F_G)=|G|\prod_p\left( 1+\sum_{k=1}^rp^{(r-k)a_{k-1}+(a_1+\ldots+a_{k-1})-1}(p^{r-k+1}-1)[a_k-a_{k-1}]_{p^{r-k}}\right) [[/math]]

where [math]a_0=0[/math] and [math]a_1\leq a_2\leq\ldots\leq a_r[/math] are such that [math]G_p=\mathbb Z_{p^{a_1}}\times\ldots\times\mathbb Z_{p^{a_r}}[/math].

Show Proof

Indeed, we know from Theorem 7.14 that we have:

[[math]] d(F_G)=|G|\delta(G) [[/math]]

The result follows then from Proposition 7.16 and Proposition 7.19.

■

As a first illustration, we can recover in this way the formula in Theorem 7.17. Indeed, assuming that [math]N=p_1^{a_1}\ldots p_s^{a_s}[/math] is the decomposition of [math]N[/math] into prime factors, we have:

[[math]] \begin{eqnarray*} d(F_N) &=&N\prod_{i=1}^s\left(1+p_i^{-1}(p_i-1)a_i\right)\\ &=&N\prod_{i=1}^s\left(1+a_i-\frac{a_i}{p_i}\right) \end{eqnarray*} [[/math]]

As a second illustration, for the group [math]G=\mathbb Z_{p^{a_1}}\times\mathbb Z_{p^{a_2}}[/math] with [math]a_1\leq a_2[/math] we obtain:

[[math]] \begin{eqnarray*} d(F_G) &=&p^{a_1+a_2}(1+p^{-1}(p^2-1)[a_1]_p+p^{a_1-1}(p-1)(a_2-a_1))\\ &=&p^{a_1+a_2-1}(p+(p^2-1)\frac{p^{a_1}-1}{p-1}+p^{a_1}(p-1)(a_2-a_1))\\ &=&p^{a_1+a_2-1}(p+(p+1)(p^{a_1}-1)+p^{a_1}(p-1)(a_2-a_1)) \end{eqnarray*} [[/math]]

Finally, let us mention that for general non-abelian groups, there does not seem to be any reasonable algebraic formula for the quantity [math]\delta(G)[/math]. As an example, consider the dihedral group [math]D_N[/math], consisting of [math]N[/math] symmetries and [math]N[/math] rotations. We have:

[[math]] \delta(D_N)=\frac{N}{2}+\delta(\mathbb Z_N) [[/math]]

Now remember the formula for [math]\mathbb Z_N[/math] established above, namely:

[[math]] \delta(\mathbb Z_N)=\prod_i(1+p_i^{-1}(p_i-1)a_i) [[/math]]

It is quite clear that the [math]N/2[/math] factor can not be incorporated in any nice way, and so, as indicated above, the quantity [math]\delta(G)[/math] remains something quite complicated.

7d. Explicit deformation

Let us discuss now, following the paper of Nicoara and White , the key fact that for the Fourier matrices the defect is “attained”, in the sense that the deformations at order 0 are true deformations, at order [math]\infty[/math]. This is something quite surprising, and non-trivial. Let us begin with some generalities. We first recall that we have:

Proposition

The unitary matrices [math]U\in U_N[/math] around [math]1[/math] are of the form

[[math]] U=e^A [[/math]]

with [math]A[/math] being an antihermitian matrix, [math]A=-A^*[/math], around [math]0[/math].

Show Proof

This is something well-known. Indeed, assuming that a matrix [math]A[/math] is antihermitian, [math]A=-A^*[/math], the matrix [math]U=e^A[/math] follows to be unitary:

[[math]] \begin{eqnarray*} UU^* &=&e^A(e^A)^*\\ &=&e^Ae^{A^*}\\ &=&e^Ae^{-A}\\ &=&1 \end{eqnarray*} [[/math]]

As for the converse, this follows either by using a dimension argument, which shows that the space of antihermitian matrices is the correct one, or by diagonalizing [math]U[/math].

■

Now back to the Hadamard matrices, we will need to rewrite a part of the basic theory of the defect, using deformations of type [math]t\to U_tH[/math]. First, we have:

Theorem

Assume that [math]H\in M_N(\mathbb C)[/math] is Hadamard, let [math]A\in M_N(\mathbb C)[/math] be antihermitian, and consider the matrix [math]UH[/math], where [math]U=e^{tA}[/math], with [math]t\in\mathbb R[/math].

• [math]UH[/math] is Hadamard when, for any [math]p,q[/math]:
  [[math]] |\sum_{rs}H_{rq}\bar{H}_{sq}(e^{tA})_{pr}(e^{-tA})_{sp}|=1 [[/math]]
• [math]UH[/math] is Hadamard at order [math]0[/math] when, for any [math]p,q[/math]:
  [[math]] |(AH)_{pq}|=1 [[/math]]

Show Proof

We already know that [math]UH[/math] is unitary, so we must find the conditions which guarantee that we have [math]UH\in M_N(\mathbb T)[/math], in general, and then at order 0.

(1) We have the following computation, valid for any unitary [math]U[/math]:

[[math]] \begin{eqnarray*} |(UH)_{pq}|^2 &=&(UH)_{pq}\overline{(UH)_{pq}}\\ &=&(UH)_{pq}(H^*U^*)_{qp}\\ &=&\sum_{rs}U_{pr}H_{rq}(H^*)_{qs}(U^*)_{sp}\\ &=&\sum_{rs}H_{rq}\bar{H}_{sq}U_{pr}\bar{U}_{ps} \end{eqnarray*} [[/math]]

Now with [math]U=e^{tA}[/math] as in the statement, we obtain:

[[math]] |(e^{tA}H)_{pq}|^2=\sum_{rs}H_{rq}\bar{H}_{sq}(e^{tA})_{pr}(e^{-tA})_{sp} [[/math]]

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

(2) The derivative of the function computed above, taken at [math]0[/math], is as follows:

[[math]] \begin{eqnarray*} \frac{\partial |(e^{tA}H)_{pq}|^2}{\partial t}_{|t=0} &=&\sum_{rs}H_{rq}\bar{H}_{sq}(e^{tA}A)_{pr}(-e^{tA}A)_{sp}{\,}_{|t=0}\\ &=&\sum_{rs}H_{rq}\bar{H}_{sq}A_{pr}(-A)_{sp}\\ &=&\sum_rA_{pr}H_{rq}\sum_s(H^*)_{qs}(A^*)_{sp}\\ &=&(AH)_{pq}(H^*A^*)_{qp}\\ &=&|(AH)_{pq}|^2 \end{eqnarray*} [[/math]]

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

■

In the Fourier matrix case we can go beyond this, and we have:

Proposition

Given a Fourier matrix [math]F_G\in M_G(\mathbb C)[/math], and an antihermitian matrix [math]A\in M_G(\mathbb C)[/math], the matrix [math]H=UF_G[/math], where [math]U=e^{tA}[/math] with [math]t\in\mathbb R[/math], is Hadamard when

[[math]] \left|\sum_s\sum_m\frac{t^m}{m!}\sum_{k+l=m}\binom{m}{l}\sum_sA^k_{p,s+n}(-A)^l_{sp}\right|=\delta_{n0} [[/math]]

for any [math]p[/math], with the indices being [math]k,l,m\in\mathbb N[/math], and [math]n,p,s\in G[/math].

Show Proof

According to the formula in the proof of Theorem 7.22 (1), we have:

[[math]] \begin{eqnarray*} |(UF_G)_{pq}|^2 &=&\sum_{rs}(F_G)_{rq}(\overline{F_G})_{sq}(e^{tA})_{pr}(e^{-tA})_{sp}\\ &=&\sum_{rs} \lt r,q \gt \lt -s,q \gt (e^{tA})_{pr}(e^{-tA})_{sp}\\ &=&\sum_{rs} \lt r-s,q \gt (e^{tA})_{pr}(e^{-tA})_{sp} \end{eqnarray*} [[/math]]

By setting [math]n=r-s[/math], can write this formula in the following way:

[[math]] \begin{eqnarray*} |(UF_G)_{pq}|^2 &=&\sum_{ns} \lt n,q \gt (e^{tA})_{p,s+n}(e^{-tA})_{sp}\\ &=&\sum_n \lt n,q \gt \sum_s(e^{tA})_{p,s+n}(e^{-tA})_{sp} \end{eqnarray*} [[/math]]

Since this quantity must be 1 for any [math]q[/math], we must have:

[[math]] \sum_s(e^{tA})_{p,s+n}(e^{-tA})_{sp}=\delta_{n0} [[/math]]

On the other hand, we have the following computation:

[[math]] \begin{eqnarray*} \sum_s(e^{tA})_{p,s+n}(e^{-tA})_{sp} &=&\sum_s\sum_{kl}\frac{(tA)^k_{p,s+n}}{k!}\,\cdot\,\frac{(-tA)^l_{sp}}{l!}\\ &=&\sum_s\sum_{kl}\frac{1}{k!l!}\sum_s(tA)^k_{p,s+n}(-tA)^l_{sp}\\ &=&\sum_s\sum_{kl}\frac{t^{k+l}}{k!l!}\sum_sA^k_{p,s+n}(-A)^l_{sp}\\ &=&\sum_s\sum_mt^m\sum_{k+l=m}\frac{1}{k!l!}\sum_sA^k_{p,s+n}(-A)^l_{sp}\\ &=&\sum_s\sum_m\frac{t^m}{m!}\sum_{k+l=m}\binom{m}{l}\sum_sA^k_{p,s+n}(-A)^l_{sp} \end{eqnarray*} [[/math]]

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

■

Following Nicoara-White , let us construct now the deformations of [math]F_G[/math]. The result here, which came a long time after the original defect paper of Tadej-\.Zyczkowski , and even more time after the early computations of Karabegov , appearing somewhat as a total surprise, puzzling all known experts at that time, is as follows:

Theorem

Let [math]G[/math] be a finite abelian group, and for any [math]g,h\in G[/math], let us set:

[[math]] B_{pq}=\begin{cases} 1&{\rm if}\ \exists k\in\mathbb N,p=h^kg,q=h^{k+1}g\\ 0&{\rm otherwise} \end{cases} [[/math]]

When [math](g,h)\in G^2[/math] range in suitable cosets, the unitary matrices

[[math]] e^{it(B+B^t)}F_G\quad,\quad e^{t(B-B^t)}F_G [[/math]]

are both Hadamard, and make the defect of [math]F_G[/math] to be attained.

Show Proof

The proof of this result, from , is quite long and technical, based on the Fourier computation from Proposition 7.23, the idea being as follows:

(1) First of all, an elementary algebraic study shows that when [math](g,h)\in G^2[/math] range in some suitable cosets, coming from the proof of Theorem 7.14, the various matrices [math]B=B^{gh}[/math] constructed above are distinct, the matrices [math]A=i(B+B^t)[/math] and [math]A'=B-B^t[/math] are linearly independent, and the number of such matrices equals the defect of [math]F_G[/math].

(2) It is also standard to check that each [math]B=(B_{pq})[/math] is a partial isometry, and that [math]B^k,B^{*k}[/math] are given by simple formulae. With this ingredients in hand, the Hadamard property follows from the Fourier computation from the proof of Proposition 7.23. Indeed, we can compute the exponentials there, and eventually use the binomial formula.

(3) Finally, the matrices in the statement can be shown to be non-equivalent, and this is something more technical, for which we refer to . With this last ingredient in hand, a comparison with Theorem 7.14 shows that the defect of [math]F_G[/math] is indeed attained, in the sense that all order 0 deformations are actually true deformations. See .

■

Finally, let us mention that the paper of Nicoara-White was written in terms of subfactor-theoretic commuting squares, which is a quite technical operator algebra notion, and with a larger class of commuting squares being actually under investigation.

We will discuss a bit the relation between Hadamard matrices and commuting squares in chapter 14 below, but in what regards the Nicoara-White theorem, which is the main known theorem regarding the geometry of the complex Hadamard matrices, this definitely remains something to be learned, from their paper and their follow-up papers, which are quite technical, and that we would like however to warmly recommend here.

General references

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