It has been observed that the mutual scalar products of the vectors in a SIC-POVM are given by algebraic units, and at least in some cases by square roots of Stark units. The full picture is somewhat more complicated, especially if non-minimal SIC-POVMs are considered. We present a mixture of exact and numerical evidence suggesting that the overlap units are always products of integral powers of square roots of Stark units from ray class fields all of which are attached to the maximal ring of integers in the base field. In the non-minimal case a lattice of such ray class fields is involved. In every second dimension (counted in a certain way) some of the overlap units equal \(\pm 1\), and we show that this follows from a special property of the ray class fields. Our observations are complementary to but consistent with the claim that the overlap units can be calculated directly from the Shintani–Faddeev modular cocycle.

1 Introduction↩︎

We will be concerned with the intersection of two unsolved problems, one of them coming from number theory and the other from quantum theory. It came as a considerable surprise that this intersection exists [1], [2].

The number theory problem is a subproblem of Hilbert’s 12th: find an elegant description of the generators of the most general abelian extension of a real quadratic number field. (This formulation is a little vague, but so was Hilbert’s [3].) A real quadratic field is an extension \(K = {\mathbb{Q}}(\sqrt{D})\) of the rational field \({\mathbb{Q}}\), where \(D\) is a positive square-free integer. In general an extension of a number field is said to be abelian if its Galois group is abelian. It has been known for a long time what the abelian extensions are. One fixes an ideal in the ring of integers of the base field, in our case \(K\), and defines what is known as a ray class field with that ideal as the finite part of its modulus. Every abelian extension is a subfield of such a ray class field. Fifty years ago Harold Stark gave a procedure that conjecturally allows us to calculate algebraic units in any such ray class field [4]. It starts from an analytic zeta function attached to the field and—provided that the numerical calculation is carried out with enough precision—ends with an exact expression for what are known as Stark units. In most but not quite all of the cases we consider they serve as generators of their ray class field. Much more can be said, a main point being that there does not at this time exist a proof of Stark’s conjectures, not even when the base field is \(K\).

A SIC—the name is a shortening of the unilluminating acronym SIC-POVM—can be projectively defined as an orbit of the Weyl–Heisenberg group in the Hilbert space \({\mathbb{C}}^d\) with a certain equiangularity property [5], [6]. We have placed the group theoretical details in Appendix 11, so as to not prolong this introduction unnecessarily for readers in the know. Once these details are accepted it is enough to consider the group elements \(\{ D_{i,j} \}_{i,j = 0}^{d-1}\), where \(D_{0,0}\) is the identity element. Choose a fiducial unit vector \(\lvert \Psi_0 \rangle\) for the group to act on. By definition the orbit is a SIC if and only if there exist phase factors \(e^{i\theta_{j,k}}\), the indices \(j\) and \(k\) not both zero, such that

\[\langle \Psi_0 \lvert D_{0,0} \lvert \Psi_0 \rangle = 1\;,\langle \Psi_0 \lvert D_{j,k} \lvert \Psi_0 \rangle = \frac{e^{i\theta_{j,k}}}{\sqrt{d+1}} \;. \label{SIC}\tag{1}\]

By taking the absolute values we see that the \(d^2\) vectors in the orbit are equiangular, and it is easy to prove that more than \(d^2\) equiangular unit vectors cannot exist in dimension \(d\). The open question is whether this upper bound can be achieved (in this way) for every choice of the dimension \(d\). Good reasons to be interested in SICs come from foundational concerns in quantum theory and from some quantum engineering applications [7], but we do not go into this here.

We focus on the quite remarkable observation that the unspecified phase factors \(e^{i\theta_{j,k}}\) that appear in the definition seem to be algebraic units as soon as \(d > 3\) [2]. For the case \(d = p\) where \(p\) is a prime equal to 2 modulo 3 Gene Kopp established in four examples that they are indeed square roots of Stark units [8]. There are other cases where Stark units enter the overlaps in a slightly different way [9], [10]. For brevity we will refer to the phase factors as overlap units.

The bridge between Hilbert space and number theory is the remarkable formula [1]

\[(d + 1)(d - 3) = f_0^2 \Delta_0 \;,\Delta_0 = \left\{ \begin{array}{ll} D & if D = 1 mod 4 \\ 4D & otherwise. \end{array} \right. \label{Eq1}\tag{2}\]

Here \(D\) is a square-free integer used to define a real quadratic field, \(\Delta_0\) is a fundamental discriminant, and \(d\) is the dimension of the Hilbert space in which the SIC exists. In our set-up, \(d\) is also the modulus defining a ray class field with the quadratic field as its base field. The prevailing conjecture, verified in all cases examined, is that this ray class field can be used to construct a SIC in dimension \(d\) [2]. The integer \(f_0\) is of interest too: its divisors allow us to define an entire lattice of abelian extensions of the minimal ray class field, and to predict the ‘spectroscopy’ of unitarily inequivalent SICs in that dimension [11], [12].

Early studies of the SIC existence problem included extensive numerical searches, leading Scott and Grassl to develop a ‘SIC phenomenology’ that tells us what symmetries the SICs enjoy in what dimensions [13], [14]. The full picture is quite intricate, but it can now be derived from the conjectures about what number fields are needed to construct the SICs. In a recent contribution Kopp has proposed a special function known as the Shintani–Faddeev modular cocycle [15] which is believed to produce the algebraic units that eventually, after the appropriate Galois transformation, appear as SIC overlaps, for every SIC [16]. The aim of the present paper is then to see exactly how these units relate to Stark units in standard ray class fields.

We take it as given that for each divisor \(f\) of \(f_0\) in equation (2 ) there exists a SIC that can be constructed in a well defined subfield of a ray class field with (finite) modulus \(fd\), and that there exists an algorithm that allows us to compute the Stark units \(S_{fd}\) in that ray class field. If \(f\) is itself composite, \(f = f_1\cdots f_n\), there is a lattice of subfields each containing its own Stark units \(S_{f_id}\). We claim that for each choice of \(d > 3\) and \(f \lvert f_0\) there exists a SIC such that:

• Every overlap unit is a product of square roots of Stark units \(S_{f_id}\). For non-minimal SICs each divisor \(f_i\) of \(f\) contributes its own square roots of Stark units to the product.

• If \(d\) admits non-trivial divisors (as an integer in \(K\)) there will be a corresponding lattice of ray class subfields, giving rise to ‘baby overlaps’ constructed from Stark units in these subfields.

• In some special ‘baby overlaps’ the square rooted Stark units appear raised to an integer exponent. This exponent is determined by a simple rule [17], equation (7 ) below, applied separately to each of the subfields defined by the divisors of \(f\).

• If either of \(d+1\) or \(d-3\) is a square the Stark units are forced to be trivial in certain subfields, giving rise to overlap units equal to \(\pm 1\).

As we shall see, these claims are not in contradiction with the standard picture that minimal SICs in even dimensions are constructed from ray class fields with modulus \(2d\) [2], or that non-minimal SICs are constructed using proper subfields of the ray class field with modulus \(fd\) [12] (or \(2fd\), as the case may be).

We will support the fourth of our claims by a theorem concerning the behaviour of ray class fields with modulus \(d-2\), assuming eq. (2 ) is still in place. In fact we will find that the moduli \(d-3\), \(d-2\), \(d-1\), and \(d\), have a very special status.

Sections 2 and 3 contain some preliminaries, and section 4 discusses the case of minimal SICs—which is governed by the divisors of the integer \(d\). Section 5 gives some details concerning the number fields needed in the non-minimal case, and Section 6 describes the non-minimal overlaps—for which we also have to consider the divisors of the integer \(f_0\). Sections 7 and 8 address two important special cases. Up to this point our paper should be regarded as nothing more than a collection of facts, but for the second of these special cases (that of ‘aligned SICs’) the theorem proved in Section 9 provides a full explanation.

2 Facts about ray class fields and Stark units↩︎

This section is a sketch of some of the number theory underlying our problem. Although we will touch on class field theory in Sections 5 and 9, we refer the interested reader to the various standard treatises [18]–[21]. For orientation, consult Figure 1. SICs that can be constructed using the fields shown there are known as minimal. In many dimensions there exist also non-minimal SICs, needing a further abelian extension of the minimal fields. We will come to them in Section 5.

2.0.0.1 The class number \(h_K\) and prime factorisation

At the bottom of the lattice sits the real quadratic field \(K = {\mathbb{Q}}(\sqrt{D})\). It is extended to the Hilbert class field \(H\). The Galois group Gal\((H/K)\) is important in the SIC problem because it acts on the minimal SIC to create a multiplet of \(h_K\) unitarily inequivalent SICs [2]. Here \(h_K\) is the class number, namely the order of the ideal class group \(\mathcal{C}_K\) of \({\mathbb{Z}}_K\), the ring of integers in \(K\). The ideal class group is isomorphic to the Galois group Gal\((H/K)\), so the class number is also the degree of the extension. If \(h_K = 1\) the extension is trivial. If \(h_K>1\) prime factorisation in \({\mathbb{Z}}_K\) fails to be unique, and we may have to express the ideal \(p{\mathbb{Z}}_K\) as a product of non-principal ideals. Ideal factorisation is unique.

Regardless of the value taken by \(h_K\) it can happen that a rational prime \(p\) does not remain prime in \({\mathbb{Z}}_K\). If \(p{\mathbb{Z}}_K\) is a prime ideal in \({\mathbb{Z}}_K\) then \(p\) is said to be inert. It splits if there exist ideals \(\partial\) and \(\overline{\partial}\) such that \(p{\mathbb{Z}}_K= \partial\overline{\partial}\), and it ramifies if it is a power of a prime ideal. For the real quadratic fields \(K = {\mathbb{Q}}(\sqrt{D})\), and for odd primes \(p\), this behaviour is determined by the value of \(D\) modulo \(p\). If \(D\) mod \(p\) is a quadratic residue the prime splits, and if it is a non-residue the prime is inert. If \(p\) divides into the discriminant it ramifies, \(p{\mathbb{Z}}_K= \mathfrak{p}^2\) where \(\mathfrak{p}\) is a unique ideal (principal or non-principal) of \({\mathbb{Z}}_K\). For odd primes this happens if and only if \(p \lvert D\).

2.0.0.2 Ray class fields

The key extension is that from Hilbert’s class field to the ray class field with modulus \(d\). This actually refers to the principal ideal \(d{\mathbb{Z}}_K\), but since \(d\) is a rational integer our notation ignores this. When \(d = 1\) we recover the Hilbert class field. To the finite modulus \(d\) one can add one or two ‘infinite places’, and this gives rise to the diamond at the top of Figure 1. The two infinite places correspond to the two different ways in which \(\sqrt{D}\) can be embedded in the real numbers. When the finite part of the modulus equals \(d\), as it appears in the key equation (2 ), the degree of the ray class field always increases by a factor of 2 when an infinite place is added [2]. This gives us four ray class fields to consider, conveniently denoted by \(K^d\), \(K^{d\infty_1}\), \(K^{d\infty_2}\), and \(K^{d\infty_1\infty_2}\).

The first of these is totally real, meaning that regardless of how it is embedded in the complex field it consists only of real numbers. The next two are isomorphic in an abstract sense, but once we choose a definite embedding of \(\sqrt{D}\) as a positive or negative real number one of them consists only of real numbers while the other contains genuinely complex numbers. In these specific situations where we know that one of the fields is real and the other is complex, we have chosen the ordering \(\infty_1,\infty_2\) in such a way that \(K^{d\infty_2}\) is real and \(K^{d\infty_1}\) complex. Moreover since we shall have need of actual fixed conjugate embeddings \(\mathfrak{j}\colon K\ \longrightarrow\ \mathbb{R}\) and \(\mathfrak{j}^\tau \colon K\ \longrightarrow\ \mathbb{R}\), we clarify the notation here. We denote by \(\tau\) the generator of the Galois group \({\mathrm{Gal}_{K/\mathbb{Q}}}\). The embedding \(\mathfrak{j}\) represents the \(\infty_2\) equivalence class in which every mapping sends \(\sqrt{D}\) to the positive real number. Similarly its \(\tau\)-conjugate \(\mathfrak{j}^\tau\) is in the \(\infty_1\) equivalence class and sends \(\sqrt{D}\) to a negative real number. This ordering was chosen to align with conventions in the Magma programme [22].

The field \(K^{d\infty_1\infty_2}\) is the largest of the four, and contains the cyclotomic field generated by the \(d\)th roots of unity as a subfield [2]. We need the roots of unity to construct the SIC because we need them to represent the Weyl–Heisenberg group, but it appears that the fiducial vector can always be chosen so that the SIC overlaps belong to the field \(K^{d\infty_1}\) (or to the field \(K^{2d \infty_1}\) if \(d\) is even, but as we will see this is irrelevant in our context).

The catch is that the Stark units that we calculate using Stark’s procedure belong to the real field \(K^{d\infty_2}\). If we know their minimal polynomial over \(K\), then they can be turned into the complex units that we need by means of the Galois transformation \(\sqrt{D} \rightarrow - \sqrt{D}\), and they will indeed be found to sit on the unit circle. The minimal polynomial is needed also to remove the sign ambiguities that arise when we take their square roots. Hence the numerical calculation of the real Stark units has to be performed with a precision high enough to identify the minimal polynomial, even if we are content to perform the rest of the calculation numerically with modest precision.

2.0.0.3 Stark units

In outline, the calculation of the Stark units proceeds as follows. Let \(A\) be a ray ideal class modulo \({\mathfrak{m}}\), where the finite part of \({\mathfrak{m}}\) is an ideal in \({\mathbb{Z}}_K\), the ring of integers in \(K\). For us, the finite part will usually be a principal ideal generated by an integer from \({\mathbb{Z}}_K\). Hence \(A\) is an element in the ray class group modulo \({\mathfrak{m}}\) (see ref. [18], chapter VI). Then we define a (partial) zeta function as

\[\zeta (s, A) = \sum_{{\boldsymbol{a}} \in A}N({\boldsymbol{a}})^{-s} \;.\]

Here \(N({\boldsymbol{a}})\) is the norm of the ideal \({\boldsymbol{a}}\). This function carries information about the real quadratic field to which it is attached. It has a pole at \(s = 1\), but we can construct a function without poles by first introducing a certain involution \(R\) and then defining

\[\delta(s,A) = \zeta (s,A) - \zeta (s,RA) \;. \label{involution}\tag{3}\]

Stark then conjectures [4] that one obtains algebraic units through the value at \(s=0\) of the derivative of this expression with respect to \(s\):

\[S_\sigma = e^{\delta^\prime (0,\sigma )} \;.\]

Here an element \(\sigma\) of the Galois group of the extension is used as a label, which is possible because of the Artin isomorphism between the ray class group modulo \({\mathfrak{m}}\) and the Galois group Gal\((K^{{\mathfrak{m}}}/K)\). In most cases these Stark units are expected to generate the number field in which they sit. For the actual calculation of the Stark units we rely on in-built Magma [22] commands for computing Hecke \(L\)-functions. These are related to the zeta function by means of a kind of glorified Fourier transformation. A short description of the Magma program we use for the calculation can be found in refs. [9], [10].

According to the Stark–Tate conjectures the square root of a Stark unit belongs to an abelian extension of the ray class field which may or may not be a trivial extension [21]. This is important here because while the overlaps belong to \(K^{d\infty_1}\) the overlap units that we aim to calculate may or may not do so. This depends on whether the factor \(\sqrt{d+1}\) that appears in equation (1 ) belongs to that field. See ref. [2] for a precise statement. Since almost all our calculations will concern squares of overlaps we can largely ignore this issue, just as we can ignore the field \(K^{2d \infty_1}\) which appears at this point if \(d\) is even.

Let us nevertheless elaborate a little: choose \(d = 5\), say, for which \(\sqrt{d+1}\) does not belong to the ray class field. If the actual overlaps do, it follows that the overlap units cannot be Stark units. They can be square roots of Stark units though, because in this case these square roots belong to the abelian extension \(K^{d\infty_1}(\sqrt{d+1})\). This provides a (weak) rationale for why we need to take square roots.

It is worth mentioning that the behaviour described in Figure 1 holds for modulus \(d\), but it is by no means universal. If the field is unaffected by the addition of an infinite place to the finite modulus then the involution that appears in equation (3 ) is absent, and the Stark units collapse to \(+ 1\). We will encounter examples of this phenomenon below, and it will be the subject of Section 9.

2.0.0.4 The unit group and the dimension towers

A key role is played by the unit group \({\mathbb{Z}^\times_K}\); that is, the multiplicative group of units in the ring of integers \({\mathbb{Z}}_K\). It has a torsion part \(\{ 1, -1\}\) and in addition there is an infinite cyclic group generated by a fundamental unit \(u_o\). We also define the totally positive unit

\[u_D = \frac{d_1-1 + \sqrt{(d_1+1)(d_1-3)}}{2} \;.\]

The subscript \(1\) in \(d_1\) refers to 4 below. This unit is totally positive in the sense that it is positive regardless of the sign we assume for the square root. If \(d_1- 3\) is a square it is the case that \(u_D = u_o^2\), in all other cases \(u_D = u_o\).

If we fix the quadratic field by fixing the square-free number \(D\) then the key equation (2 ) admits an infinite sequence of solutions \(\{d_\ell \}_{\ell = 1}^\infty\) for \(d\) [2]. In fact

\[\label{dees} d_\ell = d_\ell(D) = u_D^\ell + u_D^{-\ell} + 1 .\tag{4}\]

We refer to the resulting sequences of dimensions as AFMY towers. If we fix the quadratic field by setting \(D = 5\), for example, then the corresponding tower begins

\[\{ d_\ell \}_{\ell = 1}^\infty = \{ 4, 8, 19, 48, 124, 323, 844, 2208, 5799, 15128 , 39604, \dots \} \;. \label{tower}\tag{5}\]

The degree of the ray class field with modulus \(d\) is inversely proportional to the multiplicative order of the image of the fundamental unit \(u_o\) in \({\mathbb{Z}}_K/d{{\mathbb{Z}}_K}\), the integers taken modulo \(d\). In our case this order is always divisible by 3, and this is related to the fact that the symmetries exhibited by SICs are found to be of an order divisible by 3; something first glimpsed by Zauner [5], [13], [23]. In fact the order of the unit \(u_D\) modulo \(d_\ell{\mathbb{Z}}_K\) is \(3\ell\). Thus the symmetries of the minimal SIC increase as we climb the tower, and using this knowledge the minimal SIC has been found, for example, in exact form in all dimensions listed in (5 ), with the exception of \(d_8 = 2208\)—which is the highest dimension in which a SIC has been found by means of a numerical search [24]—and \(d_{10} = 15128\), where as yet no SIC has been found.

If \(d_\ell - 3\) is a square then the fundamental unit \(u_o\) has order \(6\ell\) modulo \(d_\ell\). Then the degree of the ray class field drops by a factor of two, and the minimal SIC responds by having anti-unitary symmetry. For reasons that are not fully understood, fiducial vectors for SICs with anti-unitary symmetry can be constructed using square roots of Stark units in a ‘small’ subfield, which means that exact SICs are available in high dimensions of this form, including some five digit ones [9], [10].

3 The lattice of divisors of \(d\)↩︎

An important fact is that if the modulus of one ray class field divides that of another then the first ray class field is a subfield of the other (see Proposition II.4.1.1 in ref. [20]). Hence the ray class field \(K^{d\infty_1}\) will have a lattice of subfields corresponding to the divisor lattice of the integer \(d\). This will manifest itself in the SIC through the occurrence of baby overlaps, that is to say some subsets of overlaps that belong to the various subfields. In this section we will lay out how the same pattern of baby overlaps arises from a consideration of the Weyl–Heisenberg group in dimension \(d\), assuming that the SIC is left invariant by a symmetry of order three. The reader may find the argument somewhat loose. At the end of the section we will discuss one possible proof strategy.

We can obtain non-trivial divisor lattices already in prime dimensions \(d = p\), because as noted in section 2 a rational prime \(p\) may be composite considered as an integer in the quadratic field \(K = \mathbb{Q}(\sqrt{D})\). If \(D\) is a quadratic residue mod \(p\) then \(p\) splits, that is to say that there exist ideals \(\partial\) and \(\bar{\partial}\) such that \(p =\partial \bar{\partial}\), and if \(p\) divides the discriminant then \(p\) ramifies, that is is to say it is a power of a prime ideal. For now we are only interested in the behaviour of primes that divide \(d\), and the discriminant is determined by \(d\) through the key formula (). We observe that

\[p \lvert d\Rightarrowf_0^2\Delta_0 = (d+1)(d-3) \equiv -3 \;mod \;p \;.\]

Let \(p\) be an odd prime, so that \(p\) divides \(\Delta_0\) if and only if \(p\) divides \(D\). Hence \(D\) will be a quadratic residue mod \(p\) if and only if \(-3\) is. But from quadratic reciprocity we know that \(-3\) mod \(p\) is a quadratic residue if and only if \(p = 1\) mod 3. As long as \(f_0\) and \(d\) are relatively prime (this can fail only if \(p = 3\)) we conclude that \(p\) is inert if \(p=2\) mod 3, it splits if \(p=1\) mod 3, and it ramifies if \(p = 3\). The resulting divisor lattices are shown in Figure 2 for prime dimensions, and in Figure 3 for prime-squared dimensions. If \(d\) has several prime factors the divisor lattices become more complicated, but they can easily be worked out case by case.

Now we turn to the Weyl–Heisenberg group. Necessary definitions can be found in Appendix 11. The first observation to be made here is that in composite dimensions \(d\) the group is a direct product of the Weyl–Heisenberg group in the prime power factors,

\[H(d) = H(p_1^{k_1}p_2^{k_2} \dots p_n^{k_n}) = H(p_1^{k_1})\times H(p_2^{k_2}) \times \dots \times H(p_n^{k_n}) \;.\]

This mirrors how prime power factors of \(d\) give rise to subfields of \(K^d\). Concretely, prime power dimensions will serve as building blocks for composite dimensions, hence we focus on prime power dimensions alone. It is helpful to be able to visualize how the maximal abelian subgroups sit inside \(H(p^k)\). In prime dimensions \(d = p\) the group can be divided into \(p+1\) cyclic subgroups having only the identity element in common. Let us refer to them as petals of a flower. See Figure 4. In prime power dimensions the cyclic subgroups become entwined with each other in a characteristic fashion. The relevant group theory is explained elsewhere [10], [25]; here we just refer to Figure 4 which illustrates the case \(d = 2^2\). In general, if \(d = p^2\) there are \(p+1\) entwined petals formed from \(p\) petals each. Each such entwined petal contains \((p-1)(p^2 +1)\) operators, including \(p-1\) operators of order \(p\). For higher powers of \(p\) the counting can be worked out with modest effort.

We refer to elements of \(H(d)\) as displacement operators \(D_{i,j}\), and recall that the symplectic group acts on the displacement operators according to equation (25 ) in Appendix 11. If we fix the representation so that the group element \(Z = D_{0,1}\) is diagonal then any symplectic operator that transforms the petal generated by \(Z\) into itself will be represented by a monomial matrix. If in addition it transforms the petal generated by \(X = D_{1,0}\) into itself it will be represented by a permutation matrix. Otherwise it will be represented by a complex Hadamard matrix whose entries are roots of unity [23]. Our key assumption is that the SIC has Zauner symmetry, that is to say that it is invariant under a unitary transformation representing a symplectic matrix of order three and trace equal to \(-1\) mod \(d\) [5], [23]. A possible choice that works in every dimension is

\[{\cal Z} = \left( \begin{array}{cc} 0 & - 1 \\ 1 & - 1 \end{array} \right) \;.\]

With a suitable choice of the fiducial vector the SIC is invariant under this transformation in the sense that

\[\langle \Psi_0 \lvert U_{\cal Z}D_{i,j}U_{\cal Z}^{-1} \lvert \Psi_0\rangle = \langle \Psi_0 \lvert D_{-j,i-j} \lvert \Psi_0\rangle = \langle \Psi_0 \lvert D_{i,j} \lvert \Psi_0 \rangle \;.\]

If the position \(\ell\) of the dimension in the AFMY towers is higher than one the symmetry may be larger, and if the dimension is of the form \(d = n^2+3\) the symmetry may include an anti-unitary symmetry as well. A different complication occurs if the dimension \(d>3\) is divisible by 3 but not by 9, because then there exists another conjugacy class of Zauner unitaries. In dimensions of the form \(d = 3(3n+1)\) this leads to SICs having a symmetry of ‘type \(F_a\)’ [13], [14]. They will be discussed in Section 7. To keep things simple we momentarily ignore all special cases, and we also focus on \(d = p\). It will then be observed that Zauner symmetry divides the \(p+1\) petals of a flower into \(k\) triplets and zero, two, or one, singlets, depending respectively upon whether \(p \equiv 2, 1 \textrm{ or } 0 \bmod 3\).

For the latter two cases it is then convenient to choose a different representative of the conjugacy class of order three matrices. For \(p = 1\) mod 3 and \(p = 3\), respectively, we choose \(\alpha\in\mathbb{Z}\) with \(2\leq\alpha\leq p-2\) and \(\gamma \in \{1,2\}\) such that

\[{\cal Z}_1 = \left( \begin{array}{cc} \alpha & 0 \\ 0 & \alpha^{-1} \end{array} \right) \;\;respectively \;\;{\cal Z}_2 = \left( \begin{array}{cc} 1 & 0 \\ \gamma & 1 \end{array} \right) \;,\]

where \(\alpha \neq 1\) satisfies \(\alpha^3 \equiv 1 \bmod p\) (such integers exist if and only if \(p \equiv 1 \bmod 3\)). The unitary transformation \(U_{{\cal Z}_1}\) transforms the petals generated by \(Z\) and \(X\) into themselves, while \(U_{{\cal Z}_2}\) transforms the petal generated by \(Z\) into itself. Hence \(U_{{\cal Z}_1}\) is a permutation matrix, and \(U_{{\cal Z}_2}\) is at least monomial. Choosing a Zauner matrix of the form \({\cal Z}_1\) means that the special petals in the flower are those generated by \(X\) and \(Z\), and we always make this choice if \(d \equiv 1 \bmod 3\). Overlap units of the form

\[\sqrt{d+1}\langle \Psi_0 \lvert Z^j \lvert \Psi_0\rangle = e^{i\theta_{0,j}} \label{Zoverlap}\tag{6}\]

will be referred to as \(Z\)-overlaps. \(X\)-overlaps are similarly defined. They will, in fact, become baby overlaps if \(d \equiv 1 \bmod 3\).

We are now ready to link the unitary geometry to the arithmetic in a more definite fashion. Let \(S\) denote the symplectic symmetry group of the SIC, and let \(M(S)\) be the centraliser of \(S\) within the group \(GL(2, {\mathbb{Z}}/d{\mathbb{Z}})\). The claim, made by Appleby et al. [2], is that

\[Gal(K^{d\infty_1}/H) \simeq M(S)/S \;.\]

The Galois group acts on the overlaps, while the group \(M(S)/S\) acts on the displacement operators according to the recipe in Appendix 11. For \(d = p \equiv 2 \bmod 3\) it turns out that the action of these groups is transitive. For \(d = p \equiv 1 \bmod 3\) the group transforms the petals generated by \(X\) and \(Z\) into themselves (assuming that we have arranged the fiducial vector suitably). The corresponding overlaps must then sit in proper subfields of the full field. If, say, \(d = p^2\) and \(d = p \equiv 1 \bmod 3\) there will be several short orbits of displacement operators, consisting of the subgroups generated by

\[\{ Z^p\}, \;\{ Z^p\}, \;\{ Z\}, \;\{ X^p, Z^p\}, \;\{ X\}, \;\{ X^p, Z\}, \;\{ X, Z^p\} \;.\]

Each such subgroup corresponds to a subfield, and indeed to an element of the second divisor lattice shown in Figure 3.

The claim then is that every element in the divisor lattice for \(d\) corresponds to baby overlaps sitting in a subfield of \(K^{d\infty_1}\). Moreover, if baby overlaps occur it is possible to arrange the fiducial vector so that the \(Z\)-overlaps, and usually also the \(X\)-overlaps, are baby overlaps. We also claim that all the overlap units are (products of) square roots of Stark units raised to the power \(n\), where the exponent is determined by the rule

\[n\times (number of overlaps in the orbit) = \lvert S \lvert \lvert Gal \lvert \;, \label{Grasslrule}\tag{7}\]

and where in turn the Galois group is that of the appropriate subfield (over \(H\)). This rule was suggested by Markus Grassl [17].

Can any of this be proved from first principles? This is a difficult matter, because at the time of writing SIC existence has been proved only in those dimensions where a SIC has been explicitly calculated. Hence we have to rely on some conjecture for a conditional proof. A particularly favourable case concerns SICs with anti-unitary symmetry, which occur whenever \(d\) is of the form \(n^2+3\) for some \(n\geq1\). In this case there exists a conjectural ‘formula’ for a SIC fiducial vector constructed from square roots of Stark units in a proper subfield of \(K^{d\infty_1}\) [9], [10]. The formula has been tested in close to a hundred cases, including (with appropriate modifications) some non-minimal SICs. The formula immediately implies that all the \(Z\)-overlaps are trivial, and that all the \(X\)-overlaps lie in the subfield containing the fiducial vector. With sufficient attention to detail the formula in fact allows one to deduce what subfields hold which baby overlaps, for all of the latter [10]. Proving that the various overlaps are indeed given by Stark units is more difficult. It has not been done [26]. Nor has it been explained why we assumed a symmetry of order three in the first place. A comment, but no answer, can be found in Appendix 12.

Hence, the situation is currently still far from understood. But the evidence to be reported below supports the claims made above.

4 Overlaps for minimal SICs↩︎

In this section we will analyse precisely how Stark units give the overlaps for minimal SICs, simply by giving the results for a set of selected examples. We begin with some known facts about prime dimensions. Suppose first that \(d = p \equiv 2\) mod 3. Then there are no non-trivial divisors of \(d\). Hence we are dealing with a single ray class field, and the order of its Galois group is (see [2])

\[\lvert Gal \lvert = \lvert Gal(K^{d\infty_1}/H) \lvert = \frac{p^2-1}{3\ell} \;.\]

The symmetry of the minimal SIC is of order \(3\ell\) where \(\ell\) is the position of \(d\) in its AFMY tower, so the order of the Galois group equals the number of distinct overlaps in the SIC. We will find that all the overlap units are square roots of Stark units in \(K^{d\infty_1}\).

If \(d = p \equiv 1\) mod 3 the prime splits over \(K\), \(p = \partial \overline{\partial}\), and it is seen in Figure 2 that we have three distinct Galois groups to consider. The details turn out to depend on whether \(p-3\) is a square or not. If \(p-3\) is not a square the orders of the Galois groups are

\[\begin{align} \lvert Gal(K^{d\infty_1}/H) \lvert = \frac{(p-1)^2}{3\ell}\nonumber \\ \\ \lvert Gal(K^{\partial \infty_1}/H) \lvert = \lvert Gal(K^{\overline{\partial} \infty_1}/H) \lvert = \frac{p-1}{3\ell} \;. \nonumber \end{align}\]

The symmetry is again of order \(3\ell\). There will be three distinct Galois orbits of overlaps, and we will arrange the fiducial vector so that the two ‘small’ orbits correspond to the cyclic subgroups of the Heisenberg group that are generated by \(X\) and by \(Z\). There are \(p-1\) \(Z\)-overlaps and equally many \(X\)-overlaps, and these overlap units are found to be square roots of Stark units in the relevant subfield of the full ray class field.

If \(d = p\) is a prime of the form \(n^2 + 3\) the situation changes. We have [9]

\[\begin{align} \lvert Gal(K^{d\infty_1}/H) \lvert = \frac{(p-1)^2}{6\ell} \;,\lvert Gal(K^{\partial \infty_1}/H) \lvert = \frac{p-1}{3\ell} \;, \nonumber \\ \label{from16} \\ K^{\overline{\partial} \infty_1} = K^{\overline{\partial}} \;.\nonumber \end{align}\tag{8}\]

The fact that \(K^{\overline{\partial} \infty_1} = K^{\overline{\partial}}\) has the consequence that Stark’s construction [4] applied to this subfield results in units that are simply equal to \(+ 1\). We arrange the fiducial vector so that these trivial units occur as \(Z\)-overlaps, while the \(X\)-overlaps belong to \(K^{\partial \infty_1}\).

The minimal SIC has anti-unitary symmetry whenever \(d = n^2+3\), not necessarily a prime, and the order of the symmetry group \(S\) is \(\lvert S \lvert = 6\ell\). An important point is that it can be proved, using pure Hilbert space arguments, that a SIC with anti-unitary symmetry must have \(d-1\) trivial baby overlap units, equal to \(+ 1\) in odd dimensions [9], [10]. With a suitable choice of the fiducial vector they are precisely the \(Z\)-overlaps.

With this preamble we can present our evidence that SICs really behave as stated. See Table 1. This is the first of many tables where \(D\) stands for a generic displacement operator, \(X\) and \(Z\) occur separately if they give rise to baby overlaps, \(\#\) is the total number of overlaps accounted for, \(\lvert Gal \lvert\) stands for the order of the Galois group Gal\((K^{{\mathfrak{m}}_0\infty_1}/H)\) for the relevant finite modulus \({\mathfrak{m}}_0\) (at least for now—this has to be modified a little when we come to the non-minimal SICs), \(\lvert S \lvert\) is the order of the symmetry group, and S\(_{{\mathfrak{m}}_0}\) is a Stark unit in the ray class field indicated by the subscript. With this information in hand it can be checked by inspection that Grassl’s rule (7 ) correctly predicts the exponent of the square rooted Stark units that give the overlaps. When \(d = p \equiv 1\) mod 3 the \(X\)-overlaps belong to \(K^{\partial \infty_1}\), and the rule tells us that the square rooted Stark units should appear raised to the power \(n=2\),

\[n\times (p-1)=6\ell \times (p-1)/3\ell\Rightarrown = 2 \;.\]

For the ‘generic’ or ‘large’ overlaps we find \(n = 1\) so we expect them to be square roots of Stark units in \(K^{d\infty_1}\), and indeed they are. The extent to which the claims of a given table has been verified is stated in its caption.

There are no surprises in prime power dimensions \(d = p^k\), at least not when \(p > 3\). We expect that \(d \neq n^2+3\) and \(\ell = 1\) for all these dimensions, where \(\ell\) is the position in the AFMY tower. If so the order of the symmetry group is always \(\lvert S \lvert = 3\). We give the pattern of unit overlaps for \(d = p^2\) in Table 2. It faithfully mirrors the lattice of divisors shown in Figure 3. When \(p = 3\) the prime 3 ramifies, \((3) = {\mathfrak{p}}^2\), and we encounter a ‘short’ orbit for which Grassl’s rule correctly gives the power \(3\) for the square rooted Stark units.

When the dimension is even we need to extend the centre of the Weyl–Heisenberg group so that it has order \(2d\) in order to obtain a clean description of the SICs [23]. But this is irrelevant for our present purposes. The Stark units that appear in the overlaps are those coming from the ray class field with finite modulus \(d\), together with its subfields. The extension to \(2d\) is needed only to take the relevant square roots. The case \(d = 2^2\) actually fits into Table 2 if we take into account that \(K^{2,\infty_1} = K^2\) so that the corresponding Stark units are equal to \(+1\). We give examples of the form \(d = 2p\) in Table 3. Note that \(d_\ell = 2p\) implies \(\ell = 1\), so \(\lvert S \lvert = 3\) for all such dimensions [10]. This is fortunate because the second row of the Table would not make sense otherwise. For \(d = 6\) we see again that a factor 3 in the dimension results in overlaps given by square rooted Stark units raised to the power 3 because one of the orbits is ‘short’.

For generic composite dimensions the lattice of divisors becomes more complicated without becoming more interesting. There are a few things to notice. First, if \(d = p_1p_2\) with neither of the primes being of the form \(n^2+3\) it can still happen that their product is. Then some trivial overlaps occur in the product dimension, an example being \(403 = 13\cdot 31\). Conversely, \(7\cdot 19 = 133\), and there are no trivial overlap phases there (or should not be—we have not checked). Two very special cases are dimensions of the form \(d = 3(3n+1)\) or of the form \(d = n(n-2)\). They will be discussed separately, in Sections 7 and 8.

A comment on the calculations is in order. For the examples in \(d = p \equiv 2\) mod 3 we constructed the exact SIC using Kopp’s method [8]. This entails calculating the Stark units numerically to enough precision so that we can determine their minimal polynomial. With this in hand we can perform the ‘flip’ indicated in Figure 1, and take the global square root if necessary. We then have to find an exact root of the resulting polynomial, act on this root with the Galois group, and place the resulting numbers in the correct order as overlap units. This will allow us to construct the SIC using the Schwinger formula (26 ), and check with an exact calculation that the resulting expression really is a projector of rank 1. The details can be found in Kopp’s paper. For \(d = 53\) the calculation is made easier by the fact that the symmetry is of order nine (\(\ell = 3\)), but it nevertheless took 4 hours to calculate the Stark units and 13 hours to find the exact root that we needed (using Magma). For most of the remaining examples we took a more pragmatic approach. We calculated the minimal polynomial of the complex Stark units in exact form, and then checked numerically to high precision that the overlap units raised to the suitable power provide its roots.

When the dimension is of the form \(d = n^2+3\) we can again resort to exact calculations, and give many more examples. For prime dimensions of this form the results were given in Table 1. For the case \(d = n^2+3 = 4p\) see Table 4, which is taken from ref. [10]. Overlap units equal to \(\pm 1\) were not included in this table. Note that \(\ell\), the position in the AFMY towers, occurs in the exponents of the Stark units in one of the rows, as predicted by the rule (7 ).

5 The lattice of divisors of \(f_0\)↩︎

We now turn to the non-minimal SICs. Our first task is to explain the role of the integer \(f_0\) that appears in the formula (2 ). Let \(f\) be any divisor of \(f_0\). Kopp and Lagarias use these integers to define subrings of the ring of integers in the quadratic field, and to define a new kind of ray class field which depends not only on the modulus \(d\) but also on the integer \(f\) [11]. They refer to \(f\) as the conductor (namely of a non-maximal ring of integers in the quadratic field). Our homemade notation for these new ray class fields is \(K_{L}^{d; f}\). Clearly \(K_{L}^{d;1} = K^d\), and as usual the degree of \(K_{L}^{d\infty_1; f}\) is twice that of the totally real field \(K_{L}^{d;f}\). It turns out that \(K_{L}^{d;f}\) is always a subfield of \(K^{fd}\). We will be interested in whether it is a proper subfield, or not. For this reason we define the excess \(e\) as the index of the extension,

\[excess = e = [ K^{fd}: K^{d;f}_L ] = \frac{degree(K^{fd})}{degree (K_{L}^{d;f})} \;. \label{excess}\tag{9}\]

If the excess equals one, then the two number fields are identical (this is also the case if we add one or two infinite places to the moduli).

Kopp and Lagarias go on to claim that for each divisor \(f\) of \(f_0\) there exists a SIC that can be constructed using their new ray class field \(K_L^{d;f}\) [12]. For \(d \leq 90\) the resulting ‘spectrum’ of geometrically inequivalent SICs agrees with the numerical findings of Scott [14], which are believed to be complete for these dimensions. Here it should be recalled that SICs are collected first into anti-unitarily equivalent multiplets based on their symmetries, and then these geometric multiplets are collected into Galois multiplets where the number of geometric multiplets in a given Galois multiplet is equal to the class number \(h_K\) of the quadratic field \(K\) if the SICs are minimal, and equal to the degree of a ring class field in the general case. The conjecture evidently has predictive content. We give the example \(d = 199\) in Figure 5. Equation (2 ) yields \(f_0 = 70\), so there is a lattice of no less than eight divisors. The SICs with \(f = 1\) and \(f = 5\) are known. The remaining cases have not yet been publicly reported.

The ray class fields with modulus \(d\) and conductor \(f\) are subfields of the ray class field with modulus \(fd\) and conductor 1, and often proper subfields. Nevertheless our thesis is that the SIC overlaps are always given by products of Stark units in the ray class field \(K^{fd \infty_1}\) and in some of its subfields. Before we present the evidence we will prove a corollary to a theorem in ref. [12]. First we define the totally positive unit

\[u = \frac{d-1 + \sqrt{(d+1)(d-3)}}{2} \;. \label{unitu}\tag{10}\]

Whether or not this is a fundamental unit in the quadratic field \(K\) depends on the position of \(d\) in the AFMY tower of dimensions connected to \(K\) (and on whether or not \(d-3\) is a square) [2]. If \(d = d_1\) it equals \(u_D\), the first totally positive power of a fundamental unit.

Next, let \(\Phi(f)\) be Euler’s totient function and

\[\Phi_K(f) = f^2\prod_{p \lvert f}\left(1 - \frac{1}{p}\right) \left( 1 - \underline{\Delta_0} \choose p \frac{1}{p}\right)\]

its generalisation to quadratic fields. Here \({\underline{\Delta_0} \choose p}\) is the Kronecker symbol, equal to the Legendre symbol if \(p\) is odd, while if \(p=2\) it equals 0 if \(\Delta_0\) is even, 1 if \(\Delta_0 = \pm 1\) mod 8, and \(-1\) if \(\Delta_0 = \pm 3\) mod 8. It tells us whether the prime \(p\) ramifies, splits, or remains inert in \(K\).

There will be a complication if \(d\) and \(f\) have a common factor. By inspection of the key equation (2 ) we see that \(\gcd(d,f) \in \{1,3\}\), and if \(\gcd(d,f) = 3\) then \(d \equiv 3\) mod 9. We will largely ignore this special case until we come to Section 7, but it will be included in the statement of the proposition.

We can now state our observation:

5.0.0.1 Proof:

We rely on a result by Kopp and Lagarias, namely that (see Theorem 6.5 in ref. [12])

\[degree(K_L^{d, \Sigma ; f}) = \left\{ \begin{array}{lcl} \frac{2^{ \lvert \Sigma \lvert }h\Phi_K(f)}{\Phi (f)}\times degree(K^{d}) & if & (d,f) = 1 \\ \\ \frac{2}{3}\frac{2^{ \lvert \Sigma \lvert }h\Phi_K(f)}{\Phi (f)}\times degree(K^{d}) & if & (d,f) = 3 \end{array} \right.\]

Here \(\lvert \Sigma \lvert\) is the number of infinite places, and \(h_K = \lvert\mathcal{C}_K\lvert\) is the class number of \(K\). For the ray class field with finite modulus \(d\) ramified at one infinite place we know (see ref. [19]), that

\[degree(K^{d\infty_1}) = \frac{2h\Phi_K(d)}{[U_K:{U_{d}^1}]} \;, \label{Lang}\tag{11}\]

where \([U_K:{U_{d}^1}]\) is the index of the subgroup of the unit group \(U_K\) consisting of units that equal 1 mod \(d\). Putting things together, and using the multiplicative property of \(\Phi_K\), we find

\[{\rm excess} = \Phi(f) \frac{[U_K:{U_{d}^1}]}{[U_K:{U_{fd}^1}]} \;, \label{index}\tag{12}\]

unless \(d\) and \(f\) have a common factor in which case we obtain

\[{\rm excess} = \frac{3}{2} \Phi(f) \frac{[U_K:{U_{d}^1}]}{[U_K:{U_{fd}^1}]} \;, \label{index3}\tag{13}\]

We need to calculate the orders of \(U_K/{U_{d}^1}\) and \(U_K/{U_{fd}^1}\). Because the torsion group \(\pm 1\) is included in \(U_K\) this is equal to twice the order of a positive fundamental unit. For the unit defined in eq. (10 ) we calculate

\[\begin{array}{lc} u^3 = 1 + \frac{d^2(d-3) + d(d-2)\sqrt{(d+1)(d-3)}}{2} \\ \\ u^6 = 1 + \frac{d^2(d+1)(d-3)(d-2)^2 + d(d-1)(d-2)(d^2-2d -2) \sqrt{(d+1)(d-3)}}{2} \;. \end{array}\]

The unit \(u\) may not be a fundamental unit, but this complication cancels between numerator and denominator in equation (12 ). By inspection we see that \(u^3 = 1\) mod \(d\) and \(u^6 = 1\) mod \(fd\). This suffices to prove the theorem as stated. 0◻

Our proposition provides a supplement to the more abstract description in ref. [12]. To compute the excess is now a question of checking whether \(u^3 \equiv 1\) mod \(fd\). We find that the excess equals 1 for \(f = 2\) whenever it occurs, namely for \(d \equiv 3\) mod 4. It may be equal to 1 also if \(\Phi (f) = 2\), namely if \(f = 3\), 4, or 6. Table 5 gives the excess in a few interesting cases.

There are two further points to make. First, for minimal SICs we observed that the Galois group Gal\((H/K)\), where \(H\) is the Hilbert class field, gives rise to multiplets of geometrically inequivalent SICs. For non-minimal SICs the role of the Hilbert class field is taken over by a ring class field defined for the particular subring of integers that is defined by the conductor \(f\).

Then we will be interested to know for what values of \(d\) and \(f\) the SIC exhibits anti-unitary symmetry. This happens only if the underlying quadratic field \(K\) has a fundamental unit of negative norm, which means that the degree of the ray class field with modulus \(d\) goes down by a factor of 2. The SIC responds to this by having an anti-unitary symmetry, so that the number of distinct overlaps decreases by a factor of 2. In general, if \(d\) is of the form \(n^2+3\) then a unit of negative norm is

\[u_0 = \frac{n + \sqrt{n^2+4}}{2} = \frac{n + f_u\sqrt{\Delta_0}}{2} \;.\]

Squaring it, we obtain the totally positive unit

\[u = u_0^2 = \frac{n^2+2 + n\sqrt{n^2+4}}{2} = \frac{d-1 + nf_u\sqrt{\Delta_0}}{2} \;,\]

where \(f_u\) is an integer. According to Kopp and Lagarias [12] non-minimal SICs are attached to a non-maximal ring of integers, and (conjecturally) non-minimal SICs have anti-unitary symmetry if and only if \(u_0\) belongs to that ring. It follows that SICs associated to a divisor \(f\) will have anti-unitary symmetry if and only if \(f \lvert f_u\), which is a more stringent condition than \(f \lvert f_0\) [16].

6 Overlaps for non-minimal SICs↩︎

We continue to build our phenomenology through the simple expedient of calculating examples. For this purpose we need to know what examples we can realistically expect to calculate. Tables 12 and 13 in Appendix 13 list low dimensional examples for which non-minimal SICs occur. The degrees of the number fields in which we are interested tend to grow (quickly) with \(d\) and \(f\), so this gives some idea of where to find accessible examples. When anti-unitary symmetry occurs we can go to high dimensions.

What we find is that the overlap units for a SIC with a given conductor \(f >1\), having divisors \(f_1, \dots , f_n\) where \(f_1 = 1\) and \(f_n = f\), are products of overlap units for SICs with conductors that divide \(f\), times a new unit coming from the field with conductor equal to \(f\) itself. If the excess \(e > 1\) this new unit cannot be a Stark unit from \(K^{fd \infty_1}\), since this would generate too large a field. Let \(K_L^{d;f}\) be the Kopp–Lagarias field in our homemade notation. The idea is that

\[\lvert Gal(K^{fd}/K_L^{d;f)} \lvert = e \;.\]

If \(e = 1\) the new unit is, as expected, a square root of a Stark unit in \(K^{fd \infty_1}\) raised to some suitable power. But suppose that the excess is \(e > 1\). What we can do then is to calculate Stark units in \(K^{fd \infty_1}\), identify the Galois group of the extension \(K^{fd \infty_1}/K_L^{d \infty_1; f}\), and act by it on one of the Stark units that we calculated. This gives an orbit consisting of \(e\) units, and we multiply all of them together so that we obtain a unit left invariant by this Galois group. In other words, we take a relative norm. We end up with a collection of \(e\)-plets of Stark units that lie in \(K_L^{d \infty_1; f}\), and these \(e\)-plets are the very units that appear in the overlaps. We furthermore find that Grassl’s rule (7 ) correctly determines all the exponents of the square rooted Stark units that appear, provided it is applied separately to the contributions from each subfield, and provided we remember that the relevant Galois group is that which keeps the ring class field fixed.

The examples support this simple picture quite consistently. A drawback is evidently that we have to perform calculations in fields of high degree. Also, in practice it may not be so easy to identify the Kopp–Lagarias field. On the bright side it may happen that even if \(e > 1\) the excess for a subfield giving some of the baby overlaps may equal 1. An example is \(d = \partial \overline{\partial}= 199\), \(f = 5\), \(e = 2\), where the subfield \(K^{5\partial, \infty_1}\) has excess 1. In this case an exact SIC fiducial vector with anti-unitary symmetry is easy to compute using the algorithm in ref. [9].

As usual dimensions \(d = p \equiv 2\) mod 3 show the simplest pattern, but unfortunately we do not find any examples with excess \(e > 1\) among those we were able to compute. For the examples in Table 6 we performed the calculations exactly using Kopp’s method [8]. Using Magma for \(d = 23\), \(f = 2\), it took 14 hours to compute the Stark units in the larger field, 12 days to find a root of the minimal polynomial, and 29 minutes to verify that we have a SIC. The first case with \(e>1\) occurs for \(d = 47\), \(f = 8\), \(e = 2\), and calculating the minimal polynomial for the Stark units in this case is beyond us. In the examples we did calculate there are non-trivial ring class fields, giving rise to two (for \(d = 11\), 17) and four (for \(d = 23\)) unitarily inequivalent SICs, respectively, in the non-minimal multiplets.

To proceed we again lower our standards. We rest content with a check that the numerical overlap units provide roots of the minimal polynomial of an appropriate combination of Stark units. If it is too time-consuming to calculate all the Stark units we rest content with analysing some of the baby overlaps, and leave some entries in the tables blank.

The case \(d = p \equiv 1\) mod 3 splits into two subcases depending on whether \(d-3\) is a square or not. See Table 6. When \(d = 31\) we come across \(f = 4\). This has two divisors, and we have an example of a non-minimal SIC whose overlaps are given by products of Stark units from three different fields. The examples for \(d-3\) being a square have anti-unitary symmetry, of order 18 for \(d = 5779\), \(f = 2\).

We need to consider some examples with excess \(e > 1\), so that the Kopp–Lagarias ray class fields are not the usual ones. Scanning Tables 12 and 13 for accessible examples we find \(d = 19\) and \(d = 28\). The non-minimal SICs with \(f > 2\) that we are about to consider have symmetry of order 3 only.

For \(d = 19\) we have \(f_0 = 8\), so there are four possible values of \(f\). See Table 6, and Table 7 for examples with \(e= 2\). When \(f = 8\) Stark units from four distinct fields appear. An example with a more interesting divisor lattice for \(f_0\) will be found in Table ¿tbl:tab:35faser? below.

An example with \(e = 4\) is \(d = 28\), \(f = 5\). We bring up this case also because repeated Stark units occur—the field \(K^{5\partial, \infty_1}\) has degree 16 over \(K\), but there are only eight distinct Stark units and they do not generate the field. This is the largest ray class field in which we have encountered this phenomenon: one which is (conjecturally) explained by extra zeroes in the L-functions and which will be the subject of a future work. One finds when working out the relevant third row in Table 8 that although the 4-plet of Stark units seemingly has too low a degree over \(K\), its degree over the ring class field—which in this case equals \({\mathbb{Q}}(\sqrt{5})\)—is the same as that of the 4-plet occurring in the first row. And this is what counts when applying the rule (7 ), which correctly predicts the exponents in all the rows. Anyway the degree over \(K\) rises to the expected value when the 4-plet is multiplied with a Stark unit squared coming from \(K^{\partial \infty_1}\).

7 Dimensions of the form \(d = 3(3n+1)\)↩︎

It was noted in Section 3 that dimensions divisible by 3 but not by 9 are special because there is more than one conjugacy class of possible order three symmetries. This happens because the order three symmetry can act like the identity in the dimension 3 factor of Hilbert space—this is allowed because the symplectic identity matrix has trace \(-1\) counted mod 3. In Section 5 we found that dimensions \(d = 3(3n+1)\) are special because it can happen that \(f_0\) admits a divisor \(f\) that also divides \(d\). The upshot of this is that dimensions of the form \(d = 3(3n+1)\) admit SICs of a different symmetry type known as type \(F_a\) [13]. The ones with an ‘ordinary’ symmetry are referred to as type \(F_z\). One finds in these dimensions that the minimal SIC is of type \(F_a\), but this is always accompanied by non-minimal SICs of type \(F_z\).

The discussion of the centraliser \(M(S)\) in Section 3 now needs amendment. Clearly any \(GL(2,{\mathbb{Z}}/3{\mathbb{Z}})\) matrix commutes with the identity matrix. To realize the isomorphism with the abelian Galois group we need to choose a maximal abelian subgroup of the centraliser in the dimension 3 factor. Up to conjugation there are three possibilities [27]. The generators of the three abelian groups are respectively

\[\begin{align} a_4:\left\langle \left( \small{ \begin{array}{cc} -1 & 0 \\ 0 & - 1 \end{array}} \right) \;, \left( \small{ \begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array}} \right) \right\rangle\nonumber \\ \\ a_6:\left\langle \left( \small{ \begin{array}{cc} -1 & 0 \\ 1 & - 1 \end{array}} \right)\right\rangle \;,a_8:\left\langle \left( \small{ \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}} \right) \right\rangle \;. \nonumber \end{align}\]

When acting on the displacement operators in the dimension 3 factor these groups divide the 8 non-trivial ones into \(2 + 2 + 4\) for subtype \({a_4}\) and into \(2 + 6\) for \({a_6}\) (as happens also for \(F_z\) SICs), while the \(a_8\) group acts transitively on all eight of them.

This will have consequences for the Galois orbits of the baby overlaps. In particular, once the Hilbert space has been expressed as \({\mathbb{C}}^3\otimes {\mathbb{C}}^{3k+1}\) it will have consequences for overlaps of the form \(\langle \Psi_0 \lvert D_{i,j}\otimes {\boldsymbol{1}} \lvert \Psi_0 \rangle\). They will be formed from Stark units in \(K^{3, \infty_1}\) and its subfields. The divisor lattices shown in Figure 2 are relevant here. If 3 is inert—that is, if it remains a prime over the quadratic field \({\mathbb{Q}}(\sqrt{D})\)—then the Galois group will act transitively on these overlaps and the subtype must be \(a_8\). If 3 splits there will be three orbits of non-trivial overlaps and the subtype must be \(a_4\). And if 3 ramifies there will be two orbits, and the subtype must be \(a_6\) [28]. This behaviour is determined by the value of \(D\), according to \[\begin{array}{lll} D = 0 \;mod \;3 & \Rightarrow \;3 ramifies & \Rightarrow \;a_6 \\ \\ D = 1 \;mod \;3 & \Rightarrow \;3 splits & \Rightarrow \;a_4 \\ \\ D = 2 \;mod \;2 & \Rightarrow \;3 is inert & \Rightarrow \;a_8 \;. \end{array}\]

cm

This is illustrated by Table 9. Note that \(d = n^2 +3 = 3\) mod 9 happens whenever \(3 \lvert n\), which means that quite a few such \(F_a\) SICs have been constructed with exact arithmetic. The subtype is then necessarily \(a_4\). See Table 10.

Dimension \(d = 12 = 3\cdot 4\) is a very special case. It is not just that ‘short’ orbits occur, so that a factor of 3 occurs in the exponents of the Stark units. It is that the Stark units S\(_{12}\) and S\(_{4\overline{\partial}}\) do not generate their fields. Indeed their degree is only one half of that of \(K^{12,\infty_1}\) and \(K^{4\bar{\partial} , \infty_1}\) respectively. However, taken jointly they do generate the full field. Grassl’s rule for the exponents holds provided the Galois group is that of the ray class field, not that of the subfield generated by the Stark units. Another special feature of \(d = 12\) is the large number of identities between Stark units in the various subfields.

Non-minimal SICs of \(F_a\) type appear (or are believed to appear) when \(f_0/f = 0\) mod 3 [16]. When \(f\) and \(d\) share a common factor of 3 there is a complication. By considering the degrees of the relevant fields one is led to expect that all non-minimal \(F_a\) SICs with \((d,f) = 3\) are of subtype \(a_6\). Were this not the case Grassl’s rule (7 ) would predict a fractional exponent. An example of such a non-minimal \(F_a\) SIC occurs when \(d = 84\), and it is indeed of subtype \(a_6\).

Whenever an \(F_a\) SIC occurs a non-minimal \(F_z\) SIC occurs as well. We then face the question of how the overlaps in the \(F_z\) SICs are related to those of the minimal \(F_a\) SIC. Table 11 gives an account of this for \(d = 39\). There are no surprises there, but note incidentally that for \(f = 3\) the excess \(e = 3\) for the fields \(K^{3d \infty_1}\) and \(K^{3\cdot 3,\infty_1}\), but \(e = 2\) for \(K^{3\cdot 13,\infty_1}\). For the degree of the Kopp–Lagarias fields \(K_L^{m,\infty_1;3}\) it matters whether the modulus \(m\) is divisible by 3 or not, so this is what one would expect given the results in Section 5. But we have no proof since the theorem on which Section 5 relies [12] covers only the case \(m = d\) explicitly.

8 SIC alignment↩︎

We now consider dimensions of the form \(d = n(n-2)\), or equivalently dimensions such that \(d+1\) is a square. This happens if and only if the position \(\ell\) of the dimension in its AFMY tower is a multiple of 2. If \(d_\ell = n\) then \(d_{2\ell} = n(n-2)\). In these dimensions the phenomenon of SIC alignment enters [29]. Actually, whenever the base dimension \(d_1 \lvert d_\ell\) in an AFMY tower, square rooted Stark units from \(K^{d_1 \infty_1}\) will appear raised to the power \(\ell\) among the baby overlaps in dimension \(d_\ell\), simply because of Grassl’s rule (7 ) and the fact that the symmetry goes up by such a factor. We saw examples (\(d = 124\), 844, and 1299) in Tables 4 and 10.

But in the case we now have in mind, more is true. It is known that from a SIC in dimension \(n\) one can construct a continuous family of configurations of vectors in dimension \(d = n(n-2)\) that share some of the properties of a SIC. If \(n\) is odd any vector \(\lvert \Psi \rangle\) in this family provably obeys [30]

\[\sqrt{d+1} \langle \Psi \lvert {\boldsymbol{1}}\otimes D \lvert \Psi \rangle = - (overlap unit from dimension n)^2 \;, \label{kvadrater}\tag{14}\]

\[\sqrt{d+1}\langle \Psi \lvert D\otimes {\boldsymbol{1}} \lvert \Psi \rangle = 1 \; , \label{ettor}\tag{15}\]

where we assumed that the Hilbert space is \({\boldsymbol{C}}^{n-2}\otimes {\boldsymbol{C}}^n\). If \(n\) is even something similar but a bit more complicated is true [31], [32]. It is expected that one can always obtain a \(d\)-dimensional SIC by specializing the parameters in this construction, and this SIC is then said to be aligned to the lower dimensional SIC. Hence there is a semi-constructive point of view on these aligned SICs.

Equation (15 ) will be our main concern in section 9. It is consistent with our overall picture only if Stark’s construction [4] gives trivial units \(+1\) for the ray class field \(K^{d-2,\infty_1}\). This will be so if

\[K^{d-2} = K^{d-2,\infty_1} = K^{d-2,\infty_2} \;. \label{Garyrule}\tag{16}\]

We will prove that, indeed, this is true for all integers \(d \geq 5\) when \(d\) is connected to the base field \(K\) by equation (2 ). That squared overlap units from the lower dimension appear in eq. (14 ) is a simple consequence of the prediction we have called Grassl’s rule (7 ), because the order of the symmetry group of the aligned SIC is a factor of two larger than that of the SIC in the lower dimension while the ray class field (for these special baby overlaps) stays the same.

As our example of a dimension where aligned SICs occur we choose \(d = 35 = 7\cdot 5\). This example is also interesting because it is the lowest dimension in which the Kopp–Lagarias inclusion lattice is non-trivial. See Figure 6. In \(d = 7 = 2^2 + 3\) there are two SICs with \(f = 1\), 2, and overlap units as given in Table 6. In Table ¿tbl:tab:35faser? the alignment makes itself felt in the first four rows, causing an otherwise unexpected behaviour among the baby overlaps there.

When \(f = 4\), \(e = 2\), there is a slight problem in the first row. In this case the Stark units \(S_{4\overline{\partial}}\) do not generate the field \(K^{4\overline{\partial},\infty_1}\), and Grassl’s rule for the exponents becomes a little ambiguous. Note also that there are a large number of identities connecting Stark units in the ‘small’ fields, so the entries in Table ¿tbl:tab:35faser? can be rewritten in various ways.

9 The moduli \({\mathfrak{m}}_0 =\) \(d-3\), \(d-2\), \(d-1\), \(d\)↩︎

Motivated by the notion of SIC alignment, we briefly develop some elementary class field theory results which would seem to be interesting in their own right, as well as in their applications to section 8. The finite part of the modulus \({\mathfrak{m}}_0\) will be a \(K\)-integral ideal of \({\mathbb{Z}}_K\) and \({\mathfrak{m}}_\infty\) an ‘infinite modulus’: a possibly empty subset of \(\{\infty_1,\infty_2\}\). The key equation (2 ) is in place, and we examine the degrees of the field extensions over the totally real ‘base’ ray class field \(K^{{\mathfrak{m}}_0}\) as we vary \({\mathfrak{m}}_\infty\).

Recall that we denote by \(\tau\) the generator of the Galois group \({\mathrm{Gal}_{K/\mathbb{Q}}}\), and that \(\mathfrak{j}\) and \(\mathfrak{j}^\tau\) will denote the two embeddings of \(K\) into \(\mathbb{R}\), chosen so that \(\mathfrak{j}(\sqrt{D}) > 0\) (meaning \(\mathfrak{j}\) is associated with the place denoted by \(\infty_2\), as explained in section 2). The function \(\mathop{\mathrm{{\mathbf{N}}}}x = \mathfrak{j}(x)\mathfrak{j}^\tau(x)\) denotes the field norm from \(K\) down to \(\mathbb{Q}\). Recall that \(u_D\) is defined to be the first totally positive power of \(u_o\), in terms of \(u_o\): \[u_D = \begin{cases} & {u_o, \textrm{ when \mathop{\mathrm{{\mathbf{N}}}}u_o= +1;}} \\ & {u_o^2, \textrm{ when \mathop{\mathrm{{\mathbf{N}}}}u_o= -1.}} \end{cases}\] The fundamental unit \(u_o\) is always assumed to have been chosen so that \(\mathfrak{j}(u_o)>1\). Hence \(0 < \mathfrak{j}^\tau(u_o) < 1\) when \(\mathop{\mathrm{{\mathbf{N}}}}u_o= 1\) and \(-1 < \mathfrak{j}^\tau(u_o) < 0\) when \(\mathop{\mathrm{{\mathbf{N}}}}u_o= -1\). By 4 the dimensions \(d_\ell(D)\geq4\) are given by \(d=d_\ell(D)=u_D^\ell+u_D^{-\ell}+1\). The main result is as follows.

In case (A) in particular, therefore, \(K^{{\mathfrak{m}}_0\infty_1\infty_2}\) is a CM field: that is to say, a totally complex field which is an extension of degree \(2\) of a totally real field. Indeed, by lemma 1 [sroots] we see that \(K^{{\mathfrak{m}}_0\infty_1\infty_2}\) is Galois over \(\mathbb{Q}\) but that it also must contain non-real roots of unity. Hence the CM assertion will be true provided that we are able to show the first part, which since \(K^{{\mathfrak{m}}_0}\) is by definition totally real, amounts to showing that \(K^{{\mathfrak{m}}_0\infty_1}\) and \(K^{{\mathfrak{m}}_0{\infty_2}}\) are also totally real. This in turn will be true (since \(K^{{\mathfrak{m}}_0}\) is a subfield of both) if and only if the respective ray class groups are isomorphic. This is how we shall prove part (A): see in particular the diagram 19 .

Part (B) is the situation described in Figure 1; because the proof is just a straightforward modification of an argument in [2], we have moved it to appendix 14, together with some technical components of the proof of part (A).

In order to put this in context, a first step is to outline what possible degrees can occur for these extensions. Let us free up \({\mathfrak{m}}_0\) to be any finite ideal of \({\mathbb{Z}}_K\) for a moment. We are interested in the sequence of degrees \[\left[ \;[ K^{{\mathfrak{m}}_0 \infty_1} \colon K^{{\mathfrak{m}}_0} ], \; [ K^{{\mathfrak{m}}_0 \infty_2} \colon K^{{\mathfrak{m}}_0} ], \;[ K^{{\mathfrak{m}}_0 \infty_1\infty_2} \colon K^{{\mathfrak{m}}_0} ] \;\right] \;. \label{degz}\tag{17}\] By standard class field theory, the first two of the indices in 17 must equal either \(1\) or \(2\), and by similar reasoning the third must be \(1\), \(2\) or \(4\); indeed \(K^{{\mathfrak{m}}_0 \infty_1\infty_2}\) is restricted in all cases to be an extension of degree either \(1\) or \(2\) of each of its subfields \(K^{{\mathfrak{m}}_0 \infty_1}\) and \(K^{{\mathfrak{m}}_0 \infty_2}\). The possible sequences are

1. \(\mathop{\mathrm{{\mathbf{N}}}}{u_o}=1\), \({\mathfrak{m}}_0\) rational or irrational: \([1,1,2], \;[2,2,4]\);

2. \(\mathop{\mathrm{{\mathbf{N}}}}{u_o}=-1\), \({\mathfrak{m}}_0\) rational: \([1,1,1], \;[1,1,2], \;[2,2,4]\);

3. \(\mathop{\mathrm{{\mathbf{N}}}}{u_o}=-1\), \({\mathfrak{m}}_0\) irrational: \([1,1,1], \;[1,1,2], \;[1,2,2], \;[2,1,2], \;[2,2,4]\).

The configurations \([1,2,2]\) and \([2,1,2]\) already played a role in section : see eq. (8 ) and ref. [9]. They can occur only for non-rational moduli, and only when the norm of the fundamental unit is \(-1\).

However, for the purposes of theorem 2, we may discard all but \([1,1,2]\) and \([2,2,4]\) since we shall always be choosing a modulus ideal which is principal over \({\mathbb{Z}}_K\) and generated by a rational integer \(\geq3\), where the Galois involution \(\tau\in{\mathrm{Gal}_{K/\mathbb{Q}}}\) cannot change the degrees of the intermediate extensions.

Considering all moduli of the form \({\mathfrak{m}}_0 = (\lambda_1 + \lambda_2 d){\mathbb{Z}}_K\), for \(\lambda_1\), \(\lambda_2 \in \mathbb{Z}\) not both zero, the ‘normal’ situation is that any such fixed form for the modulus yields a mixture of both outcomes in some ratio. So it is worth making the following observation:

The remainder of this section 9 is devoted to a proof of part (A). We rely on results in appendix 14.

For general \({\mathfrak{m}}_0\) and \({\mathfrak{m}}_\infty\), define \({U_{{\mathfrak{m}}}^1}\) to be the subgroup of \({\mathbb{Z}^\times_K}\) whose elements are simultaneously congruent to \(1\bmod{\mathfrak{m}}_0\) and positive at all of the real places inside the set \({\mathfrak{m}}_\infty\) (this is often called multiplicative congruence, and denoted by \(\bmod^\times{\mathfrak{m}}\)). The main technical ingredient for the proof is the exact sequence of global class field theory, as follows. See for example equation (2.7) of [33]. \[\label{globcft} 1 \rightarrow {U_{{\mathfrak{m}}}^1} \longrightarrow {{\mathbb{Z}^\times_K}} \xlongrightarrow{\psi } {\bigl({{\mathbb{Z}}_K/{({\mathfrak{m}}_0)}}\bigr)^\times} \times{\{\pm1\}}^{\# {\mathfrak{m}}_\infty } \xlongrightarrow{\alpha} \mathrm{Gal}(K^{\mathfrak{m}}/ K) \longrightarrow \mathcal{C}_K \rightarrow 1.\tag{18}\] As we mentioned in section 2, the Artin map \(\alpha\) establishes an isomorphism between the ray class group of modulus \({\mathfrak{m}}\), and the Galois group of the ray class field \(K^{\mathfrak{m}}\) over \(K\).

Given any \(x\) in the global units \({\mathbb{Z}^\times_K}\), the natural map \(\psi\)—whose kernel is the unit subgroup \({U_{{\mathfrak{m}}}^1}\) just defined—is given by reduction of \(x\) modulo \({\mathfrak{m}}_0\) into the first component \({\bigl({{\mathbb{Z}}_K/{({\mathfrak{m}}_0)}}\bigr)^\times}\), and then by the signs \(\pm1\) induced according to each real embedding of \(x\) into the second component \({\{\pm1\}}^{\# {\mathfrak{m}}_\infty }\). For calculations we mention that the codomain of \(\psi\) is known as the ray residue ring in Magma [22]. In particular this codomain is finite, which constrains the image of \(\psi\) also to be finite: see [19].

Just to place things in context, in section 5 we made use of the Euler generalised totient function, which ties back to this sequence in that it is the order of the term \({\bigl({{\mathbb{Z}}_K/{({\mathfrak{m}}_0)}}\bigr)^\times}\).

The varying infinite modulus in theorem 2 (A) is captured by the following commutative diagram with exact rows, built via the functoriality of the Artin isomorphism from two versions of 18 wherein we move from a given \({\mathfrak{m}}_\infty\) to a subset.

For brevity we just annotate the maps in the top row with the representative subscript \(\mathfrak{j}\) to indicate ramification being allowed at one real place, with no subscript denoting \({\mathfrak{m}}_\infty = \{\}\): \[\label{twol} \begin{adjustbox}{width=\textwidth} \begin{tikzcd} 1 \arrow[r] & {U_{{\mathfrak{m}}_0,\mathfrak{j}}^1} \arrow[r] \arrow[d,"\gamma"] & {\mathbb{Z}^\times_K}\arrow[r, "\psi_{\mathfrak{j}}"] \arrow[d,equal] & {\bigl({{\mathbb{Z}}_K/{({\mathfrak{m}}_0)}}\bigr)^\times} \times \{\pm1\} \arrow[r, "\alpha_{\mathfrak{j}}"] \arrow[d,"\rho"] & \mathrm{Gal}(K^{{\mathfrak{m}}_0,\mathfrak{j}} / K) \arrow[r] \arrow[d,"\mu"] & \mathcal{C}_K \arrow[r] \arrow[d,equal] & 1 \\ 1 \arrow[r] & {U_{{\mathfrak{m}}_0,\{\}}^1} \arrow[r] & {\mathbb{Z}^\times_K}\arrow[r, "\psi"] & {\bigl({{\mathbb{Z}}_K/{({\mathfrak{m}}_0)}}\bigr)^\times} \arrow[r, "\alpha"] & \mathrm{Gal}(K^{{\mathfrak{m}}_0,\{\}} / K) \arrow[r] & \mathcal{C}_K \arrow[r] & 1 \end{tikzcd} \end{adjustbox}\tag{19}\] The maps \(\gamma\), \(\rho\) and \(\mu\) are implicitly defined by the diagram. By the definition of the \({U_{{\mathfrak{m}}}^1}\) (or by the snake lemma) we know that \(\gamma\) is an injection, since \({U_{{\mathfrak{m}}_0{\mathfrak{m}}_\infty}^1}\) is in all cases just \({U_{{\mathfrak{m}}_0,\{\}}^1}\) under an additional compatible restriction. Similarly \(\rho\) is a projection onto the component corresponding to the finite part \({\mathfrak{m}}_0\) of the modulus and so is surjective, with kernel \(\ker\rho \cong \{\pm1\}\) representing the signs under the single real embedding \(\mathfrak{j}\). From what was said above we need to prove that \(\mu\) is an isomorphism whenever \(\mathfrak{j}\in \infty_1\) or \(\mathfrak{j}\in \infty_2\).

The key ingredient in the proof of theorem 2 (A) is the following.

Although we prove it formally below using some elementary homological algebra, it is relatively straightforward to see from diagram 19 that this defect of 2 is precisely what is needed to cancel out the extra sign module \(\{\pm1\}\) in order to ensure that the Galois groups at the right hand side—given that the map between the class groups \(\mathcal{C}_K\) is the identity—have the same order.

This completes the proof of assertion (A) of theorem 2, as follows. The first part—namely, that \(K^{{\mathfrak{m}}_0} = K^{{\mathfrak{m}}_0\infty_1} = K^{{\mathfrak{m}}_0{\infty_2}}\)—is that in our situation, the map \(\mu\) is always an isomorphism. But by 24  \(\overline{\mu}\) is injective and by 23 it was onto, so \(\overline{\mu}\) is an isomorphism; by the snake lemma applied to 22 , so is \(\mu\). The second part—namely that relative to these three identical totally real fields the full field \(K^{{\mathfrak{m}}_0{\infty_1\infty_2}}\) is indeed a proper extension of degree 2—follows from lemma 1 [sroots]. 0◻

10 Comments in closing↩︎

In closing we can say that we have verified the four claims made about overlap units in Section 1, for all our examples. The fact that the overlaps often sit in a proper subfield of \(K^{fd \infty_1}\) is no objection, since we are in effect taking relative norms to reach that subfield.

Using Hilbert space arguments it can be shown that there are cases where some squares of overlap units necessarily are equal to \(+1\). This is the exceptional case where we can support our claims with proofs, because these overlap units appear precisely when Stark’s construction results in units equal to \(+1\). The behaviour of the ray class fields with finite modulus \(d-3\), \(d-2\), \(d-1\), and \(d\), as expounded in section 9, may be of independent interest.

We did find a few examples (when \(d = 12\), 28, and 35) where the Stark units do not generate the field in which they sit. Other examples are known. This phenomenon causes no special problem for the construction of the SIC, although the rule (7 ) for the exponents of the Stark units can become a delicate one to apply.

We were able to offer only the vaguest of rationales for why it is square roots of Stark units that appear, and why one needs to include products of square roots of Stark units from several distinct fields in the non-minimal case. Indeed, we have been concerned with how Stark units enter SIC overlaps, not why they do so.

An important point to be raised is the relation between the facts we have observed on the one hand, and the conjectural identification of all the overlap units with special values of the Shintani–Faddeev modular cocycle on the other [15], [16]. In principle it should be possible to derive all our results from the main theorem (Theorem 1.1) of ref. [15]. This remains a task for the future however.

We received indispensable help from Marcus Appleby and Markus Grassl. IB is grateful to the taxpayer for support through the Digital Horizon Europeproject FoQaCiA, GA No. 101070558, funded by the European Union, NSERC (Canada), and UKRI (UK); and to the Mathematics Department at Stockholm University for being allowed to use their computers. GM is grateful to Myungshik Kim, Terry Rudolph and the QOLS group at Imperial College for their ongoing hospitality.

Appendices↩︎

11 The group↩︎

The Weyl–Heisenberg group \(H(d)\) is a finite dimensional analogue of the group that underlies Heisenberg’s uncertainty relation [35]. It is a central extension of the direct product of two cyclic groups, and can be presented as

\[ZX = \omega XZ \;,X^d = Z^d = \omega^d = {\boldsymbol{1}} \;,\]

with the understanding that \(\omega\) commutes with everything and can be represented as multiplication with a primitive root of unity. We choose

\[\omega = e^{\frac{2\pi i}{d}} \;.\]

If \(d\) is even the centre is enlarged to include also

\[\tau = - e^{\frac{\pi i}{d}} \;.\]

Once this has been agreed on there is an essentially unique irreducible unitary representation in dimension \(d\), given by

\[Z \lvert r\rangle = \omega^r \lvert r\rangle \;,X \lvert r\rangle = \lvert r + 1\rangle \;.\]

The basis vectors are indexed by integers modulo \(d\). Throughout this paper we assume that this representation is used, unless \(d = 4\) in which case a variation is useful [25]. An important fact is that if \(d = d_1d_2\) where \(d_1\) and \(d_2\) are relatively prime then there is a canonical isomorphism [25], [29]

\[H(d) = H(d_1) \times H(d_2) \;.\]

To show this the Chinese remainder theorem is used. It means that once we understand the group in prime power dimensions everything else follows.

It is useful to introduce the \(d^2\) displacement operators

\[D_{i,j} = \tau^{ij}X^iZ^j \;,\]

with this precise choice of the prefactor [23]. For them the group law reads

\[D_{i,j}D_{k,l} = \tau^{jk - il}D_{i+k,j+l} \;.\]

The symplectic form that appears in the exponent on the right hand side is the key to understanding the unitary automorphism group of the Weyl–Heisenberg group, and the symmetries exhibited by its orbits. In the representation we are using, there are outer automorphisms of the Weyl–Heisenberg group that act by conjugation, forming a representation of the symplectic group \(SL(2,{\mathbb{Z}}/d{\mathbb{Z}})\). Hence we can associate every symplectic matrix \(G\) with a unitary operator \(U_G\) according to the following scheme:

\[\left( \begin{array}{c} i' \\ j' \end{array} \right) = \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) \left( \begin{array}{c} i \\ j \end{array} \right)\leftrightarrowU_GD_{i,j}U_G^{-1} = D_{i',j'} \;. \label{symplektiskt}\tag{25}\]

If \(\alpha \delta - \beta \gamma = - 1\) mod \(d\) the operator \(U_G\) is an anti-unitary rather than a unitary operator. Moreover, the integer entries of \(G\) should be taken modulo \(2d\) if \(d\) is even; but this is not important for us [23].

The unitary or anti-unitary operators \(U_G\) transform SICs to SICs. Symmetries of SICs always include a Zauner unitary, that is to say an operator \(U_G\) coming from a symplectic matrix of order 3 and trace \(-1\). Unless \(d>3\) is divisible by 3 but not by 9 there is a unique conjugacy class of such matrices.

Another key property of the group is that it forms a unitary operator basis [36]. This means that any operator acting on \({\mathbb{C}}^d\) can be expanded as

\[A = \frac{1}{d}\sum_{i,j = 0}^{d-1} D_{i,j}TrAD_{-i,-j} \;. \label{Schwinger}\tag{26}\]

We call this the Schwinger formula for ease of reference. If we choose \(A = \lvert \Psi_0 \rangle \langle \Psi_0 \lvert\) we learn that the overlaps determine the fiducial vector uniquely up to an irrelevant phase.

Now we return to the beginning and observe that the choice \(\omega = e^{2\pi i/d}\) was arbitrary. Any primitive \(d\)th root of unity would serve. Such a change of representation is achieved by the Galois transformation \(\omega \rightarrow \omega^a\), and leads us to bring in the group \(GL(2,{\mathbb{Z}}/d{\mathbb{Z}})\) [1]. This needs an extra generator transforming the displacement operators as

\[G = \left( \begin{array}{cc} 1 & 0 \\ 0 & a \end{array} \right) \;,X \rightarrow X \;,Z \rightarrow Z^a \;,\]

where \(a\) is any integer invertible modulo \(d\). Anti-unitary transformations are a special case of this.

The reader may ask if maximal sets of equiangular vectors that are not orbits of the Weyl–Heisenberg group can exist. The answer is ‘no’ if \(d = 2\), 3 [37], [38] and ‘probably no’ if \(d = 4\), 5 [39]. In dimension \(d = 8\) an example that is an orbit of a differently defined Heisenberg group does exist [40]. Possibly this is the only such example, but the question is open.

12 Order four symmetries↩︎

The prominent role of order three symmetries in Section 3 et passim is remarkable. One can ask if, say, a symmetry of order four would lead to something of interest? Assume for simplicity that \(d = p\) is a prime. A first observation is that the centraliser of an order four symplectic matrix within \(GL({\boldsymbol{Z}}/d{\boldsymbol{Z}})\) has order \(p^2-1\) if \(d = 3\) mod 4 and order \((p-1)^2\) if \(d = 1\) mod 4. So far, so good. If we want to mimic the constructions made for SICs the next step is to ask if, given \(d\), it is possible to choose a real quadratic field \(K\) such that the ray class field \(K^{d\infty_1}\) has a degree equal to the order of the centraliser divided by 4, or possibly divided by 8 should anti-unitary symmetries occur. Will this lead to a magical formula differing from equation (2 ), and if so what kind of Weyl–Heisenberg orbits can be obtained from fiducial vectors constructed using such a field? But this seems to be a dead end. If we also insist that the degree increases by a factor of 2 when an infinite place is added to the modulus there are only a very few exceptional cases (such as \(p = 3\)) where such fields exist. This is one way to see why symmetries of order three are very special.

13 Finding accessible examples↩︎

In Tables 12–13 we give two lists of low dimensions where non-minimal SICs occur, and their excess as defined in equation (9 ). The lists are complete up to the highest dimensions cited. If an underlined entry is not mentioned in the main text it is because there were no surprises there.

14 Lemmas needed for theorem 2↩︎

In our situation we need to understand the image \(\mathop{\mathrm{im}}\psi\) in the exact sequence 18 , under certain conditions on \({\mathfrak{m}}\). It will be convenient to regard the units \({\mathbb{Z}^\times_K}\) of \(K\) as decomposing in the following standard fashion, where we denote by \(<x>\) the (possibly infinite) multiplicative cyclic group generated by an element \(x\). The torsion is just \(\pm1\) and so with our fixed choice of a fundamental unit \(u_o\) we may write: \[\label{gengen} {\mathbb{Z}^\times_K}\;\;= \;\;<u_o> \times <-1> \;\;\cong \;\;\mathbb{Z}\times C_2 .\tag{27}\] We now begin to gather up some facts about the various invariants here. We continue to regard \({\mathfrak{m}}_0\) as one of the two possibilities \((d-1){\mathbb{Z}}_K\) or \((d-2){\mathbb{Z}}_K\).

Define \(w\) to be the multiplicative order of the image of the fundamental unit \(u_o\) under reduction modulo the finite part \({\mathfrak{m}}_0\) of the modulus, irrespective of the real places. In other words, \(w\) is the lowest positive integer such that \((u_o+{\mathfrak{m}}_0)^w = u_o^w + {\mathfrak{m}}_0 = 1 + {\mathfrak{m}}_0\) in \(\left({\mathbb{Z}}_K/{\mathfrak{m}}_0\right)^\times\). In order to avoid over-burdening the reader with notation we shall use this abbreviation \(w\), assuming it known from the context whether \({\mathfrak{m}}_0\) be generated by \(d_\ell-1\) or by \(d_\ell-2\). The prevailing value of \(\ell\), of course, is also contextual.

In [2], where the finite part of the modulus \({\mathfrak{m}}_0\) is \(d_\ell\) or \(2d_\ell\), the order \(w\) of \(u_o\) modulo \({\mathfrak{m}}_0\) is a priori difficult to predict and requires a direct calculation for each case. However in the main cases of interest here—that is to say, \({\mathfrak{m}}_0 = (d_\ell-1){\mathbb{Z}}_K\) or \((d_\ell-2){\mathbb{Z}}_K\)—it is always given by the following simple rules (parts [deekm1] and [deekm2]), which we shall need in the proofs below. This is the key to why these cases give a simple hierarchy of ray class fields.

We also include some other necessary technical observations.

Lemma 2 [evev] is now the remaining technical ingredient needed to delineate the two cases which can occur for our two possible choices of finite moduli \({\mathfrak{m}}_0\), namely \((d_\ell-1){\mathbb{Z}}_K\) and \((d_\ell-2){\mathbb{Z}}_K\). Recall that we write \({\mathfrak{m}}= {\mathfrak{m}}_0{\mathfrak{m}}_\infty\).

14.0.0.1 Proof of Theorem 2 (B)

When \({\mathfrak{m}}_0 = d{\mathbb{Z}}_K\) the statement of (B) is just proposition 10 of [2], or indeed lemma 5.3 of [8]. When \({\mathfrak{m}}_0 = (d-3){\mathbb{Z}}_K\) we use the same argument. Briefly, we observe that by the definition of \(d_\ell=d_\ell(D)\), \(d_\ell-3 = u_D^\ell + u_D^{-\ell} - 2 = \frac{(u_D^\ell-1)^2}{u_D^\ell}\) and so \({\mathfrak{m}}_0 = (u_D^\ell-1)^2 {\mathbb{Z}}_K\), which means that either \(u_D^\ell \equiv 1 \bmod {\mathfrak{m}}_0\) or else that \((u_D^\ell-1)\) is nilpotent of index exactly \(2\). As it turns out, the only case for which \(u_D^\ell \equiv 1 \bmod {\mathfrak{m}}_0\) is when \(D=5\) and \(\ell=1\): that is, when \(d=4\), which is obviously a trivial case since \(d-3=1\). For every other case, we have this neat degree-2 nilpotency.

We now refer directly to the discussion around equations (15) and (16) in the proof of proposition 10 of [2]. Again invoking lemma 1, \({U_{(d-3){\mathbb{Z}}_K\{\}}^1}\) must have a generator of the form \(\pm u_D^k\) for some \(k\geq1\). Hence in order to draw the same conclusions as we did there about the modulus \(d\), we must show that the unit kernel \({U_{(d-3){\mathbb{Z}}_K\{\}}^1}\) contains no units which are totally negative. This has the consequence that all four unit kernel groups \({U_{(d-3){\mathbb{Z}}_K\{\}}^1}\), \({U_{(d-3){\mathbb{Z}}_K\infty_1}^1}\), \({U_{(d-3){\mathbb{Z}}_K\infty_2}^1}\) and \({U_{(d-3){\mathbb{Z}}_K\infty_1\infty_2}^1}\) are identical, and then by the formula 11 quoted from Lang [19] we see that the increases in the orders of the ray class groups are entirely governed by adding in successive real places to the modulus, as claimed.

So suppose to the contrary that we do in fact have a unit in \({U_{(d-3){\mathbb{Z}}_K\{\}}^1}\) which is negative at both infinite places. In the cases where \({\mathbb{Z}}_K\) has a negative fundamental unit \(u_o\), the odd powers will always have mixed signs (one positive, the other negative). Hence from the discussion just above, any totally negative unit must be of the form \(-u_D^k\) for some non-zero \(k \in\mathbb{N}\).

It is an easy consequence of the basic properties of the modified Chebyshev polynomials \(T_n^\ast(X)\)—as defined in [2] or [10]—that for every pair of positive integers \(m,n\), the following rational integer congruence holds: \[\label{dm3} (d_{mn}-3) \equiv 0 \bmod (d_n-3) .\tag{32}\] We now need to be careful about the subscripts of the \(d=d_j = d_j(D)\) for \(j=1,k,\ell\). First of all, just running through the definitions implicit in the notation, since \(d = d_\ell = d_\ell(D) = u_D^\ell + u_D^{-\ell} + 1\) for some \(\ell\), the assumption that \(-u_D^k \in {U_{(d_\ell-3){\mathbb{Z}}_K\{\}}^1}\) is equivalent to \(u_D^k \equiv -1 \bmod (d_\ell-3){\mathbb{Z}}_K\), which in turn implies \[d_k = u_D^k + u_D^{-k} + 1 \equiv -1 -1 + 1 \equiv -1 \bmod (d_\ell-3){\mathbb{Z}}_K.\] But 32 with \(m=\ell\) and \(n=1\) yields further that \[\label{drm1} d_k \equiv -1 \bmod (d_1-3){\mathbb{Z}}_K.\tag{33}\] On the other hand, again from 32 but with \(m=k\) and \(n=1\), we know that \(d_k-3\) is also divisible exactly by \(d_1-3\), and so additionally \[\label{drp3} d_k \equiv 3 \bmod (d_1-3){\mathbb{Z}}_K.\tag{34}\]

But then 34 together with 33 says that \(d_1-3\) divides into their difference, ie \(3 - (-1) = 4\), inside \({\mathbb{Z}}_K\). This leaves a very limited number of cases—that is to say, \(d_1 = 4,5,7\)—two of which (\(4\) and \(5\)) we have excluded, and the other of which may be verified by direct calculation, since for all \(k\) it is the case again by 32 that \[{U_{(d_k-3){\mathbb{Z}}_K\{\}}^1}\trianglelefteq {U_{(d_1-3){\mathbb{Z}}_K\{\}}^1} .\]

So \({U_{(d-3){\mathbb{Z}}_K\{\}}^1}\) has no totally negative units. 0◻

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