1 Introduction

In 1974, Feit showed, answering a question of Serre, that the quaternion group \(Q_8\) of order 8 is isomorphic to a subgroup of \(\textrm{GL}_2(K)\) for \(K={\mathbb Q}\bigl (\sqrt{-35}\bigr )\) but not to \(\textrm{GL}_2(R)\) where \(R={\mathfrak {O}}_K\) is the ring of integers of K. (This has been re-discovered in [1].) Of course this means that the unique faithful irreducible character \({\chi }\) of \(Q_8\) can be realized over K but not over R, that is, there is no \(RQ_8\)-lattice affording \({\chi }\) which is free over R. In three letters to Feit (published in [13]), Serre described precisely when this happens. A necessary condition is that the class number \(h_K\), the order of the ideal class group \(C_K\) of K, is even (by a result of Schur [2, Thm. 23.17]).

Throughout the paper, N will denote a square-free integer \(>3\) and \(K={\mathbb Q}\bigl (\sqrt{-N}\bigr )\); we equally well could exclude further six positive numbers N where \(h_K=1\) (Gauss–Heegner–Baker–Stark). Let \(D=\bigl ({-1,-1\over {\mathbb Q}}\bigr )\) be the standard (definite) quaternion algebra over \({\mathbb Q}\), with basis \(1, i, j, k=ij\) (\(i^2=-1, j^2=-1, ji=-k\)). Then we may identify \(Q_8=\langle i, j\rangle \), and the Hurwitz quaternions \({{\mathcal {H}}}\) generated as \({\mathbb Z}\)-module by 1, i, j, k and \((1+i+j+k)/2\) give rise to a distinguished maximal order in D (with unit group \({\mathcal {H}}^*\cong 2A_4\cong \textrm{SL}_2(3)\) containing \(Q_8\)). Let \((D)=(-1,-1)\) denote the class of D in \(Br_2({\mathbb Q})\), the 2-torsion of the rational Brauer group.

Theorem 1

The faithful irreducible character of \(Q_8\) can be realized over \(K={\mathbb Q}\bigl (\sqrt{-N}\bigr )\) if and only if \({\mathbb Q}\bigl (\sqrt{N}\bigr )\) can be embedded into a \(Q_8\)-field over the rationals (Galois with group \(Q_8\)). It can be realized over \(R={\mathfrak {O}}_K\) precisely when \({\mathbb Q}\bigl (\sqrt{2}, \sqrt{N}\bigr )\) can be embedded into such a field or N is a sum of two integer squares.

We use the results obtained by Serre [13], and a celebrated theorem of Witt [16]: A real biquadratic number field \({\mathbb Q}\bigl (\sqrt{a}, \sqrt{b}\bigr )\) can be embedded into a \(Q_8\)-field over the rationals if and only if \((-1,-1)=(-a,-b)\) in \(Br_2({\mathbb Q})\).

One knows that in \(Q_8\)-fields over the rationals at least two (finite) primes are ramified, and those where only two primes are ramified are described explicitly in [9]. In fact, either \(p=2\) and some prime \(q\equiv 1, 3\,({\textrm{mod}\,}8)\) ramify, or two distinct primes \(p\equiv q\equiv 1\,({\textrm{mod}\,}4)\). (For \(p=2, q=3\), one of the two \(Q_8\)-fields is \({\mathbb Q}\Bigl (\sqrt{(2+\sqrt{2})(3+\sqrt{3})}\,\Bigr )\) already given by Dedekind.) By genus theory (developed by Gauss [5]), the order of \(C_K/C_K^2\) equals \(2^{s-1}\) where s is the number of primes which are ramified in K.

Theorem 2

Suppose the faithful irreducible character of \(Q_8\) can be realized over \(K={\mathbb Q}\bigl (\sqrt{-N}\bigr )\) but not over \(R={\mathfrak {O}}_K\). Then at least two rational primes must ramify in K, and just two are ramified precisely when \(N=2q\) or \(N=pq\) for primes \(p\equiv 5\,({\textrm{mod}\,}8)\) and \(q\equiv 7\,({\textrm{mod}\,}8)\).

A similar example is also given when \(N=pq\) for primes \(p\equiv 3\,({\textrm{mod}\,}8)\) and \(q\equiv 7\,({\textrm{mod}\,}8)\), but then 2, p, and q are ramified in \(K={\mathbb Q}\bigl (\sqrt{-pq}\bigr )\) (and \(C_K/C_K^2\) has order 4). In his third letter, Serre [13] asked for “more complicated examples”. Candidates could be obtained via globally irreducible representations as introduced by Gross [6]. For this reason, we shall discuss the arguments given by Serre in some detail.

2 Definite rational quaternion algebras.

Let \(D=\bigl ({-a,-b \over {\mathbb Q}}\bigr )\) be a definite quaternion algebra over \({\mathbb Q}\) (\(a>0\), \(b>0\)), and let \((D)=(-a, -b)\) be its class in \(Br_2({\mathbb Q})\). This class is determined by the local Hilbert symbols \((-a,-b)_p\) over the p-adic fields \({\mathbb Q}_p\) for all rational primes p (knowing that \((-a,-b)_\infty =-1\) over \({\mathbb R}={\mathbb Q}_\infty \)). Recall that D is said to be ramified at p (including \(p=\infty \)) provided the completion \({\mathbb Q}_p\otimes D\) is a skew field, the unique non-split quaternion algebra over \({\mathbb Q}_p\), just in which case \((-a,-b)_p=-1\). By Hilbert’s reciprocity law [11, Thm. III.3], the number of such primes is finite and odd. Let \(1, i, j, k=ij\) be the usual \({\mathbb Q}\)-basis of D (\(i^2=-a\), \(j^2=-b\), \(ji=-k\)).

Following Eichler [3] (and Serre [13]), we define the signed discriminant (in German: “Grundzahl”) \(d_D=d_{(-a,-b)}\)= −(product of all primes p where D is ramified at). Here the minus sign indicates that D is ramified at \(\infty \). Eichler’s definition has the advantage that the isomorphism type of D is determined by \(d_D\). For example, \(d_{(-1,-1)}=-2\), \(d_{(-1,-3)}=-3\), and there is no definite quaternion algebra D with \(d_D=-6\).

For basic properties of (definite) quaternion algebras, we refer to Lam [7]. In particular, there is a natural involutory anti-automorphism \({\alpha }\mapsto {{\bar{{\alpha }}}}\) on D, which induces on every (imaginary) quadratic subfield the nontrivial Galois automorphism over \({\mathbb Q}\), and we may define the (reduced) trace and norm. Also, the fact that D does not split (is a skew field) may be expressed by the property that the quadratic (nondegenerate) diagonal form [a, b, ab] resulting from the pure quaternions in D is not isotropic over \({\mathbb Q}\) (does not represent zero) [7, Thm. III.2.7].

Lemma 1

The field \(K={\mathbb Q}(\sqrt{-N})\) can be embedded into D, thus splitting D, if and only if [a, b, ab] represents N, that is, if \(N=ax^2+by^2+abz^2\) for rationals x, y, z. This happens precisely when \(-N\) is not a square in \({\mathbb Q}_p^*\) for all primes p dividing \(d_D\) or, equivalently, when \((D)=(-N,-n)\) in \(Br_2({\mathbb Q})\) for some positive integer n, which may be chosen such that N and n belong to different nontrivial cosets of \({\mathbb Q}^*\) modulo squares.

Proof

We can embed K into D if and only if there exist rationals x, y, z such that \(N=ax^2+by^2+cz^2\). In fact, then we assign to \(\sqrt{-N}\) the pure quaternion \(xi+yj+zk\) and observe that \({\mathbb Q}1 \oplus {\mathbb Q}(xi+yj+zk)\) is a (commutative) subfield of D (each nonzero element being invertible as its norm is positive). Of course, one can embed K in D in many distinct ways (all resulting subfields being conjugate by the Skolem–Noether theorem), its number being related to \(h_K\) (as already observed by Gauss).

The Davenport–Cassels lemma gives a sufficient condition for the statement that N can be represented by [a, b, ab] over \({\mathbb Z}\). In particular this holds when \((D)=(-1,-1)\) where \(N=x^2+y^2+z^2\) with integers x, y, z. By a classical result of Gauss [5], this is possible if and only if \(N\not \equiv 7\,({\textrm{mod}\,}8)\) (see [11, Appendix to Chap. IV]).

Obviously [a, b, ab] represents N over \({\mathbb R}\). By the Hasse–Minkowski theorem [11, Thm. IV.8], therefore [a, b, ab] represents N over \({\mathbb Q}\) if and only if it represents N over \({\mathbb Q}_p\) for each prime p. If p is a rational prime which does not divide \(d_D\), then \({\mathbb Q}_p\otimes D\cong M_2({\mathbb Q}_p)\) is split and so [a, b, ab] is isotropic over \({\mathbb Q}_p\) and so represents N. Let p be a divisor of \(d_D\) (so that \((-a,-b)_p=-1\)). The discriminant of [a, b, ab] is a square in \({\mathbb Q}_p^*\) and the form represents N over \({\mathbb Q}_p\) if and only if either \(-N\) is not a square in \({\mathbb Q}_p^*\), or \(-N\) is a square in \({\mathbb Q}_p^*\) and \((-1,-1)_p\) agrees with the Hasse invariant of the form (see the corollary to [11, Thm. IV.6]). We have to exclude the latter case. In that case, the quadratic forms [a, b, ab] and \([1,-1,-1]\) have the same dimension, same discriminant, and the same Hasse invariant over \({\mathbb Q}_p\), so are equivalent over \({\mathbb Q}_p\) by [11, Thm. IV.7]. But then [a, b, ab] is isotropic over \({\mathbb Q}_p\) (as it is \([1,-1,-1]\)), which in turn just happens when \((-a,-b)_p=1\) (condition (8) of [7, Thm. III.2.7]). This is the desired contradiction.

For each prime p dividing \(d_D\), there is \(n_p\in {\mathbb Q}_p^*\) such that \((-N, -n_p)_p=-1\), because \(-N\not \in {{\mathbb Q}_p^*}^2\) and the Hilbert symbol over \({\mathbb Q}_p\) is nondegenerate [11, Thm. III.2]. If p divides N but not \(d_D\), choose \(n_p\in {{\mathbb Q}_p^*}^2\), and for all other (infinitely many) primes p, we have \((-N, -n_p)_p=1\) for any choice of \(n_p\in {\mathbb Q}_p^*\). By [11, Thm. III.4] (and its proof), there is a positive integer n such that \((D)=(-N,-n)\), where n may be chosen as asserted (in view of Dirichlet’s prime number theorem and the fact that the primes plus \(-1\) give rise to a basis for the \({\mathbb F}_2\)-space \({\mathbb Q}^*/{{\mathbb Q}^*}^2\)). \(\square \)

3 Genus theory.

Let again D be a definite quaternion algebra over the rationals, with class \((-a,-b)\) in \(Br_2({\mathbb Q})\). Let \(K={\mathbb Q}(\sqrt{-N})\) be embedded in D, thus splitting D, and let \(R={\mathfrak {O}}_K\) be the ring of integers of K, its unique maximal \({\mathbb Z}\)-order. Suppose L is a full R-lattice in D (of R-rank 2). Since R is a Dedekind domain and L torsion-free, up to isomorphism L is determined by its rank and its Steinitz class \((L)\in C_K\) [2, Thm. 4.13]. In particular, L is a free R-module if and only if \((L)=1\). If \((L)=({\mathfrak {a}})^2\in C_K^2\) for some fractional ideal \(\mathfrak {a}\) of K, then \(({\mathfrak {a}}^{-1}L)=(\mathfrak {a})^{-2}(L)=1\) and so \({\mathfrak {a}}^{-1}L\) is a free R-module. Thus we are led to consider the quotient group \(C_K/C_K^2\), and to apply genus theory (see Zagier [17, Sect. 12] for a nice exposition à la Gauss). The principal genus theorem tells us that an integral ideal \({\mathfrak {a}}\) of K has class \((\mathfrak {a})\in C_K^2\) if and only if \((\textrm{N}{\mathfrak {a}},-N)=1\) in \(Br_2({\mathbb Q})\) (see also [4, Prop. 2.12]). Here \(\textrm{N}{\mathfrak {a}}=|R/{\mathfrak {a}}|\) is the absolute norm of \(\mathfrak {a}\). We obtain an injective homomorphism

$$\begin{aligned} {\nu }_K: C _K/C_K^2\rightarrow Br_2({\mathbb Q}) \end{aligned}$$

which is given by representing an element \(c\in C_K/C_K^2\) by an integral ideal \(\mathfrak {a}\) of K and defining \({\nu }_K(c)=(\textrm{N}{\mathfrak {a}},-N)\). The image of \({\nu }_K\) consists of all those classes of quaternion algebras which are ramified only at those (finite) primes ramified in K. Serre [13] has proved in his last letter to Feit (Propositions 1, 3) that the following holds.

Lemma 2

(Serre). Suppose \(R\subset K\subset D\) are as above. The maximal (\(\mathbb {Z}\)-)orders of D containing R, thus being R-modules in the obvious way, give rise to the same element \(c_K\in C_K/C_K^2\), and \({\nu }_K(c_K)=(D)\cdot (d_D, -N)\).

Now let G be a finite group admitting a faithful irreducible (complex) character \({\chi }\) which is rational-valued and has the Schur index \(m_\infty ({\chi })=2\) over \({\mathbb R}={\mathbb Q}_\infty \). Then the Schur index \(m({\chi })=2\) over the rationals by the Brauer–Speiser theorem. There is an irreducible \({\mathbb Q}G\)-module V affording \(2{\chi }\), and \(D=\textrm{End}_G(V)\) is a definite quaternion algebra (as above). The degree \(m={\chi }(1)\) of \({\chi }\) is even since it is a multiple of \(m({\chi })\). Let \(K={\mathbb Q}\bigl (\sqrt{-N}\bigr )\) be a splitting field of D (and of \({\chi }\)), and let \(R={\mathfrak {O}}_K\). Suppose L and M are full RG-lattices in V. If \(L={\mathfrak {a}}M\) for some fractional ideal \(\mathfrak {a}\) of K, then the Steinitz classes \((L)=({\mathfrak {a}})^m(M)\), whence (L) and (M) belong to the same genus. Moreover, then for any prime p, the p-adic completions \(L_p={\alpha }_pM_p\) where \({\mathfrak {a}}\subseteq {\alpha }_pR_{\mathfrak {p}}\cong {\mathbb Z}_p\otimes {\mathfrak {a}}\). Here \(R_p\cong {\mathbb Z}_p\otimes R\) is a discrete valuation ring with maximal ideal \({\pi }R_p\), say, and quotient field \(K_p\cong {\mathbb Q}_p\otimes K\). In particular, \(L_p\cong M_p\) as \(R_pG\)-lattices; if this is true for all primes p, then L and M are said to be in the “same genus” (for lattices) [2, Sect. 31]. This notion is weaker, however. In order to get (global) lattices in the same genus à la Gauss, we need to know that \(L_p={\alpha }_pM_p\) for each prime p and some \({\alpha }_p\in K_p^*\), because then \(L={\mathfrak {a}}M\) where

$$\begin{aligned} {\mathfrak {a}}=K\cap \prod _p{\alpha }_pR_p. \end{aligned}$$

Such an \(\alpha _p\in K_p^*\) exists when the reduction \(L_p/{\pi }L_p \) is an irreducible \([R_p/{\pi }R_p]G\)-module, that is, when \({\chi }\) remains irreducible as a Brauer character mod p (see [12, Ex. 15.3]).

Replacing the ring R by a maximal order \(R'\) in D, the representation of G on V is called globally irreducible provided the reductions \(L/{\mathfrak {p}}L\) are irreducible for all (two-sided) maximal ideals \(\mathfrak {p}\) of \(R'\). There are quite a lot such representations; the ones known so far are listed in [15]. In particular, Tiep [14] determined all globally irreducible representations associated to basic spin characters of \(2^{\pm }S_n\) and \(2A_n\) (where we use the Atlas notation: in \(2^-S_n\), transpositions are lifted to elements of order 4). The concept of (basic) spin characters has been introduced by Schur [10]. The basic spin character of \(2^+S_n\) and \(2^-S_n\) is rational-valued when n is odd, and that of \(2A_n\) when n is even.

It should be mentioned that there are also globally irreducible representations (G, V) where the corresponding \(\textrm{End}_G(V)\) is isomorphic to \({\mathbb Q}\) (treated by Thompson; see [12, Ex. 15.4]) or to an imaginary quadratic number field.

4 Proof of Theorem 1.

The faithful irreducible character \({\chi }\) of \(Q_8\) can be realized over \(K={\mathbb Q}(\sqrt{-N})\) if and only if K splits \(D=\bigl ({-1,-1\over {\mathbb Q}}\bigr )\). Here \(d_D=-2\) and \(-N \) is not a square in \({\mathbb Q}_2^*\), that is, 2 is either inert in K or ramified (Lemma 1). Let \(R={\mathfrak {O}}_K\).

On the basis of Witt’s theorem stated in the introduction, Lemma 1 also shows that \({\chi }\) can be realized over K if and only if \({\mathbb Q}\bigl (\sqrt{N} \bigr )\) can be embeded into a \(Q_8\)-field over the rationals because there is a positive (square-free) integer n such that \({\mathbb Q}\bigl (\sqrt{N}, \sqrt{n}\bigr )\) is a biquadratic number field and \((-1,-1)=(-N,-n)\) in \(Br_2({\mathbb Q})\).

View \(Q_8\) as a subgroup of \({{\mathcal {H}}}^*\), so that \({\mathcal {H}}\) gets an \(RQ_8\)-lattice in the obvious way (\({{\mathcal {H}}}\supset R\)). By Lemma 2, the image of the Steinitz class \(({\mathcal {H}})\) in \(C_K/C_K^2\) is independent of the choice of the maximal order, and there is an R-free \(RQ_8\)-lattice in the genus of \({\mathcal {H}}\) if and only if

$$\begin{aligned} (-2,N)=(-1,-1)\cdot (-2,-N)=(D)\cdot (d_D,-N)=1 \end{aligned}$$

in \(Br_2({\mathbb Q})\). Use that \((-1,-1)=(-2,-1)\) (as \((-1,-1)_2=(-2,-1)_2\)). Hence in this case, \(N=x^2+2y^2\) with integers x, y, and \((-1,-1)=(-2,-N)\). The field \({\mathbb Q}\bigl (\sqrt{2}, \sqrt{N}\bigr )\) is biquadratic since there is an odd prime divisor of N (as \(N>3\)) which ramifies in \({\mathbb Q}\bigl (\sqrt{N}\bigr )\) but not in \({\mathbb Q}\bigl (\sqrt{2}\bigr )\). Hence Witt’s theorem shows that \({\mathbb Q}\bigl (\sqrt{2}, \sqrt{N}\bigr )\) can be embedded into a \(Q_8\)-field whenever \((-1,-1)=(-2,-N)\).

The character \({\chi }\) remains irreducible mod p for all odd primes p but not for \(p=2\). Thus the Steinitz class of any full \(RQ_8\)-lattice in D is determined by \(({\mathcal {H}})\) up to multiplication with squares and powers of \((\mathfrak {q})\), where \(\mathfrak {q}\) is the prime ideal of R above 2. If 2 is inert in K, then \({\mathfrak {q}}=2R\) is principal, and each RG-lattice affording \({\chi }\) belongs to the genus of \({\mathcal {H}}\). Otherwise 2 is ramified in K and \(\textrm{N}{\mathfrak {q}}=2\), so that \({\nu }_K(({\mathfrak {q}})C_K^2)=(2,-N)\). If \((2,-N)=1\), we get the same statement as before. Otherwise there is an \(RQ_8\)-lattice in D which is free over R precisely when

$$\begin{aligned} (-1,-1)\cdot (-2,-N)\cdot (2,-N)=1 \end{aligned}$$

in \(Br_2({\mathbb Q})\), that is, when \((-1,-1)=(-1,-N)\). This just happens when \((-1,N)=1\) and so \(N=x^2+y^2\) with integers x, y. This completes the proof.

5 Proof of Theorem 2.

By hypothesis, the faithful irreducible character \({\chi }\) of \(Q_8\) can be realized over \(K={\mathbb Q}\bigl (\sqrt{-N}\bigr )\) but not over \(R={\mathfrak {O}}_K\). By genus theory (à la Gauss), then at least two primes must ramify in K. From the proof of Theorem 1, we know that N can be neither written as \(N=x^2+y^2\) nor as \(N=x^2+2y^2\) with integers x, y. Thus in the first case, N is divisible by a prime \(q\equiv 3\,({\textrm{mod}\,}4)\), and in the second case by a prime \(q\not \equiv 1,3\,({\textrm{mod}\,}8)\). This follows from classical results due to Fermat and Euler (describing the odd prime divisors of such sums with \(gcd(x,y)=1\)). It follows that \({\chi }\) cannot be integrally realized if N is divisible by a prime \(q\equiv 7\,({\textrm{mod}\,}8)\).

Suppose that N is even. Then 2 is ramified in K and in \({\mathbb Q}(\sqrt{N})\). Let q be an odd prime such that \({\mathbb Q}\bigl (\sqrt{2}, \sqrt{q}\bigr )\) cannot be embedded into a \(Q_8\)-field over the rationals. Then \(q\not \equiv 1, 3\,({\textrm{mod}\,}8)\) by the results in [9]. By Theorem 1 and our minimality assumption, \(N=2q\). If \(q\equiv 5\,({\textrm{mod}\,}8)\), then \(N=2q\) is a sum of two integer squares and so \({\chi }\) can be realized over R, against our hypothesis. Thus \(q\equiv 7\,({\textrm{mod}\,}8)\), as required.

So let N be odd in what follows. If N is a prime, then necessarily \(N\equiv 1\,({\textrm{mod}\,}4)\) because only then 2 ramifies in K. But then N is a sum of two integer squares, which is excluded. Therefore from Theorem 1 and the results in [9], we can infer that there are two odd primes \(p\ne q\) dividing N where not both are congruent to 1 mod 4. Moreover, the assumption on K ensures that \(N=pq\). Observe that \(N\equiv 1, 3, 5\,({\textrm{mod}\,}8)\).

Assume first that \(N\equiv 1\,({\textrm{mod}\,}8)\). Then either \(p\equiv q\equiv 3\,({\textrm{mod}\,}8)\) or \(p\equiv q\equiv 5\,({\textrm{mod}\,}8)\). In the first case, p and q can be written as \(x^2+2y^2\) with integers x, y (Fermat–Euler). But then \((-2,N)=(-2,p)\cdot (-2,q)=1\) in \(Br_2({\mathbb Q})\) and so N can be written in this manner too, which is excluded. If \(p\equiv q\equiv 5\,({\textrm{mod}\,}8)\), then p, q and hence \(N=pq\) can be written as sums of two integer squares, which again is excluded.

Assume next that \(N\equiv 5\,({\textrm{mod}\,}8)\). Since not both primes are congruent to 1 (mod 4), by appropriate notation, we get that \(p\equiv 3\,({\textrm{mod}\,}8)\) and \(q\equiv 7\,({\textrm{mod}\,}8)\). Thus \({\chi }\) cannot be realized over R. However, in this case 2, p, and q ramify in K.

Let finally \(N\equiv 3\,({\textrm{mod}\,}8)\). In this case, 2 is unramified in K, thus \({\chi }\) is not realizable over R if and only if \(N=pq\) cannot be written as \(x^2+2y^2\) with integers x, y. Hence we cannot have \(p\equiv 1\,({\textrm{mod}\,}8)\), \(q\equiv 3({\textrm{mod}\,}8)\), or vice versa (Fermat–Euler). Choosing notation appropriately, we therefore have \(p\equiv 5\,({\textrm{mod}\,}8)\) and \(q\equiv 7\,({\textrm{mod}\,}8)\), as required.

6 An example.

Let us call a finite group G quaternionic provided it admits a faithful irreducible character \({\chi }\) such that the corresponding simple component of the real group algebra \({\mathbb R}G\) is isomorphic to the Hamiltonian quaternion skew field \({\mathbb H}\). We shall see that then \(G=Q_8\), \(2A_4\), or \(2^-S_3\), and that \({\chi }\) is the basic spin character of these groups. (The faithful irreducible character of \(Q_8=2V_4\) is the restriction of the basic spin character of \(2A_4\).) By assumption, \({\chi }\) is real-valued of degree \({\chi }(1)=2\) and the Schur index \(m_\infty ({\chi })=2\). In other words, the Frobenius–Schur indicator \(\textrm{ind}({\chi })=-1\), that is,

$$\begin{aligned} \sum _{g\in G}{\chi }(g^2)=-|G| \end{aligned}$$

(e.g. see [12, Prop. 39]). Since \({\chi }(1)=2\) is a divisor of |G|, there are \(t\ge 1\) involutions in G. Let r be the number of elements \(g\in G\) for which \(g^2\) is an involution with \({\chi }(g^2)=-2\) (so \(g^2\) is scalar multiplication with \(-1\)), and let s be the number of elements \(g\in G\) for which \(g^2\) has order 3 (implying that \({\chi }(g^2)=-1\)). Using that \({\chi }\) is faithful and real-valued, one gets that \(\sum _g{\chi }(g^2)\ge 0\), the sum being taken over the remaining elements \(g\ne 1\) of G. Consequently

$$\begin{aligned} |G|\le 2r+s-2(1+t). \end{aligned}$$

Thus there are no such remaining elements. It follows that G is a \(\{2,3\}\)-group and that \({\chi }\) is rational-valued.

Since the Schur index \(m({\chi })=2\) (over \({\mathbb Q}\)) and \(m_\infty ({\chi })=2\), there is an irreducible \({\mathbb Q}G\)-module V affording \(2{\chi }\) and \(D=\textrm{End}_G(V)\) is a definite quaternion algebra over \({\mathbb Q}\). Indeed \((G,{\chi })\) gives rise to a globally irreducible representation (see Gross [6, Sect. 6] and Nebe [8, Thm. 6.1]). But this was already known to Eichler [3] (and Hey). Käte Hey showed in her unpublished dissertation (Advisor: Emil Artin): “Analytische Zahlentheorie in Systemen hyperkomplexer Zahlen”, Hamburg (1929) that for any maximal order M in D, either \(M^*/\{\pm 1\}\) has order 12 (\(d_D=-2\)) or order 6 (\(d_D=-3\)), and order 2 or 3 in all other cases. One obtains the (known) result that either \((D)=(-1,-1)\) and \(G=Q_8\) or \(G=2A_4\), or \((D)=(-1,-3)\) and \(G=2^-S_3\).

It is immediate that the basic spin characters of \(2A_4\) and \(2^-S_3\) remain irreducible as Brauer characters modulo every prime. Hence letting \(K={\mathbb Q}(\sqrt{-N})\) split D, application of Lemma 2 yields the following.

The basic spin character of \(2A_4\) can be realized over \({\mathfrak {O}}_K\) if and only if \(N=x^2+2y^2\) with integers x, y.

The basic spin character of \(2^-S_3\) can be realized over \({\mathfrak {O}}_K\) if and only if \(N=x^2+3y^2\) where either x, y are integers or both are in \({1\over 2}+{\mathbb Z}\) (Davenport–Cassels).