Orthogonal determinants of characters

For an irreducible orthogonal character χ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\chi $$\end{document} of even degree, there is a unique square class det(χ)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\det ({\chi })$$\end{document} in the character field such that the invariant quadratic forms in any L-representation affording χ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\chi $$\end{document} have determinant in det(χ)(L×)2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\det ({\chi })(L^{\times })^2$$\end{document}.


Introduction
Let G be a finite group.An absolutely irreducible complex representation ρ C : G → GL n (C) fixes a non-zero quadratic form if and only if ρ C is equivalent to a real representation ρ : G → GL n (R).Then B := g∈G ρ(g)ρ(g) tr is a positive definite symmetric matrix.As ρ(g)Bρ(g) tr = B the group ρ(G) is a subgroup of the orthogonal group of B. In this case we call χ : G → R, χ(g) := trace(ρ(g)) an orthogonal character.Let K = Q(χ(g) | g ∈ G) be the character field of χ.The main result of this paper is Theorem 3.3 showing that for even character degree there is a unique square class det(χ) ∈ K × /(K × ) 2 such that for any representation ρ with character χ over some extension field L of K all non-zero ρ(G)-invariant symmetric bilinear forms B have det(B) ∈ det(χ)(L × ) 2 .The square class det(χ) is called the orthogonal determinant of χ.The proof is immediate when the Schur index of χ is one.In this case there is a representation ρ for L = K and det(χ) = det(B)(K × ) 2 for any non-zero ρ(G)-invariant form B. If the Schur index of χ is two, there is no such representation over K. Then the rational span of the matrices in ρ(G) is a central simple K-algebra A and by Remark 3.1 the adjoint involution induces an involution on A whose determinant (as defined in [5, Proposition (7.1)]) is det(χ).
In positive characteristic all Schur indices are one and the result of Theorem 3.3 holds with a direct easy proof.Therefore we restrict to characters over number fields in this short note.
In an ongoing project with Richard Parker, we aim to provide the orthogonal determinant for all irreducible orthogonal Brauer characters for all but the largest few ATLAS groups [3] over all finite fields.This is a finite (computational) problem for the primes that divide the group order.Thanks to Theorem 3.3 the infinitely many primes not dividing the group order can be treated with a characteristic zero approach as illustrated in Corollary 4.2.
Acknowledgements.I thank Richard Parker for his persisting questions and fruitful discussions that made me write this paper and Eva Bayer-Fluckiger for her interesting comments leading to Section 2.2.

Determinants of symmetric bilinear forms
Let K be a field of characteristic = 2, V a vector space of dimension n over K and B : V × V → K a symmetric bilinear form.Any choice of a basis (e 1 , . . ., e n ) of V identifies V with the row space K n .The Gram matrix of B with respect to this basis is B := ( B(e i , e j )) n i,j=1 ∈ K n×n a symmetric square matrix satisfying B(x, y) = xBy tr for all x, y ∈ K n .Base change by the matrix T ∈ GL n (K) changes the Gram matrices into T BT tr and hence the determinant of B is well defined up to squares.The bilinear form B is called non-degenerate, if det( B) ∈ K × /(K × ) 2 , i.e. det(B) = 0.

The adjoint involution
Any non-degenerate symmetric bilinear form B on V defines a K-linear involution ι B on End K (V ).For α ∈ End K (V ) the endomorphism ι B (α) is defined by B(α(x), y) = B(x, ι B (α)(y)) for all x, y ∈ V.
Identifying End K (V ) with the matrix ring K n×n by choosing a basis (e 1 , . . ., e n ) of V , the involution ι B is given by We define space of skew adjoint endomorphisms.In matrix notation we get Scaling of the bilinear form does not change the involution, E − (aB) = E − (B) for all a ∈ K × .On the other hand det(aB) = a n det(B).So we can only read off the determinant of B from the involution ι B in even dimensions.The following property of skew adjoint endomorphisms is crucial.

Proposition 2.2. E − ( B) contains invertible elements if and only
Proof.We prove the theorem in matrix notation.Let E − (I) := {X ∈ K n×n | X = −X tr } denote the space of skew symmetric matrices.It is well known that E − (I) contains an invertible matrix, if and only if n is even and then the determinant of such a matrix is a square.By Lemma 2.1 the map E − (I) → E − (B), X → BX is an isomorphism.So E − (B) contains invertible elements if and only if dim(V ) is even, and all such elements

Determinants and isometries
For any non-degenerate symmetric bilinear form B its orthogonal group is ) and denote by P the characteristic polynomial of g.Assume that P (1)P (−1) = 0.
Proof.The minimal polynomial of g divides X 2 + 1 and hence the characteristic polynomial of g is P = (X − i) a (X + i) b with P (1)P (−1) = (−2) a+b .By Proposition 2.4 (a) the degree of P is a + b = dim(V ) is even so P (1)P (−1) is a square and the statement follows from Proposition 2.4 (c).

Orthogonal representations of finite groups
Let G be a finite group and L be a field.An L-representation ρ is a group homomorphism ρ : G → GL n (L).Given a representation ρ we put

Proof of Theorem 3.3
For the proof of Theorem 3.3 we assume that we are given a field L containing the character field K of χ and a representation ρ : G → GL n (L) with character χ.We also choose some non-zero B ∈ F (ρ). Since ρ is absolutely irreducible the matrices in ρ(G) generate L n×n as a vector space over L. Also F (ρ) is one dimensional and B is non-degenerate and unique up to scalars: We now consider the Q-algebra generated by the matrices in ρ(G), It is well known that the reduced norm of a central simple algebra takes values in the center of this algebra: Lemma 3.7.For all X ∈ A we have that det(X) ∈ K.
Proof.As L is a splitting field for A the determinant is the reduced norm of the central simple K-algebra A, see for instance [6,Section 9].Reiner also shows that the reduced norm is independent of the choice of a splitting field and takes values in K.
By [5,Corollary 2.8] a central simple algebra with orthogonal involution contains invertible elements that are negated by the involution if and only if the dimension of this algebra over its center is even.In particular for our situation this yields the following proposition for which we give an independent short proof below.Proposition 3.8.E − (ρ) contains invertible elements.
Proof.The fact that E − (ρ) contains an element that is invertible in the central simple K-algebra A = ρ(g) | g ∈ G Q does not depend on the choice of the splitting field L. So without loss of generality we fix an embedding ǫ : K ֒→ R, identify K with its image ǫ(K) ⊆ R, and take L = R, one of the real completions of K.
We first choose a Now Q and hence also ǫ(K) is dense in R.So there are a i ∈ K such that ǫ(a i ) is arbitrary close to α i for all i = 1, . . ., m.Put For Y being a unit in A, it is enough to achieve that the determinant of ǫ(Y ) := m i=1 ǫ(a i )b i is non zero.As det is a polynomial, in particularly continuous, and det(X) = 0, we can find Proof.(of Theorem 3.3) By Proposition 2.2 we get det(B)(L × ) 2 = det(X)(L × ) 2 for any invertible X ∈ E − (B).Proposition 3.8 says that such an invertible element X can be chosen in E − (ρ) = E − (B) ∩ A, so in particular its determinant is an element of K by Lemma 3.7.

Some applications
4.1.An example: SL 2 (F 7 ) For illustration let G := SL 2 (F 7 ) be the special linear group of degree 2 over the field with 7 elements.The complex character table of G is given in [3].For any faithful irreducible representation ρ of G we obtain ρ( − id and hence by Corollary 2.5 all faithful irreducible orthogonal characters have determinant 1.There are six complex irreducible characters of the group L 2 (7) = PSL 2 (7), of degrees 1, 3, 3, 6, 7, and 8 giving rise to three irreducible rational representations of even degree, 3ab, 6, and 8: 3ab Restrict the representation to a Sylow-7-subgroup g of G.The eigenvalues of ρ(g) are all primitive 7th roots of unity and hence det(ρ(g) − ρ(g 6 Restriction to g as before allows to conclude that det(χ) = 7(Q × ) 2 .8 Let H = C 7 : C 3 = g : h be the normaliser in G of the Sylow-7subgroup.Then the restriction of ρ to H decomposes as 6 + 2, where g acts fixed point free on the 6-dimensional part and trivially on the 2dimensional summand.On the 2-dimensional summand, h acts faithfully.If e = 1 7 6 i=0 g i ∈ QH is the involution invariant idempotent projecting onto the fixed space of g , then Remark 4.1.Note that these techniques essentially suffice to find all orthogonal determinants for all groups SL 2 (q) as given in [2].
4.2.Orthogonal characters with rational Schur index 2 Theorem 3.3 is particularly helpful in the case that the orthogonal character is not the character of a representation over its character field.The smallest example of a simple group G in [3] is the group G = J 2 .This sporadic simple group has a complex irreducible orthogonal character χ of degree χ(1) = 336 with rational character field.By [4] (see also [7]) the rational Schur index of χ is 2. With MAGMA [8] we realise the representation as a rational representation ρ of dimension 2 • 336 = 672 (with character 2χ).Then the central simple Q-algebra is isomorphic to a matrix ring over the indefinite rational quaternion algebra Q 2,3 ramified at 2 and 3. We take three random elements g 1 , g 2 , g 3 ∈ G to achieve that x := ) ∈ E − (ρ) has full rank.To compute the reduced norm of x ∈ A, we compute the characteristic polynomial P of x which is a square P = p 2 of a unique monic polynomial p. Then the reduced norm of x is p(0).It turns out that p(0) is a rational square, so det(χ) = 1.

Split extensions G : 2
Let G be a finite group, α ∈ Aut(G) an automorphism of order 2. Then the split extension G : 2 has a pseudo-presentation Assume that there is an orthogonal character χ ∈ Irr C (G) such that χ•α = χ.
Then there is a unique character X ∈ Irr C (G : 2) such that for some δ ∈ F and det(X ) = δ χ (1) .
Proof.Let L be some extension of K and ρ : G → GL n (L) a representation affording the character χ, so n = χ(1).Then the induced representation R with character X is given by R(g) = diag(ρ(g), ρ(α(g))) for all g ∈ G and R(h) = 0 1 1 0 .
In particular R is also an L-representation and This shows that det(X )(K × ) 2 = det(χ) 2 .In particular det(X ) = 1 if K = F .Example 4.4.Let X n ∈ Irr C (J 2 : 2) be the irreducible characters of degree 2n of the automorphism group of J 2 , for n = 14, 21, 70, 189, 224 (see [3]).In all cases the character field of X n is Q and the restriction of X n to the simple group J 2 is the sum of two irreducible orthogonal characters of degree n and with character field Q[ √ 5].As J 2 : 2 is a split extension Theorem 4.3 tells us that det(X n ) = 5 n (Q × ) 2 .