On the realization of the Gelfand character of a finite group as a twisted trace

Jorge Soto-Andrade; Maria-Francisca Yáñez-Valdés

doi:10.1515/jgth-2020-0207

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On the realization of the Gelfand character of a finite group as a twisted trace

Published/Copyright: October 28, 2021

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From the journal Journal of Group Theory Volume 25 Issue 3

Abstract

We show that the Gelfand character χ G of a finite group 𝐺 (i.e. the sum of all irreducible complex characters of 𝐺) may be realized as a “twisted trace” g ↦ Tr ⁡ ( ρ g ∘ T ) for a suitable involutive linear automorphism 𝑇 of L 2 ⁢ ( G ) , where ( L 2 ⁢ ( G ) , ρ ) is the right regular representation of 𝐺. Moreover, we prove that, under certain hypotheses, we have T ⁢ ( f ) = f ∘ L ( f ∈ L 2 ⁢ ( G ) ), where 𝐿 is an involutive anti-automorphism of 𝐺. The natural representation 𝜏 of 𝐺 associated to the natural 𝐿-conjugacy action of 𝐺 in the fixed point set Fix G ⁡ ( L ) of 𝐿 turns out to be a Gelfand model for 𝐺 in some cases. We show that ( L 2 ⁢ ( Fix G ⁡ ( L ) ) , τ ) fails to be a Gelfand model if 𝐺 admits non-trivial central involutions.

1 Introduction

In the seventies, Gelfand and collaborators [3] introduced the notion of a model for the representations of a compact Lie group 𝐺, as a representation of 𝐺 where each irreducible representation of 𝐺 appears exactly once. Nowadays, these models are usually called Gelfand models.

A paradigmatic example of such a Gelfand model is the natural representation of the compact Lie group G = SO ⁢ ( 3 , R ) in the Hilbert space H = L 2 ⁢ ( S 2 ) , where S 2 denotes the unit sphere in R 3 . The spherical functions associated to the (Hilbert sum) decomposition of this representation into irreducible mutually orthogonal subspaces are given by the well-known Legendre polynomials P ℓ ( ℓ ∈ N ). More precisely, the orthogonal projectors onto the irreducible components H ℓ of 𝐻 are integral operators with kernels K ℓ ⁢ ( x , y ) = P ℓ ⁢ ( x ⋅ y ) ( x , y ∈ S 2 ).

This suggests that remarkable special functions will appear as spherical functions providing the decomposition into irreducibles of Gelfand models for particular groups.

The notion of Gelfand model makes sense, of course, for any finite group 𝐺, in which case a Gelfand model is a representation of 𝐺 isomorphic to the direct sum of all (complex) irreducible representations of 𝐺.

It is however very rarely the case (as it is for SO ⁢ ( 3 , R ) ) that a finite non-commutative group admits a Gelfand model which is just a natural representation associated to some (necessarily transitive) 𝐺-space 𝑋. In that case, we say that 𝐺 admits a natural Gelfand model.

Nevertheless, as foreseen by Gelfand, these representation often have remarkable properties. Indeed, this the case for example for G = PGL ⁢ ( 2 , F q ) , for which the main constituent of a Gelfand model is the natural representation of 𝐺 in L 2 ⁢ ( H ~ ) associated to the homographic action of 𝐺 on H ~ = F q 2 - F q , the double cover of the finite analogue of Poincaré’s upper half plane over the finite base field F q (see [12, 13]). The spherical functions providing the decomposition of L 2 ⁢ ( H ~ ) into irreducibles have turned out to play a crucial role in the construction of regular Ramanujan graphs associated to H ~ (see [9, 11]).

Note that we keep the L 2 notation in the finite case so that L 2 ⁢ ( X ) for a finite set 𝑋 denotes the complex vector space of all complex functions on 𝑋, endowed with the usual L 2 product.

In this paper, we consider the character of a Gelfand model of a finite group 𝐺, which we call the Gelfand character of 𝐺 and denote by χ G . This character, which is just the sum of all irreducible characters of 𝐺, has indeed remarkable properties.

It looks very much like a (transitive) permutation character (the character of a natural, or permutation, representation associated to a transitive 𝐺-space): its mean value is 1, and it is integer-valued. Moreover, its values are very often non-negative. In fact, this is the case for the ordinary finite classical groups, and we conjectured it to be true for every finite group until Yokonuma [17] pointed out to us that the Gelfand character of the exceptional Mathieu group M 11 took negative values at some conjugacy classes. Later, Behn [2] found a minimal counter-example for this conjecture, to wit, a group G ⁢ ( 96 ) of order 96 whose Gelfand character takes values [ 30 , 2 , - 2 , 6 , 6 , 2 , 2 , 2 ] on its 8 conjugacy classes [2].

Note that the alternating group A 4 admits a natural Gelfand model M = L 2 ⁢ ( X ) , where the 𝐺-set 𝑋 stands for the set of the six oriented edges of the regular tetrahedron, endowed with the natural action of A 4 . An interesting open question in this regard is to characterize all finite groups admitting a natural Gelfand model.

We remark that a first obstruction for this to happen is that the dimension d G of the Gelfand model of 𝐺 must divide the order of 𝐺, which is rarely the case. However, remarkably enough, for G = SL ⁢ ( 2 , F q ) we have that the dimension of its Gelfand model is q ⁢ ( q + 1 ) , which divides the order ( q - 1 ) ⁢ q ⁢ ( q + 1 ) of 𝐺, although, as we show below, there is no natural Gelfand model for 𝐺.

The realization of the Gelfand character χ G of a finite group 𝐺 is an old problem on its own [4, 12, 14, 16]. One possible approach is to try to obtain χ G by twisting the trace of some very natural representation ( V , π ) of 𝐺, like its regular representation, by a suitable linear automorphism 𝑇 of its underlying space 𝑉, so as to obtain χ G ⁢ ( g ) = Tr ⁡ ( π g ∘ T ) for all g ∈ G . Recall that twisted traces appear in many contexts in mathematics [1, 4, 7].

A classical result in the character theory of finite groups [6] is that the central function θ 1 : G → C , defined by θ 1 ⁢ ( g ) = | { h ∈ G : h 2 = g } | is a generalized character that satisfies θ 1 ⁢ ( g ) = ∑ π ∈ G ^ ν ⁢ ( π ) ⁢ χ π ⁢ ( g ) , where

ν ⁢ ( π ) = 1 | G | ⁢ ∑ g ∈ G χ π ⁢ ( g 2 )

is the Frobenius–Schur number of the character χ π of the irreducible representation 𝜋 of 𝐺. If ( π , V ) is an irreducible complex representation of a finite group 𝐺 and 𝜋 is self-contragradient, then there exists a non-degenerate bilinear form 𝐵 on 𝑉, unique up to scalar multiple, such that

B ⁢ ( v , w ) = ϵ ⁢ ( π ) ⁢ B ⁢ ( w , v ) , where ⁢ ϵ ⁢ ( π ) = ± 1 .

Frobenius and Schur [5] proved that ν ⁢ ( π ) = ϵ ⁢ ( π ) , and if ν ⁢ ( π ) = 1 , then π ⁢ ( G ) is conjugate to a subgroup of the orthogonal group O ⁢ ( n ) , and if ν ⁢ ( π ) = - 1 , then π ⁢ ( G ) is conjugate to a subgroup of the symplectic group Sp ⁢ ( n ) and 𝑛 is even. Clearly, if ν ⁢ ( π ) = 1 for all irreducible representations of 𝐺, then θ 1 is the Gelfand character χ G of 𝐺. Some groups for which this is the case are S n , D 2 ⁢ n and O ⁢ ( n , q ) for all 𝑞 (see [6, 15]).

Furthermore, Gow proved in [6] that, for G = GL ⁢ ( n , q ) , the central function θ 2 : G → C , defined by θ 2 ⁢ ( g ) = | { h ∈ G : h t ⁢ h - 1 = g } | affords χ G .

On the other hand, Kawanaka and Matsuyama [10] defined the twisted Frobenius–Schur indicator ν τ ⁢ ( χ ) = 1 | G | ⁢ ∑ g ∈ G χ ⁢ ( g ⁢ τ ⁢ ( g ) ) , where 𝜏 is an involutive automorphism of 𝐺. They proved that if the character 𝜒 is afforded by a matrix representation 𝑅 of 𝐺 such that R ∘ τ = R ¯ , then ν τ ⁢ ( χ ) = 1 . If this is not the case, but nevertheless χ ∘ τ = χ ¯ , then ν τ ⁢ ( χ ) = - 1 . Finally, if χ ∘ τ ≠ χ ¯ , then ν τ ⁢ ( χ ) = 0 . Furthermore, they get that ∑ χ ∈ G ^ ν τ ⁢ ( χ ) ⁢ χ ⁢ ( g ) = | { h ∈ G : τ ⁢ ( h ) ⁢ h = g } | .

Later, Bump and Ginzburg [4] considered an automorphism 𝜏 of 𝐺 such that τ r = 1 , the norm map N : G → G given by N ⁢ ( g ) = g ⁢ τ ⁢ ( g ) ⁢ τ 2 ⁢ ( g ) ⁢ ⋯ ⁢ τ r - 1 ⁢ ( g ) and the number M ⁢ ( g ) of solutions of the equation N ⁢ ( x ) = g with x ∈ G . They proved that M ⁢ ( g ) lies in the Z ⁢ e 2 ⁢ π ⁢ i r -algebra generated by the irreducible characters and

M ⁢ ( g ) = ∑ χ ∈ G ^ ϵ τ ⁢ ( χ ) ⁢ χ ⁢ ( g ) ¯ , where ⁢ ϵ τ ⁢ ( χ ) = 1 | G | ⁢ ∑ g ∈ G χ ⁢ ( N ⁢ ( g ) ) .

If r = 2 and 𝜏 is trivial, then ϵ τ ⁢ ( χ ) = ν ⁢ ( χ ) and M ⁢ ( g ) = θ 1 ⁢ ( g ) ; if 𝜏 is not trivial, then ϵ τ ⁢ ( χ ) = ν τ ⁢ ( χ ) and M ⁢ ( g ) = ∑ χ ∈ G ^ ν τ ⁢ ( χ ) ⁢ χ ⁢ ( g ) . If G = GL ⁢ ( n , q ) and 𝜏 is the transpose inverse, then M ⁢ ( g ) = θ 2 ⁢ ( g ) .

We prove here, in Theorem 1, that the Gelfand character χ G of a finite group 𝐺 may always be realized as a twisted trace as χ G ⁢ ( g ) = Tr ⁡ ( ρ g ∘ T ) , g ∈ G , where 𝑇 is an involutive automorphism of L 2 ⁢ ( G ) and ( L 2 ⁢ ( G ) , ρ ) is the right regular representation. Moreover, in Theorem 2, we prove that if 𝐺 admits an involutive anti-automorphism 𝐿 such that L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G , then putting L * ⁢ ( f ) = f ∘ L , we get that the central function t ⁢ ( g ) = Tr ⁡ ( ρ g ∘ L * ) of g ∈ G is equal to the number of solutions h ∈ G to the equation h - 1 ⁢ L ⁢ ( h ) = g and is given by ∑ π ∈ G ^ ε π ⁢ χ π ⁢ ( g ) , where

ε π = ν L ⁢ ( χ π ) = 1 | G | ⁢ ∑ g ∈ G χ π ⁢ ( L ⁢ ( g ) ⁢ g - 1 ) = ± 1 .

Furthermore, in Theorem 3, we prove that if the number of fixed points of 𝐿 is equal to the sum of the dimensions of all irreducible representations of 𝐺, then under the above assumptions, the central function 𝑡 is the Gelfand character of 𝐺. So we can take the involutive automorphism 𝑇 of L 2 ⁢ ( G ) above to be L * . Note that, in particular, we recover in this way the classical results of Gow in [6].

Let 𝐺 be a group for which there exists an anti-automorphism 𝐿 satisfying all the conditions above. Note that 𝐺 acts then on the fixed point set

X = { h ∈ G : L ⁢ ( h ) = h }

of 𝐿 by 𝐿-conjugacy: g ⋅ h = g - 1 ⁢ h ⁢ L ⁢ ( g ) - 1 .

Let 𝜏 be the associated natural representation of 𝐺 in L 2 ⁢ ( X ) . Although L 2 ⁢ ( X ) has the right dimension to be a Gelfand Model of 𝐺, we prove, using our twisted trace description of χ G , that this cannot be the case if 𝐺 admits non-trivial central involutions.

2 The Gelfand character χ G as a twisted trace

Let 𝐺 be a finite group, and let ( L 2 ⁢ ( G ) , ρ ) and ( L 2 ⁢ ( G ) , σ ) be the right and left regular representation of 𝐺 respectively; let ( U k , π k ) ( 1 ≤ k ≤ r ) denote the irreducible unitary representations of 𝐺 (up to isomorphism), with n k = dim ⁡ U k , and let I k ( 1 ≤ k ≤ r ) be the isotypic component of type π k of 𝜌. We denote by χ k the character of π k .

Let ( e i ⁢ j k ⁢ ( g ) ) 1 ≤ i , j ≤ n k be the matrix of the operator π k ⁢ ( g ) ( g ∈ G ) with respect to a fixed orthonormal basis of U k . Then the matrix coefficients e i ⁢ j k ( 1 ≤ i , j ≤ n k , 1 ≤ k ≤ r ) provide an orthonormal basis ℬ for the Hilbert space L 2 ⁢ ( G ) . More precisely, the matrix coefficients e i ⁢ j k ( 1 ≤ i , j ≤ n k ) provide an orthonormal basis for the isotypical component I k for 1 ≤ k ≤ r , and for each 1 ≤ i ≤ n k , the coefficients e i ⁢ j k ( 1 ≤ j ≤ n k ) afford an isomorphic copy of the irreducible representation ( U k , π k ) of 𝐺 in L 2 ⁢ ( G ) .

The matrix coefficients e i ⁢ j k satisfy the relations

(2.1) e i ⁢ j k ⁢ ( g - 1 ) = e j ⁢ i k ⁢ ( g ) ¯ ,

(2.2) e i ⁢ j k ⁢ ( g ⁢ h ) = ∑ l = 1 n k e i ⁢ l k ⁢ ( g ) ⁢ e l ⁢ j k ⁢ ( h )

for all 1 ≤ i , j ≤ n k , 1 ≤ k ≤ r . Moreover, χ k denotes the character afforded by π k .

Proposition 1

Let ( V , π 1 ) and ( V , π 2 ) be two isomorphic representations of a finite group 𝐺 in the same vector space 𝑉 such that π h 1 ∘ π g 2 = π g 2 ∘ π h 1 for all g , h ∈ G , and let 𝑆 be an involutive automorphism of 𝑉 that intertwines the representations π 1 and π 2 of 𝐺. Then the complex function 𝜃 defined on 𝐺 by θ ⁢ ( g ) = Tr ⁡ ( π g 1 ∘ S ) = Tr ⁡ ( π g 2 ∘ S ) is central and so a linear complex combination of irreducible characters of 𝐺.

Proof

For g , h in 𝐺, we have

Tr ⁡ ( π g - 1 ⁢ h ⁢ g 1 ∘ S ) = Tr ⁡ ( π g 1 ∘ S ∘ π g - 1 ⁢ h 1 ) = Tr ⁡ ( S ∘ π g 2 ∘ π g - 1 ⁢ h 1 ) = Tr ⁡ ( S ∘ π g - 1 ⁢ h 1 ∘ π g 2 ) = Tr ⁡ ( π g 2 ∘ S ∘ π g - 1 ⁢ h 1 ) = Tr ⁡ ( S ∘ π g 1 ∘ π g - 1 ⁢ h 1 ) = Tr ⁡ ( S ∘ π h 1 ) = Tr ⁡ ( π h 1 ∘ S ) . ∎

Recall that all irreducible representations π k of a finite group 𝐺 are unitarizable. Moreover, the isotypic component I π k of type ( U k , π k ) of 𝜌 is isomorphic to U k ⊗ U k ∗ .

Theorem 1

Let 𝑇 be the linear application of L 2 ⁢ ( G ) defined by T ⁢ ( e i ⁢ j k ) = e j ⁢ i k for all e i ⁢ j k ∈ B , and let σ ~ be the homomorphism from 𝐺 to Aut ⁢ ( L 2 ⁢ ( G ) ) defined by σ ~ g = T ∘ ρ g ∘ T for all g ∈ G . Then 𝑇 is a linear involutive automorphism of L 2 ⁢ ( G ) and σ ~ is a representation of 𝐺 such that

ρ g ∘ T = T ∘ σ ~ g , g ∈ G ,
ρ g ∘ σ ~ h = σ ~ h ∘ ρ g , g , h ∈ G ,
Tr ⁡ ( ρ g ∘ T ) = χ G ⁢ ( g ) , g ∈ G .

Proof

Since 𝑇 is involutive, we get from σ ~ g = T ∘ ρ g ∘ T that σ ~ g is an automorphism of L 2 ⁢ ( G ) and that ρ g ∘ T = T ∘ σ g ~ for all g ∈ G . Furthermore, since

ρ g ⁢ ( e i ⁢ j k ) = ∑ l = 1 n k e l ⁢ j k ⁢ ( g ) ⁢ e i ⁢ l k and σ ~ g ⁢ ( e i ⁢ j k ) = ∑ l = 1 n k e l ⁢ i k ⁢ ( g ) ⁢ e l ⁢ j k

for all e i ⁢ j k ∈ B , g ∈ G , we get that

( ρ g ∘ σ ~ h ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 n k e l ⁢ i k ⁢ ( h ) ⁢ ∑ m = 1 n k e m ⁢ j k ⁢ ( g ) ⁢ e l ⁢ m k = ∑ m = 1 n k e m ⁢ j k ⁢ ( g ) ⁢ ∑ l = 1 n k e l ⁢ i k ⁢ ( h ) ⁢ e l ⁢ m k = ∑ m = 1 n k e m ⁢ j ⁢ ( g ) ⁢ σ ~ h ⁢ ( e i ⁢ m k ) = ( σ ~ h ∘ ρ g ) ⁢ ( e i ⁢ j k )

for g , h ∈ G and e i ⁢ j k ∈ B , which proves (ii). Finally, since

( ρ g ∘ T ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 n k e l ⁢ i k ⁢ ( g ) ⁢ e j ⁢ l k

for all g ∈ G and e i ⁢ j k ∈ B , then

Tr ⁡ ( ρ g ∘ T ) = ∑ k = 1 r ∑ i = 1 n k e i ⁢ i k ⁢ ( g ) = ∑ k = 1 r χ k ⁢ ( g ) = χ G ⁢ ( g ) ,

which proves (iii). ∎

Next we will prove that the central function t ⁢ ( g ) = Tr ⁡ ( ρ g ∘ T ) can under certain conditions be constructed from an involutive anti-automorphism 𝐿 of the group 𝐺.

Theorem 2

Let 𝐿 be an involutive anti-automorphism of 𝐺 such that L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G , and let L ∗ be the automorphism of L 2 ⁢ ( G ) defined by L ∗ ⁢ ( f ) = f ∘ L . Then, for all g ∈ G , we have

ρ g ∘ L ∗ = L ∗ ∘ σ L ⁢ ( g ) - 1 ,
Tr ( ρ g ∘ L ∗ ) = | { h ∈ G : h - 1 L ( h ) ) = g } | ,
Tr ⁡ ( ρ g ∘ L ∗ ) = Tr ⁡ ( ρ g - 1 ∘ L ∗ ) ,
Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r ε k ⁢ χ k ⁢ ( g ) , where ε k = ± 1 .

Proof

The proof of (i) is a straightforward calculation. We obtain (ii) by computing the trace of ρ g ∘ L ∗ with respect to the canonical basis { δ g : g ∈ G } , where δ g ⁢ ( h ) = δ g , h ( h ∈ G ).

Now let A g = { h ∈ G : h - 1 ⁢ L ⁢ ( h ) = g } . Taking inverses, we see that h ∈ A g if and only if L ⁢ ( h ) ∈ A g - 1 . Then (iii) follows since 𝐿 is bijective.

To prove (iv), note first that, because of (i), the representations ( L 2 ⁢ ( G ) , ρ ) and ( L 2 ⁢ ( G ) , σ ˇ ) , where σ ˇ ⁢ ( h ) = σ L ⁢ ( h ) - 1 ( h ∈ G ), together with the involutive automorphism L ∗ satisfy all the assumptions of Proposition 1, and so it follows that the complex function t ⁢ ( g ) = Tr ⁡ ( ρ g ∘ L ∗ ) is central and

Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r λ k ⁢ χ k ⁢ ( g )

for suitable complex numbers λ k .

The anti-automorphism 𝐿 induces an anti-automorphism L ~ on the complex group algebra C ⁢ [ G ] . Since χ k ⁢ ( L ⁢ ( g ) ) = χ k ⁢ ( g ) , g ∈ G , 1 ≤ k ≤ r , L ~ acts as the identity on the centre and therefore induces an anti-automorphism L ~ k on each simple component of C ⁢ [ G ] ≅ ⨁ 1 ≤ k ≤ r M ⁢ ( n k , C ) . Due to the Skolem–Noether theorem, L ~ k is conjugate to the transposition map, i.e. there exists b k ∈ GL ⁢ ( n k , C ) such that L ~ k ⁢ ( a ) = b k - 1 ⁢ a t ⁢ b k for all a ∈ M ⁢ ( n k , C ) . Furthermore, we have that a = L ~ ⁢ ( L ~ ⁢ ( a ) ) = b k - 1 ⁢ b k t ⁢ a ⁢ ( b k t ) - 1 ⁢ b k = b k - 1 ⁢ b k t ⁢ a ⁢ ( b k - 1 ⁢ b k t ) - 1 for all a ∈ M ⁢ ( n k , C ) , so b k - 1 ⁢ b k t belongs to the centre, and therefore, b k t = ε k ⁢ b k with ε k = ± 1 since ( ( b k ) t ) t = b k .

In this way, for each representation ( U k , π k ) of 𝐺, a symmetric or an antisymmetric form b k exists, with respect to which the linear operators π k ⁢ ( g ) and L ~ ⁢ ( π k ⁢ ( g ) ) are adjoint to each other. More precisely, if we consider the bilinear form ⟨ u , v ⟩ = v t ⁢ b k ⁢ u , defined by b k , then

(2.3) ⟨ L ~ ⁢ ( π k ⁢ ( g ) ) ⁢ ( u ) , v ⟩ = v t ⁢ ( b k ⁢ L ~ ⁢ ( π k ⁢ ( g ) ) ) ⁢ u = v t ⁢ ( π k ⁢ ( g ) ) t ⁢ b ⁢ u = ⟨ u , π k ⁢ ( g ) ⁢ ( v ) ⟩ .

Assume that, for some 𝑘 ( 1 ≤ k ≤ r ), we have ε k = 1 . We choose then an orthonormal basis U k + = { u i : 1 ≤ i ≤ n k } of U k , with respect to the symmetric form b k , and we denote by e i ⁢ j k ⁢ ( g ) = ⟨ π k ⁢ ( g ) ⁢ ( u j ) , u i ⟩ the matrix coefficients of π k ⁢ ( g ) with respect to this basis. Due to equation (2.2), we have the following relations between the matrix coefficients of L ~ ⁢ ( π k ⁢ ( g ) ) = π k ⁢ ( L ⁢ ( g ) ) and π k ⁢ ( g ) :

e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = ⟨ u j , π k ⁢ ( g ) ⁢ ( u i ) ⟩ = e j ⁢ i k ⁢ ( g ) .

Let E k = ⟨ e i ⁢ j k : 1 ≤ i , j ≤ n k ⟩ , and let Tr k ⁡ ( ρ g ∘ L ∗ ) denote the trace of the restriction of ρ g ∘ L ∗ to the subspace E k of L 2 ⁢ ( G ) . In order to compute Tr k ⁡ ( ρ g ∘ L ∗ ) , we note that

[ ( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) ] ⁢ ( h ) = e i ⁢ j k ⁢ ( L ⁢ ( h ⁢ g ) ) = e i ⁢ j k ⁢ ( L ⁢ ( g ) ⁢ L ⁢ ( h ) ) = ∑ l = 1 n k e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ e l ⁢ j k ⁢ ( L ⁢ ( h ) ) = ∑ l = 1 n k e l ⁢ i k ⁢ ( g ) ⁢ e j ⁢ l k ⁢ ( h ) .

Then

( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 n k e l ⁢ i k ⁢ ( g ) ⁢ e j ⁢ l k .

Since

Tr k ⁡ ( ρ g ∘ L ∗ ) = ∑ 1 ≤ i , j ≤ n k ⟨ ( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) , e i ⁢ j k ⟩ = ∑ 1 ≤ i , j ≤ n k ∑ l = 1 n k e l ⁢ i k ⁢ ( g ) ⁢ ⟨ e j ⁢ l k , e i ⁢ j k ⟩

and ⟨ e j ⁢ l k , e i ⁢ j k ⟩ = δ ( j , l ) , ( i , j ) , we get

Tr k ⁡ ( ρ g ∘ L ∗ ) = ∑ i = 1 n k e i ⁢ i k ⁢ ( g ) = χ k ⁢ ( g ) .

Let us suppose now that 𝑘 is such that ε k = - 1 . Let n k = 2 ⁢ m k ; then we can find a symplectic basis U k - = { u i , 1 ≤ i ≤ n k } of U k such that ⟨ u i , u i + m k ⟩ = 1 and ⟨ u i , u j + m k ⟩ = ⟨ u i + m k , u j + m k ⟩ = ⟨ u i , u j ⟩ = 0 , i ≠ j ( 1 ≤ i , j ≤ m k ).

Equation (2.1) gives us now the following relations for the matrix coefficients e i ⁢ j k ⁢ ( g ) of π k ⁢ ( g ) with respect to this basis:

⟨ π k ⁢ ( g ) ⁢ ( u i ) , u j ⟩ = { - e j + m k ⁢ i k ⁢ ( g ) , j ≤ m k , e j - m k ⁢ i k ⁢ ( g ) , j > m k .

Therefore,

⟨ π k ⁢ ( L ⁢ ( g ) ) ⁢ ( u i ) , u j ⟩ = { - ⟨ π k ⁢ ( g ) ⁢ ( u j ) , u i ⟩ = e i + m k ⁢ j k ⁢ ( g ) , i ≤ m k , - e i - m k ⁢ j k ⁢ ( g ) , i > m k .

Taking into account these relations and equation (2.2), we get the following relations between the matrix coefficients of L ~ ⁢ ( π k ⁢ ( g ) ) = π k ⁢ ( L ⁢ ( g ) ) and π k ⁢ ( g ) :

e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = e j + m k ⁢ i - m k k ⁢ ( g ) , j ≤ m k , i > m k , e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = - e j - m k ⁢ i - m k k ⁢ ( g ) , j > m k , i > m k , e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = - e j + m k ⁢ i + m k k ⁢ ( g ) , j ≤ m k , i ≤ m k , e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = e j - m k ⁢ i + m k k ⁢ ( g ) , j > m k , i ≤ m k .

So we get

for i , j ≤ m k ,
( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 m k - e l + m k ⁢ i + m k k ⁢ ( g ) ⁢ e j + m k ⁢ l + m k k + ∑ l = m k + 1 n k e l - m k ⁢ i + m k k ⁢ ( g ) ⁢ e j + m k ⁢ l - m k k ,
for i ≤ m k , j > m k ,
( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 m k - e l + m k ⁢ i + m k k ⁢ ( g ) ⁢ e j - m k ⁢ l + m k k + ∑ l = m k + 1 n k e l - m k ⁢ i + m k k ⁢ ( g ) ⁢ ( - e j - m k ⁢ l - m k k ) ,
for i > m k , j ≤ m k ,
( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 m k e l + m k ⁢ i - m k k ⁢ ( g ) ⁢ ( - e j + m k ⁢ l + m k k ) + ∑ l = m k + 1 n k - e l - m k ⁢ i - m k k ⁢ ( g ) ⁢ e j + m k ⁢ l - m k k ,
and for i > m k , j > m k ,
(2.4) ( ρ g ∘ L ∗ ) ⁢ ( e i ⁢ j k ) = ∑ l = 1 m k e l + m k ⁢ i - m k k ⁢ ( g ) ⁢ e j - m k ⁢ l + m k k + ∑ l = m k + 1 n k - e l - m k ⁢ i - m k k ⁢ ( g ) ⁢ ( - e j - m k ⁢ l - m k k ) .

Note the following.

If i , j ≤ m k , then e j + m k ⁢ l + m k k ≠ e i ⁢ j k and e j + m k ⁢ l - m k k ≠ e i ⁢ j k .
If i ≤ m k , j > m k , then e j - m k ⁢ l + m k k = e i ⁢ j k if and only if j = m k + i > m k and l + m k = j > m k ; moreover, we have e j - m k ⁢ l - m k k ≠ e i ⁢ j k .
If i > m k , j ≤ m k , then e j + m k ⁢ l - m k k = e i ⁢ j k if and only if j + m k = i > m k and l - m k = j < m k ; moreover, we have e j + m k ⁢ l + m k k ≠ e i ⁢ j k .
If i > m k , j > m k , then e j - m k ⁢ l + m k k ≠ e i ⁢ j k and e j + m k ⁢ l - m k k ≠ e i ⁢ j k .

Therefore,

Tr k ⁡ ( ρ g ∘ L ∗ ) = ∑ j = m k + 1 n k - e j ⁢ j k ⁢ ( g ) + ∑ j = 1 m k - e j ⁢ j k ⁢ ( g ) = - χ k ⁢ ( g ) .

Hence, taking into account both cases, we get

Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r Tr k ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r ε k ⁢ χ k ⁢ ( g ) ,

which proves (iv). ∎

Definition 1

For any mapping 𝐿 from 𝐺 to itself, we denote by Fix G ⁡ ( L ) the fixed point set { g ∈ G : L ⁢ ( g ) = g } of 𝐿.

Theorem 3

If 𝐿 is an involutive anti-automorphism of 𝐺 such that L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G and | Fix G ⁡ ( L ) | = d G , then Tr ⁡ ( ρ g ∘ L ∗ ) = χ G ⁢ ( g ) , g ∈ G , and L ∗ = T .

Proof

From Theorem 2 (iv), we have Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r ε k ⁢ χ k ⁢ ( g ) . If we evaluate Tr ⁡ ( ρ g ∘ L ∗ ) on 𝑒, we obtain

Tr ⁡ ( L ∗ ) = ∑ k = 1 r ε k ⁢ n k .

Furthermore, from Theorem 2 (iii),

Tr ⁡ ( ρ e ∘ L ∗ ) = | { h ∈ G : h - 1 ⁢ L ⁢ ( h ) = e } | = | { h ∈ G : L ⁢ ( h ) = h } | = | Fix G ⁡ ( L ) | ,

but Tr ⁡ ( ρ e ∘ L ∗ ) = Tr ⁡ ( L ∗ ) , and by hypothesis,

| Fix G ⁡ ( L ) | = d G = ∑ k = 1 r n k .

So ∑ k = 1 r ε k ⁢ n k = ∑ k = 1 r n k ; since n k > 0 , we conclude that ε k = 1 for all 𝑘, and Tr ⁡ ( ρ g ∘ L ∗ ) = χ G ⁢ ( g ) ( g ∈ G ). Then, using equation (2.3), it follows that

L ∗ ⁢ ( e i ⁢ j k ) ⁢ ( g ) = e i ⁢ j k ⁢ ( L ⁢ ( g ) ) = e j ⁢ i k ⁢ ( g ) ,

hence L ∗ = T . ∎

Proposition 2

If 𝐿 is an involutive anti-automorphism of 𝐺 such that L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G , then

Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ 1 ≤ k ≤ r ( 1 | G | ⁢ ∑ h ∈ G χ k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ) ⁢ χ k ⁢ ( g ) .

Proof

To prove equation (2.4), we compute the Fourier coefficients λ k , 1 ≤ k ≤ r , of the central function Tr ⁡ ( ρ g ∘ L ∗ ) = Tr ⁡ ( L ∗ ∘ σ L ⁢ ( g ) - 1 ) with respect to the basis

{ χ k = ∑ i = 1 n k e i ⁢ i k : 1 ≤ k ≤ r }

of the centre 𝑍 of C ⁢ [ G ] . First we observe that

( L ∗ ∘ σ L ⁢ ( g ) - 1 ) ⁢ ( e i ⁢ j k ) ⁢ ( h ) = e i ⁢ j k ⁢ ( L ⁢ ( g ) ⁢ L ⁢ ( h ) ) = ∑ l = 1 n k e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ ( e l ⁢ j k ∘ L ) ⁢ ( h )

and

⟨ ( L ∗ ∘ σ L ⁢ ( g ) - 1 ) ⁢ ( e i ⁢ j k ) , e i ⁢ j k ⟩ = 1 | G | ⁢ ∑ h ∈ G ∑ l = 1 n k e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ e i ⁢ l k ⁢ ( L ⁢ ( h ) ) ⁢ e j ⁢ i k ⁢ ( h - 1 ) .

Therefore,

Tr ⁡ ( L ∗ ∘ σ L ⁢ ( g ) - 1 ) = ∑ k = 1 r ∑ 1 ≤ i , j ≤ n k 1 | G | ⁢ ∑ h ∈ G ∑ l = 1 n k e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ e l ⁢ j k ⁢ ( L ⁢ ( h ) ) ⁢ e j ⁢ i k ⁢ ( h - 1 ) .

Since

∑ j = 1 n k e l ⁢ j k ⁢ ( L ⁢ ( h ) ) ⁢ e j ⁢ i k ⁢ ( h - 1 ) = e l ⁢ i k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ,

we get that

Tr ⁡ ( L ∗ ∘ σ L ⁢ ( g ) - 1 ) = ∑ k = 1 r ∑ 1 ≤ i , l ≤ n k 1 | G | ⁢ ∑ h ∈ G e l ⁢ i k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ⁢ ( e i ⁢ l k ∘ L ) ⁢ ( g ) ,

and then, for 1 ≤ k ′ ≤ r ,

λ k ′ = ⟨ ∑ k = 1 r ∑ 1 ≤ i , l ≤ n k 1 | G | ⁢ ∑ h ∈ G e l ⁢ i k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ⁢ ( e i ⁢ l k ∘ L ) , χ k ′ ⟩ .

By hypothesis, χ k ⁢ ( L ⁢ ( g ) ) = χ k ⁢ ( g ) , 1 ≤ k ≤ r , so

λ k ′ = ∑ k = 1 r ∑ 1 ≤ i , l ≤ n k 1 | G | ⁢ ∑ h ∈ G e l ⁢ i k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ⁢ ∑ s = 1 n k ′ 1 | G | ⁢ ∑ g ∈ G e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ e s ⁢ s k ′ ⁢ ( L ⁢ ( g ) ) ¯ ,

But 1 | G | ⁢ ∑ g ∈ G e i ⁢ l k ⁢ ( L ⁢ ( g ) ) ⁢ e s ⁢ s k ′ ⁢ ( L ⁢ ( g ) ) ¯ = ⟨ e i ⁢ l k , e s ⁢ s k ′ ⟩ , and then

λ k ′ = ∑ s = 1 n k ′ 1 | G | ⁢ ∑ h ∈ G e s ⁢ s k ′ ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) = 1 | G | ⁢ ∑ h ∈ G χ k ′ ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) . ∎

Proposition 3

Let 𝐿 be an involutive anti-automorphism of 𝐺 such that L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G . Then the following conditions are equivalent.

| Fix G ⁡ ( L ) | = d G .
1 | G | ⁢ ∑ h ∈ G χ k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) = 1 , where 1 ≤ k ≤ r .

Proof

The equivalence follows from Theorem 2 and Proposition 2. ∎

Proposition 4

Let 𝐿 an involutive anti-automorphism of 𝐺, τ : G → G defined by τ ⁢ ( g ) = L ⁢ ( g - 1 ) , and let c τ ⁢ ( χ ) = | G | - 1 ⁢ ∑ g ∈ G χ ⁢ ( g ⁢ τ ⁢ ( g ) ) be the twisted Frobenius–Schur indicator defined by Kawanaka and Matsuyama [10]. Then

c τ ⁢ ( χ k ) = 1 | G | ⁢ ∑ h ∈ G χ k ⁢ ( h ⁢ L ⁢ ( h - 1 ) ) = ± 1 if ⁢ χ k ⁢ ( L ⁢ ( g ) ) = χ k ⁢ ( g ) .

Moreover, if the matrix representation R k corresponding to χ k satisfies the equation R k ⁢ ( L ⁢ ( g ) ) = R k ⁢ ( g ) , then c τ ⁢ ( χ k ) = 1 .

Proof

Due to g τ ( g ) ) = g L ( g - 1 ) , we have

∑ g ∈ G χ ⁢ ( g ⁢ L ⁢ ( g - 1 ) ) = ∑ g ∈ G χ ⁢ ( L ⁢ ( g ) ⁢ g - 1 ) ¯ ,

and since Tr ⁡ ( ρ g ∘ L ∗ ) = Tr ⁡ ( ρ g ∘ L ∗ ) ¯ , we can write

Tr ⁡ ( ρ g - 1 ∘ L ∗ ) = ∑ k = 1 r 1 | G | ⁢ ∑ h ∈ G χ k ⁢ ( L ⁢ ( h ) ⁢ h - 1 ) ⁢ χ k ⁢ ( g - 1 ) ¯ ,

and therefore,

Tr ⁡ ( ρ g ∘ L ∗ ) = ∑ k = 1 r 1 | G | ⁢ ∑ h ∈ G χ k ⁢ ( h ⁢ L ⁢ ( h - 1 ) ) ⁢ χ k ⁢ ( g ) .

Our proposition follows. ∎

3 Applying the twisted trace to test natural Gelfand model candidates

Let 𝐺 be a group for which there exists an involutive anti-automorphism 𝐿 satisfying all the conditions of Theorem 3. Then we can consider the 𝐺-set X = Fix G ⁡ ( L ) , endowed with the 𝐿-conjugacy action of 𝐺 given by g ⋅ h = g - 1 ⁢ h ⁢ L ⁢ ( g ) - 1 . Let us call 𝜏 the associated natural representation of 𝐺 in L 2 ⁢ ( X ) .

Note that, since the dimension of L 2 ⁢ ( X ) is equal to the Gelfand dimension d G of 𝐺, the representation ( L 2 ⁢ ( X ) , τ ) could be a Gelfand model for 𝐺.

This is indeed the case for G = PSL ⁢ ( 2 , 3 ) (see [4]), that is in fact isomorphic to A 4 , for which group a natural Gelfand model is afforded by the six edges of the regular tetrahedron, as recalled in the introduction.

To check in general whether ( L 2 ⁢ ( X ) , τ ) provides a Gelfand model for 𝐺, we can compare its character χ τ with the Gelfand character χ G that we have described as a twisted trace: χ G ⁢ ( g ) = Tr ⁡ ( ρ g ∘ L * ) , g ∈ G .

Let 𝑍 denote the centre Z ⁢ ( G ) of 𝐺; we note that if z ∈ Z , then z ∈ X because L ⁢ ( z ) is conjugate to 𝑧 by hypothesis. Furthermore, since

χ τ ⁢ ( g ) = | { h ∈ X : g ⁢ h ⁢ L ⁢ ( g ) = h } | for all ⁢ g ∈ G ,

it follows that χ τ ⁢ ( z ) = | X | for all involutions z ∈ Z . So the character χ τ does not tell central involutions from the identity. We get then the following necessary condition.

Proposition 5

Let 𝐺 be a group for which there exists an involutive anti-automorphism 𝐿 satisfying all the conditions of Theorem 3. With the notation above, a necessary condition for ( L 2 ⁢ ( X ) , τ ) to be a Gelfand model for 𝐺 is that, for all central involutions 𝑧, we have χ G ⁢ ( z ) = | X | , or equivalently, in terms of 𝐿, | { g ∈ G : g - 1 ⁢ L ⁢ ( g ) = z } | = | X | .∎

3.1 The case of G = SL ⁢ ( 2 , F q ) , 𝑞 odd

As an illustrative example, we consider here the case of G = SL ⁢ ( 2 , F q ) , 𝑞 odd. Recall that the Gelfand dimension of 𝐺, i.e. the sum of the dimensions of all its irreducible representations, is q ⁢ ( q + 1 ) .

For k = ( 1 0 0 - 1 ) , we may define L : G → G , L ⁢ ( g ) = k ⁢ g - 1 ⁢ k . We note that if g = ( a b c d ) ∈ G , then L ⁢ ( g ) = ( d b c a ) , and

X = Fix G ⁡ ( L ) = { ( a b c a ) ∈ G : a 2 - b ⁢ c = 1 } .

Proposition 6

𝐿 is an involutive anti-automorphism of 𝐺 satisfying

L ⁢ ( g ) is conjugate with 𝑔 for all g ∈ G ,
| X | = q ⁢ ( q + 1 ) ,
the action of 𝐺 on 𝑋 is transitive.

Proof

(i) This is clear since L ⁢ ( g ) has the same trace and determinant as 𝑔 and it is a scalar matrix if and only if 𝑔 is for all g ∈ G .

(ii) For ( a b c a ) ∈ X , we must have b ⁢ c = a 2 - 1 . Now, if a 2 - 1 = 0 , i.e. a = ± 1 , then we have 2 ⁢ q - 1 pairs ( b , c ) such that b ⁢ c = a 2 - 1 . On the other hand, if a 2 - 1 ≠ 0 , then there are q - 1 pairs ( b , c ) such that b ⁢ c = a 2 - 1 . So

| X | = 2 ⋅ ( 2 ⁢ q - 1 ) + ( q - 2 ) ⋅ ( q - 1 ) = q ⁢ ( q + 1 ) .

(iii) We have that | Orb G ⁡ ( e ) | = | G | / | Stab G ⁡ ( e ) | and

Stab G ⁡ ( e ) = { g ∈ G : L ⁢ ( g ) = g - 1 } .

For g = ( a b c d ) ∈ G , this means b = c = 0 and d = a - 1 . Therefore,

| Stab G ⁡ ( e ) | = | { ( a 0 0 a - 1 ) : a ∈ F q * } | = q - 1 .

So | Orb G ⁡ ( e ) | = q ⁢ ( q 2 - 1 ) q - 1 = q 2 + q = | X | , and therefore, the action is transitive. ∎

In this way, we get that 𝐿 is an involutive anti-automorphism of G = SL ⁢ ( 2 , F q ) satisfying all the conditions in Theorem 1 above. Hence, for the central involution z = ( - 1 0 0 - 1 ) , we have χ G ⁢ ( z ) = | { h ∈ G : h - 1 ⁢ L ⁢ ( h ) = z } | . But if h = ( a b c d ) ∈ G satisfies h - 1 ⁢ L ⁢ ( h ) = z , then

( d b c a ) = ( - a - b - c - d ) ,

and that is equivalent to b = c = 0 and - a 2 = 1 .

So if q ≡ 3 ⁢ mod ⁢ ( 4 ) , then χ G ⁢ ( z ) = 0 and if q ≡ 1 ⁢ mod ⁢ ( 4 ) , then χ G ⁢ ( z ) = 2 . In both cases, χ G ⁢ ( z ) ≠ χ τ ⁢ ( z ) . Therefore, ( L 2 ⁢ ( X ) , τ ) is not a Gelfand Model for 𝐺.

Motivated by the previous example, we may state the following general result.

Lemma 1

For every central element z ∈ G of order 2, we have χ G ⁢ ( z ) ≠ χ G ⁢ ( e ) .

Proof

Indeed, note that χ π ⁢ ( z ) = ± dim ⁡ ( π ) for all irreducible representations 𝜋 of 𝐺, and so for χ G ⁢ ( z ) = χ G ⁢ ( e ) to hold we need that χ π ⁢ ( z ) = dim ⁡ ( π ) = χ π ⁢ ( e ) for all irreducible representations 𝜋, but this is absurd since the regular character χ ρ tells 𝑧 from 𝑒 because χ ρ ⁢ ( e ) = | G | ≠ 0 = χ ρ ⁢ ( z ) . ∎

Proposition 7

Let 𝐺 be a finite group with an involutive anti-automorphism 𝐿 of 𝐺 such that

L ⁢ ( g ) is conjugate to 𝑔 for all g ∈ G ,
| Fix G ⁡ ( L ) | = d G .

If 𝐺 has central (non-trivial) involutions, we conclude that the natural representation ( L 2 ⁢ ( X ) , τ ) associated to the 𝐺-space X = Fix G ⁡ ( L ) cannot be a Gelfand model for 𝐺.

Proof

This follows immediately from Lemma 1 and the necessity of the condition χ G ⁢ ( z ) = χ τ ⁢ ( z ) for all central involutions 𝑧. ∎

3.2 The case of the quaternion group 𝑄

The quaternion group 𝑄 is an illustrative example of a group admitting involutive anti-automorphisms 𝐿 satisfying all the conditions of Theorem 3, whose fixed point sets X = Fix Q ⁡ ( L ) , endowed with the natural 𝐿-conjugacy action of 𝑄, are non-transitive. Of course, these 𝐺-sets do not afford then a Gelfand model for 𝑄, a fact that also follows from the existence of non-trivial central involutions in 𝐺.

Recall that the quaternion group 𝑄 may be described as

Q = { 1 , - 1 , i , - i , j , - j , k , - k } ,

where g 2 = - 1 for all g ∈ Q , g ≠ ± 1 , and

i ⁢ j = k = - j ⁢ i , j ⁢ k = i = - k ⁢ j , k ⁢ i = j = - i ⁢ k .

Every element is only conjugate to its inverse, and the centre of 𝐺 is { 1 , - 1 } . The five conjugacy class of 𝐺 are { 1 } , { - 1 } , { ± i } , { ± j } , { ± k } . The quaternion group 𝑄 has four 1-dimensional irreducible representations and one 2-dimensional irreducible representation. So d Q = 6 . The possibilities for L ⁢ ( i ) and L ⁢ ( j ) for 𝐿 to be an involutive anti-automorphism of 𝐺 such that L ⁢ ( g ) is conjugate with 𝑔, for all g ∈ Q are L ⁢ ( i ) = ± i and L ⁢ ( j ) = ± j .

If we define L ⁢ ( i ) = i and L ⁢ ( j ) = j , then L ⁢ ( k ) = - k and
Fix Q ⁡ ( L ) = { ± 1 , ± i , ± j } .
If we define L ⁢ ( i ) = i and L ⁢ ( j ) = - j , then L ⁢ ( k ) = k and
Fix G ⁡ ( L ) = { ± 1 , ± i , ± k } .
Analogously, if we define L ⁢ ( i ) = - i and L ⁢ ( j ) = j , then L ⁢ ( k ) = k and
Fix Q ⁡ ( L ) = { ± 1 , ± j , ± k } .

So, in all these cases, we get | Fix Q ⁡ ( L ) | = 6 = d Q .

Note that if we define L ⁢ ( i ) = - i and L ⁢ ( j ) = - j , then L ⁢ ( k ) = - k and

Fix Q ⁡ ( L ) = { ± 1 } .

Note that each of the first three anti-automorphisms defined above clearly satisfies all the conditions of Theorem 3. We have then, for the Gelfand character of 𝑄,

χ Q ⁢ ( g ) = | { h ∈ Q : h - 1 ⁢ L ⁢ ( h ) = g } | .

It follows that, for the non-trivial central involution −1, we have, for the first such 𝐿 (the other two cases being obtained by cyclic permutation of i , j , k ),

χ Q ⁢ ( - 1 ) = | { h ∈ G : L ⁢ ( h ) = - h } | = | { ± k } | = 2 ≠ 6 = d Q .

So χ Q is different from the character χ τ of 𝑄 associated to its natural action on Fix Q ⁡ ( L ) , given by g . x = g - 1 ⁢ x ⁢ L ⁢ ( g ) - 1 .

We remark that the action of 𝑄 on X = Fix Q ⁡ ( L ) is not transitive. There are in fact 3 orbits of 𝑄 in 𝑋, to wit,

Orb Q ⁡ ( 1 ) = { ± 1 } , Orb Q ⁡ ( i ) = { ± i } , Orb Q ⁡ ( j ) = { ± j } .

The question remains open though as to whether if a group 𝐺 with trivial centre admits an involutive anti-automorphism 𝐿 satisfying the conditions of Theorem 3 and such that the 𝐺-set X = Fix G ⁡ ( L ) is transitive, then 𝑋 affords a natural Gelfand model for 𝐺.

Funding source: Fondo Nacional de Desarrollo Científico y Tecnológico

Award Identifier / Grant number: 1140510

Funding statement: The authors have been partially supported by Fondecyt Grant 1140510.

Acknowledgements

The authors thank Pierre Cartier for helpful discussions related to this work.

Communicated by: Timothy C. Burness

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Received: 2020-12-31

Revised: 2021-08-01

Published Online: 2021-10-28

Published in Print: 2022-05-01

Articles in the same Issue

https://doi.org/10.1515/jgth-2020-0207