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TOTAL CHARACTERS AND CHEBYSHEV POLYNOMIALS
EIRINI POIMENIDOU and HOMER WOLFE
Received 8 January 2002
The total characterτof a finite groupGis defined as the sum of all the irreducible characters ofG. K. W. Johnson asks when it is possible to expressτas a polyno-mial with integer coefficients in a single irreducible character. In this paper, we give a complete answer to Johnson’s question for all finite dihedral groups. In particular, we show that, when such a polynomial exists, it is unique and it is the sum of certain Chebyshev polynomials of the first kind in any faithful irreducible character of the dihedral groupG.
2000 Mathematics Subject Classification: 20C15.
1. Introduction. AGel’fand modelMfor a groupGwas defined in [8] as any complex representation ofGwhich is isomorphic to the direct sum of all irre-ducible representations ofG. We refer to the character of such a representation as thetotal characterτofG. A Gel’fand model afforded by a (generalized) per-mutation representation is referred to as a (weakly) geometric Gel’fand model. When a Gel’fand modelMis the representation afforded by a nonnegative inte-ger linear combination of powers of a genuineG-setX, then the total character τ ofGcan be expressed as a polynomial in the characterχafforded byX. A question related to this idea was posed by Johnson [5].
Question1.1. For a finite groupG, do there necessarily exist an irreducible
characterχand a monic polynomialf (x)∈Z[x]such thatf (χ)=τ, whereτ is the total character ofG?
Johnson’s question arose in the context of character sharpness which we will briefly explain. LetGbe a finite group,χa generalized character ofGof degree n,L= {χ(g)|g=1}, andfL(x)=l∈L(x−l). It was discovered by Blichfeldt
[2] and rediscovered in a modern context by Kiyota [6] thatfL(x)∈Z[x]and
|G|dividesfL(n). In the special case thatfL(n)= |G|, the characterχis said
to besharp. Another way to characterize a sharp character is to notice that the class functionfL(χ)=l∈L(χ−l1G)=ρ, whereρis the regular character and
1Gis the trivial character ofG. In other words, every irreducible character of
Gappears as a constituent offL(χ).
We prove the following main theorem.
Theorem1.2. LetGD2nand letτbe the total character ofG.
(1) If n is odd or ifn=2m, where m≡2,3(mod 4), then there exists a unique monic polynomialP(x)∈Z[x]such thatP(χ)=τfor any faithful
χ∈Irr(G). Moreover,
P(x)= 2 m
k=0 Tk
x
2
−2Tm−1
x
2
, ifn=2m−1, n≡1,3(mod 8),
2
m−1
k=0 Tk
x
2
, ifn=2m−1, n≡5,7(mod 8),
2
m
k=0 Tk
x
2
, ifn=2m, m≡2,3(mod 4),
(1.1)
whereTk(x)is thekth Chebyshev polynomial of the first kind.
(2) Ifn=2mandm≡1(mod 4), then there exists a unique monic polyno-mialP(x)∈Z[x]such thatP(χ)=2τfor any faithfulχ∈Irr(G), where
P(x)=2
Tm+1
x 2 +2 m
k=0 Tk
x
2
−Tm−1
x
2
. (1.2)
Moreover, there does not existP(x)∈Z[x]such thatP(χ)=τ for any
χ∈Irr(G).
(3) IfGD2n, wheren≡0(mod 8), then there does not exist a polynomial P(x)∈C[x]such thatP(χ)=τfor anyχ∈Irr(G).
2. Preliminaries. The total character for all dihedral groups was computed in [7, Proposition 2.1]. A dihedral group of order 2nwithn≥3 will be pre-sented as usual as
D2n=G=a,b:an=b2=1, b−1ab=a−1. (2.1)
Using the notation in [4], we use gi and hi to denote a representative and
the size of theith conjugacy class, respectively. The character table and total characterτofD2nare given below.
Case1(nodd). The conjugacy classes ofD2n(nodd) are
{1}, ar,a−r 1≤r≤n−1
2
, asb|0≤s≤n−1. (2.2)
The character table ofD2n(nodd) and the total characterτ, where=e2πi/n,
Table2.1
gi 1 ar1≤r≤(n−1)/2 b
hi 1 2 n
χ1 1 1 1
χ2 1 1 −1
ψj1≤j≤(n−1)/2 2 jr+−jr 0
τ n+1 1 0
Case2(neven). Ifnis even, sayn=2m, then the conjugacy classes ofD2n
are
{1}, am, ar,a−r (1≤r≤m−1),
asb
|seven, asb|sodd. (2.3)
The character table ofD2n (neven,n=2m, and the total characterτ, where
=e2πi/n) is given inTable 2.2.
Table2.2
gi 1 am ar(1≤r≤m−1) b ab
hi 1 1 2 n/2 n/2
χ1 1 1 1 1 1
χ2 1 1 1 −1 −1
χ3 1 (−1)m (−1)r 1 −1
χ3 1 (−1)m (−1)r −1 1
ψj(1≤j≤m−1) 2 2(−1)j jr+−jr 0 0
τ 2(m+1) 0, modd, 0, rodd, 0 0
2, meven, 2, reven,
The nth Chebyshev polynomial of the first kind is defined as Tn(x) =
cos(ncos−1(x))for|x| ≤1. Chebychev polynomials can be expressed recur-sively asTn+2(x)=2xTn+1(x)−Tn(x), T0(x)=1, andT1(x)=x. Before we proceed with the proof ofTheorem 1.2we need the following lemmas.
Lemma2.1. IfP(ψ1)=τ for some P(x)∈Z[x], then P(ψj)=τ for any faithful characterψjofD2n.
Proof. We first observe thatψj is faithful if and only if (j,n)=1, for
Hence for(j,n)=1, the character ψj is faithful and a Galois conjugate of
ψ1with the same set of character values. Since fornoddP(ψ1[ar])=1 for all r, it follows that P(ψj[ar])=P(ψ1[ar])=1. When nis even, we have thatP(ψ1[ar])=2 whenr is even andP(ψ1[ar])=0 whenr is odd. For a faithfulψj, the set of character values is the same as those ofψ1; and sincejis necessarily odd, it follows thatP(ψj[ar])=2 whenris even, andP(ψj[ar])=
0 whenr is odd as required.
Lemma2.2. Let thenth Chebyshev polynomial of the first kind be expressed
asTn(x)=nk=1cn,kxk+cn,0. Thencn,n=2(n−1)and2(k−1)|cn,kfor1≤k≤n.
Proof. The first half follows easily by an inductive argument onn and
by using the recursive relation mentioned above. For divisibility of the coeffi-cients, observe that the result is true forn=1 andn=2. Now assume that the result is true for all Chebyshev polynomials of degree less thatnand consider thenth degree Chebyshev polynomialTn(x). We have by the recursive relation
thatcn,k=2cn−1,k−1−cn−2,k. By inductive hypothesis, 2(k−1)|2cn−1,k−1−cn−2,k
for any 1≤k≤n. Hence the result is true for all the coefficients ofTn(x)and
hence for alln.
Lemma2.3[3, 134.2]. Form∈Z+, m
k=0
coskx=12
1+sin
(m+1/2)x sin(1/2)x
. (2.4)
Lemma2.4. Letpm(x)=2mk=0Tk(x/2). The following equations hold:
pm(2)=2(m+1); (2.5)
pm
2 cos2πr 2m
=
0, ifrodd,
2, ifreven; (2.6)
pm
2 cos
2πr
(2m−1)
=1+(−1)r2 cos πr
(2m−1)
; (2.7)
pm(0)=1+sin
mπ
2
+cos
mπ
2
. (2.8)
Proof. Equation (2.5) follows immediately sinceTk(1)=1, for all k≥0.
For the other equations, we useLemma 2.3,
pm
2 cos
2πr
2m
=2
m
k=0 cos
πkr
m
=1+sin
(m
+1/2)(πr /m) sin(πr /2m)
=1+cos(πr )=
pm
2 cos
2πr
2m−1
=1+sin
(2m
−1)+2/22πr /(2m−1) sinπr /(2m−1)
=1+cos(πr )sin 2
πr /(2m−1) sinπr /(2m−1)
=1+(−1)r2 cos πr
2m−1
,
pm(0)=2 m
k=0
Tk(0)=2 m
k=0 cos
kπ
2
=1+sin
(m+1/2)(π/2) sin(π/4)
=1+sin
mπ
2
+cosmπ 2
.
(2.9)
3. Proof of main theorem. It is a well-known theorem of polynomial inter-polation that there exists a unique polynomial of degreekor less that maps k+1 distinct points in the domain to predefined points in the range. We refer the reader to [1] for a natural proof. Thus, by presenting a polynomial of de-gree less than the number of distinct values of someχ∈Irr(G), which matches the values of the total character when considered as a class function onχ, we immediately get that the polynomial is both minimal and unique.
Proof ofTheorem1.2(1). Letn=2m−1,n≡1,3(mod 8). UsingLemma 2.4, we have thatP(x)=pm(x)−2Tm−1(x/2),
Pψ1[1]=2(m+1)−2=2m=n+1=τ[1],
Pψ1
ar=1+(−1)r2 cos πr
2m−1
−2 cos
πr (m−1)
2m−1
=1=τar,
Pψ1[b]
=1+sin
mπ 2 +cos mπ 2
−2=0=τ[b].
(3.1)
Letn=2m−1,n≡5,7(mod 8). ThenP(x)=pm−1(x),
Pψ1[1]
=2(m−1)+1=2m=n+1=τ[1],
Pψ1
ar=2m− 1
k=0 cos
2πkr
2m−1
=1+ sin(πr )
sinπr /(2m−1)=1=τ
ar,
Pψ1[b]=1+sin
mπ 2 +cos mπ 2
=0=τ[b].
Letn=2m,m≡2,3(mod 4). ThenP(x)=pm(x),
Pψ1[1]
=2(m+1)=τ[1],
Pψ1
ar=
0, ifrodd, 2, ifreven,
=τar,
Pψ1[b]=Pψ1[ab]=1+sin
mπ
2
+cos
mπ
2
=0
=τ[b]=τ[ab].
(3.3)
Thus, in each of the above three cases,P(x)is minimal, and byLemma 2.2it is monic with integer entries. ByLemma 2.1, we have thatP(ψj)=τfor any
faithfulψj∈Irr(G).
Proof ofTheorem1.2(2). Letn=2mand m≡1(mod 4). ThenP(x)= 2Tm+1(x/2)+2pm(x)−2Tm−1(x/2),
Pψ1[1]=21+2(m+1)−1=2τ[1],
Pψ1ar=2
cos(m+1)πr
m +1+(−1)r−cos
(m−1)πr m
=
0, ifrodd, 4, ifreven,
=2τar,
Pψ1[b]
=Pψ1[ab]
=2
0+
1+sin
mπ
2
+cos
mπ
2
−0
=0=2τ[b]=2τ[ab].
(3.4)
Again, we have thatP(x) is minimal since it matchesm+2 distinct values with a polynomial of degreem+1. ByLemma 2.1,P(ψj)=2τfor any faithful
character ofG, and byLemma 2.2it is monic with integer entries. Thus, the unique polynomial that maps the values of a faithful χ∈Irr(G) ontoτ is (1/2)P(x)and has noninteger coefficients.
Proof ofTheorem1.2(3). Letn=2mandm≡0(mod 4). Whenr=m/2,
we have that for any faithful character,ψj[am/2]=ψj(b)=0, butτ[am/2]=2
andτ(b)=0; thus we have an inconsistent system.
Acknowledgments. Both authors wish to thank Professor Donald
also wish to thank the New College Foundation for supporting this project through a faculty/student development grant.
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[8] J. Soto-Andrade,Geometrical Gel’fand models, tensor quotients, and Weil represen-tations, The Arcata Conference on Representations of Finite Groups (Arcata, Calif., 1986), Proc. Sympos. Pure Math., vol. 47, American Mathematical So-ciety, Rhode Island, 1987, pp. 305–316.
Eirini Poimenidou: Division of Natural Sciences, New College of Florida, 5700 North Tamiami Trail, Sarasota, FL 34243, USA
E-mail address:poimenidou@ncf.edu
Homer Wolfe: Division of Natural Sciences, New College of Florida, 5700 North Tami-ami Trail, Sarasota, FL 34243, USA
