Many zeros of many characters of GL(n,q)
Patrick X. Gallagher, Michael J. Larsen, Alexander R. Miller
aa r X i v : . [ m a t h . R T ] S e p MANY ZEROS OF MANY CHARACTERS OF GL( n, q ) PATRICK X. GALLAGHER, MICHAEL J. LARSEN, AND ALEXANDER R. MILLER
Abstract.
For G = GL( n, q ), the proportion P n,q of pairs ( χ, g ) in Irr( G ) × G with χ ( g ) = 0 satisfies P n,q → n → ∞ . Introduction
A few years ago, it was shown [7] that for G = S n the proportion P n of pairs( χ, g ) in Irr( G ) × G with χ ( g ) = 0 satisfies(1) P n → n → ∞ . Here we prove the analogous statement for GL( n, q ): Theorem 1.
The proportion P n,q , in Irr(GL( n, q )) × GL( n, q ) , of pairs ( χ, g ) with χ ( g ) = 0 satisfies (2) sup q P n,q → as n → ∞ . One of the two proofs of (1) in [7] is based on the special property of S n , derivedfrom estimates due to Erd˝os and Lehner [1] and Goncharoff [4], that for large n , asuitably chosen small proportion of Cl( S n ) covers all but a small proportion of S n .For the proof of (2) for GL( n, q ), we use both conjugacy class sizes and characterdegrees. There is a general inequality, (3) below, proved in Section 3, and specialproperties (4), (5) of the degrees and sizes of almost all characters and classes ofGL( n, q ), which are proved in Section 7.To lighten the notation, for a finite group G we denote by d χ the degree χ (1) ofan (irreducible) character χ of G , by s g the size | g G | of the conjugacy class g G , andby ( d χ , s g ) the greatest common divisor of d χ and s g . Lemma A.
For each finite group G and ε > , the proportion P , in Irr( G ) × G ,of pairs ( χ, g ) with χ ( g ) = 0 satisfies (3) P ≤ Q ( ε ) + ε , with Q ( ε ) the proportion, in Irr( G ) × G , of pairs ( χ, g ) with ( d χ , s g ) /d χ ≥ ε . Lemma B.
For all δ, ε > , there exists N such that if n ≥ N , q is a prime power,and G = GL( n, q ) , then for ( χ, g ) in Irr( G ) × G , (4) ( d χ , s g ) d χ < ε, except for ( χ, g ) in a subset R ⊂
Irr( G ) × G such that (5) |R| ≤ δ | Irr( G ) × G | . ML was partially supported by the NSF grant DMS-1702152.
P. X. GALLAGHER, M. J. LARSEN, AND A. R. MILLER Proof of Theorem 1 using Lemmas A and B
For G = GL( n, q ) and ε >
0, Lemma A gives P n,q ≤ Q n,q + ε , with P n,q the proportion of pairs ( χ, g ) with χ ( g ) = 0 and Q n,q the proportionof pairs with ( d χ , s g ) /d χ ≥ ε . Lemma B gives Q n,q ≤ δ for n ≥ N . Thus for n sufficiently large, P n,q ≤ δ + ε , from which Theorem 1 follows. (cid:3) Proof of Lemma A by a device of Burnside
For each χ ∈ Irr( G ) and g ∈ G , both χ ( g ) and s g χ ( g ) /d χ are algebraic integers,so for all a, b ∈ Z , so is ( ad χ + bs g ) χ ( g ) /d χ . Choosing a and b so that ad χ + bs g isthe greatest common divisor ( d χ , s g ) of d χ and s g , this gives(6) χ ( g ) = d χ ( d χ , s g ) α χ,g , with α χ,g an algebraic integer in the cyclotomic field Q ( ζ | G | ) with ζ | G | = e πi/ | G | .From (6), for each χ ,(7) X g ∈ G (cid:0) d χ ( d χ , s g ) (cid:1) | α χ,g | = | G | . To (7), apply elements σ of the Galois group Γ = Gal( Q ( ζ | G | ) / Q ), average over Γ,and use the fact, due to Burnside, that the average over Γ of | σ ( α ) | is ≥ α ∈ Q ( ζ | G | ), [3, p. 359]. This gives, for each χ ,(8) X g ∈ G ′ (cid:0) d χ ( d χ , s g ) (cid:1) ≤ | G | , the dash meaning that the sum is over those g with χ ( g ) = 0. From (8),(9) X χ ∈ Irr( G ) X g ∈ G ′ ( d χ ( d χ , s g ) ) ≤ | Irr( G ) || G | . From (9), the proportion, in Irr( G ) × G , of pairs ( χ, g ) with both χ ( g ) = 0 and( d χ , s g ) /d χ ≤ ε is at most ε , from which (3) follows. (cid:3) Number theoretic lemmas: partitions
We denote by p ( n ) the number of partitions of a non-negative integer n . Lemma 1.
For each positive integer n , p ( n ) ≤ n − .Proof. The base case n = 1 is trivial. For n >
1, the number of partitions withsmallest part m is at most p ( n − m ), so p ( n ) ≤ p (1) + p (2) + · · · + p ( n − ≤ · · · + 2 n − = 2 n − , and the lemma follows by induction. (cid:3) Lemma 2.
Let φ := √ . Then p ( n ) ≤ φ n for all non-negative integers n . ANY ZEROS OF MANY CHARACTERS OF GL( n, q ) 3
Proof.
The partition function is non-decreasing since the number of partitions of n + 1 with a part of size 1 is p ( n ). The lemma holds for n ∈ { , } . For n ≥
2, thepentagonal number theorem implies(10) p ( n ) = p ( n −
1) + p ( n − − p ( n − − p ( n −
7) + p ( n −
12) + · · · , with sign pattern + + − − + + − − + + − − · · · and where the sum on the right-hand side terminates at the last term ± p ( n − m ), where m is the largest generalizedpentagonal number for which n ≥ m . By monotonicity, the right-hand side of (10)is at most p ( n −
1) + p ( n − n . (cid:3) Lemma 3.
There exists γ < such that if q ≥ and a and b are positive integerssuch that a ( b − ≥ N ≥ , then p ( b ) q a ( b − < γ N . Proof.
It suffices to prove the lemma for q = 2. For a = 1, we have b − ≥ N , soLemma 2 implies p ( b )2 a ( b − = p ( b )2 b − < φ/ N . For a ≥ a ( b − ≤ a − b − p ( b )2 a ( b − ≤ − ( a − b − ≤ (1 / √ N < / √ N . Therefore, we may take γ = φ/ > / √ (cid:3) Number theoretic lemmas: cyclotomic polynomials
For n a positive integer, let Φ n ( x ) denote the minimal polynomial over Q of e πi/n .Thus(11) x n − Y d | n Φ d ( x ) , so by M¨obius inversion,(12) Φ n ( x ) = Y d | n ( x n/d − µ ( d ) . For any prime ℓ , let ord ℓ ( x ) denote the largest integer e such that ℓ e divides x . Lemma 4.
Let ℓ be a prime, e a positive integer, and n an integer such that ord ℓ ( n −
1) = e . (i) If k is a positive integer prime to ℓ , then ord ℓ ( n k −
1) = e . (ii) If ℓ is odd and ord ℓ ( k ) = 1 , then ord ℓ ( n k −
1) = e + 1 .Proof. Let n = 1 + mℓ e , where ℓ ∤ m . By the binomial theorem, n k ≡ kmℓ e (mod ℓ e ) , which implies claim (i). For claim (ii), using part (i), it suffices to treat the case k = ℓ , for which we have n ℓ ≡ mℓ e +1 + m ( ℓ − ℓ e +1 (mod ℓ e ) . (cid:3) Lemma 5.
Suppose n > and a > are integers. We factor Φ n ( a ) as P n ( a ) R n ( a ) ,where P n ( a ) is relatively prime to n and R n ( a ) factors into prime divisors of n . P. X. GALLAGHER, M. J. LARSEN, AND A. R. MILLER (i)
Every prime divisor of P n ( a ) is ≡ (mod n ). (ii) If n ≥ , R n ( a ) is a square-free divisor of n . (iii) For n ≥ , P n ( a ) > √ n/ − log n − . (iv) If mℓ > n and ℓ is a prime divisor of P m ( a ) , then ord ℓ ( a n −
1) = ( ord ℓ P m ( a ) if m | n , otherwise.Proof. Fix any prime ℓ which divides Φ n ( a ). As ℓ | a n − a is not divisible by ℓ ,so it represents a class in F × ℓ . Let k be the order of this class. As a n ≡ ℓ ), k | n . Let s denote the largest square-free divisor of n/k . By (12),ord ℓ Φ n ( a ) = ord ℓ Y d | s ( a n/d − µ ( d ) . Now, if s can be written ps ′ for some prime p = ℓ ,(13) Y d | s ( a n/d − µ ( d ) = Y d | s ′ (cid:16) a n/d − a n/pd − (cid:17) µ ( d ) . Applying part (i) of Lemma 4 with k = p , the above formula implies ord ℓ Φ n ( a ) = 0,contrary to assumption. Since s is square-free, it follows that it can only be 1 or ℓ .If ℓ divides P n ( a ), then it does not divide n . That means s = 1, so the class of a has order n in a group of order ℓ −
1. This implies part (i). Conversely, if ℓ doesdivide n , it cannot be 1 (mod n ), so s = ℓ .If s = ℓ >
2, then d square-free and ord ℓ ( a n/d − > d ∈ { , ℓ } .Therefore, part (ii) of Lemma 4 implies that the left-hand side of (13) has ord ℓ equal to 1. If s = ℓ = 2, then k = 1, so we need only consider the case that n isa power of 2. For t ≥
2, Φ t ( x ) = ( x t − ) + 1, so plugging in a , the result has atmost one factor of 2. This gives claim (ii).By (12),(14) Φ n ( a ) ≥ a deg Φ n ∞ Y i =1 (1 − a − ) ≥ a φ ( n ) ∞ Y i =1 (1 − − ) ≥ φ ( n ) . As φ ( p e ) ≥ √ p e except when p e = 2, the multiplicativity of φ implies φ ( n ) ≥ p n/ R n ( a ) ≤ n , and claim (iii) follows.If ℓ divides P m ( a ), then the image of a in F × ℓ is of order m , so ℓ divides a n − n is divisible by m . In that case, P m ( a ) divides Φ m ( a ), which is a divisor of a m − a n −
1. Moreover, ℓ does not divide m , so ord ℓ P m ( a ) = ord ℓ Φ m ( a ).To prove (iv), it remains to show that a n − ℓ beyondthose in a m −
1. It suffices to prove that Φ n ′ ( a ) is not divisible by ℓ if n ′ is a divisorof n and m is a proper divisor of n ′ . Indeed, ℓ does not divide P n ′ ( a ) because a isnot of order exactly m ′ (mod ℓ ). If it divides Φ n ′ ( a ), it must divide R n ′ ( a ), so itmust divide n ′ . It does not divide m , so it must divide n ′ /m ≤ m . This is ruledout by (i). (cid:3) Irreducible characters of
GL( n, q )In what follows, G = GL( n, q ). By [2, Proposition 3.5],(15) q n ≤ | Irr( G ) | = | Cl( G ) | ≤ q n . ANY ZEROS OF MANY CHARACTERS OF GL( n, q ) 5
Denote by P the set of all integer partitions λ (including the empty partition ∅ )and by F the set of all non-constant monic irreducible polynomials f ( x ) ∈ F q [ x ]with non-zero constant term. We define the degree of ν as follows:deg( ν ) := X f ∈F deg( f ) | ν ( f ) | . By Jordan decomposition, there is a natural bijection between conjugacy classes in G and maps ν : F → P of degree n . Green [5] introduced the set G of simplices and proved (Theorem 12) that Irr( G ) has a parametrization by maps ν : G → P satisfying X f ∈G deg( f ) | ν ( f ) | = n. By fixing in a compatible way multiplicative generators of finite fields, he gavea degree-preserving bijection between F and G . We will ignore the distinctionbetween F and G henceforward. The same theorem of Green also gave a formulafor the degree of the irreducible character χ associated to ν . It can be written(16) d χ = q N ν Q ni =1 ( q i − Q f ∈F Q | ν ( f ) | i =1 ( q h ν ( f ) ,i deg( f ) − , where N ν is a certain non-negative integer, and the h λ,i are the hook lengths of thepartition λ ; in particular these are positive integers ≤ | λ | .By the support of ν , which we denote supp ν , we mean the set of f ∈ F suchthat ν ( f ) = ∅ . Lemma 6.
Let γ be defined as in Lemma 3, and let N be a positive integer. Thenthe number of degree n functions ν : F → P satisfying deg( f )( | ν ( f ) | − ≥ N forsome f is less than Nγ N (1 − γ ) q n .Proof. It suffices to prove that for each m , the number of choices of ν of degree n such that for some f ∈ F , deg( f )( | ν ( f ) | −
1) = m is less than 2 mγ m q n . Sincethere are at most m ways of expressing m as a ( b −
1) for positive integers a and b , it suffices to prove that there are less than 2 γ m q n such ν of degree n for which | ν ( f ) | = b for some f ∈ F of degree a . Since there are fewer than q a elements of F of degree a , it suffices to prove that for given f ∈ F of degree a , there are atmost 2 γ m q n − a possibilities for ν with | ν ( f ) | = b . For each partition λ of b , thefunctions ν of degree n with ν ( f ) = λ can be put into bijective correspondencewith ν ′ of degree n − ab with ν ′ ( f ) = ∅ . By (15), the number of possibilities for ν ′ and therefore for ν is at most q n − ab = q n − m − a . Summing over the possibilitiesfor λ , which by Lemma 3 number less than 2 γ m q m , we obtain less than 2 γ m q n − a possibilities for ν with | ν ( f ) | = b , as claimed. (cid:3) We define the deficiency of a character of G or of the associated ν : F → P tobe the maximum of deg( f )( | ν ( f ) | −
1) over all f ∈ F . Together, Lemma 6 and (15)imply that for all ε > N such that for all n and q , the proportionof irreducible characters of GL( n, q ) with deficiency < N is at least 1 − ε . Lemma 7.
Let m be a positive integer and ℓ a prime such that ℓm > n and ord ℓ P m ( q ) = e > . Let χ be a character whose deficiency is less than m/ . Then ord ℓ d χ = e ⌊ n/m ⌋ − e |{ f ∈ supp ν | deg( f ) ∈ m Z }| = ord ℓ | G | − e |{ f ∈ supp ν | deg( f ) ∈ m Z }| . P. X. GALLAGHER, M. J. LARSEN, AND A. R. MILLER
Proof. If f is in the support of ν and deg( f ) | ν ( f ) | < m , then f does not contributeany factor of ℓ to the denominator of (16). So we need only consider the casedeg( f ) | ν ( f ) | ≥ m , in which case deg( f )( | ν ( f ) | − ≥ m/ | ν ( f ) | ≥
2. Sincethe deficiency of χ is less than m/
2, this is impossible, which means that all f contributing factors of ℓ in (16) satisfy ν ( f ) = (1). Moreover, by Lemma 5, ℓ divides q k − m divides k , in which case ord ℓ ( q k −
1) = e . Thus, thefactors in (16) contributing to ord ℓ are q m − , q m − , . . . , q ⌊ n/m ⌋ m −
1, each ofwhich contributes e , and q deg( f ) − f ∈ supp ν of degree divisible by m ,again each contributing e . (cid:3) Lemma 8.
For any positive integer m , the number of ν : F → P of degree n forwhich there exist f ∈ F of degree m with ν ( f ) = (1) is less than q n /m .Proof. Any degree m element of F splits completely in F q m , so there are less than q m /m such elements. For each f , there is a bijective correspondence between ν ofdegree n with ν ( f ) = (1) and ν ′ of degree n − m with ν ′ ( f ) = ∅ . By (15), there areat most q n − m such ν ′ , so the total number of ν is less than q n /m . (cid:3) Lemma 9.
For all ε > , if n is sufficiently large in terms of ε , m is a sufficientlylarge positive integer, ℓ is a prime divisor of P m ( q ) , and ℓm > n , then the probabilityis at least − n − mm − ε that a random element χ chosen uniformly from Irr( G ) satisfies (17) ord ℓ d χ = ord ℓ | G | . Proof.
Choose N in Lemma 6 such that N γ N < (1 − γ ) ε/
4. By (15), the prob-ability that χ has deficiency ≥ N is less than ε . We assume m > N , so withprobability greater than 1 − ε , the deficiency of a random χ ∈ Irr( G ) is less than m/
2. By Lemma 7, this implies (17) provided that no element in the support of ν has degree a multiple of m . If f ∈ supp ν has degree km , then the deficiencycondition on ν implies ν ( f ) = (1). By Lemma 8, the probability that there existsan element in the support of ν of degree km is less than 2 /km , so the probabilitythat there is an element in the support of ν with degree in m Z is less than ⌊ n/m ⌋ X k =1 km < n − mm . (cid:3) Lemma 10.
For all δ > , if n is sufficiently large in terms of δ , m ≥ √ n , and ℓ is any prime divisor of P m ( q ) , then the probability of (17) is greater than − δ/ .Proof. By part (i) of Lemma 5, ℓ > m , so ℓm > n . Applying Lemma 9 for ε = δ/ n − mm < δ . For n ≥ m ≥ √ n , the left-hand side is less than 2 n − / log n , which goes tozero as n goes to ∞ . (cid:3) ANY ZEROS OF MANY CHARACTERS OF GL( n, q ) 7 Proof of Lemma B
Let Fact f denote the total number of factors in the decomposition of f ( x ) ∈ F q [ x ]into irreducibles. For each g ∈ GL( n, q ), let p g ( x ) denote the characteristic poly-nomial of g . Lemma 11.
There exist constants A and B such that for all m , n , and q , at most An B q − m | GL( n, q ) | elements of GL( n, q ) have a characteristic polynomial with arepeated irreducible factor of degree ≥ m .Proof. By [6, Proposition 3.3], the number of elements of GL( n, q ) with any givencharacteristic polynomial is at most ( A/ n B q n − n for some absolute constants A and B . (Actually, the statement is proven only for “classical” groups, but the prooffor GL( n, q ) is identical.) For any given f of degree m , there are q n − m polynomialsof degree ≤ n divisible by f , so there are less than q n − m polynomials of degree n with a repeated irreducible factor of degree m and less than q n − m + q n − m − + · · · < q n − m polynomials with a repeated irreducible factor of degree ≥ m . On the otherhand, by the same argument as (14), | GL( n, q ) | = n Y i =1 ( q n − q i ) > q n . The lemma follows. (cid:3)
Proof of Lemma B.
By [6, Proposition 3.4], for all δ > k such that(18) P [Fact p g > k log n ] < δ , where P denotes probability with respect to the uniform distribution on G =GL( n, q ). (Actually, the cited reference proves the analogous claim for SL( n, q ),but the proof goes through the GL( n, q ) case.) Choose k so that this holds andassume that n is large enough that(a) √ n > k log n ,(b) An B −√ n < δ , where A and B are defined as in Lemma 11,(c) p m/ > log m + 2 for all m ≥ √ n ,(d) m > /ε for all m ≥ √ n .Let X denote the set of elements g for which p g ( x ) has ≤ k log n irreduciblefactors and no repeated factor of degree ≥ √ n . By condition (a) on n , every p g with g ∈ X has a simple irreducible factor of degree ≥ √ n . By equation (18) andcondition (b), | G \X | < ( δ/ | G | . For each g ∈ X , fix an irreducible factor of degree m g ≥ √ n of p g . By condition (c) and part (iii) of Lemma 5, P m g ( q ) >
1, so foreach g , we may fix a prime divisor ℓ g of P m g ( q ). We define R to consist of all pairs( χ, g ) where g
6∈ X or where g ∈ X butord ℓ g d χ = ord ℓ g | G | . By Lemma 10, for each g ∈ X , there are at most ( δ/ | Irr( G ) | pairs ( χ, g ) ∈ R .Thus, R satisfies equation (5).For pairs ( χ, g )
6∈ R , we have g ∈ X and ord ℓ g d χ = ord ℓ g | G | . As p g ( x ) has anirreducible factor of degree m g which occurs with multiplicity 1, the centralizer of g has order divisible by q m g − ℓ g . Therefore, ord ℓ g s g < ord ℓ g | G | .This implies that ℓ g is a divisor of the denominator of ( d χ , s g ) /d χ . As ℓ g ≡ P. X. GALLAGHER, M. J. LARSEN, AND A. R. MILLER (mod m g ), we have ℓ g > m g . By condition (d) on n , m g ≥ /ε . Thus, equation (4)holds. (cid:3) References
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Duke Math. J. (1941) 335–345.2. J. Fulman and R. Guralnick, Bounds on the number and sizes of conjugacy classes in finiteChevalley groups with applications to derangements. Trans. Amer. Math. Soc. (2012)3023–3070.3. P. X. Gallagher, Degrees, class sizes and divisors of character values.
J. Group Theory (2012) 455–467.4. V. L. Goncharoff, Sur la distribution des cycles dans les permutations. C. R. (Doklady)Acad. Sci. URSS (N.S.) (1942) 267–269.5. J. A. Green, The characters of the finite general linear groups. Trans. Amer. Math. Soc. (1955) 402–447.6. M. Larsen and A. Shalev, On the distribution of values of certain word maps. Trans. Amer. Math. Soc. (2016), no. 3, 1647–1661.7. A. R. Miller, The probability that a character value is zero for the symmetric group.
Math. Z. (2014) 1011–1015.
Department of Mathematics, Columbia University, New York, NY, USA
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