aa r X i v : . [ m a t h . N T ] M a y HOOK LENGTHS AND -CORES GUO-NIU HAN AND KEN ONO
Abstract.
Recently, the first author generalized a formula of Nekrasov and Okounkovwhich gives a combinatorial formula, in terms of hook lengths of partitions, for thecoefficients of certain power series. In the course of this investigation, he conjecturedthat a ( n ) = 0 if and only if b ( n ) = 0, where integers a ( n ) and b ( n ) are defined by ∞ X n =0 a ( n ) x n := ∞ Y n =1 (1 − x n ) , ∞ X n =0 b ( n ) x n := ∞ Y n =1 (1 − x n ) − x n . The numbers a ( n ) are given in terms of hook lengths of partitions, while b ( n ) equalsthe number of 3-core partitions of n . Here we prove this conjecture. Introduction and statement of results
In their work on random partitions and Seiberg-Witten theory, Nekrasov and Okounkov[8] proved the following striking formula:(1.1) F z ( x ) := X λ x | λ | Y h ∈H ( λ ) (cid:16) − zh (cid:17) = ∞ Y n =1 (1 − x n ) z − . Here the sum is over integer partitions λ , | λ | denotes the integer partitioned by λ , and H ( λ ) denotes the multiset of classical hooklengths associated to a partition λ . In arecent preprint, the first author [3] has obtained an extension of (1.1), one which has aspecialization which gives the classical generating function(1.2) C t ( x ) := ∞ X n =0 c t ( n ) x n = ∞ Y n =1 (1 − x tn ) t − x n for the number of t -core partitions of n . Recall that a partition is a t -core if none of itshook lengths are multiples of t .In the course of his work, the first author [4] formulated a number of conjecturesconcerning hook lengths of partitions. One of these conjectures is related to classicalidentities of Jacobi. For positive integers t , he compared the functions F t ( x ) and C t ( x ).If t = 1, we obviously have that F ( x ) = C ( x ) = 1 . The second author thanks the support of the NSF, and he thanks the Manasse family.
For t = 2, by two famous identities of Jacobi, we have F ( x ) = ∞ Y n =1 (1 − x n ) = ∞ X k =0 ( − k (2 k + 1) x ( k + k ) / ,C ( x ) = ∞ Y n =1 (1 − x n ) − x n = ∞ X k =0 x ( k + k ) / . In both pairs of power series one sees that the non-zero coefficients are supported on thesame terms. For t = 3, we then have F ( x ) = ∞ X n =0 a ( n ) x n := ∞ Y n =1 (1 − x n ) = 1 − x + 20 x − x + · · · − x + 57 x + 560 x + 182 x + · · · (1.3)and C ( x ) = ∞ X n =0 b ( n ) x n := ∞ Y n =1 (1 − x n ) − x n = 1 + x + 2 x + 2 x + · · · + 2 x + 3 x + 2 x + 2 x + · · · . (1.4) Remark.
It is clear that b ( n ) = c ( n ).In accordance with the elementary observations when t = 1 and 2, one notices thatthe non-zero coefficients of F ( x ) and C ( x ) appear to be supported on the same terms.Based on substantial numerical evidence, the first author made the following conjecture. Conjecture
Assuming the notation above, we have that a ( n ) = 0 if and only if b ( n ) = 0 .Remark. The obvious generalization of Conjecture 4.6 and the examples above is not truefor t = 4. In particular, one easily finds that F ( x ) = 1 − x + 90 x − · · · + 641445 x + 1537330 x + · · · ,C ( x ) = 1 + x + 2 x + 3 x + · · · + 5 x + 8 x + 10 x + · · · . The coefficient of x vanishes in F ( x ) and is non-zero in C ( x ).Here we prove that Conjecture 4.6 is true. We have the following theorem. Theorem 1.1.
Assuming the notation above, we have that a ( n ) = 0 if and only if b ( n ) = 0 . Moreover, we have that a ( n ) = b ( n ) = 0 precisely for those non-negative n forwhich ord p (3 n + 1) is odd for some prime p ≡ .Remark. As usual, ord p ( N ) denotes the power of a prime p dividing an integer N . OOK LENGTHS AND 3-CORES 3
Remark.
Theorem 1.1 shows that a ( n ) = b ( n ) = 0 in a systematic way. The vanishingcoefficients are associated to primes p ≡ n ≡ p ( n ) is odd, then we have a (cid:18) n − (cid:19) = b (cid:18) n − (cid:19) = 0 . For example, since ord (10) = 1 ≡ a (3) = b (3) = 0.As an immediate corollary, we have the following. Corollary 1.2.
For positive integers N , we have that X λ ⊢ N Y h ∈H ( λ )) (cid:18) − h (cid:19) = 0 if and only if there are no -core partitions of N . Theorem 1.1 implies that “almost all” of the a ( n ) and b ( n ) are 0. More precisely, wehave the following. Corollary 1.3.
Assuming the notation above, we have that lim X → + ∞ { ≤ n ≤ X : a ( n ) = b ( n ) = 0 } X = 1 . Acknowledgements
The authors thank Mihai Cipu for insightful comments related to Conjecture 4.6.2.
Proofs
It is convenient to renormalize the functions a ( n ) and b ( n ) using the series A ( z ) = ∞ X n =1 a ∗ ( n ) q n := ∞ X n =0 a ( n ) q n +1 = q − q + 20 q − q + 64 q + 56 q − q − q + · · · . (2.1)and B ( z ) = ∞ X n =1 b ∗ ( n ) q n := ∞ X n =0 b ( n ) q n +1 = q + q + 2 q + 2 q + q + 2 q + q + 2 q + · · · . (2.2)Here we have that z ∈ H , the upper-half of the complex plane, and we let q := e πiz . Wemake these changes since A ( z ) and B ( z ) are examples of two special types of modularforms (for background on modular forms, see [1, 6, 7, 9]). The modularity of these twoseries follows easily from the properties of Dedekind’s eta-function(2.3) η ( z ) := q ∞ Y n =1 (1 − q n ) . GUO-NIU HAN AND KEN ONO
The proofs of Theorem 1.1 and Corollary 1.3 shall rely on exact formulas we derive forthe numbers a ∗ ( n ) and b ∗ ( n ).2.1. Exact formulas for a ∗ ( n ) . The modular form A ( z ) given by A ( z ) = η (3 z ) = ∞ X n =1 a ∗ ( n ) q n is in S (Γ (9)), the space of weight 4 cusp forms on Γ (9). This space is one dimensional(see Section 1.2.3 in [9]). Therefore, every cusp form in the space is a multiple of A ( z ).It turns out that A ( z ) is a form with complex multiplication .We now briefly recall the notion of a newform with complex multiplication (for ex-ample, see Chapter 12 of [6] or Section 1.2 of [9], [10]). Let D < K = Q ( √ D ). Let O K be the ring of integersof K , and let χ K := (cid:0) D • (cid:1) be the usual Kronecker character associated to K . Let k ≥ c be a Hecke character of K with exponent k − f c , a non-zeroideal of O K . By definition, this means that c : I ( f c ) −→ C × is a homomorphism, where I ( f c ) denotes the group of fractional ideals of K prime to f c .In particular, this means that c ( αO K ) = α k − for α ∈ K × for which α ≡ × f c . To c we naturally associate a Dirichlet character ω c defined, for every integer n coprime to f c , by ω c ( n ) := c ( nO K ) n k − . Given this data, we let(2.4) Φ
K,c ( z ) := X a c ( a ) q N ( a ) , where a varies over the ideals of O K prime to f c , and where N ( a ) is the usual ideal norm.It is known that Φ K,c ( z ) ∈ S k ( | D | · N ( f c ) , χ K · ω c ) is a normalized newform.Using this theory, we obtain the following theorem. Theorem 2.1.
Assume the notation above. Then the following are true: (1) If p = 3 or p ≡ is prime, then a ∗ ( p ) = 0 . (2) If p ≡ is prime, then a ∗ ( p ) = 2 x − xy , where x and y are integers for which p = x + 3 y with x ≡ .Remark. It is a classical fact that every prime p ≡ x + 3 y .Moreover, there is a unique pair of positive integers x and y for which x + 3 y = p .Therefore, the formula for a ∗ ( p ) is well defined. OOK LENGTHS AND 3-CORES 5
Proof.
There is a form with complex multiplication in S (Γ (9)). Following the recipeabove, it is obtained by letting k = 4, Q ( √ D ) = Q ( √−
3) and f c := ( √− p , the coefficients of q p in this form agree with the claimed formulas. Since S (Γ (9)) isone dimensional, this form must be A ( z ). (cid:3) Using this theorem, we obtain the following immediate corollary.
Corollary 2.2.
The following are true about a ∗ ( n ) . (1) If m and n are coprime positive integers, then a ∗ ( mn ) = a ∗ ( m ) a ∗ ( n ) . (2) For every positive integer s , we have that a ∗ (3 s ) = 0 . (3) If p ≡ is prime and s is a positive integer, then a ∗ ( p s ) = ( s is odd,( − s/ p s/ if s is even.(4) If p ≡ is prime and s is a positive integer, then a ∗ ( p s ) = 0 . Moreover,we have that a ∗ ( p s ) ≡ (8 x ) s (mod p ) , where p = x + 3 y with x ≡ .Proof. Since S (Γ (9)) is one dimensional and since a ∗ (1) = 1, it follows that A ( z ) is anormalized Hecke eigenform. Claim (1) is well known to hold for all normalized Heckeeigenforms.Claim (2) follows by inspection since a ∗ ( n ) = 0 if n ≡ , A ( z ) is a normalized Hecke eigenformon Γ (9), it follows, for every prime p = 3, that(2.5) a ∗ ( p s ) = a ∗ ( p ) a ∗ ( p s − ) − p a ∗ ( p s − ) . If p ≡ a ∗ ( p s ) = − p a ( p s − ) . Claim (3) now follows by induction since a ∗ (1) = 1 and a ∗ ( p ) = 0.Suppose that p ≡ a ∗ ( p ) = 0. Moreimportantly, we have that a ∗ ( p ) ≡ x (mod p ) , where p = x + 3 y with x ≡ x − xy = 2 x ( x − y ) = 2 x ( x − p − x )) ≡ x (mod p ) . Since | x | ≤ √ p and is non-zero, it follows that a ∗ ( p ) ≡ x p ). By (2.5), wethen have that a ∗ ( p s ) ≡ a ∗ ( p ) a ∗ ( p s − ) ≡ x a ∗ ( p s − ) (mod p ) . By induction, it follows that a ∗ ( p s ) ≡ (8 x ) s (mod p ), which is non-zero modulo p . Thisproves claim (4). (cid:3) GUO-NIU HAN AND KEN ONO
Example 2.3.
Here we give some numerical examples of the formulas for a ∗ ( n ).1) One easily finds that a ∗ (13) = −
70. The prime p = 13 is of the form x + 3 y where x = 1 and y = 2. Obviously, x = 1 ≡ a ∗ (13) = 2 · − · · = − a ∗ (13) = −
70 and a ∗ (16) = 64. One easily checks that a ∗ (13 ·
16) = a ∗ (208) = − ·
64 = − p = 5 and s = 3, then Corollary 2.2 (3) asserts that a ∗ (5 ) = 0. If p = 5 and s = 4,then it asserts that a ∗ (5 ) = 5 = 15625. One easily checks both evaluations numerically.4) Now we consider the prime p = 13 ≡ x = 1 and y = 2 for p = 13,Corollary 2.2 (4) asserts that a ∗ (13 s ) ≡ s (mod 13). One easily checks that a ∗ (13) = − ≡ ,a ∗ (13 ) = 2703 ≡ (mod 13) ,a ∗ (13 ) = − ≡ (mod 13) . Proof of Theorem 1.1 and Corollary 1.3.
Before we prove Theorem 1.1, werecall a known formula for b ( n ) (also see Section 3 of [2]), the number of 3-core partitionsof n . Lemma 2.4.
Assuming the notation above, we have that B ( z ) = ∞ X n =1 b ∗ ( n ) q n = ∞ X n =0 b ( n ) q n +1 = ∞ X n =0 X d | n +1 (cid:18) d (cid:19) q n +1 , where (cid:0) • (cid:1) denotes the usual Legendre symbol modulo 3.Proof. We have that B ( z ) = η (9 z ) /η (3 z ) is in M (Γ (9) , χ ), where χ := (cid:0) − • (cid:1) . Thelemma follows easily from this fact. One may implement the theory of weight 1 Eisensteinseries to obtain the desired formulas.Alternatively, one may use the weight 1 formΘ( z ) = ∞ X n =0 c ( n ) q n := X x,y ∈ Z q x + xy + y = 1 + 6 q + 6 q + 6 q + 12 q + 6 q + · · · . Using the theory of twists, we find that e Θ( z ) = X n ≡ c ( n ) q n = 6 q + 6 q + 12 q + 12 q + 6 q + 12 q + 6 q + · · · = 6 (cid:0) q + q + 2 q + 2 q + q + 2 q + q + · · · (cid:1) . By dimensionality (see Section 1.2.3 of [9]) we have that B ( z ) = 6 e Θ( z ). The claimedformulas for the coefficients follows easily from the fact that x + xy + y corresponds tothe norm form on the ring of integers of Q ( √− (cid:3) OOK LENGTHS AND 3-CORES 7
Example 2.5.
The only divisors of primes p ≡ p , and so we havethat b ∗ ( p ) = 1 + (cid:0) p (cid:1) = 1 + (cid:0) (cid:1) = 2. Proof of Theorem 1.1.
The theorem follows immediately from Theorems 2.1, 2.2 andLemma 2.4. One sees that the only n ≡ a ∗ ( n ) = 0 are those n for which ord p ( n ) is odd for some prime p ≡ b ∗ ( n ). Using the fact that a ( n ) = a ∗ (3 n + 1) and b ( n ) = b ∗ (3 n + 1) , the theorem follows. (cid:3) Proof of Corollary 1.3.
In a famous paper [11], Serre proved that “almost all” of thecoefficients of a modular form with complex multiplication are zero. This implies thatalmost all of the a ∗ ( n ) are zero. The result now follows thanks to Theorem 1.1. (cid:3) References [1] D. Bump,
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I.R.M.A., UMR 7501, Universit´e Louis Pasteur et CNRS, 7 rue Ren´e-Descartes, F-67084Strasbourg, France
E-mail address : [email protected] Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706
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