( k , j )-Colored Partitions and The Han/Nekrasov-Okounkov Hooklength Formula Emily Anible William J. Keith Michigan Technological University October 2018 Contact: eeanible@mtu.edu 1 / 23
Partition Statistics Let a partition λ of n be represented by λ = 1 ν 1 2 ν 2 . . . n ν n . From this, define the following n -dimensional vectors: ν ( λ ): i th entry is ν i in λ ν i is the multiplicity of parts of size i in λ γ ( λ ): j th entry is number of ν i = j in λ γ j is the frequency of part sizes with multiplicity j We also have ν ( n ) and γ ( n ), where each entry is the sum of all same-index entries over all partitions of n . Note: ν j ( n ) = γ ≥ j ( n ) = � λ ⊢ n γ ≥ j If we were to choose some number of these per partition, they are no longer identical. 1 e.g. � ν 2 � � γ ≥ 2 � � � � = 2 2 λ ⊢ 6 λ ⊢ 6 1 Bacher & Manivel discuss � ν k � in [4] d 2 / 23
Hooklengths The hook length h i , j of the square in the i th column and j th row of a partition λ is the number of squares to its right and directly below it (including itself). Define H k ( λ ) to be the number of hooks of length k in λ . λ = 5 1 4 1 3 2 2 2 1 1 = 11 9 6 3 1 9 7 4 1 7 5 2 6 4 1 4 2 3 1 1 e.g. h 2 , 2 = 7, and H 7 ( λ ) = 2. Note: � λ ⊢ n H k = k � λ ⊢ n ν k 3 / 23
Han/Nekrasov-Okounkov Hooklength Formula The Han/Nekrasov-Okounkov 2 hooklength formula expands the following product, giving coefficients on q n as polynomials in b , a complex indeterminate: Definition 1 (Han/Nekrasov-Okounkov). ∞ ∞ p n ( b ) q n : = � � (1 − q n ) b − 1 H ( q ) := n =0 n =1 ∞ (1 − b � q n � � = ) h 2 i , j n =0 λ ⊢ n h i , j ∈ λ 2 Named as such due to independent discovery by Guo Niu Han and Andrei Okounkov & Nikita Nekrasov 4 / 23
( k , j )-Colored Partitions Definition 2 ( ( k , j ) − Colored Partitions). j ∞ ∞ 1 � � k � � c k , j ( n ) q n = � � � (1 − q n ) j − i q in C k , j ( q ) := . j i ( q ) ∞ n =0 n =1 i =0 If we were to let k = 1 − b , and j go to infinity... ∞ ∞ ∞ � 1 − b q in q n � k � � � � � C 1 − b , ∞ ( q ) = (1 − q n ) i = 1 + 1 − q n i n =1 i =0 n =1 ∞ (1 − q n ) b − 1 = H ( q ) � = n =1 5 / 23
Truncation So, H ( q ) and C 1 − b , ∞ are equivalent. Let’s pick some truncation of H ( q ), limiting it to consider only hooks of length j or less: Definition 3 (Truncated Hooklength Formula). ∞ (1 − b � q n � � H j ( q ) := ) h 2 i , j n =0 λ ⊢ n h i , j ∈ λ h i , j ≤ j So, we are effectively interested in the following question: what term can we add to each polynomial coefficient of C 1 − b , j ( q ) to make it match H j ( q )? 6 / 23
Simplifying C 1 − b , j ( q ) To better investigate this relationship, we would like to put these formulas into a similar form, such that their expansion on a given b c can be better understood. Theorem 1 ( C 1 − b , j Simplification). ∞ ∞ � min( j , ν i ) − b � c 1 − b , j ( n ) q n = � � q n � � C 1 − b , j ( q ) := . min( j , ν i ) n =0 n =0 ν i λ ⊢ n 7 / 23
[ b c ][ q n ] Coefficients Using the previous theorem’s definition for C 1 − b , j , we can now devise a way to compare the two formulas in an easier fashion: Theorem 2 (Coefficient Expansion). The coefficients on the b c term in the coefficient of q n of C 1 − b , j ( q ) and H j ( q ) are as follows: j 1 � γ ≥ k � [ b c ] [ q n ] C 1 − b , j ( q ) = � � � k a k a k a 1 + ··· + a j = c λ ⊢ n k =1 j 1 � H k � [ b c ] [ q n ] H j ( q ) = � � � k 2 a k a 1 + ··· + a j = c λ ⊢ n k =1 8 / 23
[ b c ][ q n ] Coefficients Proof Proof. The b c coefficients in H j ( q ) are given by the binomial theorem. For C 1 − b , j ( q ), we consider its expansion and manipulate it into a similar form. [ q n ] C 1 − b , j ( q ) � γ 1 � 2 − b � γ 2 � γ ≥ j � 1 − b � j − b � = . . . 1 2 j λ ⊢ n (1 − b ) γ 1 ( 1 2!(1 − b )(2 − b )) γ 2 . . . ( 1 � j !(1 − b )(2 − b ) . . . ( j − b )) γ ≥ j = λ ⊢ n (1 1(1 − b )) γ ≥ 1 (1 2(2 − b )) γ ≥ 2 . . . (1 � j ( j − b )) γ ≥ j = λ ⊢ n Applying the binomial theorem gives us our desired form. 9 / 23
Constant and Linear Terms Theorem 3 (Constant & Linear Term Equivalence). For any j, The q i terms of H j ( q ) and C 1 − b , j ( q ) have the same constant and linear term in b. Proof. The constant term is given by � λ ⊢ n 1 = p ( n ) for both formulas. (continued) 10 / 23
Constant and Linear Terms Proof For the linear term, use the previous theorem’s expansion at c = 1: 1 1 γ ≥ 1 + · · · + 1 b 1 � � [ q n ] C 1 − b , j ( q ) = � j γ ≥ j λ ⊢ n 1 2 H 1 + · · · + 1 1 b 1 � � [ q n ] H j ( q ) = � j 2 H j λ ⊢ n Comparing these termwise, we check that for 0 ≤ i ≤ j , 1 1 ? � � i 2 H i = i γ ≥ i λ ⊢ n λ ⊢ n ? � � H i ( n ) = H i = i γ ≥ i = i ∗ ν i ( n ) λ ⊢ n λ ⊢ n The identity H k ( n ) = k ∗ ν k ( n ) is given by Bacher & Manivel in [4], so we are done. 11 / 23
Quadratic Term, j = 2 Let’s proceed to the quadratic term for j = 2. From the Theorem 2 expansion, we get: �� γ ≥ 1 � + 1 � γ ≥ 1 �� γ ≥ 2 � + 1 � γ ≥ 2 �� b 2 � � � [ q n ] C 1 − b , 2 ( q ) = 2 2 1 1 4 2 λ ⊢ n �� H 1 � + 1 � H 1 �� H 2 � + 1 � H 2 �� b 2 � � [ q n ] H 2 ( q ) = � 2 4 1 1 16 2 λ ⊢ n Observationally, these are generally not equivalent. Conjecturally, [ b 2 ][ q n ] H 2 ( q ) = [ b 2 ][ q n ]( C 1 − b , 2 ( q ) + 1 16 γ ≥ 4 ( n )). 12 / 23
� γ ≥ 1 � Quadratic Term — Piecing it Together — 2 We know the following: � γ ≥ 1 � � H 1 � � � = 2 2 λ ⊢ n λ ⊢ n as both H 1 and γ ≥ 1 count the same thing – the number of part sizes in a partition. 13 / 23
Quadratic Term — Piecing it Together — Mixed Term We can show that � γ ≥ 1 �� γ ≥ 2 � � H 1 �� H 2 � 1 1 � � = 2 1 1 4 1 1 λ ⊢ n λ ⊢ n by a 2:1 bijection, i.e. for every pair ( f 1 , f 2 ) over the partitions of n , there are two pairs of ( h 1 , h 2 ) (where f k is a multiplicity of k or more, h k is a hook of length k ). 14 / 23
Quadratic Term — Piecing it Together — Mixed Term We have two cases: Case 1: If λ ⊢ n is not self-conjugate, take a pair consisting of a corner and a repeated part in λ . Then λ ′ will have a hook of 2 that corresponds to that repeated part but as an ascent of at least 2, so it is not counted again for ( f 1 , f 2 ). 2 ′ 1 ′ 1 1 λ ′ = λ = 2 1 ′ 2 1 1 2 ( f 1 , f 2 ): (1 1 , 2) , (1 2 , 2) (1 ′ 1 , 2 ′ ) , (1 ′ 2 , 2 ′ ) ( h 1 , h 2 ): (1 1 , 2) , (1 2 , 2) Case 2: If λ is self-conjugate, consider the 2x2 square: a hook of 2 in a repeated part corresponds to one in an ascent, so the bijection holds. 15 / 23
� γ ≥ k � Quadratic Term — Piecing it Together — d Define G k ( q , u ) as in Brennan, Knopfmacher, & Wagner [1], where q marks the size of the partitions, and u marks parts with multiplicity at least k . ∞ (1 + q i + q 2 i + · · · + q ( k − 1) i + u ( q ki + q ( k +1) i + . . . )) � G k ( q , u ) = i =1 ∞ � 1 − q ki q ki � � = 1 − q i + u 1 − q i i =1 ∞ � (1 + ( u − 1) q ki ) = P ( q ) i =1 16 / 23
� γ ≥ k � Quadratic Term — Piecing it Together — d It is known that ∞ ∂ d � γ ≥ k � � = 1 � q n � � ∂ u d G k ( q , u ) . � d d ! � u =1 n =0 λ ⊢ n We could do the work by hand to find our desired result for d = 2. However, we have proved the following for all d : Theorem 4 (Closed form for choosing multiple γ ≥ k ). ∞ d ∂ d q ik � � γ ≥ k � = 1 � q n � � � ∂ u d G k ( q , u ) = P ( q ) 1 − q ik . � d d ! � u =1 n =0 λ ⊢ n i =1 17 / 23
� H k � Quadratic Term — Piecing it Together — 2 Define the following bivariate generating function: ∞ ∞ � � q n u a ( number of partitions of n with a k-hooks ) F k ( q , u ) = n =0 a =0 ∞ � (1 + q ki ( u − 1)) k = P ( q ) i =1 Han verifies this to be the case in [2]. Repeated derivatives act the same, so we have the following for d = 2: ∞ � 2 kq k kq 2 k � H k � �� � = 1 � q n � 2 P ( q ) − 2 (1 − q k ) 1 − q 2 k n =0 λ ⊢ n so we have ∞ q 4 (1 + 3 q 2 ) � H 2 � � q n � = P ( q ) (1 − q 2 )(1 − q 4 ) . 2 n =0 λ ⊢ n 18 / 23
Quadratic Equivalence From here, we can see the following: ∞ ∞ 1 � H 2 � 1 � γ ≥ 2 � ) + 1 ? � q n � � q n ( � = 16 γ ≥ 4 ( n ) 16 2 4 2 n =0 λ ⊢ n n =0 λ ⊢ n q 4 (1 + 3 q 2 ) 4 q 6 q 4 � � ? P ( q ) = P ( q ) (1 − q 2 )(1 − q 4 ) + (1 − q 2 )(1 − q 4 ) 1 − q 4 q 4 (1 + 3 q 2 ) = P ( q ) (1 − q 2 )(1 − q 4 ) . Thus, we have just proved the following: Theorem 5 (Quadratic Equivalence). For all n and j = 2 , � � � + 1 b 2 � [ q n ] H 2 ( q ) = b 2 � [ q n ] C 1 − b , 2 ( q ) 16 γ ≥ 4 ( n ) b 2 . � 19 / 23
Further Work If we increase j : Messier generating functions, counting many different hooklengths and frequencies. If we look at further order terms (increase b c ): Counting few different hooklengths and frequencies, but selecting several of each in different ways. 20 / 23
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