( k , j )-Colored Partitions and The Han/Nekrasov-Okounkov - - PowerPoint PPT Presentation

(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

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Partition Statistics

Let a partition λ of n be represented by λ = 1ν12ν2 . . . nνn. From this, define the following n-dimensional vectors: ν(λ): ith entry is νi in λ νi is the multiplicity of parts of size i in λ γ(λ): jth entry is number of νi = j in λ γj 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: νj(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.

• λ⊢6

ν2 2

• =
• λ⊢6

γ≥2 2

• 1Bacher & Manivel discuss

νk

d

• in [4]

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Hooklengths

The hook length hi,j of the square in the ith column and jth row of a partition λ is the number of squares to its right and directly below it (including itself). Define Hk(λ) to be the number of hooks of length k in λ. λ = 5141322211 = 11 9 6 3 1 9 7 4 1 7 5 2 6 4 1 4 2 3 1 1 e.g. h2,2 = 7, and H7(λ) = 2. Note:

λ⊢n Hk = k λ⊢n νk

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Han/Nekrasov-Okounkov Hooklength Formula

The Han/Nekrasov-Okounkov 2 hooklength formula expands the following product, giving coefficients on qn as polynomials in b, a complex indeterminate: Definition 1 (Han/Nekrasov-Okounkov). H(q) :=

∞

• n=0

pn(b)qn : =

∞

• n=1

(1 − qn)b−1 =

∞

• n=0

qn

λ⊢n

• hi,j∈λ

(1 − b h2

i,j

)

2Named as such due to independent discovery by Guo Niu Han and Andrei

Okounkov & Nikita Nekrasov

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(k, j)-Colored Partitions

Definition 2 ((k, j) − Colored Partitions). Ck,j(q) :=

∞

• n=0

ck,j(n)qn = 1 (q)∞

j ∞

• n=1
• j
• i=0

k i

• (1 − qn)j−iqin
• .

If we were to let k = 1 − b, and j go to infinity... C1−b,∞(q) =

∞

• n=1

∞

• i=0

k i

• qin

(1 − qn)i =

∞

• n=1
• 1 +

qn 1 − qn 1−b =

∞

• n=1

(1 − qn)b−1 = H(q)

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Truncation

So, H(q) and C1−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). Hj(q) :=

∞

• n=0

qn

λ⊢n

• hi,j∈λ

hi,j≤j

(1 − b h2

i,j

) So, we are effectively interested in the following question: what term can we add to each polynomial coefficient of C1−b,j(q) to make it match Hj(q)?

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Simplifying C1−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 bc can be better understood. Theorem 1 (C1−b,j Simplification). C1−b,j(q) :=

∞

• n=0

c1−b,j(n)qn =

∞

• n=0

qn

λ⊢n

• νi

min(j, νi) − b min(j, νi)

• .

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[bc][qn] Coefficients

Using the previous theorem’s definition for C1−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 bc term in the coefficient of qn of C1−b,j(q) and Hj(q) are as follows: [bc] [qn] C1−b,j(q) =

• λ⊢n
• a1+···+aj=c

j

• k=1

1 kak γ≥k ak

• [bc] [qn] Hj(q) =
• λ⊢n
• a1+···+aj=c

j

• k=1

1 k2 Hk ak

• 8 / 23

[bc][qn] Coefficients Proof

Proof. The bc coefficients in Hj(q) are given by the binomial theorem. For C1−b,j(q), we consider its expansion and manipulate it into a similar form. [qn] C1−b,j(q) =

• λ⊢n

1 − b 1 γ12 − b 2 γ2 . . . j − b j γ≥j =

• λ⊢n

(1 − b)γ1( 1 2!(1 − b)(2 − b))γ2 . . . ( 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 Applying the binomial theorem gives us our desired form.

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Constant and Linear Terms

Theorem 3 (Constant & Linear Term Equivalence). For any j, The qi terms of Hj(q) and C1−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)

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Constant and Linear Terms Proof

For the linear term, use the previous theorem’s expansion at c = 1:

• b1

[qn] C1−b,j(q) =

• λ⊢n

1 1γ≥1 + · · · + 1 j γ≥j

• b1

[qn] Hj(q) =

• λ⊢n

1 12 H1 + · · · + 1 j2 Hj Comparing these termwise, we check that for 0 ≤ i ≤ j,

• λ⊢n

1 i2 Hi

?

=

• λ⊢n

1 i γ≥i Hi(n) =

• λ⊢n

Hi

?

= i

• λ⊢n

γ≥i = i ∗ νi(n) The identity Hk(n) = k ∗ νk(n) is given by Bacher & Manivel in [4], so we are done.

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Quadratic Term, j = 2

Let’s proceed to the quadratic term for j = 2. From the Theorem 2 expansion, we get:

• b2

[qn] C1−b,2(q) =

• λ⊢n

γ≥1 2

• + 1

2 γ≥1 1 γ≥2 1

• + 1

4 γ≥2 2

• b2

[qn] H2(q) =

• λ⊢n

H1 2

• + 1

4 H1 1 H2 1

• + 1

16 H2 2

• Observationally, these are generally not equivalent.

Conjecturally, [b2][qn]H2(q) = [b2][qn](C1−b,2(q) + 1

16γ≥4(n)).

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Quadratic Term — Piecing it Together — γ≥1

2

• We know the following:
• λ⊢n

γ≥1 2

• =
• λ⊢n

H1 2

• as both H1 and γ≥1 count the same thing – the number of part

sizes in a partition.

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Quadratic Term — Piecing it Together — Mixed Term

We can show that

• λ⊢n

1 2 γ≥1 1 γ≥2 1

• =
• λ⊢n

1 4 H1 1 H2 1

• by a 2:1 bijection, i.e. for every pair (f1, f2) over the partitions of

n, there are two pairs of (h1, h2) (where fk is a multiplicity of k or more, hk is a hook of length k).

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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 (f1, f2). λ = 11 2 12 λ′ = 2′ 1′

2

1′

1

(f1, f2): (11, 2), (12, 2) (h1, h2): (11, 2), (12, 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.

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Quadratic Term — Piecing it Together — γ≥k

d

• Define Gk(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. Gk(q, u) =

∞

• i=1

(1 + qi + q2i + · · · + q(k−1)i + u(qki + q(k+1)i + . . . )) =

∞

• i=1

1 − qki 1 − qi + u qki 1 − qi

• = P(q)

∞

• i=1

(1 + (u − 1)qki)

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Quadratic Term — Piecing it Together — γ≥k

d

• It is known that

∞

• n=0

qn

λ⊢n

γ≥k d

• = 1

d! ∂d ∂ud Gk(q, u)

• u=1

. 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).

∞

• n=0

qn

λ⊢n

γ≥k d

• = 1

d! ∂d ∂ud Gk(q, u)

• u=1

= P(q)

d

• i=1

qik 1 − qik .

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Quadratic Term — Piecing it Together — Hk

2

• Define the following bivariate generating function:

Fk(q, u) =

∞

• n=0

∞

• a=0

qnua(number of partitions of n with a k-hooks) = P(q)

∞

• i=1

(1 + qki(u − 1))k Han verifies this to be the case in [2]. Repeated derivatives act the same, so we have the following for d = 2:

∞

• n=0

qn

λ⊢n

Hk 2

• = 1

2P(q)

• kqk

(1 − qk) 2 − kq2k 1 − q2k

• so we have

∞

• n=0

qn

λ⊢n

H2 2

• = P(q)

q4(1 + 3q2) (1 − q2)(1 − q4).

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Quadratic Equivalence

From here, we can see the following:

∞

• n=0

qn

λ⊢n

1 16 H2 2

• ?

=

∞

• n=0

qn(

• λ⊢n

1 4 γ≥2 2

• ) + 1

16γ≥4(n) P(q) q4(1 + 3q2) (1 − q2)(1 − q4)

?

= P(q)

• 4q6

(1 − q2)(1 − q4) + q4 1 − q4

• = P(q)

q4(1 + 3q2) (1 − q2)(1 − q4). Thus, we have just proved the following: Theorem 5 (Quadratic Equivalence). For all n and j = 2,

• b2

[qn] H2(q) = b2 [qn] C1−b,2(q)

• + 1

16γ≥4(n)b2.

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Further Work

If we increase j: Messier generating functions, counting many different hooklengths and frequencies. If we look at further order terms (increase bc): Counting few different hooklengths and frequencies, but selecting several of each in different ways.

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Further Work — A Route Forward

If we want to count multiple frequencies, consider the multivariate generating function GS(q, U) with S = {s1, s2, . . . , sm} and U = {u1, u2, . . . , um} s.t. ui < ui+1 and si < si+1, which marks multiplicities ≥ sk with uk: GS(q, U) =

∞

• i=1

(1 + qi + · · · + u1(qs1i + · · · + u2(qs2i + . . . ))) In the same fashion as before, differentiation with respect to uk will give us the moments of γ≥sk in the joint generating function:

∞

• n=0

qn

λ⊢n

γ≥s1 d1 γ≥s2 d2

• . . .

γ≥sm dm

• =

1 m

i=1 di!

∂d1+···+dm ∂ud1

1 . . . ∂udm m

JS(q, U)

• u∈U=1

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Further Work — A Route Forward

We have a bijection, in the smallest case, for comparing counts of multiple hooklengths and of γ≥k. (

λ⊢n H1H2 = 2 λ⊢n γ≥1γ≥2)

The question, then, is can we use similar mappings from GS(q, U) to joint generating functions of hooklengths? This would allow us to enumerate the latter.

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References

• C. Brennan, A. Knopfmacher, and S. Wagner, “The

distribution of ascents of size d or more in partitions of n,” Combinatorics, Probability and Computing, vol. 17,

• pp. 495–509, 2008.
• G. N. Han, “Discovering hook length formulas by an expansion

technique,” The Electronic Journal of Combinatorics, vol. 15,

• no. 1, p. 133, 2008.
• N. J. Fine, Basic hypergeometric series and applications.
• No. 27, 1988.
• R. Bacher and L. Manivel, “Hooks and powers of parts in

partitions,” S´

• em. Lothar. Combin., vol. 47, pp. Article B47d,

11 pp. (electronic), 2001/02.

• W. J. Keith, “Restricted k-color partitions,” The Ramanujan

Journal, vol. 40, pp. 71–92, May 2016.

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