Intro Problems
The ranks of unimodal sequences Byungchan Kim SeoulTech
KIAS Combinatorics Workshop May 31 2014
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Intro Problems
Partitions
Definition. A partition of a positive integer n is a weakly decreasing sequence of positive integers λ1 , λ2 , . . . , λk such that Pk i =1 λi = n. The λi are called the parts of the partition. Definition. p (n) = the number of partitions of n. For convenience, we define p (0) = 1. Example. p (4) = 5 because 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1.
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Intro Problems
Generating function and modularity • Euler ∞ X
p (n)qn =
X
q1n1 +2n2 +3n3 +···+knk +···+`n`
n =0
=
∞ X
q n1
n1 = 0
= =
∞ X n2 =0
1
1
1 − q 1 − q2 ∞ Y n =1
q2n2 · · ·
∞ X
qknk · · ·
nk = 0
···
1 1 − qn
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Intro Problems
Arithmetic of partition function
Theorem (Ramanujan, 1920) p (5n + 4) ≡ 0 (mod 5) p (7n + 5) ≡ 0 (mod 7) p (11n + 6) ≡ 0 (mod 11)
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Intro Problems
Arithmetic of partition function
Theorem (Ramanujan, 1920) p (5n + 4) ≡ 0 (mod 5) p (7n + 5) ≡ 0 (mod 7) p (11n + 6) ≡ 0 (mod 11)
• A natural question Can we explain these partition congruences combinatorially?
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Intro Problems
Dyson’s rank
Definition the rank of partition λ = λ1 − `(λ), where λ1 is the largest part of λ and `(λ) is the number of parts of λ.
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Intro Problems
Dyson’s rank
Definition the rank of partition λ = λ1 − `(λ), where λ1 is the largest part of λ and `(λ) is the number of parts of λ. Partitions of 4 4 3+1 2+2 2+1+1 1+1+1+1
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Intro Problems
Dyson’s rank
Definition the rank of partition λ = λ1 − `(λ), where λ1 is the largest part of λ and `(λ) is the number of parts of λ. Partitions of 4 4 3+1 2+2 2+1+1 1+1+1+1
rank 3 1 0 -1 -3
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Intro Problems
Dyson’s rank
Definition the rank of partition λ = λ1 − `(λ), where λ1 is the largest part of λ and `(λ) is the number of parts of λ. Partitions of 4 4 3+1 2+2 2+1+1 1+1+1+1
rank 3 1 0 -1 -3
rank (mod 5) 3 1 0 4 2
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Intro Problems
Dyson’s conjecture
Conjecture (1944) Let N (i , p , n) be the number of partitions of n with rank ≡ i (mod p ), then 1 p (5n + 4), 5 1 N (i , 7, 7n + 5) = p (7n + 5) 7 N (i , 5, 5n + 4) =
for all i.
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Intro Problems
Dyson’s conjecture
Conjecture (1944) Let N (i , p , n) be the number of partitions of n with rank ≡ i (mod p ), then 1 p (5n + 4), 5 1 N (i , 7, 7n + 5) = p (7n + 5) 7 N (i , 5, 5n + 4) =
for all i. Theorem (Atkin and Swinnerton-Dyer(1954)) Dyson’s conjecture is true.
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Intro Problems
Generating function and mock modularity
R (z , q) =
∞ X ∞ X
N (m, n)z m qn
n=0 m=−∞
=
∞ X
2
qn , k k k =1 (1 − zq )(1 − q /z )
Qn n=0
where N (m, n) denote the number of partitions of n with rank m.
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Intro Problems
Generating function and mock modularity
R (z , q) =
∞ X ∞ X
N (m, n)z m qn
n=0 m=−∞
=
∞ X
2
qn , k k k =1 (1 − zq )(1 − q /z )
Qn n=0
where N (m, n) denote the number of partitions of n with rank m.
• Ramanujan’s mock theta function f (q) f (q) = R (−1, q)
• R (ζn , q) is a mock modular form. 7 / 21
Intro Problems
(Strong) Unimodal sequences
• (Strong) Unimodal sequences are sequences of the form a1 < a2 < · · · < ar < c > b1 > b2 > · · · > bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
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Intro Problems
(Strong) Unimodal sequences
• (Strong) Unimodal sequences are sequences of the form a1 < a2 < · · · < ar < c > b1 > b2 > · · · > bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
• Weight 5 unimodal sequences are (5), (1, 4), (4, 1), (1, 3, 1), (2, 3), (3, 2)
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Intro Problems
(Strong) Unimodal sequences
• (Strong) Unimodal sequences are sequences of the form a1 < a2 < · · · < ar < c > b1 > b2 > · · · > bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
• Weight 5 unimodal sequences are (5), (1, 4), (4, 1), (1, 3, 1), (2, 3), (3, 2) • The rank of a unimodal sequence is s − r. Ranks are respectively are 0,-1,1,0,-1,1
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Intro Problems
Generating function and quantum modularity
U (z , q) =
∞ X ∞ X
U (m , n )z m q n
n=0 m=−∞
=
∞ X n=0
q
n+1
n Y
(1 + zqk )(1 + qk /z ),
k =1
where U (m, n) denote the number of strong unimodal sequences of weight n with rank m.
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Intro Problems
Generating function and quantum modularity
U (z , q) =
∞ X ∞ X
U (m , n )z m q n
n=0 m=−∞
=
∞ X
q
n+1
n=0
n Y
(1 + zqk )(1 + qk /z ),
k =1
where U (m, n) denote the number of strong unimodal sequences of weight n with rank m.
• Kontsevich’s strange function F (q ) =
n XY
(1 − qk )
n≥0 k =1
• F (q−1 ) = U (−1; q). 9 / 21
Intro Problems
Unimodal sequences
• Unimodal sequences are sequences of the form a1 ≤ a2 ≤ · · · ≤ ar ≤ c ≥ b1 ≥ b2 ≥ · · · ≥ bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
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Intro Problems
Unimodal sequences
• Unimodal sequences are sequences of the form a1 ≤ a2 ≤ · · · ≤ ar ≤ c ≥ b1 ≥ b2 ≥ · · · ≥ bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
• Weight 4 unimodal sequences are (4), (1, 3), (3, 1), (1, 2, 1), (2, 2), (2, 2), (1, 1, 2), (2, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1).
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Intro Problems
Unimodal sequences
• Unimodal sequences are sequences of the form a1 ≤ a2 ≤ · · · ≤ ar ≤ c ≥ b1 ≥ b2 ≥ · · · ≥ bs with weight n = c +
Pr i =1
ai +
Ps i =1
bi .
• Weight 4 unimodal sequences are (4), (1, 3), (3, 1), (1, 2, 1), (2, 2), (2, 2), (1, 1, 2), (2, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1), (1, 1, 1, 1). • The rank of a unimodal sequence is s − r.
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Intro Problems
Wright’s work (1968-1972)
• Generating functions ∞ X n=0
u (n )q n =
∞ X
qn k 2 k =1 (1 − q )
Qn n =0
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Intro Problems
Wright’s work (1968-1972)
• Generating functions ∞ X n=0
u (n )q n =
∞ X
qn k 2 k =1 (1 − q )
Qn n =0
• E. M. Wright obtained asymptotic formula for u(n). ∞ X
∞ ∞ Y X 1 qn = (−1)n qn(n+1)/2 n )2 k )2 ( 1 − q ( 1 − q k =1 n =1 n =0
Qn n =0
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Intro Problems
Unimodal rank and partial theta functions • Generating function X m,n
u(m, n)x m qn =
X
qn k k k =1 (1 − xq )(1 − q /x )
Qn n ≥0
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Intro Problems
Unimodal rank and partial theta functions • Generating function X
u(m, n)x m qn =
m,n
X
qn k k k =1 (1 − xq )(1 − q /x )
Qn n ≥0
Another generating function from Ramanujan’s lost notebook, n +1 P (−1)n x 2n+1 q( 2 ) Q∞n≥0 n n n=1 (1 − xq )(1 − q /x ) X + (1 − x ) (−1)n x 3n qn(3n+1)/2 (1 − x 2 q2n+1 )
n ≥0
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Intro Problems
Unimodal rank and partial theta functions • Generating function X
u(m, n)x m qn =
m,n
X
qn k k k =1 (1 − xq )(1 − q /x )
Qn n ≥0
Another generating function from Ramanujan’s lost notebook, n +1 P (−1)n x 2n+1 q( 2 ) Q∞n≥0 n n n=1 (1 − xq )(1 − q /x ) X + (1 − x ) (−1)n x 3n qn(3n+1)/2 (1 − x 2 q2n+1 )
n ≥0
• Partial theta functions 12 / 21
Intro Problems
unimodal sequence? • From the lost notebook, X (qn+1 )n qn n ≥0
(xq, q/x )n
= (1 − x )
X
x n qn
2
+n
n ≥0
X 2 x x 3n q3n +2n (1 − xq2n+1 ). + (xq, q/x )∞ n≥0 • (a ; q)n =
n Y
(1 − aqk −1 )
k =1
• Define v (m, n) by X m ,n
v (m, n)x m qn =
X (qn+1 )n qn n ≥0
(xq, q/x )n 13 / 21
Intro Problems
Non-negativity • q-Chu-Vandermonde summation n X ( a , q −n ) k k (c /a )n n q = a (c , q)k (c )n k =0
• From q-Chu-Vandermonde, 2 " # n X (qn+1 )n n x k qk n 1 n q =q , (xq, q/x )n (xq)k k (q/x )n−k k =0
where
" #
(q)n n = k (q)k (q)n−k
is a q-binomial coefficient. 14 / 21
Intro Problems
Unimodality 2 " # n X (qn+1 )n n x k qk n 1 n q =q . (xq, q/x )n ( xq)k k (q/x )n−k k =0
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Intro Problems
Unimodality 2 " # n X (qn+1 )n n x k qk n 1 n q =q . (xq, q/x )n ( xq)k k (q/x )n−k k =0
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Intro Problems
Symmetry
• Symmetry of ranks N (m, n) = N (−m, n), U (m, n) = U (−m, n), u(m, n) = u(−m, n)
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Intro Problems
Symmetry
• Symmetry of ranks N (m, n) = N (−m, n), U (m, n) = U (−m, n), u(m, n) = u(−m, n)
• From the generating function, v (m, n) = v (−m, n). However, a natural combinatorial conjugation is desired.
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Intro Problems
Unimodality of Ranks • Numerical evidence suggests that for n fixed and large enough, the sequences {N (m, n)}m≥0 , {u(m, n)}m≥0 , and {v (m, n)}m≥0 are decreasing.
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Intro Problems
Unimodality of Ranks • Numerical evidence suggests that for n fixed and large enough, the sequences {N (m, n)}m≥0 , {u(m, n)}m≥0 , and {v (m, n)}m≥0 are decreasing.
• Asymptotically, there are progresses in this direction. For example, r r !5/2 !7/2 1 π 2n 17 π 2n N (m, n) ∼ √ I−5/2 π I−7/2 π + √ √ 4 3 96 3 6n 6n 2π r ! !9/2 1237 m2 π 2n + − I−9/2 π √ 4608 16 3 6n 1
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Intro Problems
Combinatorial proofs? • K-Lovejoy (2014+) 2 " # n X (qn+1 )n n x k qk n 1 n q =q . (xq, q/x )n ( xq)k k (q/x )n−k k =0
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Intro Problems
Combinatorial proofs? • K-Lovejoy (2014+) 2 " # n X (qn+1 )n n x k qk n 1 n q =q . (xq, q/x )n ( xq)k k (q/x )n−k k =0
• F. Garvan (1988) ∞ X N (m, n)qn = n=0
∞ 1 X (−1)n−1 qn(3n−1)/2+mn (1 − qn ), (q)∞ n=1
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Intro Problems
Combinatorial proofs? • K-Lovejoy (2014+) 2 " # n X (qn+1 )n n x k qk n 1 n q =q . (xq, q/x )n ( xq)k k (q/x )n−k k =0
• F. Garvan (1988) ∞ X N (m, n)qn = n=0
∞ 1 X (−1)n−1 qn(3n−1)/2+mn (1 − qn ), (q)∞ n=1
• K.-Lovejoy (2014) X
u(m, n)qn =
n ≥0
δm,0 +
−1 X (−1)n+r qn(n+1)/2+r (r +1)/2+2rn+|m|r (1 − qr ). (q)2∞ r ,n≥0 18 / 21
Intro Problems
Combinatorial statistic
Theorem (Lovejoy-K.) u(1, 5, 5n + 3) = u(2, 5, 5n + 3) u(0, 5, 5n + 4) = u(2, 5, 5n + 4)
U2 (7n + 6) ≡ 0 (mod 7) Here U2 (n) =
P
m
m2 u(m, n) is the second rank moment.
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Intro Problems
Combinatorial statistic
Theorem (Lovejoy-K.) u(1, 5, 5n + 3) = u(2, 5, 5n + 3) u(0, 5, 5n + 4) = u(2, 5, 5n + 4)
U2 (7n + 6) ≡ 0 (mod 7) Here U2 (n) =
P
m
m2 u(m, n) is the second rank moment.
• An injection for the equality? • A combinatorial statistic which explains the congruence?
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Intro Problems
A positivity conjecture ∞ X
(Ak (n) − Bk (n))qn
n =1
=
∞ Y 1 + qn X
1−
n
−
qn
q2n
2
−5n+4nk −2k +2
2
−n
)(1 − q4n+2k −2 )
n =1
∞ Y 1 + qn X n
(1 − q2n
1−
qn
q2n
2
−3n+4nk
(1 − q2n
2
+n
)(1 − q4n+2k )
n =1
• Andrews-Chan-K.-Osburn conjectured that Ak (n) > Bk (n)
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Intro Problems
A positivity conjecture ∞ X
(Ak (n) − Bk (n))qn
n =1
=
∞ Y 1 + qn X
1−
n
−
qn
q2n
2
−5n+4nk −2k +2
2
−n
)(1 − q4n+2k −2 )
n =1
∞ Y 1 + qn X n
(1 − q2n
1−
qn
q2n
2
−3n+4nk
(1 − q2n
2
+n
)(1 − q4n+2k )
n =1
• Andrews-Chan-K.-Osburn conjectured that Ak (n) > Bk (n)
• More generally, find a combinatorial model for the coefficients of 1 A linear combination of partial theta functions (q)∞ 20 / 21
Intro Problems
Thank you!
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