SLIDE 1
Infinitely many new partition statistics
Greg Warrington
The University of Vermont AMS Sectional Meeting, Penn State October 25, 2009
Joint with Nick Loehr (Virginia Tech)
Infinitely many new partition statistics Greg Warrington The - - PowerPoint PPT Presentation
Infinitely many new partition statistics Greg Warrington The - - PowerPoint PPT Presentation
Infinitely many new partition statistics Greg Warrington The University of Vermont AMS Sectional Meeting, Penn State October 25, 2009 Joint with Nick Loehr (Virginia Tech) 1 P ( t, q ) = i 1 1 tq i 1 P ( t, q ) = i 1 1
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P (t, q) =
i≥1 1 1−tqi
One interpretation [tkqn]P (t, q) is the number
- f partitions of n into exactly k parts.
P (t, q) = · · · + (t + 2t2 + t3 + t4) q4 + · · · .
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P (t, q) =
i≥1 1 1−tqi
Another interpretation [tkqn]P (t, q) is the number
- f partitions of n with largest part of size k.
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Goal
Complete “[tkqn]P (t, q) is the number
- f partitions of n. . . ”
in infinitely many ways.
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Head, shoulders, knees & toes
a(c) l(c) c
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Head, shoulders, knees & toes
l(c) a(c)+1 c l(c) slope 1 = a(c)+1 slope 1 = a(c) l(c)+1
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h±
x (λ)
For x ∈ [0, ∞), y ∈ (0, ∞],
h+
x (λ) =
- c∈λ
χ
- a(c)
l(c) + 1 ≤ x < a(c) + 1 l(c)
- ,
h−
y (λ) =
- c∈λ
χ
- a(c)
l(c) + 1 < y ≤ a(c) + 1 l(c)
- .
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h+
x (3, 2) example
1/2 2 1 3 [1,3) [1/2,2) [0, ) [1, ) [0, )
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h+
0 example
1/4 1/3 1/2 2/3 1 3/2 2 3 4 x=0
- λ⊢n≥0
tℓ(λ)qn = · · · + (t + 2t2 + 2t3 + t4 + t5)q5 + · · ·
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h−
∞ example
1/4 1/3 1/2 2/3 1 3/2 2 3 4 x =
- λ⊢n≥0
tλ1qn = · · · + (t + 2t2 + 2t3 + t4 + t5)q5 + · · ·
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h+
π example
1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = pi
- λ⊢n≥0
th+
π (λ)qn = · · · + (t + 2t2 + 2t3 + t4 + t5)q5 + · · ·
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Theorem[Haiman (ca. 2000); Loehr-W]
For x ∈ [0, ∞), “[tkqn]P (t, q) is the number
- f partitions λ ⊢ n
for which h+
x (λ) = k.”
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Theorem[Haiman (ca. 2000); Loehr-W]
For x ∈ [0, ∞), y ∈ (0, ∞],
- i≥1
1 1 − tqi =
- λ⊢n≥0
tℓ(λ)qn =
- λ⊢n≥0
th+
x (λ)qn
=
- λ⊢n≥0
th−
y (λ)qn.
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A rephrasing
For x ∈ [0, ∞], δ ∈ {+, −}, Define Hδ
x(n; t) =
- λ⊢n
thδ
x(λ).
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A rephrasing
For x ∈ [0, ∞], δ ∈ {+, −}, Define Hδ
x(n; t) =
- λ⊢n
thδ
x(λ).
Show For fixed n, Hδ
x(n; t) is independent of x, δ.
SLIDE 17
h+
1 example
1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = 1
- λ⊢n≥0
th+
1 (λ)qn = · · · + (t + 2t2 + 2t3 + t4 + t5)q5 + · · ·
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h+
√ 2 example
1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = 2
- λ⊢n≥0
th+
√ 2(λ)qn = · · · + (t + 2t2 + 2t3 + t4 + t5)q5 + · · ·
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Proof: A combinatorial homotopy
H − H+ H − H+ x = 0 x = r/s x = r’/s’
= =
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Main Lemma [Loehr-W]
For all positive rational r/s and all integers n ≥ 0, H+
r/s(n; t) =
- λ⊢n
th+
r/s(λ)
=
- λ⊢n
th−
r/s(λ) = H−
r/s(n; t).
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à la J. Sjöstrand
3 4 7 10 8 6 4 7 5 1 3 1 2 3 1 2 4 2 6 3 4 5 4 = −2 = +3 5 6 4 2 7 3 1 5 10 10 8 x+y = 0
NN N EN EEE NENNE EEE NEEN EN E EE
x+y=5 x = 3/2
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Dependencies
Write h+
r/s = mr/s + c+ r/s,
h−
r/s = mr/s + c− r/s.
c−
r/s mr/s c+ r/s |λ|
Only on graph X X On Eulerian tour X X
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Arrival tree
5 4 2 7 3 1 10
NN N EN EEE NENNE EEE NEEN
8
E
6
EN
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c±
r/s → c∓ r/s involution
5 4 2 7 3 1 10
NN N EN EEE NENNE EEE NEEN
8
E
6
EN NNEE NE
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1/4 1/3 1/2 2/3 1 3/2 2 3 4 5 ∞
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Rephrasing the statistics
area(M)
= Amax(r, s, n) −
- v∈VM
⌊v/s⌋Nout(v, M);
mid(M)
= Amax(r, s, n) −
- v,w∈VM
Ein(v, M)Nin(w, M)χ(v ≥ w); ctot(M) =
- v∈VM
Ein(v, M)Nin(v, M) − (n − Ein(0, M)). Theorem For any µ ∈ Parr,s,n, we have: c+
r/s(µ) =
- v∈VM
inv(wv(µ)), c−
r/s(µ) =
- v∈VM
inv(yv(µ)); |µ| = area(M),
midr/s(µ) = mid(M),
ctotr/s(µ) = ctot(M).