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 − tq i One interpretation [ t k q n ] P ( t, q ) is the number of partitions of n into exactly k parts. ( t + 2 t 2 + t 3 + t 4 ) q 4 + · · · . P ( t, q ) = · · · +

1 P ( t, q ) = � i ≥ 1 1 − tq i Another interpretation [ t k q n ] P ( t, q ) is the number of partitions of n with largest part of size k .

Goal Complete “ [ t k q n ] P ( t, q ) is the number of partitions of n . . . ” in infinitely many ways.

Head, shoulders, knees & toes a(c) c l(c)

Head, shoulders, knees & toes a(c) 1 = slope l(c)+1 a(c)+1 c = a(c)+1 1 slope l(c) l(c)

h ± x ( λ ) For x ∈ [0 , ∞ ) , y ∈ (0 , ∞ ] , l ( c ) + 1 ≤ x < a ( c ) + 1 a ( c ) � � � h + x ( λ ) = χ , l ( c ) c ∈ λ l ( c ) + 1 < y ≤ a ( c ) + 1 a ( c ) � � � h − y ( λ ) = χ . l ( c ) c ∈ λ

h + x (3 , 2) example [1,3) [1/2,2) [0, ) [1, ) [0, ) 0 1/2 1 2 3

h + 0 example 0 1/4 1/3 1/2 2/3 1 3/2 2 3 4 x=0 t ℓ ( λ ) q n = · · · + ( t + 2 t 2 + 2 t 3 + t 4 + t 5 ) q 5 + · · · � λ ⊢ n ≥ 0

h − ∞ example 0 1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = t λ 1 q n = · · · + ( t + 2 t 2 + 2 t 3 + t 4 + t 5 ) q 5 + · · · � λ ⊢ n ≥ 0

h + π example 0 1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = pi π ( λ ) q n = · · · + ( t + 2 t 2 + 2 t 3 + t 4 + t 5 ) q 5 + · · · t h + � λ ⊢ n ≥ 0

Theorem[Haiman (ca. 2000); Loehr-W] For x ∈ [0 , ∞ ) , “ [ t k q n ] P ( t, q ) is the number of partitions λ ⊢ n for which h + x ( λ ) = k .”

Theorem[Haiman (ca. 2000); Loehr-W] For x ∈ [0 , ∞ ) , y ∈ (0 , ∞ ] , 1 � � t ℓ ( λ ) q n 1 − tq i = i ≥ 1 λ ⊢ n ≥ 0 t h + � x ( λ ) q n = λ ⊢ n ≥ 0 t h − � y ( λ ) q n . = λ ⊢ n ≥ 0

A rephrasing For x ∈ [0 , ∞ ] , δ ∈ { + , −} , t h δ � Define H δ x ( λ ) . x ( n ; t ) = λ ⊢ n

A rephrasing For x ∈ [0 , ∞ ] , δ ∈ { + , −} , t h δ � Define H δ x ( λ ) . x ( n ; t ) = λ ⊢ n Show For fixed n , H δ x ( n ; t ) is independent of x, δ .

h + 1 example 0 1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = 1 1 ( λ ) q n = · · · + ( t + 2 t 2 + 2 t 3 + t 4 + t 5 ) q 5 + · · · t h + � λ ⊢ n ≥ 0

h + 2 example √ 0 1/4 1/3 1/2 2/3 1 3/2 2 3 4 x = 2 t h + 2 ( λ ) q n = · · · + ( t + 2 t 2 + 2 t 3 + t 4 + t 5 ) q 5 + · · · � √ λ ⊢ n ≥ 0

Proof: A combinatorial homotopy H+ = H − H+ = H − x = 0 x = r/s x = r’/s’

Main Lemma [Loehr-W] For all positive rational r/s and all integers n ≥ 0 , t h + r/s ( λ ) � H + r/s ( n ; t ) = λ ⊢ n t h − r/s ( λ ) = H − � = r/s ( n ; t ) . λ ⊢ n

à la J. Sjöstrand 6 4 2 0 = −2 3 x = 3/2 4 = +3 2 0 5 3 1 10 4 2 N 7 x+y=5 5 3 1 10 8 6 4 7 5 7 NN EN 4 10 8 6 4 2 0 3 1 E EN NENNE EEE EE 0 x+y = 0 5 3 1 NEEN EEE 0

Dependencies Write h + r/s = m r/s + c + r/s , h − r/s = m r/s + c − r/s . c − r/s m r/s c + r/s | λ | Only on graph X X On Eulerian tour X X

Arrival tree 10 8 6 N E EN 7 5 NN EN 4 2 NENNE EEE 3 1 NEEN EEE 0

c ± r/s → c ∓ r/s involution 10 8 6 N E EN 7 5 NN EN NE 4 2 NENNE EEE 3 1 NEEN EEE NNEE 0

0 ∞ 5 1/4 4 1/3 3 1/2 2 2/3 3/2 1

Rephrasing the statistics � area ( M ) = A max ( r, s, n ) − ⌊ v/s ⌋ N out ( v, M ); v ∈ V M � mid ( M ) = A max ( r, s, n ) − E in ( v, M ) N in ( w, M ) χ ( v ≥ w ); v,w ∈ V M � ctot( M ) = E in ( v, M ) N in ( v, M ) − ( n − E in (0 , M )) . v ∈ V M Theorem For any µ ∈ Par r,s,n , we have: c + c − � � inv( w v ( µ )) , inv( y v ( µ )); r/s ( µ ) = r/s ( µ ) = v ∈ V M v ∈ V M | µ | = area ( M ) , mid r/s ( µ ) = mid ( M ) , ctot r/s ( µ ) = ctot( M ) .