SuperPlancherelsticexpialidocious a.k.a. The SuperPlancherel measure - PowerPoint PPT Presentation

SuperPlancherelsticexpialidocious supercharacter theory Examples 1 ( K , H ) = ( G / ∼ , Irr ( G )), trivial; 2 ( K , H ) = ( { 1 G , G \ { 1 G }} , { id , χ reg − id } ), trivial; 3 suppose A acts on G , φ : A → Aut ( G ) the superclasses are { φ ( A )([ g 1 ]) , . . . , φ ( A )([ g r ]) } but A acts also on Irr ( G ); call Ω 1 , . . . , Ω r the orbits, then the supercharacters are � χ (1) χ. χ ∈ Ω i Brauer This is a supercharacter theory

SuperPlancherelsticexpialidocious supercharacter theory (Bergeron and Thiem) A supercharacter theory for U n ( F q ) A = U n ( F q ) × U n ( F q ) × D n ( F q ) acts on U n ( F q ):

SuperPlancherelsticexpialidocious supercharacter theory (Bergeron and Thiem) A supercharacter theory for U n ( F q ) A = U n ( F q ) × U n ( F q ) × D n ( F q ) acts on U n ( F q ): φ ( g 1 , g 2 , t )( h ) = 1 + g 1 t ( h − 1) t − 1 g − 1 2

SuperPlancherelsticexpialidocious supercharacter theory (Bergeron and Thiem) A supercharacter theory for U n ( F q ) A = U n ( F q ) × U n ( F q ) × D n ( F q ) acts on U n ( F q ): φ ( g 1 , g 2 , t )( h ) = 1 + g 1 t ( h − 1) t − 1 g − 1 2   1 0 1 0 1 4 0 3 1 5 2 0 3 6 0     1 0 0 0 4 3      1 0 0 3 0    h = ,   1 0 4 0     1 0 0     1 0     1

SuperPlancherelsticexpialidocious supercharacter theory (Bergeron and Thiem) A supercharacter theory for U n ( F q ) A = U n ( F q ) × U n ( F q ) × D n ( F q ) acts on U n ( F q ): φ ( g 1 , g 2 , t )( h ) = 1 + g 1 t ( h − 1) t − 1 g − 1 2     1 0 1 0 1 4 0 3 1 0 · 0 ⋆ · · · · · · · · 1 5 2 0 3 6 0 1 ⋆         1 0 0 0 4 3 1 0 0 0 · ⋆          1 0 0 3 0   1 0 0 · 0      h = ,     1 0 4 0 1 0 ⋆ ·         1 0 0 1 0 0         1 0 1 0         1 1

SuperPlancherelsticexpialidocious supercharacter theory (Bergeron and Thiem) A supercharacter theory for U n ( F q ) A = U n ( F q ) × U n ( F q ) × D n ( F q ) acts on U n ( F q ): φ ( g 1 , g 2 , t )( h ) = 1 + g 1 t ( h − 1) t − 1 g − 1 2     1 0 1 0 1 4 0 3 1 0 · 0 ⋆ · · · · · · · · 1 5 2 0 3 6 0 1 ⋆         1 0 0 0 4 3 1 0 0 0 · ⋆          1 0 0 3 0   1 0 0 · 0      h = ,     1 0 4 0 1 0 ⋆ ·         1 0 0 1 0 0         1 0 1 0         1 1 π = 1 2 3 4 5 6 7 8

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects;

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects; 2 the supercharacters have an explicit formula;

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects; 2 the supercharacters have an explicit formula; 3 the supercharacters have rational values;

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects; 2 the supercharacters have an explicit formula; 3 the supercharacters have rational values; 4 the algebra of superclass functions is isomorphic to the algebra of symmetric functions in noncommutative variables;

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects; 2 the supercharacters have an explicit formula; 3 the supercharacters have rational values; 4 the algebra of superclass functions is isomorphic to the algebra of symmetric functions in noncommutative variables; m 13 / 24 ( x 1 , x 2 , . . . ) = x 1 x 2 x 1 x 2 + x 2 x 1 x 2 x 1 + x 1 x 3 x 1 x 3 + x 3 x 1 x 3 x 1 + x 2 x 3 x 2 x 3 + x 3 x 2 x 3 x 2 + . . . � = x i x j x i x j i � = j

SuperPlancherelsticexpialidocious supercharacter theory Why this supercharacter theory? 1 Superclasses (and supercharacters) are indexed by nice combinatorial objects; 2 the supercharacters have an explicit formula; 3 the supercharacters have rational values; 4 the algebra of superclass functions is isomorphic to the algebra of symmetric functions in noncommutative variables; m 13 / 24 ( x 1 , x 2 , . . . ) = x 1 x 2 x 1 x 2 + x 2 x 1 x 2 x 1 + x 1 x 3 x 1 x 3 + x 3 x 1 x 3 x 1 + x 2 x 3 x 2 x 3 + x 3 x 2 x 3 x 2 + . . . � = x i x j x i x j i � = j 5 Nice decomposition of the supercharacter table.

SuperPlancherelsticexpialidocious supercharacter theory Set partitions notation π = 1 2 3 4 5 6 7 8 Arcs( π ) = { (1 , 5) , (2 , 3) , (3 , 8) , (5 , 7) } ;

SuperPlancherelsticexpialidocious supercharacter theory Set partitions notation π = 1 2 3 4 5 6 7 8 Arcs( π ) = { (1 , 5) , (2 , 3) , (3 , 8) , (5 , 7) } ; a ( π ) = | Arcs( π ) | = 4;

SuperPlancherelsticexpialidocious supercharacter theory Set partitions notation π = 1 2 3 4 5 6 7 8 Arcs( π ) = { (1 , 5) , (2 , 3) , (3 , 8) , (5 , 7) } ; a ( π ) = | Arcs( π ) | = 4; dim( π ) = � j − i = 12; ( i , j ) ∈ Arcs( π )

SuperPlancherelsticexpialidocious supercharacter theory Set partitions notation π = 1 2 3 4 5 6 7 8 Arcs( π ) = { (1 , 5) , (2 , 3) , (3 , 8) , (5 , 7) } ; a ( π ) = | Arcs( π ) | = 4; dim( π ) = � j − i = 12; ( i , j ) ∈ Arcs( π ) crs ( π ) = ♯ crossings of π = 1;

SuperPlancherelsticexpialidocious supercharacter theory The dimension of a supercharacter is χ π (1) = ( q − 1) a ( π ) · q dim( π ) − a ( π ) ;

SuperPlancherelsticexpialidocious supercharacter theory The dimension of a supercharacter is χ π (1) = ( q − 1) a ( π ) · q dim( π ) − a ( π ) ; � χ π , χ π � = ( q − 1) a ( π ) · q crs ( π ) ;

SuperPlancherelsticexpialidocious supercharacter theory The dimension of a supercharacter is χ π (1) = ( q − 1) a ( π ) · q dim( π ) − a ( π ) ; � χ π , χ π � = ( q − 1) a ( π ) · q crs ( π ) ; The superplancherel measure

SuperPlancherelsticexpialidocious supercharacter theory The dimension of a supercharacter is χ π (1) = ( q − 1) a ( π ) · q dim( π ) − a ( π ) ; � χ π , χ π � = ( q − 1) a ( π ) · q crs ( π ) ; The superplancherel measure χ (1) 2 SPl G ( χ ) = 1 | G | � χ, χ �

SuperPlancherelsticexpialidocious supercharacter theory The dimension of a supercharacter is χ π (1) = ( q − 1) a ( π ) · q dim( π ) − a ( π ) ; � χ π , χ π � = ( q − 1) a ( π ) · q crs ( π ) ; The superplancherel measure ( q − 1) a ( π ) · q 2 dim( π ) − 2 a ( π ) χ (1) 2 SPl G ( χ ) = 1 1 � χ, χ � = n ( n − 1) q crs ( π ) | G | q 2

SuperPlancherelsticexpialidocious supercharacter theory Plan 1 See set partitions as objects of the same space

SuperPlancherelsticexpialidocious supercharacter theory Plan 1 See set partitions as objects of the same space (some renormalization happens);

SuperPlancherelsticexpialidocious supercharacter theory Plan 1 See set partitions as objects of the same space (some renormalization happens); 2 interpret your statistics w.r.t. this new setting;

SuperPlancherelsticexpialidocious supercharacter theory Plan 1 See set partitions as objects of the same space (some renormalization happens); 2 interpret your statistics w.r.t. this new setting; 3 let n → ∞ ;

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit →

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → →

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → → →

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → → →  1 0 · 0 ⋆ · · ·  1 ⋆ · · · · ·     1 0 0 0 · ⋆      1 0 0 · 0      1 0 ⋆ ·     1 0 0     1 0     1

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → → →  1 0 · 0 ⋆ · · ·                · · ⋆ 0 · 1 0 0 1 0 1 1 ⋆ · · · · ·   · · · · 1 0 ⋆   1 0 0 0 · ⋆   · · 1 0 0 0 1 0 0    1 0 0 · 0  → 0 ⋆ ·     1 0 ⋆ ·   ·   1 0 0   · 1 ⋆   1 0   1 0   1              

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → → → 1  1 0 · 0 ⋆ · · ·                · · ⋆ 0 · 1 0 0 1 0 1 1 ⋆ · · · · ·   · · · · 1 0 ⋆   1 0 0 0 · ⋆   · · 1 0 0 0 1 0 0    1 0 0 · 0  → → 0 ⋆ ·     1 0 ⋆ ·   ·   1 0 0   · 1 ⋆   1 0   1 0   1 0               1

SuperPlancherelsticexpialidocious supercharacter theory First step to look for a limit → → → 1  1 0 · 0 ⋆ · · ·                · · ⋆ 0 · 1 0 0 1 0 1 1 ⋆ · · · · ·   · · · · 1 0 ⋆   1 0 0 0 · ⋆   · · 1 0 0 0 1 0 0    1 0 0 · 0  → → → ? 0 ⋆ ·     1 0 ⋆ ·   ·   1 0 0   · 1 ⋆   1 0   1 0   1 0               1

SuperPlancherelsticexpialidocious supercharacter theory Our setting 1 ∆ = 0 1

SuperPlancherelsticexpialidocious supercharacter theory Our setting 1   measures µ on ∆ s.t.     ∆ = Γ = � ∆ µ ≤ 1 ( subprobability )  µ has sub-uniform marginals    0 1

SuperPlancherelsticexpialidocious supercharacter theory Our setting 1   measures µ on ∆ s.t.     ∆ = Γ = � ∆ µ ≤ 1 ( subprobability )  µ has sub-uniform marginals    0 1 1 d c 0 a b 1

SuperPlancherelsticexpialidocious supercharacter theory Our setting 1   measures µ on ∆ s.t.     ∆ = Γ = � ∆ µ ≤ 1 ( subprobability )  µ has sub-uniform marginals    0 1 1 d � y =1 � x = b d µ = b − a UNIFORM c y =0 x = a 0 a b 1

SuperPlancherelsticexpialidocious supercharacter theory Our setting 1   measures µ on ∆ s.t.     ∆ = Γ = � ∆ µ ≤ 1 ( subprobability )  µ has sub-uniform marginals    0 1 1 d � y =1 � x = b d µ = b − a UNIFORM c y =0 x = a � x =1 � y = d d µ ≤ d − c SUB-UNIFORM x =0 y = c 0 a b 1

SuperPlancherelsticexpialidocious supercharacter theory Theorem (DDS) There exists a measure Ω ∈ Γ such that µ π → Ω almost surely.

SuperPlancherelsticexpialidocious supercharacter theory Theorem (DDS) There exists a measure Ω ∈ Γ such that µ π → Ω almost surely.

SuperPlancherelsticexpialidocious supercharacter theory Theorem (DDS) There exists a measure Ω ∈ Γ such that µ π → Ω almost surely. n 1 2 3

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: a ( π ) π = 1 2 3 4 5 6 7 8

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: a ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: a ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 Each box has mass 1 n , there are a ( π ) boxes, ⇒ a ( π ) =

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: a ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 Each box has mass 1 n , there are a ( π ) boxes, ⇒ � a ( π ) = n d µ π ( x , y ) ∆

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: dim( π ) π = 1 2 3 4 5 6 7 8

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: dim( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: dim( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 � dim( π ) = j − i ( i , j ) ∈ Arcs( π )

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: dim( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 � � j − i = n 2 dim( π ) = ( y − x ) d µ π ( x , y ) ∆ ( i , j ) ∈ Arcs( π )

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: dim( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 � � j − i = n 2 dim( π ) = ( y − x ) d µ π ( x , y ) ∆ ( i , j ) ∈ Arcs( π ) y ∼ 1 x ∼ 1 nj ni each box has mass 1 n

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: crs ( π ) π = 1 2 3 4 5 6 7 8

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: crs ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: crs ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 crs ( π ) = ♯ { ( i 1 , j 1 ) , ( i 2 , j 2 ) ∈ Arcs( π ) s.t. i 1 < i 2 < j 1 < j 2 }

SuperPlancherelsticexpialidocious supercharacter theory Interpretation of statistics: crs ( π ) 1 π = µ π = 1 2 3 4 5 6 7 8 0 1 crs ( π ) = ♯ { ( i 1 , j 1 ) , ( i 2 , j 2 ) ∈ Arcs( π ) s.t. i 1 < i 2 < j 1 < j 2 } � = n 2 ∆ 2 1 [ x 1 < x 2 < y 1 < y 2 ] d µ π ( x 1 , y 1 ) d µ π ( x 2 , y 2 ) + O ( n )

SuperPlancherelsticexpialidocious supercharacter theory q 2 dim( π ) − 2 a ( π ) 1 SPl n ( χ π ) = ( q − 1) a ( π ) q crs ( π ) = n ( n − 1) q 2

SuperPlancherelsticexpialidocious supercharacter theory q 2 dim( π ) − 2 a ( π ) 1 SPl n ( χ π ) = ( q − 1) a ( π ) q crs ( π ) = n ( n − 1) q 2 � � − n 2 �� 2 + log( q − 1) 2 + n a ( π ) − 2 a ( π ) + 2 dim( π ) − crs ( π ) = exp log q log q

SuperPlancherelsticexpialidocious supercharacter theory q 2 dim( π ) − 2 a ( π ) 1 SPl n ( χ π ) = ( q − 1) a ( π ) q crs ( π ) = n ( n − 1) q 2 � � − n 2 �� 2 + log( q − 1) 2 + n a ( π ) − 2 a ( π ) + 2 dim( π ) − crs ( π ) = exp log q log q � 1 � � � − n 2 log q exp 2 − 2 I dim ( µ π ) + I crs ( µ π ) + O ( n )

SuperPlancherelsticexpialidocious supercharacter theory � 1 � � � − n 2 log q exp 2 − 2 I dim ( µ π ) + I crs ( µ π ) + O ( n )

SuperPlancherelsticexpialidocious supercharacter theory � 1 � � � − n 2 log q exp 2 − 2 I dim ( µ π ) + I crs ( µ π ) + O ( n ) H ( µ ) := 1 2 − 2 I dim ( µ ) + I crs ( µ )

SuperPlancherelsticexpialidocious supercharacter theory � 1 � � � − n 2 log q exp 2 − 2 I dim ( µ π ) + I crs ( µ π ) + O ( n ) H ( µ ) := 1 2 − 2 I dim ( µ ) + I crs ( µ ) IDEA Find Ω s.t. H (Ω) = 0 ( ⇒ Ω candidate limit shape)

SuperPlancherelsticexpialidocious supercharacter theory Playing with I dim ( µ ) = ∆ ( y − x ) d µ � Proposition φ ( µ ) has uniform marginals φ : Γ → Γ s.t.

SuperPlancherelsticexpialidocious supercharacter theory Playing with I dim ( µ ) = ∆ ( y − x ) d µ � Proposition φ ( µ ) has uniform marginals φ : Γ → Γ s.t. I dim ( φ ( µ )) ≥ I dim ( µ )

SuperPlancherelsticexpialidocious supercharacter theory Playing with I dim ( µ ) = ∆ ( y − x ) d µ � Proposition φ ( µ ) has uniform marginals φ : Γ → Γ s.t. I dim ( φ ( µ )) ≥ I dim ( µ ) I dim ( φ ( µ )) = I dim ( µ ) ⇔ µ has uniform marginals

SuperPlancherelsticexpialidocious supercharacter theory Playing with I dim ( µ ) = ∆ ( y − x ) d µ � Proposition φ ( µ ) has uniform marginals φ : Γ → Γ s.t. I dim ( φ ( µ )) ≥ I dim ( µ ) I dim ( φ ( µ )) = I dim ( µ ) ⇔ µ has uniform marginals �→ µ φ ( µ )