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Stuttering multipartitions and blocks of ArikiKoike algebras Salim Rostam Univ Rennes 17/04/2019 82nd Sminaire Lotharingien de Combinatoire and 9th Combinatorics Days Motivations 1 A theorem in combinatorics 2 Tools for the proof 3


  1. Stuttering multipartitions and blocks of Ariki–Koike algebras Salim Rostam Univ Rennes 17/04/2019 82nd Séminaire Lotharingien de Combinatoire and 9th Combinatorics Days

  2. Motivations 1 A theorem in combinatorics 2 Tools for the proof 3

  3. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n .

  4. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n . In this case, by Clifford theory the irreducible H D n -modules are H B exactly the irreducible summands in the restrictions D λ,µ  n .  � H D n The number of these irreducible summands entirely depends whether λ = µ or λ � = µ .

  5. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n . In this case, by Clifford theory the irreducible H D n -modules are H B exactly the irreducible summands in the restrictions D λ,µ  n .  � H D n The number of these irreducible summands entirely depends whether λ = µ or λ � = µ . The theory of cellular algebras gives a general framework to construct Specht and irreducible modules. The algebra H B n is cellular, with Specht modules {S λ,µ } .

  6. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n . In this case, by Clifford theory the irreducible H D n -modules are H B exactly the irreducible summands in the restrictions D λ,µ  n .  � H D n The number of these irreducible summands entirely depends whether λ = µ or λ � = µ . The theory of cellular algebras gives a general framework to construct Specht and irreducible modules. The algebra H B n is cellular, with Specht modules {S λ,µ } . To each S λ,µ corresponds a block of H B n , entirely determined by α := α ( λ, µ ). We define σ · α := α ( µ, λ ).

  7. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n . In this case, by Clifford theory the irreducible H D n -modules are H B exactly the irreducible summands in the restrictions D λ,µ  n .  � H D n The number of these irreducible summands entirely depends whether λ = µ or λ � = µ . The theory of cellular algebras gives a general framework to construct Specht and irreducible modules. The algebra H B n is cellular, with Specht modules {S λ,µ } . To each S λ,µ corresponds a block of H B n , entirely determined by α := α ( λ, µ ). We define σ · α := α ( µ, λ ). If λ = µ then σ · α = α . If σ · α = α , does there necessarily exist ν such that α = α ( ν, ν )?

  8. Motivations Let H X n be a Hecke algebra of type X ∈ { B , D } . If H B n is semisimple, its irreducible representations are indexed by the bipartitions { ( λ, µ ) } of n . In this case, by Clifford theory the irreducible H D n -modules are H B exactly the irreducible summands in the restrictions D λ,µ  n .  � H D n The number of these irreducible summands entirely depends whether λ = µ or λ � = µ . The theory of cellular algebras gives a general framework to construct Specht and irreducible modules. The algebra H B n is cellular, with Specht modules {S λ,µ } . To each S λ,µ corresponds a block of H B n , entirely determined by α := α ( λ, µ ). We define σ · α := α ( µ, λ ). If λ = µ then σ · α = α . If σ · α = α , does there necessarily exist ν such that α = α ( ν, ν )? The above problem appears when studying the cellularity of H D n .

  9. Motivations 1 A theorem in combinatorics 2 Tools for the proof 3

  10. Bipartitions Definition A partition is a finite non-increasing sequence of positive integers. We can picture a partition with its Young diagram . Example The sequence (4 , 2 , 2 , 1) is a partition and its Young diagram is .

  11. Bipartitions Definition A partition is a finite non-increasing sequence of positive integers. We can picture a partition with its Young diagram . Example The sequence (4 , 2 , 2 , 1) is a partition and its Young diagram is . Definition A bipartition is a pair of partitions. Example � � The pair (5 , 1) , (2) is a bipartition, constructed with the partitions (5 , 1) and (2).

  12. Multiset of residues Let η be a positive integer and set e := 2 η . Definition The multiset of residues of the bipartition ( λ, µ ) is the part of 0 1 2 . . . η η +1 η +2 . . . − 1 0 1 . . . η +1 . . . η − 1 η (mod e ) , − 2 − 1 0 . . . η . . . η − 2 η − 1 . . . . . . ... ... . . . . . . . . . . . . corresponding to the Young diagram of ( λ, µ ).

  13. Multiset of residues Let η be a positive integer and set e := 2 η . Definition The multiset of residues of the bipartition ( λ, µ ) is the part of 0 1 2 . . . η η +1 η +2 . . . − 1 0 1 . . . η +1 . . . η − 1 η (mod e ) , − 2 − 1 0 . . . η . . . η − 2 η − 1 . . . . . . ... ... . . . . . . . . . . . . corresponding to the Young diagram of ( λ, µ ). Example � � The multiset of residues of the bipartition (5 , 1) , (2) is given for 2 3 . 0 1 2 3 0 e = 4 by 3

  14. Residues multiplicity and shift Let e = 2 η ∈ 2 N ∗ . If ( λ, µ ) is a bipartition, write α ( λ, µ ) ∈ N e for the e -tuple of multiplicities of the multiset of residues. Example � � The multiset of residues of the bipartition (4 , 2) , (1) for e = 6 � � 3 , thus α 0 1 2 3 is (4 , 2) , (1) = (2 , 1 , 1 , 2 , 0 , 1). 5 0

  15. Residues multiplicity and shift Let e = 2 η ∈ 2 N ∗ . If ( λ, µ ) is a bipartition, write α ( λ, µ ) ∈ N e for the e -tuple of multiplicities of the multiset of residues. Example � � The multiset of residues of the bipartition (4 , 2) , (1) for e = 6 � � 3 , thus α 0 1 2 3 is (4 , 2) , (1) = (2 , 1 , 1 , 2 , 0 , 1). 5 0 Definition (Shift) For α = ( α i ) ∈ N e , we define σ · α ∈ N e by ( σ · α ) i := α η + i . We have σ · α = ( α η , α η +1 , . . . , α e − 1 , α 0 , α 1 , . . . , α η − 1 ).

  16. Stutterness Proposition We have α ( µ, λ ) = σ · α ( λ, µ ) . In particular, if α := α ( λ, λ ) then σ · α = α .

  17. Stutterness Proposition We have α ( µ, λ ) = σ · α ( λ, µ ) . In particular, if α := α ( λ, λ ) then σ · α = α . Theorem (R.) Let ( λ, µ ) be a bipartition and let α := α ( λ, µ ) ∈ N e . If σ · α = α then there exists a partition ν such that α = α ( ν, ν ) .

  18. Stutterness Proposition We have α ( µ, λ ) = σ · α ( λ, µ ) . In particular, if α := α ( λ, λ ) then σ · α = α . Theorem (R.) Let ( λ, µ ) be a bipartition and let α := α ( λ, µ ) ∈ N e . If σ · α = α then there exists a partition ν such that α = α ( ν, ν ) . Example Take e = 6. The multisets 0 3 4 5 , and 0 1 2 3 3 4 5 0 , 5 2 3 5 2 1 2 0 coincide (and α = (2 , 1 , 2 , 2 , 1 , 2)).

  19. Proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0

  20. Proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 3 , 2 , 2 , 3) 5 0 2 3 4 5 1 2

  21. Proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 3 , 2 , 2 , 3) 5 0 2 3 4 5 1 2 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 2 , 2 , 2 , 2) 5 0 2 3 4 1

  22. Proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 3 , 2 , 2 , 3) 5 0 2 3 4 5 1 2 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 2 , 2 , 2 , 2) 5 0 2 3 4 1 ↓ ↓ 0 1 2 3 4 5 α = (2 , 1 , 2 , 2 , 1 , 2) 5 0 2 3

  23. Failure of the proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0

  24. Failure of the proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 3 , 2 , 2 , 3) 5 0 2 3 4 5 1 2

  25. Failure of the proof by example We have α ( , ) = (2 , 1 , 2 , 2 , 1 , 2). 0 1 2 3 4 5 5 0 2 3 α = (3 , 2 , 3 , 3 , 2 , 3) 4 5 1 2 3 0 ↓ ↓ 0 1 2 3 4 5 α = (2 , 2 , 3 , 2 , 2 , 3) 5 0 2 3 4 5 1 2 ↓ ↓ 0 1 3 4 α = (2 , 2 , 2 , 2 , 2 , 2) 5 0 2 3 4 5 1 2

  26. Motivations 1 A theorem in combinatorics 2 Tools for the proof 3

  27. Abaci and cores To a partition λ = ( λ 1 , . . . , λ h ), we associate an abacus with e runners such that for each a ∈ N ∗ , there are exactly λ a gaps above and on the left of the bead a . Example The 3 and 4-abaci associated with the partition (6 , 4 , 4 , 2 , 2) are . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . .

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