A factorisation theorem for the number of rhombus tilings of a - PowerPoint PPT Presentation

A factorisation theorem for the number of rhombus tilings of a hexagon with triangular holes Mihai Ciucu and Christian Krattenthaler Indiana University; Universit¨ at Wien Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings — Dimer configurations Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings — Dimer configurations Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings — Dimer configurations Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings — Dimer configurations Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Prelude Rhombus tilings — Dimer configurations Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Science Fiction (Mihai Ciucu) Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Science Fiction (Mihai Ciucu) Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Science Fiction (Mihai Ciucu) Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Science Fiction (Mihai Ciucu) Let R be that region. Then M( R ) ? = M hs ( R ) · M vs ( R ) , where M( R ) denotes the number of rhombus tilings of R . Mihai Ciucu and Christian Krattenthaler A factorisation theorem

A small problem Mihai Ciucu and Christian Krattenthaler A factorisation theorem

A small problem For this region R , we have M( R ) = 6 × 6 = 36, M hs ( R ) = 6, and M vs ( R ) = 4 × 4 = 16. But, 36 � = 6 × 16 . Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? It is true for the case without holes! Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? It is true for the case without holes! Actually, this is “trivial” and “well-known”. Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence?       n  n                   2 m       Once and for all, let us fix H n , 2 m to be the hexagon with side lengths n , n , 2 m , n , n , 2 m . Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? MacMahon showed that (“plane partitions” in a given box) 2 m n n i + j + k − 1 � � � M( H n , 2 m ) = i + j + k − 2 . i =1 j =1 k =1 Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? MacMahon showed that (“plane partitions” in a given box) 2 m n n i + j + k − 1 � � � M( H n , 2 m ) = i + j + k − 2 . i =1 j =1 k =1 Proctor showed that (“transpose-complementary plane partitions” in a given box) 2 m + 2 n + 1 − i − j � M hs ( H n , 2 m ) = . 2 n + 1 − i − j 1 ≤ i < j ≤ n Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? MacMahon showed that (“plane partitions” in a given box) 2 m n n i + j + k − 1 � � � M( H n , 2 m ) = i + j + k − 2 . i =1 j =1 k =1 Proctor showed that (“transpose-complementary plane partitions” in a given box) 2 m + 2 n + 1 − i − j � M hs ( H n , 2 m ) = . 2 n + 1 − i − j 1 ≤ i < j ≤ n Andrews showed that (“symmetric plane partitions” in a given box) n 2 m + 2 i − 1 2 m + i + j − 1 � � M vs ( H n , 2 m ) = . 2 i − 1 i + j − 1 i =1 1 ≤ i < j ≤ n Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Evidence? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

How to prove such a thing? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

How to prove such a thing? By a bijection ? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

How to prove such a thing? By a bijection ? By “factoring” Kasteleyn matrices ? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

How to prove such a thing? By a bijection ? By “factoring” Kasteleyn matrices ? Maybe introducing weights helps in seeing what one can do ? Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes It is well-known that the number of rhombus tilings of the hexagon H n , 2 m is the same as the number of semistandard tableaux of rectangular shape ((2 m ) n ) with entries between 1 and 2 n . This observation connects M( H n , 2 m ) with Schur functions. Given a partition λ = ( λ 1 , . . . , λ n ), the Schur function s λ is given by � � x λ j + N − j det 1 ≤ i , j ≤ N i s λ ( x 1 , . . . , x N ) = � � x N − j det 1 ≤ i , j ≤ N i N � � x #(occurrences of i in T ) = , i T i =1 where the sum is over all semistandard tableaux of shape λ with entries between 1 and N . Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes Hence: s λ (1 , . . . , 1 ) = M( H n , 2 m ) . � �� � 2 n Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes Hence: s λ (1 , . . . , 1 ) = M( H n , 2 m ) . � �� � 2 n So, let us consider the Schur function, when not all variables are specialised to 1. Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes Computer experiments lead one to: Theorem For any non-negative integers m and n, we have s ((2 m ) n ) ( x 1 , x − 1 1 , x 2 , x − 1 2 , . . . , x n , x − 1 n )

Interlude: without holes Computer experiments lead one to: Theorem For any non-negative integers m and n, we have s ((2 m ) n ) ( x 1 , x − 1 1 , x 2 , x − 1 2 , . . . , x n , x − 1 n ) = ( − 1) mn so ( m n ) ( x 1 , x 2 , . . . , x n ) so ( m n ) ( − x 1 , − x 2 , . . . , − x n ) . Here, λ j + N − j + 1 − ( λ j + N − j + 1 2 ) 1 ≤ i , j ≤ N ( x det 2 − x ) i i so λ ( x 1 , x 2 , . . . , x N ) = N − j + 1 − ( N − j + 1 2 ) 1 ≤ i , j ≤ N ( x det 2 − x ) i i is an irreducible character of SO 2 N +1 ( C ). Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes The odd orthogonal character is “expected”, since all existing proofs for the enumeration of symmetric plane partitions use — in one form or another, directly or indirectly — the summation � so ( m n ) ( x 1 , x 2 , . . . , x n ) = ( x 1 x 2 · · · x n ) − m · s ν ( x 1 , . . . , x n ) , ν : ν 1 ≤ 2 m and, in particular, one obtains M vs ( H n , 2 m ) = so ( m n ) (1 , . . . , 1 ) . � �� � n Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes However, the appearance of so ( m n ) ( − x 1 , − x 2 , . . . , − x n ) is “unwanted”. What one would actually like to see in place of this is a symplectic character of rectangular shape, because this is what goes into all proofs of the enumeration of transpose-complementary plane partitions (in one form or another). Nevertheless, by substituting x i = − q i − 1 in the Weyl character formula, both determinants can be evaluated in closed form, and subsequently the limit q → 1 can be performed. The result is that, indeed, ( − 1) mn so ( m n ) ( − 1 , . . . , − 1 ) = M hs ( H n , 2 m ) . � �� � n Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes Proof of the theorem. By the definition of the Schur function, we have s ((2 m ) n ) ( x 1 , x − 1 1 , x 2 , x − 1 2 , . . . , x n , x − 1 n ) � � x 2 m χ ( j ≤ n )+2 n − j 1 ≤ i ≤ n i det x − (2 m χ ( j ≤ n )+2 n − t ) 1 ≤ i , j ≤ 2 n n + 1 ≤ i ≤ 2 n i − n = . denominator Now do a Laplace expansion with respect to the first n rows. This leads to a huge sum. Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes For the odd orthogonal character(s), one also starts with the Weyl character formula λ j + N − j + 1 − ( λ j + N − j + 1 2 ) 1 ≤ i , j ≤ N ( x det 2 − x ) i i so λ ( x 1 , x 2 , . . . , x N ) = . denominator Here, each entry in the determinant is a sum of two monomials. We use linearity of the determinant in the rows to expand the determinant. Also here, this leads to a huge sum. Mihai Ciucu and Christian Krattenthaler A factorisation theorem

Interlude: without holes In the end, one has to prove identities such as � � A − 1 � � ( A c ) − 1 � � A , A − 1 � � A c , ( A c ) − 1 � V ( A c ) V V ( A ) V R R A ⊆ [2 N ] | A | = N � � A − 1 � � ( A c ) − 1 � � A , ( A c ) − 1 � � A c , A − 1 � V ( A c ) V = V ( A ) V R R , A ⊆ [2 N ] where A c denotes the complement of A in [2 N ]. Here, � � � � A , B − 1 � ( x a − x − 1 := b ) , V ( A ) := ( x a − x b ) , R a ∈ A b ∈ B a , b ∈ A a < b � � A − 1 � ( x − 1 − x − 1 and V := b ) , which can be accomplished by a a , b ∈ A a < b induction. Mihai Ciucu and Christian Krattenthaler A factorisation theorem