A factorisation theorem for the number of rhombus tilings of a hexagon with triangular holes

Transcription

1 for the number of rhombus tilings of a hexagon with triangular holes Indiana University; Universität Wien

2 Prelude Rhombus tilings

3 Prelude Rhombus tilings

4 Prelude Rhombus tilings Dimer configurations

5 Prelude Rhombus tilings Dimer configurations

6 Prelude Rhombus tilings Dimer configurations

7 Prelude Rhombus tilings Dimer configurations

8 Prelude Rhombus tilings Dimer configurations

9 Science Fiction (Mihai Ciucu)

10 Science Fiction (Mihai Ciucu)

11 Science Fiction (Mihai Ciucu)

12 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.

13 A small problem

14 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,

15 Evidence?

16 Evidence? It is true for the case without holes!

17 Evidence? It is true for the case without holes! Actually, this is trivial and well-known.

18 Evidence? n n 2m Once and for all, let us fix H n,2m to be the hexagon with side lengths n,n,2m,n,n,2m.

19 Evidence? MacMahon showed that ( plane partitions in a given box) M(H n,2m ) = n n 2m i=1 j=1k=1 i +j +k 1 i +j +k 2.

20 Evidence? MacMahon showed that ( plane partitions in a given box) M(H n,2m ) = n n 2m i=1 j=1k=1 i +j +k 1 i +j +k 2. Proctor showed that ( transpose-complementary plane partitions in a given box) M hs (H n,2m ) = 1 i<j n 2m+2n+1 i j. 2n+1 i j

21 Evidence? MacMahon showed that ( plane partitions in a given box) M(H n,2m ) = n n 2m i=1 j=1k=1 i +j +k 1 i +j +k 2. Proctor showed that ( transpose-complementary plane partitions in a given box) M hs (H n,2m ) = 1 i<j n 2m+2n+1 i j. 2n+1 i j Andrews showed that ( symmetric plane partitions in a given box) n M vs 2m+2i 1 2m+i +j 1 (H n,2m ) =. 2i 1 i +j 1 i=1 1 i<j n

22 Evidence?

23 Evidence?

24 How to prove such a thing?

25 How to prove such a thing? By a bijection?

26 How to prove such a thing? By a bijection? By factoring Kasteleyn matrices?

27 How to prove such a thing? By a bijection? By factoring Kasteleyn matrices? Maybe introducing weights helps in seeing what one can do?

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

29 Interlude: without holes Hence: s λ (1,...,1) = M(H }{{} n,2m ). 2n

30 Interlude: without holes Hence: s λ (1,...,1) = M(H }{{} n,2m ). 2n So, let us consider the Schur function, when not all variables are specialised to 1.

31 Interlude: without holes In[1]:= S[la List] := Module[{L = Length[la]}, Expand[Cancel[Det[Table[x[i] (la[[j]] + L - j), {i, 1, L}, {j, 1, L}]]/ Det[Table[x[i] (L - j), {i, 1, L}, {j, 1, L}]]]]]

32 Interlude: without holes In[1]:= S[la List] := Module[{L = Length[la]}, Expand[Cancel[Det[Table[x[i] (la[[j]] + L - j), {i, 1, L}, {j, 1, L}]]/ Det[Table[x[i] (L - j), {i, 1, L}, {j, 1, L}]]]]] In[2]:= Factor[S[{2,2,0,0}]]

33 Interlude: without holes In[1]:= S[la List] := Module[{L = Length[la]}, Expand[Cancel[Det[Table[x[i] (la[[j]] + L - j), {i, 1, L}, {j, 1, L}]]/ Det[Table[x[i] (L - j), {i, 1, L}, {j, 1, L}]]]]] In[2]:= Factor[S[{2,2,0,0}]] Out[2]= x[1] x[2] + x[1] x[2] x[3] + x[1] x[2] x[3] > x[1] x[3] + x[1] x[2] x[3] + x[2] x[3] > x[1] x[2] x[4] + x[1] x[2] x[4] + 2 > x[1] x[3] x[4] + 2 x[1] x[2] x[3] x[4] > x[2] x[3] x[4] + x[1] x[3] x[4] + x[2] x[3] x[4] > x[1] x[4] + x[1] x[2] x[4] + x[2] x[4] > x[1] x[3] x[4] + x[2] x[3] x[4] + x[3] x[4]

34 Interlude: without holes In[3]:= Factor[S[{2, 2, 0, 0}] /. x[3] -> 1/x[1] /. x[4] -> 1/x[2]]

35 Interlude: without holes In[3]:= Factor[S[{2, 2, 0, 0}] /. x[3] -> 1/x[1] /. x[4] -> 1/x[2]] 2 Out[3]= ((1 - x[1] + x[1] - x[2] + 2 x[1] x[2] > x[1] x[2] + x[2] - x[1] x[2] + x[1] x[2] ) 2 2 > (1 + x[1] + x[1] + x[2] + 2 x[1] x[2] + x[1] x[2] > x[2] + x[1] x[2] + x[1] x[2] )) / (x[1] x[2] )

36 Interlude: without holes Computer experiments lead one to: Theorem For any non-negative integers m and n, we have s ((2m) 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, so λ (x 1,x 2,...,x N ) = 1 2 det 1 i,j N (xλ j+n j+ i x (λ j+n j+ 1 2 ) 1 det 1 i,j N (xn j+ 2 i ) i ) i x (N j+1 2 ) is an irreducible character of SO 2N+1 (C).

37 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 ), and, in particular, one obtains ν:ν 1 2m M vs (H n,2m ) = so (m n )(1,...,1). }{{} n

38 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,2m ). n

39 Interlude: without holes Proof of the theorem. By the definition of the Schur function, we have s ((2m) n )(x 1,x1 1,x 2,x2 1,...,x n,x 1 ( = det 1 i,j 2n n ) x 2mχ(j n)+2n j i x (2mχ(j n)+2n t) i n denominator 1 i n n+1 i 2n Now do a Laplace expansion with respect to the first n columns. This leads to a huge sum. ).

40 Interlude: without holes For the odd orthogonal character(s), one also starts with the Weyl character formula det 1 i,j N (xλ j+n j+ i x (λ j+n j+ 1 2 ) 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. 1 2

41 Interlude: without holes In the end, one has to prove identities such as A [2N] A =N = A [2N] V(A)V ( A 1) V(A c )V ( (A c ) 1) R ( A,A 1) R ( A c,(a c ) 1) V(A)V ( A 1) V(A c )V ( (A c ) 1) R ( A,(A c ) 1) R ( A c,a 1), where A c denotes the complement of A in [2N]. Here, R ( A,B 1) := a A and V ( A 1) := b B (xa 1 x 1 b a,b A a<b provide a proof of the above identity. (x a x 1 b ), V(A) := (x a x b ), a,b A a<b ). Induction on N works to

42 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem

43 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G.

44 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

45 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

46 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

47 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

48 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

49 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

50 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. a b a b a b

51 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. G + a b a b a b 1/2 1/2 G

52 Half of Science Fiction is Reality Ciucu s Matchings Factorisation Theorem Consider a symmetric bipartite graph G. G + a b a b a b 1/2 1/2 G Then M(G) = 2 #(edges on symm. axis) M(G + ) M weighted (G ).

53 Half of Science Fiction is Reality If we translate this to our situation: R + we obtain R M(R) = 2 #(rhombi on symm. axis) M(R + ) M weighted (R ).

54 Half of Science Fiction is Reality If we translate this to our situation: R + R we obtain M(R) = 2 #(rhombi on symm. axis) M(R + ) M weighted (R ). We want M(R) =? M hs (R) M vs (R).

55 The actual problem So, it only remains to prove M vs (R) = 2 #(rhombi on symm. axis) M weighted (R ).

56 The theorem The hexagon with holes H 15,10 (2,5,7)

57 The theorem Theorem For all positive integers n, m, l and non-negative integers k 1,k 2,...k l with 0 < k 1 < k 2 < < k l n/2, we have M(H n,2m (k 1,k 2,...,k l )) = M hs (H n,2m (k 1,k 2,...,k l ))M vs (H n,2m (k 1,k 2,...,k l )).

58 Sketch of proof

59 Sketch of proof First step. Use non-intersecting lattice ) paths to get a determinant for M weighted (Hn,2m (k 1,k 2,...,k l ) and a Pfaffian for M vs (H n,2m (k 1,k 2,...,k l )).

60 Sketch of proof A tiling of H n,2m (k 1,k 2,...,k l )

61 Sketch of proof Theorem (Karlin McGregor, Lindström, Gessel Viennot, Fisher, John Sachs, Gronau Just Schade Scheffler Wojciechowski) Let G be an acyclic, directed graph, and let A 1,A 2,...,A n and E 1,E 2,...,E n be vertices in the graph with the property that, for i < j and k < l, any (directed) path from A i to E l intersects with any path from A j to E k. Then the number of families (P 1,P 2,...,P n ) of non-intersecting (directed) paths, where the i-th path P i runs from A i to E i, i = 1,2,...,n, is given by det ( P(A j E i ) ), 1 i,j n where P(A E) denotes the set of paths from A to E.

62 Sketch of proof By the Karlin McGregor, Lindström, Gessel Viennot, Fisher, John Sachs, Gronau Just Schade Scheffler Wojciechowski Theorem on non-intersecting lattice paths, we obtain a determinant. Proposition ) M weighted (Hn,2m (k 1,k 2,...,k l ) is given by det(n), where N is the matrix with rows and columns indexed by {1,2,...,m,1 +,2 +,...,l + }, and entries given by N i,j = ( 2n n+j i ( 2n 2kt ( n k t i+1 2n 2kt n k ( t j+1 2n 2kt 2kˆt ) ( + 2n n i j+1), if 1 i,j m, ) ( + 2n 2kt n k t i), if 1 i m and j = t +, ) ( + 2n 2kt n k t j), if i = t + and 1 j m, ) ( n k + 2n 2kt 2kˆt t kˆt n k t kˆt 1), if i = t +, j = ˆt +, and 1 t,ˆt l.

63 Sketch of proof The left half of a vertically symmetric tiling

64 Sketch of proof Theorem (Okada, Stembridge) Let {u 1,u 2,...,u p } and I = {I 1,I 2,...} be finite sets of lattice points in the integer lattice Z 2, with p even. Let S p be the symmetric group on {1,2,...,p}, set u π = (u π(1),u π(2),...,u π(p) ), and denote by P nonint (u π I) the number of families (P 1,P 2,...,P p ) of non-intersecting lattice paths, with P k running from u π(k) to I jk, k = 1,2,...,p, for some indices j 1,j 2,...,j p satisfying j 1 < j 2 < < j p. Then we have π S p (sgnπ) P nonint (u π I) = Pf(Q),

65 Sketch of proof with the matrix Q = (Q i,j ) 1 i,j p given by Q i,j = ( P(ui I u ) P(u j I v ) P(u j I u ) P(u i I v ) ), 1 u<v where P(A E) denotes the number of lattice paths from A to E.

66 Sketch of proof Proposition M vs (H n,2m (k 1,k 2,...,k l )) is given by ( 1) (l 2) Pf(M), where M is the skew-symmetric matrix with rows and columns indexed by { m +1, m +2,...,m,1,2,...,l,1 +,2 +,...,l + }, and entries given by

67 Sketch of proof M i,j = j i ( 2n r=i j+1 i ( 2n 2kt r=i+1 i+1 r=i n+r), if m+1 i < j m, n k t+r), if m+1 i m and j = t, ( 2n 2kt n k t+r), if m+1 i m and j = t +, 0, if i = t, j = ˆt, and 1 t < ˆt l, ( ) 2n 2kt 2kˆt n k t kˆt + ( 2n 2k t 2kˆt n k t kˆt +1), if i = t, j = ˆt +, and 1 t,ˆt l, 0, if i = t +, j = ˆt +, and 1 t < ˆt l, where sums have to be interpreted according to N 1 r=m Expr(k) = N 1 r=m Expr(k) N > M 0 N = M M 1 k=n Expr(k) N < M.

68 Sketch of proof Second step.

69 Sketch of proof Second step. Lemma For a positive integer m and a non-negative integer l, let A be a matrix of the form ( ) X Y A = Y t, Z where X = (x j i ) m+1 i,j m and Z = (z i,j ) i,j {1,...,l,1 +,...,l + } are skew-symmetric, and Y = (y i,j ) m+1 i m,j {1,...,l,1 +,...,l + } is a 2m 2l matrix. Suppose in addition that y i,t = y i,t and y i,t + = y i+2,t +, for all i with m+1 i m for which both sides of an equality are defined, and 1 t l, and that z i,j = 0 for all i,j {1,...,l }. Then Pf(A) = ( 1) (l 2) det(b),

70 Sketch of proof where with ( ) X Ȳ B = 1, Z Ȳ 2 X = ( x i,j ) 1 i,j m, Ȳ 1 = (y i+1,j ) 1 i m,j {1 +,...,l + }, Ȳ 2 = ( y i,j ) i {1,...,l },1 j m, Z = (z i,j ) i {1,...,l },j {1 +,...,l + }, and the entries of X are defined by x i,j = x j i +1 +x j i x i+j 1.

71 Sketch of proof By the lemma, the Pfaffian for M vs (H n,2m (k 1,k 2,...,k l )) can be converted into a determinant, of the same size ) as the determinant we obtained for M weighted (Hn,2m (k 1,k 2,...,k l ).

72 Sketch of proof By the lemma, the Pfaffian for M vs (H n,2m (k 1,k 2,...,k l )) can be converted into a determinant, of the same size ) as the determinant we obtained for M weighted (Hn,2m (k 1,k 2,...,k l ). Third step. Alas, it is not the same determinant.

73 Sketch of proof By the lemma, the Pfaffian for M vs (H n,2m (k 1,k 2,...,k l )) can be converted into a determinant, of the same size ) as the determinant we obtained for M weighted (Hn,2m (k 1,k 2,...,k l ). Third step. Alas, it is not the same determinant. However, further row and column operations do indeed convert one determinant into the other.

74 Postlude

75 Postlude A theorem has been proved.

76 Postlude A theorem has been proved. Is the proof illuminating?

77 Postlude A theorem has been proved. Is the proof illuminating? No.

78 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation?

79 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No.

80 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon?

81 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon? I do not know.

82 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon? I do not know. Is this a theorem without applications?

83 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon? I do not know. Is this a theorem without applications? No.

84 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon? I do not know. Is this a theorem without applications? No. Is this the end?

85 Postlude A theorem has been proved. Is the proof illuminating? No. Do we understand this factorisation? No. Can this be the utmost/correct generality for this factorisation phenomenon? I do not know. Is this a theorem without applications? No. Is this the end? Yes.