A dual of MacMahon s theorem on plane partitions
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- Hollie Bridges
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1 Indiana University; Universität Wien
2 Plane Partitions Plane partitions were invented by Major Percy Alexander MacMahon around 1900.
3 Plane Partitions Plane partitions were invented by Major Percy Alexander MacMahon around They are NOT partitions of the plane, but rather planar analogues of (integer) partitions.
4 Plane Partitions Plane partitions were invented by Major Percy Alexander MacMahon around They are NOT partitions of the plane, but rather planar analogues of (integer) partitions. By definition, a plane partition is an array of non-negative integers (π i,j ) i,j such that entries along rows and along columns are non-increasing, that is, π i,j π i,j+1 and π i,j π i+1,j.
5 Plane Partitions Plane partitions were invented by Major Percy Alexander MacMahon around They are NOT partitions of the plane, but rather planar analogues of (integer) partitions. By definition, a plane partition is an array of non-negative integers (π i,j ) i,j such that entries along rows and along columns are non-increasing, that is, π i,j π i,j+1 and π i,j π i+1,j.
6 Plane Partitions - modern perspective
7 Plane Partitions - modern perspective rhombus tilings of a hexagon!
8 Plane Partitions - modern perspective rhombus tilings of a hexagon!
9 Plane Partitions - modern perspective rhombus tilings of a hexagon!
10 MacMahon s Plane Partition Theorem The number of rhombus tilings of a hexagon with side lengths a, b, c, a, b, c equals H(a) H(b) H(c) H(a + b + c) H(a + b) H(a + c) H(b + c), where the hyperfactorials H(n) are defined by H(n) := 0! 1! (n 1)!.
11 A dual?
12 A dual?
13 A dual? There are no rhombus tilings of such a thing!
14 A dual? There are no rhombus tilings of such a thing! Dualise : Count the rhombus tilings of the outside!
15 A dual? Let S (a, b, c, m) denote the exterior of an (a, b, c, m)-shamrock. Furthermore, let H N (a, b, c, m) be the hexagonal region of side-lengths alternating between N + a + b + c and N + a + b + c + m (the top side being N + a + b + c), and having the shamrock S(a, b, c, m) removed from its centre Then we define M(S (a, b, c, m)) M(H N (a, b, c, m)) M(S := lim (a + b + c, 0, 0, m)) N M(H N (a + b + c, 0, 0, m)), where M(R) denotes the number of rhombus tilings of the region R.
16 The dual MacMahon formula Theorem For any non-negative integers a, b, c and m we have M(S (a, b, c, m)) M(S (a + b + c, 0, 0, m)) = H(a) H(b) H(c) H(a + b + c + m) H(m) H(a + m) H(b + m) H(c + m) H(a + b + c) = P(a, b, m) P(a + b, c, m), where P(A, B, C) denotes the number of rhombus tilings of a hexagon with side lengths A, B, C, A, B, C.
17 The dual MacMahon formula m = a + b + c Corollary For any non-negative integers a, b, c, we have M(S (a, b, c, a + b + c)) M(S = P(a, b, c) P(a + b, b + c, c + a), (a + b + c, 0, 0, a + b + c)) where P(A, B, C) denotes the number of rhombus tilings of a hexagon with side lengths A, B, C, A, B, C.
18 m = a + b + c geometric interpretation M(S (a, b, c, a + b + c)) M(S = P(a, b, c) P(a+b, b +c, c +a) (a + b + c, 0, 0, a + b + c))
19 m = a + b + c geometric interpretation M(S (a, b, c, a + b + c)) M(S = P(a, b, c) P(a+b, b +c, c +a) (a + b + c, 0, 0, a + b + c))
20 A more general region - equal parity case The region SC x,y,z (a, b, c, m):
21 A more general region - equal parity case The region SC x,y,z (a, b, c, m): x+a+b+c z+m y+z y+m y+a+b+c x+y +a+b b m x+z x+y y+z +b+c a c x+z +a+c z+a+b+c x+m
22 The main theorem: equal parity case Theorem Let x, y, z, a, b, c and m be nonnegative integers. If x, y and z have the same parity, we have H(m) 3 H(a) H(b) H(c) M(SC x,y,z (a, b, c, m)) = H(m + a) H(m + b) H(m + c) x+y H( + m + a + b) H( x+z + m + a + c) H( y+z + m + b + c) H( x+y + m + c) H( x+z + m + b) H( y+z + m + a) H( x+y + c) H( x+z + b) H( y+z + a) H( x+y + a + b) H( x+z + a + c) H( y+z + b + c) H(x + m + a + b + c) H(y + m + a + b + c) H(x + y + m + a + b + c) H(x + z + m + a + b + c) H(z + m + a + b + c) H(x + y + z + m + a + b + c) H(y + z + m + a + b + c)
23 The main theorem: equal parity case where H( x+y+z + m + a + b + c) H( x+y + m + a + b + c) H( x+z + m + a + b + c) x+y+z H( + m + a + b + c) H( y+z + m+a+b+c ) H( y+z + m + a + b + c) H( x+y x+z y+z ) H( ) H( ) H( x H( x ) H( y ) H( x + m+a+b+c ) H( x + m+a+b+c ) H( y + m+a+b+c H( y ) H( z ) H( z ) ) H( y + m+a+b+c ) H( z + m+a+b+c ) H( z + m+a+b+c m+a+b+c H( ) H( x+y + m+a+b+c ) H( x+z + m+a+b+c ) H( x+y+z + m+a+b+c ) H( x+y+z + m+a+b+c, ) H(n) := { n 1 k=0 ) Γ(k + 1), for n a positive integer, n 1 k=0 Γ(k + 1 ), for n a positive half-integer,
24 A more general region - unequal parity case The region SC x,y,z (a, b, c, m): x+a+b+c z+m y+z y+m y+a+b+c x+y 1 +a+b b x+z 1 m a c x+y+1 y+z +b+c x+z+ 1 +a+c z+a+b+c x+m
25 The main theorem: unequal parity case Theorem Let x, y, z, a, b, c and m be nonnegative integers. If x has parity different from the parity of y and z, we have M(SC x,y,z (a, b, c, m)) = H( x+y H(m) 3 H(a) H(b) H(c) H(m + a) H(m + b) H(m + c) x+z y+z + m + a + b) H( + m + a + c) H( + m + b + c) H( x+y x+z y+z + m + c) H( + m + b) H( + m + a) H( x+y x+z y+z + c) H( + b) H( + a) H( x+y x+z y+z + a + b) H( + a + c) H( + b + c) H(x + m + a + b + c) H(y + m + a + b + c) H(x + y + m + a + b + c) H(x + z + m + a + b + c) H(z + m + a + b + c) H(x + y + z + m + a + b + c) H(y + z + m + a + b + c)
26 The main theorem: unequal parity case H( x+y H( x+y+z + m + a + b + c) x+z + m + a + b + c) H( + m + a + b + c) x+y+z H( + m + a + b + c) H( y+z + m + a + b + c) H( x ) H( x ) H( y ) H( x + m+a+b+c ) H( x + m+a+b+c ) H( y + m+a+b+c H( y ) H( z ) H( z ) ) H( y + m+a+b+c ) H( z + m+a+b+c ) H( z + m+a+b+c ) m+a+b+c H( ) H( x+y + m+a+b+c ) H( x+y + m+a+b+c ) H( x+y+z + m+a+b+c ) H( x+y+z + m+a+b+c ) H( x+y ) H( x+z + m+a+b+c ) H( x+z + m+a+b+c ) H( y+z + m+a+b+c ) H( x+z ) H( y+z ).
27 Kuo s graphical condensation Theorem (Kuo) Let G = (V 1, V, E) be a plane bipartite graph in which V 1 = V. Let vertices α, β, γ and δ appear cyclically on a face of G. If α, γ V 1 and β, δ V, then M(G)M(G {α, β, γ, δ}) = M(G {α, β})m(g {γ, δ}) + M(G {α, δ})m(g {β, γ}), where M(H) denotes the number of perfect matchings of the bipartite graph H.
28 Kuo s graphical condensation Theorem (Kuo) Let G = (V 1, V, E) be a plane bipartite graph in which V 1 = V. Let vertices α, β, γ and δ appear cyclically on a face of G. If α, γ V 1 and β, δ V, then M(G)M(G {α, β, γ, δ}) = M(G {α, β})m(g {γ, δ}) + M(G {α, δ})m(g {β, γ}), where M(H) denotes the number of perfect matchings of the bipartite graph H.
29 Kuo s graphical condensation Theorem (Kuo) Let G = (V 1, V, E) be a plane bipartite graph in which V 1 = V. Let vertices α, β, γ and δ appear cyclically on a face of G. If α, γ V 1 and β, δ V, then M(G)M(G {α, β, γ, δ}) = M(G {α, β})m(g {γ, δ}) + M(G {α, δ})m(g {β, γ}), where M(H) denotes the number of perfect matchings of the bipartite graph H. This is, as worked out by Fulmek, a combinatorial version of Dodgon s condensation formula for determinants ( = of a special case of a determinant formula of Jacobi... ).
30 The proof of the main theorem Both, the equal parity case and the unequal parity case, are simultaneously proved by induction on x + y + z.
31 The proof of the main theorem Both, the equal parity case and the unequal parity case, are simultaneously proved by induction on x + y + z. The induction base. These are the cases where x = 0, y = 0, or z = 0. Again, there are several cases. In the equal parity case, by symmetry of parameters it is sufficient to consider the case where z = 0.
32 Proof of the main theorem: z = 0 The region SC 4,4,0 (3, 1,, ):
33 Proof of the main theorem: z = 0 We need to know the number of rhombus tilings of regions of the form x+c a y+m m b c a+b+c y+c x+m
34 Proof of the main theorem: z = 0 Theorem For any nonnegative integers x, y, a, b, c and m, the number of rhombus tiling of this region equals = H(m) H(a) H(b) H(c) H(m + a + b + c) H(m + a) H(m + b) H(m + c) H(x + m + a + c) H(y + m + b + c) H(x + y + c) H(x + y + m + c) H(x + a + c) H(y + b + c) H(x + y + m + a + b + c) H(x) H(y) H(x + m + a + b + c) H(y + m + a + b + c) H(x + y).
35 Proof of the main theorem: z = 0 Theorem For any nonnegative integers x, y, a, b, c and m, the number of rhombus tiling of this region equals = H(m) H(a) H(b) H(c) H(m + a + b + c) H(m + a) H(m + b) H(m + c) H(x + m + a + c) H(y + m + b + c) H(x + y + c) H(x + y + m + c) H(x + a + c) H(y + b + c) H(x + y + m + a + b + c) H(x) H(y) H(x + m + a + b + c) H(y + m + a + b + c) H(x + y).
36 Proof of the main theorem: z = 0 Theorem For any nonnegative integers x, y, a, b, c and m, the number of rhombus tiling of this region equals = H(m) H(a) H(b) H(c) H(m + a + b + c) H(m + a) H(m + b) H(m + c) H(x + m + a + c) H(y + m + b + c) H(x + y + c) H(x + y + m + c) H(x + a + c) H(y + b + c) H(x + y + m + a + b + c) H(x) H(y) H(x + m + a + b + c) H(y + m + a + b + c) H(x + y). Proof. This can also be proved by means of Kuo s condensation.
37 Proof of the main theorem Induction step: We apply Kuo s condensation. For α, β, γ, δ we choose triangles along in the corners of the outer boundary of the hexagon.
38 Proof of the main theorem
39 Proof of the main theorem
40 Proof of the main theorem
41 Proof of the main theorem The implied recurrence: M(SC x,y,z (a, b, c, m))m(sc x,y 1,z 1 (a, b, c, m)) = M(SC y,x,z 1 (b, a, c, m))m(sc z,y 1,x (c, b, a, m)) + M(SC x 1,y,z (a, b, c, m))m(sc x+1,y 1,z 1 (a, b, c, m)).
42 Proof of the main theorem The implied recurrence: M(SC x,y,z (a, b, c, m))m(sc x,y 1,z 1 (a, b, c, m)) = M(SC y,x,z 1 (b, a, c, m))m(sc z,y 1,x (c, b, a, m)) + M(SC x 1,y,z (a, b, c, m))m(sc x+1,y 1,z 1 (a, b, c, m)). One needs to verify that the guessed formula satisfies the same recurrence.
43 Proof of the main theorem One substitutes the guessed expression in the recurrence to be verified and forms the quotient of the two sides of the recurrence. After a lot of simplification, this quotient becomes Γ(y + z + a + b + c + m) Γ(x + y + z + a + b + c + m 1) Γ(y + z + a + b + c + m 1) Γ(x + y + z + a + b + c + m) ( ) Γ x+y+z 1 + a + b + c + m Γ ( x+y+z ) + a+b+c+m Γ ( x+y+z ) ( ) + a + b + c + m Γ x+y+z 1 + a+b+c+m Γ(x + a + b + c + m + 1) Γ(x + y + z + a + b + c + m 1) + Γ(x + a + b + c + m) Γ(x + y + z + a + b + c + m) Γ ( ) ( x+1 Γ x ) + a+b+c+m Γ ( ) ( x Γ x+1 ) + a+b+c+m Γ ( x+y+z 1 Γ ( x+y+z ) + a + b + c + m Γ ( x+y+z ) + a+b+c+m ) ( ). + a + b + c + m Γ x+y+z 1 + a+b+c+m
44 Proof of the main theorem If all of x, y, z are even, this condenses down to (y + z + a + b + c + m 1) (x + y + z + a + b + c + m 1) (x + a + b + c + m) + (x + y + z + a + b + c + m 1) x/ (x + a + b + c + m)/, which indeed equals 1.
45 Proof of the main theorem If all of x, y, z are even, this condenses down to (y + z + a + b + c + m 1) (x + y + z + a + b + c + m 1) (x + a + b + c + m) + (x + y + z + a + b + c + m 1) x/ (x + a + b + c + m)/, which indeed equals 1. The other cases are similar.
46 The dual MacMahon formula For the dual of MacMahon s formula, we need to do asymptotics of hyperfactorials. By the Glaisher Kinkelin formula, which gives the asymptotics of the Barnes G-function, we have lim n 0! 1! (n 1)! n n 1 1 (π) n e 3n 4 = e 1 1 A, where A = is the so-called the Glaisher Kinkelin constant.
47 The dual MacMahon formula lim n H(n) n n 1 1 (π) n e 3n 4 = e 1 1 A, where A = is the so-called the Glaisher Kinkelin constant.
48 The dual MacMahon formula lim n H(n) n n 1 1 (π) n e 3n 4 = e 1 1 A, where A = is the so-called the Glaisher Kinkelin constant. This leads (more or less) straightforwardly to the dual MacMahon formula: Theorem For any non-negative integers a, b, c and m we have M(S (a, b, c, m)) M(S (a + b + c, 0, 0, m)) = H(a) H(b) H(c) H(a + b + c + m) H(m) H(a + m) H(b + m) H(c + m) H(a + b + c) = P(a, b, m) P(a + b, c, m).
49 A dual of MacMahon s formula
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