Symmetries in trees Olivier Bernardi (Brandeis University) First - - PowerPoint PPT Presentation

Symmetries in trees

McGill University, September 2013 Olivier Bernardi (Brandeis University)

First part is joint work with Alejandro Morales (UQAM)

1 2 3 2 5 4 2 3 3 4 6 1 1 1 2

Part I Multitype Cayley trees

Cayley trees A tree is a connected acyclic graph. A Cayley tree of size n is a tree with vertex set [n] := {1, 2, . . . , n}. 1 8 6 3 2 4 5 7

Cayley trees A tree is a connected acyclic graph. A Cayley tree of size n is a tree with vertex set [n] := {1, 2, . . . , n}. Theorem [Sylvester 1857]. The number of Cayley trees of size n is nn−2.

Cayley trees A tree is a connected acyclic graph. A Cayley tree of size n is a tree with vertex set [n] := {1, 2, . . . , n}. Theorem [Sylvester 1857]. The number of Cayley trees of size n is nn−2.

1 3 2 4 3 3 3 3 3 3 3 3 2 3 1 4 2 4 1 3 2 3 1 4 3 1 2 4 4 3 2 1 3 3 3 3 3 3 3 3 3 2 1 4 3 4 1 2 3 3 3 3 3 3 3 3 4 2 1 3 4 3 1 2 3 3 3 3 3 3 3 3 1 3 2 4 1 4 2 3

• Example. There are 16 Cayley trees of size n = 4.

3 3 3 3 3 3 3 3 1 2 3 4 1 4 3 2 3 3 3 3 3 3 3 3 1 2 4 3 1 3 4 2

Multitype Cayley trees A multitype Cayley tree of size (n1, . . . , nd) is a tree with vertex set V = {(t, i), t ∈ [d], i ∈ [nt]} .

Multitype Cayley trees A multitype Cayley tree of size (n1, . . . , nd) is a tree with vertex set V = {(t, i), t ∈ [d], i ∈ [nt]} . type 1

1 2 3 2 5 4 2 3 3 4 6

root vertex type 2 type 3

1 1

type 4

1 2

size = (4, 3, 6, 2) vertex (3, 6)

Multitype Cayley trees A multitype Cayley tree of size (n1, . . . , nd) is a tree with vertex set V = {(t, i), t ∈ [d], i ∈ [nt]} . Theorem [Knuth 68, Bousquet-M´ elou, Chapuy 12]. The number

• f multitype Cayley trees of size (n1, . . . , nd) with root-vertex of type

ρ such that the type of adjacent vertices differ by ±1 is nρ n1nd

d

• i=1

ni (ni−1 + ni+1)ni−1, with n0 = nd+1 = 0.

type 1 1 2 3 2 5 4 2 3 3 4 6 root vertex type 2 type 3 1 1 type 4 3 2 1 4

Warm up: counting Cayley trees by degrees Theorem [Cayley 1889]: The generating function of Cayley trees of size n counted according to their degrees is

• T

xdeg(1)

1

xdeg(2)

2

· · · xdeg(n)

n

= x1x2 · · · xn(x1 + x2 + . . . + xn)n−2.

Warm up: counting Cayley trees by degrees Theorem [Cayley 1889]: The generating function of Cayley trees of size n counted according to their degrees is

• T

xdeg(1)

1

xdeg(2)

2

· · · xdeg(n)

n

= x1x2 · · · xn(x1 + x2 + . . . + xn)n−2. Known proofs: Recurrences, matrix-tree theorem, Prufer’s code, Joyal’s endofunction approach, Pitman’s double counting argument.

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi. We must take α1, . . . , αn > 0 such that

• i

αi = 2n − 2.

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi.

• Claim. If βi = αi − 1, βj = αj + 1 and βk = αk for k = i, j, then

(αi − 1)|Tα| = (βj − 1)|Tβ|.

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi. Proof.

• Claim. If βi = αi − 1, βj = αj + 1 and βk = αk for k = i, j, then

(αi − 1)|Tα| = (βj − 1)|Tβ|. i i j j bijection

Tree in Tα with a marked edge incident to i not on the path i j. Tree in Tβ with a marked edge incident to j not on the path i j.

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi.

• Claim. If βi = αi − 1, βj = αj + 1 and βk = αk for k = i, j, then

(αi − 1)|Tα| = (βj − 1)|Tβ|.

• Corollary. |Tα| =

(n−2)! (α1−1)!···(αn−1)!|T(n−1,1,1,...,1)| =

• n−2

α1−1,...,αn−1

• .

1

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi.

• Claim. If βi = αi − 1, βj = αj + 1 and βk = αk for k = i, j, then

(αi − 1)|Tα| = (βj − 1)|Tβ|.

• Corollary. |Tα| =

(n−2)! (α1−1)!···(αn−1)!|T(n−1,1,1,...,1)| =

• n−2

α1−1,...,αn−1

• .

Hence,

• T

xdeg(1)

1

xdeg(2)

2

· · · xdeg(n)

n

= x1x2 · · · xn(x1 + x2 + . . . + xn)n−2.

Warm up: counting Cayley trees by degrees For α = (α1, . . . , αn) we denote by Tα the set of Cayley trees of size n where vertex i has degree αi.

• Claim. If βi = αi − 1, βj = αj + 1 and βk = αk for k = i, j, then

(αi − 1)|Tα| = (βj − 1)|Tβ|.

• Corollary. |Tα| =

(n−2)! (α1−1)!···(αn−1)!|T(n−1,1,1,...,1)| =

• n−2

α1−1,...,αn−1

• .

Hence,

• T

xdeg(1)

1

xdeg(2)

2

· · · xdeg(n)

n

= x1x2 · · · xn(x1 + x2 + . . . + xn)n−2. Rooted version:

• rooted Cayley tree

xch(1)

1

xch(2)

2

. . . xch(n)

n

= (x1 + x2 + · · · + xn)n−1, where ch(i) is the number of children of vertex i.

Main result Let n = (n1, . . . , nd) be a tuple of positive integers. Let Tρ(n) be the set of multitype Cayley trees of size n, and root-type ρ.

Main result Let n = (n1, . . . , nd) be a tuple of positive integers. Let Tρ(n) be the set of multitype Cayley trees of size n, and root-type ρ.

1 2 3 2 5 4 2 3 3 4 6 root vertex 1 1 1 2

Vertex (t, i) = (2, 3) gives weight x1,2,3 x2

3,2,3.

where chs(t, i)= number of children of type s of the vertex (t, i). The weight of a tree T is w(T) =

• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

,

Main result Let n = (n1, . . . , nd) be a tuple of positive integers. Let Tρ(n) be the set of multitype Cayley trees of size n, and root-type ρ. Theorem [B., Morales].

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 .

Cayley trees with vertex-set [d] and root-vertex ρ.

The weight of a tree T is w(T) =

• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

.

Main result Let n = (n1, . . . , nd) be a tuple of positive integers. Let Tρ(n) be the set of multitype Cayley trees of size n, and root-type ρ. Theorem [B., Morales].

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 . Lρ = (Ls,t)s,t∈[d]\{ρ}, with

• Ls,t

= −

i∈[nt] xs,t,i if s = t

Ls,s =

• t=s,i∈[nt] xs,t,i.

The weight of a tree T is w(T) =

• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

. det(Lρ)

Main result The weight of a tree T is w(T) =

• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

. Theorem [B., Morales].

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 . Example: Spanning trees of Km,n rooted on a black vertex. Set d = 2, ρ = 1, n1 = m, n2 = n, x1,1,i = x2,2,j = 0. Denote x2,1,i = xi, x1,2,j = yj.

• T ⊂Km,n
• i∈[m]

xch(bi)

i

• j∈[n]

y

ch(wj) j

= (x1 + · · · + xm)n(y1 + · · · + yn)m−1.

Sketch of proof. For α = (αs,t,i)s,t∈[d],i∈[nd] we denote by Tα the set of Cayley trees such that vertex (t, i) has αs,t,i children of type s.

• Claim. If βs,t,i = αs,t,i − 1 and βs,t,j = αs,t,j + 1 but everything else

is the same, then αs,t,i|Tα| = βs,t,j|Tβ|.

Sketch of proof. For α = (αs,t,i)s,t∈[d],i∈[nd] we denote by Tα the set of Cayley trees such that vertex (t, i) has αs,t,i children of type s. Proof.

• Claim. If βs,t,i = αs,t,i − 1 and βs,t,j = αs,t,j + 1 but everything else

is the same, then αs,t,i|Tα| = βs,t,j|Tβ|.

vertex (t, i) vertex (t, j) root-vertex vertex (t, i) vertex (t, j) root-vertex vertex (t, j) vertex (t, i) (a) (b)

• Φ
• Φ

vertex (t, i) vertex (t, j) root-vertex root-vertex

Sketch of proof. For α = (αs,t,i)s,t∈[d],i∈[nd] we denote by Tα the set of Cayley trees such that vertex (t, i) has αs,t,i children of type s.

• Claim. If βs,t,i = αs,t,i − 1 and βs,t,j = αs,t,j + 1 but everything else

is the same, then αs,t,i|Tα| = βs,t,j|Tβ|. Corollary: It suffices to count trees in which vertices not labelled 1 are leaves (the “star-trees”).

Sketch of proof. For α = (αs,t,i)s,t∈[d],i∈[nd] we denote by Tα the set of Cayley trees such that vertex (t, i) has αs,t,i children of type s.

• Claim. If βs,t,i = αs,t,i − 1 and βs,t,j = αs,t,j + 1 but everything else

is the same, then αs,t,i|Tα| = βs,t,j|Tβ|. Corollary: It suffices to count trees in which vertices not labelled 1 are leaves (the “star-trees”).

• T ∈Tρ(n)
• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

=

• T star-tree
• s,t∈[d]

nt

• i=1

xs,t,i

• chs(t,1)

.

Sketch of proof. For α = (αs,t,i)s,t∈[d],i∈[nd] we denote by Tα the set of Cayley trees such that vertex (t, i) has αs,t,i children of type s.

• Claim. If βs,t,i = αs,t,i − 1 and βs,t,j = αs,t,j + 1 but everything else

is the same, then αs,t,i|Tα| = βs,t,j|Tβ|. Corollary: It suffices to count trees in which vertices not labelled 1 are leaves (the “star-trees”).

• T ∈Tρ(n)
• s,t∈[d],i∈[nt]

xchs(t,i)

s,t,i

=

• T star-tree
• s,t∈[d]

nt

• i=1

xs,t,i

• chs(t,1)

.

Moreover, star-trees are easy to count:

• T star-tree
• s,t∈[d]

ychs(t,1)

s,t

=

• A∈Cayley

d,ρ

• (s,t)∈A

ys,t ×

d

• s=1

d

• t=1

ys,t

• ns−1

.

Adding the leaves Choosing subtree made of the d vertices labeled 1

Application 1: counting trees by edge type Thm [Bousquet, Chauve, Labelle, Leroux 03]. Let n = (n1, . . . , nd) be positive integers. The number of trees in Tρ(n) with ms,t edges of type (s, t) is

• s∈[d]

(ns − 1)!

• s,t∈[d]

nms,t

t

ms,t!

• A∈Cayley

d,ρ

• (s,t)∈A

ms,t.

Application 1: counting trees by edge type Thm [Bousquet, Chauve, Labelle, Leroux 03]. Let n = (n1, . . . , nd) be positive integers. The number of trees in Tρ(n) with ms,t edges of type (s, t) is

• s∈[d]

(ns − 1)!

• s,t∈[d]

nms,t

t

ms,t!

• A∈Cayley

d,ρ

• (s,t)∈A

ms,t. Proof: Set xs,t,i = xs,t in the main formula

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 and extract the coefficient of the monomial xmx,t

s,t .

Application 2: counting embedded trees Thm [Knuth 68]. Let n = (n1, . . . , nd) be positive integers. Let D ⊂ [d]2. The number of trees in Tρ(n) with edge-type in D is

d

• s=1

 

(s,t)∈D

nt  

ns−1

• A∈Cayley

d,ρ A⊂D

• (s,t)∈A

nt

Application 2: counting embedded trees Thm [Knuth 68]. Let n = (n1, . . . , nd) be positive integers. Let D ⊂ [d]2. The number of trees in Tρ(n) with edge-type in D is

d

• s=1

 

(s,t)∈D

nt  

ns−1

• A∈Cayley

d,ρ A⊂D

• (s,t)∈A

nt

• Proof. Set xs,t,i =
• 1 if (s, t) ∈ D,

0 otherwise in the main formula

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 .

Application 2: counting embedded trees Thm [Knuth 68]. Let n = (n1, . . . , nd) be positive integers. Let D ⊂ [d]2. The number of trees in Tρ(n) with edge-type in D is

d

• s=1

 

(s,t)∈D

nt  

ns−1

• A∈Cayley

d,ρ A⊂D

• (s,t)∈A

nt Corollary [Bousquet-M´ elou, Chapuy 12]. The number of trees in Tρ(n) such that the type of adjacent vertices differ by ±1 is nρ n1nd

d

• i=1

ni (ni−1 + ni+1)ni−1.

Application 3: injective trees and profile A tree is injective if each vertex has at most one vertex of each type. Thm [B., Morales]. Let n = (n1, . . . , nd) be positive integers. Let D ⊂ [d]2. The number of injective trees in Tρ(n) with edge-type in D is

d

• s=1

δs,ρ − 1 +

(s,t)∈D nt

ns − 1

• (ns − 1)!
• A∈Cayley

d,ρ A⊂D

• (s,t)∈A

nt.

Application 3: injective trees and profile A tree is injective if each vertex has at most one vertex of each type.

• Proof. Set xs,t,i = 0 if (s, t) /

∈ D in the main formula

• T ∈Tρ(n)

w(T) =

d

• s=1

d

• t=1

nt

• i=1

xs,t,i

• ns−

1

×  

• A∈Cayley

d,ρ

• (s,t)∈A

nt

• i=1

xs,t,i

• 

 . and extract coefficient. Thm [B., Morales]. Let n = (n1, . . . , nd) be positive integers. Let D ⊂ [d]2. The number of injective trees in Tρ(n) with edge-type in D is

d

• s=1

δs,ρ − 1 +

(s,t)∈D nt

ns − 1

• (ns − 1)!
• A∈Cayley

d,ρ A⊂D

• (s,t)∈A

nt.

Application 3: injective trees and profile Corollary [Bousquet-M´ elou, Chapuy 12]. Let a, b ≥ 0 and n = (n−a, n−a+1, . . . , nb−1, nb) be a tuple of positive numbers. The number of binary trees having shadow n is n0 n−anb

b

• s=−a

δs,0 − 1 + ns−1 + ns+1 ns − 1

• n:

1 1 3 4 3 2

Application 3: injective trees and profile Corollary [Bousquet-M´ elou, Chapuy 12]. Let a, b ≥ 0 and n = (n−a, n−a+1, . . . , nb−1, nb) be a tuple of positive numbers. The number of binary trees having shadow n is n0 n−anb

b

• s=−a

δs,0 − 1 + ns−1 + ns+1 ns − 1

• n:

1 1 3 4 3 2 (The shadow of trees is related to the Integrated SuperBrownian Excursion)

Application 4: complete degree distributions

type 1 1 2 3 2 5 4 2 3 3 4 6 root vertex type 2 type 3 1 1 type 4 1 2

The indegree of a vertex is the tuple c = (c1, . . . , cd), where cs is the number of children of type s.

Indegree c = (1, 0, 2, 0)

Application 4: complete degree distributions Thm [B., Morales]. Let n = (n1, . . . , nd) be positive integers. The number of trees in Tρ(n) with Nt,c vertices of type t and indegree c is

• t∈[d]

nt!(nt − 1)!

• t∈[d],c∈C

c!Nt,cNt,c! ×

• A∈Cayley

d,ρ

• (s,t)∈A

ms,t. where ms,t =

• c=(c1,...,cd)

csNt,c = number of edges of type (s, t). The indegree of a vertex is the tuple c = (c1, . . . , cd), where cs is the number of children of type s.

Application 4: complete degree distributions The indegree of a vertex is the tuple c = (c1, . . . , cd), where cs is the number of children of type s. Thm [B., Morales]. Let n = (n1, . . . , nd) be positive integers. The number of trees in Tρ(n) with Nt,u,c vertices of type t with parent

• f type u and indegree c is
• t∈[d]

nt!

• s,t∈[d],ms,t>0

(ms,t − 1)!

• t∈[d],c∈C

c!Nt,u,cNt,u,c! ×

• A⊂L(Kd)
• (s,t,u)∈A

ms,t,u, where ms,t= number of edges of type (s, t), and ms,t,u= number of 2-paths of type (s, t, u).

spanning tree of the line graph L(Kd)

Application 5. Multivariate Lagrange inversion formula Thm [Bender, Richmond 98]. Let G1, . . . , Gd+1, be power series in d variables with non-zero constant terms. The unique tuple (f1, . . . , fd) of power series in x1, . . . , xd satisfying ∀t ∈ [d], ft = xtGt(f1, .., fd) have coefficients given by [xn]Gd+1(f1, .., fd) = [xn]

d

• i=1

xi ni

• A∈Cayley

d + 1,ρ

d+1

• t=1

 

(s,t)∈A

∂ ∂xs  Gt(x)nt.

Part II Spanning trees of the hypercube

Hypercube The hypercube of dimension n is the graph with vertex set {0, 1}n and edges between vertices differing on one coordinate. Ceci est un hypercube de dimension 3. (1,0,0) (0,0,0) (0,0,0) (0,1,0) (1,1,0) (1,1,1) (0,1,1) (0,0,1) (1,0,1)

Counting spanning trees

• Theorem. The number of spanning trees of the n-hypercube is

2−n

• v∈{0,1}n

v=(0,0,..,0)

2|v|. 1, 4, 384, 42467328, 20776019874734407680, . . .

Counting spanning trees

• Theorem. The number of rooted spanning trees of the hypercube is
• v∈{0,1}n

v=(0,0,..,0)

2|v|.

Counting spanning trees

• Theorem. The number of rooted spanning trees of the hypercube is
• v∈{0,1}n

v=(0,0,..,0)

2|v|.

Counting spanning trees An arc (u, v) has direction i ∈ [n] and spin ǫ ∈ {0, 1} if v is obtained from u by changing the ith coordinate into ǫ.

(1,0,0) (0,0,0) (0,0,0) (0,1,0) (1,1,0) (1,1,1) (0,1,1) (0,0,1) (1,0,1)

direction 3 spin 0.

Counting spanning trees An arc (u, v) has direction i ∈ [n] and spin ǫ ∈ {0, 1} if v is obtained from u by changing the ith coordinate into ǫ.

(1,0,0) (0,0,0) (0,0,0) (0,1,0) (1,1,0) (1,1,1) (0,1,1) (0,0,1) (1,0,1)

x3,0 The weight of a rooted tree T is w(T) =

• a∈T

xdir(a),spin(a). w(T) = x1,0 x1,1 x2

2,0 x2 3,0 x3,1

Counting spanning trees Theorem [⇔ Martin, Reiner 03.]. The total weight of the rooted spanning trees of the hypercube is

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   .

Counting spanning trees Theorem [⇔ Martin, Reiner 03.]. The total weight of the rooted spanning trees of the hypercube is

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   . (x1,0 + x1,1)(x2,0 + x2,1)(x1,0 + x1,1 + x2,0 + x2,1)

Counting spanning trees Theorem [⇔ Martin, Reiner 03.]. The total weight of the rooted spanning trees of the hypercube is

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   .

• Proof. Use the matrix-tree theorem (for weighted directed graphs):

Total weight of rooted spanning trees =

λ=0 λ,

where the λ’s are the eigenvalues of the Laplacian L = (Lu,v)u,v∈V , defined by Lu,v = −w(u, v) , and Lu,u =

v w(u, v).

Counting spanning trees Theorem [⇔ Martin, Reiner 03.]. The total weight of the rooted spanning trees of the hypercube is

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   .

• Proof. Use the matrix-tree theorem (for weighted directed graphs):

Total weight of rooted spanning trees =

λ=0 λ,

where the λ’s are the eigenvalues of the Laplacian L = (Lu,v)u,v∈V , defined by Lu,v = −w(u, v) , and Lu,u =

v w(u, v).

Compute the eigenvalues of L: use either representation theory of abelian groups, or induction.

Counting spanning trees Theorem [⇔ Martin, Reiner 03.]. The total weight of the rooted spanning trees of the hypercube is

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   . Combinatorial proof?

An easy special case

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   .

(1, 1, 1) (0, 0, 0)

Proof.

• Claim. The special case xi,1 = 0 holds.

An easy special case

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   .

(1, 1, 1) (0, 0, 0)

Proof.

v = (1, 1, 0)

• v∈{0,1}n

v=(0,0,...,0)

 

i,vi=1

xi,0   x1,0 + x2,0

• Claim. The special case xi,1 = 0 holds.

The key property: independence of spins. Thm [B.]. Let S = {s1, . . . , sk} be a subset of edges in direction n. Let T be a uniformly random spanning tree conditioned to contain exactly these edges in direction n.

(1,0,0) (0,0,0) (0,0,0) (0,1,0) (1,1,0) (1,1,1) (0,1,1) (0,0,1) (1,0,1)

The key property: independence of spins. Thm [B.]. Let S = {s1, . . . , sk} be a subset of edges in direction n. Let T be a uniformly random spanning tree conditioned to contain exactly these edges in direction n. Then the spins of the edges s1, . . . , sk are uniformly random and independent. Example.

The key property: independence of spins. Thm [B.]. Let S = {s1, . . . , sk} be a subset of edges in direction n. Let T be a uniformly random spanning tree conditioned to contain exactly these edges in direction n. Then the spins of the edges s1, . . . , sk are uniformly random and independent. Example. This remains true if one further conditions on the number ni,ǫ of edges with direction i and spin ǫ, for i ∈ [n − 1], ǫ ∈ {0, 1}. n1,0 = 1, n1,1 = 0

The key property: independence of spins. Thm [B.]. Let S = {s1, . . . , sk} be a subset of edges in direction n. Let T be a uniformly random spanning tree conditioned to contain exactly these edges in direction n. Then the spins of the edges s1, . . . , sk are uniformly random and independent. This remains true if one further conditions on the number ni,ǫ of edges with direction i and spin ǫ, for i ∈ [n − 1], ǫ ∈ {0, 1}. This explains why xn,0 and xn,1 always appear together in the formula

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   . More precisely, it proves that changing the variables xn,0 and xn,1 respectively into xn,0 + xn,1 and 0 does not change the total weight. ⇒ Formula follows from independence of spins + special case.

Independence of spins in bunkbed graphs Example. A bunkbed graph is a graph of the form G × K2. G G × K2

Independence of spins in bunkbed graphs Example. A bunkbed graph is a graph of the form G × K2. G G × K2

• Remark. The hypercube is a bunkbed graph: Hn = K2 × · · · × K2.

Independence of spins in bunkbed graphs A bunkbed graph is a graph of the form G × K2. Projection: 1 1 1 2 1 Example. The projection of a tree T of G × K2 specifies:

• which vertical edges are used,
• for each arc a of G, the number of a-arcs in T.

Independence of spins in bunkbed graphs A bunkbed graph is a graph of the form G × K2. The projection of a tree T of G × K2 specifies:

• which vertical edges are used,
• for each arc a of G, the number of a-arcs in T.

Theorem [B.] Let T be a uniformly random rooted spanning tree of G × K2, conditioned to have a given projection. Then, the spins of the vertical edges are uniformly random and independent.

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. 1 1 1 2 1

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. Case (1). G′ = G − v. 1 1 1 2 1 1 1 2

2-to-1 correspondence (a) No vertical edge at v. v

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. Case (1). G′ = G − v. 1 1 1 2 1 1 1 2

2-to-1 correspondence (a) No vertical edge at v. v

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. Case (1). G′ = G − v.

(a) No vertical edge at v. (b) A vertical edge at v.

1 1 1 2 1

2-to-1 correspondence v

1 1 2 1

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. Case (1). G′ = G − v. Case (2). G′ = G − v + e.

1-to-1 correspondence

1 1 1 1 1 1

v

1 1 1 1 1 1 1 1

e e

+

Proof. Induction on # vertices of G. Claim: there is a vertex v of G such that either (1) indegree(v) = 0 in the projection, or (2) indegree(v) = 1 in the projection and there is no vertical edge at v. Case (1). G′ = G − v. Case (2). G′ = G − v + e.

1-to-1 correspondence

1 1 1 1 1 1

v

1 1 1 1 1 1 1 1

e e

+

Independence of spins and tree weights Theorem [B.] Let T be a uniformly random rooted spanning tree of G × K2, conditioned to have a given projection. Then, the spins of the vertical edges are uniformly random and independent. Corollary If wG(x0, x1) is the total weight of the rooted spanning trees of G × K2 then wG(x0, x1) = wG(x0 + x1, 0).

Independence of spins and tree weights Theorem [B.] Let T be a uniformly random rooted spanning tree of G × K2, conditioned to have a given projection. Then, the spins of the vertical edges are uniformly random and independent. Corollary. wHn =

• v∈{0,1}n

v=(0,0,...,0)

 

i, vi=1

xi,0 + xi,1   . Corollary If wG(x0, x1) is the total weight of the rooted spanning trees of G × K2 then wG(x0, x1) = wG(x0 + x1, 0).

Additional results

• Independence of spins also holds for rooted forests of G × K2.

Theorem [B.]. The total weight of the rooted forests of the hypercube (with weight t by tree) is

• v∈{0,1}n

 t +

• i, vi=1

xi,0 + xi,1   .

Additional results

• Independence of spins also holds for strong product G ⊠ K2.

G G ⊠ K2

• Independence of spins also holds for rooted forests of G × K2.

Additional results

• Independence of spins also holds for strong product G ⊠ K2.
• Another combinatorial proof of the hypercube formula.

Equivalently: for all v ∈ {0, 1}, FHn  −

• i, vi=1

xi,0 + xi,1   = 0. We want to prove FHn(t) =

• v∈{0,1}n

 t +

• i, vi=1

xi,0 + xi,1  . Use a sign reversing involution.

forest enumerator

• Independence of spins also holds for rooted forests of G × K2.

Additional results

• Independence of spins also holds for strong product G ⊠ K2.
• Another combinatorial proof of the hypercube formula.

⇒ Formula for hypercube with main diagonals. FDn(x, y; t) =

• v∈{0,1}n

 t + 2y · 1|v| odd +

• i,vi=1

xi,0 + xi,1  

• Independence of spins also holds for rooted forests of G × K2.

Conjecture Using the matrix-tree theorem one gets: FG×Kp(t) = FG(t) · FG(t + x1 + x2 . . . + xp)p−1. Conjecture. If T is a uniformly random spanning tree of G × Kp conditioned to a have a certain projection, then the weight of the subforests in the copies of Kp are independent. 2 2 3 1 1

1 1 2 1 Projection of G × K3

Thanks.

1 2 3 2 5 4 2 3 3 4 6 1 1 1 2