Mobius Functions of Posets IV: Why theCharacteristic Polynomial Factors

Bruce SaganDepartment of MathematicsMichigan State University

East Lansing, MI 48824-1027sagan@math.msu.edu

www.math.msu.edu/˜sagan

June 28, 2007

The Characteristic Polynomial

The Chromatic Polynomial

The Bond Lattice

The Connection

Outline

The Characteristic Polynomial

The Chromatic Polynomial

The Bond Lattice

The Connection

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P.

Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x .

Note thatrk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.

3. d ∈ Dn: rk d =∑

i mi where d =∏

i pmii .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

For more information on this topic, see

B. Sagan, Why the characteristic polynomial factors, Bull. Amer.Math. Soc. 36 (1999), 113–134.

Let P be a graded poset and let x ∈ P. Then all saturated 0–xchains have the same length called the rank of x and denotedrk x . Note that

rk P = rk 1.

Example. In our four running example posets:

1. i ∈ Cn: rk i = i .

2. S ∈ Bn: rk S = |S|.3. d ∈ Dn: rk d =

∑i mi where d =

∏i pmi

i .

4. π = B1/ . . . /Bk ∈ Πn: rk π = n − k .

Let P be a graded poset and let t be a variable.

Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic:

the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0

= t2(t − 1)

C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0 = t2(t − 1) C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Let P be a graded poset and let t be a variable. Thecharacteristic polynomial of P is the generating function

q(P) = q(P; t) =∑x∈P

µ(x)t rk 1−rk x .

Note that we use the corank, rk 1− rk x , as the exponent on trather than the rank so as to make q(P; t) monic: the largestpower of t is when x = 0 and µ(0) = 1.

Example.

q(C3) = t3 − t2 + 0 + 0 = t2(t − 1) C3=

s0

s1

s2

s3

1

−1

0

0

1 · t3

−1 · t2

0 · t

0 · 1

In general, q(Cn; t) = tn−1(t − 1) (easy to verify).

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1

= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).

Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t

= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(B3) = t3 − 3t2 + 3t − 1= (t − 1)3

B3 =

t∅QQ

QQQ

��

���

t t t1 2 3QQ

QQQ

t12

��

���

QQ

QQQ

t13

��

���

t23��

���

QQ

QQQ

t123

1 · t3

−1 · t2 −1 · t2 −1 · t2

1 · t 1 · t 1 · t

−1 · 1

In general q(Bn; t) = (t − 1)n (use the Binomial Theorem).Example.

q(D18) = t3 − 2t2 + t= t(t − 1)2

D18 =

t1@@

@

��

�t t2 3�

��

@@

@

��

�t t6 9�

��

@@

@t18

1 · t3

−1 · t2 −1 · t2

1 · t 0 · t

0 · 1

In general q(Dn; t) =∏

i tmi−1(t − 1) where n =∏

i pmii .

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2

= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Example.

q(Π3) = t2 − 3t + 2= (t − 1)(t − 2)

Π3 =

t1/2/3@@

@@

@@

��

��

��t t t

12/3 13/2 1/23�

��

���

@@

@@

@@t123

1 · t2

−1 · t −1 · t −1 · t

2 · 1

In general q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1). Not clear!

We would like to have a technique which would

1. prove the formula for q(Πn; t),

2. explain why these q(P; t) factor over Z≥0.

We will use a technique based on graph theory. Two othertechniques (one using the theory of hyperplane arrangementsand one using properties of posets) are given in the paper.

Outline

The Characteristic Polynomial

The Chromatic Polynomial

The Bond Lattice

The Connection

Let G be a graph with vertices V and edges E .

Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C.

Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G

while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0.

The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

u

t

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

v

t − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t

(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

vt − 1

w

t − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t(t − 1)

(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

vt − 1

wt − 1

x

t − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 1)

(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

vt − 1

wt − 1

xt − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 1)(t − 1)

= t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

vt − 1

wt − 1

xt − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let G be a graph with vertices V and edges E . Given a set C,called the color set , a coloring of G is a function κ : V → C. Acoloring is proper if for each edge uv ∈ E we have κ(u) 6= κ(v).

Example. If G =

qq q

q

JJJ

and C = {1, 2} then

q qqq

JJJ

2

2

12

is a proper coloring of G while q qqq

JJJ

1

2

12

is not.

Let t ∈ Z≥0. The chromatic polynomial of G is

p(G) = p(G; t) = # of proper κ : V → {1, . . . , t}.

Example. If G = q qqq

JJJ

ut

vt − 1

wt − 1

xt − 1

then

p(G; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 1)(t − 1) = t(t − 1)3

In general of any tree T with |E | = n: p(G; t) = t(t − 1)n.

Let KV be the complete graph having all possible edgesbetween elements of V .

If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

u

t

v

t − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

u

t

v

t − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

u

t

v

t − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

u

t

v

t − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

=

t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

v

t − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t

(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

w

t − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)

(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

wt − 2

x

t − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 2)

(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

wt − 2

xt − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

wt − 2

xt − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

wt − 2

xt − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Let KV be the complete graph having all possible edgesbetween elements of V . If |V | = n then we also write Kn for KV .

Example. If G = K4 =

r r

r r

��

��

��

@@

@@

@@

ut

vt − 1

wt − 2

xt − 3

then

p(K4; t) = # of ways to color u, then v , then w , then x

= t(t − 1)(t − 2)(t − 3)

In general p(Kn; t) = t(t − 1) · · · (t − n + 1).

We need to explain

1. why does p(G; t) always seem to be a polynomial in t?

2. why do p(T ; t) where T is a tree and p(Kn; t) seem to berelated to q(Bn; t) and q(Πn; t), respectively?

Outline

The Characteristic Polynomial

The Chromatic Polynomial

The Bond Lattice

The Connection

Let k(G) denote the number of components of G.

We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G.

If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.

A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G),

i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H),

i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z

then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G,

while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

Let k(G) denote the number of components of G. We also writeu ∼ v if u and v are in the same component of G. If we wish tobe specific about the graph in question, we will use notationsuch as V (G) or ∼G.A subgraph H ⊆ G is a bond if

1. V (H) = V (G), i.e., H is spanning.

2. u ∼H v and uv ∈ E(G) implies uv ∈ E(H), i.e., eachcomponent of H is an induced subgraph.

Example. If G =

q qJ

JJ

J

JJq q

J

JJ

q qu

v

w

x

y

z then

H =

q qJ

JJq q

q qu

v

w

x

y

z is a bond of G, while H =

q qJ

JJJ

JJq q

q qu

v

w

x

y

z is not

since u ∼H v and uv ∈ E(G) but uv 6∈ E(H).

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q q

ZZ

ZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q

��

�

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

The bond lattice of G is

L(G) = {H : H is a bond of G}

partially ordered by inclusion.

Example.

T =

qq q q

JJ

So L(T ) ∼= B3.

L(T ) =

'&

$%

qq q qZ

ZZZ

��

�

'&

$%

qq q q'&

$%

qq q qJJ

'&

$%

qq q q�

��

ZZ

ZZ

��

�

ZZ

ZZ

'&

$%

qq q qJJ

'&

$%

qq q q

'&

$%

qq q q

JJ

��

�

ZZ

ZZ

'&

$%

qq q q

JJ

In general, if T is a tree with n edges then L(T ) ∼= Bn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA�

��

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA�

��

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA

��

�

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA�

��

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA�

��

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Example.

K3 =

qq q���

AA

A

So L(K3) ∼= Π3.

L(K3) =

'&

$%

qq q

ZZ

Z

��

�

'&

$%

qq q���

'&

$%

qq q

'&

$%

qq qA

AA�

��

ZZ

Z

'&

$%

qq q���

AA

A

In general, L(Kn) ∼= Πn.

Outline

The Characteristic Polynomial

The Chromatic Polynomial

The Bond Lattice

The Connection

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1

then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q q

LemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a).

Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition.

Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G).

Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ.

So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color.

Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.

Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G)

iffE(Hκ) = ∅ iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅

iff Hκ = 0L(G).

Given a coloring κ : V → {1, . . . , t} of G, the kernel of κ is thespanning subgraph Hκ ⊆ G where

uv ∈ E(Hκ) if and only if uv ∈ E(G) and κ(u) = κ(v).

Example. If κ =

q qJ

JJ

J

JJq q

J

JJ

q q1

1

2

1

1

1 then Hκ =

q qJ

JJq q

q qLemmaFor any graph G

(a) Hκ is a bond for any κ.

(b) κ is proper iff E(Hκ) = ∅ iff Hκ = 0L(G).

Proof of (a). Hκ is spanning by definition. Now supposeu ∼Hκ

v and uv ∈ E(G). Since u ∼Hκv there is a u–v walk W

in Hκ. So κ(u) = κ(v) since every pair of adjacent vertices ofW have the same color. Thus uv ∈ E(Hκ) as desired.Proof of (b). κ is proper iff κ(u) 6= κ(v) for all uv ∈ E(G) iffE(Hκ) = ∅ iff Hκ = 0L(G).

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b).

Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,

2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,Since Hκ is always a bond, f (H) =

∑H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′).

By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0

so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).

Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′.

So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′).

Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G.

Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′

= tk(G)q(L(G); t)

Theorem (Rota, 1964)Let G be a graph with |V | = n.

(a) L(G) is a graded lattice with rk H = n − k(H).

(b) p(G; t) = tk(G)q(L(G); t).

Proof of (b). Define f , g : L(G) → Z by1. f (H) = # of κ : V (G) → {1, . . . , t} with Hκ ⊇ H,2. g(H) = # of κ : V (G) → {1, . . . , t} with Hκ = H,

Since Hκ is always a bond, f (H) =∑

H′⊇H g(H ′). By the MIT

g(0) =∑

H′⊇0 µ(H ′)f (H ′). (1)

But g(0) counts κ with Hκ = 0 so, by the lemma, g(0) = p(G; t).Also, given H ′ then Hκ ⊇ H ′ iff κ is constant on the componentsof H ′. So f (H ′) = tk(H′). Thus (1) becomes

p(G; t) =∑

H′∈L(G) µ(H ′)tk(H′). (2)

By (a), k(H ′) + rk H ′ = n = k(G) + rk G. Plugging into (2) gives

p(G; t) =∑

H′∈L(G)

µ(H ′)tk(G)+rk G−rk H′= tk(G)q(L(G); t)

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof.

For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.

So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn.

So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).

Rota: p(G; t) = tk(G)q(L(G); t).

Corollary

1. q(Bn; t) = (t − 1)n.

2. q(Πn; t) = (t − 1)(t − 2) · · · (t − n + 1).

Proof. For Bn we have L(T ) ∼= Bn for any tree T with n edges.So by Rota’s Theorem

q(Bn; t) = t−k(T )p(T ; t)

= t−1 · t(t − 1)n

= (t − 1)n.

For Πn we have L(Kn) ∼= Πn. So by Rota’s Theorem

q(Πn; t) = t−k(Kn)p(Kn; t)

= t−1 · t(t − 1) · · · (t − n + 1)

= (t − 1) · · · (t − n + 1).