[PPT] - Counting with automorphisms Lectures for CO 430 / 630 March 24 PowerPoint Presentation

SLIDE 1

Counting with automorphisms

Lectures for CO 430/630 March 24 – April 2, 2020

SLIDE 2

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§1. Counting orbits of group actions

CO 430/630 Counting with automorphisms

SLIDE 3

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Group actions

G a finite group. X a set.

Definition. An action of G on X is a map G × X → X, written

(g,x) → gx such that – 1x = x for all x ∈ X, and – (gh)x = g(hx) for all g,h ∈ G, x ∈ X. For x ∈ X:

◮ the orbit of x is

x = {gx | g ∈ G}.

◮ the stabilizer of x is Gx = {g ∈ G | gx = x}.

Orbit–Stabilizer Lemma. # x · #Gx = #G.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: Functions on G

K a set. KG = set of functions G → K. We have an action of G on KG: for α ∈ KG, g ∈ G, gα : h → α(g−1h). Special case: G = Cn ← cyclic group of order n. K ← finite set of “colours”. KCn ↔ colouring sides of a regular n-gon. Cn-action ↔ rotating the n-gon. Orbits ↔ equivalence classes of colourings up to rotation.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Figure: Colourings of the square (n = 4) with 2 colours

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Orbit counting

X = {

x | x ∈ X} ← set of orbits. For g ∈ G, the set of g-fixed points is Xg = {x ∈ X | gx = x}. Orbit Counting Lemma (a.k.a. Burnside’s Lemma) Let X be a finite set with a G-action. # X = 1 #G

g∈G

#Xg (Number of orbits = Average number of fixed points.) Proof. 1 #G

g∈G

#Xg = 1 #G

(g,x)∈G×X

gx=x

1 = 1 #G

x∈X

#Gx =

x∈X

#Gx #G =

x∈X

1 # x =

x∈

X

x∈

x

1 # x =

x∈

X

1 = # X.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: Colourings of the n-gon up to rotation

We compute the number of Cn-orbits on X = KCn, |K| = k. For each c ∈ Cn, Xc = {f : Cn → K | f(i − c) = f(c) ∀i ∈ Cn} = {f : Cn → K | f(i − d) = f(d) ∀i ∈ Cn}, d = gcd(c,n). Hence #Xc = kgcd(c,n). By the orbit counting lemma, # X = 1 n

c∈Cn

kgcd(c,n) . Let φ be the totient function: φ(m) = #{ℓ ∈ [m] | gcd(ℓ,m) = 1}. For each divisor d of n there are φ(n/d) elements c ∈ Cn such that gcd(c,n) = d. Hence # X = 1 n

d|n

φ(n/d)kd .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

An example involving generating functions

Let Sm be the set of functions f : Cn → such that

c∈Cn f(c) = m.

How many Cn-orbits? Let S = Cn (all functions Cn → ), with weight function wt : S → , wt(f) =

c∈Cn f(c).

The OGF for S is S(x) = (1 − x)−n. #Sm = [xm](1 − x)−n. For c ∈ Cn, we have a bijection Sc ↔ ( n

d)Cd, d = gcd(c,n)

(see next slide). Hence the OGF for Sc is (1 − xd)−n/d. #Sc

m = [xm](1 − xn/d)−d .

Hence the number of Cn orbits on Sm is [xm]1 n

d|n

φ(n/d)(1 − xn/d)−d .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Figure: The bijection Sc ↔ (n

d)Cd

2 5 1 2 5 1 2 0 5 1 6 15 3 Functions C12 → fixed by 4 ∈ C12 Functions C4 → 3 ↔ ↔

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Exercises

1. Let X be the set of colourings of the n-gon with two colours.

The dihedral group Dn of order 2n acts on X by rotations and

reflections. (Note: if n is even there are two different types of

reflections — we can reflect across a line through two

pposite vertices, or across a line through two opposite

edges.) For n = 4, n = 6 and n = 9, use the orbit counting lemma to compute the number of Dn-orbits.

Answer: 6, 13 and 46

2. Let Rm be the set of colourings of the n-gon with m sides

coloured red, and n − m sides coloured blue. The cyclic group Cn acts on Rm by rotation. Find a formula for the number of

rbits, as the coefficient of xm in a generating function.

Answer: [xm]

d|n φ(n/d)(1 + xn/d)d

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§2. Type generating functions

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Isomorphism types of species

Sn the symmetric group of permutations of [n]. A a species. We have an action of Sn on A[n]: for σ ∈ Sn, α ∈ A[n], σα = σ∗(α). Hence α,β ∈ A[n] are isomorphic iff α = β. The set of isomorphism types of the species A is

A =
n≥0
A[n] =
n≥0

{ α | α ∈ A[n]}, with weight function ord : A → , ord( α) = n iff α ∈ A[n]. The type generating function of the species A is the OGF for A:

A(x) =
α∈

A

xord(

α) .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Examples of type generating functions

◮ E – species of sets. One isomorphism type of each order.

E(x) =
n≥0

1xn = (1 − x)−1

◮ L• – species of rooted linear orders. L also has one

isomorphism type of each order n, but there are n non-isomorphic ways to root.

L•(x) =
n≥0

nxn = x(1 − x)−2

◮ S – species of permutations. Two permutations are isomorphic

iff they have the same cycle type (same number of cycles of each size). Isomorphism types of order n ↔ partitions of n.

S(x) =
i≥1

(1 − xi)−1

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Properties and non-properties

Proposition. Let A and B be species.
1. The type generating function of A + B is

A(x) + B(x).

2. The type generating function of A ∗ B is

A(x) B(x). Unfortunately, for other species operations we can’t determine the TGF of the output species just from the TGF of the input species. This is why TGFs are usually harder to compute than EGFs.

Example. E and L have the same TGF

, but E• and L• have different TGFs.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Type generating functions via orbit counting lemma

For σ ∈ Sn, write Aσ = {α ∈ A[n] | σ∗(α) = α} . By the orbit counting lemma, the number of Sn-orbits on A[n] is 1 n!

σ∈Sn

#Aσ . Hence

A(x) =
n≥0

1 n!

σ∈Sn

#Aσxn . This is usually difficult to use directly. Goal: Develop a generalization of this formula which will allow us to compute TGFs in a manner similar to EGFs. We will need to work with objects that are more general than formal power series.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Exercises

1. Find the first few terms of the type generating function for the

species T of trees.

Answer: T(x) = x + x2 + x3 + 2x4 + 3x5 + ...

2. Compute the type generating functions for E[E+] and L[E+].

Notice that TGFs of E and L are the same, but the TGFs of E[E+] and L[E+] are different!

Answer:

i≥1(1 − xi)−1 and (1 − x)(1 − 2x)−1

CO 430/630 Counting with automorphisms

SLIDE 17

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§3. Cycle index functions

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

A more general kind of function

Let be the set of all functions [[x]]+ → [[x]]. is a commutative -algebra under pointwise operations:

◮ For f ∈ , c ∈ , cf ∈ is the map U(x) → cf[U(x)]. ◮ For f,g ∈ , f ± g ∈ is the map U(x) → f[U(x)] ± g[U(x)]. ◮ For f,g ∈ , fg ∈ is the map U(x) → f[U(x)]g[U(x)].

We can also take pointwise limits.

◮ For a sequence f1,f2,f3 ∈ , limn→∞ fn ∈ is the map

U(x) → limn→∞ fn[U(x)] (assuming this limit exists for all U(x) ∈ [[x]]+). Under suitable conditions, we can compose functions in .

◮ Let + ⊆ be the subset of functions [[x]]+ → [[x]]+.

If f ∈ , g ∈ + then f ◦ g ∈ is the map U(x) → f[g[U(x)]].

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

A note about the notation

For a function f ∈ we are using square brackets f[·] to denote evaluations of f rather than the usual round brackets f(·). This is partly to remind ourselves that f is not a formal power series, and the expression f[U(x)] does not obey the same rules as composition of formal power series! For example, if f[x] = g[x], this does not imply f[2x] = g[2x], nor vice-versa. Even with the different notation, it is very easy to make this type of mistake.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Examples of such functions

◮ For A(x) ∈ [[x]], the left-composition (application) map is:

apA(x) ∈ , apA(x)[U(x)] = A(U(x)).

◮ For B(x) ∈ [[x]]+, the right-composition (evaluation) map is:

evB(x) ∈ +, evB(x)[U(x)] = U(B(x)).

◮ We can construct a wide variety of functions using the above

and pointwise operations. For example f = 1 + (evx+x2+x3)5 + apexp(x) ◦ evx2 1 − evx7 is the map f[U(x)] = 1 + U(x + x2 + x3)5 + exp(U(x2)) 1 − U(x7) .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

The cycle index function of a species

We associate a function ZA ∈ to every species A.

◮ For k = 1,2,3,..., let pk = evxk ∈ + (i.e. pk[U(x)] = U(xk)). ◮ For a permutation σ ∈ Sn, let cyci(σ) denote the number of

i-cycles in σ. Let pσ = pcyc1(σ)

1

pcyc2(σ)

2

···p

cycn(σ) n

.

Definition. The cycle index function of a species A is

ZA =

n≥0

1 n!

σ∈Sn

#Aσpσ . Hence ZA is the map ZA[U(x)] =

n≥0

1 n!

σ∈Sn

#Aσ

n

i=1

U(xi)cyci(σ) .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Cycle index function vs. Type generating function

Let’s compare the formulas side by side.

A(x) =
n≥0

1 n!

σ∈Sn

#Aσxn ZA =

n≥0

1 n!

σ∈Sn

#Aσpσ Not only are the formulas very similar, the TGF is obtained as an evaluation of the cycle index function.

Proposition. For any species A,

A(x) = ZA[x].

Proof. For every σ ∈ Sn, pσ[x] =

n

i=1(xi)cyci(σ) = x n

i=1 icyci(σ) = xn.

Thus ZA[x] =

n≥0

1 n!

σ∈Sn

#Aσpσ[x] =

n≥0

1 n!

σ∈Sn

#Aσxn = A(x).

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Exercises

1. Let f = exp(p2p4 − p2

3). Compute f[x + x2].

Answer: exp

(x4 − x5)2
2. Let T4 be the species of trees with 4 vertices. That is, for a

finite set X, (T4)X = TX if |X| = 4, and (T4)X = otherwise. For each permutation σ ∈ S4, compute #Tσ

4 . Hence compute ZT4.

Answer: ZT4 = 2

3p4 1 + 1 2p2 1p2 + 1 3p1p3 + 1 2p2 2

3. Let g ∈ + and let A(x) =
n≥0 anxn ∈ [[x]] be a formal

power series. Show that A(g) =

n≥0 angn ∈ is defined, and

A(g) = apA(x) ◦ g.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§4. Examples of cycle index functions

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Treating the cycle index function as a formal power series

So far, we have emphasized the fact that pi ∈ and ZA ∈ are functions [[x]]+ → [[x]]. In particular this is how we get the TGF . Note however, that the definition the cycle index function expresses ZA as a formal power series in infinitely many variables p1,p2,.... As it turns out, p1,p2,... are algebraically independent elements of . Thus the algebra of formal power series [[p1,p2,...]] is a subalgebra of . As we’ll see in the next few examples, we can often completely forget that p1,p2,... are functions, and simply regard them as independent indeterminates.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: The cycle index function of L

Linear orders have no non-trivial automorphisms. For σ ∈ Sn, #Lσ =

n!

if σ = idn

therwise.

where idn ∈ Sn denotes the identity element. Thus, ZL =

n≥0

1 n!

n!pidn
=
n≥0

pn

1 = (1 − p1)−1 .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: The cycle index function of E

Note that #Eσ = 1 for all permutations σ. Thus ZE =

n≥0

1 n!

σ∈Sn

pcyc1(σ)

1

pcyc2(σ)

2

··· Consider the mixed generating function for the species S of permutations, with weight functions cyc1,cyc2,.... S(x;t1,t2,...) =

n≥0
σ∈S[n]

tcyc1(σ)

1

tcyc2(σ)

2

...,

xn

n! Thus ZE = S(1;p1,p2,...). By the usual MGF arguments, S(x;t1,t2,...) = exp

k≥1 tk

xk k

, and so

ZE = exp

k≥1

pk k

.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: The cycle index function of S

Sn acts on S[n] by conjugation. Note: Sn and S[n] are the same set, but Sn is a group and S[n] is a set with group action. For a permutation σ with can think of σ ∈ Sn and consider its fixed point set Sσ, or think of σ ∈ S[n] and consider its stabilizer Sσ. For conjugation action, these are the same set. Thus ZS =

n≥0

1 n!

σ∈Sn

#Sσpσ =

n≥0
σ∈Sn

#Sσ n! pσ =

n≥0
σ∈Sn

1 # σpσ . The orbit σ is the set of all permutations with the same cycle type as σ. If this cycle type is the partition λ ⊢ n, then pσ = ℓ(λ)

i=1 pλi.

Thus ZS =

n≥0
λ⊢n
σ∈

σ ( σ↔λ)

1 # σ

ℓ(λ)

i=1

pλi =

n≥0
λ⊢n

ℓ(λ)

i=1

pλi =

∞

k=1

(1 − pk)−1 .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: The cycle index function of C

For α ∈ C[n], the stabilizer (Sn)α is precisely the cyclic subgroup of Sn generated by α. This subgroup has τ(ℓ) permutations of cycle type (ℓ,ℓ,...,ℓ), for all ℓ dividing n. Thus, ZC =

n≥1

1 n!

σ∈Sn

#Cσpσ =

n≥1

1 n!

α∈C[n]
σ∈(Sn)α

pσ =

n≥1

1 n!

α∈C[n]
ℓ|n

τ(ℓ)pn/ℓ

ℓ

=

n≥1

1 n!(n − 1)!

ℓ|n

τ(ℓ)pn/ℓ

ℓ

=

n≥1
ℓ|n

1 nτ(ℓ)pn/ℓ

ℓ

=

d≥1
ℓ≥1

1 ℓdτ(ℓ)pm

ℓ

=

ℓ≥1

τ(ℓ) ℓ

d≥1

pd

ℓ

d =

ℓ≥1

τ(ℓ) ℓ log(1 − pℓ)−1 .

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Exercises

1. Let A be any species. If σ,π ∈ Sn have the same cycle type,

show that #Aσpσ = #Aπpπ .

2. Suppose we ignore the fact that p1,p2,··· ∈ and view ZA

purely as formal power series ZA(p1,p2,...) ∈ [[p1,p2,...]]. Show that A(x) = ZA(x,0,0,...) is the exponential generating function of A. Verify this fact for the species L, E, S and C.

3. For α ∈ C[n], prove that the stabilizer (Sn)α is the cyclic

subgroup generated by α. (Hint: First use the orbit-stabilizer lemma, to compute #(Sn)α.)

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§5. The main theorem

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Main theorem for cycle index functions

As we saw, the cycle index function generalizes the TGF . Unlike TGFs, cycle index functions behave well under species

perations.

Main Theorem. For any species A, B: (1) ZA±B = ZA ± ZB. (2) ZA∗B = ZAZB. (3) ZA′ =

∂ ∂ p1 ZA (and ZA• = p1 ∂ ∂ p1 ZA).

(4) If B is connected, ZA◦B = ZA ◦ ZB. Corollary (TGFs under composition). If C = A ◦ B, then

C(x) = ZA[

B(x)]. Proof. C(x) = ZC[x] = ZA ◦ ZB[x] = ZA[ZB[x]] = ZA[ B(x)].

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Sketch of the proof of the Main Theorem

For a species A, define Aut(A) be the species such that Aut(A)X = {(σ,α) ∈ SX × AX | σ∗(α) = α}. We can think of ZA as a mixed generating function for Aut(A). In each case, we need to understand how to (bijectively) build Aut(·)-structures for the output species from Aut(·)-structures for the input species. For example an automorphism of a A ∗ B structure is simply an automorphism of the A-structure together with an automorphism

f a B-structure. Thus we identify Aut(A ∗ B) ≃ Aut(A) ∗ Aut(B).

This yields (2). Composition is the most complicated, but the basic idea is similar.

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Computing compositions

Although computing compositions of cycle index functions can lead to some ugly expressions, it can in principle always be done.

Theorem. In [[p1,p2,...]] ⊂ we have the following identities.

(1) pi ◦ pj = pij (2) pi ◦ f = f ◦ pi (3) f ◦ g = f(g ◦ p1,g ◦ p2,g ◦ p3,...) Here, i,j ≥ 1, and f = f(p1,p2,p3,...) and g = g(p1,p2,p3,...) are formal power series in p1,p2,p3,....

Proof. For each identity, simply check that LHS[U(x)] = RHS[U(x)]

for all U(x) ∈ [[x]]+. For (2) and (3), use the fact that f

h1,h2,...
[U(x)] = f
h1[U(x)],h2[U(x)],...
for any h1,h2,··· ∈ , provided f(h1,h2,...) is defined.

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Example: Cycle index function of species of involutions

The species of involutions is E[E1 + E2]. We know that ZE = exp(

k≥1

pk k ). By direct calcuation, ZE1 = p1

and ZE2 = 1

2p2 1 + 1 2p2.

ZE[E1+E2] = ZE ◦ (ZE1 + ZE2) = exp

k≥1

pk k

p1 + 1

2p2 1 + 1 2p2

= exp
k≥1
p1 + 1

2p2 1 + 1 2p2

pk

k

= exp
k≥1
(pk ◦ p1) + 1

2(pk ◦ p1)2 + 1 2(pk ◦ p2)

k
= exp
k≥1

pk + 1

2p2 k + 1 2p2k

k

CO 430/630

Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Colourings of the n-gon, revisited

Let K be a finite set of colours, |K| = k. C[K × X] is the species in which structures are cyclic permutations, together with an assignment of a colour for each label. Isomorphism types ↔ cyclically ordereded list of colours ↔ colourings of a regular polygon up to rotation. ZC[K×X][x] = ZC ◦ ZK×X[x] =

ℓ≥1

τ(ℓ) ℓ log(1 − pℓ)−1

kp1
[x]

=

ℓ≥1

τ(ℓ) ℓ log

1 − pℓ[kx]

−1 =

ℓ≥1

τ(ℓ) ℓ log(1 − kxℓ)−1 =

ℓ≥1
d≥1

τ(ℓ)kdxℓd ℓd =

n≥1

1 n

d|n

τ(n/d)kd

xn

Coefficient of xn counts of colourings of the n-gon up to rotation.

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Exercises

1. Let f = p2

2 − 4p4, g = p2 1 + 2p2. Compute f ◦ g.

Answer: p4

2 − 4p2 2p4 + 8p8

2. Compute the cycle index function for the species S•.

Answer: p1(1 − p1)−1ZS

3. Compute the cycle index function for the species S[E2].

Answer: ∞

k=1

1 − 1

2p2 k − 1 2p2k

−1

CO 430/630 Counting with automorphisms

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

§6. Enumeration of unlabelled trees

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§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

General formula for ZE[U(x)]

Lemma. If U(x) =
m≥1 umxm, then

ZE[U(x)] =

m≥1

(1 − xm)−um .

Proof. Since ZE = exp(
k≥1

1 kpk),

ZE[U(x)] = exp

k≥1

1 kpk[U(x)]

= exp
k≥1

1 kU(xk)

= exp
k≥1

1 k

m≥1

umxkm = exp

m≥1

um

k≥1

xkm k

= exp

m≥1

um log(1 − xm)−1 =

m≥1

(1 − xm)−um

CO 430/630 Counting with automorphisms

SLIDE 40

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Enumeration of unlabelled rooted trees

From T• ≃ X ∗ E[T•], we have

T•(x) = x · ZE[

T•(x)]. If we write T•(x) =

n≥1 tnxn, then comparing coefficients of xn on

both sides gives tn = [xn−1]

m≥1

(1 − xm)−tm which gives tn recursively in terms of t1,t2,...,tn−1.

Remark. It is also possible to obtain non-recursive formulas for tn,

using a version of LIFT for cycle index functions. However, this is not necessarily a big improvement. We get a relatively nice formula for each coeffient of ZT•, but then tn is a sum of many such

coefficients. There are no simple non-recursive formulas for tn.

CO 430/630 Counting with automorphisms

SLIDE 41

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Enumeration of unlabelled unrooted trees

There is no general theorem that relates A(x) to A•(x). However, for trees it is possible to express T(x) in terms of T•(x). Theorem. T(x) = (p1 − 1

2p2 1 + 1 2p2)

T•(x)
.
Proof. We consider two types of unlabelled trees.
I. Trees for which every automorphism has a fixed vertex.
II. Trees that have an automorphism with no fixed vertices.

Most trees are of type I. Type II trees are quite special. They are

btained by taking two isomorphic rooted trees and joining the

roots.

CO 430/630 Counting with automorphisms

SLIDE 42

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees

Now, for any unlabelled tree τ:

◮ Let vτ be the number of ways to root τ at a vertex. ◮ Let eτ be the number of ways to root τ at an edge.

For type I trees, vτ − eτ = 1. For type II trees, vτ − eτ = 0. Thus without knowing vτ and eτ, we can count unrooted trees as: #

Vertex-rooted trees
− #
Edge-rooted trees
+ #
Type II trees
(1) Vertex-rooted trees. TGF is

T•(x). (2) Edge-rooted trees. Correspond to unordered pairs of vertex rooted trees. TGF is ZE2[ T•(x)] = 1

2

T•(x)2 + 1

2

T•(x2) (3) Type II trees. These correspond to pairs of isomorphic trees. TGF is p2[ T•(x)] = T•(x2). Taking (1) − (2) + (3) gives the desired formula for T(x).

CO 430/630 Counting with automorphisms

SLIDE 43

§1. Counting orbits §2. Type generating functions §3. Cycle index functions §4. Examples §5. Main theorem §6. Unlabelled trees