Chapter 1

Algebraic Graph Theory

• Linear algebra

• Group theory (Cayley graphs, Dynkin diagrams)

1.1 Eigenvalues

Definition 1.1 (Adjacency matrix). Let G = (V,E) be a finite graph. The adjacency matrixA(G) = {0, 1}V×V is defined by

A(G)v,w =

{1 if v ∼ w

0 otherwise

Recall the characteristic polynomial of a square matrix M over C

φM(t) := det(tI −M).

Question: Is a graph discussed in the lecture (and homework) always simple?

Answer: Yes. (According to the instructor).

Definition 1.2 (Characteristic polynomial). Let G = (V,E) be a finite graph. The Char-acteristic polynomial is defined by

φG(t) := φA(G)(t),

and call the zeros (with multiplicitier) the eigenvalues of G.

Definition 1.3 (Spectrum). The spectrum of G is the multiset of its eigenvalues.

Example 1.1 (Spectrum).

A(G) =

[0 11 0

]φG(t) = det

(t −1−1 t

)= t2 − 1

spectrum : {1,−1}

1

1.1. EIGENVALUES CHAPTER 1. ALGEBRAIC GRAPH THEORY

Remark. Eigenvalues provide information about the connectivity of the graph.

Theorem 1.1. Let A ∈ Rn×n be symmetric, then A orthogonally diagonalizable over R,i.e., there exists an orthonormal basis

v(1), v(2), · · · , v(n) ∈ Rn

of eigenvectors of A corresponding to real eigenvalues λ1, λ2, · · · , λn ∈ R, we have

A =n∑i=1

λiv(i)v(i)T

Therefore the eigenvalues of any graph G are all real and we’ll denote then

λ1(G) ≥ λ2(G) ≥ · · · ≥ λn(G),

where n = |V (G)|.

Question: What is the intuition for

A =n∑i=1

λiv(i)v(i)T ?

Answer: The fact that the sum equals A we can just check explicitly: if we denote the sumby S, we compute

Sv(i) =

(n∑j=1

λjv(j)v(j)T

)v(i) =

n∑j=1

λjv(j)(v(j)Tv(i)

)= λiv

(i) ∀i = 1, · · · , n,

and so S = A. Remember that xTy is just the inner product of vectors x and y in Rn. Here weuse distributive law and the property of orthonormal basis that v(i)Tv(j) = v(j)Tv(i) = I{i=j}.

Theorem 1.2 (Perron-Forbenius). If a matrix A ∈ Rn×n has nonnegative entries, then thespectral radius of A (i.e., the maximum magnitude over all complex eigenvalues of A) is aneigenvalue of A, corresponding to an eigenvector in Rn

≥0.

Therefore for any graph G, λ1(G) is the spectral radius and corresponds to an eigenvectorwith nonnegative entries. Perron-Forbenius also implies if G is connected, then λ1(G) hasmultiplicity 1.

Definition 1.4 (disjoint union). If G = (V,E), G′ = (V ′, E ′) are graphs, their disjoint unionis the graph

G tG′ = (V t V ′, E t E ′)

and

A(G tG) =

[A(G) 0

0 A(G′)

]

2

1.2. REGULAR GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Remark. Spectrum doesn’t detect if the graph is connected.

Example 1.2.

G : V = {1, 2}, E = {{1, 2}}spectrum : {1,−1}

G tG : V = {1, 2, 3, 4}, E = {{1, 2}, {3, 4}}spectrum : {1, 1,−1,−1}

Remark. Note that the spectrum of GtG′ is the multiset union of the spectra of G and G′.

1.2 Regular Graphs

Definition 1.5. A graph G = (V,E) is called k-regular if every vertex has degree k ∈ N.

Remark. G is regular if and only if e is an eigenvector, where e =[1 1 · · · 1

]T ∈ Rn.

A(G) · e = degG(V )

Proposition 1.1. Let G = (V,E), then

Ev∈V (G)

degG(v)(i)

≤ λ1(G)(ii)

≤ maxv∈V (G)

degG(v),

where the expectation of the degree is defined to be

Ev∈V (G)

degG(v) :=1

n

∑v∈V (G)

degG(v).

We have equality in (i) if and only if G is regular. If G is connected, equality holds in (ii)if and only if G is regular.

Proof. (i) For any symmetric A ∈ Rn×n,

λmax(A) ≥ xTAx, ∀x ∈ Rn with ||x|| = 1

with equality if and only if x is an eigenvector of A with eigenvalue λmax(A).

A =∑j

λjx(j)x(j)T

x =∑j

cjx(j)

∑j

c2j = 1

xTAx =∑j

λjc2j

3

1.2. REGULAR GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Question: The first equation is from Theorem 1.1, but where do the following equa-tions come from?

Answer: The second equation comes from the fact that a vector in Rn can be repre-sented as the linear combination of an orthonormal basis. The third equation comesfrom the constraint ||x|| = 1, where the norm can be written as

√xTx = 1. The fourth

one comes from calculation

Ax =∑j

λjx(j)x(j)T

∑i

cix(i) =

∑j

cjλjx(j)

so thatxTAx =

∑i

cix(i)T∑j

cjλjx(j) =

∑i

c2iλi

Here we use distributive law and the property of orthonormal basis that x(i)Tx(j) =x(j)Tx(i) = I{i=j}.

Take A = A(G), x = 1√ne, then

xTAx =1

neTAe︸ ︷︷ ︸

sum of entries

= average degree

(ii) Let’s assume V (G) = [n] := {1, 2, · · · , n}. Recall that by Perron-Forbenius, λ1(G)corresponds to an eigenvector x ∈ Rn

≥0. Then for any i ∈ [n],

λ1(G)xi =(A(G)x

)i

=n∑j=1

A(G)i,jxj =∑j∼i

xj.

Now take i ∈ [n] so that xi is maximum. Then

λ1(G)xi =∑j∼i

xj ≤∑j∼i

xi = degG(vi)xi ≤(

maxv∈V (G)

degG(v)

)xi (*)

Soλ1(G) ≤ max

v∈V (G)degG(v).

Suppose equality holds everywhere in (ii) and G is connected, then equality holdseverywhere in (*). So xj = xi for all j ∼ i. Applying the same argument to j ∼ i, thenxh = xj = xi for any h ∼ j, etc. Since G is connected, x is a multiple of e, i.e., e is aneigenvector of G.

Corollary 1.1. The complete graph Kn is the only graph on n vertices with eigenvalue n−1.In particular, Kn is uniquely determined by its spectrum.

Remark. Not all graphs can be recovered by their spectrum, i.e., the spectrum is not a faithful(in that graph can be uniquely determined) graph invariant (in that spectrum depends ongraph only up to isomorphism).

Example 1.3. Spectrum does not tell connectivity. For example, the following two graphs(Figure 1.1) have the same spectrum (−2, 2, 0, 0, 0).

4

1.3. BIPARTITE GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Figure 1.1: Two graphs with the same spectrum

1.3 Bipartite Graphs

Recall graph is bipartite with biparts x, y if and only if

A(G) =

[0 BBT 0

]Lemma 1.1. Let B be an n× n matrix such that BTB has nonzero eigenvalues λ1, · · · , λr(with multiplicities), then the nonzero eigenvalues of

A =

[0 BBT 0

]∈ R(m+n)×(m+n)

are precisely√λ1,−

√λ1, · · · ,

√λr,−

√λr (with multiplicities).

Remark. The eigenvalues of BTB are real (from symmetry) and nonnegative since if BTBx =λx (we assume ||x|| = 1)

λ = λxTx = xT (λx) = xT (BTBx) = (Bx)T (Bx) = 〈Bx,Bx〉 ≥ 0.

Proof. Note that [tIm −B−BT tIn

] [Im B0 tIn

]=

[tIm 0−BT t2In −BTB

].

Take determinant givesφA(t)tn = tmφBTB(t2)

Question: Why does the matrix on RHS have the determinant tmφBTB(t)? Particularly,why is

det(t2In −BTB

)= φBTB

(t2)?

Answer: By definition, φBTB(x) = det(xIn−BTB). Therefore, φBTB(t2) = det(t2In−BTB)

5

1.3. BIPARTITE GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Example 1.4. Let Km,n be the complete bipartite graph with biparts of size m and n andall possible edges between them, then

A(Km,n) =

[0 BBT 0

]where B is an m× n all-one matrix. We have

BTB︸ ︷︷ ︸n×n

= mJn

where Jn is the n × n all-one matrix. Recall that Jn has exactly one nonzero eigenvaluesnamely n, so the eigenvalues of Km,n are

√mn,−

√mn and 0 (multiplicity m+ n− 2).

Example 1.5. Let C2n be a bipartite graph with adjacency matrix

A(C2n) =

[0 BBT 0

]where B ∈ Rn×n e.g.

B =

1 0 0 0 11 1 0 0 00 1 1 0 00 0 1 1 00 0 0 1 1

∈ R5×5

We can verify thatBTB = 2In + A(Cn)︸ ︷︷ ︸

0 1 0 0 11 0 1 0 00 1 0 1 00 0 1 0 11 0 0 1 0

Therefore if λ1, · · · , λn are the eigenvalues of Cn, then the eigenvalues of C2n are

±√

2 + λk, k = 1, · · · , n

This agrees with our early calculations

λk = 2 cos

(2πk

n

)with identity

1 + cos(2θ) = 2 cos2 θ

Corollary 1.2. If G is bipartite graph on n vertices then λn(G) = −λ1(G).

Remark. A converse holds for connected graphs.

Proposition 1.2. If G is connected on n vertices and λn(G) = −λ1(G), then G is bipartite.

Remark. It’s another consequence of the Perron-Frobenius theorem. We don’t prove this.

6

1.4. CARTESIAN PRODUCTS CHAPTER 1. ALGEBRAIC GRAPH THEORY

1.4 Cartesian Products

Definition 1.6 (Cartesian product). Let G and H be graphs, the Cartesian product G�Hhas vertex set V (G)× V (H) with edges of the forms

• (v, w) ∼ (v′, w) where v ∼ v′ in G and w ∈ V (H)

• (v, w) ∼ (v, w′) where w ∼ w′ in H and v ∈ V (G)

Example 1.6. Pm�Pn is the m×n rectangular lattice. P2�P2�P2 is the 1-skeleton of cube.In general, P2� · · ·�P2︸ ︷︷ ︸

n copies

is the 1-skeleton of the n-dimensional hypercube [0, 1]n ⊂ Rn

Remark. Cartesian product is associative.

Definition 1.7 (Tensor product). Let Rm ⊗ Rn denote the tensor product of Rm and Rn

which we will identify with the vector space of matrices Rm×n. Let e(i) denote the unit vector(ith entry is one and zeros elsewhere) and define the standard basis of Rm×n:

e(i)e(j)T =

j

0 0 0 0 00 0 0 0 0

i 0 0 1 0 00 0 0 0 00 0 0 0 0

, 1 ≤ i ≤ m, 1 ≤ j ≤ n

For A ∈ Rm×m, B ∈ Rn×n, let A⊗B denote the endomorphism of Rm×n given by

(A⊗B)(M) := AMBT for M ∈ Rm×n.

Lemma 1.2. Let two matrices A ∈ Rm×m, B ∈ Rn×n, then

(A⊗B)(i,j),(k,l) = Ai,kBj,l, (1 ≤ i, k ≤ m, 1 ≤ j, l ≤ n)

Proof. Recall that the entries of a matrix M ∈ Rd×d are characterized by

Me(j) =d∑i=1

Mi,je(i) for 1 ≤ j ≤ d.

(A⊗B)(e(k)e(l)T

)= A

(e(k)e(l)T

)BT

=(Ae(k)

) (Be(l)

)T=

(m∑i=1

Ai,ke(i)

)(n∑j=1

Bj,le(j)

)T

=m∑i=1

n∑j=1

Ai,kBj,le(i)e(j)T

7

1.5. ENUMERATING WALKS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Corollary 1.3. Let G and H be graphs with basis of eigenvectors x(1), · · · , x(m) ∈ Rm andy(1), · · · , y(n) ∈ Rn corresponding to eigenvalues λ1, · · · , λm and µ1, · · · , µn, then

A(G�H) = A(G)⊗ In + Im ⊗ A(H)

and G�H has eigenvectors x(i)y(j)T corresponding to eigenvalues λi + µj for 1 ≤ i ≤ m, 1 ≤j ≤ n.

Remark. Note that{x(i)y(j)T

}forms a basis of Rm×n since they are the unit matrices (ma-

trices with a 1 and 0’s elsewhere) under the basis x(1), · · · , x(m) of Rm and the basis dual toy(1), · · · , y(n) ∈ Rn. (If y(1), · · · , y(n) ∈ Rn are orthonormal, it’s self-dual).

1.5 Enumerating Walks

Definition 1.8 (Walk). Let G be a graph. A walk in G is a sequence of vertoices (v0, · · · , vl)such that v0 ∼ v1 ∼ v2 ∼ · · · ∼ vl. We call l the length of the walk. (A path is a walk with novertices repeated except possibly v0 = vl.) We call a walk closed if v0 = vl (a closed path iscalled cycle). We will consider marked closed walks, i.e., closed walks whose starting vertex isdistinguished (and an occurance of that vertex within the walk if it appears multiple times).

Example 1.7. Triangle has 6 marked closed walks:

1 2 3 1 3 22 1 3 2 3 13 1 2 3 2 1

Lemma 1.3. Let G be a graph and l ∈ N. For any v, w ∈ V (G), the number of walks oflength l from v to w in G equals(

A(G)l)v,w

, (A(G)0 = I)

Proof. (A(G)l

)v,w

=∑

u1,··· ,ul−1∈V (G)

A(G)v,u1A(G)u1,u2 · · ·A(G)ul−2,ul−1A(G)ul−1,w

=∑

u1,··· ,ul−1∈V (G)

1v∼u1∼u2∼···∼ul−1∼w

Corollary 1.4. Let G be a graph on n vertices with adjacency matrix A and l ∈ N.

(i) the number of marked closed walks in G of length l equals

∑v∈V (G)

(Al)v,v

= tr(Al)

=n∑j=1

λlj(G)

8

1.5. ENUMERATING WALKS CHAPTER 1. ALGEBRAIC GRAPH THEORY

(ii) the total number of walks in G of length l equals∑v,w∈V (G)

(Al)v,w

= sum of entries of Al

Example 1.8. Let G = Kn the complete graph on n vertices. Recall that the eigenvaluesof G are n− 1,−1,−1, · · · ,−1. Hence, the number of marked closed walks of length l is

(n− 1)l + (n− 1)(−1)l

Now let’s find the total number of walks of length l. Recall A(G) = Jn − In. Therefore bybinomial theorem

(Jn − In)l =l∑

d=0

(l

d

)(−1)l−dJdn

By J2n = nJn anmd induction,

Jdn = nd−1Jn, ∀d ≥ 1

So the sum of the entries of Jdn is nd−1n2 = nd+1 for all d ∈ N. Then the sum of entries ofA(G)l, (i.e., number of walks of length l) is

l∑d=0

(l

d

)(−1)l−dnd+1 = n(n− 1)l

Finally, note that by symmetry(A(G)l

)v,w

is the same for all distinct v, w ∈ V (G).

(A(G)l

)v,w

=sum of entries of A(G)l − sum of diagonal entries of A(G)l

# of off-diagonal entries

=n(n− 1)l −

[(n− 1)l + (n− 1)(−1)l

]n2 − n

=(n− 1)l − (−1)l

n

Corollary 1.5. The number of triangles in a graph G is

1

6A(G)3

The coefficient 16

comes from the fact that a triangle has 6 (marked closed) walks of length3.

Lemma 1.4. Let M ∈ Rn×n. The sum of the entries of M is

eTMe = tr(MJn) = det(In +MJn)− 1

where e =[1 1 · · · 1

]T ∈ Rn.

Proof. We can check explicitly that eTMe is the sum of entries.

(i)eTMe︸ ︷︷ ︸1×1

= tr(eTMe

)= tr

(MeeT

)= tr(MJn)

9

1.6. GENERATING FUNCTIONS CHAPTER 1. ALGEBRAIC GRAPH THEORY

(ii) Note that rank(MJn) ≤ 1. For any matrix N ∈ Rn×n of rank at most 1, we have theSherman-Morrison identity

det(In +N) = 1 + tr(N)

This is true since N has at most one nonzero eigenvalue λ and both sides equal to 1+λ.

1.6 Walk Generating Functions

Definition 1.9 (ordinary generating function). The ordinary generating function (ogf) ofthe sequence (al)l∈N is the formal power series

∞∑l=0

altl

It is called ordinary to distinguish from the exponential generating function

∞∑l=0

all!tl

Recall the geometric series formula

∞∑l=0

tl =1

1− t

We can think of 11−t as a rational function of t, or as the inverse of the formal power series

1− t. More generally,∞∑l=0

(tA)l = (In − tA)−1

Definition 1.10. Let G be a graph of n vertices with adjacency matrix A.

(i) For v, w ∈ V (G), let WGv,w(t) denote the ordinary generating function for number of

walks from v to w with respect to lenth.

WGv,w(t) =

∞∑l=0

(Al)v,wtl =((In − tA)−1

)v,w

(ii) Let WGclosed(t) denote the ordinary generating function for marked closed walks in G

with respect to lenth.

WGclosed(t) =

∞∑l=0

tr(Al)tl = tr((In − tA)−1

)10

1.6. GENERATING FUNCTIONS CHAPTER 1. ALGEBRAIC GRAPH THEORY

(iii) Let WGall(t) denote the ordinary generating function for all walks in G with respect to

lenth.

WGall(t) =

∞∑l=0

tr(AlJn)︸ ︷︷ ︸sum of entries of Al

tl = tr((In − tA)−1Jn

)Example 1.9. Recall

tr(A(Kn)l

)= (n− 1)l + (n− 1)(−1)l

Therefore,

WKnclosed(t) =

∞∑l=0

[(n− 1)l + (−1)l(n− 1)

]tl

=1

1− (n− 1)t+ (n− 1)

1

1− t

Theorem 1.3. Let G be a graph with n vertices.

WGclosed(t) = tr

((In − tA(G))−1

)=

n∑j=1

1

1− tλj(G)=

φ′G(1t

)tφG

(1t

)Proof. The eigenvalues of A(G) are λ1, λ2, · · · , λn, so the eigenvalues of (In − tA(G))−1 are

(1− tλ1(G))−1, · · · , (1− tλn(G))−1

Thus,tr((In − tA(G))−1

)= (1− tλ1(G))−1 + · · ·+ (1− tλn(G))−1

On the other hand,

φ′G(1t

)tφG

(1t

) =1

t

d

ds

∣∣∣∣s= 1

t

log φG(s)

=1

t

d

ds

∣∣∣∣s= 1

t

log[(s− λ1(G)) · · · (s− λn(G))]

=1

t

(1

s− λ1(G)+ · · ·+ 1

s− λn(G)

)∣∣∣∣s= 1

t

=1

1− tλ1(G)+ · · ·+ 1

1− tλn(G)

Alternative proof using the identity

tr(log(A)) = log(det(A))

as functions on matrices.

11

1.7. ASYMPTOTIC BEHAVIOR CHAPTER 1. ALGEBRAIC GRAPH THEORY

Example 1.10. Let G = Kn, then

φKn(t) = (t− (n− 1))(t+ 1)n−1

φ′Kn(t) = (t+ 1)n−2(tn− n2 + 2n)

Then

WKnclosed(t) =

(1 + 1

t

)n−2 (nt− n2 + 2n

)t(1t− (n− 1)

) (1t

+ 1)

1.7 Asymptotic Behavior

Theorem 1.4. Let G be a connected non-bipartite graph with largest eigenvalue λ1 corre-sponding to an eigenvector x with ||x|| = 1. (recall from the Perron-Frobenius theorem that|λj(G)| < λ1 for all j > 1), then

liml→∞

A(G)l

λl1= xxT

In particular, for all v, w ∈ GA(G)lv,w ∼ λl1xvxw

Proof. Let’s assume that V (G) = [n]. Let x(1), x(2), · · · , x(n) ∈ Rn be an orthonormal basisof eigenvectors of A(G) corresponding to eigenvalues λ, λ2, · · · , λn. Recall that

A(G) =n∑j=1

λjx(j)x(j)T

Since x(i)Tx(j) = δij, we get

A(G)l =n∑j=1

λljx(j)x(j)T

Therefore,A(G)l

λl1= xxT +

n∑j=2

(λjλ1

)lx(j)x(j)T

Since |λj| < λ1 for j > 1, we get

liml→∞

A(G)l

λl1= xxT

Remark. 1. Connected: λ1 has multiplicity 1.

2. Non-bipartite: −λ1 not an eigenvalue.For connected bipartite graphs, we have the following

A(G)lv,w =

{0 l 6≡ d(v, w) mod 2

∼ 2λl1xvxw l ≡ d(v, w) mod 2

12

1.7. ASYMPTOTIC BEHAVIOR CHAPTER 1. ALGEBRAIC GRAPH THEORY

Example 1.11. Consider G = Pn, which is connected and bipartite. Recall the eigenvaluesof G are

λk = 2 cos

(πk

n+ 1

)k = 1, · · · , n

We can check that an eigenvector x ∈ Rn with ||x|| = 1 corresponding to λ1 = 2 cos(

πn+1

)is

given by

xj =

√2

n+ 1sin

(πj

n+ 1

)(entries bigger if closer to the middle of the path). An eigenvector y ∈ Rn with ||y|| = 1corresponding to −λ1 is given by

yj = (−1)j−1xj j = 1, · · · , n

Hence, as l→∞, we have

(A(G)l

)ij

=

{0 l 6≡ |i− j| mod 2

∼ 2(

2 cos(

πn+1

)l) ( 2n+1

)sin(πin+1

)sin(πjn+1

)l ≡ |i− j| mod 2

Why does ||x|| = 1?n∑j=1

sin2

(πj

n+ 1

)=

1

2

2n+1∑j=0

sin2

(πj

n+ 1

)Recall

sin θ =eiθ − e−iθ

2i

Letting ρ = 2πi2n+2

,

sin

(jπ

n+ 1

)=ρj − ρ−j

2i

Thus,

1

2

2n+1∑j=0

sin2

(πj

n+ 1

)=

1

2

2n+1∑j=0

(ρj − ρ−j

2i

)

=1

2

2n+1∑j=0

(ρ2j − 2 + ρ−2j

−4

)

= −1

8

2n+1∑j=0

ρ2j︸ ︷︷ ︸=0

+2n+1∑j=0

−2︸ ︷︷ ︸=−2(2n+2)

+2n+1∑j=0

ρ−2j︸ ︷︷ ︸=0

=n+ 1

2

13

1.8. GRAPH HOMOMORPHISMS CHAPTER 1. ALGEBRAIC GRAPH THEORY

1.8 Graph Homomorphisms

Definition 1.11. Let G and H be graphs. We say that φ : V (G) → V (H) is a homomor-phism if

∀v, w ∈ V (G), v ∼G w =⇒ φ(v) ∼H φ(w).

We call φ an isomorphism if φ is a bijection and its inverse φ−1 is also a homomorphism. Ifφ is a bijection, it is an isomorphism if and only if

∀v, w ∈ V (G), v ∼G w ⇐⇒ φ(v) ∼H φ(w)

Definition 1.12 (Automorphism). An automorphism of G is an isomorphism from G to G.Note: (Aut(G), ◦) forms a group.

Let S(V ) denote the symmetric group of all permutations (i.e. bijections) of the set V(We also let Sn denote the symmetric group of permutations of [n] = {1, 2, ..., n}). Aut(G)is a subgroup of S(V (G)).

Example 1.12. Aut(Kn) = Sn, Aut(Cn) = Dn(the dihedral group of order 2n of rotationsand reflections of a regular n-gon).

Theorem 1.5 (Lie Theory). Miss this part...

Example 1.13. n ≥ 2, Pn has 2 automorphism (the identity map and the map that reflectsthe path about its center).

Definition 1.13 (m-coloring). m-coloring of a graph G is a labeling of its vertices by a colorϕ(V ) ∈ [m]. An m-coloring is called proper if no two adjacent vertices are given the samecolor, i.e.,ϕ is proper if and only if ϕ is homomorphism form G to Km.

Definition 1.14. A permutation matrix is a square {0, 1} matrix with exactly one 1 in everyrow and column. For each permutation π ∈ Sn. Let P (π) denote the n × n permutationmatrix where jth column has a 1 in row π(j) for all j ∈ [n].

Example 1.14. Let π =

(1 2 32 3 1

), σ =

(1 2 32 1 3

). Then

P (π) =

0 0 11 0 00 1 0

, P (σ) =

0 1 01 0 00 0 1

.

Note that P (π)P (σ) = P (πσ). Let Pn be the set of all n × n permutation matrices. Thenit’s easy to verify that Pn ≤ GLn(R) and Pn ∼= Sn.

Definition 1.15. Define the sign or signature sgn(π) of π ∈ Sn by sgn(π) = det(P (π)).

Lemma 1.5. Let π ∈ Sn, then

(i) If π can be represented as a product of m transpositions, then sgn(π) = (−1)m.

14

1.8. GRAPH HOMOMORPHISMS CHAPTER 1. ALGEBRAIC GRAPH THEORY

(ii) sgn(π) = (−1)# i<j:π(j)>π(i).

Proof. (i) Using multiplicativity of the determinant, it suffices to prove that every trans-position has sign −1. This is true since the permutation matrix of a transpositionis

0 1 0 · · · 01 0 0 · · · 00 0 1 · · · 0...

......

. . ....

0 0 0 · · · 1

,

up to a permutation of the basis vectors.

(ii) Check that

(a) sgn(id) = 1.

(b) Since every transposition has sign -1, note that if there is inversion ((i, j) : i <j, π(j) > π(i)) in a permutation, then there must exist an inversion between 2consecutive indices (e.g. π = 3124 contains inversion (1,2) and (1,3)). Now it isenough to note that one could reverse such an inversion using transposition. Sothe total number of transpositions needed is the number of inversions.

Example 1.15. sgn

(1 2 32 3 1

)= 1

Proposition 1.3. Let G and H be graphs on the vertex set [n] and π ∈ Sn. Then π is anisomorphism from G to H if and only if

A(G) = P (π)−1A(H)P (π).

Proof. Note that (P (π)−1A(H)P (π))i,j = A(H)π(i),π(j). Therefore we have an equivalencechain,

A(G) = P (π−1)A(H)P (π)

⇐⇒ A(G)i,j = A(H)π(i),π(j)

⇐⇒ (i ∼ j ⇐⇒ π(i) ∼ π(j))

⇐⇒ π is an isomorphism

Corollary 1.6. The automorphisms of a graph G correspond to permutation matrices Psatisfying A(G) = P−1A(G)P .

Theorem 1.6. Suppose that the eigenvalues of G are all distinct. Then every automorphismof G has order at most 2.

15

1.9. EXPANDER GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Proof. Let A = A(G) ∈ Rn×n. Every automorphism of G corresponds to a permutationmatrix P with AP = PA. We must show that P 2 = In.

Let λ1, . . . , λn ∈ R be the eigenvalues of A with eigenbasis x(1), . . . , x(n) ∈ Rn. Then fori ∈ [n], we have

AP = PA =⇒ APx(i) = PAx(i) =⇒ A(Px(i)) = λiPx(i) =⇒ Px(i) = µix

(i)

Since P n! = In, (µi)n! = 1. Thus µi = ±1. We get P 2x(i) = µ2

ix(i) = x(i) for all i ∈ [n]. Since

{x(i)} forms a basis of Rn, P 2 is In.

Remark. in fact, almost all graphs have a trivial automorphism group, i.e.

limn→∞

#graphs G with V (G) = [n] and |Aut(G)| = 1

#graphs G with V (G) = [n]= 1.

See section 2.3 of Godsil-Royle for a proof.

Remark (Graph Isomorphic Problem).

• Babai & Luks(1983) 2O(√n logn)

• Babai(2017) 2O((logn)c)

1.9 Expander Graphs

Vaguely speaking, an expander graph is a sparse graph (i.e., it has few edges) which isnonetheless highly connected. Expander graphs have applications to

• constructing error-correcting codes

• derandomize algorithms

• serve methods in number theory

• hyperbolic manifolds

Definition 1.16 (edge expansion ratio). Lety G be a graph. For S ⊆ V (G) denote the setof edges with one endpoint in S and one endpoint not in S by ∂S. Define the edge expansionratio

h(G) := minS⊆V (G)

|S|≤|V (G)|/2

|∂S||S|

Definition 1.17. A sequence of k-regular graphs (Gn)n∈N is called a family of expandergraphs if

|V (G0)| < |V (G1)| < · · ·and there exists ε > 0 such that h(Gn) ≥ 0,∀n ∈ N.

Remark. It’s relatively easy to show that such families exist by probabilistic arguments, butit’s relatively difficult to construct them explicitly.

16

1.9. EXPANDER GRAPHS CHAPTER 1. ALGEBRAIC GRAPH THEORY

Theorem 1.7. Suppose G is connected and k-regular so λ1(G) = k, then

k − λ2(G)

2≤ h(G) ≤

√2k(k − λ2(G))

Therefore a family (Gn)n∈N of k-regular graphs with |V (G0)| < |V (G1)| < · · · forms afamily of expander graphs if and only if there exists ε′ > 0 such that

k = λ2(Gn) ≥ ε′ ∀n ∈ N

The spectral gap k−λ2(G) or sometimes min{|k−λ2G|, k− |λn(G)|} measures how rapidlya random walk on G mixes.

17

Chapter 2

Tilings, Spanning Trees and ElectricalNetworks

• motivation (and some results) come from Physics, Chemistry and Computer Science.

2.1 Tilings and Perfect Matchings

We’ll prove the following theorems.

Theorem 2.1 (Kasteleyn (1961)). Let m,n ∈ N be such that mn is even. The number ofdomino tilings of an m× n board equals

T (m,n) =m∏j=1

n∏k=1

(4 cos2

(πj

m+ 1

)+ 4 cos2

(πk

n+ 1

)) 14

Example 2.1. π(2, 3) has 3 tilings.

Figure 2.1: Three tilings of π(2, 3)

Remark. If m,n both odd, there’s no domino tilings of an m× n board.

Definition 2.1 (Perfect matching). A perfect matching of a graph G is a subset of its edgeswhich meets every vertex exactly once.

Remark.

• Only graphs with even number of vertices can have a perfect matching.

• Note that domino tiling of an m × n board correspond to perfect matchings of them× n grid Pm�Pn.

Example 2.2. K4 has 3 perfect matchings. Each of them are rotated from anther.

We will present a general method for conducting perfect matchings of certain graphs dueto Kasteleyn, called Pfaffian method.

18

2.2. SKEW-SYMMETRIC MATRICES CHAPTER 2. TILINGS & SPANNING TREES

Figure 2.2: Three perfect matchings of K4

2.2 Skew-Symmetric Matrices

Definition 2.2 (skew-symmetric matrix). A matrix A ∈ Rn×n is called skew symmetric if

AT = −A.

Remark. Note that for skew-symmetric A,

det(A) = det(AT ) = det(−A) = (−1)n det(A)

Therefore if n is odd, det(A) = 0. However, if n is even, det(A) is a perfect square in theentries of A.

Example 2.3.

det

0 a b c−a 0 d e−b −d 0 f−c −e −f 0

= (af − be+ cd)2 = Pfaffian2

Definition 2.3 (Pfaffian). Let A ∈ R2n×2n be skew-symmetric. We define the Pfaffian of A

pf(A) :=∑

perfect matching of K2n

M={{i1,j1},··· ,{in,jn}}

sgn

(1 2 3 4 · · · 2n− 1 2ni1 j1 i2 j2 · · · in jn

) n∏r=1

Air,jr

Example 2.4. Let A ∈ R4×4 be skew-symmetric, then

pf(A) = sgn

(1 2 3 41 2 3 4

)A1,2A3,4

+ sgn

(1 2 3 41 3 2 4

)A1,3A2,4

+ sgn

(1 2 3 41 4 2 3

)A1,4A2,3

= A1,2A3,4 − A1,3A2,4 + A1,4A2,3

Remark. Why is the Pfaffian well-defined?

(a) the M-term does not change if we swap the order of a given edge

{ir, jr} ←→ {jr, ir}

• sgn

(· · · 2r − 1 2r · · ·· · · ir jr · · ·

)= −sgn

(· · · 2r − 1 2r · · ·· · · jr ir · · ·

)19

2.2. SKEW-SYMMETRIC MATRICES CHAPTER 2. TILINGS & SPANNING TREES

• Air,jr = −Ajr,ir

(b) the M-term does not change if we swap two consequtive edges

{ir, jr} ←→ {ir+1, jr+1}

• sgn

(· · · 2r − 1 2r 2r + 1 2r + 2 · · ·· · · ir jr ir+1 jr+1 · · ·

)• unchanged

Question: This part is not clear enough.

So we can compute

pf(A) =1

2nn!︸︷︷︸there are

(2n)!2nn!

perfect matchings

∑π∈S2n

sgn(π)2n∏i=1

Aπ(2i−1),π(2i)

Theorem 2.2 (Cayley 1849). For any skew-symmetric A ∈ R2n×2n,

det(A) = pf(A)2

Theorem 2.3. Every planar graph has a Pfaffian orientation.

Theorem 2.4 (EDIT: Kuratowski). A graph is planar if and only if there is no subdivisionisomorphic to K3,3 or K5.

Theorem 2.5. A graph has a Pfaffian orientation if and only if no K3,3.

Theorem 2.6. The number of pilings using dominos on an m× n board is

T (m,n) =m∏j=1

n∏k=1

(4 cos2(πj

m+ 1) + 4 cos2(

πk

n+ 1))

14 .

Proof. Any two domino tilings are related by a sequence of flips. Any two perfect matchings

Figure 2.3: Flip

contribute to the same sgn. Thus the number of perfect matchings of Pm�Pn is |Pf( ~A(D))| =√| det( ~A(G))|. By rescaling you get | det(A)| = | det(A′′)|, where

A′′ = Im ⊗ A(Pn)− i(A(Pm)⊗ In).

20

2.3. RHOMBIC TILINGS CHAPTER 2. TILINGS & SPANNING TREES

Recall that the eigenvalues of Pn are 2 cos( πkn+1

), k ∈ [n]. Therefore, eigenvalues of A′′ are

2 cos(πk

n+ 1)− i2 cos(

πj

m+ 1).

Then we have the number of perfect matchings

|Pf( ~A(D))| =√| det(A)|

=| det(A′′)|12

=m∏j=1

n∏k=1

|4 cos2(πk

n+ 1) + 4 cos2(

πk

n+ 1)|1/4

2.2.1 Asymptotic of T (m,n)

We have

log(T (m,n))

mn=

1

mnlog

m∏j=1

n∏k=1

(4 cos2(πj

m+ 1) + 4 cos2(

πk

n+ 1))

14

=1

4π2(π

m)(π

n)

m∑j=1

n∑k=1

log(4 cos2(πj

m+ 1) + 4 cos2(

πk

n+ 1))

m,n→∞−−−−→ =1

4π2

∫ π

0

∫ π

0

log(4 cos2 x+ 4 cos2 y)dxdy

=G

π

where G =1

4π

∫ π0

∫ π0

log(4 cos2 x+ 4 cos2 y)dxdy = 1− 1

32+

1

52+ ...

(Open Problem) is G irrational?log T (m,n) ≈ Gm,n. So T (m,n) ≈ eGm,n/π

2.3 Rhombic tilings

Theorem 2.7 (Mac Mahon (1916)).

Theorem 2.8 (Lindstorm, Gessel, Viennot).

Theorem 2.9.

N(a, b, c) =a∏i=1

b∏j=1

c∏k=1

i+ j + k − 1

i+ j + k − 2

21

2.4. EDIT: AZTEC DIAMOND CHAPTER 2. TILINGS & SPANNING TREES

Proof. We will apply Jacobi’s Identity:

det(A) det(A[2,n−1][2,n−1]) = det(A[2,n][2,n]) det(A[1,n−1][1,n−1])−det(A[2,n][1,n−1]) det(A[1,n−1][2,n])

for any n × n matrix A. This is a special case of general determinant identities calledGrassmann-pliicker relations (or Schouter(???) identities in physics). In our case, we have

N(a, b, c)N(a, b, c−2) = N(a, b, c−1)N(a, b, c−1)−N(a−1, b+1, c−1)N(a+1, b−1, c−1),

where N(a, b, 0) = 1, N(a, b, c) =(a+ba

). Now just need to check that

a∏i=1

b∏j=1

c∏k=1

i+ j + k − 1

i+ j + k − 2

satisfies the same recursion.

There is a direct way to relate determinants to enumerating perfect matchings.

2.4 EDIT: Aztec Diamond

The nth Aztec diamond is a symmetric board with rows of length 2, 4, 6, . . . , 2n, 2n, . . . , 6, 4, 2.

Theorem 2.10. The nth Aztec diamond has 2(n+12 ) domino tilings.

2.5 Spanning Trees

Theorem 2.11. Let m,n ∈ N be odd. The number of domino tilings of an m×n board withone corner square removed is

bm2c∏

j=1

bn2c∏

k=1

(4 cos2(πj

m+ 1) + 4 cos2(

πk

n+ 1))

To prove this theorem we first construct a bijection from domino tilings to spanning treesof a grid graph. Then we will count spanning trees using the matrix-tree theorem.

Definition 2.4. A spanning subgraph H of G is a graph V (H) = V (G) and E(H) ⊂ E(G).If H is a tree, we call H a spanning tree of G.

Theorem 2.12. let m,n ∈ Z>0, define a map ϕ from the set of domino tilings of a (2m−1)(2n− 1) board whose north east corner is removed to the set of spanning trees of Pm�Pnas follows:

Embed Pm�Pn inside the (2m-1)(2n-1) board. Given a domino tiling M , let ϕ(M) be thespanning tree whose edges cross only one edge of M . This ϕ is a bijection.

Proof. First we prove ϕ is well-defined. This can be seen from the fact that

22

2.6. MATRIX-TREE THEOREM CHAPTER 2. TILINGS & SPANNING TREES

1. the edges of ϕ(M) correspond to the vertices of Pm�Pn except for the northeast cornervertex. So ϕ(M) has mn− 1 edges,

2. ϕ(M) is acyclic, otherwise, the path would enclose odd number of squares, whichcannot be filled with dominoes.

Then we must construct the unique domino tiling M with ϕ(M) = T . Root T at thenortheast corner vertex and direct all edges toward the root. This allows us to fill in thedominoes along the tree.

Now we need to fill in the remainder. The remaining squares can be divided into severalconnected component H1, . . . , Hl ( of the dual graph of the board). Note that each Hi isacyclic, otherwise Hi would divide T into ≥ 2 connected components. So, the all Hi aretrees.

Similarly, each Hi intersects the boundary of the boundary of the board exactly once.Therefore each Hi has a unique root. By the same method we could tile all the area.

Theorem 2.13 (Propp).

2.6 Kirchoff’s matrix-tree theorem

Lemma 2.1 (Laplacian eigenvalues of Cartesian products).

2.6.1 Matrix Tree Theorem

Theorem 2.14 (Matrix-Tree Theorem). The number of spanning trees of a graph G equals

det(L(G)V (G)\{v},V (G)\{v}) any v in V (G)

=1

nµ1...µn−1 0, µ1, ..., µn−1 are Laplacian eigenvalue.

Proof. (This is the second part) We need to show that

det(L(G)V (G)\{v},V (G)\{v}) any v in V (G) =1

nµ1...µn−1

.

Lemma 2.2. Let A ∈ Cn×n be a matrix where eigenvalues are 0, µ1, µ2, ..., µn−1. Then

µ1µ2...µn−1 =n∑i=1

det(A[n]\{i},[n]\{i})

23

2.7. EULERIAN CYCLES CHAPTER 2. TILINGS & SPANNING TREES

Proof. Note that

ΦA(t) = t(t− µ1)...(t− µn−1)[t]ΦA(t) = (−1)n−1µ1...µn−1

on the other hand, [t]ΦA(t) = [t] det(tIn − A)

=n∑i=1

det((−A)[n]\{i},[n]\{i})

= (−1)n−1n∑i=1

det((A)[n]\{i},[n]\{i})

Apply the lemma to A = L(G).

2.6.2 Directed Matrix Tree Theorem

Definition 2.5. Let D be a directed graph, the Laplacian of D L(D) ∈ RV (D)×V (D) isdefined by

L(D)v,w =

{outdeg(v)− (loops at v) if v = w

− (edges v → w)neqw

Remark. • In our definition, we allow D with loop and multiple edges.

• Note that L(D)~e = 0.

Definition 2.6. A rooted (directed) graph is a (directed) graph with a distinguished vertex,root. A rooted in tree is the (unique) orientation of an undirected rooted tree such thatevery edge points toward the root.

Theorem 2.15 (Tutle(1948)Directed matrix tree theorem). Let D be a directed graph andv ∈ V (D). Then the number of spanning tree rooted at v is

det(L(D)V (D)\{v},V (D)\{v})

Proof. (Later...)

Remark. The directed matrix-tree theorem implies Kirchhoff’s matrix-tree theorem:Given G, let D be the directed graph with V (D) = V (G) and edges v � w for every

{v, w} of G.

2.7 Eulerian Cycles

Definition 2.7. Let d, n ∈ Z>0. A d-ary sequence of length n is a sequence of length n on thealphabet {0, 1, . . . , d−1}. A d-ary de Bruijn sequence of order n is a sequence (a1, . . . , adn) ∈{0, 1, . . . , d − 1}dn such that the subsequences (aj, aj+1, . . . , aj+n−1) (n-modular) are all thed-ary sequences of length n (each appearing exactly once.

24

2.8. DE BRUIJN SEQUENCE CHAPTER 2. TILINGS & SPANNING TREES

Example 2.5. A binary de Bruijn sequence of order 3 is

00010111 and 00011101

Why do de Bruijn sequences exist?How many of them are there?How do we construct them?Applications:

1. Position determination along a circular hallway.

2. Magic: Ask for the suits of any n consecutive cards in a deck of size 2n, etc.

Let us show that de Bruijin sequences exist, for every d, n.

Definition 2.8. Define the de Bruijn graph Dd,n as the directed graph with vertex set{0, . . . , d − 1}n−1 with edges (a1, a2, . . . , an−1) → (a2, . . . , an−1, an) for all (a1, . . . , an) ∈{0, . . . , d− 1}n.

Example 2.6. D2,3:

00 01

1110

Key observations: d-ary de Bruijn sequences of order n correspond to Eulerian cycles ofDd,n, i.e. directed cycles which use every edge (exactly once).

Theorem 2.16 (Euler). An undirected graph (without isolated vertices) is Euclian if andonly if every vertex has even degree.

Theorem 2.17 (Good (1947)).

Theorem 2.18 (“BEST”, van Aardenne-Ehrenfest, de Bruijn (1951), Smith, Tulk (1941)).

2.8 de Bruijn sequence

25

Chapter 3

Generating Functions

3.1 Catalan numbers

Definition 3.1. Let n ∈ N. A Dyck path of order n is a path in the plane from (0, 0) to(2n, 0), moving by steps up(1,1) and down(1,-1) and never passing below the x-axis.

Let Cn denote the set of Dyck paths of order n. We call Cn = |Cn|, the nth Catalannumber. e.g. C0 = 1, c1 = 1, c2 = 2, c3 = 5, c4 = 14.

Let’s prove the classical formula

Cn =1

n+ 1

(2nn

)=

(2n)!

(n+ 1)!n!

Lemma 3.1 (Catalan recurrence). The Catalan sequence (Cn)∞n=0 is characterized by therecurrence relation

Cn =n−1∑j=0

CjCn−1−j, C0 = 1

Proof. For n 6= 1, we construct a bijection

φ : Cn →n−1⊔j=0

Cj × Cn−1−j

Given a Dyck path P , let (2(j + 1), 0) be the first point where the path p touches the x-axisafter (0, 0). Decompose P to two Dyck paths- one is from (1, 1) to (2j + 1, 1) (p1), the otheris from (2(j + 1), 0) to (2n, 0) (p2). Note that p1 ∈ Cj, p2 ∈ Cn−1−j.

Set φ(p) = (p1, p2). Observe that φ has an inverse (by combining p1 and p2 in the reverseway). Therefore φ is a bijection.

Define the Catalan ordinary generating function

c(x) =∑n≥0

Cnxn.

26

3.1. CATALAN NUMBERS CHAPTER 3. GENERATING FUNCTIONS

Lemma 3.2.c(x) = xc(x)2 + 1.

Proof. Let us compare coefficients of xn on both sides, for all n ∈ N. If n = 0, both are 1.Now suppose n ≥ 1.

[xn](xc(x)2 + 1)

=[xn−1]c(x)2

=[xn−1](∑i

Cixi)(∑j

Cjxj)

=n−1∑j=0

Cn−1−jCj

=[xn]c(x)

Theorem 3.1. For n ∈ N, Cn = 1n+1

(2nn

)Proof. We use the equation xc(x)2 − x(x) + 1 = 0, so c(x) = 1±

√1−4x2x

. Since C(0) = 1, so

c(x) =1−√

1− 4x

2x

=

1−∑

j≥0

(12

j

)(−4x)j

2x

=∑j≥1

(2j − 2)!

j!(j − 1)!xj−1

Thus Cn = [xn]c(x) = (2n)!(n+1)!n!

.

3.1.1 Lagrange implicit function theorem

A systematical way to prove that C(x) = xC(x)2 + 1 is use Lagrange implicit functiontheorem.

Theorem 3.2 (Lagrange implicit function theorem). If A(x) = xφ(A(x)), where φ is formalpower series, then

[xn]A(x) =1

n[tn−1]φ(t)n.

27

3.2. CATALAN OBJECTS CHAPTER 3. GENERATING FUNCTIONS

Proof. Let φ(t) = (t+ 1)2

A(x) = c(x)− 1

c(x) = x(c(x))2 + 1

⇒ A(x) = x(A(x) + 1)2

Cn = [xn]c(x) = [xn]A(x)

=1

n[tn−1]((t+ 1)2)n

=1

n

(2n

n− 1

).

Remark. D(x) = xD(x)k + 1 giving [xn]D(x) =1

n

(knn−1

)(n ≥ 1).

3.1.2 Combinatorial proof

• Rotate Dyck path by 45◦, rescale by1√2

. We then have n× n rectangle weakly above

main diagonal.

• Add a right step at the end, we then have n× (n+ 1) rectangle.

• Given any up-right lattice path in an n× (n+ 1) rectangle, we can continue it periodi-

cally to form a path of slopen

n+ 1. If we start reading at any 2n+ 1 points in a given

period, we get distinct path since gcd(n, n+ 1) = 1.

• There is a unique path among these 2n+1 path which stays above the main diagonal.

Take a line of slopen

n+ 1and move it up until it touches our infinite path.

• So Cn =1

2n+ 1

(2n+1n

).

3.2 Catalan objects

3.3 Formal power series

• C[[x]]: ring of formal power series ∑n≥0

anxn

addition, multiplication are as for analytic functions, also a notion of composition.

• A ∈ C[[x]] is invertible if and only if A(0) 6= 0.

• A ∈ C[[x]] with A(0) = 0 has a compositional inverse if and only if [x1]A(x) 6= 0.

28

3.4. FORMAL LAURENT SERIES CHAPTER 3. GENERATING FUNCTIONS

3.4 Formal Laurent series

Definition 3.2 (formal Laurent series). Let C((x)) denote the field of formal Laurent series(in one variable), i.e., power series of the form∑

n≥k

anxn (an ∈ C)

for some k ∈ Z with the expected addition and multiplication. Note htat indeed everynonzero formal Laurent series f has an inverse: we can write

f = xk · g

where g ∈ C[[x]] is a formal power series with g(0) 6= 0. We know g−1 ∈ C[[x]], so

f−1 = x−k · g−1 ∈ C((x))

That is, C((x)) is the field of fractions of C[[x]].

3.4.1 Derivatives

Definition 3.3 (derivative). For

f =∑n≥k

fnxn ∈ C((x)),

define its derivativef ′ :=

∑n≥k

nfnxn−1 ∈ C((x)),

The key observation for proving Lagrange’s theorem is that[x−1]f ′ = 0 ∀f ∈ C((x))

Let’s see to recover the usual properties of the derivative. For example, the product rule

(fg)′ = f ′g + fg′

Since this equation is linear in f and g, it suffices to prove it where f = xl and g = xm.Then it reduces to (

xl+m)′

= xl (xm)′ +(xl)′xm

where we can verify explicitly. Applying the product rule n− 1 times n ≥ 1 we obtain

(fn)′ = nfn−1f ′

for positive n. If n is negative, we obtain the power rule from the equation

0 =(fnf−n

)′= fn

(f−n

)′+ (fn)′ f−n

This proves the power for all n ∈ Z. Note that this immediately implies the chain rule

(f(A(x)))′ = f ′(A(x))A′(x) ∀A ∈ C[[x]], A(0) 6= 0, f ∈ C((x))

This is because chain rule is linear in f and it reduces to the power rule when f = xn.

29

3.5. LAGRANGE’S THEOREM CHAPTER 3. GENERATING FUNCTIONS

3.5 Lagrange’s theorem

Theorem 3.3 (Lagrange implicit function theorem/Lagrange inversion formula). Supposethat φ ∈ C[[x]] with φ(0) 6= 0. then there exists a unique A ∈ C[[x]] with

A(x) = xφ(A(x))

for any f ∈ C((x)) and n 6= 0,

[xn] f(A(x)) =1

n

[xn−1

]f ′(x)φn(x)

If f(x) = x, this states

[xn]A(x) =1

n

[xn−1

]φn(x)

To prove this theorem, we need the following lemma.

Lemma 3.3 (Residue composition). Let A ∈ C[[x]], where

A(x) =∑n≥k

anxn where k ≥ 1 and ak 6= 0

then for any f ∈ C((x)), [x−1]f(x) =

1

k

[x−1]f(A(x)) · A′(x)

Proof of lemma. Since the equation is linear in f , it suffices to prove it when f = xn, n ∈ Z.We want to prove

1

k

[x−1]An(x) · A′(x)

{1 if n = −1

0 if n 6= −1

• If n 6= −1, then [x−1]An(x) · A′(x) =

[x−1] (An+1(x))

′

n+ 1= 0

• If n = −1, then write

A(x) = xk ·B(x) where B ∈ C[[x]], B(0) 6= 0

Then B−1(x) ∈ C[[x]]. We have[x−1]A−1(x) · A′(x) =

[x−1] (xkB(x)

)−1 (xkB(x)

)′=[x−1]x−kB−1(x)

[xkB′(x) + kxk−1B(x)

]=[x−1]

(B−1(x)B′(x)︸ ︷︷ ︸∈C[[x]]

+kx−1) = 0 + k = k

30

3.5. LAGRANGE’S THEOREM CHAPTER 3. GENERATING FUNCTIONS

Proof of Lagrange theorem. Since φ(0) 6= 0, φ−1 ∈ C[[x]]. Let

B(x) := xφ−1(x) ∈ C[[x]]

Note that the equation A(x) = xφ(A(x)) is just B(A(x)) = x. So the unique A is thecompositional inverse of B. Moreover, A(B(x)) = 0. For n 6= 0,

[xn] f(A(x)) =[x−1]x−n−1f(A(x))

=1

1

[x−1]B−n−1(x)f(A(B(x))︸ ︷︷ ︸

x

)B′(x) (Lemma 3.3 with k = 1)

=[x−1]f(x) · (B−n(x))

′

−nNote that [x−1] p(x)q(x) = 0 =⇒ [x−1] p(x)q′(x) = − [x−1] p′(x)q(x) Then,[

x−1]f(x) · (B−n(x))

′

−n=

1

n

[x−1]f ′(x)

(xφ−1(x)

)−n=

1

n

[x−1]f ′(x)x−nφn(x)

=1

n

[xn−1

]f ′(x)φn(x)

Example 3.1 (special series). Consider the formal power series with x ∈ C

E(x) :=∑n≥0

xn

n!

λ(x) :=∑n≥1

xn

n

Bα(x) :=∑n≥0

(α

n

)xn

We can show that E , λ, Bα satisfy the usual properties of ex, − log(1− x) and (1 + x)α, e.g.,

E(x+ y) = E(x)E(y)

E ′(x) = E(x)

λ(1− E(−x)) = x

Bα+β(x) = Bα(x)Bβ(x)

So we will just denote these by

ex − log(1− x) (1 + x)α

We can prove these identies directly, or use the following shortcut: if f, g analytic powerseries and there exists r > 0 such that

f(z) = g(z) |z| < r

31

3.5. LAGRANGE’S THEOREM CHAPTER 3. GENERATING FUNCTIONS

then[xn] f(x) = [xn] g(x) ∀n ∈ N

This follows by taking the derivative of f(z) = g(z) n times and set z = 0. (And this worksin multivariables.)

3.5.1 Applications of Lagrange’s Theorem

Theorem 3.4 (Cayley’s Formula). There are nn−2 trees and nn−1 rooted trees on the vectorset [n]. We can use Lagrange theorem to prove.

Proposition 3.1. Let an denote the number of rooted trees on [n],

A(x) =∑n≥1

ann!xn ∈ C[[x]](exponential generating function),

then A(x) = xeA(x).

Proof. For n ≥ 1, we have [xn]xeA(x) = [xn−1]∑m≥0

Am

m!.

n = 1⇒ a11!

n ≥ 2⇒ RHS =n−1∑m=1

1

m!

∑i1..im≥1,i1+..+im=n−1

ai1i1!...aimim!

We need to show that this equalsann!

, i.e.,

an =n−1∑1

n

m!

∑i1..im≥1,i1+..+im=n−1

(n− 1

i1, · · · , im

)ai1 ...aim(∗)

Interpretation:

• n = # vertices

• m = # neighbors of r

• m! =# ways to order the smaller rooted trees

• i1...im = # vertices of smaller trees

•(

n−1i1,··· ,im

)= # ways to choose the vertex sets of the smaller rooted trees

So we have A(x) = xφ(x), where φ(x) = ex. By Lagrange’s theorem n 6= 0,

[xn]A(x) =1

n[xn−1]φ(x)n

ann!

=1

n[xn−1]enx =

1

n

nn−1

(n− 1)!

So an = nn−1.

32

3.5. LAGRANGE’S THEOREM CHAPTER 3. GENERATING FUNCTIONS

Theorem 3.5 (Abel’s binomial theorem). For r ∈ C, we have (for n ≥ 1)

[xn]erA(x) =1

n[xn−1](erx)′(ex)n

=r

n[xn−1]erxenx

=r

n

(r + n)n−1

(n− 1)!=r(r + n)n−1

n!

[x0]erA(x) = 1

So

erA(x) =∑n≥0

r(r + n)n−1

n!xn.

Taking [xn] in the identity, e(α+β)A(x) = eαA(x)eβA(x)(α, β ∈ C).We obtain

(α + β)(α + β + n)n−1

n!=

∑i,j≥0,i+j=n

α(α + i)i−1

i!

β(β + n− i)i−1

i!

⇒ (α + β)(α + β + n)n−1 =n∑i=1

(n

i

)α(α + i)i−1β(β + n− i)n−i−1.

Theorem 3.6 (negative binomial).

(1− x)−α =∑n≥0

(α + n− 1

n

)xn α ∈ C

Proof. exercise.

Example 3.2 (quintic equation). Consider the quintic equation

t5 − t+1

2= 0

This is an example of an unsolvable quintic, i.e., we cannot find all solutions t ∈ C byradicals. However, we can solve for (one possible) t as a power series. Let’s consider moregenerally,

t5 − t+ x = 0

and regard t as a formal power series in C[[x]]. The corresponding quintic is unsolvable foralmost all value of x. We have

t(t4 − 1

)= −x =⇒ t = x

(1− t4

)−1i.e.,

t = xφ(t) where φ(x) =(1− x4

)−133

3.6. SERIES MULTISECTION CHAPTER 3. GENERATING FUNCTIONS

By Lagrange theorem for n ≥ 1,.

[xn] t =1

n

[xn−1

] (1− x4

)−nBy Theorem 3.6,

[xn] t =1

n

[xn−1

]∑i≥0

(n+ i− 1

i

)x4i =

{1

4i+1if n− 1 = 4i

0 otherwise

Thus,

t =∑i≥0

1

4i+ 1

(5i

i

)x4i+1

As an analytic power series, it has radius of convergence r = 4 × 5−54 ≈ 0.53499. So when

x = 12< 0.53, t ≈ 0.5506.

3.6 Series multisection

34