MATH 262A NOTES, 2/5/13
LATEXED BY JAY CUMMINGS
1. The Second Eigenvalue.
Suppose we have two connected graphs G and G0 on the same vertex set V (G). We want to
relate the second eigenvalues of the normalized Laplacian of these two graphs. Call these
eigenvalues λ and λ0 , respectively. Then we have the following theorem.
Theorem 1. Suppose that for each edge {x, y} ∈ E(G) there exist a path Pxy in G0 with
|Pxy | ≤ `. Also, for v ∈ V , suppose that dv ≥ ad0 v . Finally, let m be such that for all
e ∈ E(G), m ≥ me , where me is the number of paths using e. Given all this, we can
conslude that
aλ1
.
λ0 1 ≥
`m
Let’s do some examples before the proof.
Example 1. Let G = Kn and recall that λ1 = n/(n − 1). Then
a
n
n
0
λ1≥
≥
.
`m n − 1
`m(max degree)
Example 2. Let V = Zp , G = Kn , G0 = Cn . For each pair of points we pick the shortest
path, allowing us to set ` = (p − 1)/2, the diameter of G0 . Next, fix an edge. We count m
by partitioning by the paths that will use that fixed edge in a shortest path. This is seen to
be
X
p2 − 1
p−1
−i =
.
m=
2
1
0≤i≤(p−3)/2
So
λ0 1 ≥
p
p−1
2
·
p2 −1
8
·2
&
8
.
p2
But now recall that we actually know the λ1 for the cycle. It is
2
2π
2π
1
2π 2
=
= 2 .
1 − cos
p
p
2
p
We got the order of magnitude right, were just off by a constant.
1
2
LATEXED BY JAY CUMMINGS
Example 3. Let G0 = Z2 d = Qd be the d-cube, which has vertex set V = {0, 1}d . G is still
K2d , ` = d is still the diameter.
Given vertices (a1 , . . . , ad ) and (b1 , . . . , bd ). we’ll create a path as follows. Find the first
place the two differ, say ai . Then the edges are (x1 , . . . , xt−1 , ∗, xt+1 , . . . , xd ), where ∗ is 0
and then 1. This defines a single edge. Continue in this way.
Number of paths through this edge? x = ( – any – , ∗, xt+1 , . . . , xd ) and y = (x1 , . . . , xt−1 , ∗, – any – ).
So m = 2d−1 . Therefore
2
2d
λ1 ≥ d−1 = 2 .
d
d2 d
Whereas actually λ1 = d2 . So we did give up a bit with this approach.
Now let’s prove Theorem 1.
Proof. First recall that λ0 1 = RG0 (g), the Rayleigh quotient, for some Rayleigh function g.
This g has the property that
X
g(x)d0 x = 0.
Now let’s construct an
P upperbound for λ1 by finding a good instance of f . Define f :
V (G) → R. We need
f (x)dx = 0. To accomplish this, let’s set f (x) = g(x) − c. Then
X
X
X
f 2 (x)dx =
(g(x) − c)2 dx ≥ a
(g(x) − c)2 d0 x
x∈V
x∈V (G)
≥ a inf
c
x∈V
X
(g(x) − c)2 d0 x = a
x∈V
X
g 2 (x)d0 x .
x∈V
Now, for an edge e = {x, y} ∈ E(G). Define g(e) = |g(x) − g(y)|. {x, y} may not be an
edge in G0 , but we do know that there exists a path Px,y if length less
P than or equal to `
in G0 . By Cauchy-Schwarz (and telescoping), (f (x) − f (y))2 ≤ |Pxy | e∈Px,y g 2 (e). Next,
m
X
e∈E(G0
X
g 2 (e) ≥
X
{x,y}∈E(G) e∈Px,y
1
=
`
X
X
g 2 (e) ≥
{x,y}∈E(G)
(g(x) − g(y))2
`
(f (x) − f (y)2 ,
{x,y}∈E(G)
the first two inequalities by definition of m and Cauchy-Schwarz, respectively.
Then, putting everything together to get a lower bound on λ0 1 ,
P
2
e∈E(G0 ) g (e)
λ0 1 = RG0 (g) = P
2
v∈V g (v)dv
P
1
2
a
a
{x,y}∈E(G0 ) (f (x) − f (y))
m`
≥
=
RG (f ) ≥
λ1 .
1P
2
m`
m`
x∈V (G) f (x)dx
a
MATH 262A NOTES, 2/5/13
3
Homework
• Generalize the last theorem to weighted paths.
Theorem 2. Let G be a graph on n vertices. For all x, y ∈ V (G) let Px,y be the path of
length less than or equal to ` joining x and y. Define
dx dy
,
qxy =
V ol(G)
P
the weight of Pxy . Assume Q is such that Qe = Pxy ⊇e qxy ≤ Q. Then λ1 ≥ 1/(`Q).
Recall that we previously showed the Cheeger ratio to be h ≥ 1/(2q). We will now sketch
the proof to the above theorem. In it, we will use the following homework problem.
Homework
• Show that
V ol(G)
λ1 = P
P
x∼y (f (x)
{x,y} (f (x)
− f (y)2
− f (y))2 dx dy
.
Proof. We will use the homework assignment. Since
X
X
(f (x) − f (y))2 ≤ |Pxy |
f 2 (e) ≤ `
f 2 (e),
e∈Pxy
e∈Pxy
summing over all x, y gives

X
x,y
qxy (f (x)−f (y))2 ≤ `
X
x,y
qxy
X
f 2 (e) ≤ `
e∈Pxy
X

X

e∈E(G)
qxy  f 2 (e) ≤ `Q
Pxy ⊇e
X
f 2 (e).
e∈E(G)
And moreover, the left-hand side above has
X
X dx dy
qxy (f (x) − f (y))2 =
(f (x) − f (y))2
V
ol(G)
x,y
x,y
Applying the homework gives the result.