Lecture 2. (October 4, 2012)
Notes taken by Javier Garc´ıa Rodr´ıguez.
In this lecture we want to prove the following theorem.
Theorem 1. Let X be a finite, connected,k-regular graph. Then k−µ1
2 ≤h(X)≤p
2k(k−µ1).
To prove this theorem we will first need to introduce some other concepts and results. Recall thatthe Laplacian of a graphXis defined to be the operator
∆X:F(V)→ F(V), where for everyf ∈ F(V) andv∈V (∆Xf)(v) =deg(v)·f(v)−(AXf)(v).
In particular, if the graph X is k-regular we have ∆X =k·Id− AX. When it does not cause confusion we will write for clarity ∆ = ∆X.
Recall that E is a symmetric multiset of V2. An orientation on E is a multisubsetO ofE such that
E={{(x, y),(y, x)|x6=y, (x, y)∈O}} ∩ {{(x, x)∈O}}.
For each edgeefromO the first vertex ofewe denote bye− and the second by e+. Let
F(E) ={f :E→C|f((x, y)) =−f((y, x))}.
Consider the operator d:F(V)→ F(E) given by
(df)((x, y)) =f(y)−f(x), for every edge e∈E.
If we endowF(V) andF(E) with the natural scalar products hf, gi=X
v∈V
f(v)g(v), for every f, g∈ F(V), hf, gi=1
2 X
e∈E
f(e)g(e) =X
e∈O
f(e)g(e), for every f, g∈ F(E),
we can consider the adjoint operatord∗:F(E)→ F(V) characterised by hdf, giF(E)=hf, d∗giF(V).
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Observe that if we fixv∈V, we have d1v= X
e=(w,v)
1e− X
e=(v,w)
1e.
Similarly for anyg∈ F(E) we obtain (d∗g)(v) = X
e∈O,v=e+
g(e)− X
e∈O,v=e−
g(e).
Clearlyd∗gdoes not depend on the orientationO.
Proposition 1. Given a graphX we have ∆X =d∗d.
Proof. Letv∈V andf ∈ F(V). We choose an orientationO in such way that vis always positive. Following the definitions and the previous remarks we have
((d∗d)f)(v) = (d∗(df))(v) = X
e∈O,v=e+
(df)(e) = X
e∈O,v=e+
(f(v)−f(e−))
= X
e=(w,v)∈O
(f(v)−f(w)) = deg(v)·f(v)−(AXf)(v) = (∆Xf)(v)
If the graphXisk-regular, by definition of ∆Xwe have Spec ∆X=k−SpecAX
and we can write
Spec ∆X={λ0< λ1≤λ2≤ · · · ≤λn−1}
whereλi=k−µi. Hence the spectral gap ofX is given byλ1(∆X) =k−µ1. Consider the subspacel20(V)⊂ F(V) given by
l02(V) :={f ∈ F(V)| hf, ki= 0 for every constant functionk∈ F(V)}, i.e., f ∈l20(V) if and only ifP
v∈vf(v) = 0.Observe that diml20(V) =|V| −1.
We have the following proposition:
Proposition 2. For ak-regular connected graphX we have:
λ1(∆X) = inf
h∆f, fi
||f||2 :f ∈l20(V)\ {0}
= inf
||df||22
||f||22 : f ∈l02(V)\ {0}
.
Proof. SinceAX is symmetric, so is ∆X and we can find an orthonormal basis of l20(V), {h1, . . . , hn−1} such that ∆hi = λihi for every 1≤i≤n−1. Now givenf ∈l20(V) we can writef =α1h1+· · ·+αn−1hn−1 and we have
∆f =λ1α1h1+· · ·+λn−1αn−1hn−1. 2
Therefore we obtain
h∆f, fi=|α1|2λ1+· · ·+|αn−1|2λn−1≥λ1(|α1|2+· · ·+|αn−1|2) =λ1||f||22. Observe thath∆f1, f1i=hλ1f1, f1i=λ1||f1||22. Hence equality holds forf =h1 and the first equality is proved. Since we have h∆f, fi=hd∗df, fi=hdf, dfi=
||df||22, the second equality holds.
We can now show the first inequality of Theorem 1:
Proposition 3. Let X be ak-regular graph. Then h(X)≥ λ1(X)2 .
Proof. LetA⊆V. Suppose|A|=aand|V−A|=b. If we consider the function f =b1A−a1V−A, we havef ∈l20(V). Observe that||f||22=b2a+a2b=ab(a+b) and||df||22= (a+b)2|∂A|sincedf(e) =±|V|ife∈∂Aanddf(e) = 0 otherwise.
Now applying the previous proposition we obtain
||df||22
||f||22 =(a+b)2|∂A|
ab(a+b) = |∂A||V|
|A||V −A| ≥λ1(X).
If we now take Asuch that|A| ≤V /2 andh(x) =|∂A||A|, it follows that h(X) =|∂A|
|A| ≥ (k−µ1)|V −A|
|V| ≥k−µ1
2 .
Let us define the following 1-norms onF(V) andF(E) respectively:
kfk1=X
v∈V
|f(v)|, for every f ∈ F(V), kfk1=1
2 X
e∈E
|f(e)|=X
e∈O
|f(e)|, for every f ∈ F(E),
To prove the second inequality we will need the following results:
Proposition 4. Let U = n
f ∈ F(V) |f(v)≥0 ∀v∈V and|suppf| ≤ |V2|o . Then
h(X) = inf
||df||1
||f||1 |f ∈U\ {0}
.
Proof. Given f ∈U we can writef =Pn
i=1αi1Ai where 0 =α0 < α1 <· · ·<
αn andf takes the valueαi on the setAi. Clearly the setsAi are disjoint and
|Ai| ≤ |V|/2 sincef ∈U. We rewritefasf =Pn
i=1βi1Bi, whereBi=∪ik=1Ak and βi = αi−αi−1. Choose an orientation O such that if (x, y) ∈ O then
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f(y)≥f(x). Henced1Bi is non-negative one∈O. Now it follows kdfk1=kX
βid1Bik1=X
βikd1Bik1≥X
βi|∂Bi| ≥X
βih(X)k1Bik1=h(X)kfk1
which proves the proposition.
Proposition 5. For every real valuedf ∈ F(V)we havekdf2k1≤√
2kkfk2kdfk2. Proof. Applying the Cauchy-Schwarz inequality and the well known fact that (a+b)2≤2(a2−b2) for any real numbers a, b, we have
kdf2k1= X
(x,y)∈O
|f(x)2−f(y)2|= X
(x,y)∈O
|f(x)−f(y)||f(x) +f(y)|
≤
s X
(x,y)∈O
|f(x)−f(y)|2 s X
(x,y)∈O
|f(x) +f(y)|2
≤ kdfk2 s X
(x,y)∈O
2(f(x)2+f(y)2) =√
2kkdfk2kfk2.
We can now finish the proof of Theorem 1:
Letf be an eigenvector corresponding toλ1. Since λ1 is real, we may assume that f is a real valued function. Let V+ := {v ∈ V | f(v) > 0} and define f+ :=
( f(v) ifv∈V+
0 ifv /∈V+ and f− := f −f+. We may further assume that
|V+| ≤ |V|/2, otherwise replacef by−f. First observe that for everyv∈V+
∆f+(v) =k·f+(v)− X
(v,w)∈E
f+(w)≤k·f(v)− X
{v,w}∈E
f(w) = ∆f(v) =λ1f+(v).
Also if v6∈V+, then ∆f+(v)≤0 =λ1f+(v).It follows then that kdf+k22=<∆f+, f+>≤< λ1f+, f+>=λ1kf+k22. On the other hand using the last two propositions we have
λ1kf+k22≥ kdf+k22≥ kdf+2k21 2kkf+k22
= kdf+2k21kf+2k21
2kkf+k22kf+2k21 ≥ h(X)2kf+2k21
2kkf+k22 = h(X)2kf+k42
2kkf+k22 = h(X)2kf+k22
2k .
It follows thatλ1≥ h(X)2k2 as required.
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