Topology in Molecular Biology

212 R. Brooks

x= 0 x=1

y= y

y=y 1

0

Fig. 12.3.What a cusp looks like

Theorem 1.Let Sbe a Riemann surface of finite area, and let{Si}be a
family of coverings ofSgiven by graphs{Γi}.
Thenλi(Si)→ 0 asi→∞if and only ifh(Γi)→ 0 asi→∞
The basic idea of the proof is a following: ifCis a cusp ofS, thenClooks
as follows:
LetSyjbeS∩{y:y≤yj}, where we have chosen someyjfor each cusp
(Fig. 12.3).Syjis of course compact with boundary. Given our coveringsSi,
we may liftSyjtoSito get a family of coveringsSi,yjofSyj.
We may now apply our proof in the compact case to this setting, where we
replaceλ 1 withλN 1 , the first eigenvalue with Neumann boundary conditions,
and replace the Cheeger constanth(S) with the Cheeger constanthN(Si,yj)
with Neumann boundary conditions. We now have enough compactness to see
that the original argument goes through. The only point at which we have to
be careful is passing fromλ 1 (Si)toλN 1 (Si,yj).
In general, ifS∗has boundary, we have

λ 1 (S∗)=inf∫

S∗fdarea=0

∫

S∗‖grad(f)‖

2

∫

S∗f

2.

Now letfbe an eigenvalue of∆onSiwith eigenvalue

λ 1 =1/ 4 −s^2.

(Ifλ 1 is bigger than 1/4, there is nothing to prove.) We need to compare
the Rayleigh quotient offonSiwith the Rayleigh quotient offonS(i,yj).
Clearly, the numerator is less, since we are integrating the same over a small
area, but the denominator is also less, for the same reason. We want to show
that the denominator is nottoo muchless.
We must also worry about the fact that

∫

Si,yjfneed no longer be equal
to zero. We can modify this by subtracting a constant fromf, which does not
change grad(f), but we must worry that this constant is not too large.