Spectral Geometry of Riemann Surfaces These are rough notes on Spectral Geometry and their application to hyperbolic riemann surfaces. They are based on Buser s text Geometry and Spectra of Compact Riemann Surfaces and Chavel s Eigenvalues in Riemannian Geometry. 1 Preliminaries We will generally take to be an m-dimensional closed, connected manifold. The Laplacian is the differential operator most simply defined as = div grad In geodesic coordinates it can be rewritten in the following, familiar form u(p) = m 2 u () x 2 i=1 i Green s formula will be essential f gd = grad f, grad g d We will not worry too much with regularity assumptions, although these are important. 2 Heat Equation One intuitive way to open this discussion is to discuss the heat equation on. We seek a function p(x, y, t) on (, ) satisfying the following three properties. p t = xp p(x, y, t) = p(y, x, t) lim P (x, y, t)f(y)d(y) = f(x) t 1
The first condition is geometric in nature: the heat kernel p should disperse heat over time in a way that is proportional to local heat averages. The second and third conditions will be essential when using the heat kernel as the kernel of an integral transform. Theorem: For closed, connected, there exists a unique heat kernel p, and moreover for < t < 1 we have the estimate 3 Spectral Theory p(x, y, t) c t m/2 Solving the heat equation on can be accomplished by understanding the eigenvalue problem on. That is, finding smooth ϕ and constants λ for which ϕ = λϕ The Spectral Theorem promises us the existence of a complete, L 2 - orthonormal eigenfunctions ϕ, ϕ 1, and associated eigenvalues λ, λ 1,. If we arrange the eigenvalues in increasing order = λ < λ 1 λ 2 then λ n as n. oreover, p (x, y, t) = e λnt ϕ n (x)ϕ n (y) n= where the converge above is uniform on for each t >. The proof of the Spectral Theorem follows from the following classical results. Theorem (Hilbert-Schmidt): Take closed, and K : R continuous and symmetric. Define the following integral transform on f L 2 () K[f](x) = K(x, y)f(y)d(y) Then the eigenvalues problem K[ϕ] = ηϕ has a complete, orthonormal system of functions in L 2. Call these ϕ, ϕ 1, and their associated eigenvalues η, η 1,. Crucially, one can recover the kernel from these functions 2
K(x, y) = η n ϕ n (x)ϕ n (y) n= However, this theorem only guarantees this convergence in L 2. Theorem (ercer): If η i then the converge is uniform on. 4 Small Eigenvalues Before elucidating the connection between the spectral theory of and the geometry of, we offer up some strange results about the eigenvalue distribution in the case when is a compact, hyperbolic Riemann Surface S. Theorem: For any S of genus g, we have λ 4g 2 (S) > 1 4. That is, there is a genus-dependent absolute bound on the number of eigenvalues less than 1/4. Contrast this with Theorem: For any n and for ϵ > arbitrarily small, there exists a Riemann Surface S of genus g with λ n (S) 1/4 + ϵ. Here are two more theorems which we will not prove. Theorem: For any δ > there exists S of genus g with λ 2g 3 (S) δ. Theorem: There exists a universal constant c > independent of the genus such that λ 2g 2 (S) c when g 2. 5 The inimax Principles The inimax Principles give upper and lower bounds on the eigenvalues of in terms of something explicitly computable, although still analytic. Here is a general closed manifold of any dimension. (Upper Bound) Let f, f 1,, f k be smooth functions with L 2 -norm equal to 1, and whose supports intersect in measure zero sets. Then λ k () max grad f i 2 d i k 3
oreover, equality holds precisely if f is a λ k eigenfunction. Proof: Take the eigenfunctions ϕ,, ϕ k 1. Since our system of (k + 1) functions f i are linearly independent, we can find some element in their span which has norm-squared 1 and is independent from these k eigenfunctions. Write this as ext, define the Fourier coeffient α j = f = β f + + β k f k fϕ j d ow we try to subtract off the projection of f to the k through m eigenfunctions, for m > k. grad(f By Green s formula, this becomes = m α j ϕ j ) 2 d j=k grad f 2 d m αjλ 2 j Here λ j λ k and by Parseval s Theorem the sum of the squares of the Fourier coeffients is 1, From this we obtain that λ k αj 2 = 1 j=k j=k grad f 2 d Since the f i had disjoint supports this is just = k i= β 2 i grad f i 2 d Since the L 2 norm of f, which is equal to 1, is also the sum of the β 2 i, this sum is bounded by 4
max grad f i 2 d i k The case of the equality for f a λ k eigenfunction is apparent from the proof. Intuition: While the f i are not eigenfunctions, Green s theorem lets us recover some average of their constituent frequencies through the L 2 norm of their gradients. Having enough independent functions means that some combination of them f only has higher modes, and the disjointness of the f i lets us bound the L 2 norm of the gradient of f by the L 2 norm of the gradients of the f i. (Lower Bound): Let split as the union of 1,, k, compact domains with positive measure, which intersect pairwise in zero measure. Define ν() = grad f 2 d where f ranges over smooth functions for which f 2 d = 1 fd = Then λ k () min 1 i k ν( i) Proof: We work in L 2 instead of C. Let χ i be the characteristic function of i. Take the linear combination ϕ = α ϕ + + α k ϕ k that has L 2 () norm equal to 1 and that is orthogonal to these characteristic functions. By definition of ν( i ), grad ϕ 2 d ν( i ) ϕ 2 d i i Thus grad ϕ 2 d min 1 i k ν( i) At the same time by Green s Theorem 5
grad ϕ 2 d = ϕ ϕd = k λ j ()αj 2 Since the L 2 norm of ϕ on, which is the sum of the squares of the α 2 j, is equal to 1, this is bounded by λ k (). Intuition: As before, we use the L 2 norm of the gradient to obtain the eigenvalues. We can find one function ϕ to have mean zero on all the i by using enough modes: more sets i means more nodes are necessary. Green s theorem then produces the eigenvalues for the modes used on the one hand, and we can lower-bound the largest of these by using the quantities ν( i ) which are designed to serve as lower bounds to the L 2 norm of the gradient. j= 6 Cheeger s Inequality The quantity ν() that emerged in the lower bounds is sometimes called the fundamental tone of. Although we defined it earlier as an infimum, it also has a simpler analytic description, which can be seen by applying the upper-bound in the minimax principle to the subset. Since might have boundary, we consider only eigenfunctions ϕ on which vanish on the boundary. The principal implies that the infimum, and thus the fundamental tone, is obtained by an eigenfuction with the lowest eigenvalue on. In any event, the fundamental tone is still hard to compute. However, we can estimate it from below with a geometric value Cheeger s isoperimetric constant. In what follows, we assume = S is a closed, connected surface. Let S be a closed subset with positive area and piecewise smooth boundary (it is possible for = S). Cheeger s isoperimetric constant is defined as l(a) h() = inf min( B, B ) where A ranges over all finite unions of piecewise smooth curves that separate into disjoint, relatively open subsets B and B. B and B need not be connected, but we can assume that they are without affecting the constant. A small Cheeger s constant means that it is possible to cut up into two big pieces with a very short curve. Think of a dumbell with a thin 6
midbar and large weights on either end. Theorem (Cheeger s inequality): ν() (1/4)h 2 (). Proof: Let s take u to be an eigenfunction for. We compute Thus by Cauchy-Schwarz grad u 2 = 2u(grad u) ( ) 2 4 u 2 d grad u 2 d grad u 2 d For the next step in the proof we use the co-area formula grad u 2 dv = A(t)dt where A(t) is the length of the line {u 2 (x) = t}. ow h() V (t)dt where V (t) is the volume of the set {u 2 (x) > t. This inequality is clear from the definition of the Cheeger constant. Finally, we use integration by parts = h() tv (t)dt = h() u 2 d Putting all this together gives us the inequality we want. Intuition: We can estimate the fundamental tone by estimating the eigenvalues on. Cauchy-Schwarz gives us a lower bound in terms of the gradient of u 2, and the co-area formula allows us to reformulate this as a geometric integral which can be estimated using Cheeger s constant and the norm of u 2. 7 Eigenvalue Estimates We can know prove some of the eigenvalue results we mentioned earlier. 7
Theorem: For any n and for ϵ > arbitrarily small, there exists a Riemann Surface S of genus g with λ n (S) 1/4 + ϵ. Proof: We need to take S with a sufficiently large collar. Let s take a local model isometric to [ w, w] R/[τ τ + l] with metric ds 2 = dρ 2 + cosh 2 ρdτ 2. The w coordinate moves along the collar and the τ coordinate around it. Take < a < b w. Consider the function f on the annulus defined by Write f = df/dρ. We have C ab = [a, b] R/[τ τ + l] f(ρ, t) = f(ρ) = e ρ/2 sin C ab grad f 2 d = l b a π(ρ a) b a (f (ρ)) 2 cosh ρdρ We know that cosh ρ (1/2)e ρ and that for ρ a, Thus C ab grad f 2 d cosh ρ (1/2)(1 + e 2a )e ρ ( ) 1 4 + π2 (b a) 2 (1 + e 2a ) f 2 d C ab ow we can split the collar into n disjoint annuli of this kind and let a and (b a) be arbitrarily large. Applying the minimax theorem completes the proof. Theorem: For any S of genus g, we have λ 4g 2 (S) > 1 4. That is, there is a genus-dependent absolute bound on the number of eigenvalues less than 1/4. Proof: Take a Riemann Surface S and consider a canonical fundamental polygon with 4g sides. Draw diagonals from the fixed vertex p, cutting the polygon into 4g 2 triangles which descend to a decomposition of S into 4g 2 geodesic triangles. Let us apply the other half of the minimax theorem to this decomposition by first showing that ν(t ) (1/4) for any such triangle. ow we use Cheeger s inequality. We want to find some 8
number ϵ(t ) > such that for one-dimensional submanifold A cutting up T into connected B and B we have l(a) min( B, B ) 1 + ϵ(t ) There are three ways that A can cut up such a triangle, by having ends on one, two, or no sides. Let us consider the case when A has ends on both side of T. We take local coordinates ds 2 = dρ 2 + sinh 2 ρdσ 2 Let B be the domain with the distinguished vertex on its boundary, and parametrize A by α(t) = (ρ(t), σ(t)). Let R be the diameter of T, and write Then 1 + ϵ(t ) = sinh R 1 + cosh R 1 l(α(t)) = 1 sinh ρ(t) σ(t) dt (1 + (T )) ( ρ 2 (t) + (sinh 2 ρ(t)) σ 2 (t)) 1/2 1 (cosh ρ(t) 1) σ(t) dt (1 + ϵ(t )) B 9