Notes for. Spectral Geometry. by Yaiza Canzani CRM Summer School in Quebec City. Laval University.

Transcription

1 Notes for Spectral Geometry by Yaiza Canzani CRM Summer School in Quebec City. Laval University.

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3 3 Abstract I wrote these lectures using the material in set of notes Analysis on manifolds via the Laplacian available at canzani/docs/laplacian.pdf. The material there corresponds to a year long course, so it contains more topics and many more details. The following references were important sources for these notes: Eigenvalues in Riemannian geometry. By I. Chavel. Old and new aspects in Spectral Geometry. By M. Craiveanu, M. Puta and T. Rassias. The Laplacian on a Riemannian manifold. By S. Rosenberg. Local and global analysis of eigenfunctions on Riemannian manifolds. By S. Zelditch. Enjoy!!

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5 Syllabus Abstract What makes the Laplacian special? Daily life problems You want to solve the Helmholtz equation Inverse problem: Can you hear the shape of a drum? Hearing the length of a guitar string Direct problems The first eigenvalue: Rayleigh Conjecture Counting eigenvalues: Weyl s Law A hard problem: understanding the eigenfunctions The Laplacian on a Riemannian manifold Definition of the Laplacian on a manifold Definition of a manifold Definition of a Riemannian metric Definition of the Laplacian Spectral Theory for the Laplacian Spectral Theorem Eigenvalue characterization (Exercise 4) Examples of eigenfunctions and eigenvalues Circle Torus Eigenfunctions on the sphere (Exercise 3) Hearing the geometry of a manifold Heat Equation Solutions to Heat equation (Exercise 4) Definition of the fundamental solution Weyl s Law and other high energy asymptotics Isospectral manifolds Exercises Exercise 0: Recovering the eigenvalues of a guitar string Exercise 1: Weyl s Law for a rectangle Exercise 2: Weyl s law on the Torus Exericise 3: Eigenfunctions on the Sphere Exercise 4: Characterization of eigenvalues Exercise 5: Temperature decreases with time and solutions are unique Exercise 5: Nodal domains and first eigenfunctions

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7 CHAPTER 1 What makes the Laplacian special? In this chapter we motivate the study of the Laplace operator. To simplify exposition, we do this by concentrating on planar domains. 1.1 Daily life problems Let Ω R n be a connected domain and consider the operator acting on C (Ω) that simply differentiates a function ϕ C (Ω) two times with respect to each position variable: n 2 ϕ ϕ = x 2. i i=1 Figure: Pierre-Simon de Laplace

8 8 What makes the Laplacian special? Here are some examples where the Laplacian plays a key role: Heat diffusion. If you are interested in understanding how would heat propagate along Ω R n then you should solve the Heat Equation u(x, t) = 1 c u(x, t) t where c is the conductivity of the material of which Ω is made of, and u(x, t) is the temperature at the point x Ω at time t. You could also think you have an insulated region Ω (it could be a wire, a ball, etc.) and apply certain given temperatures on the edge Ω. If you want to know what the temperature will be after a long enough period of time (that is, the steady state temperature distribution), then you need to find a solution of the heat equation that be independent of time. The steady state temperature solution will be a function u(x 1,..., x n, t) such that u = 0. Wave propagation. Now, instead of applying heat to the surface suppose you cover it with a thin layer of some fluid and you wish to describe the motion of the surface of the fluid. Then you will need to solve the Wave equation u(x, t) = 1 c 2 u(x, t) t2 where c is the speed of sound in your fluid, and u(x, t) denotes the height of the wave above the point x at time t. You could also think of your domain Ω as the membrane of a drum, in which case its boundary Ω would be attached to the rim of the drum. Suppose you want to study what will happen with the vibration you would generate if you hit it. Then, you should also solve the wave equation u(x, t) = 2 u(x, t) for your drum, but this time 2 t you want to make sure that you take into account that the border of the membrane is fixed. Thus, you should also ask your solution to satisfy u(x, t) = 0 for all points x Ω. Quantum particles. If you are a bit more eccentric and wish to see how a quantum particle moves inside Ω (under the assumption that there are no external forces) then you need to solve the Schrödinger Equation 2 2m u(x, t) = i u(x, t) t where is Planck s constant and m is the mass of the free particle. Normalizing u so that u(, t) L 2 (Ω) = 1 one interprets u(x, t) as a probability density. That is, if A Ω then the probability that your quantum particle be inside A at time t is given by A u(x, t) 2 dx. Why not another operator? The Laplacian on R n commutes with translations and rotations. That is, if T is a translation or rotation then (ϕ T ) = ( ϕ) T. Something

9 1.2 You want to solve the Helmholtz equation 9 more striking occurs, if S is any operator that commutes with translations and rotations then there exist coefficients a 1,..., a m making S = m j=1 a j j. Therefore, it is not surprising that the Laplacian will be a main star in any process whose underlying physics are independent of position and direction such as heat diffusion and wave propagation in R n. 1.2 You want to solve the Helmholtz equation There are of course many more problems involving the Laplacian, but we will focus on these ones to stress the importance of solving the eigenvalue problem (also known as Helmholtz equation) ϕ = λϕ. It is clear that if one wants to study harmonic functions then one needs to solve the equation ϕ = λϕ with λ = 0. So the need for understanding solutions of the Helmholtz equation for problems such as the static electric field or the steady-state fluid flow is straightforward. In order to attack the heat diffusion, wave propagation and Schrödinger problems described above, a standard method (inspired by Stone-Weierstrass Theorem) is to look for solutions u(x, t) of the form u(x, t) = α(t)ϕ(x). For instance if you do this and look at the Heat equation then you must have ϕ(x) ϕ(x) = α (t) α(t) x Ω, t > 0. This shows that there must exist a λ R such that α = λα and ϕ = λϕ. Therefore ϕ must be an eigenfunction of the Laplacian with eigenvalue λ and α(t) = e λt. Once you have these particular solutions u k = e λ kt ϕ k you use the superposition principle to write a general solution u(x, t) = k a k e λ kt ϕ k (x) where the coefficients a k are chosen depending on the initial conditions. You could do the same with the wave equation (we do it in detail for a guitar string in Section 1.3.1) or with the Schrödinger equation and you will also find particular solutions of the form u k (x, t) = α k (t)ϕ k (x) with ϕ k = λ k ϕ k and α k (t) = e λ kt e i λ k t e iλ kt Heat eqn, Wave eqn, Schrödinger eqn.

10 10 What makes the Laplacian special? 1.3 Inverse problem: Can you hear the shape of a drum? Inverse problem: If I know (more or less) the Laplace eigenvalues of a domain, what can I deduce of its geometry? Suppose you have perfect pitch. Could you derive the shape of a drum from the music you hear from it? More generally, can you determine the structural soundness of an object by listening to its vibrations? This question was first posed by Schuster in As Berger says in his book A panoramic view of Riemannian Geometry, Already in the middle ages bell makers knew how to detect invisible cracks by sounding a bell on the ground before lifting it up to the belfry. How can one test the resistance to vibrations of large modern structures by nondestructive essays?... A small crack will not only change the boundary shape of our domain, one side of the crack will strike the other during vibrations invalidating our use of the simple linear wave equation. On the other hand, heat will presumably not leak out of a thin crack very quickly, so perhaps the heat equation will still provide a reasonable approximation for a short time... An infinite sequence of numbers determines via Fourier analysis an integrable function. It wouldn t be that crazy if an infinite sequences of eigenvalues would determine the shape of the domain. Unfortunately, the answer to the question can you hear the shape of a drum? is no. This was proved in 1992 by Gordon, Web and Wolpert. Nowadays many planar domains are known to have different shapes but exactly the same spectrum. Figure: Webb and Gordon holding isospectral drums

11 1.3 Inverse problem: Can you hear the shape of a drum? 11 k=1 Figure: Isospectral drums Not all is lost. One can still derive a lot of information of a domain by knowing its eigenvalues. Using the heat kernel, one can prove that as t 0, e λkt 1 ( area(ω) 4πt length( Ω) + 2πt ) (1 γ(ω)) 4πt 3 where γ(ω) is the number of holes of Ω and Ω is a smooth, bounded domain. The eigenvalues λ k are the ones corresponding to the Laplacian on Ω enforcing ϕ k Ω = 0. The first term was proved by Weyl in The second term was proved in 1954 by Pleijel. It shows that you can hear wether a drum is circular or not (because of the isoperimetric inequality (length( Ω)) 2 4π area(ω) whose equality is only attained by circles). The first term was found by Mark Kac in 1966 and it was rigorously justified by McKean and Singer in This means that if you know the full sequence of eigenvalues of your favorite domain Ω then you can deduce its area, its perimeter and the number of holes in it!! Figure: Mark Kac

12 12 What makes the Laplacian special? Hearing the length of a guitar string Consider an interval [0, l] with Dirichlet boundary conditions ϕ(0) = ϕ(l) = 0. The eigenfunctions are ϕ k (x) = sin ( kπ ) l x for k 1 with eigenvalues λ k = ( ) kπ 2 l for k 1. ϕ 1 ϕ 2 ϕ 3 ϕ 4 After the first half of the 18th century mathematicians such as d Alembert and Bernoulli developed the theory of a vibrational string. As one should expect, the vibrations of a string will depend on many factors such us its length, mass and tension. Jean-Baptiste d Alembert Daniel Bernoulli To simplify our exposition consider a guitar string of length l which we model as the interval [0, l]. Assume further that the density mass and the tension are constant and equal to 1. Today it comes as no surprise that the behavior of a vibrating string is described by the wave equation. That is, if we write x for a point in the string [0, l] and t for the time variable, then the height u(x, t) of the string above the point x after a time t should satisfy the wave equation u(x, t) = 2 u(x, t). t2

13 1.3 Inverse problem: Can you hear the shape of a drum? 13 There are infinitely many solutions to this problem. But we already know that there are constraints to this problem that we should take into account since the endpoints of the string are fixed and so u(x, t) must satisfy u(0, t) = 0 = u(l, t) for all time t. In addition having a unique solution to our problem depends upon specifying the initial shape of the string f(x) = u(x, 0) and its initial velocity g(x) = t u(x, 0). All in all, we are solving the system 2 u(x, t) = 2 u(x, t) x [0, l], t > 0, x 2 t 2 u(0, t) = 0 = u(l, t) t > 0, u(x, 0) = f(x) t u(x, 0) = g(x) x [0, l], x [0, l]. A general solution of this problem has the form u(x, t) = α k (t)ϕ k (x), (1.1) where k=1 ( kπ ) ( kπ ) α k (t) = a k cos l t + b k sin l t. The coefficients are a k = f, ϕ k and b k = g, ϕ k. The functions ϕ k are called harmonic modes for the string [0, l]. Since λ k = ( ) kπ 2 ( ) l is the eigenvalue of the wave ϕ k (x) = sin kπ l x, the connection between the eigenvalues λ k and the frequencies f k of the harmonic modes of the string is obvious: λk f k = 2π. Therefore, the higher the eigenvalue, the higher the frequency is. Consider the Fourier transform of a function ϕ as F(ϕ)(ξ) = ϕ(x)e ixξ dx. The Fourier transform of the function ϕ k (x) = sin( λ k x) is F(ϕ k )(ξ) = π ( δ(ξ λ k ) δ(ξ + ) λ k ). i This is because sin( λ k x) = ei λk x e i λ k x and F(e i λ k x )(ξ) = 2π δ(ξ λ k ). 2i If you pluck a guitar string then you obtain a wave of the form u(x, t) = k=1 α k(t)ϕ k (x), and by applying the Fourier transform to it you get F(u(, t))(ξ) = π ( α k (t) δ (ξ ) λ k δ (ξ + )) λ k. i k=1 So you recover all the relevant frequencies and hence all the eigenvalues. In the picture below we have the graphs of s 1 (x) = sin(x), s 2 (x) = 1 3 sin(2x) and s 3(x) = 1 8 sin(4 x).

14 14 What makes the Laplacian special? In the first picture we show the graph of the wave s 1 + s 2 + s 3, while in the second picture we show the Fourier transform of the function s 1 + s 2 + s /3 1/7 8π 7 3π 7 1π 7 1π 7 3π 7 8π Direct problems Direct problem: If I know (more or less) the shape of a domain, what can I deduce of its Laplace eigenvalues? The first eigenvalue: Rayleigh Conjecture The first eigenvalue λ 1 of the Laplacian on an interval or a region of the plane is called the fundamental tone. This is because either on a vibrating guitar string or drum membrane the first eigenvalue corresponds to the leading frequency of oscillation and it is therefore the leading tone you hear when you play one of these instruments. Seen from a heat-diffusion point of view, since the solutions of the heat equation are of the form u(x, t) = k a ke λkt ϕ k (x), it is clear that (λ 1, ϕ 1 ) give the dominant information because e λ1t ϕ 1 (x) is the mode that decays with slowest rate as time passes by. From this last point of view it is natural to expect that the geometry of Ω should be reflected on λ 1 to some extent. For instance the largest the boundary Ω is, the more quickly the heat should wear off. That is, if we consider a domain Ω and a ball B of same area as Ω, then we expect the heat on Ω to diffuse more quickly than that of B. Therefore, we should have

15 1.4 Direct problems 15 Faber-Krahn Inequality: λ 1 (Ω) λ 1 (B). This result was proved by Faber and Krahn in As expected, it extends to any dimension. Figure: Lord Rayleigh Counting eigenvalues: Weyl s Law Jeans asked once what is the energy corresponding to an infinitesimal frequency interval. In 1966 Mark Kac told this story in a very illustrating manner:...at the end of October of 1910 the great Dutch physicist H. A. Lorentz was invited to Götingen to deliver a Wolfskehl lecture... Lorentz gave five lectures under the overall title Old and new problems of physics and at the end of the fourth lecture he spoke as follows (in free translation from the original German): In conclusion, there is a mathematical problem which perhaps will arouse the interest of mathematicians who are present. It originates in the radiation theory of Jeans. In an enclosure with a perfectly reflecting surface, there can form standing electromagnetic waves analogous to tones over an organ pipe: we shall confine our attention to very high overtones. Jeans asks for the energy in the frequency interval dν. To this end, he calculates the number of overtones which lie between frequencies ν and ν +dν, and multiplies this number by the energy which belongs to the frequency ν, and which according to a theorem of statistical mechanics, is the same for all frequencies.

16 16 What makes the Laplacian special? It is here that there arises the mathematical problem to prove that the number of sufficiently high overtones which lie between ν and ν + dν is independent of the shape of the enclosure, and is simply proportional to its volume. For many shapes for which calculations can be carried out, this theorem has been verified in a Leiden dissertation. There is no doubt that it holds in general even for multiply connected regions. Similar theorems for other vibrating structures, like membranes, air masses, etc., should also hold. If we express the Lorentz conjecture in a vibrating membrane Ω, it becomes of the following form: Let λ 1 λ 2... be the Laplace eigenvalues corresponding to the problem ϕ k = λ k ϕ k ϕ k Ω = 0. Then N(λ) = #{λ k : λ k < λ} area(ω) λ as λ. 4π D. Hilbert was attending these lectures and predicted as follows: This theorem would not be proved in my life time. But, in fact, Hermann Weyl, a graduate student at that time, was also attending these lectures. Weyl proved this conjecture four months later in February of Figure: Hermann Weyl We will prove this in specific examples such as rectangles and the torus. Later on we will prove the analogue result for compact Riemannian manifolds (M, g). Let λ 0 λ 1... be the Laplace eigenvalues repeated according to its multiplicity. Then N(λ) ω n (2π) n vol g(m)λ n/2, λ where ω n is the volume of the unit ball in R n.

17 1.4 Direct problems 17 In particular, λ j (2π) 2 (ω n vol g (M)) 2/n j2/n, j Weyl s law for an interval [0, l] [ ] [ ] Note that if we scale our domain by a factor a > 0 we get λ k (0, al) = 1 λ a 2 k (0, l). Intuitively, the eigenvalue λ must balance d2, and so λ (length scale) 2. We also dx 2 note that we have the asymptotics λ k C k 2 where C is a constant independent of k. For λ > 0 consider the eigenvalue counting function N(λ) = #{eigenvalues < λ}. Proposition 1. (Weyl s law for intervals) Write λ j for the Dirichlet or Neumann eigenvalues of the Laplacian on the interval Ω = [0, l]. Then, N(λ) length(ω) λ. π Proof. Note that ω 1 = 2. { N(λ) = max{k : λ k < λ} = max k : k 2 π 2 l 2 } < λ l λ. π Weyl s Law for a rectangle (Exercise 1) Consider a rectangle Ω = [0, l] [0, m] with Dirichlet boundary conditions. exercise you will prove that N(λ) area(ω) λ. 4π In this 1. Prove that the eigenvalues are ( jπ λ jk = l ) 2 + ( kπ m ) 2 for j, k Prove that #{(j, k) : λ jk < λ} area(ω) λ as λ. 4π Hint: use the two figures below, where E λ denotes an ellipse and Ẽλ is the copy of the ellipse translated by ( 1, 1).

18 18 What makes the Laplacian special? λm π λm π E λ Ẽ λ λl π λl π 1.5 A hard problem: understanding the eigenfunctions As already mentioned, we may interpret the eigenfunction ϕ k as the probability density of a quantum particle in the energy state λ j. That is, the probability that a quantum particle in the state ϕ k belongs to the set A Ω if given by ϕ k (x) 2 dx. A High energy eigenfunctions are expected to reflect the dynamics of the geodesic flow. In the energy limit λ one should be able to recover classical mechanics from quantum mechanics. In the following picture (taken from Many-body quantum chaos: Recent developments and applications to nuclei) you can see how the dynamics of the geodesic flow for two different systems is reflected on the eigenfunctions. In the left column a cardioid billiard is represented. In the right column a ball billiard is shown. In the first line the trajectories of the geodesic flow for each system is shown. Then, from the second line to the fifth one, the graph of the functions ϕ j 2 are shown for λ k = 100, 1000, 1500, The darker the color, the higher the value of the modulus. One can see how a very chaotic system, such as the cardioid, yields a uniform distribution (chaotic) of the eigenfunctions. On the other hand, a very geometric dynamical system, such as the disc, yields geometric distributions of the eigenfunctions. Figure 4. Behavior of eigenstates. The eigenstates of the integrable circular billiard and the chaotic cardioid billiard reflect the structure of the corresponding classical dynamics. A beautiful result about the behavior of eigenfunctions takes place on manifolds with ergodic geodesic flow (like the cardioid above), including all manifolds with negative p p s 1 0 References 1. M. Robnik, Classical Dynamics of a Family of Billiards with Analytic Boundaries, J. Physics A, vol. 16, 1983, pp A. Bäcker and H.R. Dullin, Symbolic Dynamics and Periodic Orbits for the Cardioid Billiard, J. Physics A, vol. 30, 1997, pp A. Bäcker, Numerical Aspects of Eigenvalue and Eigenfunction Computations for Chaotic Quantum Systems, The Mathematical Aspects of Quantum Maps, Lecture Notes in Physics 618, M. Degli Esposti and S. Graffi, eds., Springer-Verlag, 2003, pp R. Aurich and F. Steiner, Statistical Properties of Highly Excited Quantum Eigenstates of a Strongly Chaotic System, Physica D, vol. 64, 1993, pp O. Bohigas, M.-J. Giannoni, and C. Schmit, Characterization of Chaotic Quantum Spectra and Universality of Level Fluctuation

19 1.5 A hard problem: understanding the eigenfunctions 19 constant sectional curvature. This result says that in the high energy limit eigenfunctions are equidistributed. Quantum ergodicity. If Ω is a compact and has ergodic geodesic flow, then there exists a density one subsequence of eigenfunctions {ϕ jk } k such that for any A M lim ϕ jk (x) 2 dx = vol(a) k Ω vol(ω). #{k: j By density one subsequence it is meant that inf k m} m m Schnirelman (1973) finished by Colin de Verdière (1975). = 1. This result is due to 790 G. Rong et al. Image processing. One may describe a given surface M R 3 by a function such as the normal vector. That is, to every point x M you associate the normal vector at x x (n 1 (x), n 2 (x), n 3 (x)). Each function n j : M R can be rewritten as an infinite series n j = i=1 a(j) i ϕ i by the Sturm-Liouville decomposition Theorem. You may then truncate the series and work with the approximates manifold described by the function x ( N i=1 a (1) i ϕ i (x), N i=1 a (2) i ϕ i (x), N i=1 a (3) i ϕ i (x) ). Fig. 2a f. The inverse a. M =Dragon. manifold harmonics b-f. reconstructed transform manifold for models using of N different = 100, numbers 200, 300, 500 of bases. and 900 a is eigenfunctions the original dragon model, and b f are the results of the inverse manifold harmonics transform using m 100, 200, 300, 500 and 900 bases, respectively respectively. Picture from paper Spectral mesh deformation. B ij = ( t + t This )/2 is a way of encoding the geometry of result a surface is a vector some [ x 1, extent x 2,... x using m ],witheachitem x k corresponding to each frequency basis H k as little ( ) : information as you want (at the risk of having a worse approximation). In practice the B ii = t /6. n t St(i) x k = x, H k = x i D ii H t, t i k. (7) are the two triangles that share the edge (i, j), i=1 t and t are their areas, β i, j, β i, j are the two angles opposite to the edge (i, j), andst(i) is the set of triangles The inverse MHT maps the descriptor from frequency domain onto space domain by reconstructing x at vertex i incident to vertex i. using the first m frequencies: The above formulation can be further simplified into:

20 20 What makes the Laplacian special? way people have of computing eigenfunctions on the dragon is to discretize it and work with a discretized version of the Laplacian and its corresponding eigenfunctions. For instance, the method of approximating a surface by a finite number of eigenfunctions is used to perform a change of the position of some part of the surface. Suppose you have an armadillo standing on two legs (figure a) and you wish to lift one of its legs (figure e) reducing as much as possible the amount of computations that need to be carried to get the final result. What people are doing is to compute the first 99 eigenfunctions on the (discretized) armadillo (figure b) and approximate the armadillo by them (figure c). Then, you apply the transformation to the approximate armadillo (figure d). Doing this is much cheaper -computation wise- than applying the transformation to the original armadillo. You may then add all the details to the transformed armadillo by another algorithm. 788 G. Rong et al. Fig. 1a e. The spectral mesh deformation pipeline. a The original Armadillo model, b the manifold harmonics b model reconstructed using the first m bases, d the deformed smoothed model, and e the deformed model with detai added back easily [6]. However, the size of the linear system is not reduced since the number of unknowns (vertex positions) remains the same for the smoothed surfaces. Another approach is to use topological hierarchies of coarser and coarser meshes [8]. Subdivision surfaces provide a nice coupling of the geometric and topological hierarchy, and the number of unknowns in the coarse subdivision mesh could be significantly smaller than the original mesh [2, 27]. However, automatically building subdivision surfaces for arbitrary irregular meshes is a highly non-trivial task, since the original irregular mesh may not have coherent subdivision connectivity. In this paper, we propose to use the spectrum of the Laplace Beltrami operator defined on manifold surfaces, i.e. manifold Picture harmonics, from the to compactly article Spectral encode mesh deformation. the deformation functions. The manifold harmonics can be precomputed on arbitrary irregular meshes. Compared with other subspace deformation techniques [9, 27], the computation of manifold harmonics is fully automatic, and these orthogonal bases provide a compact parametrization for the space of functions defined on the surfaces. We can use very small number (compared to the number of vertices) of frequency components to represent the geometry and motion of the smoothed model. So the number of unknowns in the linear system for the deformation can be greatly decreased, to allow interactive manipulation on and spectral methods. Section 3 introdu manifold harmonics. The details of our given in Sect. 4, and the experimental re Sect. 5. Section 6 concludes the paper w directions of future work. 2 Related work Mesh deformation is an active research graphics. There are numerous previous p Energy minimization has long been a g deform smooth surfaces [3, 22]. A varia introduced in [2] to deform subdivision serve surface details, they optimize the en tion vector field instead of the deformatio e original Armadillo model, b the manifold harmonics bases, c the smoothed ed smoothed model, and e the deformed model with details added back positions. Multiresolution mesh editing t have been developed for detail-preservin decomposing a mesh into several frequ formed mesh is obtained by first man frequency mesh and later adding back t details as displacement vectors. e linear knowns hed surand spectral methods. Section 3 introduces the concept of manifold harmonics. The details of our new algorithm are given in Sect. 4, and the experimental results are shown in Yu et al. [23] apply the widely us tion on the 3D model deformation. The before and after the deformation to be e son equation, which is a linear system

21 CHAPTER 2 The Laplacian on a Riemannian manifold 2.1 Definition of the Laplacian on a manifold Definition of a manifold A smooth manifold M of dimension n is a Hausdorff topological space with a countable basis of open sets that has a family of charts U = {(U α, ϕ α )}, with ϕ α : U α M R n so that 1. U α = M 2. If U α U β then ϕ β ϕ 1 α : ϕ α (U α U β ) ϕ β (U α U β ) are C with C inverse. In this case we say that (U α, ϕ α ) and (U β, ϕ β ) are compatible. 3. Completness property: If (V, ψ) is a chart which is compatible with every (U α, ϕ α ) U then (V, ψ) U. ϕ α ϕ β ϕ β ϕ 1 α

22 22 The Laplacian on a Riemannian manifold Examples of differential manifolds include R n, the sphere S n, the torus T n. Products of manifolds are manifolds as well Example: R n Only one chart suffices to see that R n is a manifold. The chart is the identity: Example: Circle The circle can be thought of as U = R n, φ : U R n, φ(x) = x. S 1 = {(cos(θ), sin(θ)) : θ [0, 2π)}. The circle is a manifold of dimension 1. Let U 1 = {(cos(θ), sin(θ)) : θ (0, 2π)} and ϕ 1 : U 1 (0, 2π) be defined by its inverse ψ 1 = ϕ 1 1 by ψ 1 (θ) = (cos(θ), sin(θ)). Since U 1 doesn t cover the entire circle (we are missing one point!), we need one more chart. For example, we could take U 2 = {(cos(θ), sin(θ)) : θ ( π/2, π/2)} and ϕ 2 : U 2 ( π/2, π/2) defined by ψ 2 (θ) = (cos(θ), sin(θ)) and ψ 2 = ϕ 1 2. S 1 S 1 ϕ 1 ϕ 2 ψ 1 ψ2 0 2π π 2 π Example: Sphere Here we consider the sphere S 2 = {(x, y, z) R 3 : x 2 + y 2 + z 2 = 1}. The sphere is a manifold of dimension 2. Let U 1 = {(x, y, z) R 3 : x 2 + y 2 + z 2 = 1 and x 0}

23 2.1 Definition of the Laplacian on a manifold 23 and ϕ 1 : U 1 (0, π) (0, 2π) be defined by its inverse ψ 1 (θ, φ) = (sin(θ) sin(φ), sin(θ) cos(φ), cos(θ)). These are called spherical coordinates and are illustrated below. z θ x φ y Of course, you should add more charts to include the {θ = π, φ = 2π} line. Try it! Example: Torus The torus is the surface T 2 = R 2 /Z 2. That is, we put an equivalence relation on R 2 where we say that two points (x 1, y 1 ) and (x 2, y 2 ) are the same provided x 1 y 1 Z and x 2 y 2 Z. You may think of the torus as the surface you get after identifying opposite sides of a unit square, like in the picture below In particular, we will identify the torus T 2 with the set [0, 1) [0, 1). The chart that we will usually choose on the torus for working with coordinates is φ : (0, 1) (0, 1) T 2 (0, 1) (0, 1) R 2 φ(x, y) = (x, y). Of course, this chart covers all the torus but two lines (the sides of the square). Therefore, to be rigorous, one needs to add more charts.

24 24 The Laplacian on a Riemannian manifold Definition of a Riemannian metric A metric is an inner product, and to define an inner product we need a space of vectors associated to the manifold. If the manifold lived inside R n, like the sphere, one could think of the vectors as being the velocities associated to curves on the manifold. The drawback to this is that then one had vectors that live in a space whose dimension is higher than that of the manifold. On the sphere for example, the tangent vectors live in R 3 while the manifold has dimension 2. So how do we define a notion of velocity vectors associated to the manifold? The answer is that if γ : [ 1, 1] M is a curve on M, then γ(0) is going to be defined as an operator. Indeed, γ(0) is the operator that to a function f : M R associates the derivative of f in the direction of γ evaluated at t = 0. Namely, if we let (U, ϕ) be a coordinate chart about x M induce coordinates (x 1,..., x n ), then γ(0) ( f ) = n i=1 (ϕ γ) (0) x i ϕ(γ(0)) (f ϕ 1 ). In other words, γ(0) is the operator where we have set γ(0) = Scanned by CamScanner n i=1 x i γ(0) (f) := (ϕ γ) (0) x i γ(0), x i ϕ(γ(0)) (f ϕ 1 ). More generally, let (U, ϕ) be a coordinate chart about x M. We define the derivations for x U, where { x i x : x i x f := } 1 i n f ϕ 1 (ϕ(x)). x i { } Then, we define the tangent space T x M at x as the vector space whose basis is x x i, 1 i n. This is our space of velocities!

25 2.1 Definition of the Laplacian on a manifold 25 A Riemannian Manifold is a pair (M, g) where M is a C manifold and g is a map that assigns to any x M a non-degenerate, symmetric, positive definite bilinear form, g(x) : T x M T x M R such that for all V, W derivations, the map x V, W g(x) is smooth. Notation. Let (U, ϕ) be a chart with local coordinates (x 1,..., x n ), and consider the corresponding natural basis { x x 1,..., x x n } of T x M. We adopt the following notation: g ij (x) := x i x, x x j. g(x) We will also denote by g ij the entries of the inverse matrix of (g ij ) ij. Theorem. Every manifold carries a Riemannian metric.

26 26 The Laplacian on a Riemannian manifold Example: R n The chart is (R n, φ) where φ(x) = x. The metric is g R n(x) = This means that, using coordinates (x 1,..., x n ), we have Example: Circle x x, xi xj g R n = δ ij. We use the chart ϕ defined by its inverse ψ : (0, 2π) S 1 with ψ(θ) = (cos(θ), sin(θ)). Since everything is 1-dimensional, we we define the metric as g S 1(θ) = 1. That is, we set Example: Sphere, θ θ = 1. θ θ Endow the sphere with spherical coordinates ψ : (0, π) (0, 2π) S 2 R 3

27 2.1 Definition of the Laplacian on a manifold 27 ψ(θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ). The round metric g S 2 on the 2-sphere is defined as ( 1 0 g S 2(θ, φ) = 0 sin 2 θ ) Example: Torus Identifying T 2 with [0, 1) [0, 1) and using the coordinates ψ : (0, 1) (0, 1) R 2 T 2 ψ(x, y) = (x, y), the flat Riemannian metric on the torus takes the form ( ) 1 0 g T 2(x, y) = Definition of the Laplacian Having a metric on a manifold allows us to define geodesics on the manifold. Indeed, given any two points x and y sufficiently close in M there exists a unique path that has the shortest possible length (called geodesic) joining x with y. Fix a point x M and let β 1,..., β n : [ 1, 1] M be n geodesics in M with γ i (0) = x for all i. Suppose further that the geodesics are orthogonal at x. That is, We define the Laplacian as the operator making β i (0), βj (0) = 0, i j. g : C (M) C (M) g f(x) = n (f β i ) (0). i=1 It is possible to show that if (x 1,..., x n ) are local coordinates in M, then the Laplacian on (M, g)takes the form g = Example: R n Since g R n(x) = Id, we have 1 det g n i,j=1 ( g ij det g x i x j ). gr n = 2 x x 2. n

28 28 The Laplacian on a Riemannian manifold Example: Circle Since g S 1(θ) = 1, we have Example: Sphere gs 1 = 2 θ 2. Since g S 2(θ, φ) = ( sin 2 θ ), ( = 1 gs 2 det gs 2 θ = 1 ( sin θ θ = 1 sin θ θ ( sin θ θ ( sin θ θ And so, in spherical coordinates, Example: Torus gs 2 = 1 sin θ ( g θθ det g S 2 ) + θ φ )) ) + ( 1 φ sin θ ) 1 sin 2 θ 2 φ 2. φ ( sin θ ) 1 θ θ sin 2 θ ( g φφ det g S 2 )) φ 2 φ 2. (2.1) It is clear that gt 2 = 2 x y Spectral Theory for the Laplacian Volume form. Having a Riemannian metric g allows us to have a volume form dv g on M with respect to which we integrate. Given A M, vol(a) = 1 A (x) dv g (x). M Here, 1 A (x) = 1 if x A and 1 A (x) = 0 if x / A. In local coordinates ϕ(x) = (x 1,..., x n ) the volume form is dv g (x) = det(g ij (x)) dx 1... dx n. Hilbert space. The riemannian metric induces an inner product for functions on M. Given φ, ψ : M R we set φ, ψ L 2 := φ(x) ψ(x)dv g (x). M

29 2.2 Spectral Theory for the Laplacian 29 This yields a Hilbert space L 2 (M) := {f : M R function so that f, f L 2 < }. Gradient. The metric also allows us to have a notion of gradient g. The gradient of a function f C (M) measures the direction in which the function grows the most. In local coordinates ϕ(x) = (x 1,..., x n ) the gradient is defined as g f(x) = n g ij f ϕ 1 (x) (x) x. x i x j i=1 In R n this reduces to gr n f = n f i,j=1 x i (x) x x i. Green s Identity. Let (M, g) be a compact Riemannian manifold with boundary M. Let ψ C 1 (M) and φ C 2 (M). Then, M ψ g φ dv g = g ψ, g φ g dv g + φ ν φ dσ g. M M In particular, when M =, we get that the Laplacian is self-adjoint: g φ, ψ L 2 = φ, g ψ L 2. Sobolev space. As for domains in R n one can define the Sobolev inner product between functions φ, ψ C (M) φ, ψ H 1 = M g φ, g ψ g + φψ dv g. The Sobolev space denoted by H 1 (M) is the completion of C (M) with respect to the H 1 norm.

30 30 The Laplacian on a Riemannian manifold Spectral Theorem The Hilbert space is L 2 (M) and the Sobolev space is H 1 (M). The sesquilinear form is The ellipticity follows from the fact that a(φ, ψ) := φ, ψ H 1. a(φ, φ) = φ 2 H 1 0. The sesquilinear form a is continuous and symmetric on H 1 (M). Then, there exists a basis {ϕ j } of L 2 (M) and eigenvalues {γ j } staisfying that a(ϕ j, ψ) = γ j ϕ j, ψ L 2 for all ψ H 1 (M). Since a(ϕ j, ψ) = M gϕ j, g ψ g dv g + ϕ j, ψ L 2, we get g ϕ j, g ψ g dv g = (γ j + 1) ϕ j, ψ L 2 M for all ψ H 1 (M). In particular, for all ψ H 1 (M), and so for λ j = γ j + 1. g ϕ j, ψ L 2 = (γ j + 1) ϕ j, ψ L 2 g ϕ j = λ j ϕ j Theorem 2 (Spectral Theorem). For (M, g) compact, without boundary, there exists a complete orthonormal basis {ϕ 1, ϕ 2,... } of L 2 (M) consisting of eigenfunctions of g with ϕ j having eigenvalue λ j satisfying 0 = λ 1 λ Eigenvalue characterization (Exercise 4) Let (M, g) be a compact Riemannian manifold and write λ 1 λ 2... for the eigenvalues of the Laplacian repeated according to multiplicity (for any initial problem). Write ϕ 1, ϕ 2,... for the corresponding L 2 -normalized eigenfunctions. In this exercise you will prove the following result. Theorem 3. For k N and E k (g) := {ϕ 1, ϕ 2,..., ϕ k 1 }, λ k = inf { g φ 2 } L 2 φ 2 : φ H 1 (M) E k (g). L 2 The infimum is achieved if and only if φ is an eigenfunction of eigenvalue λ k. Here, g φ 2 L 2 := M gφ 2 gdv g. To ease the exposition, define D g (φ, ψ) = M gφ, g ψ g dv g. Fix φ H 1 (M) E k (g) and assume it has expansion φ = j=1 a jϕ j.

31 2.3 Examples of eigenfunctions and eigenvalues Fix l N and prove that D g (φ, φ) l j=1 λ ja 2 j. Hint: compute D g (φ l λ j a 2 j, φ j=1 l λ j a 2 j). j=1 2. Conclude that D g (φ, φ) λ k φ 2 L 2 for all φ E k (g). 3. To prove the second statement in the theorem, assume that the equality is attained. From your previous work deduce that j=k λ ja 2 j = λ k j=k a2 j. Conclude that φ must be a linear combination of the ϕ j s. 2.3 Examples of eigenfunctions and eigenvalues Circle The eigenfunctions are 1, cos(θ), sin(θ), cos(2θ), sin(2θ),..., cos(kθ), sin(kθ),... with eigenvalues 0, 1, 1, 4, 4,..., k 2, k 2,... respectively Torus We claim that the following are a basis of eigenfunctions on the torus: ϕ y (x) := e 2πi x,y for y Z 2. If you want, you may take their real and imaginary part to get the eigenfunctions 1, sin(2πi x, y ), cos(2πi x, y ), y Z 2 with eigenvalues Let us see that the is a basis of L 2 (T 2 ). 0, 4π 2 y 2, 4π 2 y 2, y Z 2. {ϕ y : y Z 2 } We first check that these functions are linearly independent by induction: suppose ϕ y1,..., ϕ yk are linearly independent and suppose also that k+1 i=1 a iϕ yi = 0. Since ϕ yj ϕ yi = ϕ yj +y i we know that 0 = k+1 i=1 a iϕ yi y k+1 = a k+1 + k i=1 a iϕ yi y k+1, after applying the Laplacian we get 0 = k k a i 4π 2 y i y k+1 2 ϕ yi y k+1 = e 2πi x,y k+1 a i 4π 2 y i y k+1 2 ϕ yi. i=1 i=1 It then follows that a 1 = = a k = 0. Therefore a k+1 = 0.

32 32 The Laplacian on a Riemannian manifold To see that B = span{ϕ y : y Z 2 } is dense in L 2 (T 2 ) we use the Stone Weirstrass Theorem. Clearly B is a subalgebra. Let us see that it separates points. Fix two points x 1, x 2 T 2 and assume x 1 x 2 (then x 1 x 2 / Z 2 ). Suppose now that ϕ y (x 1 ) = ϕ y (x 2 ) for all y Z 2. It then follows that e 2πi x 1 x 2,y = 1 for all y Z 2 and so x 1 x 2, y Z for all y Z 2 which implies that x 1 x 2 Z 2, and this is a contradiction. Weyl s law on the Torus (Exercise 2) We continue to write ω 2 := vol(b 1 (0)) where B 1 (0) is the unit ball in R 2. Denote by N(λ) the counting function N(λ) := #{j : λ j < λ}. In this exercise you will prove that N(λ) ω 2 vol(t 2 ) (2π) 2 λ λ. 1. Define N (r) := #{Z 2 B r (0)}. Find how N(λ) and N (r) are related. Restate your goal in terms of N (r). 2. Let P (r) denote the number of Z 2 -lattice squares inside B r (0). Show that P (r) N (r) P (r + d). 3. Find upper and lower bounds for the function P (r) in terms of volumes of Euclidean balls centered at Eigenfunctions on the sphere (Exercise 3) Let ψ(θ, φ) = (ψ 1 (θ, φ), ψ 2 (θ, φ), ψ 3 (θ, φ)) represent the spherical coordinate system for the sphere S 2 R 3. Then any x R 3 can be written as rψ(θ, φ) for r > 0 and is represented by the coordinates (r, θ, φ). 1. Use that if (x 1, x 2, x 3 ) = F (ξ 1, ξ 2, ξ 3 ) is a change of coordinates then 3 i=1 (F i ) ξ j x i, to prove that in spherical coordinates g R 3(r, θ, φ) = ( r 2 g S 2(θ, φ) Hint: express r, θ, φ in terms of x1, x2, x3. 2. Deduce that in spherical coordinates 3. Define gr 3 = 1 r 2 r ). ( r 2 ) + 1 r r 2 g. S 2 P k = {homogeneous polynomials of degree k}, H k = {P P k : gr 3 P = 0}, H k = {Y = P S 2 : P H k }. ξ j = Prove that the spaces H k and H k are isomorphic. That is, find the inverse for the restriction map H k H k given by P P S 2.

33 2.3 Examples of eigenfunctions and eigenvalues Prove that H k {Y C (S 2 ) : gs 2 Y = k(k + 1)Y }. The space H k is known as the space of Spherical Harmonics of degree k. 5. Find explicit bases for H 1 and H Use separation of variables to find a basis for H k. That is, look for solutions of the form Y k (θ, φ) = P k (θ)φ k (φ). The functions P k (θ) should satisfy a second order differential equation (do not try to solve it explicitly!). Such solutions P k (θ) are called Legendre polynomials. The solutions you obtain should have the form Yk m (θ, φ) = eimφ Pk m (cos(θ)). Scanned by CamScanner

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35 CHAPTER 3 Hearing the geometry of a manifold 3.1 Heat Equation Throughout this section we assume that (M, g) is a compact Riemannian manifold without boundary. The Heat Operator g + t acts on functions in C(M (0, + )) that are C 2 on M and C 1 on (0, ). The homogeneous heat equation is { ( g + t )u(x, t) = 0 u(x, 0) = f(x) (x, t) M (0, + ). x M The function u(x, t) represents the temperature at the point x at time t assuming that the initial temperature accross the manifolds was given by the function f(x) Solutions to Heat equation (Exercise 4) 1. Prove that the temperature of the manifold decreases as time evolves. That is, show that t u(, t) L 2 is decreasing with t. 2. Using the previous part show that the solution to the Heat Equation is unique Definition of the fundamental solution We say that a fundamental solution of the heat equation is a continuous function p : M M (0, ) R which is C 2 in (x, y), and C 1 with respect to t and such that L y p = 0 and lim t 0 p(, y, t) = δ y.

36 36 Hearing the geometry of a manifold Observation. It can be shown that that the fundamental solution is symmetric: p(x, y, t) = p(y, x, t). Observation. The function u(x, t) = solves the homogeneous heat equation. M p(x, y, t)f(y) dv g (y) Fundamental solution in R n Eventhough we have been working on compact manifolds throughout this chapter, we digress briefly to inspire the form of the fundamental solution on compact manifolds. Proposition 4. The function p : R n R n (0, + ) [0, + ) 1 p(x, y, t) = (4πt) n 2 e y x 2 /4t is a fundamental solution for the heat equation on R n. Proof. It is easy to check that gr n p + t p = 0. Let us prove that p(x, y, t) δ x (y) as t 0. We first need to show that R p(x, y, t)dy = 1 for all (x, t) R n (0, + ). n Indeed, p(x, y, t)dy = p(x, x + rξ, t)r n 1 dξdr R n 0 S n 1 (x) = = 0 0 = 1 π n 2 = 1 π n 2 = 1. S n 1 (x) S n 1 (x) 0 1 (4πt) n 2 1 S n 1 (x) R n e y 2 (4πt) n 2 dy e r2 4t r n 1 dξdr e s2 s n 1 (4t) n 1 2 (4t) 1 2 dξdr e s2 s n 1 dξds Let f : R n R be a bounded and continuous function. R f(x) p(x, y, t)f(y)dy = n = p(x, y, t)(f(x) f(y))dy R n p(x, y, t) f(x) f(y) dy + p(x, y, t)(f(x) f(y))dy B 2 tr (x) R n \B 2 tr (x) sup f(x) f(y) + y B 2 tr (x) 2 p(x, x + rξ, t) f(x) f(x + rξ) r n 1 dξdr. tr S n 1 (x)

37 3.1 Heat Equation 37 We now need to choose R sufficiently large so that the second term is as small as we wish. Once R is fixed the first term can be chosen to be small since t 0. To handle the second term note that 2 tr = S n 1 (x) R R p(x, x + rξ, t) f(x) f(x + rξ) r n 1 dξdr = S n 1 (x) S n 1 (x) p(x, x + 2 ts, t)2 f (2 ts) n 1 2 t dξds e s2 s n 1 2 f dξds = 2vol(S n 1 ) f e s2 s n 1 ds. R Fundamental solution on manifolds Theorem 5. There exists ɛ > 0 such that the fundamental solution for the heat equation has expansion p(x, y, t) = e d2 g (x,y)/4t ( k ) (4πt) n/2 t j u j (x, y) + O(t k+1 ), for all x, y M with d g (x, y) ε/2. In addition, u 0 (x, x) = 1 and u 1 (x, x) = 1 6 R g(x). j=0 Formally, we have that ( + t )e t = 0 and lim t 0 e t f = f. So it should be true that e t f (x) = p(x, y, t)f(y) dv g (y). M It can be shown that the operator e t is well defined and that p(x, y, t) is indeed its kernel. Theorem 6. Let {ϕ j } j be an orthonormal basis of Laplace eigenfunctions with eigenvalues {λ j } j. Then, p(x, y, t) = e λjt ϕ j (x)ϕ j (y). j=1 Proof. Since p(x, y, t) = k=0 p(x,, t), ϕ k ϕ k (y) and it is straight forward that p(x,, t), ϕ k = e t g ϕ k (x) = e λ kt ϕ k (x), p(x, y, t) = e λkt ϕ k (x)ϕ k (y). k=0

38 38 Hearing the geometry of a manifold 3.2 Weyl s Law and other high energy asymptotics Let (M, g) be a compact boundary-less Riemannian manifold. Write 0 = λ 1 < λ 2 λ 3... for all the Laplace eigenvalues repeated according to their multiplicity. We begin this section by introducing the Zeta function Z g : (0, + ) R Z g (t) = e λjt. j=1 Since the series is uniformly convergent on intervals of the form [t 0, + ) for all t 0 > 0 we know that Z g is continuous. We also have that it is decreasing in t, that lim t 0 + Z g (t) = +, and lim t + Z g (t) = 0. Proposition 7. Proof. Z g (t) Z g (t) = j=0 = = = M 1 (4πt) n/2 (vol g(m) + O(t)) as t 0 +. e λ jt p(x, x, t) dv g (x) k t j u j (x, x) dv g (x) + O(t k+1 ) 1 (4πt) n/2 j=0 M 1 (4πt) n/2 (vol g(m) + O(t)) Let us now write 0 = ν 1 < ν 2 < ν 3 <... for all the distinct eigenvalues. Then, setting m j for the multiplicity of ν j we can rewrite Z g (t) = m j e νjt. j=1 Theorem 8. The function Z g determines all the eigenvalues and their multiplicities. Proof. Note that for µ > 0 with µ 2, lim t eµt( Z g (t) 1 ) = lim t j=2 0 if µ < ν 2, m j e (µ νj)t = + if µ > ν 2 m 2, if µ = ν 2.

39 3.2 Weyl s Law and other high energy asymptotics 39 It follows that ν 2 is the unique strictly positive real number µ such that the limit lim t e µt( Z g (t) 1 ) is a natural number. By induction, ν k is the unique strictly positive real number µ such that the limit is a natural number. m k := lim e µt( k 1 ) Z g (t) 1 m j e ν jt t j=2 Note that N(λ) = #{j : λ j < λ} = λ 0 dµ for µ = j=1 δ λ j. To understand the bahavior of µ we use the following Tauberian Theorem. Theorem 9 (Karamata). Suppose that µ is a positive measure on [0, ) and that α (0, ). Supose that Then, λ 0 0 e tx dµ(x) at α t 0. dµ(x) Let ω n be the volume of the unite ball in R n, a Γ(α + 1) λα x. ω n := Our aim is to prove the following Theorem: 2πn/2 nγ(n/2). Theorem 10 (Weyl s asymptotic formula). Let M be a compact Riemannian manifold with eigenvalues 0 = λ 1 λ 2..., each distinct eigenvalue repeated according to its multiplicity. Then for N(λ) := #{j : λ j λ}, we have In particular, N(λ) ω n (2π) n vol g(m)λ n/2, λ. λ j 2π (ω n vol g (M)) 2/n j2/n, j. Proof. For the measure µ = δ λj, Proposition 7 asserts that 0 e tλ dµ(λ) 1 vol (4π) n g (M)t n/2. 2 Using Karamata s theorem on µ and α = n/2 we obtain N(λ) = λ 0 dµ(λ) vol g (M) (4π) n/2 Γ(n/2 + 1) λn/2 = ω nvol g (M) (2π) n λ n/2.

40 40 Hearing the geometry of a manifold 3.3 Isospectral manifolds In this section we prove that if (M, g M ) and (N, g N ) are compact Riemannian manifolds which share the same eigenvalues then they must have the same dimension, same volume and same total curvature. Theorem 11. If (M, g M ) and (N, g N ) are isospectral compact Riemannian manifolds, then dim M = dim N, vol gm (M) = vol gn (N) and R gm dv gm = R gn dv gn. Proof. Let λ 0 λ 1... be the eigenvalues of both gm e λjt = j=0 1 (4πt) dim M/2 k t j j=0 u g M M M and gn. Then j (x, x) dv gm (x) + O(t k+1 ) N and e λjt = j=0 1 (4πt) dim N/2 k t j j=0 u g N M j (x, x) dv gn (x) + O(t k+1 ). It follows immediately that dim M = dim N := n. Next, note that ( 1 (4πt) n/2 M yields = 0 (x, x) dv gm (x) k ( t j u g M 1 (4πt) n/2 j=1 M M N ) u g N 0 (x, x) dv gn (x) = u g M j (x, x) dv gm (x) N ) u g N j (x, x) dv gn (x) u g M 0 (x, x) dv gm (x) = u g N 0 (x, x) dv gn (x) N + O(t k+1 ) and since u g M 0 (x, x) = u g N 0 (x, x) = 1, we have vol gm (M) = vol gn (N). Repeating the same argument it follows that if (M, g M ) and (N, g N ) are isospectral compact Riemannian manifolds, then for all j M u g M j (x, x) dv gm (x) = N u g N j (x, x) dv gn (x). In particular, since u 1 (x, x) = 1 6 R g(x) we have that (M, g M ) and (N, g N ) have the same total curvature. We next prove that in the case of compact surfaces isospectrality implies a strong result: Corollary 12. If (M, g M ) and (N, g N ) are isospectral compact Riemannian surfaces, then M and N are diffeomorphic.

41 3.3 Isospectral manifolds 41 Proof. By Gauss-Bonnet Theorem R gm dv gm = 8π(1 γ M ) M where γ M is the genus of M. The same result holds for N. We then use that R gm dv gm = M R gn dv gn N which yields γ M = γ N. The result follows from the fact that two orientable surfaces with the same genus are diffeomorphic.

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43 CHAPTER 4 Exercises 4.1 Exercise 0: Recovering the eigenvalues of a guitar string 1. Show that the vibrations of a guitar string of length l are of the form u(x, t) = ( ( ) ( )) ( ) kπt kπt kπx a k sin + b k cos sin, l l l k where u(x, t) is the height of the wave at the point x at time t. Hint: Look first for solutions of the form α(t)ϕ(x). 2. Find a k and b k if you knew that the initial conditions are u(x, 0) = f(x) and t u(x, 0) = h(x). 3. Prove that the Fourier transform of the sound wave you heard recovers all the frequencies present in the wave. Hint: decompose sin(x) as a sum of exponentials and compute F(sin)(ξ). 4.2 Exercise 1: Weyl s Law for a rectangle Consider a rectangle Ω = [0, l] [0, m] with Dirichlet boundary conditions. exercise you will prove that N(λ) area(ω) λ. 4π In this 1. Prove that the eigenvalues are ( jπ λ jk = l ) 2 + ( kπ m ) 2 forj, k 1.

44 44 Exercises 2. Prove that #{(j, k) : λ jk λ} area(ω) λ as λ. 4π Hint: use the two figures below, where E λ denotes an ellipse and Ẽλ is the copy of the ellipse translated by ( 1, 1). λm π λm π E λ Ẽ λ λl π λl π 4.3 Exercise 2: Weyl s law on the Torus We continue to write ω 2 := vol(b 1 (0)), where B 1 (0) is the unit ball in R 2. Denote by N(λ) the counting function N(λ) := #{j : λ j < λ}. In this exercise you will prove that N(λ) ω 2 vol(t 2 ) (2π) 2 λ λ. 1. Define N (r) := #{Z 2 B r (0)}. Find how N(λ) and N (r) are related. Restate your goal in terms of N (r). 2. Let P (r) denote the number of Z 2 -lattice squares inside B r (0). Show that P (r) N (r) P (r + d). 3. Find upper and lower bounds for the function P (r) in terms of volumes of Euclidean balls centered at Exericise 3: Eigenfunctions on the Sphere Let ψ(θ, φ) = (ψ 1 (θ, φ), ψ 2 (θ, φ), ψ 3 (θ, φ)) represent the spherical coordinate system for the sphere S 2 R 3. Then any x R 3 can be written as rψ(θ, φ) for r > 0 and is represented by the coordinates (r, θ, φ). Hint: express r, θ, φ in terms of x1, x2, x3. 1. Use that if (x 1, x 2, x 3 ) = F (ξ 1, ξ 2, ξ 3 ) is a change of coordinates then 3 i=1 F i ξ j x i, to prove that in spherical coordinates g R 3(r, θ, φ) = ( r 2 g S 2(θ, φ) Hint: express r, θ, φ in terms of x1, x2, x3. ). ξ j =