Spectral Geometry

Home , Heat kernel, Isospectral, Spectral geometry, Spectral theorem, Spectral theory, Weyl law

by Yaiza Canzani.

2016 CRM Summer School in Quebec City. Laval University.

Abstract

I wrote these lectures using the material in set of notes “Analysis on manifolds via the Laplacian” available at http://www.math.harvard.edu/ 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. Ras- • sias.

The Laplacian on a Riemannian manifold. By S. Rosenberg. • Local and global analysis of eigenfunctions on Riemannian manifolds. By S. Zelditch. • Enjoy!!

Syllabus

Abstract ...... 3

1 What makes the Laplacian special? 7 1.1 Daily life problems ...... 7 1.2 You want to solve the Helmholtz equation ...... 9 1.3 Inverse problem: Can you hear the shape of a drum? ...... 10 1.3.1 Hearing the length of a guitar string ...... 12 1.4 Direct problems ...... 14 1.4.1 The ﬁrst eigenvalue: Rayleigh Conjecture ...... 14 1.4.2 Counting eigenvalues: Weyl’s Law ...... 15 1.5 A hard problem: understanding the eigenfunctions ...... 18

2 The Laplacian on a Riemannian manifold 21 2.1 Definition of the Laplacian on a manifold ...... 21 2.1.1 Definition of a manifold ...... 21 2.1.2 Definition of a Riemannian metric ...... 24 2.1.3 Definition of the Laplacian ...... 27 2.2 Spectral Theory for the Laplacian ...... 28 2.2.1 Spectral Theorem ...... 30 2.2.2 Eigenvalue characterization (Exercise 4) ...... 30 2.3 Examples of eigenfunctions and eigenvalues ...... 31 2.3.1 Circle ...... 31 2.3.2 Torus ...... 31 2.3.3 Eigenfunctions on the sphere (Exercise 3) ...... 32

3 Hearing the geometry of a manifold 35 3.1 Heat Equation ...... 35 3.1.1 Solutions to Heat equation (Exercise 4) ...... 35 3.1.2 Deﬁnition of the fundamental solution ...... 35 3.2 Weyl’s Law and other high energy asymptotics ...... 38 3.3 Isospectral manifolds ...... 40

4 Exercises 43 4.1 Exercise 0: Recovering the eigenvalues of a guitar string ...... 43 4.2 Exercise 1: Weyl’s Law for a rectangle ...... 43 4.3 Exercise 2: Weyl’s law on the Torus ...... 44 4.4 Exericise 3: Eigenfunctions on the Sphere ...... 44 4.5 Exercise 4: Characterization of eigenvalues ...... 45 4.6 Exercise 5: Temperature decreases with time and solutions are unique . 46 4.7 Exercise 5: Nodal domains and ﬁrst eigenfunctions ...... 46

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

n ∞ Let Ω R 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 X ∂2ϕ ∆ϕ = . ∂x2 i=1 i

Figure: Pierre-Simon de Laplace 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 n along Ω R then you should solve the Heat Equation ⊂ 1 ∂ ∆u(x, t) = u(x, t) c ∂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(x1, . . . , xn, 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

1 ∂2 ∆u(x, t) = u(x, t) c ∂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 ∂2 should also solve the wave equation ∆u(x, t) = ∂2t u(x, t) for your drum, but this time 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 Schr¨odingerEquation

2 ~ ∂ ∆u(x, t) = i~ u(x, t) −2m ∂t where ~ is Planck’s constant and m is the mass of the free particle. Normalizing u so that u( , t) 2 = 1 one interprets u(x, t) as a probability density. That is, if k · kL (Ω) A Ω then the probability that your quantum particle be inside A at time t is given ⊂ by R u(x, t) 2dx. A | | n Why not another operator? The Laplacian on R commutes with translations and rotations. That is, if T is a translation or rotation then ∆(ϕ T ) = (∆ϕ) T . Something ◦ ◦ 1.2 You want to solve the Helmholtz equation 9 more striking occurs, if S is any operator that commutes with translations and rotations Pm j then there exist coeﬃcients a1, . . . , am making S = j=1 aj ∆ . 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 n in R .

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 Schrödingerproblems 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) α0(t) = x Ω, t > 0. ϕ(x) α(t) ∈

This shows that there must exist a λ R such that ∈ α0 = λα and ∆ϕ = λϕ. − − Therefore ϕ must be an eigenfunction of the Laplacian with eigenvalue λ and α(t) = −λt −λ t e . Once you have these particular solutions uk = e k ϕk you use the superposition principle to write a general solution

X −λkt u(x, t) = ake ϕk(x) k where the coefficients ak 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 Schrödingerequation and you will also find particular solutions of the form uk(x, t) = αk(t)ϕk(x) with

 e−λkt Heat eqn,  √ i λ t ∆ϕk = λkϕk and αk(t) = e k Wave eqn, −  eiλkt Schr¨odingereqn. 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 1882. 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 non- destructive 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 1.3 Inverse problem: Can you hear the shape of a drum? 11

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, → ∞ X 1 2πt e−λkt area(Ω) √4πt length(∂Ω) + (1 γ(Ω)) ∼ 4πt − 3 − k=1 where γ(Ω) is the number of holes of Ω and Ω is a smooth, bounded domain. The eigenvalues λk are the ones corresponding to the Laplacian on Ω enforcing ϕk ∂Ω = 0. | The first term was proved by Weyl in 1911. 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 1967.

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 What makes the Laplacian special?

1.3.1 Hearing the length of a guitar string Consider an interval [0, `] with Dirichlet boundary conditions ϕ(0) = ϕ(`) = 0.

The eigenfunctions are

kπ ϕk(x) = sin x for k 1 ` ≥ kπ 2 with eigenvalues λk = for k 1. ` ≥

ϕ1

ϕ2

ϕ3

ϕ4

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 ` which we model as the interval [0, `]. 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, `] 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 ∂2 ∆u(x, t) = u(x, t). ∂t2 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(`, 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) = ∂tu(x, 0). All in all, we are solving the system  ∂2 ∂2  2 u(x, t) = 2 u(x, t) x [0, `], t > 0,  ∂x ∂t ∈ u(0, t) = 0 = u(`, t) t > 0, u(x, 0) = f(x) x [0, `],   ∈ ∂tu(x, 0) = g(x) x [0, `]. ∈ A general solution of this problem has the form ∞ X u(x, t) = αk(t)ϕk(x), (1.1) k=1 where kπ kπ α (t) = a cos t + b sin t . k k ` k ` The coefficients are ak = f, ϕk and bk = g, ϕk . h i h i kπ 2 The functions ϕk are called harmonic modes for the string [0, `]. Since λk = ` is kπ the eigenvalue of the wave ϕk(x) = sin ` x , the connection between the eigenvalues λk and the frequencies fk of the harmonic modes of the string is obvious: √λ f = k . k 2π Therefore, the higher the eigenvalue, the higher the frequency is.

Consider the Fourier transform of a function ϕ as Z ∞ (ϕ)(ξ) = ϕ(x)e−ixξdx. F −∞

The Fourier transform of the function ϕk(x) = sin(√λkx) is π p p (ϕk)(ξ) = δ(ξ λk) δ(ξ + λk) . F i − − This is because √ √ i λkx −i λkx √ p e e i λkx p sin( λkx) = − and (e )(ξ) = 2π δ(ξ λk). 2i F − P∞ If you pluck a guitar string then you obtain a wave of the form u(x, t) = k=1 αk(t)ϕk(x), and by applying the Fourier transform to it you get ∞ π X p p (u( , t))(ξ) = αk(t) δ ξ λk δ ξ + λk . F · i − − k=1 So you recover all the relevant frequencies and hence all the eigenvalues. In the picture 1 1 below we have the graphs of s1(x) = sin(x), s2(x) = 3 sin(2x) and s3(x) = 8 sin(4 x). 14 What makes the Laplacian special?

In the ﬁrst picture we show the graph of the wave s1 + s2 + s3, while in the second picture we show the Fourier transform of the function s1 + s2 + s3.

1/3

1/7

8π 3π 1π 1π 3π 8π − 7 − 7 − 7 7 7 7

1.4 Direct problems

Direct problem: If I know (more or less) the shape of a domain, what can I deduce of its Laplace eigenvalues?

1.4.1 The ﬁrst eigenvalue: Rayleigh Conjecture

The first eigenvalue λ1 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 P −λkt the form u(x, t) = k ake ϕk(x), it is clear that (λ1, ϕ1) give the dominant infor- −λ t mation because e 1 ϕ1(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 λ1 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 1.4 Direct problems 15

Faber-Krahn Inequality:

λ1(Ω) λ1(B). ≥ This result was proved by Faber and Krahn in 1923. As expected, it extends to any dimension.

Figure: Lord Rayleigh

1.4.2 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 Götingento 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 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 λ1 λ2 ... be the Laplace eigenvalues corresponding to the ≤ ≤ problem ∆ϕk = λkϕk ϕk ∂Ω = 0. − | Then area(Ω) N(λ) = # λk : λk < λ 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 1911.

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 λ0 λ1 ... ≤ ≤ be the Laplace eigenvalues repeated according to its multiplicity. Then

ωn n/2 N(λ) volg(M)λ , λ ∼ (2π)n → ∞

n where ωn is the volume of the unit ball in R . 1.4 Direct problems 17

In particular,

2 (2π) 2/n λj 2/n j , j . ∼ (ωnvolg(M)) → ∞

1.4.2.1 Weyl’s law for an interval [0, `]

1 Note that if we scale our domain by a factor a > 0 we get λk (0, a`) = a2 λk (0, `) . 2 Intuitively, the eigenvalue λ must balance d , and so λ (length scale)−2. We also dx2 ∼ note that we have the asymptotics

2 λk C k ∼ 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 λj for the Dirichlet or Neumann eigenvalues of the Laplacian on the interval Ω = [0, `]. Then,

length(Ω) N(λ) √λ. ∼ π Proof. k2π2 ` N(λ) = max k : λk < λ = max k : < λ √λ. { } `2 ∼ π

Note that ω1 = 2.

1.4.2.2 Weyl’s Law for a rectangle (Exercise 1)

Consider a rectangle Ω = [0, `] [0, m] with Dirichlet boundary conditions. In this × exercise you will prove that area(Ω) N(λ) λ. ∼ 4π

1. Prove that the eigenvalues are

jπ 2 kπ 2 λjk = + for j, k 1. ` m ≥

2. Prove that area(Ω) # (j, k): λjk < λ λ as λ . { } ∼ 4π → ∞

Hint: use the two ﬁgures below, where Eλ denotes an ellipse and E˜λ is the copy of the ellipse translated by ( 1, 1). − − 18 What makes the Laplacian special?

√λm √λm π π

Eλ E˜λ

√λ` √λ` π π

1.5 A hard problem: understanding the eigenfunctions

As already mentioned, we may interpret the eigenfunction ϕk as the probability density of a quantum particle in the energy state λj. That is, the probability that a quantum particle in the state ϕk belongs to the set A Ω if given by ⊂ Z 2 ϕk(x) 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, 2 from the second line to the fifth one, the graph of the functions ϕj are shown for | | λk = 100, 1000, 1500, 2000. 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 A beautifulreﬂect result the structure about of the corresponding the behavior classical dynamics. of eigenfunctions takes place on manifolds with ergodic geodesic ﬂow (like the cardioid above), including all manifolds with negative 1 References 1. M. Robnik, “Classical Dynamics of a Family of Billiards with Ana- lytic Boundaries,” J. Physics A, vol. 16, 1983, pp. 3971–3986.

p 0 2. A. Bäcker and H.R. Dullin, “Symbolic Dynamics and Periodic Orbits for the Cardioid Billiard,” J. Physics A, vol. 30, 1997, pp. 1991–2020. 3. A. Bäcker, “Numerical Aspects of Eigenvalue and Eigenfunction –1 Computations for Chaotic Quantum Systems,” The Mathematical –4 0 4 Aspects of Quantum Maps, Lecture Notes in Physics 618, M. Degli s Esposti and S. Grafﬁ, eds., Springer-Verlag, 2003, pp. 91–144. 1 4. R. Aurich and F. Steiner, “Statistical Properties of Highly Excited Quantum Eigenstates of a Strongly Chaotic System,” Physica D, vol. 64, 1993, pp. 185–214.

p 0 5. O. Bohigas, M.-J. Giannoni, and C. Schmit, “Characterization of Chaotic Quantum Spectra and Universality of Level Fluctuation Laws,” Physical Rev. Letters, vol. 52, 1984, pp. 1–4. –1 6. M.V. Berry and M. Tabor, “Level Clustering in the Regular Spec- –4 0 4 trum,” Proc. Royal Soc. London A, vol. 356, 1977, pp. 375–394. s 7. M.L. Mehta, Random Matrices, Academic Press, 1991. 8. M.V. Berry, “Semiclassical Mechanics of Regular and Irregular 1 Motion,” Comportement Chaotique des Systemes Deterministes: Chaotic Behaviour of Deterministic Systems, G. Iooss, R.H.G. Helle- mann, and R. Stora, eds., North-Holland, 1983, pp. 171–271.

p 0 9. A. Bäcker, R. Schubert, and P. Stifter, “Rate of Quantum Ergod- icity in Euclidean Billiards,” Physical Rev. E, vol. 57, 1998, pp. 5425–5447. –1 10. A. Bäcker, R. Ketzmerick, and A. Monastra, “Flooding of Regular –4 0 4 s Islands by Chaotic States,” Physical Rev. Letters, vol. 94, 2005, p. 054102.

Figure 5. Mixed phase space. Eigenstates in billiards with mixed Arnd Bäcker is a scientiﬁc researcher at the Technische phase space typically either concentrate in the regular islands (ﬁrst Universität Dresden. His research interests include nonlin- two lines) or extend over the chaotic region (last line). This is most ear dynamics, quantum chaos, and mesoscopic systems. clearly seen in the quantum Poincaré Husimi representation Bäcker has a PhD in theoretical physics from the Univer- displayed in the last column for each case. sität Ulm. Contact him at [email protected].

64 COMPUTING IN SCIENCE & ENGINEERING 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 ϕj k such that for any A M { k } ⊂ Z 2 vol(A) lim ϕjk (x) dx = . k→∞ Ω | | vol(Ω)

#{k: jk≤m} By density one subsequence it is meant that infm m = 1. This result is due to Schnirelman (1973) ﬁnished by Colin de Verdi`ere(1975).

3 Image processing. One may describe a given surface M R by a function such as ⊂ the normal vector. That is, to every point x M you associate the normal vector at x ∈

x (n1(x), n2(x), n3(x)). 7→ P∞ (j) Each function nj : M R can be rewritten as an inﬁnite series nj = a ϕi by the → i=1 i Sturm-Liouville decomposition Theorem. You may then truncate the series and work with the approximates manifold described by the function

N N N ! X (1) X (2) X (3) x a ϕi(x), a ϕi(x), a ϕi(x) . 790 G. Rong et al. 7→ i i i i=1 i=1 i=1

Fig. 2a–f. The inversea. M =Dragon. manifold harmonicsb-f. reconstructed transform manifold for models using ofN different= 100, numbers200, 300, 500 of bases. and 900a 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.

Bij ( t Thist! )/ is2 a way of encoding the geometry ofresult a surface is a vector to somex1, extentx2,...x usingm ,witheachitem as little xk corres- = | |+| | ponding to each frequency[ ˜ ˜ basis˜ ] Hk: ˜ information as you want (at the risk of having a worse approximation). In practice the Bii t /6. = | | n  t St(i) k k % ∈& ' xk x, H xi Dii H . (7) ˜ = $ % = i t, t are the two triangles that share the edge (i, j), i 1  ! &= t and t are their areas, βi, j, β are the two angles op- | | | !| i!, j The inverse MHT maps the descriptor from frequency do- posite to the edge (i, j),andSt(i) is the set of triangles main 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: m D 1 Qhk λ hk, (6) k − k xi xk H . (8) − = = ˜ i k 1 where D is a diagonal matrix defined as follows: &= Figure 2 shows the inverse MHT for the Dragon model Dii Bij t /3. = = | | using 100, 200, 300, 500, and 900 bases, respectively. j t St(i) & % ∈& ' The solution to this eigenproblem yields a series of k eigenpairs (H , λk) called the manifold harmonics bases 4 Deformation process (MHB). These bases are orthogonal, i.e. the functional inner product Hi, H j 0ifi j.Wealsoensurethatthe The manifold harmonics transform can help us to trans- MHB is orthonormal,$ % = by dividing&= each basis vector Hk by fer the geometric representation of a surface model and its functional norm Hk, Hk .ByusingtheArnoldimethod, its deformation functions from the space domain to the it is possible to compute$ eigenvectors% band-by-band utiliz- frequency domain. When reconstructing geometric and ing the shift-invert spectral transform. A detailed derivation deformation information using only the first m frequen- of these formulae can be found in [21]. cies, we are in fact conducting a low-pass signal filter- Using the MHB, we can define the manifold harmonics ing. The high frequencies (details) are filtered, and we transform (MHT) to convert the geometry of the mani- get a smoother approximation of the original model. In fold surface M into the spectral domain. The geometry x this paper we use a linear Laplacian-based variational al- (resp. y, z)ofthetriangulatedsurfacecanbeconsidered gorithm for mesh deformation. It is well known that the as a piecewise linear function defined over the hat func- gradient-based transformation cannot handle the local ro- n tions φi x i 1 xiφi where xi denotes the x coordinate tation of details [5]. We can overcome this problem by at vertex: i.TheMHTistheprojectionof= = x onto the or- using a spectral multi-resolution approach. The multi- thonormal MHB( via the functional inner product, and the resolution here means different levels of resolution in 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.

788 G. Rong et al.

Fig. 1a–e. The spectral mesh deformation pipeline. a The original Armadillo model, b the manifold harmonics bases, c the smoothed model reconstructed using the ﬁrst m bases, d the deformed smoothed model, and e the deformed model with details added back

added back easily [6]. However, the size of the linear and spectral methods. Section 3 introduces the concept of system is not reduced since the number of unknowns manifold harmonics. The details of our new algorithm are (vertex positions) remains the same for the smoothed sur- given in Sect. 4, and the experimental results are shown in faces. Another approach is to use topological hierarchies Sect. 5. Section 6 concludes the paper with some possible of coarser and coarser meshes [8]. Subdivision surfaces directions of future work. 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 ori- 2 Related work ginal mesh [2, 27]. However, automatically building subdivision surfaces for arbitrary irregular meshes is a highly Mesh deformation is an active research area in computer non-trivial task, since the original irregular mesh may not graphics. There are numerous previous papers on this topic. have coherent subdivision connectivity. Energy minimization has long been a general approach to In this paper, we propose to use the spectrum of the deform smooth surfaces [3, 22]. A variational approach is Laplace–Beltrami operator defined on manifold surfaces, introduced in [2] to deform subdivision surfaces. To pre- Fig. 1a–e. The spectral mesh deformation pipeline. a The original Armadilloi.e. model, manifoldPicture harmonics,b the from manifold the to compactly article Spectral harmonics encode mesh deformation the defor- bases,. servec the surface smoothed details, they optimize the energy of a deforma- mation functions. The manifold harmonics can be pre- tion vector field instead of the deformation energy of vertex model reconstructed using the first m bases, d the deformed smoothed model,computed and e onthe arbitrary deformed irregular model meshes. with Compared details with addedpositions. back Multiresolution mesh editing techniques [11, 28] other subspace deformation techniques [9, 27], the com- have been developed for detail-preserving deformations by putation of manifold harmonics is fully automatic, and decomposing a mesh into several frequency bands. A de- these orthogonal bases provide a compact parametrization formed mesh is obtained by first manipulating the low- for the space of functions defined on the surfaces. We frequency mesh and later adding back the high frequency can use very small number (compared to the number of details as displacement vectors. added back easily [6]. However, the size of the linear and spectralvertices) of methods. frequency components Section to represent 3 introduces the geom- theYu concept et al. [23] apply of the widely used Poisson equa- system is not reduced since the number of unknowns manifoldetry and harmonics. motion of the smoothed The details model. So of the our number newtion algorithm on the 3D model are deformation. They set the gradients of unknowns in the linear system for the deformation can before and after the deformation to be equal to get a Pois- (vertex positions) remains the same for the smoothed sur- givenbe in greatly Sect. decreased, 4, and to the allow experimental interactive manipulation results on son are equation, shown which in is a linear system. The solutions of large triangle meshes. The process of our algorithm is de- this equation give the deformed model. This class of al- faces. Another approach is to use topological hierarchies Sect. 5.scribed Section as the following 6 concludes steps: the paper withgorithm some is possible called gradient domain deformation.Gradient domain mesh deformation techniques [1, 9, 12, 17, 23, 24, of coarser and coarser meshes [8]. Subdivision surfaces directions1. compute of future manifold work. harmonics bases for the original 26, 27] have been intensively investigated for mesh edit- model with n vertices (Fig. 1b); provide a nice coupling of the geometric and topological ing. The main challenge is to handle nontrivial transform- 2. perform the inverse manifold harmonics transform ations which include rotations (especially large rotations) using the first m (m n)frequenciestogetasmoothed hierarchy, and the number of unknowns in the coarse sub- while preserving as much as possible the visual character- model (Fig. 1c); ! istic of the shape at interactive rates. The most important division mesh could be significantly smaller than the ori- 3. perform the deformation on the smoothed model by 2 Related work idea is to factor out the rotation from the deformation. For solving a linear system with m unknowns (Fig. 1d); ginal mesh [2, 27]. However, automatically building sub- shape editing, the factorization and shape definition have 4. add the details back to the deformed smoothed model to be solved simultaneously [17] since the target shape is division surfaces for arbitrary irregular meshes is a highly Mesh deformationto get the final result is (Fig. an 1e). active research area in computer not explicitly given. Instead of factoring out the rotation, non-trivial task, since the original irregular mesh may not graphics.The There rest of this are paper numerous is organized previous as follows: Sect. papers 2 abettersolutionistorepresenttheshapewithrotation- on this topic. have coherent subdivision connectivity. Energyreviews minimization some previous work has related long to mesh been deformation a generalinvariant approach coordinates to [12]. Zayer et al. [24] use a harmonic In this paper, we propose to use the spectrum of the deform smooth surfaces [3, 22]. A variational approach is Laplace–Beltrami operator defined on manifold surfaces, introduced in [2] to deform subdivision surfaces. To pre- i.e. manifold harmonics, to compactly encode the defor- serve surface details, they optimize the energy of a deforma- mation functions. The manifold harmonics can be pre- tion vector field instead of the deformation energy of vertex computed on arbitrary irregular meshes. Compared with positions. Multiresolution mesh editing techniques [11, 28] other subspace deformation techniques [9, 27], the com- have been developed for detail-preserving deformations by putation of manifold harmonics is fully automatic, and decomposing a mesh into several frequency bands. A de- these orthogonal bases provide a compact parametrization formed mesh is obtained by first manipulating the low- for the space of functions defined on the surfaces. We frequency mesh and later adding back the high frequency can use very small number (compared to the number of details as displacement vectors. vertices) of frequency components to represent the geom- Yu et al. [23] apply the widely used Poisson equa- etry and motion of the smoothed model. So the number tion on the 3D model deformation. They set the gradients of unknowns in the linear system for the deformation can before and after the deformation to be equal to get a Pois- be greatly decreased, to allow interactive manipulation on son equation, which is a linear system. The solutions of large triangle meshes. The process of our algorithm is de- this equation give the deformed model. This class of al- scribed as the following steps: gorithm is called gradient domain deformation.Gradient domain mesh deformation techniques [1, 9, 12, 17, 23, 24, 1. compute manifold harmonics bases for the original 26, 27] have been intensively investigated for mesh edit- model with n vertices (Fig. 1b); ing. The main challenge is to handle nontrivial transform- 2. perform the inverse manifold harmonics transform ations which include rotations (especially large rotations) using the first m (m n)frequenciestogetasmoothed while preserving as much as possible the visual character- model (Fig. 1c); ! istic of the shape at interactive rates. The most important 3. perform the deformation on the smoothed model by idea is to factor out the rotation from the deformation. For solving a linear system with m unknowns (Fig. 1d); shape editing, the factorization and shape definition have 4. add the details back to the deformed smoothed model to be solved simultaneously [17] since the target shape is to get the final result (Fig. 1e). not explicitly given. Instead of factoring out the rotation, The rest of this paper is organized as follows: Sect. 2 abettersolutionistorepresenttheshapewithrotation- reviews some previous work related to mesh deformation invariant coordinates [12]. Zayer et al. [24] use a harmonic CHAPTER 2

The Laplacian on a Riemannian manifold

2.1 Deﬁnition of the Laplacian on a manifold

2.1.1 Deﬁnition of a manifold

A smooth manifold M of dimension n is a Hausdorﬀ topological space with a countable n basis of open sets that has a family of charts = (Uα, ϕα) , with ϕα : Uα M R U { } ⊂ → so that S 1. Uα = M

−1 ∞ ∞ 2. If Uα Uβ = then ϕβ ϕ : ϕα(Uα Uβ) ϕβ(Uα Uβ) are C with C ∩ 6 ∅ ◦ α ∩ → ∩ inverse. In this case we say that (Uα, ϕα) and (Uβ, ϕβ) are compatible.

3. Completness property: If (V, ψ) is a chart which is compatible with every (Uα, ϕα) ∈ then (V, ψ) . U ∈ U

ϕα ϕβ

1 ϕ ϕ− β ◦ α 22 The Laplacian on a Riemannian manifold

n n n Examples of diﬀerential manifolds include R , the sphere S , the torus T . Products of manifolds are manifolds as well.

n 2.1.1.1 Example: R n Only one chart suﬃces to see that R is a manifold. The chart is the identity: n n U = R , φ : U R , φ(x) = x. → 2.1.1.2 Example: Circle The circle can be thought of as

S1 = (cos(θ), sin(θ)) : θ [0, 2π) . { ∈ } The circle is a manifold of dimension 1. Let

U1 = (cos(θ), sin(θ)) : θ (0, 2π) { ∈ } −1 and ϕ1 : U1 (0, 2π) be deﬁned by its inverse ψ1 = ϕ by → 1 ψ1(θ) = (cos(θ), sin(θ)).

Since U1 doesn’t cover the entire circle (we are missing one point!), we need one more chart. For example, we could take U2 = (cos(θ), sin(θ)) : θ ( π/2, π/2) and { ∈ −−1 } ϕ2 : U2 ( π/2, π/2) deﬁned by ψ2(θ) = (cos(θ), sin(θ)) and ψ2 = ϕ . → − 2 S1 S1

ϕ ϕ 1 ψ1 2 ψ2

0 2π π π − 2 2

2.1.1.3 Example: Sphere Here we consider the sphere

2 3 2 2 2 S = (x, y, z) R : x + y + z = 1 . { ∈ } The sphere is a manifold of dimension 2. Let

3 2 2 2 U1 = (x, y, z) R : x + y + z = 1 and x = 0 { ∈ 6 } 2.1 Definition of the Laplacian on a manifold 23

and ϕ1 : U1 (0, π) (0, 2π) be deﬁned by its inverse → ×

ψ1(θ, φ) = (sin(θ) sin(φ), sin(θ) cos(φ), cos(θ)).

These are called spherical coordinates and are illustrated below.

z θ

φ x y

Of course, you should add more charts to include the θ = π, φ = 2π line. Try it! { }

2.1.1.4 Example: Torus

2 2 2 2 The torus is the surface T = R /Z . That is, we put an equivalence relation on R where we say that two points (x1, y1) and (x2, y2) are the same provided x1 y1 Z − ∈ and x2 y2 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.

0 1

2 In particular, we will identify the torus T with the set [0, 1) [0, 1). The chart that we × will usually choose on the torus for working with coordinates is

2 2 φ : (0, 1) (0, 1) T (0, 1) (0, 1) R × ⊂ → × ⊂

φ(x, y) = (x, y).

Of course, this chart covers all the torus but two lines (the sides of the square). There- fore, to be rigorous, one needs to add more charts. 24 The Laplacian on a Riemannian manifold

2.1.2 Definition of a Riemannian metric A metric is an inner product, and to define an inner product we need a space of vectors n associated to the manifold. If the manifold lived inside R , 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 3 in R 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, thenγ ˙ (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 ofγ ˙ evaluated at → t = 0. Namely, if we let (U, ϕ) be a coordinate chart about x M induce coordinates ∈ (x1, . . . , xn), then

n X ∂ γ˙ (0)f = (ϕ γ)0(0) (f ϕ−1). ◦ ∂x ϕ(γ(0)) ◦ i=1 i

In other words,γ ˙ (0) is the operator

n X 0 ∂ Scannedγ˙ by(0) CamScanner = (ϕ γ) (0) , ◦ ∂x γ(0) i=1 i where we have set ∂ ∂ (f) := (f ϕ−1). ∂xi γ(0) ∂xi ϕ(γ(0)) ◦ More generally, let (U, ϕ) be a coordinate chart about x M. We deﬁne the derivations ∈ ∂ : 1 i n ∂xi x ≤ ≤ for x U, where ∈ ∂ ∂f ϕ−1 f := ◦ (ϕ(x)). ∂xi x ∂xi

n ∂ o Then, we deﬁne the tangent space TxM at x as the vector space whose basis is , 1 i n . ∂xi x ≤ ≤ This is our space of velocities! 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 ∈ , : TxM TxM R such that for all V,W derivations, the map h· ·ig(x) × →

x V,W 7→ h ig(x) is smooth.

Notation. Let (U, ϕ) be a chart with local coordinates (x1, . . . , xn), and consider the ∂ ∂ corresponding natural basis ,..., of TxM. We adopt the following nota- { ∂x1 x ∂xn x} tion:

∂ ∂ gij(x) := , . ∂xi x ∂xj x g(x)

ij We will also denote by g the entries of the inverse matrix of (gij)ij.

Theorem. Every manifold carries a Riemannian metric. 26 The Laplacian on a Riemannian manifold

n 2.1.2.1 Example: R n The chart is (R , φ) where φ(x) = x. The metric is

 1 0 0 0  ···  0 1 0 0   . . ··· . .  n  ......  gR (x) =  . . . . .  .    0 0 1 0  ··· 0 0 0 1 ···

This means that, using coordinates (x1, . . . , xn), we have

D ∂ ∂ E , = δij. x x g n ∂xi ∂xj R

2.1.2.2 Example: Circle We use the chart ϕ deﬁned by its inverse ψ : (0, 2π) S1 with ψ(θ) = (cos(θ), sin(θ)). → Since everything is 1-dimensional, we we deﬁne the metric as

gS1 (θ) = 1.

That is, we set ∂ ∂ , = 1. ∂θ θ ∂θ θ

2.1.2.3 Example: Sphere Endow the sphere with spherical coordinates

2 3 ψ : (0, π) (0, 2π) S R × → ⊂ 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 gS2 on the 2-sphere is deﬁned as

1 0 g 2 (θ, φ) = . S 0 sin2 θ

2.1.2.4 Example: Torus

2 Identifying T with [0, 1) [0, 1) and using the coordinates × 2 2 ψ : (0, 1) (0, 1) R T × ⊂ → ψ(x, y) = (x, y), the ﬂat Riemannian metric on the torus takes the form 1 0 g 2 (x, y) = . T 0 1

2.1.3 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 β1, . . . , βn :[ 1, 1] M be n geodesics in M with γi(0) = x ∈ − → for all i. Suppose further that the geodesics are orthogonal at x. That is,

β˙i(0) , β˙j(0) = 0, i = j. h i 6 We deﬁne the Laplacian as the operator

∞ ∞ ∆g : C (M) C (M) → making n X 00 ∆gf(x) = (f βi) (0). ◦ i=1

It is possible to show that if (x1, . . . , xn) are local coordinates in M, then the Laplacian on (M, g)takes the form n 1 X ∂ ijp ∂ ∆g = p g det g . det g ∂xi | |∂xj | | i,j=1

n 2.1.3.1 Example: R

Since gRn (x) = Id, we have ∂2 ∂2 ∆g n = + + . R 2 2 ∂x1 ··· ∂xn 28 The Laplacian on a Riemannian manifold

2.1.3.2 Example: Circle

Since gS1 (θ) = 1, we have ∂2 ∆g = . S1 ∂θ2

2.1.3.3 Example: Sphere Since 1 0 g 2 (θ, φ) = , S 0 sin2 θ

1 ∂ θθp ∂ ∂ φφp ∂ ∆ = g det g 2 + g det g 2 gS2 p S S det g 2 ∂θ | |∂θ ∂φ | |∂φ | S | 1 ∂ ∂ ∂ 1 ∂ = sin θ + sin θ ∂θ ∂θ ∂φ sin θ ∂φ 1 ∂ ∂ 1 ∂2 = sin θ . sin θ ∂θ ∂θ − sin2 θ ∂φ2

And so, in spherical coordinates,

1 ∂ ∂ 1 ∂2 ∆g = sin θ . (2.1) S2 sin θ ∂θ ∂θ − sin2 θ ∂φ2

2.1.3.4 Example: Torus It is clear that ∂2 ∂2 ∆g 2 = + . T ∂x2 ∂y2

2.2 Spectral Theory for the Laplacian

Volume form. Having a Riemannian metric g allows us to have a volume form dvg on M with respect to which we integrate. Given A M, ⊂ Z vol(A) = 1A(x) dvg(x). M

Here, 1A(x) = 1 if x A and 1A(x) = 0 if x / A. In local coordinates ϕ(x) = ∈ ∈ (x1, . . . , xn) the volume form is q dvg(x) = det(gij(x)) dx1 . . . dxn.

Hilbert space. The riemannian metric induces an inner product for functions on M. Given φ, ψ : M R we set → Z φ, ψ L2 := φ(x) ψ(x)dvg(x). h i M 2.2 Spectral Theory for the Laplacian 29

This yields a Hilbert space

2 L (M) := f : M R function so that f, f L2 < . { → h i ∞}

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) = (x1, . . . , xn) the gradient is deﬁned as

n −1 X ij ∂f ϕ ∂ gf(x) = g (x) ◦ (x) . ∇ ∂x ∂x x i=1 i j

n Pn ∂f ∂ In R this reduces to g n f = i,j=1 (x) . ∇ R ∂xi ∂xi x Green’s Identity. Let (M, g) be a compact Riemannian manifold with boundary ∂M. Let ψ C1(M) and φ C2(M). Then, ∈ ∈ Z Z Z ψ ∆gφ dvg = gψ, gφ g dvg + φ ∂νφ dσg. M · − M h∇ ∇ i ∂M ·

In particular, when ∂M = , we get that the Laplacian is self-adjoint: ∅

∆gφ, ψ 2 = φ, ∆gψ 2 . h iL h iL n Sobolev space. As for domains in R one can deﬁne the Sobolev inner product between functions φ, ψ C∞(M) ∈ Z φ, ψ H1 = gφ, gψ g + φψ dvg. h i M h∇ ∇ i

The Sobolev space denoted by H1(M) is the completion of C∞(M) with respect to the 1 norm. k · kH 30 The Laplacian on a Riemannian manifold

2.2.1 Spectral Theorem The Hilbert space is L2(M) and the Sobolev space is H1(M). The sesquilinear form is

a(φ, ψ) := φ, ψ 1 . h iH The ellipticity follows from the fact that

2 a(φ, φ) = φ 1 0. k kH ≥ The sesquilinear form a is continuous and symmetric on H1(M). Then, there exists a 2 basis ϕj of L (M) and eigenvalues γj staisfying that { } { }

a(ϕj, ψ) = γj ϕj, ψ 2 h iL 1 R for all ψ H (M). Since a(ϕj, ψ) = gϕj, gψ gdvg + ϕj, ψ 2 , we get ∈ M h∇ ∇ i h iL Z gϕj, gψ gdvg = (γj + 1) ϕj, ψ L2 M h∇ ∇ i h i for all ψ H1(M). In particular, ∈

∆gϕj, ψ 2 = (γj + 1) ϕj, ψ 2 h− iL h iL for all ψ H1(M), and so ∈ ∆gϕj = λjϕj − for λj = γj + 1. Theorem 2 (Spectral Theorem). For (M, g) compact, without boundary, there exists 2 a complete orthonormal basis ϕ1, ϕ2,... of L (M) consisting of eigenfunctions of ∆g { } with ϕj having eigenvalue λj satisfying

0 = λ1 λ2 ... . ≤ ≤ → ∞ 2.2.2 Eigenvalue characterization (Exercise 4)

Let (M, g) be a compact Riemannian manifold and write λ1 λ2 ... for the eigenval- ≤ ≤ ues of the Laplacian repeated according to multiplicity (for any initial problem). Write 2 ϕ1, ϕ2,... for the corresponding L -normalized eigenfunctions. In this exercise you will prove the following result.

⊥ Theorem 3. For k N and Ek(g) := ϕ1, ϕ2, . . . , ϕk−1 , ∈ { } 2 n gφ L2 o λk = inf k∇ k : φ H1(M) Ek(g) . φ 2 ∈ ∩ k kL2

The inﬁmum is achieved if and only if φ is an eigenfunction of eigenvalue λk. Here, 2 R 2 gφ 2 := gφ dvg. k∇ kL M k∇ kg R 1 To ease the exposition, deﬁne Dg(φ, ψ) = M gφ, gψ gdvg. Fix φ H (M) Ek(g) P∞ h∇ ∇ i ∈ ∩ and assume it has expansion φ = j=1 ajϕj. 2.3 Examples of eigenfunctions and eigenvalues 31

P` 2 1. Fix ` N and prove that Dg(φ, φ) λja . Hint: compute ∈ ≥ j=1 j ` ` X 2 X 2 Dg(φ λja , φ λja ). − j − j j=1 j=1

2 2. Conclude that Dg(φ, φ) λk φ 2 for all φ Ek(g). ≥ k kL ∈ 3. To prove the second statement in the theorem, assume that the equality is at- P 2 P∞ 2 tained. From your previous work deduce that j=k λjaj = λk j=k aj . Conclude that φ must be a linear combination of the ϕj’s.

2.3 Examples of eigenfunctions and eigenvalues

2.3.1 Circle The eigenfunctions are

1, cos(θ), sin(θ), cos(2θ), sin(2θ),..., cos(kθ), sin(kθ),... with eigenvalues 0, 1, 1, 4, 4, . . . , k2, k2,... respectively.

2.3.2 Torus We claim that the following are a basis of eigenfunctions on the torus:

2πihx,yi 2 ϕy(x) := e for y Z . ∈ If you want, you may take their real and imaginary part to get the eigenfunctions

2 1, sin(2πi x, y ), cos(2πi x, y ), y Z h i h i ∈ with eigenvalues 2 2 2 2 2 0, 4π y , 4π y , y Z . | | | | ∈ Let us see that the 2 ϕy : y Z { ∈ } 2 2 is a basis of L (T ).

We ﬁrst check that these functions are linearly independent by induction: suppose Pk+1 ϕy1 , . . . , ϕyk are linearly independent and suppose also that i=1 aiϕyi = 0. Since Pk+1 Pk ϕy ϕy = ϕy +y we know that 0 = aiϕy −y = ak+1 + aiϕy −y , after j ◦ i j i i=1 i k+1 i=1 i k+1 applying the Laplacian we get

k k X 2 2 −2πihx,yk+1i X 2 2 0 = ai4π yi yk+1 ϕy −y = e ai4π yi yk+1 ϕy . | − | i k+1 | − | i i=1 i=1

It then follows that a1 = = ak = 0. Therefore ak+1 = 0. ··· 32 The Laplacian on a Riemannian manifold

2 2 2 To see that = span ϕy : y Z is dense in L (T ) we use the Stone Weirstrass B { ∈ } Theorem. Clearly is a subalgebra. Let us see that it separates points. Fix two points 2 B 2 x1, x2 T and assume x1 = x2 (then x1 x2 / Z ). Suppose now that ϕy(x1) = ϕy(x2) ∈ 2 6 2πihx −−x ,y∈i 2 for all y Z . It then follows that e 1 2 = 1 for all y Z and so x1 x2, y Z ∈ 2 2 ∈ h − i ∈ for all y Z which implies that x1 x2 Z , and this is a contradiction. ∈ − ∈

Weyl’s law on the Torus (Exercise 2) We continue to write ω2 := vol(B1(0)) where 2 B1(0) is the unit ball in R . Denote by N(λ) the counting function N(λ) := # j : λj < { λ . In this exercise you will prove that } 2 ω2 vol(T ) N(λ) λ λ . ∼ (2π)2 → ∞

∗ 2 ∗ 1. Deﬁne N (r) := # Z Br(0) . Find how N(λ) and N (r) are related. Restate { ∩ } your goal in terms of N ∗(r).

∗ 2 2. Let P (r) denote the number of Z -lattice squares inside Br(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 Eu- clidean balls centered at 0.

2.3.3 Eigenfunctions on the sphere (Exercise 3)

Let ψ(θ, φ) = (ψ1(θ, φ), ψ2(θ, φ), ψ3(θ, φ)) represent the spherical coordinate system for 2 3 3 the sphere S R . Then any x R can be written as rψ(θ, φ) for r > 0 and is ⊂ ∈ represented by the coordinates (r, θ, φ).

1. Use that if (x , x , x ) = F (ξ , ξ , ξ ) is a change of coordinates then ∂ = 1 2 3 1 2 3 ∂ξj P3 ∂(Fi) ∂ , to prove that in spherical coordinates i=1 ∂ξj ∂xi

1 0 gR3 (r, θ, φ) = 2 . 0 r gS2 (θ, φ)

Hint: express ∂r, ∂θ, ∂φ in terms of ∂x1 , ∂x2 , ∂x3 . 2. Deduce that in spherical coordinates 1 ∂ 2 ∂ 1 ∆g 3 = r + ∆g 2 . R r2 ∂r ∂r r2 S

3. Deﬁne

k = homogeneous polynomials of degree k , P { } k = P k : ∆g 3 P = 0 , H { ∈ P R } Hk = Y = P 2 : P k . { |S ∈ H }

Prove that the spaces k and Hk are isomorphic. That is, ﬁnd the inverse for the H restriction map k Hk given by P P 2 . H → 7→ |S 2.3 Examples of eigenfunctions and eigenvalues 33

4. Prove that ∞ 2 Hk Y C (S ) : ∆g Y = k(k + 1)Y . ⊂ { ∈ S2 − } The space Hk is known as the space of Spherical Harmonics of degree k.

5. Find explicit bases for H1 and H2.

6. Use separation of variables to ﬁnd a basis for Hk. That is, look for solutions of the form Yk(θ, φ) = Pk(θ)Φk(φ). The functions Pk(θ) should satisfy a second order diﬀerential equation (do not try to solve it explicitly!). Such solutions Pk(θ) are called Legendre polynomials. The solutions you obtain should have the form

m imφ m Yk (θ, φ) = e Pk (cos(θ)).

Scanned by CamScanner

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 C2 on M and C1 on (0, ). × ∞ ∞ The homogeneous heat equation is ( ( ∆g + ∂t)u(x, t) = 0 (x, t) M (0, + ) − ∈ × ∞ . u(x, 0) = f(x) 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).

3.1.1 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) 2 is decreasing with t. 7→ k · kL 2. Using the previous part show that the solution to the Heat Equation is unique.

3.1.2 Deﬁnition of the fundamental solution We say that a fundamental solution of the heat equation is a continuous function 2 1 p : M M (0, ) R which is C in (x, y), and C with respect to t and such that × × ∞ → Lyp = 0 and lim p( , y, t) = δy. t→0 · 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 Z u(x, t) = p(x, y, t)f(y) dvg(y) M solves the homogeneous heat equation.

n 3.1.2.1 Fundamental solution in R 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. n n Proposition 4. The function p : R R (0, + ) [0, + ) × × ∞ → ∞ 1 −ky−xk2/4t p(x, y, t) = n e (4πt) 2 n is a fundamental solution for the heat equation on R .

Proof. It is easy to check that ∆g n p + ∂tp = 0. Let us prove that p(x, y, t) δx(y) − R R n → as t 0. We ﬁrst need to show that n p(x, y, t)dy = 1 for all (x, t) R (0, + ). → R ∈ × ∞ Indeed, Z Z ∞ Z p(x, y, t)dy = p(x, x + rξ, t)rn−1dξdr Rn 0 Sn−1(x) Z ∞ Z 2 1 − r n−1 = n e 4t r dξdr 0 Sn−1(x) (4πt) 2 Z ∞ Z 1 −s2 n−1 n−1 1 = n e s (4t) 2 (4t) 2 dξdr 0 Sn−1(x) (4πt) 2 Z ∞ Z 1 −s2 n−1 = n e s dξds π 2 0 Sn−1(x) Z 1 −kyk2 = n e dy π 2 Rn = 1.

n Let f : R R be a bounded and continuous function. → Z

f(x) p(x, y, t)f(y)dy = − Rn Z

= p(x, y, t)(f(x) f(y))dy Rn −

Z Z p(x, y, t) f(x) f(y) dy + p(x, y, t)(f(x) f(y))dy ≤ √ | − | n √ − B2 tR(x) R \B2 tR(x) Z ∞ Z n−1 sup f(x) f(y) + √ p(x, x + rξ, t) f(x) f(x + rξ) r dξdr. ≤ √ | − | n−1 | − | y∈B2 tR(x) 2 tR S (x) 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

Z ∞ Z n−1 √ p(x, x + rξ, t) f(x) f(x + rξ) r dξdr = 2 tR Sn−1(x) | − | Z ∞ Z n−1 p(x, x + 2√ts, t)2 f ∞(2√ts) 2√t dξds ≤ R Sn−1(x) k k Z ∞ Z −s2 n−1 = e s 2 f ∞dξds R Sn−1(x) k k Z ∞ n−1 −s2 n−1 = 2vol(S ) f ∞ e s ds. k k R

3.1.2.2 Fundamental solution on manifolds Theorem 5. There exists > 0 such that the fundamental solution for the heat equation has expansion 2 k −dg(x,y)/4t e X j k+1 p(x, y, t) = n/2 t uj(x, y) + O(t ) , (4πt) j=0 1 for all x, y M with dg(x, y) ε/2. In addition, u0(x, x) = 1 and u1(x, x) = Rg(x). ∈ ≤ 6 t∆ t∆ Formally, we have that ( ∆ + ∂t)e = 0 and limt→0 e f = f. So it should be true − that Z t∆ e f (x) = p(x, y, t)f(y) dvg(y). M It can be shown that the operator et∆ is well deﬁned and that p(x, y, t) is indeed its kernel.

Theorem 6. Let ϕj j be an orthonormal basis of Laplace eigenfunctions with eigen- { } values λj j. Then, { } ∞ X −λj t p(x, y, t) = e ϕj(x)ϕj(y). j=1 P∞ Proof. Since p(x, y, t) = p(x, , t), ϕk ϕk(y) and k=0h · i

t∆g −λkt p(x, , t), ϕk = e ϕk(x) = e ϕk(x), h · i it is straight forward that

∞ X −λkt p(x, y, t) = e ϕk(x)ϕk(y). k=0 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 = λ1 < λ2 λ3 ... ≤ ≤ for all the Laplace eigenvalues repeated according to their multiplicity. We begin this section by introducing the Zeta function Zg : (0, + ) R ∞ → ∞ X −λj t Zg(t) = e . j=1

Since the series is uniformly convergent on intervals of the form [t0, + ) for all t0 > 0 we ∞ know that Zg is continuous. We also have that it is decreasing in t, that limt→0+ Zg(t) = + , and limt→+∞ Zg(t) = 0. ∞ Proposition 7.

1 + Zg(t) (volg(M) + O(t)) as t 0 . ∼ (4πt)n/2 → Proof.

∞ X −λj t Zg(t) = e j=0 Z = p(x, x, t) dvg(x) M   k Z 1 X j k+1 = n/2  t uj(x, x) dvg(x) + O(t ) (4πt) j=0 M 1 = (volg(M) + O(t)) (4πt)n/2

Let us now write 0 = ν1 < ν2 < ν3 < . . . for all the distinct eigenvalues. Then, setting mj for the multiplicity of νj we can rewrite

∞ X −νj t Zg(t) = mje . j=1

Theorem 8. The function Zg determines all the eigenvalues and their multiplicities. Proof. Note that for µ > 0 with µ = 2, 6  0 if µ < ν , ∞  2 µt X (µ−νj )t lim e Zg(t) 1 = lim mje = + if µ > ν2 t→∞ − t→∞ j=2  ∞ m2, if µ = ν2. 3.2 Weyl’s Law and other high energy asymptotics 39

It follows that ν2 is the unique strictly positive real number µ such that the limit µt limt→∞ e Zg(t) 1 is a natural number. By induction, νk is the unique strictly − positive real number µ such that the limit

k−1 µt X −νj t mk := lim e Zg(t) 1 mje t→∞ − − j=2 is a natural number.

R λ P∞ Note that N(λ) = # j : λj < λ = dµ for µ = δλ . To understand the { } 0 j=1 j 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 ∈ ∞ Z ∞ e−tx dµ(x) at−α t 0. 0 ∼ → Then, Z λ a dµ(x) λα x . 0 ∼ Γ(α + 1) → ∞ n Let ωn be the volume of the unite ball in R , 2πn/2 ωn := . nΓ(n/2) Our aim is to prove the following Theorem: Theorem 10 (Weyl’s asymptotic formula). Let M be a compact Riemannian manifold with eigenvalues 0 = λ1 λ2 ... , each distinct eigenvalue repeated according to its multiplicity. ≤ ≤ Then for N(λ) := # j : λj λ , we have { ≤ } ωn n/2 N(λ) volg(M)λ , λ . ∼ (2π)n → ∞ In particular,

√2π 2/n λj 2/n j , j . ∼ (ωnvolg(M)) → ∞ Proof. P For the measure µ = δλj , Proposition 7 asserts that Z ∞ −tλ 1 −n/2 e dµ(λ) n volg(M)t . 0 ∼ (4π) 2 Using Karamata’s theorem on µ and α = n/2 we obtain Z λ volg(M) n/2 ωnvolg(M) n/2 N(λ) = dµ(λ) n/2 λ = n λ . 0 ∼ (4π) Γ(n/2 + 1) (2π) 40 Hearing the geometry of a manifold

3.3 Isospectral manifolds

In this section we prove that if (M, gM ) and (N, gN ) 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, gM ) and (N, gN ) are isospectral compact Riemannian manifolds, then Z Z

dim M = dim N, volgM (M) = volgN (N) and RgM dvgM = RgN dvgN . M N

Proof. Let λ0 λ1 ... be the eigenvalues of both ∆g and ∆g . Then ≤ ≤ M N   ∞ k Z X −λj t 1 X j gM k+1 e = dim M/2  t uj (x, x) dvgM (x) + O(t ) j=0 (4πt) j=0 M and   ∞ k Z X −λj t 1 X j gN k+1 e = dim N/2  t uj (x, x) dvgN (x) + O(t ) . j=0 (4πt) j=0 M

It follows immediately that dim M = dim N := n. Next, note that Z Z 1 gM gN n/2 u0 (x, x) dvgM (x) u0 (x, x) dvgN (x) = (4πt) M − N   k Z Z 1 X j gM gN k+1 =  t u (x, x) dvgM (x) u (x, x) dvgN (x) + O(t ) n/2 j − j (4πt) j=1 M N yields Z Z gM gN u0 (x, x) dvgM (x) = u0 (x, x) dvgN (x) M N

gM gN and since u0 (x, x) = u0 (x, x) = 1, we have volgM (M) = volgN (N). Repeating the same argument it follows that if (M, gM ) and (N, gN ) are isospectral compact Rieman- nian manifolds, then for all j Z Z gM gN uj (x, x) dvgM (x) = uj (x, x) dvgN (x). M N

1 In particular, since u1(x, x) = 6 Rg(x) we have that (M, gM ) and (N, gN ) have the same total curvature.

We next prove that in the case of compact surfaces isospectrality implies a strong result:

Corollary 12. If (M, gM ) and (N, gN ) are isospectral compact Riemannian surfaces, then M and N are diﬀeomorphic. 3.3 Isospectral manifolds 41

Proof. By Gauss-Bonnet Theorem Z

RgM dvgM = 8π(1 γM ) M − where γM is the genus of M. The same result holds for N. We then use that Z Z

RgM dvgM = RgN dvgN M N which yields γM = γN . The result follows from the fact that two orientable surfaces with the same genus are diﬀeomorphic.

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 ` are of the form

X kπt kπt kπx u(x, t) = a sin + b cos sin , k ` k ` ` 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 ak and bk if you knew that the initial conditions are u(x, 0) = f(x) and ∂tu(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 (sin)(ξ). F

4.2 Exercise 1: Weyl’s Law for a rectangle

Consider a rectangle Ω = [0, `] [0, m] with Dirichlet boundary conditions. In this × exercise you will prove that area(Ω) N(λ) λ. ∼ 4π

1. Prove that the eigenvalues are

jπ 2 kπ 2 λjk = + forj, k 1. ` m ≥ 44 Exercises

2. Prove that area(Ω) # (j, k): λjk λ λ as λ . { ≤ } ∼ 4π → ∞

Hint: use the two ﬁgures below, where Eλ denotes an ellipse and E˜λ is the copy of the ellipse translated by ( 1, 1). − −

√λm √λm π π

Eλ E˜λ

√λ` √λ` π π

4.3 Exercise 2: Weyl’s law on the Torus

2 We continue to write ω2 := vol(B1(0)), where B1(0) is the unit ball in R . Denote by N(λ) the counting function N(λ) := # j : λj < λ . In this exercise you will prove that { } 2 ω2 vol(T ) N(λ) λ λ . ∼ (2π)2 → ∞

∗ 2 ∗ 1. Deﬁne N (r) := # Z Br(0) . Find how N(λ) and N (r) are related. Restate { ∩ } your goal in terms of N ∗(r).

4.4 Exericise 3: Eigenfunctions on the Sphere

1. Use that if (x , x , x ) = F (ξ , ξ , ξ ) is a change of coordinates then ∂ = 1 2 3 1 2 3 ∂ξj P3 ∂Fi ∂ , to prove that in spherical coordinates i=1 ∂ξj ∂xi

1 0 gR3 (r, θ, φ) = 2 . 0 r gS2 (θ, φ)

Hint: express ∂r, ∂θ, ∂φ in terms of ∂x1 , ∂x2 , ∂x3 . 4.5 Exercise 4: Characterization of eigenvalues 45

2. Deduce that in spherical coordinates 1 ∂ 2 ∂ 1 ∆g 3 = r + ∆g 2 . R r2 ∂r ∂r r2 S

3. Deﬁne

k = homogeneous polynomials of degree k , P { } k = P k : ∆g 3 P = 0 , H { ∈ P R } Hk = Y = P 2 : P k . { |S ∈ H }

Prove that the spaces k and Hk are isomorphic. Hint: ﬁnd the inverse for the H restriction map k Hk given by P P 2 . H → 7→ |S 4. Prove that ∞ 2 Hk Y C (S ) : ∆g Y = k(k + 1)Y . ⊂ { ∈ S2 − } The space Hk is known as the space of Spherical Harmonics of degree k.

5. Find explicit bases for H1 and H2.

6. Use separation of variables to ﬁnd a basis for Hk. That is, look for solutions of the form Yk(θ, φ) = Pk(θ)Φk(φ). The functions Pk(θ) should satisfy a second order diﬀerential equation (don’t try to solve it explicitly!). Such solutions Pk(θ) are called Legendre polynomials. The solutions you obtain should have the form

m imφ m Yk (θ, φ) = e Pk (cos(θ)).

4.5 Exercise 4: Characterization of eigenvalues

⊥ Theorem 13. For k N and Ek(g) := ϕ1, ϕ2, . . . , ϕk−1 , ∈ { } 2 n gφ L2 o λk = inf k∇ k : φ H1(M) Ek(g) . φ 2 ∈ ∩ k kL2 The inﬁmum is achieved if and only if φ is an eigenfunction of eigenvalue λk. Here, 2 R 2 gφ 2 := gφ dvg. k∇ kL M k∇ kg R 1 To ease the exposition, deﬁne Dg(φ, ψ) = M gφ, gψ gdvg. Fix φ H (M) Ek(g) P∞ h∇ ∇ i ∈ ∩ and assume it has expansion φ = j=1 ajϕj.

P` 2 1. Fix ` N and prove that Dg(φ, φ) λja . Hint: compute ∈ ≥ j=1 j ` ` X 2 X 2 Dg(φ λja , φ λja ). − j − j j=1 j=1 46 Exercises

4.6 Exercise 5: Temperature decreases with time and solutions are unique

1. Prove that the temperature of the manifold decreases as time evolves. That is, show that t u( , t) 2 is decreasing with t. 7→ k · kL 2. Using the previous part show that the solution to the Heat Equation is unique.

4.7 Exercise 5: Nodal domains and ﬁrst eigenfunctions

1. Prove the Courant nodal domain Theorem: Write λ1 λ2 ... for the eigenval- ≤ ≤ ues of the Laplacian repeated according to multiplicity. Write ϕ1, ϕ2,... for the corresponding L2-normalized eigenfunctions. Then, the number of nodal domains of ϕk is strictly smaller than k + 1.

Follow the next steps:

(a) Suppose that ϕk has at least k+1 nodal domains D1,...,Dk+1,... and deﬁne ( ϕk Dj on Dj, ψj = | . 0 else

Pk Show that there exists φ = ajψj H1(M) orthogonal to ϕ1, . . . , ϕk−1. j=1 ∈ 2 k∇gφkg (b) Explain why λk 2 . ≤ kφkg 2 Pk k∇gφkg (c) Using that φ, φ g = ij=1 aiaj ∆gψi, ψj g show that λk 2 . h∇ ∇ i h i ≥ kφkg (d) Conclude that φ is an eigenfunction of eigenvalue λk. The theorem then follows from a result known as “Unique Continuation” which says that eigenfunctions cannot vanish on an open set unless they are equal to zero every- where.

2. Conclude that ϕ1 always has constant sign.

3. Conclude that the multiplicity of λ1 is 1.

4. Conclude that ϕ2 has precisely two nodal domains and ϕk has at least two nodal domains for all k 2. ≥