Eigenvalues of the Laplace-Beltrami operator on a compact Riemannnian manifold












11












$begingroup$


Let $(M,g)$ be a compact Riemannian manifold, and let $Delta_g$ be its Laplace-Beltrami operator. A "well-known fact" is that the eigenvalues of $Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this?










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:04






  • 1




    $begingroup$
    To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
    $endgroup$
    – Sönke Hansen
    May 18 at 14:29






  • 1




    $begingroup$
    @Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
    $endgroup$
    – Max Schattman
    May 18 at 14:30








  • 1




    $begingroup$
    @SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:57






  • 1




    $begingroup$
    @MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
    $endgroup$
    – Nate Eldredge
    May 18 at 15:09


















11












$begingroup$


Let $(M,g)$ be a compact Riemannian manifold, and let $Delta_g$ be its Laplace-Beltrami operator. A "well-known fact" is that the eigenvalues of $Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this?










share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:04






  • 1




    $begingroup$
    To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
    $endgroup$
    – Sönke Hansen
    May 18 at 14:29






  • 1




    $begingroup$
    @Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
    $endgroup$
    – Max Schattman
    May 18 at 14:30








  • 1




    $begingroup$
    @SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:57






  • 1




    $begingroup$
    @MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
    $endgroup$
    – Nate Eldredge
    May 18 at 15:09
















11












11








11


2



$begingroup$


Let $(M,g)$ be a compact Riemannian manifold, and let $Delta_g$ be its Laplace-Beltrami operator. A "well-known fact" is that the eigenvalues of $Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this?










share|cite|improve this question











$endgroup$




Let $(M,g)$ be a compact Riemannian manifold, and let $Delta_g$ be its Laplace-Beltrami operator. A "well-known fact" is that the eigenvalues of $Delta_g$ have finite multiplicity and tend to infinity. What is the easiest/simplest way to estblish this fact? By simplest, I mean an abstract, formal, approach to the problem, not too heavily based on technical calculation. In the same line of thought, is there a philosophical/intuitive reason to see why the spectrum should behave like this?







dg.differential-geometry riemannian-geometry sp.spectral-theory differential-operators






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 18 at 13:23







Max Schattman

















asked May 18 at 11:37









Max SchattmanMax Schattman

13710 bronze badges




13710 bronze badges








  • 4




    $begingroup$
    I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:04






  • 1




    $begingroup$
    To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
    $endgroup$
    – Sönke Hansen
    May 18 at 14:29






  • 1




    $begingroup$
    @Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
    $endgroup$
    – Max Schattman
    May 18 at 14:30








  • 1




    $begingroup$
    @SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:57






  • 1




    $begingroup$
    @MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
    $endgroup$
    – Nate Eldredge
    May 18 at 15:09
















  • 4




    $begingroup$
    I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:04






  • 1




    $begingroup$
    To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
    $endgroup$
    – Sönke Hansen
    May 18 at 14:29






  • 1




    $begingroup$
    @Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
    $endgroup$
    – Max Schattman
    May 18 at 14:30








  • 1




    $begingroup$
    @SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
    $endgroup$
    – Nate Eldredge
    May 18 at 14:57






  • 1




    $begingroup$
    @MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
    $endgroup$
    – Nate Eldredge
    May 18 at 15:09










4




4




$begingroup$
I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
$endgroup$
– Nate Eldredge
May 18 at 14:04




$begingroup$
I guess the route I think of is like this. You show that the resolvent (or the semigroup) of $Delta_g$ maps $L^2$ into an appropriate Sobolev space. The latter is compactly embedded in $L^2$ by Rellich's theorem, so the resolvent is a compact operator. So the eigenvalues of the resolvent (with multiplicity) converge to zero, and hence the eigenvalues of the Laplacian converge to infinity.
$endgroup$
– Nate Eldredge
May 18 at 14:04




1




1




$begingroup$
To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
$endgroup$
– Sönke Hansen
May 18 at 14:29




$begingroup$
To Nate Eldredge's route, which is a standard one, I would add: $Delta_g$ is self-adjoint and, by elliptic regularity, its domain is the Sobolev space $H^2$. So the resolvent factors through $H^2$, hence is compact by Rellich's Theorem.
$endgroup$
– Sönke Hansen
May 18 at 14:29




1




1




$begingroup$
@Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
$endgroup$
– Max Schattman
May 18 at 14:30






$begingroup$
@Nate: So I guess this changes the question into a request for an abstract/formal approach to the Rellich embedding theorem :) I've never actually seen a proof for a general Riemannian manifold . . . might an easy proof exist for spaces with enough symmetries. For example, what does Rellich look like for a compact homogeneous space?
$endgroup$
– Max Schattman
May 18 at 14:30






1




1




$begingroup$
@SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
$endgroup$
– Nate Eldredge
May 18 at 14:57




$begingroup$
@SönkeHansen: I was thinking about that, but I seem to recall that the proof of elliptic regularity goes through the Sobolev embedding argument itself, so I wasn't sure off the top of my head whether that is circular.
$endgroup$
– Nate Eldredge
May 18 at 14:57




1




1




$begingroup$
@MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
$endgroup$
– Nate Eldredge
May 18 at 15:09






$begingroup$
@MaxSchattman: I think you just use a partition of unity. When $M$ is compact, can find a finite number of smooth functions $psi_k$, each compactly supported in a coordinate chart, whose sum is 1. Now if I have a sequence of functions $f_n$ in my Sobolev space, the functions $f_n psi_1$ are in the corresponding Sobolev space of the coordinate chart. Apply the Euclidean version of Rellich to pass to an $L^2$ convergent subsequence, and repeat iteratively for $psi_2, psi_3, dots, psi_k$. Adding them up, you have a subsequence of the original $f_n$ converging in $L^2$.
$endgroup$
– Nate Eldredge
May 18 at 15:09












2 Answers
2






active

oldest

votes


















9












$begingroup$

A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators.



This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912.
(see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works.



An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain
meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity.
(Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.)



Three remarks should be made:



a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned!



b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem.



c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in
1637. This started a long story of research about these eigenvalues.






share|cite|improve this answer











$endgroup$





















    6












    $begingroup$

    What follows is only an answer to the philosophical/intuitive question and follows only one specific point of view (there are probably many others).



    Eigenvalue problem for the Laplace operator on a Riemannian manifold $(M,g)$ is a quantization of the problem of the classical motion of a particle "freely moving", i.e. following geodesics, on $(M,g)$. More precisely, the phase space of this classical mechanics problem is the symplectic manifold $T^{*}M$ (cotangent bundle, with its standard symplectic form) and the Hamiltonian is the function on $T^{*}M$ given by $H=|p|_g^2/2$, where $p$ is linear form on cotangent fibers.



    Eigenspaces of the Laplace operator with eigenvalues less than E are quantization of the part of the phase space with $H <E$. If M is compact, then $H<E$ is a subset of $T^{*}M$ of finite volume and so its quantization will produce a finite dimensional vector space (of dimension roughly the volume in units of Planck constant $hbar$): in particular, there will be finitely many eigenvalues below $E$, all with finitely many multiplicities. When $E$ goes to infinity, the volume of $H<E$ goes to infinity and so eigenvalues go to infinity.



    In fact, this picture tells you that the number of eigenvalues less than $E$ should be of the order of the volume of the set $H<E$, which is of order $vol(M,g) E^{dim(M)/2}$, where $vol(M,g)$ is the volume of $(M,g)$, and $dim(M)$ is the dimension of $n$. This is indeed true (Weyl law).



    (Maybe a trivial comment, but one never knows: it is probably helpful to think about the case of the circle, and more generally flat tori, where everything is trivial Fourier analysis, before thinking about more general situations).






    share|cite|improve this answer











    $endgroup$









    • 4




      $begingroup$
      Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
      $endgroup$
      – Victor Protsak
      May 18 at 18:00














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "504"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f331870%2feigenvalues-of-the-laplace-beltrami-operator-on-a-compact-riemannnian-manifold%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators.



    This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912.
    (see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works.



    An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain
    meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity.
    (Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.)



    Three remarks should be made:



    a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned!



    b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem.



    c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in
    1637. This started a long story of research about these eigenvalues.






    share|cite|improve this answer











    $endgroup$


















      9












      $begingroup$

      A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators.



      This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912.
      (see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works.



      An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain
      meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity.
      (Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.)



      Three remarks should be made:



      a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned!



      b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem.



      c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in
      1637. This started a long story of research about these eigenvalues.






      share|cite|improve this answer











      $endgroup$
















        9












        9








        9





        $begingroup$

        A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators.



        This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912.
        (see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works.



        An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain
        meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity.
        (Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.)



        Three remarks should be made:



        a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned!



        b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem.



        c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in
        1637. This started a long story of research about these eigenvalues.






        share|cite|improve this answer











        $endgroup$



        A modern "simple philosophical" explanation is that this problem can be restated as an eigenvalue problem for a compact operator in an appropriate Hilbert space, whose eigenvalues are reciprocal to those of the Laplacian. Then two properties that you stated follow from the general properties of compact operators.



        This approach is due to Hilbert. He wrote a long series of papers on the subject in 1904-1912.
        (see also his book with Courant, Methods of mathematical physics). Modern expositions are usually based on Hilbert's ideas. The notions of Hilbert space and compact operator were essentially distilled from these works.



        An earlier philosophy of Poincaré interprets these eigenvalues as poles of certain
        meromorphic function in the plane (in modern language it is essentially the resolvent), and the poles of a meromorphic function are isolated and tend to infinity. Poincaré was the first to prove under general conditions the existence of an infinite sequence of eigenvalues tending to infinity.
        (Sur les équations de la physique mathématique, Rend. Circ. mat. Palermo, 1894 8, 57-155.)



        Three remarks should be made:



        a) At the time of Poincaré and Hilbert, the modern formal notion of compact Riemannian manifold did not exist. (The notion of compact was introduced by Aleksandrov and Urysohn in 1924, and the notion of manifold by Weyl 1913, and only for dimension 2). Even in the classical book on the subject by Titchmarsh, Eigenfunction expansions..., 1958, the words "manifold" and "compact" are not mentioned!



        b) There was a very large number of problems about vibrations which were solved "explicitly" in 18th and 19th century. So Hilbert and Poincaré had a lot of "empirical material" to generalize. Fourier should be mentioned: his work inspired Hilbert and Poincare. He solved many concrete eigenvalue problems but had no tools to attack the general problem.



        c) As a physical fact, existence of infinite discrete spectrum was first discovered (for the case of a string) by a music theorist Marin Mersenne in
        1637. This started a long story of research about these eigenvalues.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 19 at 2:02

























        answered May 18 at 21:21









        Alexandre EremenkoAlexandre Eremenko

        53.3k6 gold badges151 silver badges272 bronze badges




        53.3k6 gold badges151 silver badges272 bronze badges























            6












            $begingroup$

            What follows is only an answer to the philosophical/intuitive question and follows only one specific point of view (there are probably many others).



            Eigenvalue problem for the Laplace operator on a Riemannian manifold $(M,g)$ is a quantization of the problem of the classical motion of a particle "freely moving", i.e. following geodesics, on $(M,g)$. More precisely, the phase space of this classical mechanics problem is the symplectic manifold $T^{*}M$ (cotangent bundle, with its standard symplectic form) and the Hamiltonian is the function on $T^{*}M$ given by $H=|p|_g^2/2$, where $p$ is linear form on cotangent fibers.



            Eigenspaces of the Laplace operator with eigenvalues less than E are quantization of the part of the phase space with $H <E$. If M is compact, then $H<E$ is a subset of $T^{*}M$ of finite volume and so its quantization will produce a finite dimensional vector space (of dimension roughly the volume in units of Planck constant $hbar$): in particular, there will be finitely many eigenvalues below $E$, all with finitely many multiplicities. When $E$ goes to infinity, the volume of $H<E$ goes to infinity and so eigenvalues go to infinity.



            In fact, this picture tells you that the number of eigenvalues less than $E$ should be of the order of the volume of the set $H<E$, which is of order $vol(M,g) E^{dim(M)/2}$, where $vol(M,g)$ is the volume of $(M,g)$, and $dim(M)$ is the dimension of $n$. This is indeed true (Weyl law).



            (Maybe a trivial comment, but one never knows: it is probably helpful to think about the case of the circle, and more generally flat tori, where everything is trivial Fourier analysis, before thinking about more general situations).






            share|cite|improve this answer











            $endgroup$









            • 4




              $begingroup$
              Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
              $endgroup$
              – Victor Protsak
              May 18 at 18:00
















            6












            $begingroup$

            What follows is only an answer to the philosophical/intuitive question and follows only one specific point of view (there are probably many others).



            Eigenvalue problem for the Laplace operator on a Riemannian manifold $(M,g)$ is a quantization of the problem of the classical motion of a particle "freely moving", i.e. following geodesics, on $(M,g)$. More precisely, the phase space of this classical mechanics problem is the symplectic manifold $T^{*}M$ (cotangent bundle, with its standard symplectic form) and the Hamiltonian is the function on $T^{*}M$ given by $H=|p|_g^2/2$, where $p$ is linear form on cotangent fibers.



            Eigenspaces of the Laplace operator with eigenvalues less than E are quantization of the part of the phase space with $H <E$. If M is compact, then $H<E$ is a subset of $T^{*}M$ of finite volume and so its quantization will produce a finite dimensional vector space (of dimension roughly the volume in units of Planck constant $hbar$): in particular, there will be finitely many eigenvalues below $E$, all with finitely many multiplicities. When $E$ goes to infinity, the volume of $H<E$ goes to infinity and so eigenvalues go to infinity.



            In fact, this picture tells you that the number of eigenvalues less than $E$ should be of the order of the volume of the set $H<E$, which is of order $vol(M,g) E^{dim(M)/2}$, where $vol(M,g)$ is the volume of $(M,g)$, and $dim(M)$ is the dimension of $n$. This is indeed true (Weyl law).



            (Maybe a trivial comment, but one never knows: it is probably helpful to think about the case of the circle, and more generally flat tori, where everything is trivial Fourier analysis, before thinking about more general situations).






            share|cite|improve this answer











            $endgroup$









            • 4




              $begingroup$
              Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
              $endgroup$
              – Victor Protsak
              May 18 at 18:00














            6












            6








            6





            $begingroup$

            What follows is only an answer to the philosophical/intuitive question and follows only one specific point of view (there are probably many others).



            Eigenvalue problem for the Laplace operator on a Riemannian manifold $(M,g)$ is a quantization of the problem of the classical motion of a particle "freely moving", i.e. following geodesics, on $(M,g)$. More precisely, the phase space of this classical mechanics problem is the symplectic manifold $T^{*}M$ (cotangent bundle, with its standard symplectic form) and the Hamiltonian is the function on $T^{*}M$ given by $H=|p|_g^2/2$, where $p$ is linear form on cotangent fibers.



            Eigenspaces of the Laplace operator with eigenvalues less than E are quantization of the part of the phase space with $H <E$. If M is compact, then $H<E$ is a subset of $T^{*}M$ of finite volume and so its quantization will produce a finite dimensional vector space (of dimension roughly the volume in units of Planck constant $hbar$): in particular, there will be finitely many eigenvalues below $E$, all with finitely many multiplicities. When $E$ goes to infinity, the volume of $H<E$ goes to infinity and so eigenvalues go to infinity.



            In fact, this picture tells you that the number of eigenvalues less than $E$ should be of the order of the volume of the set $H<E$, which is of order $vol(M,g) E^{dim(M)/2}$, where $vol(M,g)$ is the volume of $(M,g)$, and $dim(M)$ is the dimension of $n$. This is indeed true (Weyl law).



            (Maybe a trivial comment, but one never knows: it is probably helpful to think about the case of the circle, and more generally flat tori, where everything is trivial Fourier analysis, before thinking about more general situations).






            share|cite|improve this answer











            $endgroup$



            What follows is only an answer to the philosophical/intuitive question and follows only one specific point of view (there are probably many others).



            Eigenvalue problem for the Laplace operator on a Riemannian manifold $(M,g)$ is a quantization of the problem of the classical motion of a particle "freely moving", i.e. following geodesics, on $(M,g)$. More precisely, the phase space of this classical mechanics problem is the symplectic manifold $T^{*}M$ (cotangent bundle, with its standard symplectic form) and the Hamiltonian is the function on $T^{*}M$ given by $H=|p|_g^2/2$, where $p$ is linear form on cotangent fibers.



            Eigenspaces of the Laplace operator with eigenvalues less than E are quantization of the part of the phase space with $H <E$. If M is compact, then $H<E$ is a subset of $T^{*}M$ of finite volume and so its quantization will produce a finite dimensional vector space (of dimension roughly the volume in units of Planck constant $hbar$): in particular, there will be finitely many eigenvalues below $E$, all with finitely many multiplicities. When $E$ goes to infinity, the volume of $H<E$ goes to infinity and so eigenvalues go to infinity.



            In fact, this picture tells you that the number of eigenvalues less than $E$ should be of the order of the volume of the set $H<E$, which is of order $vol(M,g) E^{dim(M)/2}$, where $vol(M,g)$ is the volume of $(M,g)$, and $dim(M)$ is the dimension of $n$. This is indeed true (Weyl law).



            (Maybe a trivial comment, but one never knows: it is probably helpful to think about the case of the circle, and more generally flat tori, where everything is trivial Fourier analysis, before thinking about more general situations).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 18 at 16:54

























            answered May 18 at 16:48









            user25309user25309

            5,21224 silver badges42 bronze badges




            5,21224 silver badges42 bronze badges








            • 4




              $begingroup$
              Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
              $endgroup$
              – Victor Protsak
              May 18 at 18:00














            • 4




              $begingroup$
              Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
              $endgroup$
              – Victor Protsak
              May 18 at 18:00








            4




            4




            $begingroup$
            Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
            $endgroup$
            – Victor Protsak
            May 18 at 18:00




            $begingroup$
            Even in the "trivial" flat $m$-torus case, where the eigenvalues are given by an explicit formula, nontrivial analytic number theory is needed to understand the asymptotics the spectral function for $mgeq 2$. The principal term is of course controlled by the (co)volume, but already in the simplest case of square torus, estimating the error term is the Gauss circle problem.
            $endgroup$
            – Victor Protsak
            May 18 at 18:00


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to MathOverflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f331870%2feigenvalues-of-the-laplace-beltrami-operator-on-a-compact-riemannnian-manifold%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

            He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

            Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029