For a compact Riemannian manifold $M$, $L^2(M)$ is spanned by the eigenfunctions of the Laplacian. The Next CEO of Stack OverflowWhy is the laplacian positive-definiteprojection onto the nullspace of the Laplacian on a conformally compact surfaceKernel of the Laplacian on a compact manifoldEigenfunctions and spectrum of $T:H to H^*$ where $H$ is a Hilbert spaceSelf-adjoint extension of the LaplacianDiagonalizable operators on compact Riemannian manifoldsUniform estimate for Laplacian eigenfunctionsGeneral questions on the eigenfunctions of Laplacian and Dirac operatorsLocal form of $p$-Laplacian operator in Riemannian manifoldLaplacian of the position vector
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For a compact Riemannian manifold $M$, $L^2(M)$ is spanned by the eigenfunctions of the Laplacian.
The Next CEO of Stack OverflowWhy is the laplacian positive-definiteprojection onto the nullspace of the Laplacian on a conformally compact surfaceKernel of the Laplacian on a compact manifoldEigenfunctions and spectrum of $T:H to H^*$ where $H$ is a Hilbert spaceSelf-adjoint extension of the LaplacianDiagonalizable operators on compact Riemannian manifoldsUniform estimate for Laplacian eigenfunctionsGeneral questions on the eigenfunctions of Laplacian and Dirac operatorsLocal form of $p$-Laplacian operator in Riemannian manifoldLaplacian of the position vector
$begingroup$
In some paper I read the following statement:
For a compact Riemannian manifold $M$ and the corresponding Laplace-Beltrami operator $Delta$ on $M$ we have, that $$L^2(M) = widehatbigoplus_lambda operatornameEigbig( Delta , lambda big) .$$
So the $L^2$-space of $M$ is spanned by the eigenfunctions of the laplacian.
Now the question is: do you know a good book where to look the proof up or from where to cite this result?
Thanks a lot in advance!
functional-analysis reference-request spectral-theory laplacian eigenfunctions
edited Mar 21 at 0:57
Andrews
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
$endgroup$
add a comment |
$begingroup$
In some paper I read the following statement:
For a compact Riemannian manifold $M$ and the corresponding Laplace-Beltrami operator $Delta$ on $M$ we have, that $$L^2(M) = widehatbigoplus_lambda operatornameEigbig( Delta , lambda big) .$$
So the $L^2$-space of $M$ is spanned by the eigenfunctions of the laplacian.
Now the question is: do you know a good book where to look the proof up or from where to cite this result?
Thanks a lot in advance!
functional-analysis reference-request spectral-theory laplacian eigenfunctions
edited Mar 21 at 0:57
Andrews
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
$endgroup$
1
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55
add a comment |
$begingroup$
In some paper I read the following statement:
For a compact Riemannian manifold $M$ and the corresponding Laplace-Beltrami operator $Delta$ on $M$ we have, that $$L^2(M) = widehatbigoplus_lambda operatornameEigbig( Delta , lambda big) .$$
So the $L^2$-space of $M$ is spanned by the eigenfunctions of the laplacian.
Now the question is: do you know a good book where to look the proof up or from where to cite this result?
Thanks a lot in advance!
functional-analysis reference-request spectral-theory laplacian eigenfunctions
edited Mar 21 at 0:57
Andrews
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
$endgroup$
In some paper I read the following statement:
For a compact Riemannian manifold $M$ and the corresponding Laplace-Beltrami operator $Delta$ on $M$ we have, that $$L^2(M) = widehatbigoplus_lambda operatornameEigbig( Delta , lambda big) .$$
So the $L^2$-space of $M$ is spanned by the eigenfunctions of the laplacian.
Now the question is: do you know a good book where to look the proof up or from where to cite this result?
Thanks a lot in advance!
functional-analysis reference-request spectral-theory laplacian eigenfunctions
functional-analysis reference-request spectral-theory laplacian eigenfunctions
edited Mar 21 at 0:57
Andrews
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
edited Mar 21 at 0:57
Andrews
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
edited Mar 21 at 0:57
Andrews
1,2812422
edited Mar 21 at 0:57
Andrews
1,2812422
edited Mar 21 at 0:57
Andrews
1,2812422
1,2812422
asked Mar 20 at 10:56
TargonTargon
677
asked Mar 20 at 10:56
TargonTargon
677
asked Mar 20 at 10:56
TargonTargon
677
677
1
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55
add a comment |
1
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55
1
1
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55
add a comment |
0
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gedemfIowow,XT,ulE9x0O erT1iJ4QjjT0Qo0ojdfB4QEcx7GxMW85PIs
1
$begingroup$
I don't have a good reference at hand, but the argument goes as follows: (1) For every self-adjoint operator with compact resolvent, the underlying Hilbert spaces decomposes as direct sum of its eigenspaces. This is a direct consequence of the spectral theorem for compact operators. (2) The Laplace-Beltrami operator on a compact Riemannian manifold has compact resolvent. This follows from the compact the Rellich–Kondrachov embedding theorem.
$endgroup$
– MaoWao
Mar 20 at 14:39
$begingroup$
Thank you very much! In the meantime I did also find a reference were they proof it for the case of compact hyperbolic surfaces, which is enough for me and might be enough for some other people: The Spectrum of Hyperbolic Surfaces by Nicolas Bergeron
$endgroup$
– Targon
Mar 20 at 17:28
$begingroup$
And I'd say locally in some chart $Delta_M$ looks like $fmapsto Delta_BbbR^n (f circ psi)$ for some uniformly smooth $psi$ so the resolvent is compact for almost every $z$ and the same holds globally because $M$ is compact thus covered by finitely many charts / bounded domains
$endgroup$
– reuns
Mar 21 at 2:55