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Two general facts: $L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues in your example are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent via the stereographic projection.

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$. E.g. adding a point at infinity will give you a sphere as well with a slightly different metric though.

I think this question on Mathexchange is well in the spirit of your question: http://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum

Edit due to question in the comments: Wikipedia http://en.wikipedia.org/wiki/Clifford_analysis#Cayley_transform_.28stereographic_projection.29 tells me that the Laplace operator on the plane is then the usual one. When you consider the circle, and compare it with the line, there you have definitely a lot more eigenfunctions $e^{\lambda x}$ or not? The same thing happens for the 2-sphere.

Also in the example $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, you definitely get a more flexibility, since usually the growth conditions rule out certain series of Bessel function as solutions. So there the growth conditions does not come for free as well.

Two general facts: $L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues in your example are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent via the stereographic projection.

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$. E.g. adding a point at infinity will give you a sphere as well with a slightly different metric though.

I think this question on Mathexchange is well in the spirit of your question: https://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum

Edit due to question in the comments: Wikipedia http://en.wikipedia.org/wiki/Clifford_analysis#Cayley_transform_.28stereographic_projection.29 tells me that the Laplace operator on the plane is then the usual one. When you consider the circle, and compare it with the line, there you have definitely a lot more eigenfunctions $e^{\lambda x}$ or not? The same thing happens for the 2-sphere.

Also in the example $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, you definitely get a more flexibility, since usually the growth conditions rule out certain series of Bessel function as solutions. So there the growth conditions does not come for free as well.

Two general facts: $L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues in your example are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent via the stereographic projection.

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$. E.g. adding a point at infinity will give you a sphere as well with a slightly different metric though.

I think this question on Mathexchange is well in the spirit of your question: http://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum

Two general facts: $L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues in your example are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent via the stereographic projection.

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$. E.g. adding a point at infinity will give you a sphere as well with a slightly different metric though.

I think this question on Mathexchange is well in the spirit of your question: http://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum

Edit due to question in the comments: Wikipedia http://en.wikipedia.org/wiki/Clifford_analysis#Cayley_transform_.28stereographic_projection.29 tells me that the Laplace operator on the plane is then the usual one. When you consider the circle, and compare it with the line, there you have definitely a lot more eigenfunctions $e^{\lambda x}$ or not? The same thing happens for the 2-sphere.

Also in the example $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, you definitely get a more flexibility, since usually the growth conditions rule out certain series of Bessel function as solutions. So there the growth conditions does not come for free as well.

$L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$.

I think this question on Mathexchange is well in the spirit of your question: http://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum

Two general facts: $L^2$ spaces do not see, when you remove one point. Laplace operators on compact smooth manifolds have always discrete spectrum. So if you are not interested in smooth or continuous eigenfunctions, then this gives you the answer yes. All eigenvalues in your example are preserved, because the $L^2$ spaces together with the Laplace operator are unitary equivalent via the stereographic projection.

The situation changes drastically, if you consider e.g. the Laplace Beltrami on $SL(2, \mathbb{Z}) \backslash \mathbb{H}$, which has finite volume, but fails to be compact. Hence it admits continouous and discrete spectrum. All naive compactifications at "$\infty$" will destroy all of the eigenvalues, except for the trivial eigenvalue $0$. E.g. adding a point at infinity will give you a sphere as well with a slightly different metric though.

I think this question on Mathexchange is well in the spirit of your question: http://math.stackexchange.com/questions/46919/changing-the-manifold-preserving-the-discrete-spectrum