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I would like to find the spectrum of a circle. I know that the Laplace-Beltrami eigenvalue problem for unit circle is equivalent to the regular Laplacian eigenvalue problem for the interval of the length $2\pi$ with periodic Boundary conditions.

In a sense, the Laplace-Beltrami operator (i.e. $∆ =\frac{1}{\sqrt{G}} \sum_{i,j=1}^n \frac{\partial}{\partial_i} (\sqrt{G} g^{ij} \frac{\partial}{\partial_i})$) on circle can be viewed as a Laplacian of a function depending on the arc length.

I am not familiar with the use of the Laplace-Beltrami operator and Helmholtz equation. Some things I know is in the definition of page $1$ of pdf. The first formula of the pdf, the definition of the $Δ$ operator in manifolds in term of local coordinates $x_i$ and the metric $g$ and I'd say that if the manifold is a $1$-manifold homeomorphic to the circle, we can choose the coordinates such that $g$ is constant and $=1$, and it reduces again to the eigenvalue equation $h″=λh$ with $h:ℝ→ℂ$ and periodic.

Is there anyone could solve it in using these tools?

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  • $\begingroup$ I'm not sure what your question is. You say you do know the problem is equivalent to the regular Laplacian eigenvalue problem for the interval of length $2\pi$ with periodic Boundary condition, and then you also explain how to reduce the problem to $h^{\prime\prime} = \lambda h$. So if you put this together you should be done, at least regarding the eigenvalue problem -- the differential equation is easy enough to solve. (This or very similar questions popped up several times during the last days. Is there some course where this has just been given as an exercise?) $\endgroup$
    – Thomas
    May 21, 2016 at 15:04
  • $\begingroup$ @Thomas This is some fact some people explain to me, but honestly I don't know how works the LB operator, and I would like that someone explain to me rigorously with this example. It is maybe a silly question and maybe evident for some people, but it is not clear for me. $\endgroup$
    – user316765
    May 21, 2016 at 16:04

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I'll try to put something together to help you a bit here:

If you are working with manifolds, you usually only have local descriptions of a manifold $M$, given by coordinate systems $x:M\cap U\rightarrow \mathbb{R}^n$ (here $U$ is some neighbourhood on $M$ diffeomorphis to a ball in Euclidean space).

On a manifold you can do differential calculus by doing it in the image of the coordinate systems. With this approach, differential operators must be shown to be well defined, which basicall means that they behave in the same way regardless of the choice of coordinate system.

If you, in addition, have a Riemannian metric on the manifolds, several constructions known from differential calculus carry over to manifolds. One of these is the Laplacian, which is defined as the trace of the Hessian (read: second derivative, whatever this means precisely). It turns out that the resulting expression of the Laplacian (also called Laplace Beltrami Operator) is given by the expression you have written down -- there is a typo in there, the formula shoud read $$\Delta f = \frac{1}{\sqrt{G}}\sum_{i,j} \frac{\partial}{\partial x^i}\left( \sqrt{G}g^{ij}\frac{\partial f}{\partial x^j }\right)$$ Here $g^{ij}$ is the inverse of the metric tensor $g_{ij} =\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\rangle $

As mentioned earlier, usually you have this represention only locally, and it looks rather complicated. In some cases you can find an easier reprentation (e.g. when the metric tensor is diagonal, or even better, a function times the identity). In the one dimensional case you have a curve and the metric tensor is just a function. If you choose the inverse of an arclength parametrization as coordinate system, the metric tensor becomes (as you indicated in your post) a constant, and the operator reduces to the ordinary second derivative. In case of the circle you can choose a parametrization which covers all of the manifold with the exception of a point, which allows you to identify functions on the circle with periodic functions on the interval. If you put all this together the equation $\Delta f = \lambda f$ reduces to $f^{\prime\prime} = \lambda f$. (The sign of the operator varies among authors).

That's about it. In order to grasp the details and fully understand them in a general setup you will have to dig into Differential Geometry.

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  • $\begingroup$ I need I little clarification on the meaning of $g_{ij} =\langle \frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\rangle$. Is it an operator that we have to apply at our coordinate system $x$, i.e. $g_{ij}(x) =\langle \frac{\partial}{\partial x^i}(x), \frac{\partial}{\partial x^j} (x) \rangle$? $\endgroup$
    – user316765
    May 21, 2016 at 17:41
  • $\begingroup$ @george It's the scalar product of the basis vectors in the coordinate system. This way you determine the components of the metric tensor. $\endgroup$
    – Thomas
    May 21, 2016 at 19:26
  • $\begingroup$ I am not use to with this notation. For me, when you displayed $\frac{\partial}{\partial x^i}$, it means the usual differential operator. Could you compute the metric tensor for our coordinate system? I imagine as we are on the circle, our coordinate system is simply $(r \cos \theta, r \sin \theta)$. $\endgroup$
    – user316765
    May 21, 2016 at 23:34
  • $\begingroup$ @george In Differential Geometry it is common to use $\frac{partial}{\partial xî}$ as notation for the $i-$th coordinate basis vector. If you don't like that replace it by some other notation you are comfortable with. In our example it's just the unit tangent vector to the circle. $\endgroup$
    – Thomas
    May 22, 2016 at 5:51

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