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Green's function of a differential operator contains a lot of information of that operator. In particular, if we have a differential operator on a compact manifold with discrete spectrum, then Green's function satisfies

$$G(x,x',\lambda)=\sum_n\frac{\psi_n^*(x)\psi_n(x)}{\lambda-\lambda_n}$$

where $\lambda_n$ are the eigenvalues of the differential operator, $\psi_n$ the corresponding eigenfunctions, and $\psi_n^*$ the complex conjugate.

In theory, we can look where the poles are in a Green's function and find the spectrum, and then calculate the residue of the poles to get the eigenfunctions. However, when an eigenvalue has multiplicity, then the residue might not give us the eigenfunctions in a straight-forward way. In this situation, is there any way to work around this difficulty and calculate the eigenfucntions?

More specifically, how can we write down a basis of the eigenspace associated to an eigenvalue, say 6, of the Laplacian on $S^3$ in terms of the Green's function of this operator?

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I assume that your differential operator is linear unbounded with compact resolvent. Eigenvalues of higher multiplicity have eigenspaces: any basis of the eigenspace form the eigenfunctions for this eigenvalue. They are not unique! But the expression in the Greens function is independent of the choice of an orthonormal basis of the eigenspace.

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  • $\begingroup$ I should be more explicit I think... I am interested in writing down explicitly the eigenfunctions of the Laplacian associated to a given eigenvalue using the Green's function. As you mentioned, there are in general an eigenspace around. I am wondering if I can write down a basis of the eigenspace using Green's function. $\endgroup$
    – Qijun Tan
    Commented Nov 13, 2014 at 22:43
  • $\begingroup$ No, you cannot, as I explained. $\endgroup$
    – Peter Michor
    Commented Nov 13, 2014 at 22:58
  • $\begingroup$ @QijunTan: I think your question is way too vague. Of course you can in principle "write down a basis of the eigenspace using Green's function" for the trivial reason that you can recover the operator from $G$, so you can answer all questions about it (this may involve choices, as Peter explained). Obviously, this is not what you're intending to ask, and it doesn't really become clear what your question is then. $\endgroup$
    – Christian Remling
    Commented Nov 14, 2014 at 1:34
  • $\begingroup$ @ChristianRemling Thanks for your comment. I have revised my question so that it is now very concrete. $\endgroup$
    – Qijun Tan
    Commented Nov 14, 2014 at 15:15

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