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Consider the Beltrami-Laplacian $\Delta$ on $\mathbb{S}^n$ with standard metric. One can define a family of operators $A(z):H^1(\mathbb{S}^n)\to H^1(\mathbb{S}^n)$ as the following $$A(z)=\Delta+z$$ Then $A$ is holomorphic on $z$. From the Fredholm theory of elliptic operator, $A(z)$ is invertible only $z$ is not the eigenvalue of $-\Delta$ which are $k(k+n-1)$. However, it is also known that $A(z)^{-1}$ is actually meromorphic near these eigenvalues.

So my question is what is the order of the pole near each eigenvalue? Can we find Laurent series of it?

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    $\begingroup$ $\langle f, (A-z)^{-1} f\rangle$ has positive imaginary part in the upper half plane for any self-adjoint $A$, so possible real poles must be of order $1$. $\endgroup$ – Christian Remling May 17 '18 at 23:05
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    $\begingroup$ If you replace $\Delta$ by some finite dimensional matrix, not necessarily symmetric/hermitian, then higher order poles of $A(z)^{-1}$ correspond to higher order Jordan blocks in the canonical form of $\Delta$. If you do the calculation explicitly for $\Delta$ being a single Jordan block, you'll see where the higher order singularities appear. $\endgroup$ – Igor Khavkine May 18 '18 at 5:36
  • $\begingroup$ Using a resolvent integral along a simple closed curve containing some eigenvalues in the interior you can reduce the question to the finite dimensional matrix case. So Igor Khavkine's comment precisely answers your question. $\endgroup$ – Peter Michor May 19 '18 at 8:04
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At least on the circle, we have $\lambda_{n}=n^2$. Thus you want to work with $$ (z^{2}-\Delta)^{-1}=z^{-2}(1-\frac{1}{z^{2}}\Delta)^{-1}=z^{-2}\sum_{k=0}^{\infty} (-\frac{1}{z^2}\Delta)^{k} $$ If we let $f(x)=e^{in x}$, then we have $$ z^{-2}\sum(-\frac{n^2}{z^2})^{k}\rightarrow z^{-2}(1-w)^{-1}, w=\frac{n^2}{z^2} $$ And we know that this has a pole of order one at $w=1$ ($z=n$). So we have a simple pole for every $\lambda_{n}$ of order $1$, with residue being $\frac{1}{n^2}$. I suspect the higher dimensional situation is similar.

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  • $\begingroup$ Are you saying we only have simple pole? This will contradict with the comment of Igor $\endgroup$ – Slm2004 May 19 '18 at 15:13

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