# Inverse of holomorphic elliptic differential operator

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?

• $\langle f, (A-z)^{-1} f\rangle$ has positive imaginary part in the upper half plane for any self-adjoint $A$, so possible real poles must be of order $1$. May 17, 2018 at 23:05
• 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. May 18, 2018 at 5:36
• 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. May 19, 2018 at 8:04

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.