Let $(M^n,g)$ be a closed Riemannian manifold. Let $\Delta$ be the Laplace–Beltrami operator acting on scalar functions defined on $M$, and let $\lambda_1 < \lambda_2 \leq \cdots$ be its spectrum.
Say we are now instead looking at complex-valued scalar functions defined on $M$. It seems like for such a function $u: M \to \mathbf{C}$, a natural extension of $\Delta$ would act on $u$ as \begin{equation} \Delta u := \Delta \operatorname{Re} u + \mathrm{i} \Delta \operatorname{Im} u. \end{equation} Its real eigenvalues exactly coincide with the set $\{ \lambda_i \mid i \geq 1 \}$. However for a complex eigenvalue $\lambda = a + \mathrm{i} b$ with eigenfunction $u = u_1 + \mathrm{i} u_2$ the equation becomes the less transparent \begin{equation} \Delta u_1 + (au_1 - bu_2) = 0 \quad \text{and} \quad \Delta u_2 + (a u_2 + b u_1) = 0. \end{equation}
What is the (complex) spectrum of this operator? Can it be expressed in terms of $(\lambda_i \mid i \geq 1)$?
- I suspect the answer is well-known (or directly follows from, say, Fourier analysis), but I haven't been able to find it on the web.
- I am ultimately interested in the same question for Schrodinger-type operators like $L = \Delta + q$ with real-valued coefficient $q: M \to \mathbf{R}$, but it seemed to me that adding $q$ would not materially change the answer.
