Mathematics

Disreteness of the spectrum for the Schrödinger operator and metric transformation on manifold

A. V. Svetlov

Volgograd State University

Abstract: In this paper we proof the conservation property for the discreteness of the spectrum for the Schrödinger operator on the simple warped products of order $k$ with the special kind of quasi-isometric transformation of the metric.
Let's consider a complete noncompact Riemannian manifold $D$, which is isometric to the product ${\mathbb R}_+ \times \mathrm{S}_1\times \mathrm{S}_2\times\cdots\times \mathrm{S}_k$ (where ${\mathbb R}_+=(0,+\infty)$, and $\mathrm{S}_i$ are compact Riemannian manifolds without boundary) with metric
$$ds^2=dr^2+q_1^2(r)d\theta_1^2+\cdots+q_k^2(r)d\theta_k^2,$$
where $d\theta_i^2$ is the metric on ${\mathrm S}_i$ and $q_i(r)$ is a smooth positive function on ${\mathbb R}_+$. We assume $\dim{\mathrm S}_i=n_i$ and denote $s(r)=q_1^{n_1}(r)\cdots q_k^{n_k}(r).$
Metric transformation on this manifold is determined by the following matrix $\sigma(r)$.
$$\|\sigma(r)\|= \left\|
\begin{array}{c|ccc} \delta_0^2(r) & 0 & \ldots & 0\\ \hline 0 & \delta_1^2(r)E_{n_1} & \ldots & 0 \\ \hline \vdots & & \ddots & \vdots \\ \hline 0 & 0 & \ldots & \delta_k^2(r)E_{n_k} \end{array}
\right\|. $$
The coefficients of this matrix are $C^1$-smooth, and let's $\Sigma(r)$ will stand for its determinant. Actually, we can easily calculate it:
$$\Sigma(r)=\det\|\sigma(r)\|=\delta_0^2(r)\delta_1^{2n_1}(r)\cdots\delta_k^{2n_k}(r).$$

On the manifold $D$ we study the Laplace–Beltrami operator
$$-\Delta=-\mathrm{div}\nabla$$
and the Schrödinger operator
$$-\Delta=-\mathrm{div}\nabla+c(r).$$

With the mentioned metric transformation the Laplace–Beltrami operator will change to
$$ -\widetilde{\Delta}=-\frac 1 {\sqrt{\Sigma}}\mathrm{div}(\sqrt{\Sigma}\sigma^{-1}\nabla). $$

Transformed Schrödinger operator we write as $\widetilde{L}=-\widetilde{\Delta}+c(r)$. Also we put
$$ F(r)=c(r)+\left(\frac{s'(r)}{2s(r)}\right)' +\left(\frac{s'(r)}{2s(r)}\right)^2, $$

$$ \Phi(r)=\left(\frac{\delta'(r)}{2\delta(r)}\right)'+\frac{s'(r)\delta'(r)}{2s(r)\delta(r)} +\left(\frac{\delta'(r)}{2\delta(r)}\right)^2, $$
where $\delta(r)=\frac{\sqrt{\Sigma(r)}}{\delta_0(r)}$.
Then we get the following theorem.
Theorem. Let's $F(r)+\Phi(r)>-C \ (C=\mathrm{const}>0)$. The spectrum of the Schrödinger operator $\widetilde{L}$ on the manifold $D$ is discrete if and only if
$$\forall \omega>0\quad\lim_{r\to\infty}\int\limits_r^{r+\omega}(F(r)+\Phi(r))dr=+\infty.$$

And next we come to the following corollary.
Corollary. If the Schrodinger operator $L$ on manifold $D$ has discrete spectrum, and we transform the metric of $D$ with some diagonal matrix $\|\sigma(r)\|$, and $\Phi(r)>\mathrm{const}$, then the Schrödinger operator $\widetilde{L}$ has discrete spectrum too. The same way non-discrete spectrum holds this characteristic.

Keywords: spectrum discreteness, Schrödinger operator, Riemannian manifolds, quasimodel manifolds, warped products.

UDC: 517.984
BBK: 22.162

DOI: 10.15688/jvolsu1.2016.5.9