Irreducibility of the Fermi variety for discrete periodic Schrödinger operators and embedded eigenvalues

Abstract

Let $H_0$ be a discrete periodic Schrödinger operator on $\ell ^2(\mathbb {Z}^d)$:

$$\begin{aligned} H_0=-\Delta +V, \end{aligned}$$

where $\Delta$ is the discrete Laplacian and $V:\mathbb {Z}^d\rightarrow \mathbb {C}$ is periodic. We prove that for any $d\ge 3$, the Fermi variety at every energy level is irreducible (modulo periodicity). For $d=2$, we prove that the Fermi variety at every energy level except for the average of the potential is irreducible (modulo periodicity) and the Fermi variety at the average of the potential has at most two irreducible components (modulo periodicity). This is sharp since for $d=2$ and a constant potential V, the Fermi variety at V-level has exactly two irreducible components (modulo periodicity). We also prove that the Bloch variety is irreducible (modulo periodicity) for any $d\ge 2$. As applications, we prove that when V is a real-valued periodic function, the level set of any extrema of any spectral band functions, spectral band edges in particular, has dimension at most $d-2$ for any $d\ge 3$, and finite cardinality for $d=2$. We also show that $H=-\Delta +V+v$ does not have any embedded eigenvalues provided that v decays super-exponentially

Appendix A. Proof of Claim 1

Proof

Otherwise, ${\mathcal {P}}_1(z,\lambda )$ has two non-trivial polynomial factors f(z) and g(z) such that both $\{z\in \mathbb {C}^d: f(z)=0\}$ and $\{z\in \mathbb {C}^d: g(z)=0\}$ contain $z_1=z_2=\cdots =z_d=0$. Let

$$\begin{aligned} {\tilde{f}}(z)=f\big (z_1^{q_1},z_2^{q_2},\ldots ,z_d^{q_d}\big ), {\tilde{g}}(z)=g\big (z_1^{q_1},z_2^{q_2},\ldots ,z_d^{q_d}\big ). \end{aligned}$$

Let ${\tilde{f}}_1(z)$ (${\tilde{g}}_1(z)$) be the component of the lowest degree of ${\tilde{f}}(z)$ (${\tilde{g}}(z)$). Since both $\{z\in \mathbb {C}^d: f(z)=0\}$ and $\{z\in \mathbb {C}^d: g(z)=0\}$ contain $z_1=z_2=\cdots =z_d=0$, one has that ${\tilde{f}}_1(z)$ and ${\tilde{g}}_1(z)$ are non-constant.

Since both ${\tilde{f}}(z)$ and ${\tilde{g}}(z)$ are polynomials of $z_1^{q_1},z_2^{q_2},\ldots , z_d^{q_d}$, we have ${\tilde{f}}_1(z)$ and ${\tilde{g}}_1(z)$ are also polynomials of $z_1^{q_1},z_2^{q_2},\ldots , z_d^{q_d}$ and hence there exist $f_1(z)$ and $g_1(z)$ such that

$$\begin{aligned} {\tilde{f}}_1(z)=f_1\big (z_1^{q_1},z_2^{q_2},\ldots ,z_d^{q_d}\big ), {\tilde{g}}_1(z)=g_1\big (z_1^{q_1},z_2^{q_2},\ldots ,z_d^{q_d}\big ). \end{aligned}$$

By (28) and (29), one has

$$\begin{aligned} {\tilde{f}}_1(z) {\tilde{g}}_1(z)={\tilde{h}}_1(z) \end{aligned}$$

and hence

$$\begin{aligned} {f}_1(z) {g}_1(z)= {h}_1(z). \end{aligned}$$

This is impossible since $h_1(z)$ is irreducible. $\square$