Spectral asymptotics for the Landau Hamiltonian on cylindrical surfaces

Abstract

For a given discrete subgroup $\Gamma =\alpha \mathbb Z$ of $(\mathbb C,+)$, we consider the Landau Hamiltonian acting on the space of $(L^2,\Gamma )$-automorphic functions, perturbed by an electric potential with compact support on $\mathbb C/\Gamma$. We investigate the asymptotic behaviour of the discrete spectrum near Landau levels.

Appendix A. Boundedness of \(\mathcal P_\Gamma \)

Lemma A.1

Let $\Gamma$ be a discrete subgroup of $(\mathbb C,+)$ ($\Gamma =\alpha \mathbb Z$; $\alpha \in \mathbb C\setminus \{0\}$ or $\Gamma =\mathbb Z+\tau \mathbb Z$; $\Im (\tau )>0$). Then, the operator $P_{\Gamma }$ is bounded from $\mathcal E(q)$ to $\mathcal E_\Gamma (q)$.

Proof

We begin by studying the operator $\mathcal P_\Gamma$ on the Fock space $\mathcal E(0)$. Recall the following well-known atomic decomposition for $f\in \mathcal E(0)$

$$\begin{aligned} \left| f(a) e^{-B \mid a\mid ^2 / 2}\right| \leqslant \frac{C}{r^2} \int _{D(a, r)}\left| f(z) e^{-\frac{B}{2}|z|^2 }\right| d A(z). \end{aligned}$$

We choose $r_\gamma =|\gamma |^7+1$. Thus, for $f\in \mathcal E(0)$, using the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \left| \mathcal P_\Gamma (f)(z)\right| e^{-\frac{B}{2}|z|^2}&\leqslant C\sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^2} \int _{D(z+\gamma ,r_\gamma )}\left| f(w) e^{-\frac{B}{2}|w|^2}\right| d A(w) \\&\leqslant C\sum _{\gamma \in \Gamma } \frac{1}{r_\gamma } \left( \int _{D(z+\gamma ,r_\gamma )}|f(w)|^{2} e^{-B|w|^2} d A(w)\right) ^{\frac{1}{2}}. \end{aligned}$$

Applying again the Cauchy–Schwarz inequality, we get

$$\begin{aligned} \left| \mathcal P_\Gamma (f)(z)\right| ^2 e^{-B|z|^2}\le C \left( \sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^{2/3}} \right) \left( \sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^{4/3}}\left( \int _{D(z+\gamma ,r_\gamma )}|f(w)|^{2} e^{-B|w|^2} d A(w)\right) \right) \end{aligned}$$

Noting that the series $\sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^{2/3}}$ converges in both cases $\Gamma =\mathbb Z$ and $\Gamma =\mathbb Z+\tau \mathbb Z$. We integrate over a fundamental domain, which can be chosen as $\Lambda (\Gamma )=[0,1]\times \mathbb R$ if $\Gamma =\mathbb Z$ and as $\Lambda (\Gamma )=[0,1]\times [0,\Im (\tau )]$ if $\Gamma =\mathbb Z+\tau \mathbb Z$. We get

$$\begin{aligned} {\left\| {\mathcal P_\Gamma (f)}\right\| }^{2}_\Gamma&\le C\int _\mathbb C|f(w)|^{2}e^{-B|w|^2}\sum _{\gamma \in \Gamma } \left( \frac{1}{r_\gamma ^{4/3}}\int _{\Lambda (\Gamma )}\chi _{D(w-\gamma ,r_\gamma )}(z)d A(z)\right) d A(w). \end{aligned}$$

Noting that we have $\int _{\Lambda (\Gamma )}\chi _{D(w-\gamma ,r_\gamma )}(z)d A(z)\le 2r_\gamma ,$ we can conclude that

$$\begin{aligned} \sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^{4/3}}\int _{\Lambda (\Gamma )}\chi _{D(w-\gamma ,r_\gamma )}(z)d A(z)\le 2\sum _{\gamma \in \Gamma } \frac{1}{r_\gamma ^{1/3}}. \end{aligned}$$

The series in the right-hand side converges in both cases, and hence, $\mathcal P_\Gamma (f)$ is bounded on $\mathcal E(0)$.

Now, since the $L^2$ convergence implies uniform convergence in a holomorphic setting, the boundedness of $\mathcal P_\Gamma$ implies the uniform convergence of $\mathcal P_\Gamma (f)(z)$ on every compact subset of the fundamental domain $\Lambda (\Gamma )$. To conclude, it suffices to notice that for every compact subset K of $\mathbb C$, we can find a compact subset $K_0 \subset \Lambda (\Gamma )$ and a finite sequence $\gamma _0,\cdots ,\gamma _n$ such that $K \subset \bigcup \nolimits _{j=0}^{n} (K_0 + \gamma _j)$. Hence, $\mathcal P_\Gamma (f)$ is a holomorphic function on $\mathbb C$. Finally, it is clear that $\mathcal P_\Gamma$ satisfies the functional equation, and hence, $\mathcal P_\Gamma$ is a bounded operator from $\mathcal E(0)$ into $\mathcal E_\Gamma (0)$.

The space $\mathcal E(q)$ (resp. $\mathcal E_\Gamma (q)$) is in connection with $\mathcal E(0)$ (resp. $\mathcal E_\Gamma (0)$) via the annihilation and creation operators. Precisely, let $A=\partial _{{\bar{z}}}$ be the annihilation operator. Its adjoint operator is $A^*=\partial _{ z}-B{\bar{z}}$, the creation operator. In addition, let $T_a$ be the Weil representation defined by

$$\begin{aligned} T_a(f)(z)=e^{-Bz{\bar{a}}-\frac{B}{2}|a|^2}f(z+a);\quad a\in \mathbb C. \end{aligned}$$

These operators satisfy the following equations:

$$\begin{aligned}{}[A,A^*]=-B,\quad AT_a=T_aA,\quad A^*T_a=T_aA^*,\quad \Delta _B=AA^*. \end{aligned}$$

(A.1)

Furthermore, the creation operator gives an isometry between $\mathcal E(0)$ (resp. $\mathcal E_\Gamma (0)$) abd $\mathcal E(q)$ (resp. $\mathcal E_\Gamma (q)$). Precisely, we have that

$$\begin{aligned} (B^qq!)^{-1/2}(iA^*)^q:\mathcal E(0)\rightarrow \mathcal E(q), \quad (\text {resp.}:\mathcal E_\Gamma (0)\rightarrow \mathcal E_\Gamma (q)), \end{aligned}$$

(A.2)

is an isometry from $\mathcal E(0)$ to $\mathcal E(q)$ (resp. from $\mathcal E_\Gamma (0)$ to $\mathcal E_\Gamma (q)$). Using (A.1) as well as (A.2), we see that $\mathcal P_\Gamma$ is a bounded operator from $\mathcal E(q)$ to $\mathcal E_\Gamma (q)$. This finishes the proof. $\square$