\section{Proof of Theorem~\ref{thr:2}}
\label{sec:proof-theor-refthr:1}

In our previous work~\cite{Fi-Kl:04,Fi-Kl:04c}, we proved the absolute
continuity of the spectrum of the Schr{\"o}dinger operator where the
properties of the potential were similar to those imposed on
permittivity $\varepsilon$ and the permeability $\mu$ in
Theorem~\ref{thr:1} and~\ref{thr:2}. The scheme of the proof of
Theorem~\ref{thr:2} is globally the same as that of Theorem~1.1
in~\cite{Fi-Kl:04,Fi-Kl:04c}; so, we will omit some details.

\smallpagebreak First, basing on the relation (\ref{eq:13}), we
construct a convenient representation of the resolvent $({\mathcal M}
-\la)^{-1}$ (see Lemma~\ref{Lem2.3} below).

\smallpagebreak First of all we need to define some notations. Let
$\<x\>= \sqrt{x^2+1}$. For $a\in\R$, introduce the spaces

\begin{equation*}
  L_{p,a}=\{f: e^{a\<x\>}f\in L_p(\OO)\},\qquad
  H^l_a=\{f:e^{a\<x\>}f\in H^l(\OO)\},
\end{equation*}

where $1\le p\le \infty$ and $H^l(\OO)$ is the standard Sobolev space.
Introduce the function spaces in $\Omega$ with quasi-periodic boundary
conditions

\begin{equation*}
  H^l_a(k):=\left\{f\in H^l_a:(D^\alpha f)\mid_{y_j=2\pi}=e^{2\pi
  ik_j}(D^\alpha f)\mid_{y_j=0},\quad|\alpha|\le l-1\right\}\text{ and
  } H^l(k):=H^l_0(k).
\end{equation*}

Finally, for $X$ and $Y$ Banach spaces, $B(X,Y)$ is the space of all
bounded operators from $X$ to $Y$, and $B(X)=B(X,X)$, both endowed
with their natural norm topology.

\smallpagebreak Due to the Bloch-Floquet-Gelfand transformation, the
Maxwell operator ${\mathcal M}$ is unitary equivalent to the direct
integral $\int_{[0, 1)^d}^\oplus {\mathcal M} (k) dk$, where
${\mathcal M}$ is the operator given by the differential expression
(\ref{eq:4}) on the domain $\dom {\mathcal M} (k) = H^1(k)$.  The
Laplace operator on the
domain $H^2(k)$ will be denoted by $\Delta(k)$.\\

In~\cite{Fi-Kl:04,Fi-Kl:04c}, we essentially proved the following result

\begin{Le}
  \label{Lem2.1}
  Assume that the pair $(k_0, \la_0)\in \RR^{d+1}$ satisfies
  \begin{equation}
    \label{eq:2.1}
    (k_0 + n)^2 \not= \varepsilon_0 \mu_0 \la_0, \qquad \forall n \in \ZZ^d.
  \end{equation}
  Then, there exist numbers $\delta>0$, $a>0$, an open set
  $\Xi_0 \subset \C^{d+1}$ such that
                                
  \begin{equation*}
    \left( B_\de (k_0) \cup \{ k(\tau) \}_{\tau
      \in \RR} \right) \times B_\de (\la_0) \subset \Xi_0,
  \end{equation*}
                                
  where $B_\de (k_0)$ is a ball in real space
  \begin{equation*}
    B_\de (k_0) = \{ k \in \RR^d : |k - k_0| < \de \},
  \end{equation*}
  and $k (\tau) = (\tilde k_1 + i \tau, \tilde k')$ with fixed $\tilde
  k \in B_\de (k_0)$, and there exists an analytic
  $B(L_{2,a},H^2_{-a})$-valued function $R_0$, defined in $\Xi_0$,
  having the properties
  \begin{itemize}
  \item for $(k,\lambda)\in\Xi_0$, $k\in\R^d$, $\im\lambda>0$, $U\in
    L_{2,a}$, one has
                                
    \begin{equation*}
      R_0(k,\lambda)U=(-\Delta(k)-\varepsilon_0\mu_0\lambda)^{-1}U;
    \end{equation*}
  \item
    \begin{equation}
      \label{eq:2.2}
      \|R_0(k(\tau),\lambda)\|_{B(H^2_a,\,H^2_{-a})}\leq C|\tau|^{-1};
    \end{equation}
  \item $R_0(k,\lambda)L_{2, a}\subset H^2_{-a}(k)$.
\end{itemize}
\end{Le}

\noindent This lemma is proved in~\cite{Fi-Kl:04} (see Theorem 3.1)
except for the fact that estimate~\eqref{eq:2.2} is replaced with

\begin{equation}
  \label{eq:2}
  \|R_0(k(\tau),\lambda)\|_{B(L_{2, a},L_{2, -a})}\leq C|\tau|^{-1}.
\end{equation}

The proof of estimate~(\ref{eq:2.2}) is exactly the same as that
of~\eqref{eq:2}.

\smallpagebreak Clearly, in Lemma~\ref{Lem2.1}, we can replace
$\Delta(k)$ with $\Delta_8(k)$ (defined in~\eqref{eq:8}) at
the expense of changing the constants; the resolvent of
$\Delta_8(k)$ (and its analytic extension) will henceforth
be denoted by $R_{\mathcal{M}}^0(k,\la)$. So

\begin{equation*}
  R_{\mathcal{M}}^0(k,\la)=R_0(k,\la)\,\text{Id}_{\C^8}.
\end{equation*}

\smallpagebreak To deal with the potential, we prove

\begin{Le}
  \label{Lem2.2}
  Let $\varepsilon$, $\mu$ satisfy hypothesis (H1)-(H4), ${\mathcal
    A}$ be defined by (\ref{eq:5}), and $(k_0, \lambda_0)$ satisfy
  (\ref{eq:2.1}). Then, there exist $\delta>0$, $a>0$, an open set
  $\Xi \subset \C^{d+1}$ with $B_\de (k_0) \times B_\de (\lambda_0) \subset
  \Xi$, a function $h:\ \Xi\to\C$ analytic in $\Xi$ with the property
                                
  \begin{equation}
    \label{eq:2.3}
    \forall\lambda\in B_\delta(\lambda_0),\quad\exists k\in B_\delta(k_0)\quad
    \text{such that}\quad h(k,\lambda)\not= 0,
  \end{equation}
                                
  and there exists an analytic $B(L_{2,a},H^2_{-a})$-valued function
  $Z$, defined in
                                
  \begin{equation*}
    \Xi_1 := \{ (k, \lambda) \in \Xi : h (k, \lambda)\not= 0\},
  \end{equation*}
                                
  such that, for $(k,\lambda)\in\Xi_1$, $k\in\R^d$, $\im\lambda^2>0$,
  $U\in H^2_a(k)$, one has
                                
  \begin{equation}
    \label{eq:2.4}
    Z(k, \lambda) \left(-\Delta_8 + \mathcal{V} -
      \varepsilon\mu\mathcal{J}^{-1}
      (\lambda \mathcal{A} + \lambda^2)\mathcal{J}\right) U = U
  \end{equation}
                                
  and
                                
  \begin{equation*}
    Z(k, \lambda) L_{2, a} \subset H^2_{-a} (k).
  \end{equation*}
\end{Le}

\begin{proof} Note that
                                
  \begin{equation*}
    \mathcal{V} - \varepsilon\mu\mathcal{J}^{-1}
    (\lambda \mathcal{A} + \lambda^2)\mathcal{J} =
    -\er_0 \mu_0 \la^2 + \mathcal{W} (\la),
  \end{equation*}
                                
  where, by assumptions (H2)-(H3), $\lambda\mapsto{\mathcal
    W}(\lambda)$ is an entire function valued in $L_{\infty,b}$ for
  any $b\in\R$. Set
                                
  \begin{equation*}
    Z(k,\lambda)=\left(I+R_{\mathcal{M}}^0(k,\la^2)\mathcal{W}(\la)
    \right)^{-1}R_{\mathcal{M}}^0(k,\la^2).
  \end{equation*}
                                
  The operator of multiplication by ${\mathcal W}$ is bounded as an
  operator from $H^2_{-a}$ to $H^2_a$, and is compact as an operator
  from $H^2_{-a}$ to $L_{2, a}$. It remains to use the estimation
  (\ref{eq:2.2}) and the analytic Fredholm alternative in the Hilbert
  space $H^2_{-a}$ (see e.g.~\cite{Ka:80,MR58:12429c}) to complete the
  proof of Lemma~\ref{Lem2.2}.
\end{proof}

\noindent In the following lemma, we construct an analytic extension
of the resolvent of Maxwell operator to the non-physical sheet. Set

\begin{equation*}
  Q(\la) = \varepsilon\mu\mathcal{J}^{-1}
(\mathcal{M} + \mathcal{A} + \lambda).
\end{equation*}

Then, for any $b\in\R$, $Q$ is an entire function with values in
$B(H^1_b,L_{2, b})$.
The next result we need is

\begin{Le}
  \label{Lem2.3}
  Under the assumptions of Lemma~\ref{Lem2.2}, on the set $\Xi_1$, we
  define the operator-function
                                
  \begin{equation*}
    (k,\lambda)\mapsto R_\mathcal{M}(k,\la):=\mathcal{J}Z(k,\lambda)
    (I -\mathcal{F}Z(k,\lambda))Q(\la).
  \end{equation*}
  Then, one has
                                
  \begin{enumerate}
  \item $(k,\lambda)\mapsto R_\mathcal{M}(k,\lambda)$ is analytic in
    $\Xi_1$ with values in $B(H^1_a, H^2_{-a}))$;
  \item for $(k,\lambda)\in\Xi_1$, $k\in\R^d$, $\im\lambda^2>0$, there
    exists $\mathcal{H}(k)\subset H^1_a(k)$ such that
    $\overline{{\mathcal H}(k)}=L_2(\Omega)$ and for
    $U\in\mathcal{H}(k)$,
                                
    \begin{equation*}
       R_\mathcal{M} (k, \la) U = (\mathcal{M} (k) - \la)^{-1} U
    \end{equation*}
  \end{enumerate}
\end{Le}

\begin{proof}
  The first property is true because ${\mathcal F}$ is a bounded
  operator from $H^1_{-a}$ to $L_{2, a}$.\\
                                
  To prove the second one, pick $(k,\lambda)\in\Xi_1$ such that
  $k\in\R^d$ and $\im\lambda^2>0$; define
                                
  \begin{equation*}
    {\mathcal H}(k)=({\mathcal M}(k)-\la)H^2_a (k).
  \end{equation*}
                                
  That $\mathcal{H}(k)$ is dense in $L^2(\Omega)$ is a consequence of
  the self-adjointness of $\mathcal{M}$ and the fact that
  $\lambda\not\in\R$.\\
                                
  Let $W\in H^2_a (k)$ and $U=({\mathcal M}-\la){\mathcal J} W$.
  Then, one computes
  \begin{equation}
    \label{eq:38}
    \begin{split}
      R_\mathcal{M}(k,\la)U&=\mathcal{J}Z(k,\lambda)
      (I-\mathcal{F}Z(k,\lambda))Q(\la)({\mathcal M}-\la){\mathcal J}W\\
      &=\mathcal{J}(Z(k,\lambda)-Z(k,\lambda)\mathcal{F}Z(k,\lambda))
      \left(-\Delta_8+\mathcal{V}- \varepsilon\mu\mathcal{J}^{-1}
        (\lambda\mathcal{A}+\lambda^2)\mathcal{J}+\mathcal{F}\right)W\\
      &=\mathcal{J}\left[W+Z(k,\lambda)\mathcal{F}W\right.\\
      &\hskip2cm\left.-Z(k,\lambda)\mathcal{F}Z(k,\lambda)
        \left(-\Delta_8+\mathcal{V}-
          \varepsilon\mu\mathcal{J}^{-1}(\lambda\mathcal{A}+
          \lambda^2)\mathcal{J}+\mathcal{F}\right)W\right]\\
      &=\mathcal{J} (W - Z(k,\lambda)\mathcal{F}
      Z(k,\lambda)\mathcal{F} W)
    \end{split}
  \end{equation}
                                
  where we used (\ref{eq:13}) and (\ref{eq:2.4}).
  Furthermore, one can check that
  \begin{equation*}
   \mathcal{F} Z(k, \la)\mathcal{F} = 0.
  \end{equation*}
  Plugging this into~\eqref{eq:38}, we obtain
                                
  \begin{equation*}
    R_\mathcal{M}(k,\la)U=\mathcal{J}W=(\mathcal{M}(k)-\la)^{-1}U.
  \end{equation*}
                                
  This completes the proof of Lemma~\ref{Lem2.3}.
\end{proof}

\begin{Rem}
  \label{rem:1}
  One presumably has ${\mathcal H} (k) = H^1_a (k)$.
\end{Rem}

\begin{Le}
  \label{Lem2.4}
  Let $G_0$ and $G$ be two Hilbert spaces, $G_0\subset G$, and $G_0^*$
  be a dual space to $G_0$ with respect to the scalar product in $G$.
  Let $B$ be a self-adjoint operator in $G$.  Suppose that $R_B$ is an
  analytic function defined in a complex neighborhood of an interval
  $[\al,\be]$ except at a finite number of points
  $\{\mu_1,\dots,\mu_N\}$, that the values of $R_B$ are in $B (G_0,
  G_0^*)$ and that
                                
  \begin{equation*}
    R_B (\la) \ph = (B - \la)^{-1} \ph \quad \text{if } \im \la > 0, \ph
  \in {\mathcal H}
  \end{equation*}
                                
  where ${\mathcal H}\subset G_0$ is dense in $G$.  Then, the spectrum
  of $B$ in the set $[\al,\be]\setminus\{\mu_1, \dots, \mu_N \}$ is
  absolutely conti\-nuous.  If $\La \subset [\al, \be]$, $\mes\La = 0$
  and $\mu_j \not\in \La$, $j = 1, \dots, N$, then $E_B (\La) = 0$,
  where $E_B$ is the spectral projector of $B$.
\end{Le}

\noindent This lemma is an immediate consequence of Proposition 2 and
equation (18) in section 1.4.5 of~\cite{MR94f:47012}.\\

Now, let $G$ be a Hilbert space, and let $(H(k))_{k\in\C^d}$ be an
analytic family of self-adjoint operators on $G$.  On
$\mathcal{G}=L^2([0,1)^d,G)$, following~\cite{MR58:12429c}, one
defines the self-adjoint operator
                                
\begin{equation*}
  H=\int_{[0,1)^d}^\oplus H (k) dk.
\end{equation*}
                                
The following abstract theorem on the spectrum of the fibered operator
$H$ is based on the Lemma \ref{Lem2.4}.  Its proof repeats the proof
of Theorem 1.1 in~\cite{Fi-Kl:04} although this explicit formulation
is not given there.

\begin{Th}
  \label{thr:4}
  Suppose that there exists a sequence
  
  of analytic functions $f_m:\C^{d+1}\to\C$ such
  that
                                
  \begin{equation*}
    \forall\la\quad\exists k\quad \text{such that}\quad f_m
    (k,\la)\not= 0,
  \end{equation*}
                                
  and the set of real points $(k,\la)$ where $f_m(k,\la)\not=0$ for
  all $m$ can be represented as
                                
  \begin{equation*}
    \R^{d+1}\setminus \bigcup_{m=1}^\infty \{(k,\lambda) : f_m (k,\la)=
    0\} = \bigcup_{j=1}^\infty B_{\er_j} (k_j) \times B_{\er_j} (\la_j).
  \end{equation*}
                                
  Suppose moreover that, for every $j$, there exist
                                
  \begin{itemize}
  \item an analytic scalar function $h_j$ defined in a complex
    neighborhood of $\overline{B_{\er_j} (k_j) \times B_{\er_j} (\la_j)}$
    satisfying property (\ref{eq:2.3});
  \item a Hilbert space $G_j (k)\subset G$, its dual $G^*_j (k)$ with
    respect to the scalar product in $G$, and a set ${\mathcal
      H}_j(k)$ such that
   $${\mathcal H}_j (k)\subset G_j (k)\subset G,\quad
      \overline{{\mathcal H}_j (k)}=G;$$
  \item an analytic $B(G_j,G_j^*)$-valued function $R_j$ defined
    on the set $\{(k,\lambda);\ h_j(k,\la)\not=0\}$ such that for
    $k\in\R^d$, $\im\la>0$, $f\in {\mathcal H}_j(k)$,
    \begin{equation*}
      R_j (k, \la) f = (H(k) - \la)^{-1} f.
    \end{equation*}
  \end{itemize}
                                
  Then, the spectrum of $H$ is purely absolutely continuous.
\end{Th}

\noindent The spectral theory of a class of analytically fibered
operators has been studied in~\cite{MR99b:47030}; their definition of
an analytically fibered operator cannot be used in the present case as
they require the resolvent of the fiber operators to be compact.

\smallpagebreak Theorem~\ref{thr:4} completes the proof of
Theorem~\ref{thr:2} if we take

\begin{equation*}
  G = L_2 (\Omega), \quad H(k) = {\mathcal M} (k),
  \quad H = {\mathcal M}, \quad f_n (k, \la) = (k+n)^2 - \er_0\mu_0 \la^2,
\end{equation*}

use Lemma~\ref{Lem2.3} in a neighborhood of each pair $(k,\la)$ for
which $f_n$ does not vanish, and set

\begin{equation*}
  {\mathcal H}_j(k) = ({\mathcal M} - \la) H^2_{a_j} (k), \quad
  G_j (k) = H^1_{a_j} (k), \quad  H^2_{-a_j} (k)\subset G^*_j (k) =
  H^{-1}_{-a_j} (k), \quad R_j = R_{\mathcal M}.
\end{equation*}
{"config": "arxiv", "file": "math-ph0406045/abs-max.tex"}