\geq 0}B(x)dx.
\]
The following two assertions are immediate consequences of
Theorems~\ref{SMSp} and \ref{SMsing} (see also Remark~\ref{d2}).
\begin{theorem}\label{ABStil}
Let $A(x)$ admit representation $(\ref{eq:B3})$ where
$A_\infty(x)$ is function $(\ref{eq:A'})$ $($for sufficiently large $|x|)$
and $A_r(x)$
obeys estimates $(\ref{eq:H1})$ with $\tilde{\rho}>1$.
Let the function $f(\omega)$ be defined by formula $(\ref{eq:B5})$,
and let $\gamma_+=-\min f(\omega)$ and $\gamma_-=-\max f(\omega)$.
Then
for all $\lambda>0$ relation $(\ref{eq:AhBo})$ holds if $\gamma_+-\gamma_-
<2\pi$,
and relation $(\ref{eq:SMT})$ holds if $\gamma_+-\gamma_-\geq 2\pi$.
\end{theorem}
\begin{theorem}\label{ABStilde}
Let assumptions of Theorem~$\ref{ABStil}$ hold.
Let $S_{0}$ be the
integral operator on $L_2(\S)$ with kernel
\begin{equation}\label{eq:SAB}
{\rm s}_{0} (\omega,\omega')= e^{i (f(\omega^{(-)})-f(\omega^{(+)}))/2}
\Bigl( \cos(\Phi/2)\delta(\omega,\omega')
+\pi^{-1}\sin(\Phi/2)
{\rm P.V.}|\omega - \omega'|^{-1}{\rm sgn}\{\omega,\omega'\}\Bigr).
\end{equation}
Then estimate $(\ref{eq:ABboundd})$ is satisfied.
\end{theorem}
It follows from formula (\ref{eq:SAB}) that, up to the phase
factor, the singular part of the SM is determined by
the magnetic flux $\Phi$ only. In particular, if
$\Phi\in 2\pi{\Bbb Z}$, then, according to (\ref{eq:B5fl}),
the SM acts as multiplication
by the function $e^{i f(\omega^{(-)})}$, where
\[
f(\omega^{(-)})= \int_{\S(-\omega ,\omega )} a(\psi)d\psi.
\]
As we shall see in the next section,
this situation is typical for dimensions $d\geq 3$.
Note also that if $a(\hat{x})$ is even, then, according to (\ref{eq:B5fl}),
$f(\omega)=\Phi/2$ and hence the first factor in the right-hand side
of (\ref{eq:SAB}) equals $1$.
As a concrete example, let us consider the function
$$
a(\hat{x})=\alpha + ,\quad \alpha\in{\Bbb R},\quad p\in{\Bbb R}^2.
$$
Then $\Phi=2\pi\alpha$ and function (\ref{eq:B5}) equals
$$
f(\omega)=\pi\alpha+ 2
,
$$
so that the conclusion of Theorem~\ref{ABStil} is true with
$\gamma_+= 2|p| - \pi\alpha$ and $\gamma_2=-2|p|-\pi\alpha$.
In particular, relation (\ref{eq:SMT}) holds if, for example, $2|p|\geq
\pi$ or, to the
contrary, it consists of the two points
$\exp(-\pi i \alpha)$ and $\exp(\pi i \alpha)$ if $p=0$.
The phase factor in (\ref{eq:SAB}) equals $ \exp(2 i < p,\omega^{(-)}>)$.
In the case $d=2$ equality (\ref{eq:GT}) implies that
\begin{equation}\label{eq:aaw}
\tilde{a}(\hat{x})=a(\hat{x})+\phi_0^\prime(\hat{x}).
\end{equation}
In particular, the magnetic fluxes for potentials $\tilde{\Phi}$ and $\Phi$
are the same, i.e.,
$\tilde{\Phi}=\Phi$. Conversely, if $\tilde{\Phi}=\Phi$, then defining
$\phi_0 (\omega)$
by the formula
\begin{equation}\label{eq:aaw1}
\phi_0 (\omega)=\int_{\S(\omega_0 ,\omega )} (\tilde{a}(\psi)-a(\psi))d \psi
\end{equation}
(the point $\omega_0\in\S$ is arbitrary but fixed) we obtain that, due to
the condition
$\tilde{\Phi}=\Phi$, function (\ref{eq:aaw1}) is single-valued on the unit
circle. Obviously,
equality (\ref{eq:aaw}) is also satisfied. Relations (\ref{eq:GT1SA}) and
(\ref{eq:aaw1}) show
that, for a given $\Phi$, it suffices to prove Theorem~\ref{ABStilde} only
for one function $a$
satisfying (\ref{eq:B5fl}).
In particular, we can choose $ a (\psi)=(2\pi)^{-1}\Phi$, which reduces the
proof of
Theorem~\ref{ABStilde} to the case of a constant function $a$.
\section{dimensions larger than two}
The three-dimensional case
${\cal H}=L_2({\Bbb R}^3)$ (or of a higher dimension $d\geq 3$)
is qualitatively different from the case $d = 2$. On the one hand,
the topological obstruction created
by a non-integer (modulo $2\pi$) magnetic flux disappears
since the exterior of any ball $|x|\leq R$ is simply connected. On the
other hand, for
dimensions $d\geq 3$, all magnetic potentials satisfying Assumption~\ref{MPTC}
cannot be described by a simple formula of type (\ref{eq:A'}). Actually, we
distinguish
two qualitatively different cases: the regular one when condition
(\ref{eq:BBad})
is fulfilled and the singular one when $B(x)$ behaves as $b(\hat{x})| x|^{-2}$
at infinity. As always, in this section Assumption~\ref{MPTC}
is supposed to be fulfilled.
\bigskip
{\bf 5.1.}
Let us start with the regular case.
\begin{lemma}\label{Baux}
Let condition $(\ref{eq:BBad})$ for $B(x)={\rm curl}\,A(x)$
be satisfied.
Then the function $\Theta_\infty(x,\xi)=:\Theta_\infty(\xi)$
does not depend on $x$ provided $=0$, $|x|\geq 1/2$.
\end{lemma}
{\it Proof.} --
By virtue of Lemma~$\ref{CWO}$,
it suffices to check that, for any $\hat{x}_1,\hat{x}_2\in \Pi_\xi$,
\begin{equation}\label{eq:indep}
\int_{-\infty}^\infty
( -
) dt
\end{equation}
tends to zero as
$R\rightarrow\infty$.
Let us apply the Stokes theorem
to a part of the lateral surface of the cylinder
$R\theta+ t\xi$, where $\theta\in (\hat{x}_1,\hat{x}_2) \subset {\Bbb S} $,
$t\in{\Bbb R}$. Then
the difference (\ref{eq:indep})
can be rewritten as
\begin{equation}\label{eq:St}
-R \int_{-\infty}^\infty dt \int_{\theta\in(\hat{x}_1,\hat{x}_2)} d \theta
<{\rm curl}\,A_\infty (R\theta + t\hat{\xi}),\theta>
\end{equation}
where we have used the cylindrical coordinates in the three-dimensional
subspace
spanned by the vectors $\hat{x}_1,\hat{x}_2$ and $\xi$.
By condition (\ref{eq:BBad}), the integrand here is bounded by
$o((t^2+R^2)^{-1})$, so that expression (\ref{eq:St}) tends to zero as
$R\rightarrow\infty$.
$\quad\Box$
Combining this result with
Theorem~\ref{SMsing}, we obtain
\begin{theorem}\label{ABdaux}
Let $d\geq 3$ be arbitrary.
Under assumption $(\ref{eq:BBad})$
the function $ \Theta_\infty(\psi,\omega)$
does not depend on $\psi$, and
the SM is, up to a compact term $($whose kernel is bounded by
$C |\omega-\omega^\prime|^{ -d+1+\nu}$ for any
$\nu<\nu_0=\min\{\tilde{\rho}-1,1\})$,
the operator of
multiplication by the function $\exp(i\Theta_\infty(\omega))$.
In this case the essential spectrum of the SM coincides with values
of the function $\exp(i\Theta_\infty(\omega))$ for all $\omega\in{\Bbb
S}^{d-1}$.
Moreover, the differential cross section satisfies estimate
\[
\Sigma_{diff}(\omega ;\omega_0,\lambda)= O(|\omega -\omega_0|^{-2d+2+\nu}).
\]
\end{theorem}
We emphasize that according to (\ref{eq:symm})
$\Theta_\infty(-\omega)=-\Theta_\infty(\omega)$.
Theorem~\ref{ABdaux} shows that under assumption (\ref{eq:BBad})
scattering has short-range nature although the magnetic potential decays as
$|x|^{-1}$ at infinity.
\medskip
We can get a slightly different expression for the function
$\Theta_\infty(\omega)$
if $B \in C_0^\infty({\Bbb R}^d)$.
Then ${\rm curl}\,A_\infty(x)=0$ for sufficiently large $ |x|$, say $
|x|\geq R$.
Let us standardly define (for $ |x|\geq R$) the function $U(x)$ as a
curvilinear integral
\begin{equation}\label{eq:Curv}
U(x)= \int_{\Gamma_x} < A_\infty(y), dy>
\end{equation}
taken between some fixed point $x_0$ and a variable point $x$. It is
required that
$\Gamma_x$ lies outside the ball $|x|\leq R$,
so that by the Stokes theorem $U(x)$ does not depend on a choice of $\Gamma_x$
(here the condition $d\geq 3$ is used).
Clearly,
\begin{equation}\label{eq:GG}
A_\infty(x)={\rm grad}\, U(x),\quad |x|\geq R.
\end{equation}
Moreover, $U(x)$ is homogeneous of degree $0$. Indeed, if $x_2=\gamma
x_1$, $\gamma>1$,
then, by transversal gauge condition (\ref{eq:H3XX}),
$A_\infty(x)$ for $x\in(x_1,x_2)$ is orthogonal to the line $(x_1,x_2)$
and hence $U(x_1)=U (x_2)$. We extend $U(x)$ as a homogeneous function
to all $|x|\geq 1/2$.
It follows from equalities (\ref{eq:GG2}) where the role of $\phi_0$ is
played by $U$ that
\[
\Theta_\infty(x,\xi)= U(\xi)-U(-\xi).
\]
In particular, we see again that the function $\Theta_\infty(x,\xi)$
does not depend on $x$.
Thus, we obtain (cf. Proposition~\ref{GItr}) the following
\begin{theorem}\label{ABd}
If $B \in C_0^\infty({\Bbb R}^d)$, $d\geq 3$, and $U$ is defined by
$(\ref{eq:Curv})$, then the
conclusion of Theorem~$\ref{ABdaux}$ remains true with the function
$\Theta_\infty(\omega)=U(\omega)-U(-\omega) $.
\end{theorem}
\begin{Remark}\label{ABd2}
If $d=2$, then of course these arguments do not work since,
by the Stokes theorem, the integral $(\ref{eq:Curv})$ over the circle
$|x|=R$ tends as $R\rightarrow\infty$ to the total flux $\Phi$ of the
magnetic field. This
difficulty is inessential if $\Phi\in 2\pi{\Bbb Z}$. Indeed, if $d=2$, then
Assumption~\ref{MPTC} implies that $B(x)=O(|x|^{-\tilde{\rho}-1})$ where
$\tilde{\rho}>1$. Therefore, again by the Stokes theorem,
\[
\Theta_\infty(x,\xi)-\Theta_\infty(-x,\xi)
=\lim_{R\rightarrow\infty}
\Bigl(\Theta_\infty(R\hat{x},\xi)-\Theta_\infty(-R\hat{x},\xi)\Bigr)
\in 2\pi{\Bbb Z}.
\]
It follows that the function $\exp(i\Theta_\infty(\psi,\omega))
=\exp(i\Theta_\infty(\omega))$ does not depend on $\psi$ such that
$<\psi,\omega>=0$.
In this case the SM acts, up to compact terms,
as multiplication by $\exp(i\Theta_\infty(\omega))$,
so that we recover a result of subs.~4.2 (see formula (\ref{eq:SAB}))
for the case $\Phi\in 2\pi{\Bbb Z}$.
\end{Remark}
\bigskip
{\bf 5.2.}
Let us consider a concrete example of a toroidal solenoid
${\bf T}$ in the space ${\Bbb R}^3$ symmetric with respect to rotations
around the axis $x_3$. We do not assume that
the section of ${\bf T}$
by a half-plane passing through the $x_3$-axis is a disc.
Suppose
(which looks quite realistic) that
\begin{equation}\label{eq:GB}
B(x_1,x_2,x_3)= \alpha (-x_2, x_1,0), \quad \alpha={\rm const},
\end{equation}
inside of ${\bf T}$ and is zero outside.
Then ${\rm div}\,B(x)=0$. Of course such a field is not smooth,
but we neglect this circumstance. According to formula (\ref{eq:BB2})
in this case we have for large $|x|$
\begin{equation} \left.\begin{array}{lcl}
A_1(x)&=&-x_1 x_3|x|^{-3} g (z),
\nonumber\\
A_2(x)&=&-x_2 x_3|x|^{-3} g (z),
\nonumber\\
A_3(x)&=&(x_1^2+ x_2^2) |x|^{-3} g (z),
\end{array}\right\}
\label{eq:lagrA}\end{equation}
where the function $g \in C_0^\infty({\Bbb R})$,
depends on parameters of the
solenoid and $z=z(x)=x_3 (x_1^2+ x_2^2)^{-1/2}$. Indeed, plugging (\ref{eq:GB})
into (\ref{eq:BB2}) and taking into account the rotational symmetry, we
find that for
sufficiently large $|x|$
\[
A_1(x)= \alpha x_1 x_3 |x|^{-3}\int_{c_-(z)}^{c_+(z)}s^2ds,\quad c_+(z)>c_-(z),
\]
where $c_\pm(z)$ are the points where the half-line $s(1,0,z)$, $s\in{\Bbb
R}_+$,
intersects the torus. Components $A_2(x)$ and $A_3(x)$ can be found quite
similarly.
Thus, we can set
\[
g (z)=-3^{-1}\alpha(c_+^3(z) -c_-^3 (z) )
\]
which determines the function $g$ in (\ref{eq:lagrA}). Clearly, $\pm g\geq
0$ if $\mp\alpha\geq 0$.
Let ${\rm supp}\, g=[-z_0,z_0]$. Of course, for any function $g$ in
(\ref{eq:lagrA}), we have that $=0$ and ${\rm curl}\, A(x)=0$
for sufficiently large $|x|$. Note also that
$A(x)=0$ if $|z(x)|\geq |z_0|$.
In case (\ref{eq:lagrA}) a function $U(x)$ satisfying (\ref{eq:GG})
can be constructed by the explicit formula
\[
U(x)=G(x_3 (x_1^2+ x_2^2)^{-1/2}),
\]
where
$$
G^\prime (z)=g(z)(z^2+1)^{-3/2}.
$$
Clearly, $G(z) $ increases (decreases) if $\alpha<0$ ($\alpha>0$) and it is
a constant
if $|z|\geq |z_0|$. We set $G(z)=0 $ for $z\leq -z_0$. Then $G(z)=G_0$ for
$z\geq z_0$
where $-G_0$ is the magnetic flux over a section of
our toroidal solenoid. Indeed, by the Stokes
theorem it equals the circulation of the magnetic potential
$A(x)$, for
example, over the closed contour formed by the $x_3$-axis and the line
$x_1+\xi t$,
$t\in{\Bbb R}$, where $\xi$ is directed along the $x_3$-axis. We remark
that $x_1$
is sufficiently large here and the integrals over pieces of
lines connecting these infinite parallel lines tend to zero because
$A(x)=O(|x|^{-1})$.
As was already noted, the integral over the $x_3$-axis is zero.
By (\ref{eq:GG2}), the integral over the line $x_1+\xi t$ equals
$U(\xi)- U(-\xi)=G_0$. Let us formulate the results obtained.
\begin{proposition}\label{exT}
Let a magnetic potential be given $($for large $|x|)$ by equations
$(\ref{eq:lagrA})$.
Then, up to a compact term $($whose kernel is bounded by $C
|\omega-\omega^\prime|^{-d+\mu}$
for any $\mu<2)$, the SM
is the operator of multiplication by the function
$$
\exp\Bigl(iG(\omega_3 (1- \omega_3^2)^{-1/2})- iG(-\omega_3 (1-
\omega_3^2)^{-1/2}) \Bigr).
$$
If the magnetic flux $-G_0$ over a section of our toroidal solenoid
satisfies $|G_0| <\pi$, then
the essential spectrum of the SM coincides with the arc
$[\exp(-i|G_0|),\exp(i|G_0|)]$.
If $|G_0|\geq\pi$, then relation $(\ref{eq:SMT})$ holds.
The points $\exp(\pm iG_0)$ are always eigenvalues of infinite multiplicity.
\end{proposition}
\bigskip
{\bf 5.3.}
Let us pass to the singular case.
Here we consider two examples of
magnetic potentials $A(x)$, $x=(x_1,x_2,x_3)\in{\Bbb R}^3$,
homogeneous (as always, for large $|x|$) of degree $-1$ and satisfying
transversal gauge condition
(\ref{eq:H3}). However, the corresponding
magnetic fields $B(x)$ will decay only as $|x|^{-2}$ at infinity, which will
change completely the conclusion of Theorem~\ref{ABdaux}.
We define the first of these potentials by the equation
\begin{equation}\label{eq:EX1}
A(x)=|x|^{-3}(\alpha_1 x_2 x_3,\alpha_2 x_3 x_1 , \alpha_3 x_1 x_2),
\quad |x|\geq 1,
\end{equation}
where $\alpha_j$ are constants and
\begin{equation}\label{eq:EX1a}
\alpha_1+\alpha_2 + \alpha_3 =0.
\end{equation}
Let us calculate function (\ref{eq:EABAB}). According to (\ref{eq:EX1})
\begin{eqnarray*}
=(x^2+t^2 \xi^2)^{-3/2}
\\
\times\Bigl(\alpha_1 \xi_1 (x_2+t\xi_2) (x_3+t\xi_3)
+\alpha_2 \xi_2 (x_3+t\xi_3) (x_1+t\xi_1)
+\alpha_3 \xi_3 (x_1+t\xi_1) (x_2+t\xi_2)\Bigr).
\end{eqnarray*}
This function is a polynomial of the second degree in $t$. The sum of terms
containing
$t^2$ is zero due to condition (\ref{eq:EX1a}). The integral over ${\Bbb R}$
of terms containing $t$ is zero since the corresponding function is odd. So
integrating
$(x^2+t^2 \xi^2)^{-3/2}$, we find that
function (\ref{eq:EABAB}) equals
\begin{equation}\label{eq:EX2}
\Theta_\infty(x,\xi)= 2 |\xi|^{-1} |x|^{-2}
(\alpha_1\xi_1 x_2 x_3+\alpha_2 \xi_2 x_3 x_1+ \alpha_3\xi_3
x_1 x_2).
\end{equation}
Let us find the essential spectrum of the corresponding SM.
According to Theorem~\ref{SMSp} we have to find the maximum of function
(\ref{eq:EX2}) restricted to the set $T_1^\ast {\Bbb S}^2$ where $
|x|^2=|\xi|^2=1$ and condition
\begin{equation}\label{eq:EX2a}
x_1 \xi_1+ x_2\xi_2 + x_3\xi_3 =0
\end{equation}
is satisfied.
We fix $x$ and consider first (\ref{eq:EX2})
as a function of $\xi$ only. The method of Lagrange multiplies
gives us the equations
\begin{equation} \left.\begin{array}{lcl}
\alpha_1 x_2 x_3=\mu x_1+\nu \xi_1,
\nonumber\\
\alpha_2 x_3 x_1=\mu x_2 +\nu \xi_2,
\nonumber\\
\alpha_3 x_1 x_2=\mu x_3 +\nu \xi_3.
\end{array}\right\}
\label{eq:lagr}\end{equation}
Let us multiply these equations by $x_1, x_2, x_3$, respectively,
and take their sum. It follows from conditions (\ref{eq:EX1a}) and
(\ref{eq:EX2a}) that $\mu=0$ and hence
\begin{equation}\label{eq:EX3}
\nu^2(x)= (\alpha_1 x_2 x_3)^2 +(\alpha_2 x_3 x_1)^2 +(\alpha_3 x_1 x_2)^2.
\end{equation}
Multiplying (\ref{eq:lagr}) by $\xi_1, \xi_2,
\xi_3$, respectively, we find that
\begin{equation}\label{eq:tth}
\Theta_\infty(x,\xi)=2\nu(x),
\end{equation}
which is the maximum (if $\nu(x)>0$) of the function
$\Theta_\infty(x,\xi)$ for fixed $x$. Next we have to find
the maximum of function (\ref{eq:EX3}) on the sphere
$|x|=1$. It is an easy exercise to check that the function
$\nu^2(x)$ attains the maximum $\alpha_1^2/4$ at the points $x_1=0,
x_2^2=x_3^2=2^{-1}$
and, similarly, for other permutations of indices. It follows from
(\ref{eq:tth}) that
\begin{equation}\label{eq:EX4a}
\max \Theta_\infty(x,\xi)=\max \{|\alpha_1|,|\alpha_2|,|\alpha_3|\}
=:\alpha_0.
\end{equation}
Using now Theorem~\ref{SMSp}, we find the spectrum of the SM.
\begin{proposition}\label{ex1}
Let a magnetic potential be given $($for large $|x|)$ by equation
$(\ref{eq:EX1})$
where the constants $\alpha_j$ satisfy condition $(\ref{eq:EX1a})$,
and let $\alpha_0$ be defined by $(\ref{eq:EX4a})$.
Then the essential spectrum of the SM
coincides with the arc $[\exp(-i\alpha_0),\exp(i\alpha_0)]$
if $\alpha_0<\pi$ and relation $(\ref{eq:SMT})$ holds if $\alpha_0\geq\pi$.
\end{proposition}
The function (\ref{eq:EX2}) depends on variables
$x_1, x_2, x_3$, so that it can be expected that
the second term in representation (\ref{eq:DD})
of the corresponding SM is non-trivial.
Set $\omega =(\omega_1,
\omega_2,\omega_3)$ and
\[
n = (\omega_1^2+\omega_2^2)^{-1/2}(-\omega_2,\omega_1,0),\quad
m=\omega\times n =
(\omega_1^2+\omega_2^2)^{-1/2}(-\omega_1\omega_3,-\omega_2\omega_3,
\omega_1^2 +\omega_2^2).
\]
An arbitrary point $x\in{\Bbb S}_\omega={\Bbb S}^2 \cap \Pi_\omega$
can be written as $x=x(\theta)=n\cos\theta+m\sin\theta$, $\theta\in[0,2\pi)$,
so that
\begin{eqnarray}\label{eq:xxx}
x_1 = -(\omega_1^2+\omega_2^2)^{-1/2}(\omega_2\cos\theta+\omega_1
\omega_3\sin\theta ),
\nonumber\\
x_2 = (\omega_1^2+\omega_2^2)^{-1/2}(\omega_1\cos\theta -\omega_2
\omega_3\sin\theta ),
\quad
x_3 = (\omega_1^2+\omega_2^2)^{1/2} \sin\theta.
\end{eqnarray}
Plugging these expressions in (\ref{eq:EX2}), we find that
\begin{eqnarray}\label{eq:asssix}
\Theta_\infty(x,\omega)=(\omega_1^2+\omega_2^2)^{-1}
\Bigl(-2\alpha_3 \omega_1\omega_2\omega_3\cos(2\theta)
\nonumber\\
+(\alpha_1(\omega_1^2-\omega_2^2\omega_3^2)
-\alpha_2(\omega_2^2-\omega_1^2\omega_3^2))\sin(2\theta)
\Bigr) = {\cal A}(\omega)\sin(2\theta+\theta_0(\omega))
\end{eqnarray}
where
\[
{\cal A} (\omega)=(\omega_1^2+\omega_2^2)^{-1}
\Bigl(4\alpha_3^2 \omega_1^2\omega_2^2\omega_3^2
+(\alpha_1(\omega_1^2-\omega_2^2\omega_3^2)
-\alpha_2(\omega_2^2-\omega_1^2\omega_3^2))^2
\Bigr)^{1/2}
\]
and
\[
\tan \theta_0(\omega)=-2\alpha_3 \omega_1 \omega_2 \omega_3
(\alpha_1(\omega_1^2-\omega_2^2\omega_3^2)
-\alpha_2(\omega_2^2-\omega_1^2\omega_3^2))^{-1}.
\]
Now we can calculate function (\ref{eq:DD0}). Using expression
(\ref{eq:asssix}), we see that
\[
p^{(0)}(\omega)=(2\pi)^{-1}\int_0^{2\pi}\Bigl(\cos ({\cal
A}(\omega)\sin(2\theta+\theta_0(\omega))+i
\sin ({\cal A}(\omega)\sin(2\theta+\theta_0(\omega))\Bigr)d\theta.
\]
The integral of the second term is zero, and the first one can be simplified
by the change of variables $2\theta+\theta_0(\omega)\mapsto \theta$.
Therefore Theorem~\ref{SMsing} implies
\begin{proposition}\label{ex1s}
Under the assumptions of Proposition~$\ref{ex1}$ the singular part of the
scattering amplitude is given by expression $(\ref{eq:DD})$
where
\[
p^{(0)}(\omega)=(2\pi)^{-1}\int_0^{2\pi}\cos ({\cal
A}(\omega)\sin\theta)d\theta,
\]
\begin{equation}\label{eq:ex1sb}
q(\omega ,\tau)=-(2\pi )^{-2}\int_0^{2\pi}\Bigl( \exp (i{\cal A}(\omega)
\sin(2\theta+\theta_0(\omega)))-
p^{(0)}(\omega)\Bigr)(-i0)^{-2}d\theta
\end{equation}
and the vector $x(\theta)$ is defined by formulas
$(\ref{eq:xxx})$.
Estimate $(\ref{eq:ABboundd})$ holds for any $\nu<1$.
\end{proposition}
\medskip
As another example of a
magnetic potential $A(x)$ homogeneous (as always, for large
$|x|$) of degree $-1$ and satisfying transversal gauge condition
(\ref{eq:H3}),
we choose a modification of the Aharonov-Bohm potential
\begin{equation}\label{eq:EXA2}
A(x)=\alpha |x|^{-2}(- x_2, x_1,0),
\quad |x|\geq 1,\quad x=(x_1,x_2,x_3)\in{\Bbb R}^3.
\end{equation}
Once again the corresponding
magnetic fields $B(x)$ decays only as $|x|^{-2}$ at infinity, so that
the conclusion of Theorem~\ref{ABdaux} is violated.
Calculating function (\ref{eq:EABAB}),
we find that
\begin{equation}\label{eq:EXA21}
\Theta_\infty(x,\xi)= \pi\alpha |\xi|^{-1} |x|^{-1} (x_1\xi_2-\xi_1 x_2).
\end{equation}
Let us first find the essential spectrum of the corresponding SM.
According to Theorem~\ref{SMSp},
we have to find the image of function (\ref{eq:EXA21}) restricted to
the set
$T_1^\ast {\Bbb S}^2$ where $
|x|^2=|\xi|^2=1$ and condition (\ref{eq:EX2a}) is satisfied. Easy
calculations yield
\begin{proposition}\label{ex22}
Let a magnetic potential be given by
equation
$(\ref{eq:EXA2})$. Then the essential spectrum of the SM
coincides with the arc $[\exp(-i\pi|\alpha|),\exp(i\pi|\alpha|)]$
if $|\alpha|< 1$ and relation $(\ref{eq:SMT})$ holds if $|\alpha|\geq 1$.
\end{proposition}
To find the singularity of the scattering amplitude,
we repeat, with significant simplifications,
the construction of Proposition~\ref{ex1s}. In particular, it follows from
(\ref{eq:xxx}) and (\ref{eq:EXA21}) that
\[
\Theta_\infty(x,\omega)=-\pi\alpha(1-\omega_3^2)^{1/2}
\cos \theta.
\]
Therefore Theorem~\ref{SMsing} implies
\begin{proposition}\label{ex22si}
Under the assumptions of Proposition~$\ref{ex22}$ the singular part
of the scattering amplitude is given by expression $(\ref{eq:DD})$ where
\[
p^{(0)}(\omega)=(2\pi)^{-1}\int_0^{2\pi}\cos ( \pi
(1-\omega_3^2)^{1/2}\cos\theta)d\theta,
\]
\begin{equation}\label{eq:ex22sb}
q(\omega ,\tau)=-(4\pi )^{-1}\int_0^{2\pi}
\Bigl( \exp (- \pi i\alpha (1-\omega_3^2)^{1/2}\cos\theta)
- p^{(0)}(\omega)\Bigr)(-i0)^{-2}d\theta
\end{equation}
and the vector $x(\theta)$ is defined by formulas
$(\ref{eq:xxx})$.
Estimate $(\ref{eq:ABboundd})$ holds for any $\nu<1$.
\end{proposition}
We emphasize that both terms (\ref{eq:ex1sb}) and (\ref{eq:ex22sb}) are not
zero, so that the
singular part of the SM in the examples of this subsection does not reduce
to an operator of
multiplication.
\section{electric potentials}
The SM for
electric potentials $V(x)$ which are odd functions and behave at infinity as
homogeneous functions of
order $-1$ possess the same properties
as for magnetic potentials considered above.
Suppose that a potential $V(x)$
satisfies condition $(\ref{eq:H1})$ with $\rho =1$ and is such that
\[
V(x)=V_\infty(x)+V_r (x), \quad |x|\geq 1,
\]
where
\[
V_\infty(x)=v(\hat{x})|x|^{-1},\quad v(-\hat{x})=-v(\hat{x}),
\]
and
$V_r(x)$ obeys estimates (\ref{eq:H1})
with some $ \tilde{\rho}>1$. We assume additionally
that a magnetic potential $A(x)=0$.
Now the SM is defined in terms of wave operators (\ref{eq:MWO}) where
$J_\pm$ are PDO
(\ref{eq:W1}) and $\Phi_\pm$ is integral (\ref{eq:EAB}) with $A=0$.
Using again Theorem~\ref{SM4}, we see that the singular part of the SM is
determined by the
operator $S_0(\lambda)$. Since $V(x)=-V(-x)$, function
(\ref{eq:EABq}) now equals
\[
\Theta (x,\xi)= - 2^{-1} \int_{0}^\infty
\Bigl(V(x+ t\xi)+ V(x- t\xi) \Bigr) dt.
\]
As always, we assume $=0$ and
replace here $V $ by $ V_\infty$, since
the arising error is $O(|x|^{-\epsilon})$,
$\epsilon=\tilde{\rho}-1>0$. This gives the expression
\[
\Theta (x,\xi)= |\xi|^{-1} \Theta_\infty (x,\xi)+ O(|x|^{-\epsilon}),
\]
where
\[
\Theta_\infty (x,\xi)= - 2^{-1} \int_{0}^\infty
\Bigl(v\Bigl(\frac{\hat{x}+ t\hat{\xi}}{(t^2+1)^{1/2}}\Bigr) +
v\Bigl(\frac{\hat{x}- t\hat{\xi}}{(t^2+1)^{1/2}}
\Bigr)\Bigr)(t^2+1)^{-1/2} dt.
\]
Making here the change of variables $t=\tan \theta$, we can rewrite
the last expression as
\begin{equation}\label{eq:EABqx3}
\Theta_\infty (x,\xi)= - 2^{-1} \int_{0}^{\pi/2}
\Bigl(v(\cos\theta\hat{x}+ \sin\theta\hat{\xi}) +
v(\cos\theta\hat{x} - \sin\theta\hat{\xi})
\Bigr)(\cos\theta)^{-1} d\theta.
\end{equation}
The last integral converges at the upper limit $\theta=\pi/2$ by
virtue of the condition $v(-\hat{x})=-v(\hat{x})$. Note also that
$$
\Theta_\infty (-x,\xi)=
-\Theta_\infty (x,\xi).
$$
Using Propositions~\ref{SpHom} and \ref{SpHomXY}, we can now formulate
analogues of
Theorems~\ref{SMSp} and \ref{SMsing}.
\begin{theorem}\label{SMSpV}
Under the
assumptions above,
let the function $\Theta_\infty$ and the numbers $\gamma$, $\gamma_+$,
$\gamma_-$ be defined by
equations
$(\ref{eq:EABqx3})$ and $(\ref{eq:gamma})$, $(\ref{eq:gamma12})$,
respectively. Then
the assertion of Theorem~$\ref{SMSp}$ remains true if the numbers
$\gamma$, $\gamma_+$,
$\gamma_-$ are replaced by the numbers $\lambda^{-1/2}\gamma$,
$\lambda^{-1/2}\gamma_+$,
$\lambda^{-1/2}\gamma_-$, respectively.
The assertion of Theorem~$\ref{SMsing}$ also remains true if
the function $\Theta_\infty(\psi,\omega)$
in definitions $(\ref{eq:DD0})$ and $(\ref{eq:DD1})$
is replaced by function $\lambda^{-1/2} \Theta_\infty(\psi, \omega)$
defined by $(\ref{eq:EABqx3})$.
\end{theorem}
As a concrete example, let us consider the function
$$
v(\hat{x})=2 ,\quad p\in{\Bbb R}^d,\quad d\geq 2.
$$
Calculating integral (\ref{eq:EABqx3}), we find that
$$
\Theta_\infty (x,\xi)=- |\xi|^{-1} \pi
,\quad |x|\geq 1/2.
$$
Thus, the essential spectrum of the SM $S(\lambda)$
coincides with the arc $[\exp(-\pi i |p| \lambda^{-1/2}),\\
\exp(\pi i |p| \lambda^{-1/2})]$ if
$|p| < \lambda^{1/2}$, and it
covers the whole unit circle if
$|p| \geq \lambda^{1/2}$.
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\end{document}