On the spectrum of a multidimensional periodic magnetic Shrödinger operator with a singular electric potential

We prove absolute continuity of the spectrum of a periodic $n$-dimensional Schrödinger operator for $n\geqslant 4$. Certain conditions on the magnetic potential $A$ and the electric potential $V+\sum f_j\delta _{S_j}$ are supposed to be fulfilled. In particular, we can assume that the following conditions are satisfied.
(1) The magnetic potential $A\colon{\mathbb{R}}^n\to {\mathbb{R}}^n$ either has an absolutely convergent Fourier series or belongs to the space $H^q_{\mathrm {loc}}({\mathbb{R}}^n;{\mathbb{R}}^n)$, $2q>n-1$, or to the space $C({\mathbb{R}}^n;{\mathbb{R}}^n)\cap H^q_{\mathrm {loc}}({\mathbb{R}}^n;{\mathbb{R}}^n)$, $2q>n-2$.
(2) The function $V\colon{\mathbb{R}}^n\to \mathbb{R} $ belongs to Morrey space ${\mathfrak L}^{2,p}$, $p\in \big( \frac {n-1}2, \frac n2\big] $, of periodic functions (with a given period lattice), and
$$ \lim\limits_{\tau\to+0} \sup\limits_{0<r\leqslant\tau}\sup\limits_{x\in{\mathbb{R}}^n}r^2\bigg(\big(v(B^n_r)\big)^{-1}\int_{B^n_r(x)}|{\mathcal{V}}(y)|^p dy\bigg)^{1/p}\leqslant C, $$
where $B^n_r(x)$ is a closed ball of radius $r>0$ centered at a point $x\in{\mathbb{R}}^n$, $B^n_r=B^n_r(0)$, $v(B^n_r)$ is volume of the ball $B^n_r$, $C=C(n,p;A)>0$.
(3) $\delta_{S_j}$ are $\delta$-functions concentrated on (piecewise) $C^1$-smooth periodic hypersurfaces $S_j$, $f_j\in L^p_{\mathrm {loc}}(S_j)$, $j=1,\dots ,m$. Some additional geometric conditions are imposed on the hypersurfaces $S_j$, and these conditions determine the choice of numbers $p\geqslant n-1$. In particular, let hypersurfaces $S_j$ be $C^2$-smooth, the unit vector $e$ be arbitrarily taken from some dense set of the unit sphere $S^{n-1}$ dependent on the magnetic potential $A$, and the normal curvature of the hypersurfaces $S_j$ in the direction of the unit vector $e$ be nonzero at all points of tangency of the hypersurfaces $S_j$ and the lines $\{x_0+te\colon t\in\mathbb{R}\}$, $x_0\in{\mathbb{R}}^n$. Then we can choose the number $p>\frac {3n}2-3$, $n\geqslant 4$.