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\begin{document}
\author{O.A. Veliev \and \c S. Somali}
\title{The Periodic Multidimensional Schr\"odinger Operator, Part 4, An
Algorithm
for Finding the Potential by Spectral Invariants }
\date{{\small \textit{Dokuz Eyl\"ul Univ., Fen-Ed. Fak.,}}\\
{\small \textit{Dept. of Math., Tinaztepe Kamp\"us\"u, Buca, 35160, Izmir,
TURKEY}}\\
{\small \textit{\ \ E-mail: oktay.veliev@deu.edu.tr, sennur.somali@deu.edu.tr%
}}}
\maketitle
\begin{abstract}
In this paper, we give an algorithm for finding the potential of the
three-dimensional Schr\"odinger operator by spectral invariants.
\end{abstract}
\section{Introduction}
We investigate the Schr\"odinger Operator
\[
L(q)=-\Delta +q(x),\text{ }x\in R^d,\text{ }d\geq 2
\]
with a real periodic (relative to the Lattice $\Omega $) potential $q(x)\in
W_2^l(F),$ where $l\geq 6(3^d(d+1)^2)+d$, and $F$ is the fundamental domain $%
R^d/\Omega $ of $\Omega .$ The spectrum of $L(q)$ is the union of the
spectra of the operators $L_t(q)$, for $t\in F^{*}\equiv R^d/\Gamma $
generated by $-\Delta u+q(x)u$ and the conditions ( see [1])
\[
u(x+\omega )=e^{i\langle t,\omega \rangle }u(x),\ \forall \omega \in \Omega
,
\]
where $\Gamma \equiv \{\delta \in R^d:\langle \delta ,\omega \rangle \in
2\pi \Bbb{Z},\forall \omega \in \Omega \}$ is the lattice dual to $\Omega $
and $\langle .,.\rangle $ is the inner product in $R^d.$ The eigenvalues $%
\Lambda _1(t)\leq \Lambda _2(t)\leq ...$of $L_t$ define functions $\Lambda
_n(t)$ $(n=1,2,..)$ of $t$ that are called the band functions or Floquet
spectrum of $L$.
In paper [5] by the Floquet spectrum of $L$ we constructively determined the
following spectral invariants
\begin{eqnarray}
&&\ \int_F\left| q^a(x)\right| ^2dx, \\
&&\ \int_F\left| q_{a,\beta }(x)\right| ^2q^a(x)dx,
\end{eqnarray}
where $a$ is the maximal element of $\Gamma ,$ that is, $a$ is the element
of $\Gamma $ of minimal norm belonging to the line $aR,$ the $%
q^a(x)=\sum_{n\in Z}q_{n\delta }e^{in\langle a,x\rangle }$ is directional
(one dimensional) potential $q_b\equiv (q(x),e^{i\langle b,x\rangle })$ is
the Fourier coefficient of $q(x).$ We assume that $q_0=0.$ The function $%
q_{a,\beta }(x)$ is defined as
\[
q_{a,\beta }(x)=\sum_{b\in S(a,\beta )\backslash aR}\frac b{\langle \beta
,b\rangle }q_be^{i\langle b,x\rangle },
\]
where $S(a,\beta )=P(a,\beta )\cap \Gamma $, and $P(a,\beta )$ is the plane
containing $a$, $\beta $ and $0$. Here $\beta \in \Gamma _a,$ where $\Gamma
_a$ is the lattice dual to $\Omega _a$ and $\Omega _a$ is the sublattice $%
\{d\in \Omega :\langle d,a\rangle =0\}$ of $\Omega $ in the hyperplane $H_a=$
$\{x\in R^d:\langle x,a\rangle =0\}$. If
\begin{equation}
q^a(x)=q_ae^{i\langle a,x\rangle }+q_{-a}e^{-i\langle a,x\rangle }
\end{equation}
then besides (1), (2) we found the invariant
\begin{equation}
\int_F|q_{a,\beta }(x)|^2(q_a^2e^{i2\langle a,x\rangle
}+q_{-a}^2e^{-i2\langle a,x\rangle })dx.
\end{equation}
In this paper fixing the translations $q(x)\rightarrow q(-x),\
q(x)\rightarrow q(x+\tau ),\ \tau \in R^3,$ we give an algorithm for finding
the potential $q(x)$ ot the three- dimensional Schr\"odinger operator $L(q)$
by the invariants (1), (2) and (4).
Note that the operators $L(q(x)),\ L(q(-x)),\ L(q(x+\tau ))$ have the same
band functions, i.e., $q(x),\ q(-x),\ q(x+\tau )$ are isospectral
potentials. Eskin \textit{et al.} [2], [3] proved the following result for
the two-dimensional periodic Schr\"odinger operator:\newline
\textit{For $\Omega \subset R^2$ satisfying the condition: if $\left| \omega
\right| =\left| \omega ^{\prime }\right| $ for $\omega ,\ \omega ^{\prime
}\in \Omega ,$ then $\omega ^{\prime }=\pm \omega $; there is a set $%
\{M_\alpha \}$ of manifolds of potentials such that }
\textit{a) $\{M_\alpha \}$ is dense in the set of smooth periodic potentials
in }$\Bbb{C}^\infty -$\textit{topology; }
\textit{b) for each $\alpha $ there is a dense open set $Q_\alpha \subset
M_\alpha $ such that for $q\in Q_\alpha $ the set of real analytic $%
\widetilde{q}$ satisfying $Spec(L_0(q)=Spec(L_0(\widetilde{q}))$}
\textit{and the set of $\widetilde{q}\in $}$\Bbb{C}^6(R^n/L)$\textit{\
satisfying $Spec(L_t(q)=Spec(L_t(\widetilde{q}))\ \forall t\in R^n$ are
finite modulo translations}.
For the case of the two-dimensional difference periodic Schr\"odinger
operator Gieseker \textit{et al.}[4] proved that:\newline
\textit{There is a Zariski-open and dense subset $P$ of $L^2$ such that for
any $V\in P,\ V^{\prime }\in L^2$ equality $B(V)=B(V^{\prime })$ of the
Bloch varieties implies that $V$ and $V^{\prime }$ coincide up to
symmetries, where $L^2=L^2($}$\Bbb{Z}^2/a\Bbb{Z}\oplus b\Bbb{Z}\Bbb{)}$,%
\textit{\ $a$ and $b$ are distinct odd primes,}\
\[
B(V)=\{(\xi _1,\xi _2,\lambda )\in \Bbb{C^{\star }\times C^{\star }\times
C\mid }\ there\ is\ a\ complex-valued
\]
\[
function\ \Psi \ on\ \Bbb{Z}^2\ satisfying
\]
\[
\Psi (m+1,n)+\Psi (m-1,n)+\Psi (m,n+1)+\Psi (m,n-1)+V(m,n)\Psi (m,n)=
\]
\[
\lambda \Psi (m,n)\ and\ \Psi (m+a,n)=\xi _1\Psi (m,n),
\]
\[
\Psi (m,n+b)=\xi _2\Psi (m,n)\ for\ all\ (m,n)\ in\ \Bbb{Z}^2\}.
\]
In Section 3 and 4 we give an algorithm for finding the potential $q(x)$ by
the invariants (1), (2) and (4) under the assumption that $q(x)$ is a
trigonometric polynomial. In Theorem 10 we construct a dense in $L_2(F)$ set
$E_0$ of trigonometric polynomials, such that every $q\in E_0$ can be found
by this algorithm. In Theorem 11 we give a scheme for construction of smooth
potential. Then we give an application of Theorem 11 (see Corollary).
\section{On Invariants for Polynomials}
We assume that $q(x)$ is a trigonometric polynomial of the following form
\begin{equation}
q(x)=\sum_{a\in Q_{N,M,S}}q_ae^{i\langle a,x\rangle },\ \ q_a\neq 0,
\end{equation}
where $Q_{N,M,S}=\{n\gamma _1+m\gamma _2+s\gamma _3:|n|\leq N,\ |m|\leq M,\
|s|\leq S\}\backslash \{0\}$ for some basis $\gamma _1,\gamma _2,\gamma _3$
of $\Gamma $. Here $N$, $M$ and $S$ are prime numbers satisfying
\begin{equation}
S>3M+2,\ \ M>2N,\ \ N\gg 1.
\end{equation}
We introduce the following notation
\[
q_a=z(a)=x(a)+iy(a)=r(a)e^{i\alpha (a)},\ \ \ \alpha (a)\in [-\pi ,\pi ).
\]
If $a=n\gamma _1+m\gamma _2+s\gamma _3$, then sometimes we shall write $%
(n,m,s),\ z(n,m,s)$ instead of $a$ and $z(a)$ respectively. We say that the
Fourier coefficient $z(n,m,s)$ corresponds to $(n,m,s)$. For simplicity, we
write $Q$ instead of $Q_{N,M,S}$.
\begin{definition}
A maximal vector $a\in \Gamma $ is said to be long maximal (with respect to $%
Q$) if $a\in Q$, but $2a\notin Q$, i.e., $sa\in Q$\ if\ and\ only\ if\ $%
s=\mp 1.$
\end{definition}
If $a$ is long maximal, then the directional potential $q^a$ has form (3).
Therefore the invariant (1) is
\begin{equation}
\Vert q^a\Vert ^2\equiv 2|q_a|^2.
\end{equation}
Let us consider the invariants (2) and (4) in this case. Let $\beta \in
\Gamma _a\backslash 0$ and $Q(a,\beta )=Q\cap P(a,\beta )$. Putting the
Fourier decompositions (3) of directional potentials $q^a(x)$ into (2), (4)
and using the introduced notation, we get the invariants
\begin{equation}
\mathbf{Re}z(-a)(\sum\limits_{b\in Q(a,\beta )\backslash aR}\frac{%
\left\langle a-b,b\right\rangle }{-(\left\langle b,\beta \right\rangle )^2}%
z(a-b)z(b)),
\end{equation}
\begin{equation}
\mathbf{Re}z^2(-a)(\sum\limits_b\frac{\left\langle a+b,a-b\right\rangle }{%
-(\left\langle b,\beta \right\rangle )^2}z(a+b)z(a-b)),
\end{equation}
where the last sum is taken under the conditions $(a+b),\ (a-b)\in Q(a,\beta
)\backslash aR$. Here $\left\langle b,\beta \right\rangle \neq 0$, since the
line orthogonal to $\beta $ in plane $P(a,\beta )$ is $aR$, but $b\not \in
aR $. If $b\in \Gamma \cap P(a,\beta )$ then the plane $P(a,\beta )$
coincides with the plane $P(a,b)$. Moreover by (12) of [5] every vector $%
b\in \Gamma $ has decomposition $b=\beta +ca,$ where $\beta \in \Gamma
_a,c\in R.$ Hence for every plane $P(a,b)$, where $b\in \Gamma ,$ there
exists the plane $P(a,\beta ),$ where $\beta \in \Gamma _a,$ which coincides
with $P(a,b).$ Therefore for each long maximal $a\in \Gamma $ and for each
plane $P(a,b),$ where $b\in \Gamma ,$ we have the invariants (8), (9).
\begin{lemma}
Let $a$ be a long maximal element of $Q$. If the plane $P(a,b),$ where $b\in
\Gamma ,$ does not contain any vector $c$ except $\pm b$ and $\pm (a-b)$,
such that $c,\ (a-c)\in Q\backslash aR,\ \left\langle c,\ a-c\right\rangle
\neq 0,$ then the invariant (8) for $a$ and the plane $P(a,b)$ has form
\begin{equation}
\mathbf{Re}z(-a)z(a-b)z(b)
\end{equation}
\end{lemma}
\begin{lemma}
If $a$ is long maximal and the plane $P(a,b),$ where $b\in \Gamma ,$ does
not contain any vector $c$ except $\pm b$, such that $c,\ (a-c)\in
Q\backslash aR,\ \left\langle c,\ a-c\right\rangle \neq 0,$ then the
invariant (9) for $a$ and the plane $P(a,b)$ has form
\begin{equation}
\mathbf{Re}z^2(-a)z(a+b)z(a-b).
\end{equation}
\end{lemma}
Finding the planes satisfying the conditions of these lemmas and using the
simple invariants (7), (10), (11), we shall determine the Fourier
coefficients $z(a)$ of polynomial (5) in Sections 3 and 4.
\section{Finding the Fourier Coefficients Corresponding to the Boundary
Points of $Q$}
In this section, we find the Fourier coefficients $z(n,m,s)$ if either $%
n=N,-N$, or $m=M,-M$, or $s=S,-S$ using the following two observations:
1) All boundary points $a=(n,m,s)$ except points $(\pm N,0,0),\ (0,\pm M,0)$%
, $(0,0,\pm S)$ and cases $|n|=|m|=|s|=N$, are long maximal, since $N,M,S$
are distinct prime numbers. Hence, $r(a)$ is known by (7).
2) If $a$ is a boundary point of $Q$, then there are a lot of vectors $b$
such that the plane $P(a,b)$ satisfies the conditions of Lemma 2. That is,
we have the simple invariant (11). So from (7) and (11), we get the value of
$cos(-2\alpha (a)+\alpha (a+b)+\alpha (a-b)).$ In other words, for all $a$
and $b$ satisfying the conditions of Lemma 2, we have the equation
\begin{equation}
-2\alpha (a)+\alpha (a+b)+\alpha (a-b)=d(a,b)e(a,b)(mod2\pi ),
\end{equation}
where $e(a,b)$ is known number in $[0,\pi ]$, $d(a,b)$ is either 1 or -1 and
$c=d(mod2\pi )$ means that $d-c=2k\pi $ for some integer $k$. For simplicity
of notation of $e(a,b)$ and $d(a,b)$ we shall write $e_i$, $d_i$. For
different pairs $(a,b)$ we write different indices (different numbers
1,2,.., as well as, different letters $i,j,l,...$). So we can write a lot of
linear equations with respect to the argument of Fourier coefficients. If we
know the sign of $d(a,b)$ and the values of two summands in the left side of
(12), then we can find the value of third term in the left side of this
equation. To use these equations we need to know the values of argument of
some Fourier coefficients. Three of them can be determined by fixing the
translation $q(x)\rightarrow q(x+\tau )$. It means that we take one of the
functions $q(x+\tau )$, i.e., we fix one value of $\tau $ in $F$. The
conditions
\begin{equation}
\alpha _\tau (N-1,M,S)=\alpha _\tau (N,M-1,S)=\alpha _\tau (N,M,S-1)=0,
\end{equation}
\begin{equation}
\alpha _\tau (N,M,S)\in [0,\frac{2\pi }{N+M+S-1}),
\end{equation}
where $\alpha _\tau (a)=\arg (q(x+\tau ),\ e^{i\langle a,x\rangle })=\langle
a,\tau \rangle +\alpha (a)$, determine a unique value of $\tau $. Indeed,
every $\tau \in F$ has the decomposition $\tau =c_1\omega _1+c_2\omega
_2+c_3\omega _3;$ $c_1,c_2,c_3\in [0,1)$ and the conditions (13) are
equivalent to
\[
2\pi ((N-1)c_1+Mc_2+Sc_3)=-\alpha (N-1,M,S)(mod2\pi ),
\]
\[
2\pi (Nc_1+(M-1)c_2+Sc_3)=-\alpha (N,M-1,S)(mod2\pi ),
\]
\[
2\pi (Nc_1+Mc_2+(S-1)c_3)=-\alpha (N,M,S-1)(mod2\pi ).
\]
This system of equations with respect to the unknowns $c_1,\ c_2,\ c_3\in
[0,1)$ has $N+M+S-1$ solutions $(c_{1,l},\ c_{2,l},\ c_{3,l})$ satisfying
\[
c_{j,l+1}-c_{j,l}=\frac 1{N+M+S-1},\ \ j=1,2,3.\ \ \ l=1,2,...,N+M+S-2.
\]
Hence we obtain $\tau _1,\ \tau _2,...,\ \tau _{N+M+S-1}$ such that $\tau
_{l+1}-\tau _l=\frac{\omega _1+\omega _2+\omega _3}{N+M+S-1}.$ Then
\begin{equation}
\alpha _{\tau _{l+1}}(N,M,S)-\alpha _{\tau _l}(N,M,S)=\frac{2\pi }{N+M+S-1}.
\end{equation}
This together with (14) gives us the unique value of $\tau $. Thus we can
take
\begin{equation}
\alpha (N-1,M,S)=\alpha (N,M-1,S)=\alpha (N,M,S-1)=0,
\end{equation}
\begin{equation}
\alpha (N,M,S)\in [0,\frac{2\pi }{N+M+S-1}).
\end{equation}
On the other hand, the invariants (7) determine the modulus of
\begin{equation}
z(N-1,M,S),\ z(N,M-1,S),\ z(N,M,S-1).
\end{equation}
So the Fourier coefficients (18) are known. Now using (18), we find all
Fourier coefficients $z(n,m,s)$ of $q(x)$ step by step.
\textit{Step 1:} In this step we fix the translation $q(x)\rightarrow q(-x)$
and find
\begin{equation}
z(N,M,s),\ z^2(N,M-1,s),\ z(N,M-2,s)\ \ \forall s,
\end{equation}
considering the following lemma
\begin{lemma}
The plane $P(a,b)$ satisfies the conditions of Lemma 2 in each of the
following cases:
1) $a=(N,M-1,s)$, $b=(0,\pm 1,p)$, where $s+p,\ s-p\in [-S,S]$,
2) $a=(N,m,S-1)$, $b=(0,q,\pm 1)$, where $m+q,\ m-q\in [-M,M]$,
3) $a=(N,m,s)$, $b=(0,\pm 1,p)$, where $m\in [-M+1,M-1]$, $s+p,s-p\in [-S,S]$%
, but $s-2p\not \in [-S,S]$.
\end{lemma}
\textit{Proof:} Let $c=(n,m,l)$ be any nonzero vector satisfying $(a+c),\
(a-c)\in Q$. By definition of $Q$, this condition implies that $n=0$ for all
cases 1-3. Hence $c$ belongs to intersection of planes $P(a,b)$ and $%
\{x_1=0\}\equiv\{(x_1,x_2,x_3)\in R^3:x_1=0\}$. It means that, $c=kb$ for
some nonzero integer $k$. Clearly, if $k$ is not $\pm1$, then either $a+c$
or $a-c$ is not in $Q$ in all cases $\blacksquare$
>From Lemma 2 and Lemma 3, it follows that we have the equation (12) for $%
a=(N,M-1,s)$, $b=(0,1,p)$. By taking $p=1,0,-1$, we get the equations
\[
-2\alpha (N,M-1,s)+\alpha (N,M,s+1)+\alpha (N,M-2,s-1)=d_ie_i(mod2\pi ),
\]
\begin{equation}
-2\alpha (N,M-1,s)+\alpha (N,M,s)+\alpha (N,M-2,s)=d_je_j(mod2\pi ),
\end{equation}
\[
-2\alpha (N,M-1,s)+\alpha (N,M,s-1)+\alpha (N,M-2,s+1)=d_ke_k(mod2\pi ).
\]
Suppose the constants $d_i$, $d_j$, $d_k$ are already determined. Then from
these equations, we can find
\begin{equation}
-2\alpha (N,M-1,l),\ \alpha (N,M,l),\ \alpha (N,M-2,l)\ \mbox{ for }\ l=s-1,
\end{equation}
if (21) are known for $l=s+1,s$. We find (21) for $l=S,S-1,S-2$ as follows.
Substituting $a=(N,M,S-1),\ b=(0,0,1);\ a=(N,M-1,S),\ b=(0,1,0);$%
\[
a=(N,M-1,S-1),\ b=(0,1,-1);\ a=(N,M-1,S-1),\ b=(0,1,0);
\]
\[
a=(N,M-1,S-1),\ b=(0,0,1);\ a=(N,M-1,S-1),\ b=(0,1,1)
\]
into (12) and using (16) we obtain the equations
\[
\alpha +\alpha (N,M,S-2)=d_1e_1(mod2\pi ),
\]
\[
\alpha +\alpha (N,M-2,S)=d_2e_2(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-1)+\alpha (N,M,S-2)+\alpha (N,M-2,S)=d_3e_3(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-1)+\alpha (N,M-2,S-1)=d_4e_4(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-1)+\alpha (N,M-1,S-2)=d_5e_5(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-1)+\alpha +\alpha (N,M-2,S-2)=d_6e_6(mod2\pi ),
\]
where we denote the $\alpha (N,M,S)$ by $\alpha $ and for a while consider
it as known. Here we have used the fact that the planes $P(a,b)$ satisfy the
conditions of Lemma 2 for all above listed values of $a$ and $b$ (see Lemma
3). From these equations, we obtain
\[
\alpha (N,M,S-2)=(d_1e_1-\alpha )(mod2\pi ),
\]
\[
\alpha (N,M-2,S)=(d_2e_2-\alpha )(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-1)=(d_3e_3-d_2e_2-d_1e_1+2\alpha )(mod2\pi ),
\]
\[
\alpha (N,M-2,S-1)=(d_4e_4+d_2e_2+d_1e_1-d_3e_3-2\alpha )(mod2\pi ),
\]
\[
\alpha (N,M-1,S-2)=(d_5e_5+d_2e_2+d_1e_1-d_3e_3-2\alpha )(mod2\pi ),
\]
\[
\alpha (N,M-2,S-2)=(d_6e_6+d_2e_2+d_1e_1-d_3e_3-3\alpha )(mod2\pi ).
\]
Combining these with the equations
\[
-2\alpha (N,M-2,S-1)+\alpha (N,M-2,S)+\alpha (N,M-2,S-2)=d_7e_7(mod2\pi ),
\]
\[
-2\alpha (N,M-1,S-2)+\alpha (N,M,S-2)+\alpha (N,M-2,S-2))=d_8e_8(mod2\pi )
\]
obtained from (12) by taking $b=(0,0,1)$, $a=(N,M-2,S-1)$ and $b=(0,1,0)$, $%
a=(N,M-1,S-2)$, we get
\begin{equation}
f_1(V)=0(mod2\pi ),\ f_2(U)=d_6e_6+d_3e_3(mod2\pi ),
\end{equation}
where $V=(d_1,d_3,d_4,d_6,d_7)$, $f_1(V)\equiv
d_7e_7-(d_6e_6-2d_4e_4+d_3e_3-d_1e_1)$, $U=(d_8,d_5,d_2)$, $f_2(U)\equiv
d_8e_8+2d_5e_5+d_2e_2$ and $d_i$ is either $1$ or $-1$. Hence the vector $V$
takes 32 distinct values $V_1,V_2,...,V_{16}$ and $-V_1,-V_2,...,-V_{16}$.
Then the function $f_1(V)$ takes 32 values $f_1(V_k)$, $f_1(-V_k)$ $%
(k=1,2,...,16)$. Similarly, the vector $U$ takes 8 distinct values $%
U_1,U_2,...,U_8$ and the function $f_2(U)$ takes 8 values $f_2(U_k)$ $%
(k=1,2,...,8)$.
Assume that, for $k\neq j$
\begin{equation}
f_1(V_k)-f_1(Vj),\ \ d_6e_6+d_3e_3,\ f_2(U_k)-f_2(U_j)\neq 0(mod2\pi ).
\end{equation}
(Note that these conditions mean that we have the conditions on invariants
(11), i.e., corresponding Fourier coefficients.) Then there are only one
index $k$ and two values $V_k,\ -V_k$ of $V$ satisfying $f_1(V_k)\equiv
-f_1(-V_k)=0(mod2\pi )$. On the other hand, the argument of Fourier
coefficients of $q(-x)$ takes the opposite value of the argument of Fourier
coefficients of $q(x)$. Therefore, for fixing the translation
\begin{equation}
q(x)\longrightarrow q(-x),
\end{equation}
we take one of these two remaining values $V_k,\ -V_k$ of $V$. Thus, we have
determined the signs of $d_1,\ d_3,\ d_4,\ d_6,\ d_7$ from first inequality
in (23). Since the signs of $d_3$ and $d_6$ are already known we find $d_8$,
$d_5$ and $d_2$ from second and third inequalities in (23). So we found
\begin{equation}
\alpha (N,M-1,S-2)
\end{equation}
and (21) for $l=S,S-1,S-2$. They satisfy the formulae
\[
-2\alpha (N,M-1,S-p)=E_i+2p\alpha ,
\]
\begin{equation}
\alpha (N,M,S-p)=E_j-(p-1)\alpha ,
\end{equation}
\[
\alpha (N,M-2,S-p)=E_k-(p+1)\alpha
\]
for $p=0,1,2$. Further by $E{i},\ E{j},...,$ we denote the linear
combinations of $e_k$ with known integer coefficients. These formulae for
all $p$ can be easily obtained from (20) by induction. Here the signs of $%
d_i $, $d_j$, $d_k$ ( see (20)) can be found as above using the equations
\[
-2\alpha (N,M-1,s)+\alpha (N,M,s+2)+\alpha (N,M-2,s-2)=d_ne_n(mod2\pi ),
\]
\[
-2\alpha (N,M-1,s)+\alpha (N,M,s-2)+\alpha (N,M-2,s+2)=d_me_m(mod2\pi )
\]
obtained from (12) by taking $b=(0,1,2)$, $a=(N,M-1,s)$ and $b=(0,1,-2)$, $%
a=(N,M-1,s)$. In the same way, we obtain the formulae
\begin{equation}
\alpha (N,M-p,S)=E_l-(p-1)\alpha ,
\end{equation}
\begin{equation}
\alpha (0,M,-S)=E_q-(2S+N-1)\alpha .
\end{equation}
Now let us check that the vector $a=(N,M,0)$ and the plane $P(a,b)$, where $%
b=(N,0,S)$ satisfy the conditions of Lemma 1. For this we prove that this
plane contains only the following vectors of $Q$: $0,\ \pm (N,M,0),\ \pm
(N,0,S),\ \pm (0,M,-S).$ In fact, every element $(n,m,s)$ of this plane
satisfies the equation
\begin{equation}
S(nM-mN)=sNM.
\end{equation}
First let us consider the case $s=0$, i.e., the case $nM=mN$. Since $N$ and $%
M$ are distinct prime numbers and $-N\leq n\leq N$, $-M\leq m\leq M$, it
follows that either $n=\pm N$, $m=\pm M$ or $n=m=0$. Now we consider the
case $s\neq 0$. Then the left side of (29) is multiple of $S$. Therefore
taking into account that $S$ is prime number (see (6)) and $-S\leq s\leq S$,
we have $s=\pm S$. This together with (29) gives us the relation $(n\pm
N)M=mN$ and arguing as above, we obtain that either $(n=\pm N),\ m=0$ or $%
n=0,\ m=\pm M$. Hence, using Lemma 1 and formulas (26), (27), (28), from
invariants (7), (10) for $a=(N,M,0)$, $b=(N,0,S)$ we get the value of
$cos(-\alpha (a)+\alpha (b)+\alpha (a-b))=cos((N+M+S-1)\alpha +E_n).$
Similarly, taking $a=(N,M,0),\ b=(N,0,-S)$, we find $cos((N+M+S-1)\alpha
+E_m).$ By these two values of $cos$, we determine $(N+M+S-1)\alpha $ under
the condition
\begin{equation}
E_n\neq E_m(mod\pi )
\end{equation}
This together with (17) gives us the unique value of $\alpha $ and we found
the (19).
\textit{Step 2:} In this step we find
\begin{equation}
z(N,M-1,s)\ \ \forall s.
\end{equation}
These Fourier coefficients are known up to sign (see (19)), i.e., $%
z(N,M-1,s)=d_se_s,$ where $e_s$ is known and $d_s$ is either $1$ or $-1$.
Moreover, $d_s$ is known for $s=S,S-2$ (see (16), (25)). To find it for $%
s=S-2p$, where $p=2,3,...,$ we use the invariant (9) for $a=(N,M-1,S-p)$ and
the plane $P(a,b)$, where $b=(0,0,1)$. Clearly, if $a+c,\ a-c\in P(a,b)\cap
Q $, then $c$ has form $(0,m,s)$. Hence $c$ has to belong to $P(a,b)\cap
\{x_1=0\}\equiv bR$. It means that $c=(0,0,q)$ for some integer $q$.
Therefore the invariant (9) for $a$ and the plane $P(a,b)$ has form
\begin{equation}
\mathbf{Re}\sum\limits_qc_qd_{S-p+q}e_{S-p+q}d_{S-p-q}e_{S-p-q}\equiv f_3(V),
\end{equation}
where $c_q$ are known constants, $V=(h_1,h_2,...,h_N),\
h_q=d_{S-p+q}d_{S-p-q}$. Clearly, $h_q$ is either $1$ or $-1$. Assume that
the function $f_3$ takes distinct nonzero values at distinct points $V$.
Then (32) determine $h_q$ and we find $d_{S-p+q}d_{S-p-q}$. Taking $p=q$, we
find $d_Sd_{S-2p}$. Since $d_S$ is known we find $d_{S-2p}$ and we determine
$z(N,M-1,S-2p)$.
To find $z(N,M-1,S-2p+1)$ we use the following two lemmas.
\begin{lemma}
The plane $P(a,b)$ satisfies the conditions of Lemma 1 if
\[
a=(N,M-1,s),\ b=(0,-1,N),\ S-N\frac N2$, then the plane $P((n,m,s),(0,pn,1))$
contains only the following vectors of $Q$: $q(0,pn,1),\ \pm
(n,m,s)+q(0,pn,1),$ where $q\in Z$ and $p=1,2$.
\end{lemma}
\textit{Proof:} $(a)$ Any element $(n_1,m_1,s_1)$ of this plane satisfy
$S(nm_1-n_1m)=s_1Mn.$ If $(nm_1-n_1m)=0$, then $s_1=0$. If $(nm_1-n_1m)\neq
0 $, then $s_1=\pm S$ since $S$ is prime number which is greater than $n$
and $M$ (see (6)). Hence, $(n_1,m_1,s_1)$ belongs either to $\{x_3=0\}$ or
to $\{x_3=\pm S\}$. To complete the proof, it suffices to apply (34).
($b$) The elements $(n_1,m_1,s_1)$ of this plane satisfy
\begin{equation}
n(pns_1-m_1)=n_1(pns-m).
\end{equation}
The relation $gcd(n,m)=1$ implies that $pns-m\neq 0,\ gcd(n,pns-m)=1$.
Therefore, it follows from inequalities $\left| n\right| >\frac N2$, $\left|
n_1\right| \leq N$ and equality (39) that either $n_1=0$ or $n_1=\pm n$.
This together with (34) gives us the proof $\blacksquare $
\textit{Step 1:} In first step of this section, we find
\begin{equation}
z(n,m,0),\ z(n,0,s),\ z(0,m,s)\ \ \forall n,m,s.
\end{equation}
Let us find $z(n,m,0)$. Without loss of generality it can be assumed that $%
m\geq 0$ and $n>0$. Let $p$ be greatest integer satisfying $pn\leq N$, $%
pm\leq M$. Clearly $q(n,m,0)\in Q$ if and only if $-p\leq q\leq p$. Moreover
$(0,-M,S)+q(n,m,0)\in Q$ if $0\leq q\leq p$ and $(0,M,S)+q(n,m,0)\in Q$ if $%
-p\leq q\leq 0$. It follows from Lemma 7 (see part $a$) that the invariant
(8) for $a_q=(0,-M,S)+q(n,m,0)$ and the plane $P((0,-M,S),(n,m,0))$ has the
form
$\mathbf{Re}z(-(a_q))\sum\limits_lc_lz(l(n,m,0))z(a_q-l(n,m,0)),$ where $%
q=1,2,...,p.$
Similarly, we have $\mathbf{Re}z(-(b_q))\sum%
\limits_ld_lz(l(n,m,0))z(b_q-l(n,m,0)),$ where $b_q=(0,M,S)+q(n,m,0)$; $%
q=-1,-2,...,-p$ and $c_l$, $d_l$ are known constants. Since the Fourier
coefficients $z(a_q),\ z(a_q-l(n,m,0)),\ z(b_q),\ z(b_q-l(n,m,0))$ are known
from Section 2, we have $2p$ linear form with respect to $2p$ unknown $%
x(q(n,m,0))$ and $y(q(n,m,0))$. As a result we find these unknowns if the
corresponding determinant $D(2p)$ is not zero. Let us show that the
determinant is not identically zero. Let $x(q(n,m,0))$ be the $q$-th and $%
y(q(n,m,0))$ be the $(p+q)$-th unknown of the system, where $q=1,2,...,p.$
Similarly let the $q$-th linear form ot the first system be the $q$-th
linear form of the joint system and the $q$-th linear form ot the second
system be the $(p+q)$-th linear form of the joint system . Then the
determinant $D(2p)$ can be written in the form
\[
\left|
\begin{array}{cc}
A & B \\
C & D
\end{array}
\right| ,
\]
where $A=(a_{q,l}),$ $B=(b_{q,l}),$ $C=(c_{q,l}),$ $D=(d_{q,l}),$ $%
q,l=1,2,...,p;$%
\[
a_{q,l}=x(a_q)(c_lx(a_{q-l})+c_{-l}x(a_{q+l}))+y(a_q)(c_ly(a_{q-l})+c_{-l}y(
a_{q+l})),
\]
$%
b_{q,l}=x(a_q)(c_ly(a_{q-l})-c_{-l}y(a_{q+l}))+y(a_q)(c_lx(a_{q-l})-c_{-l}x(
a_{q+l})),
$%
\[
c_{q,l}=x(b_{-q})(d_lx(b_{-q-l})+d_{-l}x(b_{-q+l}))+y(b_{-q})(d_ly(b_{-q-l})
+d_{-l}y(b_{-q+l})),
\]
\[
d_{q,l}=x(b_{-q})(d_ly(b_{-q-l})-d_{-l}y(b_{-q+l}))+y(b_{-q})(d_lx(b_{-q-l})
-d_{-l}x(b_{-q+l})).
\]
The $q$-th and $(p+q)$-th diogonal elements ($a_{q,q}$ and $d_{q,q}$) of the
determinant contain the summand $x(a_q)c_qx(a_0)$ and $x(b_{-q})d_{-q}y(b_0)$
respectively. The nondiogonal elements do not contain neither $x(a_0)$ nor $%
y(b_0).$ Therefore the determinant contain the summand $\sqcap
_{q=1,2,...,p}(x(a_q)c_qx(a_0)x(b_{-q})d_{-q}y(b_0))$ which can not be
canseled by the other summand of the determinant. Note that this determinant
is a homogeneous polynomial of $x(a_q),y(a_q),x(b_{-q}),y(b_{-q}),$ for $%
q=1,2,...,p,$ of degree $4p.$ Therefore the zero set of the determinant is
either zero or a union of conical surfaces.
In the same way we find the other term of (40).
\textit{Remark 1:} Taking into account that the cases $n=0, \pm N;\ \ m=0,
\pm M;\ \ s=0, \pm S$ have been considered and $z(-n,-m,-s)$ is the complex
conjugate of $z(n,m,s)$, further we assume that $n\neq 0,\pm N;\ m\neq 0,
\pm M;\ \ 0\frac N2.
\end{equation}
Since the $z(n,m,s)$ is known for $s=S$ (see Section 2), we find $z(n,m,s)$
for all $(n,m,s)$ satisfying (41) by induction. Assume that these Fourier
coefficients are already known for $s>k>0$. If
\begin{equation}
z(q(0,n,1))=0,\ \ \mbox{for}\ \left| n\right| >\frac N2\ \mbox{ and }\
\left| q\right| >1,
\end{equation}
then by Lemma 7 (see part $(b)$) the invariant (8) for $a=(n,n+m,k+1)$ and
the plane $P((n,m,k),(0,n,1))$ has the form
\[
\mathbf{Re}z(-a)(z(n,m,k)z(0,n,1)+cz(n,m+2n,k+2)z(0,-n,-1)).
\]
Similarly, the invariant (8) for $a=(n,2n+m,k+1)$ and the plane
$P((n,m,k),(0,2n,1))$ has the form
\[
\mathbf{Re}z(-a)(z(n,m,k)z(0,2n,1)+dz(n,m+4n,k+2)z(0,-2n,-1)),
\]
where $c,d$ are known constants. Here all Fourier coefficients are known
except $z(n,m,k)$, since its either were found in step 1 or are known by
induction assumption. Moreover if $m$ and $n$ have different signs then the
condition (6) shows that $-M\frac N2\mbox{ or }gcd(m,s)=1,\ \left|
m\right| >\frac M2.
\]
Here instead the condition (42) we have to use the conditions:
\[
z(q(0,1,n))=0,\ z(q(1,0,m))=0\mbox{ for }|n|>\frac N2,\ \ |q|>1,\
|m|>\frac M2.
\]
\textit{Step 3:} In this step, we find $z(n,m,s)$ satisfying
\begin{equation}
\gcd (n,m)=1,\ \ \left| n\right| <\frac N2,\ \left| \frac mn\right| <\frac{%
M+1}{N+1}.
\end{equation}
For this we need to analyze some property of maximal elements of $Z^2$.
Since $nq-pm=\langle (n,m),(q,-p)\rangle $ from the definition of maximal
elements of $Z^2$ we obtain that $(n,m)$ is maximal if and only if for each $%
l\in Z$ there is $(p,q)\in Z^2$ such that $nq-pm=l$. Denote $%
A_l(n,m)=\{(p,q)\in Z^2:nq-pm=l\},$
$Q(3)=\{(n,m):\left| n\right| \leq N,\ \left| m\right| \leq M\}.$
\begin{definition}
A maximal vector $(n,m)$ is said to be long maximal (with respect to $Q(3)$)
if $(n,m)\in Q(3)$, but $2(n,m)\notin Q(3)$, i.e., if $(n,m,0)$ is long
maximal vector of $Q$. In other words a vector $(n,m)$ of $Q(3)$ is long
maximal if $gcd(n,m)=1$ and at least one of the following inequalities holds
$\left| n\right| >\frac N2$, $\left| m\right| >\frac M2$.
\end{definition}
\begin{lemma}
Let $(n^{\prime },m^{\prime })$ and $(n^{\prime \prime },m^{\prime \prime })$
be longest element of $A_1(n,m)\bigcap Q(3)$ and $A_{-1}(n,m)\bigcap Q(3)$
forming an acute angle with $(n,m)$, where $(n,m)$ is maximal.
(a) The vector $(n^{\prime },m^{\prime })+p(n,m)$ is maximal for all $p\in Z$%
.
(b) If $p$ is positive integer, then $(n^{\prime },m^{\prime })+p(n,m)\not
\in Q(3)$, i.e., at least one of the following inequalities holds $\left|
n^{\prime }+pn\right| >N$,$\left| m^{\prime }+pm\right| >M$.
(c) If $(n,m)$ satisfies (43), then $\left| n^{\prime }+n\right| >N$ hence, $%
\left| n^{\prime }\right| >\frac N2$.
(d) If $\left| n\right| <\frac N2$ and $\left| m\right| <\frac M2$, then $%
(n^{\prime },m^{\prime })$ is long maximal. All statements (a)-(d) are hold,
if we replace $n^{\prime }$ and $m^{\prime }$ by $n^{\prime \prime }$ and $%
m^{\prime \prime }$.
\end{lemma}
\textit{Proof:} $(a)$ By definition of $(n^{\prime },m^{\prime })$ we have
\begin{equation}
n(m^{\prime }+pm)-m(n^{\prime }+pn)=nm^{\prime }-mn^{\prime }=1.
\end{equation}
If the vector $(n^{\prime },m^{\prime })+p(n,m)$ is not maximal for some $p$%
, then $n^{\prime }+pn=lk$ and $m^{\prime }+pm=lq$ for some $l>1$, where $%
k,q\in Z$. Therefore it gives a contradiction in (44) as $l(nq-mk)=1$.
$(b)$ Suppose to the contrary that there is a positive integer $p$ such that
$(n^{\prime },m^{\prime })+p(n,m)\in Q(3)$. Then $(n^{\prime },m^{\prime
})+p(n,m)$ belongs to $A_1(n,m)$, (see (44)) longer than $(n^{\prime
},m^{\prime })$ and forms acute angle with $(n,m)$. However it contradicts
to the definition $(n^{\prime },m^{\prime })$.
$(c)$ If the inequality $\left| n^{\prime }+n\right| >N$ is not true, then
part $(b)$ of this Lemma gives $\left| n^{\prime }+n\right| \leq N$ and $%
\left| m^{\prime }+m\right| >M$. Using these inequalities and (44) for $p=1$%
, we obtain
\[
\left| \frac nm\right| =\left| \frac{n+n^{\prime }}{m+m^{\prime }}+\frac
1{m(m+m^{\prime })}\right| \leq \frac{N+1}{M+1},
\]
which contradicts to (43). Thus $\left| n^{\prime }+n\right| >N$.
$(d)$ follows from $(b)$ by taking $p=1$. Similarly we prove these
statements for $n^{\prime\prime}$ and $m^{\prime\prime}$ $\blacksquare$
Now let us find $z(n,m,s)$ satisfying (43). The invariant (8) for $a^{\prime
}=(n^{\prime },m^{\prime },s-S)$ and the plane $P((n,m,s),a^{\prime })$ has
form
\begin{equation}
\mathbf{Re}z(-a^{\prime })(z(n,m,s)z(n^{\prime }-n,m^{\prime
}-m,-S)+\sum\limits_{a_1}c(a_1)z(a_1)z(a^{\prime }-a_1)).
\end{equation}
Similarly (8) for $a^{\prime \prime }=(n^{\prime \prime },m^{\prime \prime
},s-S)$ and the plane $P((n,m,s),a^{\prime \prime })$ has form
\begin{equation}
\mathbf{Re}z(-a^{\prime \prime })(z(n,m,s)z(n^{\prime \prime }-n,m^{\prime
\prime }-m,-S)+\sum\limits_{a_2}c(a_2)z(a_2)z(a^{\prime \prime }-a_2)),
\end{equation}
where $a_i=(n_i,m_i,s_i)$ and $c(n_i,m_i,s_i)$, ($i=1,2$) are known
constants. Here the sums are taken under the conditions that
\[
(n_1,m_1,s_1),(n^{\prime }-n_1,m^{\prime }-m_1,s-S-s_1)\in
P((n,m,s),(n^{\prime },m^{\prime },s-S))\bigcap Q,
\]
\[
(n_2,m_2,s_2),(n^{\prime \prime }-n_2,m^{\prime \prime }-m_2,s-S-s_2)\in
P((n,m,s),(n^{\prime \prime },m^{\prime \prime },s-S))\bigcap Q,
\]
and
\begin{equation}
(n_1,m_1,s_1)\neq 0,(n,m,s),(n^{\prime },m^{\prime },s-S),(n^{\prime
}-n,m^{\prime }-m,-S),
\end{equation}
\begin{equation}
(n_2,m_2,s_2)\neq 0,(n,m,s),(n^{\prime \prime },m^{\prime \prime
},s-S)(n^{\prime \prime }-n,m^{\prime \prime }-m,-S).
\end{equation}
Let us show that all terms under the sign of the sum in (45) is zero. Since $%
(n,m),\ (n^{\prime },m^{\prime })$ are maximal and $nm^{\prime }-mn^{\prime
}=1$, they form a basis in $Z^2$. Therefore
\begin{equation}
(n_1,m_1)=l(n^{\prime },m^{\prime })+p(n,m),
\end{equation}
for some integers $l$ and $p$. Then the condition
$(n_1,m_1,s_1)\in P((n,m,s),(n^{\prime },m^{\prime },s-S))$ is equivalent to
\begin{equation}
s_1=(s-S)l+sp.
\end{equation}
Without loss of generality, it can be assumed that $l\geq 0$. There are the
cases: $l=1$, $l>1$, $l=0$.
\textit{Case 1.} $(l=1)$ In this case $(n_1,m_1)=(n^{\prime },m^{\prime
})+p(n,m)$. By part $(b)$ of Lemma 8 we need to consider $p\leq 0$. If $p=0$%
, then by (49) and (50), we have $s_1=s-S$ and $(n_1,m_1,s_1)=(n^{\prime
},m^{\prime },s-S)$ which contradicts to (47). If $p=-1$, then again by (49)
and (50), we have $(n_1,m_1,s_1)=(n^{\prime }-n,m^{\prime }-m,-S)$ which
again contradicts to (47). If $p<-1$, then by (50) and Remark 1 we have $%
s_1=-S+(1+p)s<-S$ and $z(n_1,m_1,s_1)=0$.
\textit{Case 2.} ( $l>1$) By (49), we have
\begin{equation}
n_1=ln^{\prime }+pn=l(n^{\prime }+n)+(p-l)n.
\end{equation}
By part $(c)$ of Lemma 8, $\left| n+n^{\prime }\right| >N$. We shall
consider the case $n+n^{\prime }>N$ (The case $n+n^{\prime }<-N$ is
similar). Then both $n$ and $n^{\prime }$ are positive. Hence by (43) and
Remark 1 we have the inequalities
\begin{equation}
n+n^{\prime }>N;\ \ 0<2n1;\ \ 0lN+(p-l)n,\ (l-p)n>(l-1)N,\ l-p>2l-2,
\]
\[
l+p\leq 1,\ s(l+p)\frac 23S$, satisfying
\begin{equation}
gcd(n,m)=1,\ \ \left| n\right| <\frac N2,\ \ \left| \frac mn\right| <\frac{%
M-N}{N+1},
\end{equation}
where $q=1,2,...,q_0$ and the integer $q_0$ is taken so that $q_0(n,m)\in
Q(3)$, but $(1+q_0)(n,m)\not \in Q(3)$. For $q=1$ this Fourier coefficient
is known from step 3, since the condition (53) implies (43). Therefore we
take $2\leq q\leq q_0$. By part $(c)$ of Lemma 8 $\left| n+n^{\prime
}\right| >N$. We shall consider the case $n+n^{\prime }>N$. (The case $%
n+n^{\prime }<-N$ is similar) Then both $n$ and $n^{\prime }$ are positive.
The relation $q_0(n,m)\in Q(3)$ implies that $q_0n\leq N$ and by (53), we
have the inequalities
\begin{equation}
n+n^{\prime }>N;\ \ 0<2n2$
and $l=2$.
\textit{Case 1.} $(l=1)$ In this case $(n_1,m_1)=(n^{\prime },m^{\prime
})+p(n,m)$. By part $(b)$ of Lemma 8 we need to consider $p\leq 0$. If $p<-q$%
, then we have $s_1<-S$ by (59). If $p=-q$, then $s_1=-S$ and $%
(n_1,m_1,s_1)=(n^{\prime }-qn,m^{\prime }-qm,-S)$ which contradicts to (57).
The case $p=0$ also contradicts to (57). So we consider the case $0>p>-q$.
By (44), we have
\begin{equation}
\frac{m^{\prime }+pm}{n^{\prime }+pn}=\frac mn+\frac 1{n(n^{\prime }+pn)}.
\end{equation}
Here $n\neq 0$ by Remark 1. Combining the inequality $-pN-qn\geq
N-q_0n\geq 0$. Now from (60) and (53), we get
\[
\left| \frac{m^{\prime }+pm}{n^{\prime }+pn}\right| <\frac{M+1}{N+1}.
\]
So if $\left| n^{\prime }+pn\right| <\frac N2$, then $z(n_1,m_1,s_1)\equiv
z(n^{\prime }+pn,m^{\prime }+pm,s)$ is known from step 3 (see part $(a)$ of
Lemma 8). If $\left| n^{\prime }+pn\right| >\frac N2$ , then this Fourier
coefficient is known from step 2. Since $-p2$) By (49) we have $-pn=ln^{\prime }-n_1$. Taking into
account that $n_1\leq N$ and using (54) we get $-pn\geq ln^{\prime
}-N>l(N-n)-N=N(l-1)-ln$ and $-p\geq (l-1)q_0-l+1=(l-1)(q_0-1).$ This
together with (59) implies
\begin{equation}
-s_1\geq Sl-sl+\frac{s(l-1)(q_0-1)}q.
\end{equation}
Since $(l-1)\geq 2$, $q_0\geq 2$ and $0S$ and $z(n_1,m_1,s_1)=0$.
\textit{Case 3.} ($l=2$) In this case (61) has form $-s_1\geq 2S+(\frac{q_0-1%
}q-2)s.$ If $q\leq q_0-1$, then $-s_1>S$ due to $sS$, i.e., in these cases $z(n_1,m_1,s_1)=0$%
.
Thus we proved that in all cases except $q=q_0$ and $s>\frac 23S$, all terms
under the sum in (55) (similarly in(56)) either is zero or is known.
Moreover $z(-(n^{\prime },m^{\prime },s-S)),\ z(-(n^{\prime \prime
},m^{\prime \prime },s-S))$ are known from step 2 of this section (see parts
$(a)$, $(c)$ of Lemma 8) and $z(n^{\prime \prime }-qn,m^{\prime \prime
}-qm,-S),\ z(n^{\prime }-qn,m^{\prime }-qm,-S)$ are known from Section 3. By
Lemma 5 the invariants (55), (56) give us $z(qn,qm,s)$ for all cases except $%
q=q_0$ and $s>\frac 23S$.
\textit{Remark 4:} Using Remark 3, replacing the role of second coordinate
by the role of third coordinate and arguing as above we find all $z(qn,m,qs)$%
, except $z(nq_1,m,sq_1)$ for $\left| m\right| >\frac 23M$, satisfying
\[
gcd(n,s)=1,\ \ \left| n\right| <\frac N2,\ \ \left| \frac sn\right| <\frac{%
S-N}{N+1},
\]
where $q=1,2,...,q_1$ and the integer $q_1$ is taken so that $q_1(n,s)\in
Q(2)\equiv \{(n,s):\left| n\right| \leq N,\ \left| s\right| \leq S\}$, but $%
(1+q_1)(n,s)\not \in Q(2).$
Similarly replacing the role of first coordinate by the role of second
coordinate we find all $z(n,qm,qs)$, except $z(n,mq_2,sq_2)$ for $\left|
n\right| >\frac 23N$, satisfying
\[
gcd(m,s)=1,\ \ \left| m\right| <\frac M2,\ \ \left| \frac sm\right| <\frac{%
S-M}{M+1},
\]
where $q=1,2,...,q_2$ and the integer $q_2$ is taken so that $q_2(m,s)\in
Q(1)\equiv \{(m,s):\left| m\right| \leq M,\ \left| s\right| \leq S\}$, but $%
(1+q_2)(m,s)\not \in Q(1).$ Since $S-M>2(M+1)$ (see (6)) we have found $%
z(n,m,s)$ satisfying $\left| s\right| \leq \left| 2m\right| $ and $\left|
n\right| <\frac{2N}3.$
\textit{Step 5:} In this step we find $z(n,m,s)$ satisfying $\left| m\right|
>\frac M2,\ \ gcd(n,m)=1.$
The case $\left| n\right| >\frac N2$ is considered in step 2.Therefore we
consider the case $\left| n\right| <\frac N2$.By Remark 4, these Fourier
coefficients are known for $\left| s\right| \leq 2\left| m\right| $. So we
study the case $s>2\left| m\right| $ (The case $s<-2\left| m\right| $ is
similar). We use the following lemma.
\begin{lemma}
If $(n,m)$ is maximal then for each integer $s$ there are integers $%
s^{\prime }$ and $s^{\prime \prime }$ from interval $[s-\left| m\right| ,s)$
such that $ns^{\prime }-s=mp^{\prime }+1,\ \ ns^{\prime \prime
}-s=mp^{\prime \prime }-1,$ where $p^{\prime }$, $p^{\prime \prime }$ are
integers.
\end{lemma}
\textit{Proof:} Since $(n,m)$ is maximal, there are integers $p$ and $q$
such that $nq-mp=s+1$. Moreover for every integer $k$ we have $%
n(q+km)-m(p+kn)=s+1$. Therefore there is an integer $k^\prime$ such that $%
q+k^\prime m$ lies in $[s-\left|m\right|,s)$ and $n(q+k^\prime
m)-m(p+k^\prime n)=s+1$. Now taking $p^\prime=p+nk^\prime$, $%
s^\prime=q+mk^\prime$ we obtain the first equality of this Lemma. To obtain
the second equality, we take $s-1$ instead of $s+1$ and repeat this argument
$\blacksquare$
Thus in this step we have the inequalities
\begin{equation}
s>2\left| m\right| >M;\ \ \left| m\right| M/2\mbox{ and
}|q|>1.
\end{equation}
Then the invariant (8) for $(n,m,s)$ and the plane $P((n,m,s),(1,0,s^{\prime
}))$ has form
$\mathbf{Re}z(-(n,m,s))(z(n-1,m,s-s^{\prime })z(1,0,s^{\prime })+\
cz(n+1,m,s+s^{\prime })z(-1,0,-s^{\prime })).$
Similarly we have
$\mathbf{Re}z(-(n,m,s))(z(n-1,m,s-s^{\prime \prime })z(1,0,s^{\prime \prime
})+\ dz(n+1,m,s+s^{\prime \prime })z(-1,0,-s^{\prime \prime })).$
Since $|s-s^{\prime }|,\ |s-s^{\prime \prime }|\leq |m|$ (see Lemma 9),and $%
\left| n-1\right| <\frac{2N}3$ the Fourier coefficients $z(n-1,m,s-s^{\prime
})$, $z(n-1,m,s-s^{\prime \prime })$ are known (see the end of Remark 4).
Moreover $z(1,0,s^{\prime })$, $z(1,0,s^{\prime \prime })$ are known from
step 1. If $s=S-1$, then $s^{\prime }+s,\ s+s^{\prime \prime }>S$ ( see(62))
and $z(n+1,m,s+s^{\prime }),\ \ z(n+1,m,s+s^{\prime \prime })=0$. Hence in
case $s=S-1$ all Fourier coefficients in above invariants except $z(n,m,s)$
are known. Therefore it follows from Lemma 5 that these invariants determine
$z(n,m,S-1)$. Now we find $z(n,m,s)$ by induction method. Indeed, if we
assume that $z(n,m,S-2)$, $z(n,m,S-3)$,..., $z(n,m,s+1)$ are known, then we
see that all Fourier coefficients in these invariants, except $z(n,m,s)$,
are known.
\textit{Remark 5:} The combination of step 2 and step 5 gives: The Fourier
coefficient $z(n,m,s)$ is known if $(n,m)$ is long maximal.
\textit{Step 6:} In this step we find all Fourier coefficients. The vectors
of $Q$ can be written in the form $(qn,qm,s)$, where $gcd(n,m)=1$; $%
q=1,2,...,q_0$ and the integer $q_0$ is taken so that $q_0(n,m)\in Q(3)$,
but $(1+q_0)(n,m)\not \in Q(3)$. If $q_0=1$, then $(n,m)$ is long maximal
and $z(n,m,s)$ is known by Remark 5. Therefore we take $q_0\geq 2$. Let us
find $z(qn,qm,s)$ by induction. For $q=1$ we find these Fourier coefficients
repeating step 3. The step 3 is based on step 2. In step 2 $z(n,m,s)$ is
found for all long maximal $(n,m)$ satisfying $|n|>N/2$. Now we repeat step
3 using Remark 5, where $z(n,m,s)$ is found for all long maximal $(n,m)$
without condition $|n|>N/2$. So we again consider the invariants (45), (46).
By Lemma 8 the vectors $(n^{\prime },m^{\prime }),\ (n^{\prime \prime
},m^{\prime \prime })$ are long maximal and $z(n^{\prime },m^{\prime
},s-S),\ z(n^{\prime \prime },m^{\prime \prime },s-S)$ are known (see Remark
5). The Fourier coefficients $z(n^{\prime }-n,m^{\prime }-m,-S),\
z(n^{\prime \prime }-n,m^{\prime \prime }-m,-S)$ are known from Section 3.
To find $z(n,m,s)$ from (45), (46) it suffices to prove that all terms under
the sign of the sums in (45), (46) are zeros. For this we repeat case 1 and
case 3 word by word. Consider case 2. If $|n+n^{\prime }|>N$, then we again
repeat it word by word. If $\left| n+n^{\prime }\right| \leq N$, then we
have $\left| m+m^{\prime }\right| >N$ by part $(b)$ of Lemma 8. Now we
replace the roles of first coordinates $n,\ n^{\prime },\ N$ by the roles of
second coordinates $m,\ m^{\prime },\ M$. More precisely, instead the
inequalities (52) we use
\begin{equation}
m+m^{\prime }>M;\ \ 0<2m1;\ 0\frac 23S$). We only note the following concerning the
repetition of step 4. In case 1, we have $(n_1,m_1)=(n^{\prime },m^{\prime
})+p(n,m)$. By part $(a)$ of Lemma 8 $(n_1,m_1)$ is maximal and $%
z(n_1,m_1,s_1)$ is known. The repetition of remained part of step 4 is
similar to the repetition of step 3. So we have found all Fourier
coefficients, exsept $z(nq_0,mq_0,s)$ for $S>\left| s\right| >\frac 23S$. To
find its ,i.e., $z(a)$ for
\[
a\in Q_1\equiv \{(nq_0,mq_0,s):S>\left| s\right| >\frac 23S,gcd(n,m)=1\}
\]
we consider the invariants (12) in the plane $P((n,m,0),(0,0,1))$ for the
directional potential $q^\delta (x)$, where $\delta $ lies on line $%
(nq_0,mq_0,l)R.$ It follows from (34) that this plane contains only the
vectors $(0,0,s)$ and $\pm (nq,mq,s)$ for $s,q\in Z$. Therefore these
invariants have the form
\[
\mathbf{Re}z(-(nq_0,mq_0,l))(\sum%
\limits_{s_1}z(nq_0,mq_0,s_1)z(0,0,l-s_1)c(s_1)
\]
\[
+\sum%
\limits_{(n_1,m_1,s_1)}z(n_1,m_1,s_1)z(nq_0-n_1,mq_0-m_1,l-s_1)c(n_1,m_1,s_1
)),
\]
where $c(s_1),\ d(s_1),\ c(n_1,m_1,s_1)$ are known constants. Here $%
z(n_1,m_1,s_1)$ and $z(nq_0-n_1,mq_0-m_1,s-s_1)$ are known since $%
(n_1,m_1,s_1)$ and $(nq_0-n_1,mq_0-m_1,s-s_1)$ belong to the plane $%
P((n,m,0),(0,0,1))$ and clearly, have the form $(qn,qm,k),$ where $01,\left| n\right| >\frac
N2,\left| m\right| >\frac M2\}.
\]
Moreover we had the conditions (23), (30) and $f_3(V_j)-f_3(V_k)\neq 0$, for
$j\neq k$ (see (32)), on the argument of the Fourier coefficients.Then we
had conditions (36) and (38). In Section 4 we assumed that some determinants
$D(j)$ of order $j$ are not zero. If $j=2$ then these assumptions are just
(36). In Step 1 and Step 6 of that section we had the conditions $D(2p)\neq
0 $ and $D(4[\frac 13S])\neq 0.$ All the listed conditions, exsept $%
D(4[\frac 13S])\neq 0,$(We call these conditions as Condition A) are related
to the $x(a)$, $y(a)$ for $\Vert a\Vert >\frac N2$. It means that if we
consider the set $E(N,M,S)$ of the polynomials of the form
\begin{equation}
q(x)=\sum_{a\in G(N,M,S)}z(a)e^{i\langle a,x\rangle },
\end{equation}
where $G(N,M,S)=Q_{N,M,S}\setminus (Q_1\cup Q_2)$ (see Step 6) then for $%
\Vert a\Vert <\frac N2$ we have only the condition $z(a)\neq 0.$ Since the
set $G(N,M,S)$ contains the ball $\{\mid \mid (n,m,s)\mid \mid <\frac N2\}$,
the polynomials (66) can be used for appoximation of smooth functions.
Denote by $E_0(N,M,S)$ the set of all elements of $E(N,M,S)$ satisfying
Condition A.
Now consider a triple sequence $\{(N_k,M_k,S_k)\}$ such that for all $k$ the
triple $(N_k,M_k,S_k)$ satisfies the conditions which are satisfied for $%
(N,M,S)$ (see (6)). So $N_k,\ M_k,\ S_k$ are prime numbers satisfying
\[
M_k>2N_k,\ S_k>3M_k+2,\ N_{k+1}>2N_k,\ M_{k+1}>2M_k,\ S_{k+1}>2S_k,\ N_1\gg
1.
\]
Denote by $E(N_k,M_k,S_k)$ and $E_0(N_k,M_k,S_k)$ the set obtained from $%
E(N,M,S)$ and $E_0(N,M,S)$ by substitution $(N_k,M_k,S_k)$ for $(N,M,S)$.
Let
\[
E_0=\bigcup\limits_{k=1}^\infty E_0(N_k,M_k,S_k).
\]
Obviously the set $E_0$ is dense in $L_2(F)$.
\begin{theorem}
Suppose that $q(x)\in E_0$. If the invariants (1), (2), (4) are known, then
one can determine constructively and uniquely ( in $E_{0\text{ }}$modulo
translations $q(x)\rightarrow q(-x),\ q(x)\rightarrow q(x+\tau )$) the
potential $q(x)$.
\end{theorem}
\textit{Proof}: The vector $(N_k,1,0)$ is maximal for each $N_k>1$, i.e.,
the invariants $\Vert q^{(N_k,1,0)}(x)\Vert $ ( see (1)) are given. By
definition of the set $E_0$, the number
\[
k\equiv \{\max k:\ \mbox{ for which }\Vert q^{(N_k,1,0)}(x)\Vert \neq 0\}
\]
is finite. Therefore, $q(x)\in E_0(N_k,M_k,S_k)$. The statement of this
Theorem is proved for this set in Sections 3, 4 $\blacksquare $
Now consider the set $C^p(F)$ of $p$ times differentiable periodic function,
where $p\gg 1$. Assign to every function $q(x)$ of $L_2(F)$ the vector $%
V(q)=(x(a_1),y(a_1),x(a_2),y(a_2),...,)$ of $l_2$, where $x(a_j)$ and $%
y(a_j) $ are real and imaginary part of the Fourier coefficient $z(a_j)$ of $%
q(x)$. Here the vectors $a_1,\ a_2,...$ of the lattice $\Gamma $ are
numerated in non-decreasing order of $\Vert a_j\Vert $. Since $z(-a_j)$ is
the complex conjugate of $z(a_j)$, we include one of the vector $a_j,-a_j$
to the list $a_1,\ a_2,...$. Hence we defined the isometric mapping $V$ from
$L_2(F)$ onto $l_2$ taking $q(x)$ to $V(q)$. For fixing the translations $%
q(x)\rightarrow q(-x),\ q(x)\rightarrow q(x+\tau )$ we set
\begin{equation}
\alpha (a_i)=0,\ \alpha (a_4)\in (0,\pi ),\ or\ \ y(a_i)=0,y(a_4)>0
\end{equation}
where $i=1,2,3$, $a_1=(1,0,0),\ a_2=(0,1,0),\ a_3=(0,0,1),\ a_4=(1,1,0)$.
Denote by $\widetilde{C}^p(F)$, $\widetilde{E}(N_k,M_k,S_k)$ and $\widetilde{%
E}_0(N_k,M_k,S_k)$ the set of all function of $C^p(F)$, $E(N_k,M_k,S_k)$ and
$E_0(N_k,M_k,S_k)$ satisfying (67). In other words $\widetilde{A}$ denotes
the set $A$ with fixed translations $q(x)\rightarrow q(-x),\ q(x)\rightarrow
q(x+\tau )$. Let
\[
\widetilde{l}_2=V(\widetilde{C}^p(F)),\ \widetilde{l}_2(p_k)=V(\widetilde{E}%
(N_k,M_k,S_k)),\ \widetilde{l}_2(p_k,0)=V(\widetilde{E}_0(N_k,M_k,S_k)),
\]
where $p_k$ is the number of elements of $G(N_k,M_k,S_k)$. The sets $%
\widetilde{l}_2(p_k,0)$ and $\widetilde{l}_2(p_k)$ are $(p_k-3)$-dimensional
(see (67)) subsets of $l_2$. The Condition A shows that the set $\widetilde{l%
}_2(p_k,0)$ is obtained from $\widetilde{l}_2(p_k)$ by eliminating a finite
number of conical surfaces $A_1$, $A_2$,..., $A_q$ of dimension less than $%
p_k-3$.
In [5] ( see introduction of [5]) by given Floquet spectrum we
constructively determined the spectral invariants
$J_k(a,\beta ,q)=\int_F|q_{a,\beta }(x)|^2A_k(x)dx,k=0,1,2,...,$ where $%
A_k(x)$ is expressed via the $q^a(x)=Q(s)$ ( $s=\langle x,a\rangle $) and is
defined from the decomposition
$\left| \varphi _{n,v}^a(s)\right| ^2=A_0+\frac{A_1(s)}n+\frac{A_2(s)}{n^2}%
+....$
Here $\varphi _{n,v}^a(s)$ is eigenfunction of the boundary value problem
$-\mid a\mid ^2y^{\prime \prime }(s)+Q(s)y(s)=\mu y(s),$ $y(s+2\pi
)=e^{i2\pi v}y(s).$
In case (3) we have $A_0(x)=1,A_1(x)=0,$ $A_2=\frac{q^a(x)}2+c_1\left|
q_a\right| ^2,$ $A_3=c_2q^a(x)+c_3\left| q_a\right| ^2,$ $%
A_4(x)=c_4q^a(x)+c_5(q_a^2e^{i2\langle a,x\rangle }+q_{-a}^2e^{-i2\langle
a,x\rangle })+c_6$ where $c_1,c_2,...,c_6$ are the known constants.
For $q(x)$ of $\widetilde{C}^{p}(F)$ we denote by $B(q(x))$ the set of all
invariants
\begin{equation}
I_{j}(a,\beta ,q),\ \Vert q^{a}(x)\Vert ,\ \mbox{for}\ j=2,3,4;\ a\in S;\
\beta \in \Gamma _{a},
\end{equation}
where $S$ is the set of all maximal elementsof $\Gamma .$ The set $B(q(x))$
can be considered as the vector of the space $l_{2}$. To determine a
polynomial of $E_{0}(N_{k},M_{k},S_{k})$ we use some invariants of (68). Let
$O_{1}$ and $O_{2}$ be the set of vectors $a$ and pairs $\{a,\beta \}$ such
that the invariants $\Vert q^{a}(x)\Vert $ and $I_{j}(a,\beta )$ $(j=2,3,4)$
are used for finding the elements of $E_{0}(N_{k},M_{k},S_{k})$. For every
function $q(x)$ of $\widetilde{C}^{p}(F)$ we denote by $B_{k}(q(x))$ the set
of invariants $\Vert q^{a}(x)\Vert ,$ for $a\in O_{1},$ and $I_{j}(a,\beta
,q)$ $(j=2,3,4)$ for $\{a,\beta \}\in O_{2}$. Denote by $E_{0}^{\prime
}(N_{k},M_{k},S_{k})$ the set of all vectors $B_{k}(p(x))$, for $p(x)\in
E_{0}(N_{k},M_{k},S_{k})$. This set is a $p_{k}^{\prime }$-dimensional
subset of $l_{2}$ . Clearly $p_{k}^{\prime }\sim p_{k},$ $p_{k}From definition of invariants (68) we see that $B(q(x))$ continuously
depends on $q$. Therefore the mapping $B_k$ taking $q(x)$ to $B_k(q(x))$ is
a continuous mapping from $\widetilde{C}^p(F)$ to $R^{p_k^{^{\prime }}}$.
Its restriction $B_{k,0}$ on $\widetilde{E}_0(N_k,M_k,S_k)$ is a continuous
mapping from $\widetilde{E}_0(N_k,M_k,S_k)$ onto $E_0^{\prime }(N_k,M_k,S_k)$%
. In Sections 3, 4, we proved that the inverse mapping $B_{k,0}^{-1}$
exists. It is continuous on every compact subset of $E_0^{\prime
}(N_k,M_k,S_k)$. (Moreover every coordinate of $V(q)$ was obtained from
coordinates of $B_k(q(x))$ by a finite number of arithmetical operations)
Hence the mapping $B_{k,0}^{-1}$ from $E_0^{\prime }(N_k,M_k,S_k)$ onto $%
\widetilde{E}_0(N_k,M_k,S_k)$ is continuous. Thus for every $q(x)\in
\widetilde{E}_0(N_k,M_k,S_k)$ and $\epsilon _k>0$ there is $\delta _k(q)$
such that
\begin{equation}
\Vert B_k(q)-B_k(\widetilde{q})\Vert <\delta _k(q)\ \Rightarrow \Vert q(x)-%
\widetilde{q}(x)\Vert <\epsilon _k
\end{equation}
for all $\widetilde{q}(x)\in \widetilde{E}_0(N_k,M_k,S_k)$. By continuity of
$B_k$ on $\widetilde{C}^p(F)$ for every $q_k(x)\in \widetilde{E}%
_0(N_k,M_k,S_k)$ there is $\delta _k^{\prime }(q_k)$ such that
\begin{equation}
\Vert q_k(x)-q(x)\Vert <\delta _k^{\prime }(q_k)\ \mbox{ implies }\Vert
B_k(q_k)-B_k(q)\Vert <\frac 12\delta _k(q_k)
\end{equation}
for all $q(x)\in \widetilde{C}^p(F)$. Let $\widetilde{C}_0^p(F)$ be the set
of all functions $q(x)$ of $\widetilde{C}^p(F)$ such that
\begin{equation}
q_k(x)\equiv \sum_{a\in G(N_k,M_k,S_k)}(q(x),e^{i\langle a,x\rangle
})e^{i\langle a,x\rangle }
\end{equation}
is contained in $\widetilde{E}_0(N_k,M_k,S_k)$ and at least one of the
inequalities $\Vert q(x)-q_k(x)\Vert <\delta _k^{\prime }(q_k(x))$ ,
\begin{equation}
\Vert B_k(q_k)-B_k(q)\Vert <\frac 12\delta _k(q_k)
\end{equation}
holds for all $k=1,2,...,$. Suppose $\epsilon _k$ tends to zero as $k$ tends
to infinity.
Denote by $\widetilde{E}_1(N_k,M_k,S_k)$ the set of all $q(x)\in \widetilde{E%
}_0(N_k,M_k,S_k)$ such that (69) is satisfied for
\[
\delta _k(q)=\frac 1{N_k^{p-2}},\ \epsilon _k=\frac 1{\ln (\ln (N_k))}
\]
and for all $\widetilde{q}(x)\in \widetilde{E}_0(N_k,M_k,S_k)$. Let $%
\widetilde{C}_1^p(F)$ be the set of all functions $q(x)$ of $\widetilde{C}%
^p(F)$ such that the corresponding partial sum (71) is contained in $%
\widetilde{E}_1(N_k,M_k,S_k)$ for all $k=1,2,...,$. Direct calculations show
that for all $q(x)\in C^p(F)$ we have
\begin{equation}
\Vert q(x)-q_k(x)\Vert <\frac 1{N_k^p},\ \Vert B_k(q)-B_k(q_k)\Vert <\frac
1{2N_k^{p-2}}.
\end{equation}
\begin{theorem}
If $q(x)\in \widetilde{C}_0^p(F)$ or $q(x)\in \widetilde{C}_1^p(F)$ then $%
q(x)$ can be constructed by $B(q)$, hence by the Floquet spectrum of the
operator $L(q)$. If $q(x),\ \widetilde{q}(x)\in \widetilde{C}_0^p(F)$ or $%
q(x),\ \widetilde{q}(x)\in \widetilde{C}_1^p(F)$ and $B(q)=B(\widetilde{q})$%
, (say, the operators $L(q)$, $L(\widetilde{q})$ have the same Floquet
spectrum) then $q(x)=\widetilde{q}(x)$.
\end{theorem}
\textit{Proof}: Assume that $q(x)\in \widetilde{C}_0^p(F)$ and $B(q)$) is
given. Using definition of $\widetilde{C}_0^p(F)$ and (70), (71), ( 72) we
obtain that there is an element $b_k$ of $E_0^{\prime }(N_k,M_k,S_k)$ such
that
\begin{equation}
\Vert B_k(q)-b_k\Vert <\frac 12\delta _k(q_k).
\end{equation}
By $b_k$, we construct the polynomial $p_k(x)=B_{k,0}^{-1}(b_k)$. Then by
(69), (70) and (74) we have
\begin{equation}
\Vert B_k(q_k)-B_k(p_k)\Vert <\delta _k(q_k),\ \Vert q_k(x)-p_k(x)\Vert
<\epsilon _k.
\end{equation}
Hence $q(x)$ is the limit of the sequence $\{p_k(x)\}$.
Now let $q(x)$ be in $\widetilde{C}_1^p(F)$. Then by (71) and (73) there is
an element $d_k$ of $E_1^{\prime }(N_k,M_k,S_k)$ such that
\begin{equation}
\Vert B_k(q)-d_k\Vert <\frac 1{2N_k^{p-2}},\ \ \Vert B_k(q_k)-d_k\Vert
<\frac 1{N_k^{p-2}}
\end{equation}
By $d_k$ we can construct a function $\widetilde{p}_k(x)=B_{k,0}^{-1}(d_k)$.
Then the inequalities (69), (76) imply that (see definition of $\widetilde{E}%
_1(N_k,M_k,S_k)$)
\[
\Vert q_k(x)-\widetilde{p}_k(x)\Vert <\frac 1{\ln (\ln (N_k))}.
\]
Hence $q(x)$ is the limit of the sequence $\{\widetilde{p}_k(x)\}$.
If $\widetilde{q}(x),\ q(x)$ are in $\widetilde{C}_0^p(F)$ or in $\widetilde{%
C}_1^p(F)$ and $B(q)=B(\widetilde{q})$, then applying this reasoning to $%
\widetilde{q}(x)$ and $q(x)$, we see that both $\widetilde{q}(x)$ and $q(x)$
are the limit of $g_k(x)$ or $\widetilde{p}_k(x)$, where $%
g_k=B_{k,0}^{-1}(e_k)$ and $e_k\in E_0^{\prime }(N_k,M_k,S_k)$ is chosen so
that $\Vert B_k(q)-e_k\Vert <\frac 12\min (\delta _k(q_k),\delta _k(%
\widetilde{q}_k))$. Hence $\widetilde{q}(x)=q(x)$ $\blacksquare $
In this paper, we have shown how the potential can be constructed by
spectral invariants. We do not analyse the sets $\widetilde{C}_0^p(F)$ and $%
\widetilde{C}_1^p(F)$. Analysis of these sets is very technical. It requires
the additional investigations. We shall do it in following papers. Now we
only note the following:
\textit{Remark 6:} It can be proved that the set $\widetilde{E}%
_1(N_k,M_k,S_k)$ has asymptotically full measure on $\widetilde{E}%
(N_k,M_k,S_k)$. It means that the ratio of measures of the sets
\begin{equation}
V(\widetilde{E}_1(N_k,M_k,S_k))\cap \{x\in R_{+}R^{p_k-4}:\Vert x\Vert
N_1.$ Note that there exsist such a number for
every $q(x)\in C^p(F).$
b) $N_k\gg P_{k-1},$ where $k=2,3,...,$and $P_{k-1}$ is a number satisfying $%
P_{k-1}>S_{k-1},P_{k-1}^{5-p}<\frac 12\delta _{k-1}(q_{k-1}).$ Here $\delta
_{k-1}(q_{k-1})$ is defined as $\delta _k(q)$ in (69) , i.e.,
\[
\Vert B_k(q_{k-1})-B_k(\widetilde{q}_{k-1})\Vert <\delta _{k-1}(q_{k-1})\
\Rightarrow \Vert q_{k-1}(x)-\widetilde{q}_{k-1}(x)\Vert <\epsilon _k
\]
for all $\widetilde{q}_{k-1}(x)\in \widetilde{E}_0(N_{k-1},M_{k-1},S_{k-1})$.
Since $\Gamma =G(N_1,M_1,S_1)\cup (\cup _{k=2}^\infty
(G(N_k,M_k,S_k)\backslash G(N_{k-1},M_{k-1},S_{k-1}))$ the Fourier series of
$q(x)$ can be written in the form
\[
q(x)=q_1(x)+\sum_{k=2}^\infty (q_k(x)-q_{k-1}(x))=q_1(x)+\sum_{k=2}^\infty
(q_{k,1}(x)+q_{k,2}(x))
\]
where $q_{k,j}(x)=\sum_{a\in G_{k,j}}z(a)e^{i(a,x)},j=1,$
$G_{k,1}=\{a:\mid a\mid \varepsilon >0,$ $q^a(x)$ and $q_{k-1}^a(x)$ are directional
potentials for $q(x)$ and $q_{k-1}(x)$ respectively. If $a$ is long maximal
with respect $Q(N_{k-1},M_{k-1},S_{k-1}),$ then
\[
\mid q^a(x)-(z(a)e^{i\langle a,x\rangle }+z(-a)e^{i\langle -a,x\rangle
})\mid