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     \def\({\big(}       \def\al{\alpha}           \def\lee{\leqslant}
     \def\){\big)}      \def\e{\varepsilon}    \def\gee{\geqslant}
     \def\[{\left[}       \def\la{\lambda}
     \def\]{\right]}      \def\ffi{\varphi}
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     \def\<{\langle}                 \def\wh{\widehat}
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\def\tr{\operatorname{trace}\,}
                              \def\id{\operatorname{id}\,}
                   \def\det{\operatorname{det}\,}

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\def\Gr{\operatorname{Gr}}
\def\AP{\operatorname{AP}}
\def\BAP{\operatorname{BAP}}
\def\N{\operatorname{N}}
\def\I{\operatorname{I}}
\def\id{\operatorname{id}}
\def\L{\operatorname{L}}
\def\QN{\operatorname{QN}}
\def\J{\operatorname{J}}
\def\R{\operatorname{R}}
\def\dim{\operatorname{dim}}
\def\dual{\operatorname{dual}}
                     \def\sbs{\subset}

            \def\co{\operatorname{co}\,}
\def\span{\operatorname{span}}
        \def\op{\operatorname}
            \def\sptr{\op{sp\,tr\,}}
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                     \newtheorem{defi}{Definition}[section]
                     \newtheorem{theor}{Theorem}[section]
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\title{On $Z_d$-symmetry of spectra of some nuclear operators}
%{}
\author{Oleg Reinov}
\address{St. Petersburg State University,
Saint Petersburg, RUSSIA.}
\email{orein51@mail.ru}

\thanks{%${ }^\maltese$
AMS Subject Classification 2010: 47B06 Riesz operators; eigenvalue distributions;
approximation numbers, s-numbers, Kolmogorov numbers, entropy numbers, etc. of operators.
47B10 Operators belonging to operator ideals (nuclear, p-summing, in the
Schatten-von Neumann classes, etc.).
46B28 Spaces of operators; tensor products; approximation properties.
}
\thanks{${ }$ Key words:  nuclear operator; tensor product; approximation property; eigenvalue.}

\begin{document}

  \maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
     \begin{abstract}
It was shown by M. I. Zelikin (2007) that the spectrum of a nuclear operator in
a Hilbert space is central-symmetric iff the traces of all
odd powers of the operator equal zero.
B. Mityagin (2016) generalized Zelikin's criterium to the case of compact operators
(in Banach spaces) some of which powers are nuclear,
considering even  a notion of so-called $\Bbb Z_d$-symmetry of spectra introduced by him.
We study $\al$-nuclear operators generated by the tensor elements
of so-called $\al$-projective tensor products of Banach spaces,
introduced in the paper ($\al$ is a quasi-norm).
We give exact generalizations of Zelikin's theorem to
the cases of $\Bbb Z_d$-symmetry of spectra of $\al$-nuclear operators (in particular,
for $s$-nuclear and for $(r,p)$-nuclear operators).
We show that the results are optimal.
             \end{abstract}

                  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%   INTRODUCT:
                \section{Introduction} \label{S:intro}
 It is well known that every nuclear (= trace class) operator on a Hilbert space has
 the absolutely summable sequence of eigenvalues \cite{Weyl}. Moreover,
 the famous Lidski\v{\i} theorem \cite{Lid} says that for such an operator its trace
 is equal to the sum of all its eigenvalues (written in according to their algebraic multiplicities).

 It is clear that if the spectrum of such an operator is central-symmetric, then
 its trace equals zero. Moreover, since every power of a nuclear operator $T$
 is nuclear too and has a central-symmetric spectrum  if $T$ has, we see that, for such  $T,$
 $\tr T^k=0$ for every odd natural number $k.$

 M.I. Zelikin has noticed that for an finite dimensional spaces the converse is also true
 (see \cite[Theorem 1]{Zel}), and then he proved the corresponding theorem for any nuclear
 operator in a separable Hilbert space (\cite[Theorem 2]{Zel}).
 At the same time, his proof was rather complicated. We are going to present,
 in particular, a more simple proof below.

 %Recall a definition from \cite[]{Zel}.
Recall that the spectrum of a compact operator is central-symmetric,
if together with any eigenvalue $\la\neq0$ it has the
eigenvalue $-\la$ of the same multiplicity. Thus,  M.I. Zelikin has proved:
The spectrum of a nuclear operator $A$
acting on a separable Hilbert space is central-symmetric iff
$\tr A^{2n - 1} = 0, \, \forall\,  n \in \mathbf N.$

Let us mention that this theorem can not be extended to the case of general
Banach spaces: it follows from Grothendieck-Enflo-Davie results \cite{Gr, Enf, Dav} that
there exists a nuclear operator $T$ in the space $l_1$ of absolutely summable sequences
such that $T^2=0$ but $\tr T=1$ (the operator can be chosen even in such a way,
that it is $s$-nuclear for every $s\in (2/3, 1];$ see Definition 1 below and
\cite[10.4.5]{PiOP}).

A right generalization of  Zelikin's theorem was found by B. Mityagin \cite{Mit}.
He introduced a notion of so-called $Z_d$-symmetry of the spectra of compact
operators in Banach spaces and gave a criterium for
the spectra of an operator (some of which power is nuclear) to be $Z_d$-symmetric.
For $d=2,$ this gives a generalization of the criterium of M.I. Zelikin.
We will use this notion of the $Z_d$-symmetry to formulate and to prove
an {\it exact}\, generalization of Zelikin's theorem for the case
of subspaces of quotients of $L_p$-spaces (thus getting, in a simpler way,
Zelikin's result putting $p=2$ and $d=2).$ However, we will have to consider
so-called $s$-nuclear operators instead of nuclear ones in Zelikin's theorem.
To formulate our main result, let us recall the definitions of $s$-nuclearity
of operators and of $Z_d$-symmetry of spectra.

  \begin{defi}[A. Grothendieck]
An operator $T: X\to Y$ is {\it $s$-nuclear}\, $(0<s\le1),$ if
$$\exists\ (x'_k)\subset X^*, (y_k)\subset Y:\
\sum_{k=1}^\infty ||x'_k||^s\, ||y_k||^s<\infty,
$$
$$
 T(x)= \sum_{k=1}^\infty x'_k(x) y_k,\ \, \forall\ x\in X.
$$
For $s=1,$ they say that $T$ is nuclear.
            \end{defi}

Let us note that A. Grothendieck in \cite{Gr} called such operators
"applications  de puissance p.\'eme sommable".

 \begin{defi}[B. Mityagin]
Let $T$ be an operator in $X,$
all non-zero spectral values of which
are eigenvalues of finite
 multiplicity and have no limit point except possibly zero.
 For a fixed $d=2, 3, \dots$ and for the operator $T,$ the spectrum of $T$
 is called $\mathbb Z_d$-symmetric,
if $0\neq\lambda\in \operatorname{sp}\, (T)$ implies $t\lambda\in \operatorname{sp}\, (T)$ for every $t\in\sqrt[d]{1}$
and of the same multiplicity.
                                \end{defi}

Our generalization of the Zelikin's theorem is:

\begin{theor} \label{T:main0}
%Theorem 1 is optimal with respect to $p$ and $r:$\,
Let $Y$ be a subspace of a quotient
(or a quotient of a subspace) of some
        $L_p(\mu)$-space, $1\le p\le\infty$ and
        $1/r=1+|1/2-1/p|.$
If $T: Y\to Y$ is r-nuclear, then $\tr T$ is well-defined.
For a fixed $d=2, 3, \dots,$ the spectrum of $T$
 is $\mathbb Z_d$-symmetric iff
     $$\operatorname{trace}\, T^{kd+j}=0\, \text{ for all }\, k=0, 1, 2, \dots \,
     \text{ and } \, j= 1, 2, \dots, d-1.$$
%$trace\,  T^{2n - 1} = 0, \, n \in \mathbf N.$
In particular, if $\tr T\neq0,$ then $T^2\neq0.$
\end{theor}

Note that if $d=2,$ we obtain an exact generalization of Zelikin's theorem on
the central symmetry.

Also, we present some sharp (optimal in $r,p)$ generalizations of Zelikin's
theorem to the case of so-called $(r,p)$-nuclear and dually $(r,p)$-nuclear
 operators (see Theorem \ref{T:Zel1}).

 Theorems \ref{T:main0} and \ref{T:Zel1} are optimal with respect to $p$ and $r:$
 \smallskip

\begin{theor} \label{T:sharp}
  Let  $p\in [1,\infty],  p \neq2, $ $1/r= 1+|1/2-1/p|.$
There exists a nuclear operator $V$ in $l_p$  $($in $c_0$ for $p=\infty)$ such that

$1)$\,
$V$ is $s$-nuclear for each $s\in (r, 1];$

$2)$\,
$V$ is not $r$-nuclear;

$3)$\,
$\operatorname{trace}\, V=1$ and $V^2=0.$
\end{theor}


         % \smallskip

Note that for $p=\infty$ or $p=1$ we have $r=2/3$  and
for $p=2$ we have $r=1.$ The proofs will be given in Sections \ref{last2}, \ref{last3}
\smallskip

Let us note that some of the implications of our results on $\Bbb Z$-symmetry
of spectra are the consequences of Mityagin's theorem. But it seems that our
proofs are shorter. Besides, our aim was to obtain the {\it exact}\,
generalizations of Zelikin's theorem in an independent way.


                %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     NOTATIONS:

           % \bigskip


           \section{Content}

In Section \ref{S:nota}, we present the general notations concerning
Banach spaces, spaces of operators, tensor products, vector-valued sequence
spaces.

In first Subsection of Section \ref{S:main4},
we give a definition of projective tensor quasi-norms $\al$ and
introduce the $\al$-projective tensor products of Banach spaces.
We show that these tensor  products are continuously imbedded in the projective
products of A. Grothendieck. For complete $\al$-projective tensor products,
we define $\al$-nuclear operators in a natural way (as elements of corresponding
factor spaces). Also in a natural way, we define a notion of the approximation property
$AP_\al,$ give a simple characterization of Banach spaces with this property
and present some main examples.

In second Subsection of Section \ref{S:main4},
we consider some properties of the $\al$-projective tensor products of spectral type
$l_1$ (so that all $\al$-nuclear operators have absolutely summable sequences
of their eigenvalues). In particular, we are interested in the question of when the
trace formulas are true. In the end, examples are given.

 In third Subsection of Section \ref{S:main4},
we introduce so-called $\al$-extension and $\al$-lifting properties for a
projective tensor quasi-norms $\al.$ W are interested here in connection between
the $AP_\al,$ trace formulas and the statements of type
"$\tr T=1 \implies T^2\neq 0$".

 In forth Subsection of Section \ref{S:main4},
we proof one of the main our theorem on $\Bbb Z_d$-symmetry of spectra of
$\al$-nuclear operators.
We apply the results to some concrete quasi-normed
tensor products, getting a generalization of Zelikin's theorem to
the case of $(r,p,q)$-nuclear operators in general Banach spaces.

 In fifth Subsection of Section \ref{S:main4},
 a proof of Theorem \ref{T:main0} is given.

 Finally, in sixth (last) Subsection of Section \ref{S:main4},
 we show that the main results of the previous subsections
 are sharp. Maybe, it it worthwhile to mention a new result on the
 asymptotically Hilbertian spaces (the last theorem in the paper).

   %%
            \section{Notation and preliminaries} \label{S:nota}
%\centerline{\bf 0. Notation, preliminaries}\,
 %                                         \bigskip
  %\subsection{General notation}
Throughout, we denote by $X,Y,E,F,W\dots$ Banach spaces over a field $\mathbb K$
(which is either $\mathbb R$ or $\mathbb C);$ $X^*, Y^*,\dots$ are
Banach dual to $X, Y, \dots.$ By $x, y, x',\dots$ (maybe with indices) we denote elements of
$X, Y, \dots, Y^*\dots$ respectively. $\pi_Y: Y\to Y^{**}$ is a natural isometric imbedding.
By a subspace of a Banach space we mean a closed linear subspace.

 Notations
 $l_p, (0<p\le\infty, n=1,2,\dots), $ $c_0$ are standard;   $e_k$
 $(k=1,2,\dots)$ is the k-th unit vector in $l_p$ or $c_0$
 (when we consider the unit vectors as the linear functionals, we use
 notation $e'_k$).
We use  $\id_X$ for the identity map in $X.$
                                 %  $BAP, MAP, AP, C-MAP$     %!!!

It is denoted by $F(X,Y)$ a vector space of all linear continuous finite  rank mappings
from $X$ to $Y.$ By $X\ot Y$ we denote the algebraic tensor product of the spaces
$X$ and $Y.$
  $X\ot Y$ can be considered as a subspace of the vector space $F(X^*, Y)$
(namely, as a vector space of all linear weak${^*}$-to-weak continuous
finite rank operators). We can identify also the tensor product (in a natural way)
with a corresponding subspace of $F(Y^*,X).$ If $X=W^*,$ then
$W^*\ot Y$ is identified with $F(W,Y).$
If $z\in X\ot Y,$ then $\wt z$ is the corresponding finite rank operator.
If $z\in X^*\ot X$ and e.g. $z=\sum_{k=1}^n x'_k\ot x_k,$ then
   $\tr z:= \sum_{k=1}^n \<x'_k, x_k\>$ does not depend on representation
   of $z$ in $X^*\ot X.$
 $L(X,Y)$ is a Banach space of all linear continuous mappings ("operators")
 from $X$ to $Y$ equipped with the usual operator norm.
 %We have:
%$X\ot Y\subset F(X^*, Y)\subset L(X^*,Y);$ a completion of the algebraic
%tensor product $X\ot Y$ with respect to the operator norm is denoted by
%$X\wh\ot Y$ ("")


  If $A\in L(X,W)$ and $B\in L(Y,G),$ then  a linear map
  $A\ot B: X\ot Y\to W\ot G$ is defined by
  $A\ot B(x\ot y):= Ax\ot By$ (and then extended by linearity). Since
 $\wt{A\ot B(z)}= B\wt z A^*$ for $z\in X\ot Y,$ we will use notation
$B\circ z \circ A^*\in W\ot G$ for $A\ot B(z).$
In the case where $X$ is a dual space, say $F^*,$ and $T\in L(W,F)$
(so, $A=T^*: F^*\to W^*),$
one considers a composition $B\wt z T;$ in this case $T^*\ot B$ maps
$F^*\ot Y$ into $W^*\ot Y$ and
we use notation
$B\circ z\circ T$ for $T^*\ot B (z).$

A projective tensor product $X\wh\ot Y$ of Banach spaces $X$ and $Y$ is defined
as a completion of $X\ot Y$ with respect to the norm $||\cdot||_{\land}:$
if $z\in X\ot Y,$ then
$
 ||z||_\land:= \inf \sum_{k=1}^n ||x_k||\, ||y_k||,
$
where infimum is taken over all representation of $z$ as $\sum_{k=1}^n x_k\ot y_k.$
We can try to consider
    $X\wh\ot Y$ also as operators $X^*\to Y$ or $Y^*\to X,$ but this
    correspondence is, in general, not one-to-one.
  However, the natural map $(X\ot Y, ||\cdot||_{\land})\to L(X^*, Y)$
is continuous and can be extended to the completion $X\wh\ot Y;$ for
a tensor element $z\in X\wh\ot Y,$ we still denote by $\wt z$ the corresponding
operator.
Note that $X\wh\ot Y= Y\wh\ot X$ in a sense.
If $z\in X\wh\ot Y, \e>0,$ then one can represent $z$ as
$z=\sum_{k=1}^\infty x_k\ot y_k$ with $\sum_{k=1}^\infty ||x_k||\, ||y_k||<||z||_\land+\e.$
For $z\in X^*\wh\ot X$ with a "projective representation" $z=\sum_{k=1}^\infty x'_k\ot x_k,$
 $\tr z:=\sum_{k=1}^\infty \<x_k, y_k\>$ does not depend of representation of $z.$
The Banach dual $(X\wh\ot Y)^*$ equals $L(Y,X^*)$ (with duality $\<T, z\>=\tr T\circ z.)$

One more notation:
If $\frak A$ is an operator ideal \cite{PiOP} then
we often use the notation $\frak A(X)$ for the space $\frak A(X,X).$

Finally,
      $$l_p(X):=\{(x_i)\subset X:\ ||(x_i)||_p:=\(\sum ||x_i||^p)^{1/p}<\infty\},$$
    $$l_\infty(X):=\{(x_i)\subset X:\ ||(x_i)||_\infty:=\sup_i ||x_i||<\infty\},$$
   $$l^w_p(X):=\{(x_i)\subset X:\ ||(x_i)||_{w,p}:= \sup_{||x'||\le1}\(\sum |\<x', x_i\>|^p)^{1/p}<\infty\},$$
   $$l^w_\infty(X):=\{(x_i)\subset X:\ ||(x_i)||_{w,\infty}:=\sup_i ||x_i||<\infty\}.$$
 Note that if $p\le q,$ then $||\cdot||_q\le  ||\cdot||_p$ and $||\cdot||_{w,q}\le  ||\cdot||_{w,p}.$
If $0<p\le\infty,$ then $p'$ is a conjugate exponent:
$1/p+1/p'=1$ if $p\ge1$ and $p'=\infty$ if $p\in (0,1].$




%%%%%%%%%%%%%%%%%%%%%
\section{Quasi-normed tensor products and approximation properties} \label{S:main4}

               \subsection{Projective quasi-norms and approximation properties} \label{S:AP}
    %{\bf 1.1}\
Let $\al$ be a function on a vector space $E,$ $\al: E\to \wh{\Bbb R}.$ We say that $\al$ is
a {\it quasi-norm} on $E$ if
1)\,
$\al(E)\subset [0,+\infty]$ and
$\al(x)=0$ implies $x=0;$
2)\,
there exists a constant $C>0$ such that
$\al(x+y)\le C\, [\al(x)+\al(y)]$ for $x,y\in E;$
3)\,
$\al(ax)=|a|\, \al(x)$ for $a\in \Bbb K, x\in E.$

 \begin{defi}[]
{\rm(i)}\
Given a pair $(E, \al),$ where $\al$ is a quasi-norm on a vector space $E,$
a quasi-normed space associated with the pair $(E, \al)$ is
the quasi-normed vector space
$$
E_\al:= \{x\in E:\ \al(x)<\infty\}.
$$
{\rm(ii)}\,
The quasi-normed space $E_\al$ is complete (= a quasi-Banach space), if
every Cauchy sequence in $E_\al$\, $\al$-converges to an element of $E_\al.$
 \end{defi}

 Note that $E_\al$ is a quasi-normed vector space in the sense of \cite[p. 159]{KOT}
 and we may generate the corresponding topology (see \cite[p. 159-160]{KOT},
 \cite[p. 445]{BenLi}).

\begin{rem}
{\rm 1)\,
 It may be $E_\al=E.$

2)\,
   It is well known \cite[p. 445]{BenLi}
  that if $E_\al$ is a quasi-normed space,
   then there are a number $\beta\in (0,1]$ and a $\beta$-norm $||\cdot||$
   on $E_\al$ which is equivalent to the quasi-norm $\al.$
   Recall that a $\beta$-norm on a vector space $F$
   is a quasi-norm $||\cdot||: F\to \Bbb R$ such that
   for all $x,y\in F$ one has the following $\beta$-triangle
   inequality:
   $||x+ y||^\beta\le ||x||^\beta+||y||^\beta.$ }
\end{rem}

Now, let $\al$ be a quasi-norm on a projective tensor product $X\wh\ot Y$ such that
$\al(x\ot y)=||x||\, ||y||$ for $x\in X, y\in Y.$
The associated quasi-normed tensor product (which will be denoted by $X\wh\ot_\al Y$
and called "$\al$-projective tensor product")
is the $\al$-closure of $X\ot Y$ in $(X\wh\ot Y)_\al$
(in the concrete cases we will use some specific notations).
Thus,
$$
X\wh\ot_\al Y:= \{u\in X\wh\ot Y:\ \al(u)<\infty \ \text{ and }\
   \exists\, (u_n)\subset X\ot Y:\ \al(u-u_n)\underset{n\to\infty}\to 0\}.
$$
More generally:
\begin{defi}[]
{\rm (i)}\,
Let $\wh\ot$ denotes the class of all tensor elements of the projective
tensor products of arbitrary Banach spaces.
A projective tensor quasi-norm $\al$ is a map from $\wh\ot$ to $\wh{\Bbb R}$
such that $\al$ is a quasi-norm on each component $X\wh\ot Y$ with the properties:

 $(Q_1)$\,
$\al(x\ot y)=||x||\, ||y||$ for $x\in X, y\in Y.$

$(Q_2)$\,
There exists a constant $C>0$ such that
$\al(u_1+u_2)\le C\, [\al(u_1)+\al(u_2)]$
for all $X, Y$ and $u_1, u_2\in X\wh\ot Y.$

$(Q_3)$\,
If $u\in X\wh\ot Y,$\, $A\in L(X, E)$ and $B\in L(Y,F),$ then
$\al(A\ot B (u))\le ||A||\, \al(u)\, ||B||.$

%$(Q_4)$\,
%For all $X, Y$ the tensor product $X\ot Y$ is dense in $X\wh\ot_\al Y.$

{\rm (ii)}\,
A projective tensor quasi-norm $\al$ ia said to be complete, if every
$\al$-projective tensor product $X\wh\ot_\al Y$ is complete, that is
quasi-Banach.
\end{defi}

For every projective tensor quasi-norm $\al$ there exist $\beta\in (0,1]$
and an equivalent  $\beta$-norm $||\cdot||_\beta$ on $\wh\ot$
so that $X\wh\ot_\alpha Y= X\wh\ot_{||\cdot||_\beta} Y$
(i.e. there exists a quasi-norm $||\cdot||_\beta$ with $\beta$-triangle inequality such
that for some positive constants $C_1, C_2$ and for all projective tensor elements
$u$ the inequalities
$C_1 \al(u)\le ||u||_\beta\le C_2 \al(u)$ hold). Thus, we may assume, if needed,
that a priori $\al$ is a $\beta$-norm.

We are not going to consider here in detail the properties of just introduced objects.
But we need below the fact that   %!!! Engl- OK?
  %{\bf TO PROVE? --- to proof all anal.PiOP p 70h-89 or In separate paper?} SEE BELOW!!!
the inclusions $X\wh\ot_\alpha Y\hookrightarrow X\wh\ot Y$ are continuous for all Banach spaces $X, Y$
(in the main Example \ref{E:rpq} below this will be automatically fulfilled).

\begin{prop} \label{P:cont}
Let $\al$ be a complete projective tensor norm.
 The natural injections $X\wh\ot_\alpha Y\to X\wh\ot Y$ are continuous for all
 Banach spaces $X$ and $Y.$ Moreover, there is a constant $d= d(\al)$ such that
 for all $X,$ $Y$ and $u\in X\wh\ot_\al Y$ we have: $||u||_\land \le d\,\al(u).$
\end{prop}

\begin{proof}
Suppose the last assertion is not true and there exist the sequences
$(X_n), (Y_n)$ and $(u_n)$ with $u_n\in X_n\wh\ot_\al Y_n$
so that $\al(u_n)\le 1/(2C)^{n}$ and $||u_n||_\land\ge n.$
Put $X:= \(\sum X_n\)_{l_2}$ and $Y:= \(\sum Y_n\)_{l_2}.$ %$u:= \(\sum u_n\)_{l_2}$
Let $i_n: X_n\to X$ and $j_n: Y_n\to Y$ be the natural injections.
Consider the sequence $(z_N):= \(\sum_{k=1}^N (i_k\ot j_k)(u_k)\).$ For any natural
numbers $K$ and $m,$ we have:
$$
  \al\(\sum_{k=K+1}^{K+m} (i_k\ot j_k)(u_k)\)\le
  \sum_{k=1}^{m} C^k\,\al\((i_{K+k}\ot j_{K+k})(u_{K+k})\)\le
  \sum_{k=1}^{\infty} \frac{C^k}{(2C)^{K+k}}\le \frac1{(2C)^K}.
  $$
Hence, $(z_N)$ is a Cauchy sequence in $X\wh\ot_\al Y$ and,
by the completeness of $\al,$ converges to an element
$u:= \sum_{k=1}^\infty (i_k\ot j_k)(u_k)\in X\wh\ot_\al Y.$
On the other hand,
if $P_n: X\to X_n$ and $Q_n: Y\to Y_n$ are the natural "projections", then
$||u||_{\land} \ge||(P_n\ot Q_n)(u)||_{\land} =||u_n||_\land \ge n.$
\end{proof}

Since $X\wh\ot_\al Y$ ia a linear subspace of $X\wh\ot Y,$
the space $L(Y,X^{*})$ separates points of $X\wh\ot_\al Y.$
If $u\in X\wh\ot_\al Y,$ then $u=0$ iff $\tr U\circ u=0$ for
every $U\in L(Y,X^{*}).$ In particular, the dual space $(X\wh\ot_\al Y)^*$
separates points of $X\wh\ot_\al Y.$
      %{\bf DENSITY OF F-R TENSORS THERE???}

It is clear that every tensor element $u\in X\wh\ot_\al Y$ generates
a nuclear operator $\wt u: X^*\to Y.$ If $X$ is a dual space, say $E^*,$
then we get a canonical mapping $j_\al: E^*\wh\ot_\al Y\to L(E, Y).$
The image of $j_\al$ is denoted here by $N_\al (E,Y),$ and we equip it
with an "$\al$-nuclear" quasi-norm $\nu_{\al}:$ This is a quasi-norm
induced from $E^*\wh\ot_\al Y$ via the quotient map $E^*\wh\ot_\al Y\to N_{\al}(E,Y).$
If the projective tensor quasi-norm $\al$ is complete, then $N_\al(E,Y)$
is a quasi-Banach space.

\begin{defi}[]
  Let $\al$ be a complete projective tensor quasi-norm. We say that
  a Banach space $X$ has the approximation property $AP_\al,$ if
  for every Banach space $E$ the canonical map $E^*\wh\ot_\al X\to N_\al(E,X)$
  is one-to-one (in other words, if $E^*\wh\ot_\al X= N_\al(E,X)$).
\end{defi}

Note that if $\al=||\cdot||_\land,$ then we get the classical approximation property $AP$ of
A. Grothendieck \cite{Gr}. It must be clear that the $AP$ implies the $AP_\al,$
for any projective tensor quasi-norm.

We will need below the following

\begin{lem} \label{L:AP}
A Banach space $X$ has the $AP_\alpha$ iff
the canonical map $X^*\widehat\otimes_\alpha X\to L(X)$ is one-to-one.
\end{lem}

\begin{proof}
It is enough to repeat (word for word with same notations) the proof
of \cite[Proposition 6.1]{Trend}.
\end{proof}

\begin{exam} \label{E:rpq}
{\rm
   Let $0<r, s\le1,$ $0< p,q \le\infty$ and $1/r+1/p+1/q=1/\beta\ge1.$
 We define a tensor product $X\wh\ot_{r,p,q} Y$ as a linear subspace of
 the projective tensor product $X\wh\ot Y,$ consisting of all tensor elements $z$
 which admit representations of type
 \begin{equation}
  z=\sum_{k=1}^\infty \al_k x_k\ot y_k,\
  (\al_k)\in l_r,\, (x_k)\in l_{w,p}(X),\, (y_k)\in l_{w,q}(Y);      \label{E0}
\end{equation}
 we equip it with the quasi-norm $||z||_{r,p,q}:= \inf ||(\al_k)||_r\,
  ||(x_k)||_{w,p}\, ||(y_k)||_{w,q},$ where the infimum is taken over all
  representations of $z$ in the above form (\ref{E0}).
 Note that this tensor product is $\beta$-normed (cf. \cite{166}, where it is
 considered a "finite-sums-representation" version of the above tensor product). %!!! Engl - OK?
 It is quasi-Banach (for the completeness, see the author's preprint
"Approximation properties associated with quasi-normed operator ideals
of $(r,p,q)$-nuclear operators"\footnote{
 http://www.mathsoc.spb.ru/preprint/2017/index.html\#08}).
 The corresponding quasi-normed operator ideal $N_{r,p,q}$ is the quasi-Banach ideal of
 $(r,p,q)$-nuclear operators (cf. \cite{PiOP, 166}).              %!!!ref
 In particular cases where one or two of the exponents $p,q$
are $\infty,$ we will use the notations close to those from \cite{RQ, Trend}
(here we change $p', q'$ to $p, q):$
 We denote              %!!!ref
$N_{r,\infty,\infty}$ by $N_r,$\,
$N_{r,\infty,q}$ by $N_{[r,q]},$ \,
$N_{r,p,\infty}$ by $N^{[r,p]},$\,
$\wh\ot_{r,\infty,\infty}$ by $\wh\ot_{r},$\,
$\wh\ot_{r,\infty,q}$ by $\wh\ot_{[r,q]},$\,
$\wh\ot_{r,p,\infty,}$ by $\wh\ot^{[r,p]}.$

The corresponding notations are used also for the $AP_{r,p,q}:$

(i)\,
For $p=q=\infty,$ we get the $AP_r$ from \cite{Trend}.              %!!!

(ii)\,
For $p=\infty,$ we get the $AP_{[r,q]}$ from \cite{RQ, Trend}.              %!!!

(iii)\,
For $q=\infty,$ we get the $AP^{[r,p]}$ from \cite{RQ, Trend}.

    \begin{comment}
 1.\,
 For Banach spaces $X$ and $Y,$ let $X\wh\ot_s Y$ be a subspace of the projective tensor product
$X\wh\ot Y$ consisting of the tensors $u\in X\wh\ot Y,$ which admit representations of the form
\begin{equation}
u= \sum_{n=1}^\infty \la_n x_n\ot y_n,      \label{E1}
\end{equation}
where $(\la_n)\in l_s,$\, $(y_n)$ and $(x_n)$ are bounded sequences from $Y$ and $X$
respectively. With a natural quasi-norm
$||u||=\inf ||(\la_n)||_{l_s}\, \sup_n||y_n||\, \sup_n||x_n||,$
it is a projective quasi-Banach tensor product.

2.\,  Consider (\ref{E1})
       \end{comment}
}
\end{exam}

We will need some known facts concerning the approximation properties from
Example \ref{E:rpq}. Let us collect them in

\begin{lem} \label{L:Lp}
{\rm 1)} \cite[Corollary 10]{RQ14}\,
  Let $s\in (0,1],$ $p\in [1,\infty]$ and $1/s=1+|1/p-1/2|.$
    If  a Banach space $Y$ is isomorphic to a
      subspace of a quotient (or to a quotient of a subspace)
        of an $L_p$-space then it has the property $AP_s.$

{\rm2)} \cite[Corollary 4.1]{RQ}, \cite[Theorem 7.1]{Trend}\,
 Let   $1/r-1/p=1/2.$ Every Banach space has the properties $AP_{[r,p']}$
and $AP^{[r,p']}.$
\end{lem}

A proof of the assertion 2) can be found below (see Example \ref{E:main}).
 See also \cite{Trend} for some other results in this direction.

 \begin{rem} \label{R:AP}  {\rm
   As a matter of fact, a proof of the assertion that every Banach space has the $AP^{[1,2]}$
is contained implicitly in \cite{PiOP}. It was obtained also there that
this assertion (after applying some results of Complex Analysis)
implies the Grothendieck-Lidski\v{i} type trace formulas for operators from $N^{[1,2]}$
\cite[27.4.11]{PiOP} (and this implies the Lidski\v{i} trace formula for trace-class
operators in Hilbert spaces and the Grothendieck trace formula for $N_{2/3}$ as well).
On the other hand, there is a very simple way
to get these results on  $AP^{[1,2]}$ and $N^{[1,2]}$ from the Lidski\v{i} theorem
(see the proofs of \cite[Theorems 7.1-7.3]{Trend} for $p=2).$
 }
 \end{rem}




 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \subsection{Spectral type $l_1$}


 Let $T$ be an operator in $X,$
all non-zero spectral values of which
are eigenvalues of finite
 multiplicity and have no limit point except possibly zero.
 Put $\la(T)= \{\la\in \text{eigenvalues}\,(T)\setminus\{0\}\}$ (the eigenvalues of
 $T$ are taken in according to their multiplicities).
 We say that an operator $T\in L(X,X)$ is {\it of spectral type $l_1,$}
 if the sequence of all eigenvalues $\la(T):= (\la_k(T))$ is
 absolutely summable. In this case, we can define the {\it spectral trace of}\, $T:$\,
 $\sptr(T):= \sum \la_k(T).$
 We say that a subspace $L_1(X,X)\subset L(X,X)$ is {\it of spectral type $l_1,$}
 if every operator $T\in L_1(X,X)$ is of spectral type $l_1.$
 Recall that an operator ideal $\frak A$ is of spectral type $l_1,$
 if every its component $\frak A(X,X)$ is of spectral type $l_1.$

\begin{defi}
Let $\al$ be a projective tensor quasi-norm.
The tensor product $X\wh\ot_\al X$ is of spectral type $l_1,$ if
the space $N_\al(X,Y)$ is of spectral type $l_1.$
The projective tensor quasi-norm $\al$ (or the tensor product $\wh\ot_\al)$
is of spectral type $l_1,$
if the corresponding operator ideal $N_\al$ is of spectral type $l_1.$
\end{defi}

\begin{exam} \label{E:sptr} {\rm
  $N_1(H)$\, ($= N_{[1,2]}(H)= N^{[1,2]}(H)= S_1(H),$
  trace class operators in a Hilbert space) is of spectral type $l_1$ \cite{Weyl}.
  $\wh\ot_{2/3}$ and $N_1\circ N_1$ are of spectral type $l_1$ \cite{Gr}.
  $N^{[1,2]}$ is of spectral type $l_1$ (see \cite[see 27.4.9, end of the proof]{PiOP}).
  $N_{[1,2]}$ is of spectral type $l_1$ (see \cite[Theorem 7.2 for $p=2$]{Trend};
  it follows also from the previous assertion). More general, if $1/r-1/p=1/2,$
  then $\wh\ot_{[r,p]}= N_{[r,p]},$ $\wh\ot^{[r,p]}= N^{[r,p]}$ and they are
  of spectral type $l_1$ (see \cite[Theorems 7.1-7.3]{Trend}).
  }
\end{exam}

Let us note that in all cases in Example
\ref{E:sptr} the trace formula for corresponding operators
(say, $T)$ is valid: $\tr T=\sptr T.$
A general result in this direction is

\begin{prop} \label{P:sptype}
Let $\al$ be a complete projective tensor quasi-norm of spectral type $l_1.$
For every Banach space $X$ with the $AP_\al$ and every $T\in N_\al(X),$
one has: $\tr T=\sptr T.$ Conversely, if for a Banach space $X$ and for
every $z\in X^*\wh\ot_\al X$ the equality $\tr z= \sptr \wt z$ holds, then $X$
possesses the $AP_\al.$
   \end{prop}

   \begin{proof}
    % Зам: к-м не предполаг плотны в $\wh\ot_\al$        - what if???
Let $X$ has the $AP_\al.$
Since the ideal $N_\al$ is quasi-Banach and of spectral type $l_1,$
by White's theorem \cite[Theorem 2.2]{Whit} the spectral trace is linear
and continuous on $N_\al.$ On the other hand,
by Proposition \ref{P:cont} the usual (nuclear) trace is continuous
on $X^*\wh\ot_\al X,$ which can be identified with $N_{\al}(X)$
by assumption about $X.$ Since the tensor product $X^*\ot X$ is dense
in $X^*\wh\ot_\al X,$ we obtain that $\tr T= \sptr T.$

Now, suppose that $X$ does not have the $AP_\al.$ By Lemma \ref{L:AP},
there exists an element $z\in X^*\wh\ot_\al X$ such that
$\tr z=1$ and $\wt z=0.$ By assumptions, $\sptr \wt z=\tr z=1.$
Contradiction.
     \end{proof}

\begin{exam} \label{E:main}           {\rm
Let $0<r\le1,$ $1\le p\le2,$ $1/r=1/2+1/p.$

1)\,
If $T\in N_{[r,p']}(X)$ (see Example \ref{E:rpq}),
then $T$ admits a factorization
$$
T=BA: X\overset{A}\to l_p\overset{B}\to X,\ \, A\in N_r(X, l_p),\, B\in L(l_p, X).
$$
The complete systems of eigenvalues of $T=BA$ and $AB$ are the same.
But $AB\in N_r(l_p,l_p).$ Therefore, $AB$ is of spectral type $l_1,$
as any $r$-nuclear operator in $l_p$ \cite[Theorem 7]{Kon}.
It follows from this that $N_{[r,p']}$ is of spectral type $l_1.$
It is easy to see that if $z\in X^*\wh\ot_{[r,p']} X$ such that $\wt z=T,$
then $\tr z=\tr AB$ (recall that $l_p$ has the $AP).$ But $\tr AB=\sptr AB$
(it was shown, e.g., in \cite{RQ1, Trend} and follows also from
Proposition \ref{P:sptype}). Hence, for each $z\in X^*\wh\ot_{[r,p']} X$
we have: $\tr z= \sptr \wt z.$ By the second part of Proposition \ref{P:sptype},
every Banach space has the property $AP_{[r,p']}$ $( = AP_{r,\infty,p'},$\,
see Example \ref{E:rpq}; thus, we gave a proof of Lemma \ref{L:Lp}, 2) for
the case of $AP_{[r,p']}).$

2)\,
If $T\in N^{[r,p']}(X)$ (see Example \ref{E:rpq}),
then $T$ admits a factorization
$$
T=BA: X\overset{A}\to l_p\overset{B}\to X,\ \, A\in L(X, l_p),\, B\in N_r(l_p, X).
$$
As in 1), we see that
for each $z\in X^*\wh\ot^{[r,p']} X$
we have: $\tr z= \sptr \wt z.$ Furthermore,
by the second part of Proposition \ref{P:sptype},
every Banach space has the property $AP^{[r,p]}$ $( = AP^{r,\infty,p'},$\,
see Example \ref{E:rpq}; thus, we have a proof of Lemma \ref{L:Lp}, 2) for
the case of $AP^{[r,p']}).$               }
\end{exam}


 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \subsection{$\al$-extension property and $\al$-lifting property}
We give now two definitions, which will be of use below. Let us note that these
definitions can be generalized in many different ways.

\begin{defi} \label{D:ext}
 Let $\al$ be a complete projective tensor quasi-norm.
 A Banach space $X$ has the $\al$-extension
property, if for any subspace $X_0\subset X$ and every tensor element
$z_0\in X_0^*\wh\ot_\al X_0$ there exists an extension $z\in X^*\wh\ot_\al X_0$
(so that $z\circ i= z_0$ and $\tr i\circ z= \tr z_0,$ where $i: X_0\to X$ is the natural injection).
A Banach space $X$ has the $\al$-lifting property, if for every subspace
$X_0$ and every tensor element $z_0\in (X/X_0)^*\wh\ot_\al X/X_0$ there exists a lifting
$z\in (X/X_0)^*\wh\ot_\al X$ (so that $Q\circ z=z_0,$ where $Q$ is a quotient map from $X$
onto $X/X_0,$ and $\tr z\circ Q= \tr z_0).$
\end{defi}

    \begin{exam} \label{E:ext}    {\rm
   For instance, every Banach space has the $||\cdot||_{r,\infty,q}$-extension property and
$||\cdot||_{r,p,\infty}$-lifting property (see Example \ref{E:rpq}).
For the tensor products $(\wh\ot_s, ||\cdot||_{s,\infty,\infty}),$ $s\in (0,1],$
all Banach spaces have both the $||\cdot||_{s,\infty,\infty}$-extension
and $||\cdot||_{s,\infty,\infty}$-lifting properties.
This follows from Hahn-Banach theorem and from definition of Banach quotients.
 }
             \end{exam}

\begin{prop} \label{P:S2}
 Let $\al$ be a complete projective tensor quasi-norm and $X$ have the $\al$-extension
property. Suppose that $X$ possesses the $AP_\al,$ but there exists a subspace
$X_0\subset X$ without the $AP_\al.$ There exists an operator $S\in N_\al(X)$
such that $\tr S=1$ and $S^2=0.$
\end{prop}

\begin{proof}
Take $z_0\in X_0\wh\ot_\al X_0$ with $\tr z_0=1$ and $\wt z_0=0$
(we use Lemma \ref{L:AP}). By assumption, there exists $z\in X^*\wh\ot_\al X_0$
such that $z_0= z\circ i$ and $\tr i\circ z=1,$ where $i: X_0\to X$ is an inclusion. Here is
a diagram for the operators:
\begin{equation}
X_0\overset{i}\to X\overset{\wt z}\to X_0\overset{i}\to X.    \label{Eq:dia}
 \end{equation}
Now, $X$ has the $AP_\al.$ Therefore, we can identify the operator $S:=\wt{i\circ z}$ with
the tensor element $i\circ z.$ It is clear that $\tr S=1$ and $S^2=0.$
\end{proof}

The following proposition is a strengthening of Proposition \ref{P:sptype}
in an important case.

 \begin{prop} \label{P:ext}
Let $\al$ be a complete projective tensor quasi-norm of spectral type $l_1$
and $X$ have the $\al$-extension property. If for
every $z\in X^*\wh\ot_\al X$ the equality $\tr z= \sptr \wt z$ holds,
then every subspace $X_0$ of $X$ possesses the $AP_\al.$ Consequently, for
every $T\in N_\al(X_0),$ one has: $\tr T=\sptr T.$
 \end{prop}

\begin{proof}
Firstly, note that by Proposition \ref{P:sptype} $X$ has the $AP_\al.$
Let $X_0$ be a subspace of $X,$ $i: X_0\to X$ be an inclusion map and
$z_0\in X_0^*\wh\ot_\al X_0$ with $\tr z_0=1.$  Take an extension $z\in X^*\wh\ot _\al X_0$
(as in Definition \ref{D:ext}; hence, $\wt z|_{X_0}=\wt z_0$ and $\tr i\circ z= \tr z_0)$
and consider the operators $\wt{i\circ z}: X\to X$ and  $\wt{z\circ i}: X_0\to X_0$
(see the  diagram (\ref{Eq:dia})). By the principle of related operators \cite[27.3.3]{PiOP},
$\sptr \wt{i\circ z}=\sptr \wt{z\circ i}.$ By assumption, $\sptr \wt{i\circ z}= \tr {i\circ z}$
Now, since $X$ has the $AP_\al,$ it follows from the equality $\tr i\circ z= \tr z_0$ that
$$1= \tr z_0= \sptr \wt{i\circ z}= \sptr \wt{z\circ i}= \sptr \wt{z_0}.
$$
Therefore, $\wt z_0\neq 0.$
By Lemma \ref{L:AP}, $X_0$ has the $\AP_\al.$
The last statement follows from the first part of Proposition \ref{P:sptype}.
 \end{proof}

 The following propositions are in a sense dual the previous ones.

\begin{prop} \label{P:S2l}
 Let $\al$ be a complete projective tensor quasi-norm and $X$ have the $\al$-lifting
property. Suppose that $X$ possesses the $AP_\al,$ but there exists a factor space
$X/X_0$ $(X_0\subset X)$ without the $AP_\al.$ There exists an operator $S\in N_\al(X)$
such that $\tr S=1$ and $S^2=0.$
\end{prop}

\begin{proof}
Take $z_0\in X/X_0\wh\ot_\al X/X_0$ with $\tr z_0=1$ and $\wt z_0=0$
%(we use Lemma \ref{L:AP}).
By assumption, there exists $z\in (X/X_0)^*\wh\ot_\al X$
such that $Q\circ z=z_0,$ where $Q$ is a factor map from $X$
onto $X/X_0,$ and $\tr z\circ Q= \tr z_0=1.$
 Here is a diagram for the operators:
\begin{equation}
X\overset{Q}\to X/X_0\overset{\wt z}\to X\overset{Q}\to X/X_0\overset{\wt z}\to X.    \label{Eq:dia2}
 \end{equation}
Now, $X$ has the $AP_\al.$ Therefore, we can identify the operator $S:=\wt{z\circ Q}$ with
the tensor element $z\circ Q.$ It is clear that $\tr S=1$ and $S^2=0.$
\end{proof}

 \begin{prop}  \label{P:lif}
Let $\al$ be a complete projective tensor quasi-norm of spectral type $l_1$
and $X$ have the $\al$-lifting property. If for
every $z\in X^*\wh\ot_\al X$ the equality $\tr z= \sptr \wt z$ holds,
then every quotient $X/X_0$ of $X$ possesses the $AP_\al.$ Consequently, for
every $T\in N_\al(X/X_0),$ one has: $\tr T=\sptr T.$
 \end{prop}

\begin{proof}
By Proposition \ref{P:sptype}, $X$ has the $AP_\al.$
Let $X_0$ be a subspace of $X,$ $Q: X\to X/X_0$ be a factor map and
$z_0\in (X/X_0)^*\wh\ot_\al X/X_0$ with $\tr z_0=1.$  Take a lifting $z\in (X/X_0)^*\wh\ot _\al X$
(as in Definition \ref{D:ext}; hence, $Q\circ z=z_0,$ and $\tr z\circ Q= \tr z_0)$
and consider the operators $\wt{z\circ Q}: X\to X$ and  $\wt{Q\circ z}: X/X_0\to X/X_0$
(see the  diagram (\ref{Eq:dia2})). By the principle of related operators \cite[27.3.3]{PiOP},
$\sptr \wt{z\circ Q}=\sptr \wt{Q\circ z}.$ By assumption, $\sptr \wt{z\circ Q}= \tr {z\circ Q}.$
Now, since $X$ has the $AP_\al,$ it follows from the equality $\tr z\circ Q= \tr z_0$ that
$$1= \tr z_0= \sptr \wt{z\circ Q}= \sptr \wt{Q\circ z}= \sptr \wt{z_0}.
$$
Therefore, $\wt z_0\neq 0.$
By Lemma \ref{L:AP}, $X_0$ has the $\AP_\al.$
The last statement follows from the first part of Proposition \ref{P:sptype}.
 \end{proof}

An immediate consequence of Propositions \ref{P:ext} and \ref{P:lif} is

\begin{prop}
Let $\al$ be a complete projective tensor quasi-norm of spectral type $l_1$ such that
every Banach space has both the $\al$-extension property and the $\al$-lifting property.
If for every $z\in X^*\wh\ot_\al X$ the equality $\tr z= \sptr \wt z$ holds,
then every quotient of any subspace of $X$ (= every subspace of any quotient of $X$)
possesses the $AP_\al.$ Consequently, for $X_0\subset X_1\subset X,$ $Y= X_1/X_0$
(or for $X_0\subset X,$ $Y\subset X/X_0)$ and for
every $T\in N_\al(Y)$ one has: $\tr T=\sptr T.$
\end{prop}

\begin{proof}
 Apply in different orders Propositions \ref{P:ext} and \ref{P:lif}.
\end{proof}



 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
    \subsection{Applications. $\Bbb Z_d$-symmetry for $N_{[r,p]}$ and $N^{[r,p]}$} \label{last1}
One of our main result (in context of the $\Bbb Z_d$-symmetry of the spectra
of nuclear operators) is

\begin{theor}  \label{T:main}
Let $\al$ be a complete projective tensor quasi-norm of spectral type $l_1$
and let a Banach space $X$ have the $AP_\al.$
For a fixed $d=2, 3, \dots,$ the spectrum of an operator
$T\in N_\al(X)$ is $\Bbb Z_d$-symmetric if and only if
   $$\operatorname{trace}\, T^{kd+j}=0\, \text{ for all }\, k=0, 1, 2, \dots \,
     \text{ and } \, j= 1, 2, \dots, d-1.$$
In particular, if $\tr T\neq0,$ then $T^2\neq0.$
\end{theor}

\begin{proof}
Let the spectrum of an operator $T\in N_\al(X)$ is $\Bbb Z_d$-symmetric.
The traces $\tr T^n$\, $(n\in \Bbb N)$ are well defined since $T^n\in N_\al(X)$
and $X$ has the $AP_\al.$ Take an integer $l:= kd+j$ with $0<j<d.$
The eigenvalue sequences of $T$ and $T^l$ can be arranged in such a way that
$\{\lambda_n(T)^l\}= \{\lambda_n(T^l)\}$ (see \cite[3.2.24, p. 147]{PiEig}).
Since the spectrum of ${T}^l$ is absolutely summable,
$\operatorname{trace}\, T^l= \sum_{\lambda\in \operatorname{sp}\, (T^l)} \lambda,$ \,
$\sum_{t\in\sqrt[d]{1}} t=0$
 and we may assume that $\{\lambda_m(T^l)\}= \{\lambda_m(T)^l\},$ we get that
 $\operatorname{trace}\, T^{kd+j}=0.$

 To prove the converse, we need some information from Fredholm Theory.
Let $u$ be an element of the projective tensor product $Y^*\widehat\otimes Y,$
where $Y$ is an arbitrary Banach space.
 Recall that the Fredholm determinant $\operatorname{det}\, (1-wu)$ of $u$
 (see \cite[Chap II, \S 1, $n^{\circ} 4,$ p. 13]{Gr}, \cite{Gr56}, \cite{PiOP} or \cite{PiEig}
 %[2, Chap II, \S 1, $n^{\circ} 4$, p. 13] or [3], [8], [9])
  is an entire function
$$
 \operatorname{det}\, (1-wu)= 1-w\, \operatorname{trace}\, u+ \dots + (-1)^n w^n\alpha_n(u) +  \dots,
$$
all zeros of which are exactly (according to their multiplicities) the inverses of nonzero eigenvalues
of  the operator $\widetilde{u},$ associated with the tensor element $u.$
By \cite[Chap II, \S 1, $n^{\circ} 4$, Corollaire 2, pp. 17-18]{Gr},
this entire function is of the form
$$
  \operatorname{det}\, (1-wu) =
  e^{-w\operatorname{trace}\, u} \prod_{i=1}^\infty (1- w w_i)\, e^{ww_i},
$$
where
$\{w_i=\lambda_i(\widetilde u)\}$ is a complete sequence of all eigenvalues of the operator $\widetilde{u}.$
Hence, there exists a $\delta>0$ such that for all $w, |w|\le\delta,$ we have

\begin{equation}
     \operatorname{det}\, (1-wu) =
  \exp {\(\sum_{n=1}^\infty c_n\, w^n \operatorname{trace}\, u^n\)}         \label{(1)}
\end{equation}
(see \cite[p. 350]{Gr56}; cf. \cite[Theorem I.3.3, p. 10]{Goh}).  %[1, Theorem I.3.3, p. 10]).

  Now, let $\operatorname{trace}\, T^{kd+r}=0$ for all $k=0, 1, 2, \dots$ and $r= 1, 2, \dots, d-1.$
 By (\ref{(1)}),
  $\operatorname{det}\, (1-wT) =
  \exp {(\sum_{m=1}^\infty c_{md}\, w^{md} \operatorname{trace}\, T^{md})}$
   in a neighborhood $V$ of zero.
   Hence, for the analytic function $f(w):= \operatorname{det}\, (1-wT),$ we have:
 there exists a $\delta>0$ such that for all $w, |w|\le\delta,$
 $f(tw)=f(w)$ for every $t\in \sqrt[d]{1}.$ By the uniqueness theorem,
   the complete system of eigenvalues of $T$ is $\Bbb Z_d$-symmetric.
\end{proof}

Applying Theorem \ref{T:main} to the tensor products $\wh\ot_{[r,p']}$ and
$\wh\ot^{[r,p']}$ and using Example \ref{E:main}, we get the following
generalizations of Zelikin's theorem:

\begin{theor} \label{T:Zel1}
Let $0<r\le1,$ $1\le p\le2,$ $1/r=1/2+1/p$ and $d=2, 3, \dots.$
For any Banach space $X$ and every operator $T\in N_{[r,p']}(X)$
(or $T\in N^{[r,p']}(Z)$) we have:
The spectrum of an operator
$T\in N_\al(X)$ is $\Bbb Z_d$-symmetric if and only if
   $$\operatorname{trace}\, T^{kd+j}=0\, \text{ for all }\, k=0, 1, 2, \dots \,
     \text{ and } \, j= 1, 2, \dots, d-1.$$
In particular, if $\tr T\neq0,$ then $T^2\neq0.$
\end{theor}

We obtain Zelikin's theorem, if we put $d=2,$ $r=1, p=2$ and $X=H$ (a Hilbert space),
since $N_1(H)= S_1(H)=N_{[1,2]}(H)=N^{[1,2]}(H).$



\subsection{Proof of Theorem \ref{T:main0}} \label{last2}
  Here it is

\begin{proof} %[of Theorem \ref{T:main0}]
Let $T\in N_r(Y).$
Under the conditions of the theorem we have:
every quotient of every subspace of an $L_p$-space
has the $AP_r,$ $\la(T)\in l_1$ and the trace of $T$ is well defined and equals the sum
of the eigenvalues of $T$ (written in according to their multiplicities;
see, e.e., \cite{RQ14, Trend}).
%Since the spectra of the operators ${T}^l, l=1, 2, \dots,$  are absolutely summable,

Supposing that the spectrum of $T$ is $\mathbb Z_d$-symmetric,
we can proceed as in the beginning of the proof of Theorem \ref{T:main} to obtain
that
$\operatorname{trace}\, T^{kd+j}=0\, \text{ for all }\, k=0, 1, 2, \dots \,
     \text{ and } \, j= 1, 2, \dots, d-1.$

To proof the converse, we repeat word for word the second part of the proof
of Theorem \ref{T:main}.
\end{proof}

%%%%%%%%%

\subsection{Sharpness of main results} \label{last3}
We need the following auxialary result:

\begin{lem} \label{L:3}
If $r\in [2/3, 1), q\in (2,\infty]$ and $1/r=3/2-1/q,$
then there exist a subspace $Y_q\subset l_q$ $(c_0$ for $q=\infty)$
and a tensor element $w_q\in Y_q^*\wh\ot Y_q$ so that
$w_q\in  Y_q^*\wh\ot^{[s,q]} Y_q$ for every $s>r,$
$\tr w_q=1$ and $\wt w_q=0.$
Moreover, $w_q$ can be chosen in such a way that
$w_q=\sum_{k=1}^\infty e'_k|_{Y_q}\ot y_k,$ where $(e'_k)$ is a sequence of the
linear functionals on $l_q$ generated by the unit vectors from $l_{q'}$ and
$(y_k)$ is in $l_s(Y_q)$ for all $s>r.$
\end{lem}

\begin{proof}
Let us look at the proof from \cite[Example 2]{RJMAA} and take the space $Y_q$
and the tensor element $w_q$ from that proof.
We have: $Y_q$ is isometrically imbedded into $l_q,$
$w_q=\sum_{k=1}^\infty e'_k|_{Y_q}\ot y_k,$ where $(e'_k)$ and $(y_k)$ are
as above.
\end{proof}

The following two theorems show that Theorem \ref{T:Zel1} is optimal.
%%
\begin{theor}
  Let $r\in [2/3, 1),$  $q\in (2,\infty], $ $1/r= 3/2-1/q.$
There exists a nuclear operator $V$ in $l_q$  $($in $c_0$ for $q=\infty)$ such that

$1)$\,
$V\in N^{[s,q]}(l_q)$ for each $s\in (r, 1];$

$2)$\,
$V$ is neither in $N^{[r,q]}(l_q)$ nor $r$-nuclear;

$3)$\,
$\operatorname{trace}\, V=1$ and $V^2=0.$
\end{theor}

 \begin{proof}
Take a pair $(Y_q, w_q)$ from Lemma \ref{L:3} and let $i: Y_q\to l_q$ be
an isometric imbedding. Define $v\in l_q^*\wh\ot l_q$ by
$v= \sum_{k=1}^\infty e'_k\ot iy_k$ and put $V:= \wt v.$
This operator possesses the properties 1)--3)
(we have to mention only that $N^{[r,q]}(l_q)\subset N^r(l_q)$
%
%$AP^{[r,q]}$ (Example \ref{E:main})
and that if $T\in N_r(l_q)$ with $\tr z=1,$ then $T^2\neq0$
by Theorem \ref{T:main0}).
 \end{proof}


%%
 \begin{theor}
  Let $r\in [2/3, 1),$  $p\in [1,2), $ $1/r= 1/2+1/p.$
There exists a nuclear operator $U$ in $l_p$ such that

$1)$\,
$U\in N_{[s,p']}(l_p)$ for each $s\in (r, 1];$

$2)$\,
$U$ is neither in $N_{[r,p']}(l_p)$ nor $r$-nuclear;

$3)$\,
$\operatorname{trace}\, V=1$ and $V^2=0.$
\end{theor}

\begin{proof}
Consider $U:=V^*,$ where $V$ is from the previous theorem.
\end{proof}
%%


Now, {\it Theorem \ref{T:sharp} follows from the above theorems,
since, e.g., $N^{[s,q]}\subset N^s$}.

\smallskip

One more auxiliary fact:

\begin{lem} \label{L:4}
Let $r\in(2/3,1], q\in[2,\infty), 1/r=3/2-1/q.$ One can find
the number sequences $(q_k)$ and $(n_k)$ with $q_n>q, q_n\to q$ and
$k_n\to\infty$ for which the following statement is true:
There exist a Banach space $Y_0$ and a tensor element
$w\in Y_0^*\widehat\otimes_r Y_0$ so that
$Y_0\subset Y:= \left(\sum_k l_{q_k}^{n_k}\right)_{l_q},$\,
$w\neq0, \tilde w=0,$ the space $Y_0$ (as well as $Y_0^*)$
has the $AP_s$ for every $s<r$
(but does not have the  $AP_{r,\bar q}$ for any $\bar q\in(q,\infty]).$
Moreover, $w$ can be chosen in such a way that
$w=\sum_{k=1}^\infty \sum_{m=1}^{n_k} e'_{mk}|_{Y_0}\ot y_{mk},$ where $(e'_{mk})$ is a
weakly $\bar q$-summable $(\forall\,\bar q>q)$ sequence of the
linear functionals on $Y$ generated by the unit vectors from $Y^*$
and
$(y_{mk})$ is in $l_r(Y_0)\setminus \cup_{s<r} l_s(Y_0).$
\end{lem}

\begin{proof}
It is enough to take the space $Y_0$
and the tensor element $w$ from the proof of \cite[Example 1]{RJMAA} and put
$n_k:= 3\cdot 2^k$ in that proof. After this we get exactly the desired Banach space
and tensor element.
We have also: $Y_0\subset Y\subset l_{\bar q}$ for every $\bar q>q.$ Hence,
the sequence $(e'_{mk}|_{Y_0})$ is weakly $\bar q$-summable $(\forall\,\bar q>q).$
\end{proof}
%%



            %   \begin{comment}
\begin{theor} \label{T:l3}
Let $r\in(2/3,1], q\in[2,\infty), 1/r=3/2-1/q.$
One can find
the number sequences $(q_k)$ and $(n_k)$ with $q_k>q, q_k\to q$ and
$n_k\to\infty$ for which the following statement is true:

There exists a nuclear operator $U$ in $Y:=\left(\sum_k l_{q_k}^{n_k}\right)_{l_q}$
 such that

$1)$\,
$U\in N^{[r,\bar q]}(Y)$ for each $\bar q>q.$

$2)$\,
$U$ is not in $N^{[r,q]}(Y).$

$3)$\,
$\operatorname{trace}\, U=1$ and $U^2=0.$
\end{theor}
     %    \end{comment}

\begin{proof}
Take a pair $(Y_0, w)$ from Lemma \ref{L:4} and let $j: Y_0\to Y$ be
an injection map. Define $u\in Y^*\wh\ot Y$ by
$u= \sum_{k=1}^\infty \sum_{m=1}^{n_k} e'_{mk}\ot jy_{mk}$ and put $U:= \wt u.$
This operator possesses the properties 1)--3)
(we have to mention only %that every Banach space has the property
%$AP^{[r,q]}$ (Example \ref{E:main}),
that if $T\in N_{[r,q]}(Y)$ with $\tr z=1,$ then $T^2\neq0$
by Theorem \ref{T:Zel1}).
\end{proof}
%%

\begin{theor} \label{T:l4}
  Let $r\in (2/3, 1],$  $p\in (1,2], $ $1/r= 1/2+1/p.$
One can find
the number sequences $(p_k)$ and $(n_k)$ with $p_k<p, p_k\to p$ and
$n_k\to\infty$ for which the following statement is true:

There exists a nuclear operator $V$ in $E:=\left(\sum_k l_{p_k}^{n_k}\right)_{l_p}$
 such that

$1)$\,
$V\in N_{[r,\bar q]}(E)$ for each $\bar q>q.$

$2)$\,
$V$ is not in $N_{[r,q]}(E).$

$3)$\,
$\operatorname{trace}\, V=1$ and $V^2=0.$
\end{theor}

\begin{proof}
Consider $V:=U^*,$ where $U$ is from the previous theorem.
\end{proof}


%%
Let us emphasize an important particular case of Theorems
\ref{T:l3} and \ref{T:l4}, namely, the case of so-called
"asymptotically Hilbertian spaces" (see, e.g., \cite{CGJ}
for a definition):

\begin{theor}
There exist an asymptotically Hilbertian space
$Y_2:=\left(\sum_k l_{q_k}^{n_k}\right)_{l_2}$
$(q_k\to 2$ and $n_k\to\infty)$
and a nuclear operator $U$ in this space so that

$1)$\,
$U\in N^{[1,2+\e]}(Y_2)$ for each $\e>0.$

$2)$\,
$U$ is not in $N^{[1,2]}(Y_2)$.

$3)$\,
$\operatorname{trace}\, U=1$ and $U^2=0.$

The corresponding statements hold for the adjoint operator $U^*.$
 \end{theor}

As we know, the last theorem is the best strengthening of related results from
\cite{CGJ}, \cite{RFA} and \cite{RJMAA}.



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