%%%%%%%%%% % %%%%% 7ECM Oct 31, 2016 20:36 %%%%%%%%%%%%%% %%%% %%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%% !!! %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[12pt,oneside]{amsproc} \usepackage[T2A]{fontenc} %\usepackage[cp1251]{inputenc} \usepackage[english]{babel} %\usepackage[russian]{babel} \usepackage{amssymb} \usepackage{amsthm} \textheight=22cm \textwidth=15cm \oddsidemargin=5mm \topmargin=-5mm %%%%%%%%%%%%%%% %%%%%%%%%%% \title{Some remarks on spectra of nuclear operators } \author{Oleg I. Reinov} \address{Saint Petersburg State University } \email{orein51@mail.ru} \begin{document} \maketitle %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{abstract} We give criteria for the spectra of some nuclear operators in subspaces of quotients of $L_p$-spaces to be central-symmetric, as well as for the spectra of linear operators in Banach spaces to be $Z_d$-symmetric in the sense of B. Mityagin. Also, we present a short proof of a generalization of Mityagin's theorem. \end{abstract} %%%%%%%%%%%%%%%%%%%%% {\bf 1.\, Introduction}.\ It was shown by M. I. Zelikin in [16] that the spectrum of a nuclear operator in a separable Hilbert space is central-symmetric iff the spectral traces of all odd powers of the operator equal zero. Recall that the spectrum of every nuclear operator in a Hilbert space consists of non-zero eigenvalues of finite algebraic multiplicity, which have no limit point except possibly zero, and maybe zero. This system of all eigenvalues (written according to their multiplicities) is absolutely summable, and the spectral trace of any nuclear operator is, by definition, the sum of all its eigenvalues (taken according their multiplicities). The space of nuclear operators in a Hilbert space may be defined as the space of all trace-class operators (see [14, ñ. 77]); % [3, Theorem 8.1]; %!!! check eng pages in this case we say about the "nuclear trace" of an operator). Trace-class operators in a Hilbert space can be considered also as the elements of the completion of the tensor product of the Hilbert space and its Banach dual with respect to the greatest crossnorm on this tensor product [14, ñ. 119]. The well known Lidski\v{\i} theorem [5] %(see also [3, Theorem 8.4]) %!!! eng pages and V.\,B.~Lidski\v{\i} says that the nuclear trace of any nuclear operator in a Hilbert space (or, what is the same, of the corresponding tensor element) coincides with its spectral trace. Thus, Zelikin's theorem [16] can be reformulated in the following way: the spectrum of a nuclear operator in a separable Hilbert space is central-symmetric iff the nuclear traces of all odd powers of the corresponding tensor element are zero. One of the aim of our notes is to give an exact generalization of this result to the case of tensor elements of so-called $s$-projective tensor products of subspaces of quotients of $L_p(\mu)$-spaces. In particular, we get as a consequence Zelikin's theorem (taking $p=2).$ Another problem which is under consideration in our notes is concentrated around the so called $\mathbb Z_d$-symmetry of the spectra of the linear operators. The notion of $\mathbb Z_d$-symmetry of the spectra was introduced by B.S. Mityagin in a preprint [6] %!!!REF and in his paper [7]. %!!!REF He is interested there in a generalization of the result from [16] %!!!REF in two directions: to extend Zelikin's theorem to the case of general Banach spaces and to change the property of a compact operator to have central-symmetric spectrum to the property of the one to have $\mathbb Z_d$-symmetric spectrum. Roughly speaking, $\mathbb Z_d$-symmetry of a spectrum of a compact operator $T$ means that for any non-zero eigenvalue $\lambda$ of $T$ the spectrum contains also as eigenvalues of the same algebraic multiplicities all "$d$-shifted"\, numbers $t\lambda$ for $t\in \sqrt[d]{1}.$ B.S. Mityagin has obtained a very nice result, showing that the spectrum of a compact operator $T$ in an arbitrary Banach space, some power $T^m$ of which is nuclear, is $\mathbb Z_d$-symmetric iff for all large enough integers of type $kd+r$ $(0< r < d)$ the nuclear traces of $T^{kd+r}$ are zero. We present some thoughts around this theorem, giving, in particular, a short (but not so elementary as in [6, 7]) proof for the case where the operator is not necessarily compact. Let us mention, however, that the proof from [6, 7] %!!!REF can be adapted for this situation too. Some words about the content of the paper. In Section 2, we introduce some notations, definitions and terminology in connection with so called $s$-projective tensor products, $s$-nuclear operators and the approximation properties of order $s,$\, $s\in (0, 1].$ We formulate here two auxiliary assertions from the author's paper [13]; they give us generalized Grothendieck-Lidski\v{\i} trace formulas which will be useful in the next section. Section 3 contains an exact generalization of Zelikin's theorem. In this section we present a criterium for the spectra of $s$-nuclear operators in subspaces of quotients of $L_p$-spaces to be central-symmetric. Results of Section 4 show that the criterium of the central symmetry, obtained in the previous section, is optimal. In particular, we present here (Theorem 2) sharp examples of $s$-nuclear operators $T$ in the spaces $l_p,$\, $1\le p\le+\infty,$\, $p\neq2,$ for which $\operatorname{trace}\, T=1$ and $T^2=0.$ Finally, Section 5 is devoted to the study of Mityagin's $\mathbb Z_d$-symmetry of the spectra of linear operators. Our aim here is to give a short (but using Fredholm Theory) proof of Mityagin's theorem [6-7] for arbitrary linear continuous (Riesz) operators. Firstly, we consider a $\mathbb Z_2$ situation (central symmetry) to clarify an idea which is to be used then in the general case. We finish the paper with a short proof of the theorem from [6-7] for continuous (not necessarily compact) operators and with some simple examples of applications. \bigskip %%%%% {\bf 2.\, Preliminaries}.\ We denote by $X, Y, \dots$ Banach spaces, $L(X, Y)$ is a Banach space of all linear continuous operators from $X$ to $Y;$ $L(X):= L(X, X).$ For a Banach dual to a space $X$ we use the notation $X^*.$ If $x\in X$ and $x'\in X^*,$ then $\langle x',x\rangle$ denotes the value $x'(x).$ By $X^*\widehat\otimes X$ we denote the projective tensor product of the spaces $X^*$ and $X$ [2] (see also [11, 12]). %% !!! It is a completion of the algebraic tensor product $X^*\otimes X$ (considered as a linear space of all finite rank continuous operators $w$ in $X)$ with respect to the norm $$ ||w||_{\land}:= \inf \{\left(\sum_{k=1}^N ||x'_k||\, ||x_k||\right):\ w=\sum_{k=1}^N x'_k\otimes x_k\}. $$ Every element $u$ of the projective tensor product $X^*\widehat\otimes X$ can be represented in the form $$ u=\sum_i \lambda_i x'_i\otimes x_i, $$ where $(\lambda_i)\in l_1$ è $||x'_i||\le1,$ $||x_i||\le1$ [2]. %!!! More generally, if $01,$ the nuclear trace $\operatorname{trace}\, T^n$ is well defined (see [2, Chap II, \S 1, $n^{\circ} 4,$ Corollaire 2, p. 16]) and equal the sum of all its eigenvalues (according their multiplicity) [2, Chap II, \S 1, $n^{\circ} 4$, Corollaire 1, p. 15]. Therefore, if the spectrum of a nuclear operator $T: X\to X$ is central-symmetric, then for each odd $m=3, 5, 7, \dots$ the nuclear trace of the operator $T^m$ is equal to zero. This follows from the fact that the eigenvalue sequences of $T$ and $T^m$ can be arranged in such a way that $\{\lambda_n(T)^m\}= \{\lambda_n(T^m)\}$ (see, e.g., [9, 3.2.24, p. 147]). \smallskip Let us formulate and prove the central result of this section. \smallskip {\bf Theorem 1}.\. Let $Y$ be a subspace of a quotient (or a quotient of a subspace) of an $L_p$-space, $1\le p\le\infty,$ and $u\in Y^*\widehat\otimes_s Y,$ where $1/s=1+|1/2-1/p|,$ The spectrum of the operator $\widetilde u$ is central-symmetric iff $\operatorname{trace}\, u^{2n - 1} = 0, n =1,2,\dots.$ \smallskip {\it Proof}.\ If the spectrum of $\widetilde u$ is central-symmetric, then, by Lemma 2, $\operatorname{trace}\, u= \operatorname{trace}\, \widetilde u= \sum_{n=1}^\infty \lambda_n(T)=0;$ also, by Remark 1, $\operatorname{trace}\, u^m= \sum_{n=1}^\infty \lambda_n(T^m)=0$ for $m=3, 5, \dots.$ To prove the inverse, we need some information from Fredholm Theory. Let $u$ be an element of the projective tensor product $X^*\widehat\otimes X,$ where $X$ is an arbitrary Banach space. Recall that the Fredholm determinant $\operatorname{det}\, (1-zu)$ of $u$ (see [2, Chap II, \S 1, $n^{\circ} 4$, p. 13] or [3], [8], [9]) is an entire function $$ \operatorname{det}\, (1-zu)= 1-z\, \operatorname{trace}\, u+ \dots + (-1)^n z^n\alpha_n(u) + \dots, $$ all zeros of which are exactly (according to their multiplicities) the inverses of nonzero eigenvalues %$(\mu_k)$ of the operator $\widetilde{u},$ associated with the tensor element $u.$ By [2, Chap II, \S 1, $n^{\circ} 4$, Corollaire 2, pp. 17-18], this entire function is of the form $$ \operatorname{det}\, (1-zu) = e^{-z\operatorname{trace}\, u} \prod_{i=1}^\infty (1- z z_i)\, e^{zz_i}, $$ where $\{z_i=\lambda_i(\widetilde u)\}$ is a system of all eigenvalues of the operator $\widetilde{u}$ (written according their algebraic multiplicities). Also, there exists a $\delta>0$ such that for all $z, |z|\le\delta,$ we have %$$ % \operatorname{det}\, (1-zu) = % \exp {(-\sum_{n=1}^\infty \frac1n\, z^n \operatorname{trace}\, u^n)} % $$ \begin{equation} \operatorname{det}\, (1-zu) = \exp {(-\sum_{n=1}^\infty \frac1n\, z^n \operatorname{trace}\, u^n)} \label{(1)} \end{equation} (see [3, p. 350]; cf. [1, Theorem I.3.3, p. 10]). %!!!Ref Now, let $u\in Y^*\widehat\otimes_s Y$ be as in the formulation of our theorem and suppose that $\operatorname{trace}\, u^{2n - 1} = 0, n =1,2,\dots.$ By (\ref{(1)}), we get: for a neighborhood $U= U(0)$ of zero in $\mathbb C,$ $\operatorname{det}\, (1-zu)= \operatorname{det}\, (1+zu)$ for $z\in U.$ Therefore, the entire function $\operatorname{det}\, (1-zu)$ is even. By definition of $\operatorname{det}\, (1-zu),$ the sequence of zeros of this function is exactly the eigenvalue sequence of $\widetilde u.$ Hence, the spectrum of $\widetilde u$ is central-symmetric. \smallskip Since under the conditions of Theorem 1 the space $Y$ has the $AP_s,$ the tensor product $Y^*\widehat\otimes_s Y$ can be identified naturally with the space of all $s$-nuclear operators in $Y.$ Hence, the statement of Theorem 1 may be reformulated in the following way: \smallskip {\bf Corollary 1}. {\it Let $s\in [2/3,1], p\in[2,\infty], 1/s=1+|1/2-1/p|,$ $Y$ be a subspace of a quotient (or a quotient of a subspace) of an $L_p$-space, $T$ be an $s$-nuclear operator in $Y.$ The spectrum of $T$ is central-symmetric iff $\operatorname{trace}\, T^{2n - 1} = 0, \, n \in \mathbf N.$ } \smallskip {\bf Corollary 2}\, [16]. {\it %!!! The spectrum of a nuclear operator $T,$ acting on a Hilbert space, is central-symmetric iff $\operatorname{trace}\, T^{2n - 1} = 0, \, n \in \mathbf N.$ } For a proof, it is enough to apply Theorem 1 for the case $p=2.$ \smallskip {\bf Corollary 3}. {\it The spectrum of a 2/3-nuclear operator $T,$ acting on an arbitrary Banach space, is central-symmetric iff $\operatorname{trace}\, T^{2n - 1} = 0, \, n \in \mathbf N.$ } For a proof, it is enough to apply Theorem 1 for the case $p=\infty,$ taking into account the fact that every Banach space is isometric to a subspace of an $L_\infty(\mu)$-space. \smallskip In connection with Corollary 3, let us pay attention again to the nuclear operator from Example 1. \medskip %%%%%%%%%%%%%% {\bf 4.\, Sharpness of results of Section 3}.\ %%%%%%%%%% $1/s=1+|1/2-1/p|,$ Now we will show that the statement of Theorem 1 is sharp and that the exponent $s$ is optimal if $p$ is fixed (if of course $p\neq2,$ i.e. $s\neq1).$ Consider the case $2< p\le\infty.$ In this case $1/s=1+|1/2-1/p|= 3/2-1/p.$ In a paper of the author [12, Example 2] %!!! the following result was obtained (see a proof in [12]): %!!! ${(\star)}$\, {\it Let $r\in [2/3,1), p\in(2,\infty], 1/r=3/2-1/p.$ There exist a subspace $Y_p$ of the space $l_p$ $(c_0$ if $p=\infty)$ and a tensor element $w_p\in Y_p^*\widehat\otimes_1 Y_p$ such that $w_p\in Y_p^*\widehat\otimes_s Y_p$ for every $s>r,$ $\operatorname{trace}\, w_p=1, \tilde w_p=0$ and the space $Y_p$ (as well as $Y_p^*)$ has the $AP_r$ $($but evidently does not have the $AP_s$ if $1\ge s>r).$ Moreover, this element admits a nuclear representation of the form $$ w_p=\sum_{k=1}^\infty \mu_k\, x'_k \otimes x_k,\ \text{ where }\ ||x'_k||=||x_k||=1,\, \sum_{k=1}^\infty |\mu_k|^s<\infty\ \forall \, s>r. $$ } %\smallskip Evidently, we have for a tensor element $u:=w_p$ from the assertion ${(\star)}:$ $\operatorname{trace}\, u=1$ and the spectrum of the operator $\widetilde u$ equals $\{0\}.$ The case where $2< p\le\infty$ can be considered analogously (with an application of the assertion ${(\star)}$ to a "transposed" tensor element $w_p^t\in Y_p \widehat\otimes_1 Y_p^*.)$ As was noted above (Example 1), there exists a nuclear operator $U$ in $l_1$ such that $U^2=0$ and $\operatorname{trace}\, U=1.$ The following theorem is an essential generalization of this result and gives us the sharpness of the statement of Corollary 1 (even in the case where $Y=l_p).$ \smallskip {\bf Theorem 2}.\, Let $p\in [1,\infty], p \neq2,$ $1/r= 1+|1/2-1/p|.$ There exists a nuclear operator $V$ in $l_p$ (in $c_0$ if $p=\infty)$ such that 1)\, $V \in N_s(l_p)$ for each $s\in (r, 1];$ 2)\, $V\notin N_r(l_p);$ 3)\, $\operatorname{trace}\, V=1$ and $V^2=0.$ \smallskip {\it Proof}.\, Suppose that $p>2.$ Consider the tensor element $w:= w_p$ from the assertion $(\star)$ and its representation $ w=\sum_{k=1}^\infty \mu_k\, x'_k \otimes x_k,$ where $||x'_k||=||x_k||=1$ è $\sum_{k=1}^\infty |\mu_k|^s<\infty$ for each $s, s>r.$ Let $l: Y:= Y_p\to l_p$ be the identity inclusion. Let $y'_k$ be an extension of the functional $x'_k$\, $(k=1, 2, \dots)$ from the subspace $Y$ to the whole space $l_p$ with the same norm and set $v:= \sum_{k=1}^\infty \mu_k\, y'_k \otimes l(x_k).$ Then $v\in l_{p'} \widehat\otimes_s l_p$ \, $(1/p+1/p'=1)$\, for each $s\in (r, 1],$ $\operatorname{trace}\, v= \sum \mu_k\, \langle y'_k, l(x_k)\rangle=1$ and $\widetilde v(l_p)\subset l(Y)\subset l_p.$ On the other hand, we have a diagram: $$ Y \overset{l}\to l_p \overset{\widetilde v_0}\to Y \overset{l}\to l_p \overset{\widetilde v_0}\to Y \overset{\widetilde v_0}\to l_p, $$ where $\widetilde v_0$ is an operator generated by $\widetilde v,$ $\widetilde v = l\widetilde v_0$ and $\widetilde v_0 l= \widetilde w=0.$ Put $V:= \widetilde v.$ Clearly, $\operatorname{trace}\, V=1$ and the spectrum $\operatorname{sp}\, \,V^2=\{0\}.$ Let us note that $V\notin N_r(l_p)$ (by Lemma 2). If $p\in [1,2),$ then it is enough to consider the adjoint operator. \smallskip It follows from Theorem 2 that the assertion of Corollary 1 is optimal already in the case of the space $Y=l_p$ (which is, by the way, has the Grothendieck's approximation property). \bigskip %%%%%%%%%%%%%% {\bf 5.\, Generalizations: around Mityagin's theorem}.\ Recall that if $T\in L(X)$ and, for some $m\in \mathbb N,$ $T^m$ is a Riesz operator (see, e.g., [4, p. 943] for a definition), then $T$ is a Riesz operator too (see, e.g., [9, 3.2.24, p. 147]). In particular, if $T^m$ is compact, then all non-zero spectral values $\lambda(T)\in \operatorname{sp}\, (T)$ are eigenvalues of finite (algebraic) multiplicity and have no limit point except possibly zero. Also, in this case the eigenvalue sequences of $T$ and $T^m$ can be arranged in such a way that $\{\lambda_n(T)^m\}= \{\lambda_n(T^m)\}$ (see [9, 3.2.24, p. 147]). Recall that in this case we denote by $\operatorname{sp}\, (T)$ (resp., by $\operatorname{sp}\, (T^m)$ the sequence $\{\lambda_n(T)^m\}$ (resp., $\{\lambda_n(T^m)\}).$ % \smallskip We are going to present a short proof of the theorem of B. Mityagin from [6, 7]. %!!!REF and below To clarify our idea of the proof, let us consider firstly the simplest case where $d=2.$ \smallskip {\bf Theorem 3.}\ Let $X$ be a Banach space and $T\in L(X).$ Suppose that some power of $T$ is nuclear. The spectrum of $T$ is central-symmetric iff there is an integer $K\ge0$ such that for every $l> K$ the value $\operatorname{trace}\, T^{l}$ is well defined and $\operatorname{trace}\, T^{2l+1}=0$ for all $l> K.$ \smallskip {\it Proof}.\ Suppose that $T\in L(X)$ and there is an $M\in \mathbb N$ so that $T^M\in N(X).$ Fix an odd $N_0, N_0>M,$ with the property that $T^{N_0}\in N_{2/3}(X)$ (it is possible since a product of three nuclear operators is 2/3-nuclear) and $\operatorname{trace}\, T^{N_0+2k}=0$ for all $k= 0, 1, 2, \dots.$ By Corollary 3, the spectra of all $T^{N_0+2k}$ are central-symmetric (since, e.g., $\operatorname{trace}\, T^{N_0}= \operatorname{trace}\, (T^{N_0})^3 = \operatorname{trace}\, (T^{N_0})^5= \dots = 0$ by assumption). Assume that the spectrum of $T$ is not central-symmetric. Then there exists an eigenvalue $\lambda_0\in \operatorname{sp}\, (T)$ such that $-\lambda_0\notin \operatorname{sp}\, (T).$ Now, $\lambda_0^{N_0}\in \operatorname{sp}\, (T^{N_0}),$ so $-\lambda_0^{N_0}\in \operatorname{sp}\, (T^{N_0}).$ Hence, there exist $\mu_{N_0}\in \operatorname{sp}\, (T)$ and $\theta_{N_0}$ so that $|\theta_{N_0}|=1,$ $\mu_{N_0}^{N_0}=-\lambda_0^{N_0}$ and $\mu_{N_0}=\theta_{N_0} \lambda_0,$\, $\theta_{N_0}\neq -1.$ Analogously, $\lambda_0^{N_0+2}\in \operatorname{sp}\, (T^{N_0+2}),$ so $-\lambda_0^{N_0+2}\in \operatorname{sp}\, (T^{N_0+2}).$ Hence, there exist $\mu_{N_0+2}\in \operatorname{sp}\, (T)$ and $\theta_{N_0+2}$ so that $|\theta_{N_0+2}|=1,$ $\mu_{N_0+2}^{N_0+2}=-\lambda_0^{N_0+2}$ and $\mu_{N_0+2}=\theta_{N_0+2} \lambda_0,$ \, $\theta_{N_0+2}\neq -1$ etc. By induction we get the sequences $\{\mu_{N_0+2k}\}_{k=0}^\infty$ and $\{\theta_{N_0+2k}\}_{k=0}^\infty$ with the properties that $\mu_{N_0+2k}\in \operatorname{sp}\, (T),$\, $|\theta_{N_0+2k}|=1,$ $\mu_{N_0+2k}^{N_0+2k}=-\lambda_0^{N_0+2k}$ and $\mu_{N_0+2k}=\theta_{N_0+2k} \lambda_0,$ \, $\theta_{N_0+2k}\neq -1.$ Since $\mu_{N_0+2k}\in \operatorname{sp}\, (T)$\, and $|\mu_{N_0+2k}|=|\lambda_0|>0,$ the sequence $\{\mu_{N_0+2k}\}$ is finite as a set, i.e., we have that $\mu_{N_0+2K}=\mu_{N_0+2K+2}=\dots$ for some $K>1.$ It follows that $\theta_{N_0+2K}=\theta_{N_0+2K+2}=\dots.$ But $\theta_{N_0+2k}^{N_0+2k}=-1$ for all $k.$ Thus $\theta_{N_0+2K}^l=-1$ for every odd $l\ge N_0+2K.$ Therefore, $\theta_{N_0+2K}=-1.$ Contradiction \medskip Now we are going to consider a general case of a notion of $\mathbb Z_d$-symmetry of a spectra, introduced and investigated by B. Mityagin in [6, 7]. %!!!REF Let $T$ be an operator in $X,$ all non-zero spectral values of which %$\lambda(\widetilde{v})\in \operatorname{sp}\, (\widetilde{v})$ are eigenvalues of finite multiplicity and have no limit point except possibly zero. Recall that we denote by $\operatorname{sp}\, (T)$ the corresponding unordered eigenvalue sequence for $T$ (possibly, including zero). For a fixed $d=2, 3, \dots$ and for the operator $T,$ the spectrum of $T$ is called $\mathbb Z_d$-symmetric, if $\lambda\in \operatorname{sp}\, (T)$ implies $t\lambda\in \operatorname{sp}\, (T)$ for every $t\in\sqrt[d]{1}.$ Let $r\in (0,\infty],$ \, $\mathbb D:= \{z\in \mathbb C:\ |z|1.$ If $u\in\Phi(X),$ then the Fredholm determinant $\operatorname{det}\, (1-zu)$ is $d$-even iff the eigenvalue sequence of $\widetilde u$ is $\mathbb Z_d$-symmetric iff $\operatorname{trace}\, u^{kd+r}=0$ for all $k=0, 1, 2, \dots$ and $r= 1, 2, \dots, d-1.$ \smallskip {\it Proof}.\, If the function $\operatorname{det}\, (1-zu)$ is $d$-even, then the eigenvalue sequence of $\widetilde u$ is $\mathbb Z_d$-symmetric, since this sequence coincides with the the sequence of all zeros of $\operatorname{det}\, (1-zu)$ (according their multiplicities). If the eigenvalue sequence of $\widetilde u$ is $\mathbb Z_d$-symmetric, then $\operatorname{trace}\, u= \sum_{\lambda\in\operatorname{sp}\, (\widetilde{u})} \lambda=0$ (since $\Phi(X)$ is of spectral type $l_1$ and $\sum_{t\in\sqrt[d]{1}} t=0).$ Also, by the same reason $\operatorname{trace}\, u^{kd+r}=0$ for all $k=0, 1, 2, \dots$ and $r= 1, 2, \dots, d-1,$ since the spectrum of $\widetilde{u}^l$ is absolutely summable for every $l\ge2$ and we may assume that $\{\lambda_m(\widetilde{u}^l)\}= \{\lambda_m(\widetilde{u})^l\}$ (for every fixed $l)$ [9, 3.2.24, p. 147]. Now, let $\operatorname{trace}\, u^{kd+r}=0$ for all $k=0, 1, 2, \dots$ and $r= 1, 2, \dots, d-1.$ By (\ref{(1)}), $\operatorname{det}\, (1-zu) = \exp {(-\sum_{m=1}^\infty \frac1{md}\, z^{md} \operatorname{trace}\, u^{md})}$ in a neighborhood $U$ of zero. Therefore, this function is $d$-even in the neighborhood $U.$ By the uniqueness theorem $\operatorname{det}\, (1-zu)$ is $d$-even in $\mathbb C.$ \smallskip {\bf Corollary 4.}\ %!!! For any Banach space $X$ and for every $u\in X^*\widehat\otimes_{2/3} X$ the conclusion of Lemma 3 holds. %!!! \smallskip {\bf Corollary 5.}\ Let $Y$ be a subspace of a quotient of an $L_p$-space, $1\le p\le\infty.$ For any $u\in Y^*\widehat\otimes_s Y,$ where $1/s=1+|1/2-1/p|,$ the conclusion of Lemma 3 holds. \medskip Now we are ready to present a short proof of the theorem of B. Mityagin [6, 7]. %!!!REF and below Note that the theorem in [6, 7] is formulated and proved for compact operators, but the proof from [6, 7] can be easily adapted for the general case of linear operators. \smallskip {\bf Theorem 4.}\ Let $X$ be a Banach space and $T\in L(X).$ Suppose that some power of $T$ is nuclear. The spectrum of $T$ is $\mathbb Z_d$-symmetric iff there is an integer $K\ge0$ such that for every $l> Kd$ the value $\operatorname{trace}\, T^{l}$ is well defined and $\operatorname{trace}\, T^{kd+r}=0$ for all $k=K, K+1, K+2, \dots$ and $r= 1, 2, \dots, d-1.$ \smallskip {\it Proof}.\ Fix $N_0\in\mathbb N$ such that $T^{N_0}$ is $2/3$-nuclear (it is possible by a composition theorem from [2, Chap II, \S 1, $n^{\circ} 3$, Theorem 3, p. 10]). Note that, by A. Grothendieck, the trace of $T^l$ is well defined for all $l\ge N_0.$ Suppose that the spectrum of $T$ is $\mathbb Z_d$-symmetric. Take an integer $l:= kd+r\ge N_0$ with $0N_0.$ Consider an infinite increasing sequence $\{p_m\}$ of prime numbers with $p_1> (K+1)d.$ Assuming that $\operatorname{trace}\, T^{kd+r}=0$ if $k=K, K+1, K+2, \dots$ and $r= 1, 2, \dots, d-1,$ for a fixed $p_m$ we get from Lemma 3 (more precisely, from Corollary 4) that the function $\operatorname{det}\, (1-zT^{p_m})$ is $d$-even. %and for all small enough $z$ we have (by (\ref{(1)}) % $$ % \operatorname{det}\, (1-zT^{p_m}) = %\exp {(-\sum_{n=1}^\infty \frac1n\, z^n \operatorname{trace}\, (T^{p_m})^n)} = % \exp {(-\sum_{j=1}^\infty \frac1{dj}\, z^{dj} \operatorname{trace}\, T^{p_m dj})} %$$ Suppose that the spectrum of $T$ is not $\mathbb Z_d$-symmetric. Then there exist an eigenvalue $\lambda_0\in \operatorname{sp}\, (T)$ and a root $\theta\in\sqrt[d]{1}$ so that $\theta \lambda_0\notin \operatorname{sp}\, (T).$ On the other hand, again by Lemma 3, the spectrum of $T^{p_m}$ is $\mathbb Z_d$-symmetric. Since $\lambda_0^{p_m}\in \operatorname{sp}\, (T^{p_m}),$ there exists $\mu_m\in \operatorname{sp}\, (T)$ such that $\mu_m^{p_m}=\theta^{p_m} \lambda_0^{p_m}\in \operatorname{sp}\, (T^{p_m});$ hence, $\mu_m = \theta_m \lambda_0$ for some $\theta_m$ with $|\theta_m|=1.$ But $|\lambda_0|>0.$ Therefore, the set $\{\mu_m\}$ is finite and it follows that there is an integer $M>1$ such that $\theta_M=\theta_{M+1}=\theta_{M+2}=\dots.$ Hence, $\theta^{p_m}= \theta_M^{p_m}$ for all $m\ge M.$ Thus, $\theta_M=\theta.$ Contradiction. \smallskip Let us give some examples in which we can apply Theorem 4, but the main result of [6, 7] does not work. \smallskip {\bf Example 2.}\ Let $\Pi_p$ be the ideal of absolutely $p$-summing operators $(p\in [1,\infty);$ see [8] for a definition and related facts). Then for some $n$ one has $\Pi_p^n\subset N.$ In particular, $\Pi_2^2(C[0,1])\subset N(C[0,1]),$ but not every absolutely 2-summing operator in $C[0,1]$ is compact. Another interesting example: $\Pi_3^3$ is of spectral type $l_1$ [15]. %!!!REF We do not know (maybe, it is not known to everybody), whether the finite rank operators are dense in this ideal. However, Theorem 4 may be applied. Moreover, it can be seen that, for example, the spectrum of an operator $T$ from $\Pi_3^3$ is central symmetric iff {\it the spectral traces}\, of the operators $T^{2k-1}$ are zero for all $k>0.$ \bigskip {\bf Acknowledgement.}\, I would like to express my gratitude to B. S. Mityagin for a two-days helpful discussion on the topic of this paper and on the problems related to the notion of nuclear operators at Aleksander Pe{\l}czy{\'n}ski Memorial Conference (13 July -- 19 July, 2014, B\c{e}dlewo, Poland) as well as for very useful information which B. S. Mityagin gave me in letters. Also, I am very grateful to him for drawing my attention to the paper of M. I. Zelikin [16]. %!!!REF %%%%%%%% refrnces \begin{thebibliography}{99} \bibitem{1} I. Gohberg, S. 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