\documentstyle{SibMatJ} % \TestXML \topmatter \Author Efimov \Initial S. \Initial V. \Gender he \Email EfimovSergei1960\@mail.ru \AffilRef 1 \endAuthor \Affil 1 \Division Department of Information Security \Organization North Caucasus Branch of Moscow Technical University of Communication and Information Technology \City Rostov-on-Don \Country Russia \endAffil \Origin \Journal \VMZh \Year 2024 \CopyrightYear 2024 \Volume 26 \Issue 4 \Pages 66--77 \DOI 10.46698/a3118-8799-1112-n \endOrigin \datesubmitted April 1, 2024\enddatesubmitted \dateaccepted April 25, 2025\enddateaccepted %\? \UDclass 517.9 % 47A53, 47G10, 45E05 \endUDclass \title The Index of a~Bisingular Operator with an Involutive Shift \endtitle \abstract In the theory of singular operators with an involutive shift, the issues of Noether (Fredholm) property and the index of an operator of the form $A+VB$ are fully studied, where $A$ and $B$ are singular operators, and $V$ is an operator of an involutive shift in the space of $p$-summable functions on a~simple closed contour of the Lyapunov type. Together with the operator $A+VB$ we consider the corresponding matrix singular operator without shift $M=\pmatrix A&{VBV}\\B&{VAV}\endpmatrix$. It is well known that the operators $A+VB$ and $M$ are Noetherian operators or not simultaneously, and their indices are related as $1:2$. Similar questions about simultaneous Noetherian property and proportionality of indices arise for bisingular operators with an involutive shift $A+WB$ and their corresponding matrix operators $M=\pmatrix A&{WBW}\\B&{WAW}\endpmatrix$, where $A$ and $B$ are bisingular operators, and $W$ is an operator of an involutive shift in the space of $p$-summable functions on the direct product of simple closed contours of the Lyapunov type. In this paper, we study bisingular operators with an involutive shift that decomposes into one-dimensional components. Two types of such shifts are considered coordinatewise and cross. In these cases, the corresponding matrix operators are matrix bisingular operators without shift. The simultaneous Noetherian property of the bisingular operator with a shift and the corresponding matrix bisingular operator without shift is obtained. The proportionality of the indices of bisingular operators with a coordinatewise shift and the corresponding matrix operators is established; namely, it is proved that the indices of these operators are related as 1:2. In a special case, the same result about the indices is obtained for the cross shift. \endabstract \keywords Noether operator, operator index, bisingular operator, involutive shift \endkeywords \endtopmatter \head 1. Introduction and Setting of the Problem \endhead Let $\Gamma_1$ and $\Gamma_2$ be simple closed contours of the Lyapunov type in the complex plane, oriented counterclockwise, and let $\Gamma$ be any of these contours, ${10$) there is a neighborhood $u$ ($\subset\Gamma_1$) of the point $t_1$ such that $$ \|(A-B)(P_u\otimes I)\|<\varepsilon, $$ where $P_u$ is the operator of multiplication by the characteristic function of the neighborhood $u$, acting in $L_{p}(\Gamma_1)$. Briefly, this fact will be indicated as follows: $A\otimes{1,t_1}\sim B$. We symmetrically introduce a~partial local equivalence relation with respect to the second variable at the point $t_2$ ($\in\Gamma_2$), marked as $A\otimes{2,t_2}\sim B$. It should be noted [11] that \Item (1) if $K_i\in\goth{K}_{p}^i$, then $K_i\overset{i,t_i}\to{\sim}O$ ($i=1,2$ and $t_i\in\Gamma_i$); \Item (2) if $A,B\in\goth{B}_{p}$, $C\in\goth{A}_{p}$, and $A\overset{1,t_1}\to{\sim}B$, where $t_1$ is a certain point of the contour~$\Gamma_1$, then ${AC\overset{1,t_1}\to{\sim}BC}$ (a~similar property holds for partial local equivalence with respect to the second variable). If $A\in\goth{A}_{p}$, then for any $t_1$ ($\in\Gamma_1$) and $t_2$ ($\in\Gamma_2$) there are [11] operators ${A_1^{\pm}(t_1)}$ ($\in\goth{A}_{p}(\Gamma_2)$) and $A_2^{\pm}(t_2)$ ($\in\goth{A}_{p}(\Gamma_1)$) such that $$ \align &A\overset{1,t_1}\to{\sim}P_{+}\otimes A_1^{+}(t_1)+P_{-}\otimes A_1^{-}(t_1), \\ &A\overset{2,t_2}\to{\sim}A_2^{+}(t_2)\otimes P_{+}+A_2^{-}(t_2)\otimes P_{-}. \endalign $$ In this case, a family of singular operators $A_i^{\pm}(t_i)$ ($i=1,2$ and $t_i\in\Gamma_i$) is defined by the bisingular operator~$A$ in a~unique way and is called the symbol of %\? the bisingular operator~$A$. On the other hand, a~bisingular operator is defined by its symbol in a unique way up to a compact term from $\goth{K}_{p}$. Let us also note that the operator functions $A_1^{\pm}$ and~$A_2^{\pm}$ are continuous in the uniform operator topology, and the four correspondences $A\mapsto A_1^{\pm}$ and $A\mapsto A_2^{\pm}$ are homomorphisms of Banach algebras. \proclaim{Theorem~2 \rm [11]}\Label{T2} The Noetherian property of a bisingular operator $A$ from the algebra $\goth{A}_{p}$ is equivalent to the invertibility of all operators of its symbol $A_i^{\pm}(t_i)$ $(i=1,2$, with $t_i\in\Gamma_i)$. In this case, the regularizer of the operator $A$ belongs to the algebra $\goth{A}_{p}$ and the symbol of the regularizer consists of the operators $(A_i^{\pm}(t_i))^{-1}$. \endproclaim In the space $L_{p}(\Gamma_1\times\Gamma_2)$, we introduce the four projections ${P_{\pm+}=P_{\pm}\otimes P_{+}}$ and ${P_{\pm-}=P_{\pm}\otimes P_{-}}$. If the function $a$ of class $C(\Gamma_1\times\Gamma_2)$ does not have zero values, then by ${\operatorname{mind}_{1}a}$ and ${\operatorname{id}_{2}a}$ we will denote the partial indices of the function~$a$ for the first and second variables, respectively. \proclaim{Theorem~3 \rm [11--13]}\Label{T3} Let $a_{\pm+},a_{\pm-}\in C(\Gamma_1\times\Gamma_2)$. Then for the bisingular operator $A=a_{++}P_{++}+a_{+-}P_{+-}+a_{-+}P_{-+}+a_{--}P_{--}$ to be Noetherian it is necessary and sufficient that the following conditions are fulfilled: \Item (i) %\?(1) the functions $a_{\pm+}$ and $a_{\pm-}$ do not have zero values; \Item (ii) %\?(2) $\operatorname{ind}_{1}a_{+\pm}=\operatorname{ind}_{1}a_{-\pm}$ and $\operatorname{ind}_{2}a_{\pm +}=\operatorname{ind}_{2}a_{\pm -}$. In this case, the index of the operator~$A$ is calculated by the formula $$ \operatorname{Ind}\,A=(\operatorname{ind}_{1}a_{++}-\operatorname{ind}_{1}a_{--})(\operatorname{ind}_{2}a_{+-}-\operatorname{ind}_{2}a_{-+})= \operatorname{ind}_{1}\frac{a_{++}}{a_{--}}\cdot\operatorname{ind}_{2}\frac{a_{++}}{a_{--}}. $$ \endproclaim Let us now consider matrix bisingular operators. If $\goth{R}$ is some ring or algebra, then by $\goth{R}^{(2)}$ we denote the ring or, respectively, the algebra of all second-order square matrices with the elements from~$\goth{R}$. In what follows, the condition $${\pmatrix a&c\\b&d\endpmatrix \in\goth{R}^{(2)}} $$ will mean by default that $a,b,c,d\in\goth{R}$. If $\goth{R}$ is some algebra of linear operators in a~linear space~$L$, then the matrices from the algebra $\goth{R}^{(2)}$ are naturally considered as linear operators in the linear space ${L\oplus L}$. The operators from the algebra $(\goth{A}_{p})^{(2)}$ act in the space ${L_{p}(\Gamma_1\times\Gamma_2)\oplus L_{p}(\Gamma_1\times\Gamma_2)}$ and called {\it matrix bisingular operators}. If $M=\pmatrix A&C\\B&D\endpmatrix \in(\goth{A}_{p})^{(2)}$, then the symbol of the operator $M$ will be understood as the {\it family of operators\/} $M_i^{\pm}(t_i)= \pmatrix {A_i^{\pm}(t_i)}&{C_i^{\pm}(t_i)}\\{B_i^{\pm}(t_i)}&{D_i^{\pm}(t_i)}\endpmatrix $ ($\in(\goth{A}_{p}(\Gamma_{3-i}))^{(2)}$), where $i=1,2$ and $t_i\in\Gamma_i$. Obviously, the matrix bisingular operator from the algebra $(\goth{A}_{p})^{(2)}$ is defined by its symbol in a~unique way up to a~compact term from $(\goth{K}_{p})^{(2)}$. For matrix bisingular operators, the following statement is true, the proof of which is in no way fundamentally different from the proof of \Par{T2}{Theorem~2}. \proclaim{Theorem~4}\Label{T4} The Noetherian property of the matrix bisingular operator $M$ from the algebra ${(\goth{A}_{p})^{(2)}}$ is equivalent to the invertibility of all operators of its symbol $M_i^{\pm}(t_i)$ $({i=1,2}$, ${t_i\in\Gamma_i})$. In this case, the regularizer of the operator $M$ belongs to $(\goth{A}_{p})^{(2)}$ and the symbol of the regularizer consists of the operators~$(M_i^{\pm}(t_i))^{-1}$. \endproclaim Let $\goth{A}_{p,W}$ be the set of all operators in $L_{p}(\Gamma_1\times\Gamma_2)$ of the form $A+WB$, where ${A,B\in\goth{A}_{p}}$. We call the operators from $\goth{A}_{p,W}$ {\it bisingular operators with shift\/} $\alpha$. \proclaim{Lemma~2 \rm[1, p.~35; 2, p.~39]}\Label{L2} If $\varphi$ $(\Gamma_i\rightarrow\Gamma_j)$ is a one-dimensional shift $(i,j=1,2)$, then $$ V_{\varphi}S_{\Gamma_j}V_{\varphi}^{-1}-\gamma S_{\Gamma_i}\in\goth{K}_{p}(\Gamma_i), $$ where $\gamma=+1$ when $\varphi$ preserves orientation, and $\gamma=-1$ when $\varphi$ changes orientation. \endproclaim \Par{L2}{Lemma~2} implies that if $A\in\goth{A}_{p}$ then so does $WAW\in\goth{A}_{p}$. Then the set ${\goth{A}_{p,W}}$ is the algebra generated by the operator $W$ and all bisingular operators from the algebra $\goth{A}_{p}$. Together with the operator~$A+WB$, where $A,B\in\goth{A}_{p}$, we will consider the accompanying operator $A-WB$ and the corresponding matrix operator $\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $ belonging to $(\goth{A}_{p})^{(2)}$. Let us note one well-known elementary algebraic fact that establishes a connection between these operators (see, for instance, [1, p.~398]). \proclaim{Lemma~3}\Label{L3} Let $\goth{R}$ be some algebra with the unity $e$ and zero $o$ over the field of complex numbers, $a,b,w\in\goth{R}$ and $w^2=e$. Then there is the following similarity in $\goth{R}^{(2)}$: $$ \pmatrix a&{wbw}\\b&{waw}\endpmatrix =G\pmatrix {a+wb}&o\\o&{a-wb}\endpmatrix G^{-1}, $$ where $G=\frac{1}{\sqrt{2}}\pmatrix e&e\\w&{-w}\endpmatrix $ and $\biggl(G^{-1}= \frac{1}{\sqrt{2}}\pmatrix e&w\\e&{-w}\endpmatrix \biggr)$. \endproclaim We will also need a property related to the regularization of the operator ${\pmatrix A&{WBW}\\B&{WAW}\endpmatrix }$, where $A,B\in\goth{A}_{p}$. Let us first consider an auxiliary statement. \proclaim{Lemma~4}\Label{L4} Let $\goth{R}$ be some associative ring with the unity $e$. Also, let ${a,b,w\in\goth{R}}$ and $w^2=e$. Then if the matrix $\mu=\pmatrix a&{wbw}\\b&{waw}\endpmatrix $ is invertible in $\goth{R}^{(2)}$, then the inverse matrix has the form $\mu^{-1}=\pmatrix c&{wdw}\\d&{wcw}\endpmatrix $, where $c$ and $d$ are some elements of the ring $\goth{R}$. \endproclaim \demo{Proof} We have $\mu^{-1}=\pmatrix c&g\\d&h\endpmatrix $, where $c,d,g,h\in\goth{R}$. Consider the matrix $\rho=\pmatrix c&{wdw}\\d&{wcw}\endpmatrix $. Since $\mu\mu^{-1}=\pmatrix e&o\\o&e\endpmatrix $, where $o$ is the zero of the ring $\goth{R}$, we have $ac+wbwd=e$ and $bc+wawd=o$. Multiplying these equalities on the left and right by $w$, we get two more equalities ${bwdw+wacw=e}$ and ${awdw+wbcw=o}$. However, in this case we have $\mu\rho=\pmatrix e&o\\o&e\endpmatrix $. Thus, $\mu^{-1}=\rho$. \qed\enddemo \proclaim{Theorem~5}\Label{T5} A Noetherian operator of the form $\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $, where $A,B\in\goth{A}_{p}$, has a regularizer of a similar type $\pmatrix C&{WDW}\\D&{WCW}\endpmatrix $, where $C$ and $D$ are some operators from~$\goth{A}_{p}$. \endproclaim \demo{Proof} Let $A,B\in\goth{A}_{p}$ and let the operator $M=\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $ be Noetherian. By \Par{L4}{Lemma~4}, the operator~$M$ has a regularizer of the form $R=\pmatrix C&{WDW}\\D&{WCW}\endpmatrix $, where~$C,D\in\goth{B}_{p}$. However, $M\in(\goth{A}_{p})^{(2)}$, and then, by \Par{T4}{Theorem~4}, the operator $R$ also belongs to~$(\goth{A}_{p})^{(2)}$. Therefore, ${C,D\in\goth{A}_{p}}$. \qed\enddemo \head 3. Simultaneous Noetherian Property of a Bisingular Operator with a~Shift and the Corresponding Matrix Operator \endhead Hereafter, if $A$ and $B$ are linear bounded operators in a certain Banach space, then the notation $A\sim B$ means that $A-B$ is a compact operator. \proclaim{Theorem~6}\Label{T6} Let $A,B,C,D\in\goth{A}_{p}$ and $A+WB=C+WD$. Then $A\sim C$ and $B\sim D$. \endproclaim \demo{Proof} Let us first assume that $\alpha$ is the shift \Tag(1). %\Par{s(1)}(i). It suffices to prove that $A\sim C$. To this end, it is enough to make sure that the symbols of the operators $A$ and $C$ are the same. Let us prove, for instance, that ${A_1^{\pm}(t_1)=C_1^{\pm}(t_1)}$ ($t_1\in\Gamma_1$). It is important to note here that the shift $\alpha_1$ ($\Gamma_1\rightarrow\Gamma_1$) is involutive. Then (1) $\alpha_1$ is not identical mapping; (2) if $\alpha_1$ preserves orientation on $\Gamma_1$, then $\alpha_1$ has no fixed points: ${\alpha_1(t_1)\neq t_1}$ at each point $t_1$ of the contour $\Gamma_1$; (3) if $\alpha_1$ changes orientation on $\Gamma_1$, then $\alpha_1$ has exactly two fixed points where the condition $\alpha_1(t_1)=t_1$ is satisfied, and at all other points of the contour $\Gamma_1$ the condition $\alpha_1(t_1)\neq t_1$ is fulfilled. Let us fix an arbitrary point $t_1^0$ on $\Gamma_1$ at which $\alpha_1(t_1^0)\ne_1^0$. Then there is a~function $h$ of class $C(\Gamma_1)$ such that $h(t_1^0)=1$ and $h(\alpha_1(t_1^0))=0$. We multiply ${A+WB=C+WD}$ on the left by the operator $(h\otimes 1)I$: $$ (h\otimes 1)\,A+W((h\circ\alpha_1)\otimes 1)\,B=(h\otimes 1)\,C+W((h\circ\alpha_1)\otimes 1)\,D. $$ It follows from the properties of partial local equivalence described in \Sec{Sec2}{Section~2} that $$ \align &(h\otimes 1)A+W((h\circ\alpha_1)\otimes 1)B\overset{1,t_1^0}\to{\sim}h(t_1^0)A+Wh(\alpha_1(t_1^0))B=A, \\ &(h\otimes 1)C+W((h\circ\alpha_1)\otimes 1)D\overset{1,t_1^0}\to{\sim}h(t_1^0)C+Wh(\alpha_1(t_1^0))D=C. \endalign $$ Then $A\overset{1,t_1^0}\to{\sim}C$. We have $$A\overset{1,t_1^0}\to{\sim}P_{+}\otimes C_1^{+}(t_1^0)+P_{-}\otimes C_1^{-}(t_1^0). $$ But then it follows from the uniqueness of the symbol of operator $A$ that $A_1^{\pm}(t_1^0)=C_1^{\pm}(t_1^0)$. So, at each~$t_1$ of the contour $\Gamma_1$, where $\alpha_1(t_1)\neq t_1$, the condition $A_1^{\pm}(t_1)=C_1^{\pm}(t_1)$ is satisfied. Since the operator functions $A_1^{\pm}$ and $C_1^{\pm}$ are continuous in the uniform operator topology, the condition $A_1^{\pm}(t_1)=C_1^{\pm}(t_1)$ must also be fulfilled at fixed points of the shift $\alpha_1$, if any such point exists. If $\alpha$ is the shift \Tag(2), %\Par{s(2)}(2), then the statement of the theorem is immediate from [4,\,5]. \qed\enddemo It follows from \Par{T6}{Theorem~6} that for every operator from the algebra $\goth{A}_{p,W}$ its accompanying and corresponding matrix operators are determined in a unique way up to a compact term. \proclaim{Theorem~7}\Label{T7} Let $A,B\in\goth{A}_{p}$. Then the operators $A+WB$, $A-WB$, and ${\pmatrix A&{WBW}\\B&{WAW}\endpmatrix }$ are Noetherian or not simultaneously. \endproclaim \demo{Proof} Let us turn to the symbolic calculus of bisingular operators with a~shift~$\alpha$ which are constructed in [3--5]. In [3], the symbol of an operator with shift~\Tag(1) %\Par{s(1)}(i) (and even of a more general operator) is constructed, and in [4,\,5], with shift~\Tag(2). %\Par{s(2)}(2). The symbol of a~bisingular operator with a shift is a family of some matrix operators acting in the spaces $L_p(\Gamma_1)\oplus L_p(\Gamma_1)$ and $L_p(\Gamma_2)\oplus L_p(\Gamma_2)$. It is proved in [3--5] that the Noetherian property of a bisingular operator with a shift is equivalent to the invertibility of all operators of its symbol. It is easy to see that the symbols of the operators $A\pm WB$ are related by a~similarity relation; namely, the symbol of the operator $A-WB$ consists of all possible operators of the form $$ \pmatrix I&O\\O&{-I}\endpmatrix H\pmatrix I&O\\O&{-I}\endpmatrix , $$ where $H$ belongs to the symbol of the operator ${A+WB}$. Then the operators $A\pm WB$ are Noetherian or not at the same time. It remains to note using \Par{L3}{Lemma~3} that the Noetherian property of the operator $\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $ is equivalent to the Noetherian property of both operators $A\pm WB$. \qed\enddemo \demo{Remark} If $\alpha$ is the shift \Tag(1), %\Par{s(1)}(i), then the condition $V_{\alpha_1}\neq I$, $V_{\alpha_2}\neq I$, inherent in the definition of \Tag(1), %\Par{s(1)}(i), is essential for the simultaneous Noetherian property of a bisingular operator with shift~\Tag(1), %\Par{s(1)}(i) and the corresponding matrix operator. Indeed, let us consider a shift operator $W_1=V_{\alpha_1}\otimes I$, where $\alpha_1$ is an involutive one-dimensional shift $\Gamma_1\rightarrow\Gamma_1$, and the bisingular operators $A=I\otimes I$ and $B=P_0\otimes I$, where $P_0$ is a~one-dimensional projection in~${L_p(\Gamma_1)}$ whose image consists of constant functions. Then the operator ${A+W_1 B=(I+P_0)\otimes I}$ is invertible and therefore Noetherian. However, the operator $A-W_1 B=(I-P_0)\otimes I$ has an infinite-dimensional kernel, and therefore this operator cannot be Noetherian. Then, by \Par{L3}{Lemma~3}, the matrix operator ${\pmatrix A&{W_1 BW_1}\\B&{W_1 AW_1}\endpmatrix }$ is not a Noetherian operator. \endproclaim In the sequel, we will need the following: \proclaim{Theorem~8}\Label{T8} The algebra $\goth{A}_{p,W}$ contains regularizers of all its Noetherian operators. In this case, the Noetherian mutually accompanying operators $A\pm WB$, where ${A,B\in\goth{A}_{p}}$, will have mutually accompanying regularizers $C\pm WD$, respectively, where $C$ and $D$ are some operators from the algebra~$\goth{A}_p$. \endproclaim \demo{Proof} Let $A,B\in\goth{A}_{p}$ and let the operator $A+WB$ be Noetherian. Then, by \Par{T7}{Theorem~7}, the operators ${A-WB}$ and $\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $ are also Noetherian. By \Par{T5}{Theorem~5}, there exist operators $C$ and $D$ from the algebra $\goth{A}_p$, such that the operator $\pmatrix C&{WDW}\\D&{WCW}\endpmatrix $ is a regularizer of the operator ${\pmatrix A&{WBW}\\B&{WAW}\endpmatrix }$. But then it follows from \Par{L3}{Lemma~3} that the operators $C\pm WD$ are regularizers of the operators $A\pm WB$, respectively. \qed\enddemo \head 4. Indices of Bisingular Operators with a~Shift \endhead \specialhead 4.1. The case of a coordinatewise shift \endspecialhead \proclaim{Lemma~5}\Label{L5} Let $\alpha$ be the shift~$(1)$, $K,T\in\goth{K}_{p}^1$ and let the operator $I+K+WT$ be Noetherian. Then $$ \operatorname{Ind}(I+K+WT)=\operatorname{Ind}\,(I+K-WT). $$ \endproclaim \demo{Proof} By \Par{T7}{Theorem~7}, together with the operator $H=I+K+WT$, the operator ${N=I+K-WT}$ will be Noetherian as well. Let us select an invertible operator $U$ from the algebra $\goth{A}_{p}(\Gamma_2)$ such that ${UV_{\alpha_2}\sim -V_{\alpha_2}U}$: \Item (1) if $\alpha_2$ preserves orientation on $\Gamma_2$, then $\alpha_2$ has no fixed points and the operator $U$ can be defined as the operator of multiplication by the function $\alpha_2(t_2)-t_2$ ($t_2\in\Gamma_2$); \Item (2) if $\alpha_2$ changes orientation on $\Gamma_2$, then by \Par{L2}{Lemma~2} we can introduce $U=S$. Applying \Par{L1}{Lemma~1}, we obtain $(I\otimes U)H(I\otimes U^{-1})\sim N$. Consequently, ${\operatorname{Ind}\,H=\operatorname{Ind}\,N}$. \qed\enddemo \proclaim{Theorem~9}\Label{T9} Let $\alpha$ be the shift~$(1)$, $A,B\in\goth{A}_{p}$ and let the operator $A+WB$ be Noetherian. Then $$ \operatorname{Ind}\,(A+WB)=\operatorname{Ind}\,(A-WB)=\frac{1}{2}\,\operatorname{Ind}\pmatrix A&{WBW}\\B&{WAW}\endpmatrix . $$\endproclaim \demo{Proof} By \Par{T7}{Theorem~7}, the operators $A-WB$ and $\pmatrix A&{WBW}\\B&{WAW}\endpmatrix $ are Noetherian. By \Par{T8}{Theorem~8}, the operators $A\pm WB$ have regularizers $C\pm WD$, respectively, where $C$ and $D$ are some operators from the algebra $\goth{A}_{p}$. As in the proof of \Par{L5}{Lemma~5}, we select an invertible operator~$U$ in the algebra $\goth{A}_{p}(\Gamma_1)$ such that $UV_{\alpha_1}\sim -V_{\alpha_1}U$. Then by \Par{L1}{Lemma~1} $$ \cases {(U\otimes I)(A-WB)(U^{-1}\otimes I)=A+K+WB+WT},\\ {(U\otimes I)(A+WB)(U^{-1}\otimes I)=A+K-WB-WT}, \endcases $$ where $K,T\in\goth{K}_{p}^1$. It follows from here that $$ \cases {(U\otimes I)(A-WB)(U^{-1}\otimes I)\sim(A+WB)(I+(C+WD)(K+WT))},\\ {(U\otimes I)(A+WB)(U^{-1}\otimes I)\sim(A-WB)(I+(C-WD)(K-WT))}. \endcases $$ Then $$ \cases {(U\otimes I)(A-WB)(U^{-1}\otimes I)\sim(A+WB)(I+K_1+WT_1)},\\ {(U\otimes I)(A+WB)(U^{-1}\otimes I)\sim(A-WB)(I+K_1-WT_1)}, \endcases $$ where $K_1,T_1\in\goth{K}_{p}^1$. The operators $I+K_1\pm WT_1$ are Noetherian and, by \Par{L5}{Lemma~5}, their indices are equal to one and the same number~$n$. Then $$ \cases {\operatorname{Ind}(A-WB)=\operatorname{Ind}(A+WB)+n},\\ {\operatorname{Ind}(A+WB)=\operatorname{Ind}(A-WB)+n}. \endcases $$ Adding these equalities, we get that $n=0$ and $$\operatorname{Ind}(A+WB)=\operatorname{Ind}(A-WB). $$ It remains to note that by \Par{L3}{Lemma~3} $$ \operatorname{Ind}\pmatrix A&{WBW} \\B&{WAW}\endpmatrix =\operatorname{Ind}(A+WB)+\operatorname{Ind}(A-WB). \qed $$\enddemo Here is an example of calculating the index of a bisingular operator with shift (1). \demo{Example~1} Let $\Gamma_0$ be a unit circle on the complex plane centered at the origin and let $\Gamma_1=\Gamma_2=\Gamma_0$. Consider the shift~\Tag(1) by $\Gamma_0\times\Gamma_0$ of the form ${\alpha(t_1;t_2)=(-t_1;-t_2)}$ ($t_1,t_2\in\Gamma_0$) and bisingular operators $${A=\sum\limits_{x,y=\pm}a_{xy}P_{xy}} \quad\text{and}\quad {B=\sum\limits_{x,y=\pm}b_{xy}P_{xy}} $$ in the space $L_p(\Gamma_0\times\Gamma_0)$, where $a_{xy},b_{xy}\in C(\Gamma_0\times\Gamma_0)$, under the additional condition ${a_{xy}\circ\alpha=a_{xy}}$, ${b_{xy}\circ\alpha=b_{xy}}$ ($x,y=+$). Note that $$ WAW=A\quad\text{and}\quad WBW=B. $$ Then, by \Par{T7}{Theorem~7}, the Noetherian property of the operator $A+WB$ is equivalent to the Noetherian property of the operator ${\pmatrix A&B\\B&A\endpmatrix }$, which is similar to the operator $\pmatrix {A+B}&O\\O&{A-B}\endpmatrix $ according to \Par{L3}{Lemma~3}. Thus, the Noetherian property of the operator $A+WB$ is equivalent to the Noetherian property of two bisingular operators $$ A+B=\sum\limits_{x,y=\pm}(a_{xy}+b_{xy})P_{xy} \quad\text{and}\quad A-B=\sum\limits_{x,y=\pm}(a_{xy}-b_{xy})P_{xy}. $$ Then, by \Par{T3}{Theorem~3}, for the operator $A+WB$ to be Noetherian it is necessary and sufficient that the following conditions be fulfilled: \Item (1) $a_{xy}+b_{xy}$ and $a_{xy}-b_{xy}$ do not have zero values ($x,y=\pm$); \Item (2) $\operatorname{ind}_{1}(a_{+\pm}+b_{+\pm})=\operatorname{ind}_{1}(a_{-\pm}+b_{-\pm})$ and $\operatorname{ind}_{2}(a_{\pm +}+b_{\pm +})=\operatorname{ind}_{2}(a_{\pm -}+b_{\pm -})$; \Item (3) $\operatorname{ind}_{1}(a_{+\pm}-b_{+\pm})=\operatorname{ind}_{1}(a_{-\pm}-b_{-\pm})$ and $\operatorname{ind}_{2}(a_{\pm +}-b_{\pm +})=\operatorname{ind}_{2}(a_{\pm -}-b_{\pm -})$. Moreover, it follows from \Par{T9}{Theorems~9} and~\Par{T3}{3} that $$ \align \operatorname{Ind}\,(A+WB) &=\frac{1}{2}\,\operatorname{Ind}\pmatrix A&B\\B&A\endpmatrix = \frac{1}{2}\,\operatorname{Ind}\pmatrix {A+B}&O\\O&{A-B}\endpmatrix \\ &=\frac{1}{2}\,(\operatorname{Ind}(A+B)+\operatorname{Ind}(A-B)) \\ &=\frac{1}{2}\,\Bigl(\operatorname{ind}_{1}\frac{a_{++}+b_{++}}{a_{--}+b_{--}}\cdot\operatorname{ind}_{2}\frac{a_{++}+b_{++}}{a_{--}+b_{--}} +\operatorname{ind}_{1}\frac{a_{++}-b_{++}}{a_{--}-b_{--}}\cdot\operatorname{ind}_{2}\frac{a_{++}-b_{++}}{a_{--}-b_{--}}\Bigr). \endalign $$ \specialhead 4.2. The case of cross-shift \endspecialhead \proclaim{Lemma~6}\Label{L6} Let $\alpha$ be the shift \Tag(2), %\? $A\in\goth{A}_{p}$, $K\in\goth{K}_{p}^1$, $T\in\goth{K}_{p}^2$, and let the operator ${A+W(K+T)}$ be Noetherian. Then the operators $A\pm W(K+T)$ are homotopic in the class of Noetherian operators. \endproclaim \demo{Proof} Let us consider the homotopy $$ H_\xi=A+W(e^{i\pi\xi}K+e^{-i\pi\xi}T),\quad 0\leq\xi\leq 1, $$ from $H_0=A+W(K+T)$ to $H_1=A-W(K+T)$. Let us construct the symbol [4,\,5] of~$H_\xi$, taking into account that $WKW\in\goth{K}_{p}^2$ and $WTW\in\goth{K}_{p}^1$: $$ \align (H_\xi)^{\pm}(t_1) &=\pmatrix A_1^\pm(t_1)&(W(e^{i\pi\xi}K+e^{-i\pi\xi}T)W)_1^\pm(t_1) \\ (e^{i\pi\xi}K+e^{-i\pi\xi}T)_1^\pm(t_1)&(WAW)_1^\pm(t_1)\endpmatrix \\ &=\pmatrix A_1^\pm(t_1)&e^{i\pi\xi}\cdot(WKW)_1^\pm(t_1)\\ e^{-i\pi\xi}\cdot T_1^\pm(t_1)&(WAW)_1^\pm(t_1)\endpmatrix \\ &=\pmatrix I&O\\O&e^{-i\pi\xi}I\endpmatrix \pmatrix A_1^\pm(t_1)&(WKW)_1^\pm(t_1)\\ T_1^\pm(t_1)&(WAW)_1^\pm(t_1)\endpmatrix \pmatrix I&O\\O&e^{i\pi\xi}I\endpmatrix \\ &=\pmatrix I&O\\O&e^{-i\pi\xi}I\endpmatrix (H_0)^{\pm}(t_1) \pmatrix I&O\\O&e^{i\pi\xi}I\endpmatrix ,\quad t_1\in\Gamma_1. \endalign $$ Since operator $H_0$ is Noetherian; therefore (see [4,\,5]), all operators of its symbol $(H_0)^{\pm}(t_1)$ are invertible in ${L_{p}(\Gamma_2)\oplus L_{p}(\Gamma_2)}$ ($t_1\in\Gamma_1$). But, together with them, all operators ${(H_\xi)^{\pm}(t_1)}$ ($0\leq\xi\leq1$ and $t_1\in\Gamma_1$) are also invertible in $L_{p}(\Gamma_2)\oplus L_{p}(\Gamma_2)$. Then (see [4,\,5]) all operators $H_\xi$ are Noetherian ($0\leq\xi\leq1$). \qed\enddemo The next statement follows from \Par{T7}{Theorem~7} and \Par{L3}{Lemmas~3} and~\Par{L6}{6}. \proclaim{Theorem~10}\Label{T10} Let $\alpha$ be the shift $(2)$, $A\in\goth{A}_{p}$, $K\in\goth{K}_{p}^1$, $T\in\goth{K}_{p}^2$, and let the operator ${A+W(K+T)}$ be Noetherian. Then $$ \operatorname{Ind}\,(A+W(K+T))=\operatorname{Ind}\,(A-W(K+T))=\frac{1}{2}\,\operatorname{Ind}\pmatrix A&{W(K+T)W} \\ {K+T}&{WAW}\endpmatrix . $$ \endproclaim However, in the general case, there is no proportion between the indices of bisingular operators with shift \Tag(2) %\Par{s(2)}(2) and the indices of the corresponding matrix operators. This is shown by the following example, in which the relation of the indices established in the conditions of \Par{T10}{Theorem~10} becomes impossible. \demo{Example~2}\Label{E2} Let $\Gamma_0$ be a unit circle on the complex plane centered at the origin and $\Gamma_1=\Gamma_2=\Gamma_0$. We introduce the function $e(t)=t$ ($t\in\Gamma_0$) and the functions ${e_{01}=1\otimes e}$, $e_{10}=e\otimes 1$, and $e_{11}=e\otimes e$ on $\Gamma_0\times\Gamma_0$. Consider the shift (2) of the form $$ \alpha(t_1;t_2)=(t_2;t_1),\quad t_1,t_2\in\Gamma_0, $$ and the bisingular operators $$ \align &A=\Bigl(e_{11}-\frac{1}{2}\Bigr)P_{++}+\Bigl(e_{01}-\frac{1}{2}\Bigr)P_{+-}+\Bigl(e_{10}-\frac{1}{2}\Bigr)P_{-+}+P_{--}, \\ &B=P_{++}+P_{+-}+P_{-+}. \endalign $$ Note that $WAW=A$ and $WBW=B$. Then, to the bisingular operator with shift ${A+WB}$, there corresponds the matrix operator $$ M=\pmatrix A&WBW\\B&WAW\endpmatrix =\pmatrix A&B\\B&A\endpmatrix . $$ We need to check that the operators $A+WB$ and $M$ are Noetherian, however, ${\operatorname{Ind}(A+WB)\neq \frac{1}{2}\operatorname{Ind}M}$. By \Par{L3}{Lemma~3}, operator $M$ is similar to the operators $$\pmatrix A+WB&O\\O&A-WB\endpmatrix \quad\text{and}\quad {\pmatrix A+B&O\\O&A-B\endpmatrix }. $$ The operators $$ \align &A+B=\Bigl(e_{11}+\frac{1}{2}\Bigr)P_{++}+\Bigl(e_{01}+\frac{1}{2}\Bigr)P_{+-}+\Bigl(e_{10}+\frac{1}{2}\Bigr)P_{-+}+P_{--}, \\ &A-B=\Bigl(e_{11}-\frac{3}{2}\Bigr)P_{++}+\Bigl(e_{01}-\frac{3}{2}\Bigr)P_{+-}+\Bigl(e_{10}-\frac{3}{2}\Bigr)P_{-+}+P_{--} \endalign $$ are Noetherian operators by \Par{T3}{Theorem~3}. But then operators $M$ and $A\pm WB$ are also Noetherian. Moreover, by \Par{T3}{Theorem~3} $\operatorname{Ind}\,(A+B)=1$ and $\operatorname{Ind}\,(A-B)=0$. Then ${\operatorname{Ind}\,M=1}$. Therefore, in this case, $\operatorname{Ind}\,(A+WB)$ cannot be equal to $\frac{1}{2}\,\operatorname{Ind}\,M$. \demo{Remark} We can check that in the conditions of \Par{E2}{Example~2} $\operatorname{ker}(A+B)$ and ${\operatorname{ker}(A-WB)}$ coincide with the one-dimensional linear space generated by the function ${1\otimes \frac{1}{e+\frac{1}{2}}-\frac{1}{e+\frac{1}{2}}\otimes 1}$, while $\operatorname{ker}(A-B)$, $\operatorname{ker}(A+WB)$, $\operatorname{coker}(A\pm B)$ and ${\operatorname{coker}(A\pm WB)}$ are trivial. 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