Riesz Decomposition Relative to C*-Algebra Homomorphisms

TL;DR

This paper investigates the Riesz decomposition relative to C*-algebra homomorphisms, establishing conditions under which T-Riesz elements can be decomposed into almost T-null and quasi-nilpotent parts, and discusses polynomial and generalized decompositions.

Contribution

It introduces new conditions for T-Riesz element decomposition relative to C*-algebra homomorphisms and explores polynomial and generalized decompositions.

Findings

T-Riesz elements can be decomposed into almost T-null and quasi-nilpotent elements under certain conditions

Discussion of polynomial T-Riesz and generalized T-Riesz decompositions

Provides theoretical framework for Riesz decomposition in C*-algebra homomorphisms

Abstract

The aim of the present work is to study the Riesz decomposition relative to a C*-algebra homomorphism T : A --> B. We prove that under some conditions on T, T-Riesz elements can be decomposed into the sum of almost T-null element and quasi-nilpotent element. Also the so-called polyno- mial T-Riesz and generalized T-Riesz decompositions will be discussed.

Equations

$$T^{-1}{\mathcal{B}}^{-1}=\{a\in \mathcal{A}:Ta\in {\mathcal{B}}^{-1}\},$$

$$T^{-1}(0)+{\mathcal{A}}^{-1}=\{c+d:c\in T^{-1}(0)andd\in {\mathcal{A}}^{-1}\},$$

$$T^{-1}(0){+}_{com}{\mathcal{A}}^{-1}=\{c+d:c\in T^{-1}(0),d\in {\mathcal{A}}^{-1}andcd=dc\},$$

$$and\{a\in \mathcal{A}:a\in T^{-1}{\mathcal{B}}^{-1}and0\notin acc{\sigma }_{\mathcal{A}}(a)\}.$$

$$Invertible\Rightarrow almostinvertibleT-Fredholm\Rightarrow T-Browder\Rightarrow T-Weyl\Rightarrow T-Fredholm.$$

$${\sigma }_T(a)$$

$${\omega }_T(a)$$

$${\beta }_T(a)$$

$$and{\sigma }_T(a)\cup acc{\sigma }_A(a)$$

$${\sigma }_T(a)\subseteq {\omega }_T(a)\subseteq {\beta }_T(a)\subseteq {\sigma }_T(a)\cup acc{\sigma }_{\mathcal{A}}(a)\subseteq {\sigma }_{\mathcal{A}}(a).$$

$$\partial H\subseteq K\subseteq H⟹\partial H\subseteq \partial K\subseteq K\subseteq H\subseteq \eta K=\eta H.$$

$$\eta K=\eta H⟹Hfinite\iff Kfinite.$$

$$a\in T^{-1}(0)⟹acc({\sigma }_{\mathcal{A}}(a))\subseteq \{0\}.$$

$$\partial {\sigma }_{\mathcal{A}}(a)\subseteq {\sigma }_T(a)\cup iso({\sigma }_{\mathcal{A}}(a)),$$

$${\beta }_T(a)={\sigma }_T(a)\cup acc({\sigma }_{\mathcal{A}}(a)).$$

$$aisT-Riesz⟺Taisquasi-nilpotentin\mathcal{B}.$$

$${\mathcal{R}}_T(\mathcal{A})\subseteq T^{-1}QN(\mathcal{B})$$

$$T^{-1}QN(\mathcal{B})=T^{-1}(0)+QN(\mathcal{A}).$$

$$P_j=\frac{1}{2\pi i}{\int }_{{\Gamma }_j}(\lambda e_{\mathcal{A}}-a{)}^{-1}d\lambda$$

$$T{P}_j=\frac{1}{2\pi i}{\int }_{{\Gamma }_j}(\lambda e_{\mathcal{A}}-Ta{)}^{-1}d\lambda =0(\text{mbox}since{\lambda }_j\notin {\sigma }_{\mathcal{B}}(Ta)foreveryj=1,2,...)$$

$$r_{\mathcal{A}}((\lambda e-a+d_n{)}^{-1}(d-d_n))$$

$$\le \Vert h_n\Vert \Vert (\lambda -Z{)}^{-1}\Vert$$

$$<1.$$

$${\sigma }_{\mathcal{C}(\mathcal{H})}(\pi (T))={\sigma }_{\pi }(T)=\{\lambda \in \mathbb{C}:T-\lambda eisnotFredholm\}.$$

$$\mathcal{R}(\mathcal{H})={\pi }^{-1}(0)+\mathcal{Q}\mathcal{N}(\mathcal{H}).$$

$${\sigma }_T(a)={\omega }_T(a)={\beta }_T(a)=P^{-1}(0),$$

$$a\in pol{y}^{-1}{T}^{-1}(QN(\mathcal{B}))⟺{\omega }_T(a)isfinite$$

$$acc({\sigma }_{\mathcal{A}}(a))\subseteq {\omega }_T(a).$$

$$P_i=\frac{1}{2\pi i}{\int }_{\partial N_i}(\lambda e_{\mathcal{A}}-a{)}^{-1}d\lambda .$$

$$a=\underset{i=1}{\overset{n}{⨁}}(a_i+{\lambda }_i{e}_i)\text{mbox}andTa=\underset{i=1}{\overset{n}{⨁}}(T{a}_i+{\lambda }_i{e}_i)$$

$$a$$

$$=c+d+f$$

$$P(a)$$

$$=\prod _{i=1}^n(\underset{j=1}{\overset{n}{⨁}}(c_j+d_j+{\lambda }_j{e}_j)-{\lambda }_i{e}_i)$$

$$=\underset{j=1}{\overset{n}{⨁}}(\prod _{i=1}^n(c_j+d_j+{\lambda }_j{e}_j-{\lambda }_i{e}_i))$$

$$=L+R$$

Full text

1. Introduction and Preliminaries

Atkinson’s and Ruston’s characterizations in classical Fredholm theory have motivated several authors to introduce a general Fredholm theory in Banach algebra relative to a closed inessential ideal, this theory has been used by Smyth to generalize the well-known West decomposition in a $C^{\star}$-algebra, for more details of these decompositions we refer the reader to [3, 4, 11, 12] as well as some characterizations of polynomially Riesz operators are given by authors in [5, 6, 13].

Let $\mathcal{A}$, $\mathcal{B}$ two complex Banach algebras with identity $e\neq 0$, $T:\mathcal{A}\rightarrow\mathcal{B}$ be a homomorphism, $\mathcal{A}^{-1}$ the set of invertible elements in $\mathcal{A}$, $QN(\mathcal{A})$ the set of quasinilpotent elements in $\mathcal{A}$, $T^{-1}(0)$ the null space of the homomorphism $T$ and $Rad(\mathcal{B})$ the Jacobson radical of $\mathcal{B}$. It is clair that $T\mathcal{A}^{-1}\subset\mathcal{B}^{-1}$ and $TQN(\mathcal{A})\subset QN(\mathcal{A})$.

For any $a\in\mathcal{A}$ denotes $\sigma_{\mathcal{A}}(a)$, $\sigma_{\mathcal{B}}(Ta)$, $r_{\mathcal{A}}(a)$, $iso\sigma_{\mathcal{A}}(a)$ and $acc(\sigma_{\mathcal{A}}(a))=\sigma_{\mathcal{A}}(a)\backslash iso(\sigma_{\mathcal{A}}(a))$, the spectrum of a in $\mathcal{A}$, the spectrum of $Ta$ in $\mathcal{B}$, the spectral radius of a in $\mathcal{A}$, the set of isolated points and the accumulation points of $\sigma_{\mathcal{A}}(a)$ respectively. Recall that an ideal $\mathcal{J}$ is inessential ideal if for every $x\in\mathcal{J}$ the spectrum $\sigma_{\mathcal{A}}(a)$ is either finite or is a sequence converging to zero.

From Harte [8] the set of $T$-Fredholm elements, $T$-Weyl elements, $T$-Browder elements and almost invertible $T$-Fredholm elements are define respectively by:

[TABLE]

Clearly we have

[TABLE]

For every $a\in\mathcal{A}$, let us define $T$-Fredholm spectrum, the $T$-Weyl spectrum, the $T$-Browder spectrum and the almost invertible $T$-Fredholm spectrum as follows respectively by:

[TABLE]

Evidently

[TABLE]

On the other hand, let $K\subseteq\mathbb{C}$ be a compact subset, $\partial K$ the topological boundary of $K$ and $\eta K$ the connected hull of $K$ with $\mathbb{C}\backslash\eta K$ is the unique unbounded component of $\mathbb{C}\backslash K$, a hole of K is a component of $\eta K\backslash K$, thus $\eta K$ is the union of K and its holes.

It is known [7] that for compact subsets $H,K\subseteq\mathbb{C}$

[TABLE]

If $K\subseteq\mathbb{C}$ is finite then $\eta K=K$. Consequently for compact subsets $H,K\subseteq\mathbb{C}$

[TABLE]

The homomorphism $T$ is said to has the Riesz property if for every $a\in\mathcal{A}$ we have

[TABLE]

Equivalently, $T$ satisfies the Riesz property if and only if $T^{-1}(0)$ is inessential ideal.

And $T$ has the strong Riesz property, if

[TABLE]

Therefore $T$ satisfies the strong Riesz property if and only if $\sigma_{\mathcal{A}}(a)\subseteq\eta\sigma_{T}(a)\cup iso(\sigma_{\mathcal{A}}(a))$.

Recall that if $T$ is a bounded homomorphism and has the Riesz property then almost invertible $T$-Fredholm are $T$-Browder, in this case for an arbitrary $a\in\mathcal{A}$:

[TABLE]

According to [10, 14], $a\in\mathcal{A}$ is $T$-Riesz element if $\lambda e-a$ is $T$-Fredholm for every $\lambda\in\mathbb{C}\backslash\{0\}$. We have the following characterization of $T$-Riesz elements:

[TABLE]

The set of all $T$-Riesz elements is denoted by $T^{-1}QN(\mathcal{B})$.

Now, let $T$ be a bounded homomorphism which has the Riesz property and

$\pi:\mathcal{A}\rightarrow\mathcal{A}/T^{-1}(0)$ the quotient map.

Write $\mathcal{R}_{T}(\mathcal{A})=\{a\in\mathcal{A}:\>\pi(a)\,\,\,is\,\,\,quasi-nilpotent\,\,\,in\,\,\,\mathcal{A}/T^{-1}(0)\}$ the set of Riesz elements relative to $T^{-1}(0)$. It is easy to show that

[TABLE]

Furthermore if $a\in T^{-1}QN(\mathcal{B})$ then $acc(\sigma_{\mathcal{A}}(a))\subseteq\{0\}$.

This work focus on Fredholm theory relative to Banach algebra homomorphism introduced by Harte [8, 9, 10, 14] in the special case of $C^{\star}$-algebras to extend West and Smyth decompositions also the decomposition of polynomially Riesz operators will be generalized .

2. Main results

Throughout the rest of this paper $\mathcal{A}$ and $\mathcal{B}$ are two $C^{\star}$-algebras however proofs are due to Gelfand-Naimark’s theorem [[1], theorem 6.2.20] which states that every $C^{\star}$-algebra is isometrically $\star$-isomorph to a closed self-adjoint subalgebra of $\mathcal{B}(\mathcal{H})$ for some Hilbert space $\mathcal{H}$.

Definition 2.1**.**

[14]** Let $T:\mathcal{A}\longrightarrow\mathcal{B}$ be a homomorphism, an element $a\in\mathcal{A}$ is said $T$-null if $Ta=0$ and almost $T$-null if $Ta\in Rad(\mathcal{B})$.

Because $C^{\star}$-algebras are semi-simple then $Rad(\mathcal{B})=0$, in this case $T$-null and almost $T$-null elements are the same.

Now, we are in position to establish the following theorem, the technique used for dealing with the proof of [[3], $C^{\star}$.2.3 Theorem] can also be used for showing the following theorem.

Theorem 2.1**.**

Let $T:\mathcal{A}\rightarrow\mathcal{B}$ be a bounded homomorphism which has Riesz property. Then every $a\in T^{-1}QN(\mathcal{B})$ has the decomposition $a=c+d$ with d is a normal element of $T^{-1}(0)$ and $c\in QN(\mathcal{A})$, i.e:*

[TABLE]

**Proof:    

**It is clair that $T^{-1}(0)+QN(\mathcal{A})\subseteq T^{-1}QN(\mathcal{B})$ so it remains to show the other inclusion, in fact let $a\in T^{-1}QN(B)$, if $\sigma_{\mathcal{A}}(a)=\{0\}$ or $\sigma_{\mathcal{A}}(a)$ is finite then the result is trivial. To avoid triviality let assume that $\sigma_{\mathcal{A}}(a)$ has an infinite number of isolated points, set $\sigma_{\mathcal{A}}(a)=(\lambda_{k})_{k\geq 1}\cup\{0\}$ with $\{|\lambda_{k}|\}_{1}^{\infty}$ is a decreasing sequence of non zero isolated points of $\sigma_{\mathcal{A}}(a)$ such that $\lim\lambda_{k}=0$.

Consider $P_{j}$ the spectral projection of $a$ associated to the spectral set $\{\lambda_{j}\}$ given by:

[TABLE]

where $\Gamma_{j}$ is a circle of center $\lambda_{j}$ which contains no other points of $\sigma_{\mathcal{A}}(a)$. By [[10], theorem 1.1] we have

[TABLE]

Therefore $P_{j}\in T^{-1}(0)$ for every j=1,2,… , set $S_{n}=\sum_{k=1}^{n}{P_{k}}$ and let $q_{n}$ be a self-adjoint projection in $T^{-1}(0)$ such that $q_{n}(\mathcal{H})=S_{n}(\mathcal{H})$.

Write $E_{i}=q_{i}-q_{i-1}$ with $q_{0}=0$ and set $d_{n}=\sum_{k=1}^{n}{\lambda_{k}E_{k}}$, then $d_{n}\in T^{-1}(0)$. Since $\{E_{i}\}_{1}^{n}$ is a sequence of self-adjoint projections then $d_{n}$ is a normal element of $T^{-1}(0)$ for every n and $d_{n}\longrightarrow d\in T^{-1}(0)$ when $n\rightarrow\infty$.

Considering the decomposition $\mathcal{H}=q_{n}(\mathcal{H})\bigoplus(1-q_{n})(\mathcal{H})$, the matrix formes associated to this decomposition are:

d_{n}=\left(\begin{array}[]{c c c}k_{n}\>\>\>0\\ \\ 0\>\>\>\>\>0\end{array}\right) with $\sigma(d_{n})=\sigma(k_{n})=\{\lambda_{k}\}_{1}^{n}$

and a=\left(\begin{array}[]{c c c}k_{n}\>\>\>*\\ \\ 0\>\>\>\>\>Z\end{array}\right) with $\sigma(a)=\sigma(k_{n})\cup\sigma(Z)$, consequently $\sigma(Z)=\sigma(a)\setminus\{\lambda_{k}\}_{1}^{n}.$

Now, it suffice to show that $c=a-d$ is quasi-nilpotent, to show this let $\lambda\in\mathbb{C}\backslash\{0\}$. Without lose generality we may assume that $\lambda\notin\sigma(a-d_{n})$, then $\lambda e-a+d_{n}$ is invertible.

We have a-d_{n}=\left(\begin{array}[]{c c c}0\>\>\>\>\>*\\ \\ 0\>\>\>\>\>Z\end{array}\right) then $\sigma(a-d_{n})=\sigma(Z)=\sigma(a)\setminus\{\lambda_{k}\}_{1}^{n}$

Observe that (\lambda e-a+d_{n})^{-1}=\left(\begin{array}[]{c c c}\frac{1}{\lambda}I_{n}\>\>\>\>\>\>\>\>\>\>\>\>*\\ \\ \>\>0\>\>\>\>\>\>\>(\lambda-Z)^{-1}\end{array}\right) and d-d_{n}=\left(\begin{array}[]{c c c}0\>\>\>\>\>0\\ \\ 0\>\>\>\>\>h_{n}\end{array}\right)

where $I_{n}$ is the identity matrix and $\sigma(h_{n})=\sigma(d)\setminus\{\lambda_{k}\}_{1}^{n}$.

Thus (\lambda-a+d_{n})^{-1}(d-d_{n})=\left(\begin{array}[]{c c c}0\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>*\\ \\ \>\>0\>\>\>\>\>\>\>h_{n}(\lambda-Z)^{-1}\end{array}\right).

Assume that $\|h_{n}\|<\|(\lambda-Z)^{-1}\|^{-1}$ then

[TABLE]

Therefore $1+(\lambda e-a+d_{n})^{-1}(d-d_{n})$ is invertible, which means that $\lambda e-a+d$ is invertible and hence $\lambda\notin\sigma(a-d)$, consequently $\sigma(c)={0}$.         $\blacksquare$

Remarks 1**.**

If in addition $T$ is onto then $\mathcal{R}_{T}(\mathcal{A})=T^{-1}QN(\mathcal{B})$, in this case the decomposition in theorem 2.1 follows immediately from [11].

Example 1**.**

*Let $\mathcal{H}$ be a Hilbert space, $\mathcal{B}(\mathcal{H})$ the algebra of bounded linear operators on $\mathcal{H}$, $\mathcal{K}(\mathcal{H})$, $\mathcal{QN}(\mathcal{H})$ and $\mathcal{R}(\mathcal{H})$ the closed ideal of compact operators, the class of quasi-nilpotent operators and the class of Riesz operators respectively. Consider $\mathcal{C}(\mathcal{H})=\mathcal{B}(\mathcal{H})/\mathcal{K}(\mathcal{H})$ the Calkin algebra and $\pi:\mathcal{B}(\mathcal{H})\longrightarrow\mathcal{C}(\mathcal{H})$ the canonical bounded homomorphism.

It is clear that $\pi$ has the Riesz property since $\pi^{-1}(0)=\mathcal{K}(\mathcal{H})$ is an inessential ideal.

In the other hand if $T\in\mathcal{B}(\mathcal{H})$, the Atkinson’s characterization said that $T$ is Fredholm if and only if $\pi(T)$ is invertible in the Calkin algebra, thus*

[TABLE]

According to Ruston characterization, $T$ is Riesz operator if and only if $\pi(T)$ is quasi-nilpotent in the Calkin algebra, so by West [12] we have the following decomposition

[TABLE]

Definition 2.2**.**

[14*]** Let $a\in\mathcal{A}$, $a$ is called polynomial $T$-Riesz element if there exist a non-zero polynomial P such that $P(a)$ is $T$-Riesz.

Note $poly^{-1}T^{-1}(QN(\mathcal{B}))$ the set of all polynomial $T$-Riesz elements.*

Lemma 2.1**.**

Let $a\in poly^{-1}T^{-1}(QN(\mathcal{B}))$ with $T$ is a bounded homomorphism which has the strong Riesz property then*

[TABLE]

For a polynomial P of minimal degree, and $acc(\sigma_{\mathcal{A}}(a))\subseteq\omega_{T}(a)$.

**Proof:    

**By [[14], theorem 11.1] (1) is hold,

In this case we have $\omega_{T}(a)=\sigma_{T}(a)\cup acc\sigma_{\mathcal{A}}(a)$ then obviously $acc(\sigma_{\mathcal{A}}(a))\subseteq\omega_{T}(a)$.

The following theorem gives the decomposition of polynomially $T$-Riesz element in $C^{\star}$-algebra.

Theorem 2.2**.**

Let T be a bounded homomorphism which has the strong Riesz property then

[TABLE]

In this case the following decomposition $a=c+d+f$ is hold with $c\in T^{-1}(0)$, $d\in QN(\mathcal{B})$ and $\sigma(f)=\omega_{T}(a).$

**Proof:    

**Assume that $\omega_{T}(a)$ is finite, claim that $acc\sigma_{\mathcal{A}}(a)\subseteq\omega_{T}(a)$.

In fact, let $\lambda\in\sigma_{\mathcal{A}}(a)\backslash\omega_{T}(a)$, then $\lambda e-a$ is $T$-Fredholm but not invertible, thus $\lambda\in\partial\sigma_{\mathcal{A}}(a)$, if it is not then $\lambda\in int(\sigma_{\mathcal{A}}(a))$, in this case $\sigma_{T}(a)$ would be infinite which is impossible.

So $\lambda\in\partial\sigma_{\mathcal{A}}(a)\backslash\sigma_{T}(a)\subseteq iso\sigma_{\mathcal{A}}(a)$, and therefore

[TABLE]

Set $\sigma_{T}(a)=\omega_{T}(a)=\{\lambda_{1},\lambda_{2},...\lambda_{n}\}$, so $acc\sigma_{\mathcal{A}}(a)\subseteq\{\lambda_{1},\lambda_{2},...\lambda_{n}\}$.

Let now define a collection $\{\triangle_{1},\triangle_{2},...\triangle_{n}\}$ of subsets of $\sigma_{\mathcal{A}}(a)$ in the following way:

$(i)\>\>\>\bigcup_{i=1}^{n}\triangle_{i}=\sigma_{\mathcal{A}}(a),$

$(ii)\>\>\>\triangle_{i}\cap\triangle_{j}=\emptyset\>\>\>if\>i\neq j,$

$(iii)\>\>\>\lambda_{i}\in\triangle_{i}\>\>\>for\>\>i=1,2,...,n,$

For every i=1,2,..,n let $N_{i}$ be a neighborhood of $\triangle_{i}$ which contains no other points of $\sigma_{\mathcal{A}}(a)$ than $\lambda_{i}$ and let $P_{i}$ the spectral projection of $a$ associated to $\triangle_{i}$ for i=1,2,…,n

[TABLE]

We have the decomposition $\mathcal{H}=R(P_{1})\bigoplus R(P_{2})\bigoplus...\bigoplus R(P_{n})$, thus we have

[TABLE]

Therefore $\sigma_{\mathcal{B}}(Ta)=\bigcup_{i=1}^{n}\sigma_{\mathcal{B}}(Ta_{i}+\lambda_{i}e_{i})$ and consequently $\sigma_{\mathcal{B}}(Ta_{i})=\sigma_{T}(a_{i})=\{0\}$ for every $i=1,2,...,n$. Then $a_{i}$ is $T$-Riesz for every $i=1,2,...,n$.

Next, using theorem 2.1 we have $a_{i}=c_{i}+d_{i}$ where $c_{i}\in QN(\mathcal{A})$ and $d_{i}\in T^{-1}(0)$. Consequently

[TABLE]

where $c=\bigoplus_{i=1}^{n}c_{i}$, $d=\bigoplus_{i=1}^{n}d_{i}$ and $f=\bigoplus_{i=1}^{n}\lambda_{i}e_{i}$. so $c\in QN(\mathcal{A})$, $d\in T^{-1}(0)$

and $\sigma(f)=\omega_{T}(a)$.

Now, set $P(\lambda)\equiv\prod_{i=1}^{n}(\lambda-\lambda_{i})$ thus

[TABLE]

With $L\in T^{-1}(0)$ and $R=c_{j}^{n}+\mu_{n-1}c_{j}^{n-1}+...+\mu_{1}c_{j}$ for some $\mu_{i}\in\mathbb{C}$       $i=1,2,...,n-1$.

It is clear that $R\in QN(\mathcal{A})$ and

[TABLE]

Hence $P(a)\in T^{-1}QN(\mathcal{B})$, thus the proof is complete.       $\blacksquare$

Similarly as in [[5], definition 1.2] we have the following

Definition 2.3**.**

Let $a\in\mathcal{A}$, $a$ is said generalized $T$-Riesz element if there exist a finite subset $\Omega\subseteq\mathbb{C}$ such that:

[TABLE]

Write $GT^{-1}QN(\mathcal{B})$ the set of generalized $T$-Riesz elements.

To conclude this section we have the following result and remark.

Corollary 2.1**.**

Let $T:\mathcal{A}\longrightarrow\mathcal{B}$ be a bounded homomorphism which has the Riesz property then every $a\in GT^{-1}QN(\mathcal{B})$ has the decomposition $a=d+c$ with $d$ is normal element of $T^{-1}(0)$ and $\sigma(c)=\Omega$.

Remarks 2**.**

*1)- Obviously $T^{-1}QN(B)\subseteq poly^{-1}T^{-1}QN(\mathcal{B})$ and if $a\in\mathcal{A}$ is polynomial $T$-Riesz for a polynomial $P$ which has no zero in $\sigma_{\mathcal{A}}(a)$, then a is $T$-Riesz.

2)- By lemma 2.1 and theorem 2.1 we have*

[TABLE]