Semi-Fredholm Theory on Hilbert C*-modules
Stefan Ivkovic
TL;DR
This paper extends the classical semi-Fredholm theory to Hilbert C*-modules, defining semi-Fredholm operators and proving their properties, including one-sided invertibility modulo compact operators and openness of the set.
Contribution
It introduces a new semi-Fredholm theory on Hilbert C*-modules, generalizing existing Fredholm theory and establishing foundational properties of semi-Fredholm operators in this context.
Findings
Semi-Fredholm operators are characterized as one-sided invertible modulo compact operators.
The set of proper semi-Fredholm operators is shown to be open.
The theory generalizes classical results to the setting of Hilbert C*-modules.
Abstract
In this paper we establish the semi-Fredholm theory on Hilbert C*-modules as a continuation of Fredholm theory on Hilbert C*-modules established by Mishchenko and Fomenko. We give a definition of a semi-Fredholm operator on Hilbert C*-module and prove that these semi-Fredholm operators are those that are one-sided invertible modulo compact operators, that the set of proper semi-Fredholm operators is open and many other results that generalize their classical counterparts.
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††Copyright 2019 The Mathematical Institute of the Serbian Academy of Sciences and Arts, Kneza Mihaila 36, 11000 Beograd, p.p. 367, Serbia, [email protected]
Semi-Fredholm theory on Hilbert -modules
Stefan Ivković
The Mathematical institute of the Serbian Academy of Sciences and Arts
Acknowlegedgement. I am especially grateful to my main supervisor Professor Vladimir M. Manuilov for careful reading of the first version of this paper and for detailed comments and suggestions which led to the improved presentation of the paper. I am also grateful to my supervisor Professor Dragan S. Đorđević for suggesting the research topic of this paper and for introducing to me the relevant reference books. In addition, I am also grateful to my external supervisor Professor Camillo Trapani for reading my paper and for inspiring and useful comments and advice. Finally, I am grateful to the Refere for careful reading of my paper and for the inspiring and useful comments and suggestions that led to the final, improved presentation of the paper.
Abstract.
In this paper we establish the semi-Fredholm theory on Hilbert -modules as a continuation of Fredholm theory on Hilbert -modules established by Mishchenko and Fomenko. We give a definition of a semi-Fredholm operator on Hilbert -module and prove that these semi-Fredholm operators are those that are one-sided invertible modulo compact operators, that the set of proper semi-Fredholm operators is open and many other results that generalize their classical counterparts.
Key words and phrases:
Hilbert -module, semi--Fredholm Operator, Calkin algebra.
2010 Mathematics Subject Classification:
Primary 47A53; Secondary 46L08.
1. Introduction
The Fredholm and semi-Fredholm theory on Hilbert and Banach spaces started by studying the certain integral equations which was done in the pioneering work by Fredholm in 1903 in [2]. After that the abstract theory of Fredholm and semi-Fredholm operators on Banach spaces was during the time developed in numerous papers. Some recent results in the classical semi-Fredholm theory can be found in [17]. Now, Fredholm theory on Hilbert -modules as a generalization of Fredholm theory on Hilbert spaces was started by Mishchenko and Fomenko in [8]. They have elaborated the notion of a Fredholm operator on the standard module and proved the generalization of the Atkinson theorem. Our aim is to study more general operators than the Fredholm ones, namely a generalization of semi-Fredholm operators. In this paper we give the definition of those and establish several properties as an analogue or a generalized version of the properties of the classical semi-Fredholm operators on Hilbert and Banach spaces.
Recall that if is a Hilbert space, then is a semi-Fredholm operator on , denoted by if and is closed, that is if there exists a decomposition
[TABLE]
with respect to which has the matrix \left[\begin{array}[]{ll}F_{1}&0\\ 0&0\\ \end{array}\right], where is an isomorphism, and either
[TABLE]
If then is called an upper semi-Fredholm operator on denoted by whereas if then is called a lower semi-Fredholm operator on denoted by If is both an upper and lower semi-Fredholm operator on , then is said to be a Fredholm operator on denoted by In the case when , the index of is defined as Now, Hilbert -modules are a natural generalization of Hilbert spaces when the field of scalars is replaced by an arbitrary -algebra. Some recent results in the theory of Hilbert -modules can be found in [3], [5], [6], [9], [14]. In [8] one considers a standard Hilbert -modul over a unital -algebra , denoted by and one defines an -Fredholm operator on as generalization of a Fredholm operator on Hilbert space in the following way: [8, Definition] A (bounded linear) operator is called -Fredholm if
-
it is adjointable;
-
there exists a decomposition of the domain and the range, , where are closed -modules and have a finite number of generators, such that has the matrix from
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right]
with respect to these decompositions and is an isomorphism.
It is then proved in [8] that some of the main results from the classical Fredholm theory on Hilbert spaces also hold when one considers this generalization of Fredholm operator on Hilbert -module. The idea in this paper was to go further in this direction, to give, in a similar way, a definition of semi-Fredholm operators on to investigate and prove generalized version in this setting of significantly many results from the classical semi-Fredholm theory on Hilbert and Banach spaces.
In the second section, inspired by [8, Definition], we define upper and lower semi--Fredholm operators on We say that F is an upper semi--Fredholm operator on if all the conditions in the definition above hold except that does not need to be finitely generated and we say that is a lower semi--Fredholm operator on if all the conditions in the definition above hold except that in this case does not need to be finitely generated. Then we show that the classes of upper and lower semi--Fredholm operators on denoted respectively by and coincide with the inverse images of the sets of the left and right invertible elements respectively in the -algebra under the quotient map, where denotes the -algebra of all bounded, adjointable operators on Semi--Fredholm operators on have been considered in [1] and [4]. In [1] they define semi--Fredholm operators to be those that are one-sided invertible modulo compact operators. However, in this paper we give another definition of semi--Fredholm operators on and then prove that these operators are exactly those that are one-sided invertible modulo compact operators. Moreover, we prove an analogue or generalized versions of main results in [18, Section 1.2] and [18, Section 1.3], as well as some additional new results. We wish to remark that if one considers the classes of operators in the sense of Definition 2.1 in this paper, which are given in terms of the decompositions for given it is not obvious that as we have in the classical semi-Fredholm theory on Hilbert and Banach spaces. This is due to the fact that these decompositions may not be unique for an whereas in the classical case one always considers the decomposition
[TABLE]
where is a Hilbert space and Key arguments in proving that is the equivalence between property and the left invertibility in the -algebra and the equivalence between -property and the right invertibility in
This is also the main argument in proving that
[TABLE]
are semigroups, as we have in the classical semi-Fredholm theory on Banach spaces. On the other hand, Lemma 2.16 and Lemma 2.17 are very important as well and they are also used in proving several other fundamental results in the paper, where we meet the challenge with non uniqueness of the decomposition.
In the third section in this paper we prove analogue results of [18, Lemma 1.4.1] and [18, Lemma 1.4.2]. More precisely, we give generalizations on Hilbert -modules of the results from the classical semi-Fredholm theory on Banach spaces connected with the Schechter’s characterization of operators on Banach spaces given in [13] .
In the fourth section, we prove that and are open as an analogue of the well known result in the classical semi-Fredholm theory which states that the sets and are open, where is a Banach space, the result which was proved in [12]. Here again Lemma 2.16 and Lemma 2.17 from the second chapter are one of the main arguments in the proof. Also, we prove an analogue version on of [18, Corollary 1.6.10] and [18, Lemma 1.6.11].
In the fifth section we give first a generalization on Hilbert -modules of and operators on Hilbert spaces. A natural generalization on of the class as defined in [18, Definition 1.2.1], where is a Banach space would be the following: Let Then, if there exists a decomposition
[TABLE]
with respect to which has the matrix \left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right], where is an isomorphism, is finitely generated and that is is isomorphic to a closed submodule of If that is if is an ordinary Hibert space, then this definition would concide with [18, Definition 1.2.1] of the class of operators on Hilbert spaces. However, if is a Hilbert or a Banach space, then when we consider the [18, Definition 1.2.1] of classes and whereas if we consider this „generalized“ definition of the class it is not true in general that
[TABLE]
This is due to the fact that given a finitely generated closed submodule of and a countably, but not finitely generated closed submodule of it is not true in general that that is isomorphic to a closed submodule of Therefore, we define the class to be the class of operators in which satisfy the conditions of the suggested „generalized“ definition above of Moreover, we define be the subclass of , where in the decomposition above we have in addition that is finitely generated. Then we set
[TABLE]
In a similar way we define the classes and set
[TABLE]
We also then prove that
[TABLE]
In addition we show that if and
[TABLE]
is any decomposition with respect to which has the matrix \left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right], where is an isomorphism and are finitely generated, then if satisfies cancellation property, we have Similar conclusion yields for operators in the class only in this case In addition, we show that the classes
[TABLE]
are open. In the rest of this section we work with these classes of operators and prove analogy or generalized versions of almost all results in [18, Section 1.9].
The generalized versions on of the results from the classical semi-Fredholm theory on Banach and Hilbert spaces, which are presented here in this paper, demand different proofs from the proofs in the classical case, however the techniques used in these proofs are to a certain extent inspired by the techniques used in the proofs of some of the results in [8]. Moreover, the first section and especially the fourth section, where we introduce new, additional classes of operators as and contain some new results and not just generalizations on of results from the classical semi-Fredholm theory on Banach spaces.
2. Semi--Fredholm operators on
In this section we define semi--Fredholm operators on the standard module and prove some of the main properties and results concerning these operators. Throughout this paper we let be a unital -algebra, be a standard module over and we let denote the set of all bounded , adjointable operators on According to [10, Definition 1.4.1], we say that a Hilbert -module over is finitely generated if there exists a finite set such that equals the linear span (over C and ) of this set.
Definition 2.1**.**
Let We say that is an upper semi--Fredholm operator if there exists a decomposition
[TABLE]
with respect to which has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism are closed submodules of and is finitely generated. Similarly, we say that is a lower semi--Fredholm operator if all the above conditions hold except that in this case we assume that ( and not ) is finitely generated.
Set
is upper semi--Fredholm
is lower semi--Fredholm
is -Fredholm operator on Then obviously . We are going to show later in this section that actually ”=” holds.
Notice that if are two arbitrary Hilbert modules -modules, the definition above could be generalized to the classes and .
Recall that by [10, Definition 2.7.8], originally given in [8], when and
[TABLE]
is an decomposition for , then the index of is definited by index where and denote the isomorphism classes of and respectively. „By [10, Definition 2.7.9], the index is well defined and does not depend on the choice of decomposition for . As regards the -group , it is worth mentioning that it is not true in general that [M]=[N] implies that for two finitely generated submodules of . If satisfies the property that [N]=[M] implies that for any two finitely generated, closed submodules of , then is said to satisfy ”the cancellation property”, see [16, Section 6.2].
Theorem 2.2**.**
*Let The following statements are equivalent
-
-
There exists such that for some *
Proof.
2) 1) If 2) holds, then by [10, Lemma 2.7.12]. Let be a decomposition w.r.t which DF has the matrix
\left[\begin{array}[]{ll}(DF)_{1}&0\\ 0&(DF)_{4}\\ \end{array}\right],
where is an isomorphism and are finitely generated. We wish to show that is closed and we will do it by showing that is bounded below. Suppose that this is not the case. Then there exists a sequence s.t. for all and as Since is bounded, we must have that as But this would mean that is not bounded below on as for all This is a contradiction since is an isomorphism. Hence we must have that is bounded below on which means that is closed.
Now, by [10, Theorem 2.7.6], the result which was originally proved in [15], we may assume that is orthogonally complementable in . Hence is adjointable, so by [10, Theorem 2.3.3], which was originally proved in [7] , is orthogonally complementable in .
Hence With respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}F_{1}&F_{2}\\ 0&F_{4}\\ \end{array}\right], where is an isomorphism. If we let
U=\left[\begin{array}[]{ll}1&-{F_{1}}^{-1}{F_{2}}\\ 0&1\\ \end{array}\right]
with respect to the decomposition
[TABLE]
then is an isomorphism and with respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}F_{1}&0\\ 0&\tilde{F}_{4}\\ \end{array}\right]. Since is finitely generated, is finitely generated also, hence
Let
[TABLE]
be a decomposition with respect to which has the matrix \left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right], where is an isomorphism and is finitely generated. Since is finitely generated, it is orthogonally complementable in by [10, Lemma 2.3.7] which was originally proved in [7]. Then, by the proof of [10, Theorem 2.7.6] , we can deduce that is an isomorphism onto . Now, , where denotes the orthogonal projection onto . Since and is closed being isomorphic to by [10, Theorem 2.3.3], it follows that is orthogonally complementable. With respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}{\tilde{F}_{1}}&{\tilde{F}_{2}}\\ 0&{\tilde{F}_{4}}\\ \end{array}\right], where is an isomorphism. Clearly and are then adjointable.
Let be the operator which has the matrix \left[\begin{array}[]{ll}{\tilde{F}_{1}}^{-1}&0\\ 0&0\\ \end{array}\right] with respect to the decomposition
[TABLE]
Then and DF=\left[\begin{array}[]{ll}1&{\tilde{F}_{1}}^{-1}{\tilde{F}_{2}}\\ 0&0\\ \end{array}\right] with respect to the decomposition
[TABLE]
Let K=\left[\begin{array}[]{ll}0&{\tilde{F}_{1}}^{-1}\tilde{F_{2}}\\ 0&-1\\ \end{array}\right] with respect to the same decomposition. Since is finitely generated, we have . Moreover, ∎
Theorem 2.3**.**
*Let Then the following statements are equivalent:
-
-
There exist s.t. *
Proof.
Let
[TABLE]
be an decomposition for As in the proof of Theorem 2.2, we deduce that is closed and orthogonally complementable in
With respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}D_{1}&D_{2}\\ 0&D_{4}\\ \end{array}\right], where is an isomorphism, as in the proof of Theorem 2.2, part we deduce thet has the matrix \left[\begin{array}[]{ll}D_{1}&0\\ 0&\tilde{D}_{4}\\ \end{array}\right] with respect to the decomposition
[TABLE]
where is an isomorpism. Since is finitely generated, it follows that
Let
[TABLE]
be an decomposition for ( so that is finitely generated ). Since is finitely generated, it is orthogonally complementable by [10, Lemma 2.3.7]. Now, since
[TABLE]
we have that is an isomorphism from onto , where denotes the orthogonal projection onto Since has the matrix \left[\begin{array}[]{ll}D_{1}&0\\ 0&D_{4}\\ \end{array}\right] with respect to the decomposition
[TABLE]
where is an isomorphism, it follows that , and which is closed. By [10, Theorem 2.3.3], is orthogonally complementable, so Hence is an isomorphism from onto , where denotes the projection onto along . Therefore, is an isomorphism from onto But since and it follows that
[TABLE]
Hence is an isomorphism from onto being a composition of isomorphisms, so with respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}\tilde{D_{1}}&0\\ \tilde{D_{3}}&\tilde{D_{4}}\\ \end{array}\right], where is an isomorphism.
Let F=\left[\begin{array}[]{ll}(\tilde{D_{1}})^{-1}&0\\ 0&0\\ \end{array}\right] with respect to the decomposition
[TABLE]
Then and DF=\left[\begin{array}[]{ll}1&0\\ \tilde{D}_{3}\tilde{D}_{1}^{-1}&0\\ \end{array}\right] with respect to the decomposition
[TABLE]
Since is finitely generated, it follows that if we let the operator
K=\left[\begin{array}[]{ll}0&0\\ D_{3}D_{1}^{-1}&-1\\ \end{array}\right] w.r.t the decomposition above , the . Moreover ∎
Recall that is a -algebra and is a closed two sided ideal in Hence is also -algebra, equipped with the quotient norm. We will call this algebra the ”Calkin” algebra.
Corollary 2.4**.**
**
Proof.
It suffices to show By Theorem 2.2, consists of all elements that are left invertible in the ”Calkin” algebra, whereas consists of all elements that are right invertible in the ”Calkin” algebra by Theorem 2.3. Now by [10, Theorem 2.7.14] and also by the proof of [10, Lemma 2.7.15], we have that consists of all elements that are invertible in the ”Calkin” algebra. The corollary follows. ∎
Corollary 2.5**.**
* and are semigroups under multiplication.*
Proof.
Follows directly from Theorem 2.2 and Theorem 2.3, as consists of all elements that are left invertible in the ”Calkin” algebra whereas consists of all elements that are right invertible in the ”Calkin” algebra. ∎
Corollary 2.6**.**
*Let If then If then
Proof.
Suppose that By Theorem 2.2 there exists some s.t. Again, by Theorem 2.2 it follows that The proof of the second statement of Corollary 2.6 is similar.
∎
Corollary 2.7**.**
Let If and then If and then
Proof.
Suppose that and By Theorem 2.2 there exist some s.t. as by assumption. Moreover, since by the proof of [10, Lemma 2.7.15] there exist some s.t. Hence and Therefore, So Since by Theorem 2.2 it follows that The proof of the second statement of Corollary 2.7 is similar. ∎
Corollary 2.8**.**
Let If and then If and then
Proof.
Let and Since by the proof of [10, Lemma 2.7.15] there exists some such that By Theorem 2.3, we have then that So But, by Corollary 2.4, so The proof of the second statement of Corollary 2.8 is similar. ∎
Corollary 2.9**.**
If and , then . If and then
Proof.
Suppose that and Since by the proof of [10, Lemma 2.7.15] there exist some s.t
Moreover, since by the proof of [10, Lemma 2.7.15] there exist some s.t. Hence and Thus Hence But so we obtain So Since by Theorem 2.3 we have that Now, since by Corollary 2.6 it follows that also. Hence
[TABLE]
by Corollary 2.5. The proof of the second statement of Corollary 2.9 is similar. ∎
Corollary 2.10**.**
* and are two sided ideals in and respectively. In particular, they are semigroups under multiplication.*
Proof.
Let and suppose first that
Since is a semigroup by Corollary 2.5 Now, if by Corollary 2.8 we have Then, by Corollary 2.9, it would follow that which is a contradiction. Thus we must have that Suppose next that Again, if then, since by Corollary 2.8 we would have that which is impossible. So, also in this case, we must have that
Similarly one can prove the statement for ∎
Corollary 2.11**.**
Let Then if and only if Moreover, if then and
Proof.
Observe that it follows from the proofs of Theorems 2.2 and 2.3 part which could be generalized to the case when (and not only when that if then for and consequently for there exist decompositions
with respect to which and have matrices
\left[\begin{array}[]{ll}F_{1}&F_{2}\\ 0&F_{4}\\ \end{array}\right],
\left[\begin{array}[]{ll}F_{1}^{*}&0\\ F_{2}^{*}&F_{4}^{*}\\ \end{array}\right],
respectively, where are isomorphisms and is finitely generated. Using the technique of diagonalization as in the proof of [10, Lemma 2.7.10], we deduce that as is finitely generated. The proof is analogue when only in this case is finitely generated. If in addition is in , then both and will be finitely generated. Using again the tecnique of diagonalization, one deduces easily that in this case and , , so ∎
Lemma 2.12**.**
Let M be a closed submodule of s.t. for some finitely generated submodule Let , be the inclusion map from into and suppose that Then
Proof.
Consider a decomposition with respect to which
FJ_{M}=\left[\begin{array}[]{ll}(FJ_{M})_{1}&0\\ 0&(FJ_{M})_{4}\\ \end{array}\right],
where is an isomorphism and is finitely generated. Then has the matrix
\left[\begin{array}[]{ll}F_{1}&F_{2}\\ 0&F_{4}\\ \end{array}\right]
with respect to the decomposition
[TABLE]
where is an isomorphism. Using the technique of diagonalization as in the proof of [10, Lemma 2.7.10] and the fact that is finitely generated as both and are so, we deduce that ∎
Suppose now that and that is closed. Then, again by [10, Theorem 2.3.3 ], is orthogonally complementable in , so .
Lemma 2.13**.**
Suppose that and is closed. Then
Proof.
Let be the decomposition with respect to which has the matrix
\left[\begin{array}[]{ll}(DF)_{1}&0\\ 0&(DF)_{4}\\ \end{array}\right],
where is an isomorphism and is finitely generated.
Since , that is , we get that
[TABLE]
To see this, let , then , for some and for some .
Hence , so . As also, we have . Thus . Since and we deduce that
[TABLE]
as was arbitrary. With respect to the decomposition
[TABLE]
has the matrix
\left[\begin{array}[]{ll}(DJ_{{\rm ran}F})_{1}&0\\ 0&(DJ_{{\rm ran}F})_{4}\\ \end{array}\right],
where is an isomorphism. Now, since has the matrix
\left[\begin{array}[]{ll}(DF)_{1}&0\\ 0&(DF)_{4}\\ \end{array}\right]
with respect to the decomposition
[TABLE]
it is easily seen that which is finitely generated. We are done. ∎
Corollary 2.14**.**
Let V be a finitely generated Hilbert submodule of and suppose that , where is the orthogonal projection onto along Then
Proof.
Passing to the adjoints and using Lemma 2.12 together with Corollary 2.11, one obtains the result. ∎
Corollary 2.15**.**
Let and suppose that is closed. If , then
Proof.
Observe that since is closed, then by the proof of [10, Theorem 2.3.3], we have that Hence Passing to the adjoints and using Lemma 2.13 together with Corollary 2.11, one deduces the corollary. ∎
Lemma 2.16**.**
Let and suppose that there are two decompositions
[TABLE]
[TABLE]
with respect to which F has matrices
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],* \left[\begin{array}[]{ll}F_{1}^{\prime}&0\\ 0&F_{4}^{\prime}\\ \end{array}\right],*
respectively, where are isomorphisms, are closed, finitely generated and is just closed. Then is finitely generated also.
Proof.
Since are finitely generated, by [10, Theorem 2.7.5], there exist an such that
and
, ,
Then
[TABLE]
consequently , are isomorphisms from onto and from onto respectively, where , denote the restrictions of projections onto and along and restricted to and respectively. Since and and it follows that
[TABLE]
where denotes the projection onto along . Hence is an isomorphism as and are so. Similarly, is an isomorphism, where denotes the projection onto along .
We get then that has the matrices
\left[\begin{array}[]{ll}\tilde{F_{1}}&0\\ \tilde{F}_{3}&F_{4}\\ \end{array}\right], \left[\begin{array}[]{ll}\tilde{F_{1}^{\prime}}&0\\ \tilde{F}_{3}^{\prime}&F_{4}^{\prime}\\ \end{array}\right] with respect to the decompositions
[TABLE]
[TABLE]
respectively, where , are isomorphisms. As in the proof of [10, Lemma 2.7.11], we let
V=\left[\begin{array}[]{ll}1&0\\ -\tilde{F_{3}}\tilde{F_{1}}^{-1}&1\\ \end{array}\right], V^{\prime}=\left[\begin{array}[]{ll}1&0\\ -\tilde{F^{\prime}_{3}}\tilde{F^{\prime}_{1}}^{-1}&1\\ \end{array}\right],
with respect to the decomposition
[TABLE]
[TABLE]
Then has the matrices
\left[\begin{array}[]{ll}\tilde{\tilde{F_{1}}}&0\\ 0&\tilde{\tilde{F_{4}}}\\ \end{array}\right], \left[\begin{array}[]{ll}\tilde{\tilde{F_{1}}}^{\prime}&0\\ 0&\tilde{\tilde{F_{4}}}^{\prime}\\ \end{array}\right]
with respect to the decompositions
[TABLE]
[TABLE]
respectively, where are isomorphism. Again, as in the proof of [10, Lemma 2.7.11], we change these decompositions into
[TABLE]
[TABLE]
and with respect to these decompositions has matrices
\left[\begin{array}[]{ll}\tilde{\tilde{F_{1}}}&0\\ 0&\tilde{\tilde{F_{4}}}\\ \end{array}\right], \left[\begin{array}[]{ll}\tilde{\tilde{F_{1}}}^{\prime}&0\\ 0&\tilde{\tilde{F_{4}}}^{\prime}\\ \end{array}\right], respectively, where are isomorphisms.
As
[TABLE]
clearly we have
[TABLE]
Now, and are finitely generated since , are isomorphisms and are finitely generated. Hence is finitely generated, consequently is finitely generated being isomorphic to a finitely generated submodule . Therefore, is finitely generated, since it is generated by the images of the generators of under the projection onto along . But is an isomorphism, hence must be finitely generated. ∎
Lemma 2.17**.**
Let and let
[TABLE]
be a decomposition with respect to which F has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],**
where is an isomorphism, is finitely generated and is just closed. Then is finitely generated.
Proof.
By Corollary 2.11, we have since and moreover by the proof of Theorem 2.3, are orthogonally complementable. With respect to the decomposition
[TABLE]
has the matrix
\left[\begin{array}[]{ll}{\tilde{F}_{1}}&0\\ {\tilde{F}_{3}}&{F_{4}}\\ \end{array}\right],
where is an isomorphism and hence with respect to the decomposition
[TABLE]
has the matrix
\left[\begin{array}[]{ll}\tilde{F}_{1}^{*}&\tilde{F}_{3}^{*}\\ 0&\tilde{F}_{4}^{*}\\ \end{array}\right].
Clearly, is an isomorphism as is so. Then has the matrix
\left[\begin{array}[]{ll}\tilde{F}_{1}^{*}&0\\ 0&F_{4}^{*}\\ \end{array}\right] with respect to the decomposition
[TABLE]
where is an isomorphism. But since , is an isomorphism and is finitely generated (as is finitely generated by assumption ), we can use the previous lemma to deduce that is finitely generated. ∎
Corollary 2.18**.**
Let and let
[TABLE]
[TABLE]
be two decompositions for . Then there exists some finitely generated submodules and s.t.
Proof.
Statement follows from the proof of Lemma 2.16. ∎
Corollary 2.19**.**
Let and let
[TABLE]
[TABLE]
be two decompositions for Then there exists some finitely generated,closed submodules and s.t.
Proof.
Statement follows from the proof of Lemma 2.17. ∎
Lemma 2.20**.**
Let and suppose that is closed. If
[TABLE]
[TABLE]
are two decomposition for then are closed finitely generated projective modules and
[TABLE]
in
Proof.
First of all, it is obvious that
[TABLE]
Now, since is closed (by assumption), we have that are closed and also finitely generated as are so. Then, by [10, Lemma 2.3.7]
[TABLE]
for some closed submodules of respectively.
Thus and . Since are finitely generated, from [10, Theorem 2.7.5] it follows that are projective. Moreover, again by [10, Theorem 2.7.5], we may assume that there exists some such that
[TABLE]
and
[TABLE]
where is the projection onto along and are projective, finitely generated A-modules.
Set and . Note that . By the arguments similar to the proof of [10, Theorem 2.7.9], we deduce that
[TABLE]
[TABLE]
[TABLE]
Hence
[TABLE]
∎
3. Generalized Schechter characterization of operators on
In this section we investigate the classes , and prove an analogue of some results concerning the classes , (where is a Banach space) in [13].
Lemma 3.1**.**
Let Then if and only if there exists a closed, orthogonally complementable submodule such that is bounded below and is finitely generated.
Proof.
If such exists, then is closed in Moreover, as is orthogonally complementable, is adjointable. By [10, Theorem 2.3.3], is orthogonally complementable in . Then with respect to the decomposition
[TABLE]
has the matrix
\left[\begin{array}[]{ll}F_{1}&F_{2}\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism.Using the technique of diagonalization as in the proof of [10, Lemma 2.7.10] and the fact that is finitely generated, we deduce that . On the other hand if , by the similar arguments as in the proof of [10, Theorem 2.7.6 ] we may assume that there exists a decomposition
[TABLE]
with respect to which F has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism and is finitely generated. ∎
Lemma 3.2**.**
*Let . Then there exists a sequence and an increasing sequence s.t.
[TABLE]
and
[TABLE]
Proof.
Since , there exists an
[TABLE]
because is then not bounded below by the previous lemma. As
[TABLE]
there exists an such that (here again denotes the orthogonal projection onto along ). Hence
[TABLE]
Set , then
[TABLE]
and . Suppose so that there exists
[TABLE]
such that the hypothesis of the lemma holds. By previous lemma, is not bounded below on hence we can find an such that and
[TABLE]
Again, since
[TABLE]
there exits an such that
[TABLE]
Then
[TABLE]
Set . We then have (because ),
[TABLE]
By induction, the lemma follows. ∎
4. Openness of the set of semi--Fredholm operators on
In this section we prove that the sets and are open in the norm topology, as an analogue of the result in [12]. Also, we derive some consequences. Recall that is open in the norm topology by [10, Lemma 2.7.10].
Theorem 4.1**.**
The sets and are open in where is equipped with the norm topology.
Proof.
Let Then there exists a decomposition
[TABLE]
with respect to which F has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism, is closed finitely generated and is closed, but not finitely generated. If such that then for small enough we may by the same arguments as in the proof of [10, Lemma 2.7.10] find isomorphisms such that has the matrix
\left[\begin{array}[]{ll}(F+D)_{1}&0\\ 0&(F+D)_{4}\\ \end{array}\right]
with respect to the decomposition
[TABLE]
where is an isomorphism. Since is an isomorphism and is not finitely generated, it follows that is not finitely generated. Now, as has the matrix
\left[\begin{array}[]{ll}(F+D)_{1}&0\\ 0&(F+D)_{4}\\ \end{array}\right]
with respect to the decomposition above, where is an isomorphism, is finitely generated whereas is not finitely generated, it follows by Lemma 2.16 that
[TABLE]
(because, by that lemma, if was -Fredholm, then would be finitely generated, which is a contradiction). The first part of the theorem follows, whereas the second part can be proved in the analogue way or can be deduced directly from the first part by passing to the adjoints and using Corollary 2.11. ∎
Corollary 4.2**.**
If belongs to the boundary of in then
Proof.
Follows by the same arguments as in the proof of [18, Corollary 1.6.10] since
[TABLE]
is open in ∎
Corollary 4.3**.**
*Let be continuous and assume that Then the following statments hold:
-
If then
-
If then
-
If then and *
Proof.
We have that is a disjoint union of
and The first two sets are open by previous theorem whereas is open by [10, Lemma 2.7.10]. Moreover, by assumption in the corollary, we have that Since is continuous by assumption, must be connected in hence must be completely contained in one of these three sets
[TABLE]
or (otherwise we would get a separation of which is impossible).
Thus and the first part of folows. For the second part of use [10, Lemma 2.7.10] together with the proof of [18, Lemma 1.6.1]. ∎
5. and operators on
In this section we construct certain classes of operators on as a generalizations of classes , (where is a Hilbert space). Then we investigate then and prove several properties concerning these new classes of operators on .
Definition 5.1**.**
Let . We say that if there exists a decomposition
[TABLE]
with respect to which has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism, are closed, finitely generated and that is is isomorphic to a closed submodule of . We define similarly the class , the only difference in this case is that . Then we set
[TABLE]
and
[TABLE]
Lemma 5.2**.**
Suppose that satisfies ”the cancellation property”. If then for any decomposition
[TABLE]
with respect to which has the matrix
\left[\begin{array}[]{ll}F_{1}^{\prime}&0\\ 0&F_{4}^{\prime}\\ \end{array}\right],**
where is an isomorphism, are finitely generated, we have Similarly if
Proof.
Given choose a decomposition for
[TABLE]
as described in the definition above. Then for some closed submodule of . Since is finitely generated, so is , therefore, , is orthogonally complementable in So for some closed submodule of
Hence
[TABLE]
Thus
[TABLE]
Taking the inverses on both sides of the equality in , we get
[TABLE]
so
[TABLE]
Since
[TABLE]
it follows that
[TABLE]
as satisfies ”the cancellation property”.
Let be the isomorphism, then since is a closed submodule of , it follows that is a closed submodule of . Thus . But , so . One treats analogsly the case when ∎
Lemma 5.3**.**
* and are semigroups under multiplication.*
Proof.
Let Then there exist decompositions
[TABLE]
[TABLE]
with respect to which have matrices \left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right], \left[\begin{array}[]{ll}D_{1}&0\\ 0&D_{4}\\ \end{array}\right], respectively, where are isomorphisms, are finitely generated and moreover By the proof of [10, Lemma 2.7.11], with respect to the decomposition
[TABLE]
has the matrix \left[\begin{array}[]{ll}(DF)_{1}&0\\ 0&(DF)_{4}\\ \end{array}\right], where is an isomorphism,
[TABLE]
for some , and are isomorphisms. Since is isomorphic to a closed submodule of and it follows that is isomorphic to a closed submodule of (here we consider the direct sums of modules in the sense of [10, Example 1.3.4]). But since there are natural isomorphisms between ( ) and (), between () and (), it follows that is isomorphic to a closed submodule of (). As is an isomorphism, it follows that is isomorphic to a closed submodule of Now, is isomorphic to so is isomorphic to a closed submodule of Next, using that and that is isomorphic to a closed submodule of by the same arguments as above (considering direct sums of modules), we can deduce that is isomorphic to a closed submodule of so Thus Similarly one can show that is a semigroup. ∎
Lemma 5.4**.**
* and are semigroups under multiplication.*
Proof.
Let By definition, so
then. By Corollary 2.5 Now, if by the previous lemma it follows that If then as is a semigroup by Corollary 2.10. If and then in particular as by definition. By Corollary 2.9, it follows that can not be in as Since we get that If it is clear that can not be an element of Indeed, if then by Corollary 2.6 we would get that Hence which is a contradiction as by Corollary 2.4. Collecting all these arguments together, we deduce that is a semigroup. Similarly one can show that is a semigroup. ∎
Lemma 5.5**.**
* and are open.*
Proof.
Given , let
[TABLE]
be a decomposition, with respect to which
F=\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism, are finitely generated and . By the proof of [10, Lemma 2.7.10], there exists an s.t. if , then there exists a decomposition
[TABLE]
with respect to which
D=\left[\begin{array}[]{ll}D_{1}&0\\ 0&D_{4}\\ \end{array}\right],
where is an isomorphism, and moreover
[TABLE]
Let
[TABLE]
be these isomorphisms. Since , there exists an isomorphism from onto some closed submodule . Then is an isomorphism from onto which is a closed submodule of
Thus ( and also are finitely generated as are so). Therefore, . Similarly we can show that is open. ∎
Definition 5.6**.**
Let We say that if there exists a decomposition
[TABLE]
with respect to which
F=\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism, is closed, finitely generated and Similarly, we define the class , only in this case , is finitely generated and
Proposition 5.7**.**
[TABLE]
Proof.
By definition of the inclusion is obvious. Let us show the other inclusion. To this end, choose some Since there exists a decomposition
[TABLE]
with respect to which has the matrix \left[\begin{array}[]{ll}D_{1}&0\\ 0&D_{4}\\ \end{array}\right], where is an isomorphism, is finitely generated and On the other hand, since there exists a decomposition
[TABLE]
with respect to which D=\left[\begin{array}[]{ll}\tilde{D}_{1}&0\\ 0&\tilde{D}_{4}\\ \end{array}\right], where is an isomorphism, are finitely generated. By Lemma 2.16, must be then finitely generated. Hence Similarly, using Lemma 2.17, one can show that
[TABLE]
∎
Remark 5.8*.*
and and on Hilbert spaces ”=” holds due to that given any finite dimensional subspace and infinite dimensional subspace , then is isomorphic to a closed subspace of
Lemma 5.9**.**
The sets and are open. Moreover, if and , then
[TABLE]
If and , then
[TABLE]
Proof.
Suppose and choose a decomposition.
[TABLE]
such that as described in the Definition 5.6. Then, again by the proof of [10, Lemma 2.7.10], we have that there exists an such that if then there exists a decomposition
[TABLE]
with respect to which has the matrix
\left[\begin{array}[]{ll}D_{1}&0\\ 0&D_{4}\\ \end{array}\right]
where is an isomorphism and . Therefore, by the same arguments as in the proof of Lemma 5.5, we have as . Thus is in also, so is open. Next, let . By the proof of [10, Lemma 2.7.13] there exists an such that has the matrix
\left[\begin{array}[]{ll}(F+K)_{1}&0\\ 0&(F+K)_{4}\\ \end{array}\right]
with respect to the decomposition
[TABLE]
where is an isomorphism, , and for some closed, finitely generated submodule (here are as given above). Now, since is isomorphic to a closed submodule of then clearly is isomorphic to a closed submodule of as . Therefore,
[TABLE]
Since are isomorphisms, then
[TABLE]
so . Similarly one proves the statments for ∎
Theorem 5.10**.**
*Let . The following statements are equivalent
-
-
There exist such that is bounded below and *
Proof.
)
Let and let
[TABLE]
be a decomposition as given in the Definition 5.6, so that is finitely generated, and is an isomorphism. Since is finitely generated, by the proof of [10, Theorem 2.7.6], we may assume that Let be the isomorphism from onto a closed submodule Set , where is the orthogonal projection onto Then is in and in addition Since is an isomorphism from onto , is an isomorphism from onto and it follows that is bounded below being an isomorphism of onto which is a closed submodule of Moreover and is compact. Note that is indeed adjointable: Since and is self-dual being finitely generated, then by [10, Proposition 2.5.2], the result which was originally proved in [11], is adjointable. Moreover, since is finitely generated being isomorphic to it follows that is an orthogonal direct summand in by [10, Lemma 2.3.7]. Hence the inclusion is adjointable. Also is adjointable, so
If is bounded below, then obviously As by previous lemma we get that ∎
Proposition 5.11**.**
*1)
-
-
*
Proof.
- Let , choose a decomposition
[TABLE]
w.r.t which has the matrix
\left[\begin{array}[]{ll}F_{1}&0\\ 0&F_{4}\\ \end{array}\right],
where is an isomorphism, and is finitely generated. Again, by the proof of [10, Theorem 2.7.6], we may assume that W.r.t. the decomposition
[TABLE]
has the matrix
\left[\begin{array}[]{ll}\tilde{F}_{1}&\tilde{F}_{2}\\ 0&\tilde{F}_{4}\\ \end{array}\right],
where is an isomorphism and are adjointable, so
F^{*}=\left[\begin{array}[]{ll}\tilde{F}_{1}^{*}&0\\ \tilde{F}_{2}^{*}&\tilde{F}_{4}^{*}\\ \end{array}\right]
w.r.t. the decomposition
[TABLE]
Moreover, since is an isomorphism, has the matrix
\left[\begin{array}[]{ll}\tilde{\tilde{F}_{1}^{*}}&0\\ 0&\tilde{F}_{4}^{*}\\ \end{array}\right]
w.r.t. the decomposition
[TABLE]
where
V=\left[\begin{array}[]{ll}1&0\\ -\tilde{F}_{2}^{*}(\tilde{F}_{1}^{*})^{-1}&1\\ \end{array}\right]
w.r.t the decomposition
[TABLE]
so that is an isomorphism and also is an isomorphism. Now, since is an isomorphism and there exists an isomorphism (as ), we get that is an isomorphism, so . Moreover, finitely generated, as is so. Therefore, . Conversely, if let
[TABLE]
be an decomposition for , then and is finitely generated. By the proof of Theorem 2.3 part has the matrix
\left[\begin{array}[]{ll}\tilde{F}_{1}&0\\ \tilde{F}_{3}&\tilde{F}_{4}\\ \end{array}\right]
w.r.t. the decomposition
[TABLE]
where are adjointable and is an isomorphism. Then has the matrix
\left[\begin{array}[]{ll}\tilde{F}_{1}^{*}&\tilde{F}_{2}^{*}\\ 0&\tilde{F}_{4}^{*}\\ \end{array}\right]
w.r.t. the decomposition
[TABLE]
and is an isomorphism. Hence
F^{*}=\left[\begin{array}[]{ll}\tilde{F}_{1}^{*}&0\\ 0&\tilde{\tilde{F}_{4}^{*}}\\ \end{array}\right]
w.r.t. the decomposition
[TABLE]
where
U=\left[\begin{array}[]{ll}1&-\tilde{F_{1}^{*}}^{-1}(\tilde{F}_{3}^{*})\\ 0&1\\ \end{array}\right]
w.r.t the decomposition
[TABLE]
so that is an isomorphism. Since is an isomorphism, then is also an isomorphism, so . Thus
- Use 1) together with the fact that
[TABLE]
by Corollary 2.11 and the fact that
[TABLE]
[TABLE]
by Proposition 5.7.
- Use 2) together with the fact that
[TABLE]
by Corollary 2.11 and the fact that
[TABLE]
[TABLE]
by Definition 5.1. ∎
Definition 5.12**.**
We set is bounded below.
Lemma 5.13**.**
*Let Then
if and only if *
Proof.
Let By the proof of [10, Theorem 2.3.3], as is closed in this case, we have that is also closed. Moreover, by the proof of [10, Theorem 2.3.3] since is closed, we also have . Since it follows that
Conversely, if then so is injective. Moreover, since is closed, then is closed also, (again by the proof of [10, Theorem 2.3.3]). By the Banach open mapping theorem, it follows that is an isomorphism from onto its image. Thus is bounded below. ∎
Corollary 5.14**.**
*Let The following statements are equivalent:
-
-
There exist s.t. *
Proof.
Follows from Theorem 5.10, Proposition 5.11 part 1) and Lemma 5.13 by passing to the adjoints. ∎
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