Equivalence after extension and Schur coupling for Fredholm operators on Banach spaces

Schur coupling (SC) and equivalence after extension (EAE) are important relations for bounded operators on Banach spaces. It has been known for 30 years that the former implies the latter, but only recently Ter Horst, Messerschmidt, Ran and Roelands disproved the converse by constructing a pair of Fredholm operators which are EAE, but not SC. Motivated by this result, we investigate when EAE and SC coincide for Fred\-holm operators. Fredholm operators which are EAE have the same Fredholm index. Surprisingly, we find that for each integer $k$ and every pair of Banach spaces $(\mathcal{X},\mathcal{Y})$, either no pair of Fredholm operators of index~$k$ acting on $\mathcal{X}$ and $\mathcal{Y}$, respectively, is SC, or every pair of this kind which is EAE is also SC. Consequently, the question whether EAE and SC coincide for Fredholm operators of index~$k$ depends only on the geometry of the underlying Banach spaces $\mathcal{X}$ and $\mathcal{Y}$, not on the properties of the operators themselves. We quantify this finding by introducing two numerical indices which capture the coincidence of EAE and SC, and provide a number of examples illustrating the possible values of these indices. Notably, this includes an example showing that the above-mentioned result of Ter Horst et al, which is based on a pair of essentially incomparable Banach spaces, does not extend to projectively incomparable Banach spaces.


Introduction
Equivalence after extension (EAE), matricial coupling (MC) and Schur coupling (SC) are three relations for bounded operators on Banach spaces that originate in the study of Wiener-Hopf integral operators [5] and have found numerous applications since.A key feature in many of these applications is that the three relations coincide.This observation led Bart and Tsekanovskiȋ [9] to ask whether this is always true.They already knew that EAE and MC are equivalent [5,7] and that SC implies EAE [8,9], so their precise question was whether EAE implies SC.
Despite numerous results confirming this implication in special cases [8,9,6,31,51,27,30], recently Ter Horst, Messerschmidt, Ran and Roelands [29] showed that EAE does not in general imply SC.Their counterexample uses bounded operators defined on a pair of Banach spaces which is essentially incomparable, in which case EAE (and hence SC) can occur only for Fredholm operators, while SC additionally requires that the operators have index zero.By contrast, it is known that EAE and SC coincide for Fredholm operators acting on isomorphic Banach spaces [28,Proposition 6.1(iv)].
These results motivated the present work, in which we study EAE and SC for Fredholm operators without imposing any restrictions on the underlying Banach spaces.Further justification for focussing on Fredholm operators comes from the prominent role this class plays in many applications of the theory, as the following studies from the last decade evidence: diffraction problems [15,48]; Wiener-Hopf factorization [46,26] and invertibility of Wiener-Hopf plus Hankel operators [16]; truncated Toeplitz operators [14,35]; Riemann-Hilbert problems [13]; Helmholtz and Sylvester equations [45,47] and [17], respectively; completeness theorems for integral and differential operators [33]; and problems concerning electrical networks [10].
Before we state our main results, let us introduce some notation and terminology, most of which is standard.We follow the convention that N = {1, 2, 3, . ..} and N 0 = {0, 1, 2, . ..}.Let X and Y be Banach spaces, either real or complex, with K ∈ {R, C} denoting the scalar field.The term "operator" means a bounded linear map between Banach spaces, and B(X , Y) denotes the Banach space of all operators from X to Y.As usual, B(X , X ) is abbreviated B(X ).This convention applies whenever we consider sets of operators: Once a subset Σ(X , Y) of B(X , Y) has been defined, we write Σ(X ) instead of Σ(X , X ).
The identity operator on a Banach space X is denoted by I X , while the kernel and the range of an operator T are denoted by ker T and ran T , respectively.Two Banach spaces X and Y are isomorphic, written X ∼ = Y, if B(X , Y) contains a bijection, called an isomorphism.The Banach Isomorphism Theorem ensures that the inverse of an isomorphism is automatically bounded.Definition 1.1.Let U ∈ B(X ) and V ∈ B(Y).We say that: (i) U and V are equivalent after extension, abbreviated EAE, if there exist Banach spaces X 0 and Y 0 and isomorphisms E ∈ B(Y ⊕ Y 0 , X ⊕ X 0 ) and (ii) U and V are Schur coupled, abbreviated SC, if there exist isomorphisms A ∈ B(X ) and D ∈ B(Y) and operators B ∈ B(Y, X ) and C ∈ B(X , Y) such that As noted above, whenever U and V are SC, they are also EAE.Motivated by many applications in which the converse implication holds, Bart and Tsekanovskiȋ asked the following question in [9]: Question 1.2.Under which conditions on the Banach spaces X and Y and/or on the operators U and V is it true that whenever the operators U ∈ B(X ) and V ∈ B(Y) are EAE, they are also SC?
We shall address this question in the case where U and V are Fredholm operators.Before doing so, let us recall some basic facts about this class of operators.An operator T ∈ B(X , Y) is called a Fredholm operator if the quantities α(T ) = dim ker T and β(T ) = dim Y/ ran T are both finite.The latter condition implies that ran T is closed.As usual, we write Φ(X , Y) for the subset of B(X , Y) consisting of Fredholm operators.The index of a Fredholm operator T is defined by and for k ∈ Z, Φ k (X , Y) denotes the set of T ∈ Φ(X , Y) such that i(T ) = k.
In the 1990s, Bart and Tsekanovskiȋ gave the following characterization of equivalence after extension for Fredholm operators; see [8,Theorem 4], and also [6,Theorem 6,page 211].
Theorem 1.3.Let U ∈ B(X ) and V ∈ B(Y) for some Banach spaces X and Y.
(i) Suppose that U and V are EAE.Then U is a Fredholm operator if and only if V is a Fredholm operator.(ii) Suppose that U and V are Fredholm operators.Then U and V are EAE if and only if In particular, Fredholm operators which are EAE have the same index.
As a consequence, the following sets, defined for every k ∈ Z and every pair of Banach spaces (X , Y), provide the natural setting in which to study Question 1.2 for Fredholm operators: ( In view of Theorem 1.3(ii), the former set can alternatively be written as (1.4) These sets are useful in our investigation because they allow us to express the statement that EAE and SC are equivalent for every pair of Fredholm operators of index k on X and Y, respectively, in the concise form SC k (X , Y) = EAE k (X , Y), where we note that the inclusion SC k (X , Y) ⊆ EAE k (X , Y) is always true because SC implies EAE.
Using this notation, we can state easily two important results that motivated our work.First, the answer to Question 1.2 is always affirmative for Fredholm operators of index 0 (see [8,Theorem 3] and [6,Theorem 5]).In the above notation, this simply means that SC 0 (X , Y) = EAE 0 (X , Y) for every pair of Banach spaces (X , Y).Second, we can state the seminal result of Ter Horst, Messerschmidt, Ran and Roelands [29] showing that there are pairs of Fredholm operators which are EAE, but not SC.This requires the following piece of terminology.Definition 1.4.A pair of Banach spaces (X , Y) is essentially incomparable if I X − ST ∈ Φ(X ) for every S ∈ B(Y, X ) and T ∈ B(X , Y).
(i) Let (X , Y) be a pair of essentially incomparable Banach spaces.Then U ∈ B(X ) and V ∈ B(Y) are SC if and only if (U, V ) ∈ EAE 0 (X , Y).
(ii) There exists a pair of essentially incomparable Banach spaces (X , Y) such that EAE k (X , Y) = ∅ for every k ∈ Z.
Hence EAE and SC are not equivalent for Fredholm operators of non-zero index on such Banach spaces.An example is given by X = ℓ p and Y = ℓ q for 1 p < q < ∞.
The significance of analyzing whether SC and EAE are equivalent for each value k of the Fredholm index separately will become clear from the next result, which is the first main outcome of our work.To state it concisely, we introduce a numerical index eae(X , Y) as follows: (1.6) Theorem 1.6.Let k ∈ Z, and let X and Y be Banach spaces.
The most remarkable part of Theorem 1.6 is the implication ⇐ in (ii) which, when written out, states that as soon as one pair of operators (U, In other words, equivalence of EAE and SC for Fredholm operators depends only on the geometry of the underlying Banach spaces X and Y and on the Fredholm index k, not on the Fredholm operators themselves. In view of Theorem 1.6(ii), it would be of great interest to establish a counterpart of Theorem 1.6(i) for SC.In analogy with (1.6), we introduce the set and the associated index , our main questionwhether EAE and SC are equivalent for every pair of Fredholm operators on X and Y, respectively -boils down to whether I SC (X , Y) = eae(X , Y)Z.We address this question in the following proposition.
Proposition 1.7.Let X and Y be Banach spaces.
In general, sc(X , Y) = n eae(X , Y) for some n ∈ N 0 , and the following chain of inclusions holds: (1.9) Remark 1.8.The first part of Proposition 1.7 implies that the second inclusion in (1.9) is an equality if and only if sc(X , Y) = eae(X , Y), in which case the first inclusion is also an equality.In fact, we do not know any instances where the first inclusion in (1.9) is proper, and we conjecture that it may always be an equality; see Section 5 for a more detailed discussion of this question.
We conclude this overview of our main findings with some results that illustrate the values which the numerical indices eae(X , Y) and sc(X , Y) can take when various incomparability conditions are imposed on the Banach spaces X and Y.This work is motivated by, and closely related to, the seminal result of Ter Horst, Messerschmidt, Ran and Roelands that we stated in Theorem 1.5.We begin with a result whose first part is simply a restatement of Theorem 1.5(i), while its second part contains Theorem 1.5(ii) as a special case, corresponding to k 0 = 1.Theorem 1.9.
(i) Let (X , Y) be a pair of essentially incomparable Banach spaces.Then sc(X , Y) = 0. (ii) For every k 0 ∈ N 0 , there exists a pair of essentially incomparable Banach spaces (X , Y) such that eae(X , Y) = k 0 .Theorem 1.9(i) immediately raises the question whether we can weaken the hypothesis that the pair (X , Y) is essentially incomparable without losing the conclusion that sc(X , Y) = 0.The most obvious, very modest weakening would be to assume that X and Y are projectively incomparable in the following sense.Definition 1.10.A pair of Banach spaces (X , Y) is projectively incomparable if no infinite-dimensional, complemented subspace of X is isomorphic to a complemented subspace of Y.
However, it turns out that this hypothesis is too weak to imply that sc(X , Y) = 0, as our next result will show.It also contains some information about the possible values of the indices eae(X , Y) and sc(X , Y).
(i) For every k 0 ∈ N, there exists a pair of projectively incomparable Banach spaces (X , Y) such that sc(X , Y) = eae(X , Y) = k 0 .(ii) For every k 0 ∈ N, there exists a pair of projectively incomparable Banach spaces (X , Y) such that eae(X , Y) = 1 and sc(X , Y) = k 0 .
Theorem 1.11 is highly surprising because the difference between essential and projective incomparability is very subtle, as evidenced by the fact that it took nearly 30 years to find an example which distinguishes them.Indeed, Tarafdar [49,50] asked in 1972 whether projective incomparability implies essential incomparability, having noted that the converse is true, but it was not until 2000 that Aiena and González answered this question by giving a counterexample (see [2,Proposition 3.7]).It relied on a sophisticated Banach space constructed by Gowers and Maurey [25].To the best of our knowledge, no simpler examples have subsequently been found.We shall discuss the relationship between essential and projective incomparability in more detail in Section 2.
In view of Theorem 1.11, let us consider what may happen when X and Y are not projectively incomparable.Then they contain isomorphic, infinite-dimensional complemented subspaces; that is, X and Y admit decompositions of the form where X 2 and Y 2 are infinite-dimensional.The next proposition answers the question when is eae(X , Y) = sc(X , Y) in the case where X 1 , X 2 and Y 1 are pairwise essentially incomparable?Its statement involves the greatest common divisor (gcd) of two quantities that could potentially both be 0, in which case the gcd is not defined.We fix this issue by adopting the convention that gcd(0, 0) = 0.
Proposition 1.12.Let X and Y be Banach spaces that decompose as in (1.10), and suppose that each of the pairs Suppose additionally that the pair (1.12) In particular, EAE and SC coincide for all pairs of Fredholm operators on X and Y, i.e., eae(X , Y) = sc(X , Y), if and only if eae(X 2 , Y 2 ) divides eae(X 1 , Y 1 ).
Note that we do not demand that the isomorphic subspaces X 2 and Y 2 are infinite-dimensional in Proposition 1.12.Therefore part (i) of Theorem 1.9 appears as a special case of Proposition 1.12 corresponding to X 2 = Y 2 = {0}, while the case where X and Y are isomorphic is obtained by taking In analogy with Theorem 1.11, it turns out that the second part of Proposition 1.12 is no longer true if we replace the hypothesis that the pair (X 1 , Y 1 ) is essentially incomparable with the weaker hypothesis that it is projectively incomparable.
Remark 1.14.Let us compare and contrast Theorem 1.13 with Proposition 1.12.
To align notation, note that X 2 = Y 2 = Z.Theorem 1.13(2) implies that (1.11) holds true.However, (1.12) fails for the pair (X , Y) in both parts (i) and (ii) of Theorem 1.13 because they satisfy sc(X , Y) = k 0 = 0 = eae(Z, Z).This difference is due to the fact that (1.12) requires that the pair (X 1 , Y 1 ) is essentially incomparable, but we only know that it is projectively incomparable in Theorem 1.13.

Conclusion.
Prior to this paper, at the level of general Banach spaces, the only conclusive results regarding the question whether EAE and SC coincide for Fredholm operators were that they do if the underlying spaces are isomorphic, and that there exist examples where they do not if the underlying spaces are essentially incomparable.We have shown that the result for essentially incomparable spaces does not carry over to projectively incomparable spaces (see Theorem 1.11), despite the fact that the difference between these two incomparability notions is very subtle.
In the case where the Banach spaces X and Y admit decompositions of the form (1.10) in which the subspaces X 1 , Y 1 and X 2 ( ∼ = Y 2 ) are pairwise essentially incomparable, the question whether EAE and SC coincide for Fredholm operators is completely resolved in Proposition 1.12; the answer can be expressed in terms of the values of the Fredholm index of operators on the subspaces X 1 , Y 1 and X 2 .Theorem 1.13 shows that, once again, this result does not carry over to projectively incomparable spaces; see Remark 1.14 for details.
The question that remains is whether one can always find a decomposition of the form (1.10) in which the subspaces X 1 , Y 1 and X 2 are pairwise essentially incomparable.Unfortunately, this is not possible, even if we replace "essentially incomparable" with "projectively incomparable", as we shall see in Corollary 2.8 and Proposition 2.10.
The above results rely on the remarkable observation in Theorem 1.6 that, for Banach spaces X and Y and k ∈ Z, either no pair of operators (U, is SC, or a pair of this kind which is SC exists, in which case the set of all such pairs that are SC is the same as the set of all such pairs that are EAE.This means that the question whether EAE and SC coincide for Fredholm operators on X and Y reduces to determining the sets of indices for which pairs of Fredholm operators on X and Y with these particular indices that are EAE or SC, respectively, exist. Our analysis of these sets led us to define the numerical indices eae(X , Y) and sc(X , Y) which satisfy that eae(X , Y) = sc(X , Y) if and only if EAE and SC coincide for all Fredholm operators on X and Y.We have computed their values in various cases; see Theorems 1.9, 1.11 and 1.13.
Organization.We conclude this introduction with a brief outline of how the remainder of this paper is organized.It consists of six sections, including the present.
In Section 2 we elaborate on the incomparability notions for Banach spaces introduced in Definitions 1.4 and 1.10, focussing on their connections with certain classes of operators.Section 3 contains a characterization of when the set EAE k (X , Y) is non-empty for Banach spaces X and Y and k ∈ Z, and also the proof of Equation (1.11).It turns out to be much more complicated to obtain a similar characterization for the non-emptiness of the set SC k (X , Y), and only a partial analogue is obtained in Section 5, where we also prove Proposition 1.7 and the remainder of Proposition 1.12.These results rely strongly on a novel characterization of the existence of Schur-coupled Fredholm operators of a given index that we establish in Section 4. This characterization may be viewed as the fundamental new insight of the paper.Theorem 1.6 is also proved in Section 4.
Finally, in Section 6 we use some of the "exotic" Banach spaces constructed by Gowers and Maurey, together with ideas from subsequent work of Aiena, González and Ferenczi, to prove Theorems 1.9, 1.11 and 1.13.

Incomparability notions for Banach spaces and their connection to operator theory
The notions of essential and projective incomparability of a pair of Banach spaces will play a key role in the final section of this paper, where Theorems 1.9, 1.11 and 1.13 are proved.However, there are certain related notions and results that we shall require beforehand.For that reason, we survey the relevant material at this point.
The formal definitions of essential and projective incomparability were already given in Definitions 1.4 and 1.10, respectively.We refer to [1, Section 7.5] for a much more comprehensive treatment of them than we can give here.Indeed, we shall consider only the aspects that we require later, namely their relationship and certain connections to operator theory.This will involve a third incomparability notion, which is stronger, older and arguably more "natural" than the other two.It is defined as follows.
Definition 2.1.A pair of Banach spaces (X , Y) is totally incomparable if no closed, infinite-dimensional subspace of X embeds isomorphically into Y.
Totally incomparable Banach spaces are clearly projectively incomparable, and it is well known and not hard to see that the converse fails; for instance, ℓ 2 embeds into L 1 [0, 1], but not complementably, so L 1 [0, 1] and ℓ 2 are projectively incomparable without being totally incomparable.
However, more is true, namely that essential incomparability lies between these two properties, in the sense that total incomparability implies essential incomparability, which in turn implies projective incomparability.The easiest way to explain this goes via the following two operator-theoretic notions, which will also be useful elsewhere in this work.Definition 2.2.Let X and Y be Banach spaces.An operator T ∈ B(X , Y) is: (i) strictly singular if, for every ε > 0, every infinite-dimensional subspace W of X contains a unit vector w such that T w ε.In other words, the restriction of T to W is not an isomorphism onto its range.(ii) inessential if I X − ST ∈ Φ(X ) for every operator S ∈ B(Y, X ).
We write S (X , Y) and E (X , Y) for the collections of strictly singular and inessential operators, respectively, from X to Y.They generalize the ideal K (X , Y) of compact operators in several ways.The following remark lists the main properties that we require.

Remark 2.3.
(i) Every compact operator is strictly singular, and every strictly singular operator is inessential (see for instance [1,Theorems 7.36 and 7.44] or [43, §26.7.3]).(ii) The assignments S and E are closed operator ideals in the sense of Pietsch (see for instance [1, page 388 and Theorem 7.5] or [43, Theorem 1.9.4,Proposition 4.2.7 and Section 4.3]).(iii) For every k ∈ Z, the class of Fredholm operators of index k is stable under inessential perturbations in the sense that Comparing Definitions 1.4 and 2.2(ii), we see that a pair of Banach spaces (X , Y) is essentially incomparable if and only if every operator from X to Y is inessential.Both definitions display an obvious lack of symmetry, which raises the question what happens if we replace the condition that in either of them?It turns out that it makes no difference because the following well-known elementary lemma implies that I X − ST ∈ Φ(X ) if and only if Using Definition 2.2 and Remark 2.3(i), we can explain the relationship between total, essential and projective incomparability of a pair of Banach spaces (X , Y) as follows.
(i) If X and Y are totally incomparable, then clearly every operator between X and Y is strictly singular and therefore inessential, so X and Y are essentially incomparable.(ii) If X and Y are essentially incomparable, then they are also projectively incomparable.Indeed, suppose contrapositively that X and Y contain isomorphic, complemented infinite-dimensional subspaces.Then it is easy to construct operators S ∈ B(Y, X ) and T ∈ B(X , Y) such that I X − ST is a projection with infinite-dimensional kernel and therefore not a Fredholm operator.Hence X and Y are not essentially incomparable.We refer to [1,Theorem 7.69] and the paragraph following Definition 7.102 for further details.
The fact that S and E are operator ideals has the following important consequence, which we shall use repeatedly without further reference.Suppose that we express an operator T : X 1 ⊕ X 2 → Y 1 ⊕ Y 2 between two direct sums of Banach spaces as an operator-valued matrix in the usual way, that is, Then T is strictly singular (respectively, inessential) if and only if T 11 , T 12 , T 21 and T 22 are strictly singular (respectively, inessential).
We conclude this section by answering a natural question about a pair of Banach spaces (X , Y) which is not projectively incomparable.This material will not play any role in the remainder of the paper; we have included it simply because the question is very natural in our context.Negating the definition of projective incomparability, we see that X and Y admit decompositions of the form (1.10), where the isomorphic subspaces X 2 and Y 2 are infinite-dimensional.If the subspaces X 1 and Y 1 fail to be projectively incomparable, then they contain isomorphic, complemented, infinite-dimensional subspaces.One may wonder whether all such subspaces can somehow be "transferred" to X 2 and Y 2 , respectively, leading to the following question: Question 2.6.Let X and Y be Banach spaces which are not projectively incomparable.Is it always possible to find decompositions of the form (1.10), where the subspaces X 1 and Y 1 are projectively incomparable?
The answer to this question is "no".We shall present two short examples showing this.In the first, we consider Banach spaces which are c 0 -direct sums of certain sequences of finite-dimensional Banach spaces.The formal definition is as follows.
endowed with the pointwise vector-space operations and with the norm given by (x n ) = sup n∈N x n .Bourgain, Casazza, Lindenstrauss and Tzafriri [12, §8] classified the complemented subspaces of , and let W be a complemented, infinite-dimensional subspace of Z. Then either In particular X 1 and Y 1 both contain a complemented subspace isomorphic to c 0 , so they are not projectively incomparable.
Proof.This follows immediately from Theorem 2.7 because X and Y are not isomorphic to each other or to c 0 , so we must have Our second example is similar, but uses only reflexive Banach spaces.It relies on the following well-known, important properties of Proof.(i).This is shown in [4].
, and suppose that X and Y are decomposed as in (1.10), with In particular X 1 and Y 1 both contain a complemented subspace isomorphic to L 2 [0, 1], so they are not projectively incomparable. Proof.
However, the latter is impossible by Theorem 2.9(ii) because The main purpose of this section is to prove the following characterization of the integers k for which the set EAE k (X , Y) defined in (1.3) is non-empty.This is a natural starting point for our investigation because when EAE k (X , Y) is empty, it is obviously equal to its subset SC k (X , Y). Proposition 3.1.The following three conditions are equivalent for every pair of Banach spaces (X , Y) and every k ∈ Z : The proof of Proposition 3.1 is relatively simple, using only (1.4) and two lemmas that follow from basic Fredholm theory.Both of these lemmas are almost certainly known to specialists; we include their (short) proofs for completeness.
Proof.We consider three cases: for every finite-rank operator R ∈ B(X , Y). Lemma 3.4.For every Banach space X , the set Proof.The result is clear if X is finite-dimensional because I Φ (X ) = {0} in this case.In general the set I Φ (X ) contains 0 because I X ∈ Φ 0 (X ).Moreover, it is closed under addition and under multiplication by positive integers because the Index Theorem implies that ST ∈ Φ k+m (X ) and 2 implies that Φ k (X ) contains a surjection T .Since ker T is finite-dimensional, it follows that T has a right inverse, which must be a Fredholm operator of index −k.A similar argument works for k 0, except that we find that Φ k (X ) contains an injection which has a left inverse.
Corollary 3.5.Let X and Y be Banach spaces.Then Remark 3.6.Elaborating on these ideas, we obtain an alternative formula for eae(X , Y), which will be useful later.Indeed, for every Banach space X , the ideal I Φ (X ) has a unique non-negative generator, which we shall denote by γ(X ); in other words, γ(X ) ∈ N 0 is the unique number such that I Φ (X ) = γ(X )Z.
It follows from elementary number theory that, for every pair of Banach spaces X and Y, the ideal I Φ (X )∩I Φ (Y) = γ(X )Z∩γ(Y)Z is generated by the lowest common multiple (lcm) of γ(X ) and γ(Y), provided that we set lcm(n, 0) = lcm(0, n) = 0 for every n ∈ Z. Combining this result with Corollary 3.5, we see that eae(X , Y) = lcm(γ(X ), γ(Y)). (3.1) Proof of Proposition 3.1.Conditions (i) and (iii) are equivalent because where the final bi-implication follows from Corollary 3.5.We complete the proof by showing that conditions (i) and (ii) are also equivalent.Here, the implication (ii) Indeed, the implication ⇐ is immediate from the definition of eae(X , Y).Conversely, suppose that Φ k (X ) = ∅ and Φ m (Y) = ∅ for some k, m ∈ N. Then the Index Theorem implies that Φ n (X ) = ∅ and Φ n (Y) = ∅ for every common multiple n ∈ N of k and m, so eae(X , Y) 1. (Alternatively, this follows from (3.1) because lcm(γ(X ), γ(Y)) = 0 if and only if γ(X ) = 0 or γ(Y) = 0.) Proposition 3.1 naturally raises the question: What are the possible values of the index eae(X , Y) ∈ N 0 ?Obviously, eae(X , Y) = 0 if X or Y is finite-dimensional.Theorems 1.9(ii), 1.11(i) and 1.13(i) show that all non-negative integers can be realized as eae(X , Y) for infinite-dimensional Banach spaces X and Y satisfying various additional conditions.At this stage, let us use the index γ(X ) from Remark 3.6 to verify that all non-negative integers can be realized as eae(X , X ).Although this example may appear simpler than the three above-mentioned theorems, ultimately they all rely on the same family of "exotic" Banach spaces constructed by Gowers and Maurey in [25].
To verify this claim for k 0 = 0, we require an infinite-dimensional Banach space on which all Fredholm operators have index 0. Gowers and Maurey constructed such a Banach space in [24].We shall encounter another space with this property in Theorem 6.8.
For k 0 = 1, any Banach space which is isomorphic to its hyperplanes satisfies the claim.Virtually every infinite-dimensional Banach space known prior to 1990 has this property.
Finally, to see that the claim is true for every k 0 2, we use a family of Banach spaces which Gowers and Maurey constructed in [25, §(4.3)].These Banach spaces will play a key role in Section 6, where Theorem 6.1 summarizes their main properties; we refer to (iv) for the particular result required at this point.
We conclude this section with two easy observations.The first is that the index γ(X ), and therefore the associated quantity eae(X , Y), is an isomorphic invariant in the following precise sense.Lemma 3.9.Let X 1 and X 2 be isomorphic Banach spaces.Then γ(X 1 ) = γ(X 2 ).
Consequently, if This proves that I Φ (X 1 ) ⊆ I Φ (X 2 ).The opposite inclusion follows by interchanging X 1 and X 2 .Hence the ideals I Φ (X 1 ) and I Φ (X 2 ) are equal, so they must have the same non-negative generator; that is, γ(X 1 ) = γ(X 2 ).The final clause is an immediate consequence of (3.1).
Our second easy observation will be the key ingredient in the proof of the first part of Proposition 1.12, as we shall show immediately after it.Lemma 3.10.Let X = X 1 ⊕ X 2 be a Banach space.
(ii) Suppose that X 1 and X 2 are essentially incomparable.Then Proof.(i).Suppose that k j ∈ I Φ (X j ) for j ∈ {1, 2}, and take T j ∈ Φ kj (X j ).Then we have . This shows that Recall that we have defined gcd(0, 0) = 0.This ensures that the formula mZ+nZ = gcd(m, n)Z holds true for all values of m, n ∈ Z.Using it, we can rewrite the identity (3.3) as gcd(γ(X 1 ), γ(X 2 ))Z ⊆ γ(X )Z, which proves (i).
(ii).Suppose that the subspaces X 1 and X 2 are essentially incomparable.Then, for each k ∈ I Φ (X ), we can take where T 12 and T 21 are inessential.In view of Remark 2.3(iii), this implies that which in turn means that T 11 ∈ Φ(X 1 ) and T 22 ∈ Φ(X 2 ) with Hence I Φ (X 1 ) + I Φ (X 2 ) = I Φ (X ), and (3.2) follows.

4.
The key technical theorem and the deduction of Theorem 1.6 from it The main aim of this section is to prove the following theorem, which will be the key tool in the remainder of our investigation, and is arguably the most important new insight in the paper, despite the somewhat technical nature of conditions (i)-(ii).Notably, Theorem 1.6 is an easy consequence of it (using also Proposition 3.1), as we shall show at the end of this section.The second lemma can be viewed as a technical refinement of Lemma 3.2.Its proof is an adaption of the proof of [28,Lemma 5.10].
Lemma 4.3.Let X and Y be Banach spaces, and suppose that Proof.We begin by observing that X must be infinite-dimensional because it admits a Fredholm operator of non-zero index.We can therefore apply Lemma 3.2 to find a finite-rank operator R ∈ B(X ) such that Take a finite-dimensional subspace W of X such that ran R ⊆ W and dim W = n|k| for some n ∈ N, and let R 0 ∈ B(X , W) denote the operator R regarded as a map into W.Moreover, let J ∈ B(W, X ) be the inclusion map.Lemma 2.4 shows that and Hence it suffices to consider the case k = 0. Suppose that Suppose that (U, V ) ∈ EAE k (X , Y), so that α(U ) = α(V ) by (1.4).Call this number m, and note that m k.Our strategy is to modify the operators S 1 and T 1 to obtain a pair for which we can construct isomorphisms M ∈ B(X ) and N ∈ B(Y) such that (4.1) is satisfied.
We begin by applying Lemma 4.3 to find operators S 2 ∈ B(Y, X ) and T 2 ∈ B(X , Y) such that S 1 T 1 − S 2 T 2 has finite rank and ) and ran U are closed subspaces of the same finite codimension in X , and therefore we can take an isomorphism A ∈ B(X ) such that , and observe that these operators satisfy This implies that α(I X − S 3 T 3 ) = α(I X − S 2 T 2 ) = m = α(U ), which is finite, so we can take an isomorphism M 1 ∈ B(ker(I X − S 3 T 3 ), ker U ). Choose closed subspaces X 1 and X 2 of X such that X = ker(I X − S 3 T 3 ) ⊕ X 1 and X = ker U ⊕ X 2 , and let be the restrictions of I X − S 3 T 3 and U , respectively.The choices of X 1 and X 2 imply that R and U 0 are isomorphisms.Using (4.5) and (4.3), we see that ran(I X − S 3 T 3 ) = ran U , so we can define an isomorphism M 2 ∈ B(X 1 , X 2 ) by M 2 = U −1 0 R, and therefore , and combining the identity with (4.4), we deduce that ran(I Y − T 3 S 3 ) = ran V .Therefore we can repeat the constructions from the previous paragraphs to obtain an isomorphism N ∈ B(Y) such that V N = I Y − T 3 S 3 .Now the conclusion that U and V are SC follows from Lemma 4.2.
The implication (iii)⇒(iv) is clear.(iv)⇒(i).Suppose that (U, V ) ∈ SC k (X , Y).Then Lemma 4.2 implies that there are operators S ∈ B(Y, X ) and T ∈ B(X , Y) and an isomorphism M ∈ B(X ) such that U M = I X − ST .We have U M ∈ Φ k (X ) because U ∈ Φ k (X ) and M is an isomorphism, and consequently (i) is satisfied.
Proof of Theorem 1.6.(i) is simply a restatement of the equivalence of conditions (ii) and (iii) in Proposition 3.1.
We conclude this section by showing how Theorem 4.1 can be deduced from results obtained in [28].To this end, take (U, V ) ∈ EAE k (X , Y). Translating the conclusion of [28,Proposition 5.9] about what is called "the Banach space operator problem" in [28] to the setting of EAE and SC, as explained in [28, Section 3], we see that U and V are SC if and only if there exist operators B 1 ∈ B(ran U, ran V ) and B 2 ∈ B(ran V, ran U ) such that Alternative proof of Theorem 4.1.As before, Lemma 2.4 shows that conditions (i) and (ii) are equivalent, and the implication (iii)⇒(iv) is trivial.
(ii)⇒(iii).Suppose that (ii) is satisfied, so that I Y − T S ∈ Φ k (Y) for some operators S ∈ B(Y, X ) and T ∈ B(X , Y), and take (U, V ) ∈ EAE k (X , Y).We must show that (U, V ) ∈ SC k (X , Y), which by the result from [28] stated above amounts to finding operators B 1 ∈ B(ran U, ran V ) and B 2 ∈ B(ran V, ran U ) which satisfy (4.6).Take finite-dimensional subspaces X 1 and Y 1 of X and Y, respectively, such that and decompose the operators S and T accordingly; that is, where Since X 1 and Y 1 are finite-dimensional, the operators S 2 = S − S 1 and T 2 = T − T 1 have finite rank, and where which in turn implies that I ran V − T 11 S 11 ∈ Φ k (ran V ).Therefore the operators B 1 = T 11 and B 2 = S 11 satisfy (4.6).
(iv)⇒(ii).Suppose that SC k (X , Y) = ∅, and take (U, V ) ∈ SC k (X , Y).Then, by the result from [28] stated above, we can find operators B 1 ∈ B(ran U, ran V ) and B 2 ∈ B(ran V, ran U ) which satisfy (4.6).As before, take finite-dimensional subspaces X 1 and Y 1 of X and Y, respectively, such that (4.7) is satisfied, and define Then which shows that (ii) is satisfied.
5. Non-emptiness of SC k (X , Y) and the proofs of Propositions 1.7 and 1.12 The aim of this section is to investigate the set of integers k for which there exist Schur-coupled operators U ∈ Φ k (X ) and V ∈ Φ k (Y); that is, SC k (X , Y) = ∅.We follow a similar strategy to the one successfully employed in Section 3, beginning with a partial analogue of Lemma 3.4 for the set ).As we shall see, the situation for SC is considerably more complicated than for EAE, primarily due to the difficulty of analyzing the technical conditions (i)-(ii) in Theorem 4.1.In particular, we have been unable to obtain an exact counterpart of Lemma 3.4 for SC because we do not know if the set I SC (X , Y) is always closed under addition.
Lemma 5.1.The set I SC (X , Y) has the following properties for every pair of Banach spaces (X , Y) : Proof.The first two properties are easy to verify.Indeed, (i) follows from the fact that (I X , I Y ) ∈ SC 0 (X , Y), while (ii) follows from (i) if sc(X , Y) = 0, and otherwise from the definition (1.8) of sc(X , Y).However, the proof of (iii) requires more work.Take k ∈ I SC (X , Y) and m ∈ Z.By (i), we may suppose that k = 0 and m = 0. Since SC k (X , Y) = ∅, Theorem 4.1 implies that we can find operators S 1 ∈ B(Y, X ) and T 1 ∈ B(X , Y) such that (5.1) Once we have established this claim, the conclusion will follow from another application of Theorem 4.1.
We prove the claim by considering three different cases: m 2, m = −1 and m −2.(Note that the case m = 1 is already covered by the choice of S 1 and T 1 .) Case 1.For m 2, we can apply the Binomial Theorem because I X and S 1 T 1 commute.It shows that so the Index Theorem implies that the operators S m = S 1 ∈ B(Y, X ) and T m = T 1 m j=1 m j (−S 1 T 1 ) j−1 ∈ B(X , Y) satisfy (5.1).Case 2. For m = −1, we consider the cases k > 0 and k < 0 separately.For k > 0, Lemma 4.3 implies that we can find operators U ∈ B(Y, X ) and V ∈ B(X , Y) such that S 1 T 1 − U V is a finite-rank operator and is a surjective Fredholm operator, and therefore it has a right inverse R ∈ Φ −k (X ).Consequently The argument for k < 0 is very similar.In this case, we can apply Lemma 4. Second, suppose that eae(X , Y) = sc(X , Y), and call this number k 0 .Then, by (ii), (iii) and Corollary 3.5, we have Proof of Proposition 1.7.We begin by verifying the chain of inclusions (1.9), which we restate here for ease of reference: The first inclusion is immediate from Lemma 5.1(ii)-(iii), while the equality of the second and third set in the first line follows by combining the definition of I SC (X , Y) with the equivalence of conditions (iii) and (iv) in Theorem 4.1.Proposition 3.1 shows that the three sets in the second line are equal (the equality of the first and last of these sets was also recorded in Corollary 3.5), and finally the inclusion at the beginning of the second line follows because the final set in the first line is trivially contained in the second set in the second line.
Next, to prove the first claim of Proposition 1.7, suppose that EAE k (X , Y) = SC k (X , Y) for every k ∈ Z. Then where the final equality follows from (1.9).Hence the definitions (1.6) and (1.8) show that eae(X , Y) = sc(X , Y).
Conversely, suppose that eae(X , Y) = sc(X , Y), and call this number k 0 .Then the inclusions in (1.9) are in fact equalities, so EAE k (X , Y) = SC k (X , Y) ( = ∅) for every k ∈ I SC (X , Y) = k 0 Z.On the other hand, Theorem 1.6(iii) shows that the identity EAE k (X , Y) = SC k (X , Y) (= ∅) is also true for every k / ∈ eae(X , Y)Z = k 0 Z.
Remark 5.2.To illustrate the applicability of our work thus far, let us explain how it leads to an explicit algorithm for deciding whether EAE and SC are equivalent for all pairs of Fredholm operators on a given pair of Banach spaces (X , Y).

and EAE and SC
are equivalent for all pairs of Fredholm operators on X and Y by Proposition 1.7.More precisely, we have (iii) Otherwise choose any pair of Fredholm operators (U, V ) ∈ Φ k0 (X )×Φ k0 (Y) with α(U ) = α(V ) (and hence β(U ) = β(V )), and decide whether U and V are SC.
(1) If U and V are SC, then Proposition 1.7 implies that EAE and SC are equivalent for all pairs of Fredholm operators on X and Y, and (2) Otherwise EAE and SC are evidently not equivalent for all pairs of Fredholm operators on X and Y, as (U, V ) is a concrete example of a pair which is EAE, but not SC.
Our next lemma is the counterpart of Lemma 3.9 for SC, showing that the set I SC (X , Y), and hence the associated index sc(X , Y), is an isomorphic invariant.
) which are SC, so that there exist isomorphisms A ∈ B(X 1 ) and D ∈ B(Y 1 ) and operators , and it is easy to check that they are SC, using the operators . The opposite inclusion follows by interchanging X 1 and X 2 , and Y 1 and Y 2 .
The final statement is immediate from the definition (1.8) of sc.
As another consequence of Lemma 5.1, we obtain the following variant of Proposition 3.1 for SC.Proof.The implication ⇐ is obvious because the set sc(X , Y)Z is closed under addition.Conversely, suppose that I SC (X , Y) is closed under addition.Then, in view of Lemma 5.1(i) and (iii), it is an ideal of Z, so I SC (X , Y) = mZ for some m ∈ N 0 .Combining this identity with the definition (1.8) of sc(X , Y), we conclude that m = sc(X , Y).
Corollary 5.4 is not entirely satisfactory because we have been unable to answer the following question.Question 5.5.Is the set I SC (X , Y) closed under addition for every pair of Banach spaces (X , Y)?
We know that the answer to this question is "yes" in certain cases because Lemma 5.1(iv) shows that I SC (X , Y) is closed under addition if sc(X , Y) = 0 or sc(X , Y) = eae(X , Y).We can also obtain a positive answer to it by imposing suitable conditions on the Banach spaces X and Y.To state this result precisely, we require the following additional notation and terminology.Definition 5.6.Let X and Y be Banach spaces.
(ii) We say that a subset Σ of B(X , Y) is essentially closed under addition if, for every pair of operators U, V ∈ Σ, there exists an inessential operator Proposition 5.7.Let X and Y be Banach spaces, and suppose that at least one of the sets G Y (X ) and G X (Y) is essentially closed under addition.Then I SC (X , Y) is closed under addition.
Proof.Suppose that G Y (X ) is essentially closed under addition, and take k 1 , k 2 ∈ I SC (X , Y).Then by Theorem 4.1, we can find operators S j ∈ B(Y, X ) and T j ∈ B(X , Y) such that I X − S j T j ∈ Φ kj (X ) for j = 1, 2. The Index Theorem shows that Both of the operators S 1 T 1 (I X − S 2 T 2 ) and S 2 T 2 belong to G Y (X ), so by the hypothesis, we can find operators and therefore SC k1+k2 (X , Y) = ∅ by another application of Theorem 4.1.This shows that k 1 + k 2 ∈ I SC (X , Y), as required.
The case where G X (Y) is essentially closed under addition is similar, just using condition (ii) in Theorem 4.1 instead of condition (i).
Remark 5.8.The set G Y (X ) is closed under addition (without the need for any inessential perturbations) if the Banach space Y contains a complemented subspace isomorphic to Y ⊕ Y.This result is "folklore"; it can for instance be found in [ There are infinite-dimensional Banach spaces Y which do not contain any complemented subspaces isomorphic to Y ⊕ Y. James' quasi-reflexive Banach space, which will feature prominently in the next example, was the first such example.
Example 5.9.The purpose of this example is to show that the converse of Proposition 5.7 fails; that is, we shall construct Banach spaces X and Y such that I SC (X , Y) is closed under addition, but neither G Y (X ) nor G X (Y) are essentially closed under addition.This construction relies heavily on the quasi-reflexive James spaces J p for 1 < p < ∞.These Banach spaces originate in James' paper [32], where only the case p = 2 was considered.Subsequently, Edelstein and Mityagin [18] observed that James' methods and results carry over to arbitrary p ∈ (1, ∞).We require the following specific facts about this family of Banach spaces: (i) J p is isomorphic to its hyperplanes for every p ∈ (1, ∞), so γ(J p ) = 1.
Berkson and Porta [11, page 18] for p = 2 and Edelstein and Mityagin [18] for general p ∈ (1, ∞) observed that W (J p ) has codimension 1 in B(J p ), so we have a unital algebra homomorphism ϕ : B(J p ) → K with ker ϕ = W (J p ).We shall in fact require the amplification of this homomorphism to the 2 × 2 matrices, that is, the unital algebra homomorphism .
We are now ready to begin our construction: Choose distinct numbers p, q ∈ (1, ∞), and set X = J p ⊕ J p ⊕ J q and Y = J p ⊕ J q ⊕ J q .First, we observe that I SC (X , Y) is closed under addition.This follows immediately from the fact that I SC (X , Y) = Z.Indeed, for each k ∈ Z, (i) implies that we can take R ∈ Φ k (J p ). Then the operators satisfy 1, and therefore k ∈ I SC (X , Y), as desired.Second, we shall show the set G Y (X ) is not essentially closed under addition.Assume the contrary, and consider the operators They both belong to G Y (X ) as the indicated factorizations show.Therefore, by the hypothesis, we can find operators S ∈ B(Y, X ) and T ∈ B(X , Y) such that U + V − ST ∈ E (X ).By (ii), we can write S = S 1 + S 2 and T = T 1 + T 2 , where and S 2 ∈ E (Y, X ) and T 2 ∈ E (X , Y).Since E is an operator ideal, we deduce that Hence, applying the algebra homomorphism ϕ 2 from (iii), we obtain However, this is impossible because the diagonal entries imply that ϕ(S 11 ), ϕ(T 11 ), ϕ(S 21 ) and ϕ(T 12 ) are all non-zero, but then the off-diagonal entries ϕ(S 11 )ϕ(T 12 ) and ϕ(S 21 )ϕ(T 11 ) are also non-zero.This contradiction proves that G X (Y) cannot be essentially closed under addition.Finally, a similar argument with X and Y interchanged shows that the set G X (Y) is not essentially closed under addition.
Remark 5.10.The purpose of this remark is to summarize our knowledge about the values of k ∈ Z for which the equation SC k (X , Y) = EAE k (X , Y) holds true for a given pair of Banach spaces (X , Y), and explain how this problem is related to Question 5.5.Recall from (1.9) and Theorem 1.6(iii) that ( We now split in two cases, beginning with the case where the answer to Question 5.5 is affirmative, so that the set I SC (X , Y) is closed under addition.Then Corollary 5.4 shows that I SC (X , Y) = sc(X , Y)Z and where we have applied Proposition 3.1 to conclude that EAE k (X , Y) = ∅.Together with (5.3), this covers all possible values of k ∈ Z.It follows in particular that EAE and SC coincides for all pairs of Fredholm operators (U, V ) ∈ Φ(X ) × Φ(Y) if and only if eae(X , Y) = sc(X , Y), as we have already seen in Proposition 1.7.
Otherwise, when the answer to Question 5.5 is negative, so that I SC (X , Y) fails to be closed under addition, Proposition 1.7 implies that sc(X , Y) = n eae(X , Y) for some n 2, and is the smallest positive number for which SC k (X , Y) = ∅, and I SC (X , Y) is closed under sign changes.However, since I SC (X , Y) fails to be closed under addition, there must be some number k = m eae(X , Y), where m ∈ N ∩ (n, ∞) \ nN, for which We conclude this section with a couple of results about direct sums which will be useful in the proof of Proposition 1.12, as well as in the next section.Lemma 5.11.Let X and Y be Banach spaces, and suppose that Y is isomorphic to a complemented subspace of X .Then and consequently sc(X , Y) = eae(X , Y).
Proof.In view of Lemmas 3.9 and 5.3, we may suppose that X = Y ⊕ Z for some Banach space Z.
The inclusion Conversely, for k ∈ I Φ (Y), we can take R ∈ Φ k (Y).Then the operators Lemma 5.12.Let X = X 1 ⊕ X 2 and Y = Y 1 ⊕ Y 2 be Banach spaces.Then: 2) is satisfied.Then it is easy to see that the operators are Schur-coupled via is inessential, and therefore, using Remark 2.3(iii), we obtain ) by another application of Theorem 4.1.
We can now complete the proof of Proposition 1.12.
6.The Gowers-Maurey-Aiena-González-Ferenczi cycle of ideas and the proofs of Theorems 1.9, 1.11 and 1.13 The Banach space that Aiena and González used in [2] to show that projective incomparability does not imply essential incomparability is the so-called "shift space" constructed by Gowers and Maurey in [25, §(4.2)].Refining the approach of Aiena and González, Ferenczi [20,Section 4] has more recently used this space to prove that there is no largest proper operator ideal, thereby solving a famous open problem going back to Pietsch's monograph [43].
The proofs of Theorems 1.11 and 1.13 are inspired by this body of work, notably the proof of Lemma 6.3(ii).However, the shift space itself will not suffice for our purposes; we need to work with a larger family of "higher-order shift spaces" which Gowers and Maurey also constructed in [25].We shall now give a brief introduction to this family.
Following the terminology introduced in [25, page 549], for two infinite subsets where (e n ) n∈N denotes the standard unit vector (Hamel) basis for c 00 .Let k 0 ∈ N 0 , and set proper set of spreads" as defined in [25, page 549], but since we do not need the precise definition of this term in the sequel, we omit the details.The important point is that, by [25,Theorem 5], S k0 induces a Banach space, which we shall call the k 0 -fold Gowers-Maurey shift space and denote by X GM (k 0 ).(It is denoted X(S k0 ) in [25].)As already mentioned, Gowers and Maurey defined and investigated this family of Banach spaces in [25].More precisely, they studied the space X GM (0) in [25, §(4.1)],X GM (1) in [25, §(4.2)] and X GM (2) in [25, §(4.3)], before outlining the general case of X GM (k 0 ) for k 0 3 in the final paragraph of [25, §(4.3)].The following theorem summarizes the results from [25] that we require, together with the necessary notation and terminology.(i) The Banach space X GM (k 0 ) has a normalized Schauder basis (e n ) n∈N which admits an isometric k 0 -fold right shift operator R k0 ∈ B(X GM (k 0 )) given by R k0 e n = e n+k0 for every n ∈ N, with left inverse L k0 ∈ B(X GM (k 0 )) given by L k0 e n = 0 for n k 0 and L k0 e n = e n−k0 for n > k 0 .(ii) The Banach space X GM (k 0 ) satisfies a lower f -estimate for the function f (t) = log 2 (t + 1); that is, x k for every n ∈ N and vectors x 1 , . . ., x n ∈ X GM (k 0 ) which are consecutive in the sense that there are integers 0 The index γ introduced in Remark 3.6 is given by γ(X GM (k 0 )) = k 0 , and X GM (k 0 ) is not isomorphic to any of its subspaces of infinite codimension.Therefore a closed subspace W of The Banach space X GM (k 0 ) contains no unconditional basic sequences.
Proof.Parts (i) and (ii) follow from [25,Theorem 5] and the definitions and conventions that it relies on.
For Corollary 6.2.Let Y be a Banach space with an unconditional basis.Then, for every k 0 ∈ N 0 , X GM (k 0 ) and Y are totally incomparable.
Proof.As explained in the comment after [39, Problem 1.d.5], every closed, infinitedimensional subspace of Y contains an unconditional basic sequence.Hence Theorem 6.1(v) implies that no such subspace embeds isomorphically into X GM (k 0 ).
Using these results, we can easily prove Theorem 1.9.
Proof of Theorem 1.9.(i).We must show that I SC (X , Y) = {0}.This was already proved in [29, Theorem 2.1(2)], but we would like to point out that it is also an almost immediate consequence of Theorem 4.1.Indeed, the essential incomparability of X and Y means that I X − ST ∈ Φ(X ) for every S ∈ B(Y, X ) and T ∈ B(X , Y), and Remark 2.3(iii) shows that i(I X − ST ) = 0. Therefore condition (i) in Theorem 4.1 is satisfied only for k = 0, so SC k (X , Y) = ∅ for every k ∈ Z \ {0}.
(ii).Set X = X GM (k 0 ), and let Y be a Banach space which has an unconditional basis and is isomorphic to its hyperplanes, so that γ(Y) = 1.(For instance, Y = ℓ 2 has these properties.)Corollary 6.2 shows that X and Y are totally incomparable and therefore essentially incomparable.Moreover, we have γ(X ) = k 0 by Theorem 6.1(iv), so eae(X , Y) = lcm(k 0 , 1) = k 0 by (3.1).
While the above proof did not involve any ideas from [2] or [20], the proofs of Theorems 1.11 and 1.13 will, namely in the shape of part (ii) of the following lemma.Lemma 6.3.Let k 0 ∈ N.
(i) Suppose that X 1 and Y 1 are essentially incomparable Banach spaces with unconditional bases and that Y 2 is a closed, infinite-dimensional and infinitecodimensional subspace of X GM (k 0 ).Then the Banach spaces Remark 6.4.Lemma 6.3(i) is also true for X 1 = {0} (even though it may be debatable whether this space has an unconditional basis).This observation will be important in the proofs of Theorems 1.11(i) and 1.13(i).The conscientious reader can check that the proof which we are about to present remains valid for X 1 = {0}.
Proof of Lemma 6.3(i).To unify notation, set X 2 = X GM (k 0 ).The proof is by contradiction, so assume that X 1 ⊕ X 2 contains an infinite-dimensional, complemented subspace W which is isomorphic to a complemented subspace Z of Y 1 ⊕ Y 2 .Corollary 6.2 shows that each of the pairs (X 1 , X 2 ) and (Y 1 , Y 2 ) is totally incomparable, so a theorem of Edelstein and Wojtaszczyk (see [19,Theorem 3.5], or [39, Theorem 2.c.13] for an exposition) implies that where W j and Z j are complemented subspaces of X j and Y j , respectively, for j ∈ {1, 2}.Take an isomorphism The hypothesis that X 1 and Y 1 are essentially incomparable implies that the operator U 11 is inessential because essential incomparability clearly passes to complemented subspaces.Moreover, U 12 and U 21 are strictly singular and therefore inby Corollary 6.2 and Remark 2.5(i).Consequently is an inessential perturbation of the isomorphism U and hence a Fredholm operator.This implies that U 22 is a Fredholm operator and that W 1 is finite-dimensional, so W 2 must be infinite-dimensional.Since it is complemented in X 2 , Theorem 6.1(iii) shows that W 2 has finite codimension in X 2 .
Choose a closed subspace W 3 of finite codimension in W 2 such that Then W 3 ∼ = X 2 by Theorem 6.1(iv), and the restriction of U 22 to W 3 is an isomorphic embedding into Z 2 ⊆ Y 2 .However, another application of Theorem 6.1(iv) shows that no such embedding exists because Y 2 has infinite codimension in X 2 .
In order to prove the second part of Lemma 6.3, we require two lemmas.The statement of the first of these involves the following standard piece of terminology.An operator T ∈ B(X , Y) (where X and Y can be any Banach spaces) is bounded below if there exists ε > 0 such that T x ε x for every x ∈ X .This is equivalent to saying that T is injective and has closed range, or in other words that T is an isomorphic embedding.
Finally, we observe that Y 2 must be infinite-dimensional because otherwise T would be a finite-rank operator, in which case i(I XGM(k0) − ST ) = 0 = k 0 .Remark 6.7.Lemma 6.6 does not follow from the statement of [37,Theorem 5.4] itself.In fact, Lebow Schechter could have concluded their proof of [37, Theorem 5.4] after its first four lines by invoking the well-known fact that if β(A) < ∞ for an operator A between Banach spaces, then A has closed range.
However, as we have seen, the remainder of their proof is very useful for our purposes because it establishes the stronger conclusion stated in Lemma 6.6 that we required to prove Lemma 6.3(ii).More precisely, what we needed was that we can perturb the operator A by an inessential operator B to obtain that ran(A − B) has infinite codimension in Y.We did not need that the perturbation B can be chosen to be nuclear and have arbitrarily small norm; we chose to state those facts simply because they follow automatically from the proof.
We remark that both Aiena-González and Ferenczi cite [37,Theorem 5.4] in their work, but as far as we can see, that result does not suffice to give their conclusions.Like us, they appear to rely on the stronger statement given in Lemma 6.6.
Proof of Theorem 1.11.(i).Set X = X GM (k 0 ) and Y = Y 1 ⊕ Y 2 , where Y 1 is a Banach space which is isomorphic to its hyperplanes and has an unconditional basis (so for instance we can take Y 1 = ℓ 2 or Y 1 = c 0 ), and Y 2 is the closed, infinitedimensional and infinite-codimensional subspace of X constructed in Lemma 6.3(ii).Then X and Y are projectively incomparable, as observed in Remark 6.4.Moreover, Theorem 6.1(iv) shows that γ(X ) = k 0 , while γ(Y) = 1 because Y 1 being isomorphic to its hyperplanes implies that the same is true for Y. Therefore eae(X , Y) = lcm(k 0 , 1) = k 0 by (3.1).
In view of Proposition 1.7, this means that sc(X , Y) is a multiple of k 0 .Hence, to show that sc(X , Y) = k 0 , it will suffice to show that SC k0 (X , Y) = ∅, which follows by combining Lemma 5.12(i) with the fact that SC k0 (X , Y 2 ) = ∅.
(ii).This proof is a slightly more elaborate variant of the proof of (i) that we have just given.We begin by choosing two distinct spaces X 1 and Y 1 from the family {ℓ p : 1 p < ∞} ∪ {c 0 }, so that X 1 and Y 1 are isomorphic to their hyperplanes, have unconditional bases and are totally incomparable (as observed in [39, page 75], for instance).Set X 2 = X GM (k 0 ), and let Y 2 be the subspace of X 2 constructed in Lemma 6.3(ii), as in the first part of the proof.Then X = X 1 ⊕ X 2 and Y = Y 1 ⊕ Y 2 are projectively incomparable by Lemma 6.3(i), and γ(X ) = γ(Y) = 1 because X 1 and Y 1 are isomorphic to their hyperplanes, so eae(X , Y) = 1 by (3.1).
The proofs of the two parts of Theorem 1.13 are somewhat more complicated variants of the proofs of the corresponding parts of Theorem 1.11 given above.They involve one additional ingredient, namely Gowers' solution to Banach's hyperplane problem, which was the first infinite-dimensional Banach space shown not to be isomorphic to its hyperplanes (see [23], as well as [25, §(5.1)] for further results).The following result summarizes the properties of this space that we require.Theorem 6.8 (Gowers).There exists an infinite-dimensional, reflexive Banach space X G with an unconditional basis such that X G fails to be isomorphic to any proper subspace of itself.
Proof.The only part of this statement that Gowers did not prove explicitly in [23] is that X G is reflexive.We believe that this fact is known to specialists, but as we have been unable to locate a proof of it in the literature, we outline one here.Since X G is not isomorphic to its hyperplanes, it cannot contain any complemented subspace which is isomorphic to its hyperplanes, so in particular no complemented subspace of X G is isomorphic to c 0 or ℓ 1 .Hence, no subspace of X G is isomorphic to c 0 by Sobczyk's Theorem (see, e.g., [39, Theorem 2.f.5]) or to ℓ 1 by a much more recent theorem of Finol and Wójtowicz [21].(This result was previously stated without proof in [38].)Therefore, a classical result of James (see [32], or [39, Theorem 1.c.12(a)] for an exposition) shows that X G is reflexive.
Proof of Theorem 1.13.(i).Following the same approach as in the proof of Theorem 1.11(i), but using different notation, we define X 1 = X GM (k 0 ) and Y 1 = c 0 ⊕Y 2 , where Y 2 is the subspace of X 1 constructed in Lemma 6.3(ii).Then, as shown in the proof of Theorem 1.11(i), X 1 and Y 1 are projectively incomparable, so (1) holds, and eae(X 1 , Y 1 ) = k 0 .( Let Z = X G be the Banach space from Theorem 6.8.Then γ(Z) = 0, so eae(Z, Z) = 0, which verifies (3).Moreover, Corollary 6.2 shows that Z is totally incomparable with X 1 , and therefore also with Y 2 .Since every closed subspace of c 0 contains an isomorphic copy of c 0 , while Z is reflexive, Z and c 0 are also totally incomparable, and therefore Z and Y 1 are essentially incomparable.This shows that (2) is satisfied.
(ii).As above, let Z = X G be the Banach space from Theorem 6.8 and set Y 1 = c 0 ⊕ Y 2 , where Y 2 is the subspace of X GM (k 0 ) from Lemma 6.3(ii), but now define X 1 = ℓ 1 ⊕ X GM (k 0 ).Then X 1 and Y 1 are projectively incomparable by Lemma 6.3(i).We showed in the first part of the proof that Y 1 and Z are essentially incomparable; a similar argument gives the same conclusion for X 1 and Z. Hence conditions (1)-(3) are satisfied.
Arguing as before, we see that the Banach spaces satisfy γ(X ) = γ(Y) = 1, so that eae(X , Y) = 1, and we also have SC k0 (X , Y) = ∅, which implies that k 0 Z ⊆ I SC (X , Y).To verify the opposite inclusion, we observe that each of the pairs (ℓ 1 , c 0 ), (ℓ 1 , Y 2 ⊕ Z) and (c 0 , X GM (k 0 ) ⊕ Z) is essentially incomparable, so Lemma 5.12(ii) shows that where the final equality follows from Corollary 6.2 and Lemma 3.10(ii).Hence we have I SC (X , Y) = k 0 Z, and therefore sc(X , Y) = k 0 .

Lemma 2 . 4 .
Let S ∈ B(Y, X ) and T ∈ B(X , Y) for some Banach spaces X and Y. Then ker(I X − ST ) ∼ = ker(I Y − T S).Moreover, if ran(I X − ST ) and ran(I Y − T S) are closed, then X / ran(I X − ST ) ∼ = Y/ ran(I Y − T S).Proof.The earliest mention of the first part of this result that we know of is [42, Satz 1], while both parts can be found in [41, Sätze 2.4-A-2.5-A].Alternatively, the result follows from [6, page 211, properties 1 and 6], see also [7, Proposition 1], because the operators I X − ST and I Y − T S are Schur coupled via A = I X , B = S, C = T and D = I Y .

Case 1 :
If α(T ) = m, then we can simply take R = 0. Case 2: If α(T ) > m, set n = α(T )−m ∈ N, and note that β(T ) = α(T ) − i(T ) n, so we can find an n-dimensional subspace Z of Y such that Z ∩ ran T = {0}.Take an operator A ∈ B(ker T, Y) with range Z and a bounded linear projection P of X onto ker T , and define R = AP ∈ B(X , Y).Then ker(T + R) = ker A, which has dimension α(T ) − dim ran A = m.Case 3: If α(T ) < m, choose an m-dimensional subspace W of X such that ker T ⊆ W, and let P ∈ B(X ) be a projection onto W. Then ker T (I X − P ) = W, so R = −T P has the required property.Remark 3.3.The condition that m i(T ) is necessary in Lemma 3.2 because

Theorem 4 . 1 .
The following four conditions are equivalent for every pair of Banach spaces (X , Y) and every k ∈ Z :(i) There exist operatorsS ∈ B(Y, X ) and T ∈ B(X , Y) such that I X − ST ∈ Φ k (X ).(ii) There exist operators S ∈ B(Y, X ) and T ∈ B(X , Y) such that I Y − T S ∈ Φ k (Y).(iii) EAE k (X , Y) = SC k (X , Y), and this set is non-empty.(iv) SC k (X , Y) = ∅.The proof of Theorem 4.1 involves two lemmas.The first of these reformulates SC in a way that is much closer to condition (i) above.Lemma 4.2.Let U ∈ B(X ) and V ∈ B(Y) for some Banach spaces X and Y. Then U and V are SC if and only if there are isomorphisms M ∈ B(X ) and N ∈ B(Y) and operators S ∈ B(Y, X ) and T ∈ B(X , Y) such that U M = I X − ST and V N = I Y − T S. (4.1) Proof.This is a straightforward verification.On the one hand, if the operators A, B, C and D satisfy (1.2), then M = A −1 , N = D −1 , S = BD −1 and T = CA −1 satisfy (4.1), and on the other, if M , N , S and T satisfy (4.1), then A = M −1 , B = SN −1 , C = T M −1 and D = N −1 satisfy (1.2).
and Φ k (Y) is also non-empty by Lemma 2.4, so EAE k (X , Y) is non-empty by Proposition 3.1.

Corollary 5 . 4 .
Let X and Y be Banach spaces.Then the set I SC (X , Y) is closed under addition if and only if I SC (X , Y) = sc(X , Y)Z.
36, the paragraph following Definition 3.6].Most "classical" Banach spaces Y satisfy the even stronger condition that Y ∼ = Y ⊕ Y.The two conditions are not equivalent because Gowers and Maurey [25, §(4.4)] have constructed a Banach space Y which is isomorphic to its cube Y ⊕ Y ⊕ Y, but not to its square Y ⊕ Y. Hence Y contains a complemented subspace isomorphic to Y ⊕ Y without being isomorphic to it.
we define the associated spread S A,B to be the linear map on c 00 determined byS A,B e j = e b k if j = a k for some k ∈ N, 0 otherwise,
ii).Suppose that k ∈ I SC (X , Y).By Theorem 4.1, we can find operators X − ST ∈ Φ k (X ).The hypothesis implies that S 11 , T 11 , S 12 , T 12 , S 21 and T 21 are inessential.Since E is an operator ideal, it follows that the operator