SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS

VICTOR I. LOMONOSOV, HEYDAR RADJAVI, AND VLADIMIR G. TROITSKY

Abstract. An algebra of operators on a Banach space X is said to be transitive if X has no nontrivial closed subspaces invariant under every member of the algebra. In this paper we investigate a number of conditions which guarantee that a transitive algebra of operators is “large” in various senses. Among these are the conditions of algebras being localizing or sesquitransitive. An algebra is localizing if there exists a closed ball B ∌ 0 such that for every sequence (x_{n}) in B there exists a subsequence (x_{n_{k}}) and a bounded sequence (A_{k}) in the algebra such that (A_{k}x_{n_{k}}) converges to a non-zero vector. An algebra is sesquitransitive if for every non-zero z ∈ X there exists C > 0 such that for every x linearly independent of z, for every non-zero y ∈ X, and every ε > 0 there exists A in the algebra such that ∥Ax − y∥ < ε and ∥Az∥ ⩽ C∥z∥. We give an algebraic version of this definition as well, and extend Jacobson’s density theorem to algebraically sesquitransitive rings.

1. Introduction, preliminaries and notation

Throughout this paper, X will be a real or complex Banach space, and L(X) will denote the space of all continuous linear operators on X. If T ∈ L(X), we say that T has an invariant subspace if there exists a closed non-zero proper subspace Y of X such that T (Y) ⊆ Y. We say that a subspace Y is hyperinvariant for T if Y is invariant under every operator in {T }′. Here S′ is the commutant of a set S ⊆ L(X), that is, S′ = {A ∈ L(X) | ∀S ∈ S AS = SA}. A subset S ⊆ L(X) is said to be transitive if Sx is dense in X for every non-zero x ∈ X, where Sx = {Ax | A ∈ S}. The symbol A will usually stand for a subalgebra of L(X). We will write B_{X} and B_{A} for the closed unit balls of X and A respectively. It can be easily verified that A is transitive iff it has no common invariant subspaces. Furthermore, A is transitive iff \overline{A}^{WOT} is transitive, where \overline{A}^{WOT} stands for the closure of A in the weak operator topology (WOT).

It was proved in [Lom73] that if T ∈ L(X) commutes with a non-zero compact operator, then T has an invariant subspace. If, in addition, X is a complex Banach space and T is not a multiple of the identity operator then T has a hyperinvariant subspace. Hooker [Hoo81] proved that in the real case T would still have a hyperinvariant subspace provided that, in addition, T doesn’t satisfy a real-irreducible quadratic equation.

Integral Equations and Operator Theory, 60(3), 2008, 405-418.

1 2 V. I. LOMONOSOV, H. RADJAVI, AND V. G. TROITSKY

However, in general there exist operators on real and complex Banach spaces with no invariant subspaces; see [Enf76, Read84].

Note that for an operator T, its commutant {T }′ is a WOT-closed algebra, and T has no hyperinvariant subspaces iff {T }′ is transitive. This naturally leads to the study of transitive algebras. It follows from [Enf76, Read84] that there exist transitive algebras of operators on Banach spaces which are not WOT-dense. However, there are several known conditions which, together with transitivity, guarantee that the algebra is WOTdense. For example, every strictly transitive algebra (see Section 4 for the definition) is WOT-dense [Yood49, Ric50]. In the finite-dimensional case, the Burnside Theorem asserts that M_{n}(ℂ) contains no proper transitive subalgebras. Also an algebraic version of [Lom73] (see, e.g., [RR03]) asserts that if A is a transitive algebra of operators on a complex Banach space such that A contains a compact operator then \overline{A}^{WOT} = L(X).

In this paper we study several conditions on an operator algebra A which, although do not necessarily imply that \overline{A}^{WOT} = L(X), provide some information about the size of A by ensuring that A′ is small (e.g., finite-dimensional). We also introduce several new conditions on algebras of operators.

Definition 1.1. We will say that an algebra A of operators on a Banach space X is localizing if there exists a closed ball B in X such that 0 ∉ B and for every sequence (x_{n}) in B there is a subsequence (x_{n_{i}}) and a sequence (S_{i}) in A such that ∥S_{i}∥ ≤ 1 and (S_{i}x_{n_{i}}) converges in norm to a nonzero vector.

It is easy to see that if T is an injective compact operator, then {T }′ is localizing. The following theorem was obtained in [Tro04] using the method of minimal vec-

tors [AE98, And03, CPS04].

Theorem 1.2. Let T be a quasinilpotent operator on a Banach space X. If {T }′ is localizing then T has a hyperinvariant subspace.

Theorem 1.2 easily extends to algebras of operators as follows.

Theorem 1.3. Suppose that X is a Banach space and A is a transitive localizing subalgebra of L(X). Then A′ contains no non-zero quasinilpotent operators.

Proof. Suppose T ∈ A′ is non-zero and quasinilpotent. It follows from A ⊆ {T }′ that {T }′ is localizing, so that T has a hyperinvariant subspace. This contradicts transitivity of A. □

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 3

In Section 2 we investigate SC-algebras, i.e., the algebras where the unit ball is relatively compact in the strong operator topology (SOT). In particular, we show that if {T }′ is an SC-algebra and {T^{∗}}′ is localizing then T^{∗} has an invariant subspace.

In Section 3 we introduce quasi-localizing algebras by replacing the condition ∥S_{i}∥ ≤ 1 in Definition 1.1 with inequalities ∥S_{i}z_{n_{i}}∥ < C∥z_{n_{i}}∥ for a subsequence of a given sequence (z_{n}). We show that Theorem 1.3 remains valid for quasi-localizing algebras.

Motivated by the quasi-localizing property, in Section 4 we define an algebra to be sesquitransitive if for every non-zero z ∈ X there exists C > 0 such that for every x linearly independent of z, for every non-zero y ∈ X, and every ε > 0 there exists A in the algebra such that ∥Ax − y∥ < ε and ∥Az∥ ⩽ C∥z∥. We say that A is uniformly sesquitransitive if C can be chosen to be independent of z. We prove that sesquitransitive algebras have trivial commutant. We show in Section 5 that the Burnside theorem and [Lom73] remain valid in the real case if transitivity is replaced with sesquitransitivity.

2. SC-algebras with localizing adjoint

In this section we make use of the following fixed point theorem due to Ky Fan [Fan52]. Recall that if Ω is a topological space and C: Ω → P(Ω) is a point-to-set map from Ω to the power set of Ω, then C is said to be upper semi-continuous if for every x_{0} ∈ Ω and every open set U such that C(x_{0}) ⊆ U there is a neighborhood V of x_{0} such that C(x) ⊆ U whenever x ∈ V.

Theorem 2.1 ([Fan52]). Let K be a compact convex set in a locally convex space, and suppose that C is an upper semi-continuous point-to-set map from K to closed convex non-empty subsets of K. Then there is x_{0} ∈ K with x_{0} ∈ C(x_{0}).

Recall that the original proof of the main result in [Lom73] involved the following fact.

Lemma 2.2 ([Lom73]). Let X be a real or complex Banach space, S a convex transitive subset of L(X), and K a non-zero compact operator. Then there exists A ∈ S such that AK has a non-zero fixed vector.

The following theorem goes along the same lines. Suppose that X is a dual Banach space, i.e., X = Y^{∗} for some Banach space Y. The weak* operator topology on L(X) is defined as follows: a net (A_{α}) converges to A in W*OT if ⟨(A_{α} − A)x, ξ⟩ → 0

4 V. I. LOMONOSOV, H. RADJAVI, AND V. G. TROITSKY

for all x ∈ X and ξ ∈ Y. It is known that the norm closed unit ball B_{L(X)} of L(X) is W*OT-compact. It follows easily that if A is a W*OT-closed subalgebra of L(X), then B_{A} is also W*OT-compact. If B a is a subalgebra of L(Y), we write B^{∗} = {A^{∗} | A ∈ B} and call it the algebra of adjoints of B.

Following [Lom80] we say that an algebra of operators is an SC-algebra if its unit ball is SOT-relatively compact. It is easy to see that if A is an SC-algebra then the map A ∈ A ↦ Ax ∈ X is compact for every x ∈ X. It was shown in [Lom80] that if T is an essentially normal operator on a Hilbert space such that neither {T }′ nor {T^{∗}}′ is an SC-algebra, then T has an invariant subspace. However, in the following theorem we use the SC condition in order to prove existence of invariant subspaces.

Theorem 2.3. Suppose that X is a dual complex Banach space, X = Y^{∗}, and A is a transitive localizing W*OT-closed algebra in L(X) such that A = B^{∗} for some SCalgebra B in L(Y). Then for every non-zero adjoint operator T in A′ there exists A ∈ A such that AT has a non-zero fixed vector. Furthermore, T has an invariant subspace.

Proof. Let T be a non-zero operator in A′ such that T = S^{∗} for some S ∈ L(Y). Since A is transitive and ker T and \overline{Range T} are A-invariant, T is one-to-one and has dense range.

Let B be a ball as in Definition 1.1. We claim that there exists r > 0 such that for every x ∈ B we have rB_{A}(T x) ∩ B ≠ ∅, that is, there exists A ∈ A such that ∥A∥ ⩽ r and AT x ∈ B. Indeed, if this were false, then for every n we would find x_{n} ∈ B such that ∥A∥ ⩾ n whenever A ∈ A and AT x_{n} ∈ B. We can choose a subsequence (x_{n_{i}}) and a sequence of contractions (S_{i}) in A such that S_{i}x_{n_{i}} → w ≠ 0. It follows that S_{i}T x_{n_{i}} = T S_{i}x_{n_{i}} → T w. Since A is transitive, we can find R ∈ A such that RT w ∈ Int B. It follows that for all sufficiently large i we have RS_{i}T x_{n_{i}} ∈ Int B, so that ∥RS_{i}∥ ⩾ n_{i} → +∞ by our choice of x_{n_{i}}. But ∥RS_{i}∥ ⩽ ∥R∥ is bounded, contradiction.

Define a set function C: B → P(B) via x ∈ B ↦ C(x) = B ∩ rB_{A}(T x). By the preceding argument C(x) is non-empty. Clearly, C(x) is convex. Observe also that C(x) is weak* closed for every x ∈ B because B_{A}(T x) is weak* compact as the image of the W*OT-compact set B_{A} under the map A ∈ L(X) ↦ AT x ∈ X which is W*OT-w*-continuous.

We will show that C is weak* upper semi-continuous. Suppose not; then there exists x_{0} ∈ B and a weak* open set U such that C(x_{0}) ⊆ U, but for every weak*

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 5

neighborhood α of x_{0} there exists x_{α} ∈ α such that C(x_{α}) is not contained in U. Pick any y_{α} ∈ C(x_{α}) \ U. Let Λ be the set of all weak* neighborhoods of x_{0}, ordered by the reverse inclusion. The collections (x_{α})_{α∈Λ} and (y_{α})_{α∈Λ} can be viewed as nets indexed by Λ, and x_{α} ⟶^{w^{∗}} x_{0}. Since B is weak* compact, by passing to a sub-net if necessary we can assume that y_{α} ⟶^{w^{∗}} y_{0} for some y_{0} ∈ B. Also, y_{α} ∉ U for every α implies y_{0} ∉ U. Note that y_{α} ∈ C(x_{α}) implies that there exists A_{α} ∈ rB_{A} such that y_{α} = A_{α}T x_{α}. For every α we have A_{α} = F_{α}^{∗} for some F_{α} in B. Since B is an SC-algebra, we can assume that F_{α} ⟶^{SOT} F_{0} for some F_{0} ∈ L(Y). It follows that A_{α} ⟶^{W^{∗}OT} A_{0}, where A_{0} = F_{0}^{∗}, so that A_{0} ∈ rB_{A}. Let ξ ∈ Y. We have SF_{α}ξ → SF_{0}ξ in norm (recall that T = S^{∗}), so that

⟨y_{α}, ξ⟩ = ⟨A_{α}T x_{α}, ξ⟩ = ⟨x_{α}, SF_{α}ξ⟩ → ⟨x_{0}, SF_{0}ξ⟩ = ⟨A_{0}T x_{0}, ξ⟩.

It follows that y_{0} = A_{0}T x_{0}, so that y_{0} ∈ C(x_{0}) ⊆ U, contradiction.

Since the map C: B → P(B) is upper semi-continuous in the weak* topology, it has a fixed point by Ky Fan’s Theorem, i.e., there exists x ∈ B such that x ∈ C(x). That is, there exists A ∈ rB_{A} such that x = AT x.

Thus, the fixed space of AT, defined by F = ker(I − AT), is non-trivial and closed. If T is not invertible then F is proper and we are done. If T is invertible, pick any λ ∈ σ(T) and put S = λI −T. Then S is not invertible, so that the preceding reasoning yield that S has an invariant subspace. Clearly, it will be also invariant under T. □

Now we can prove a version of Theorem 1.2 for non-quasinilpotent operators.

Corollary 2.4. Suppose that T is an adjoint operator on a dual complex Banach space. If {T }′ is localizing and is the algebra of adjoints of an SC-algebra, then T has an invariant subspace.

Proof. If T has a hyperinvariant subspace then there is nothing to prove. Otherwise, {T }′ is transitive, and we get the result by Theorem 2.3. □

Note that the hypotheses of Corollary 2.4 are satisfied for a one-to-one compact operator on a reflexive Banach space.

Remark 2.5. Observe that Theorem 2.3 and Corollary 2.4 remain valid for real spaces provided that T doesn’t satisfy an irreducible quadratic equation. Indeed, the problem with T occurs only in the last statement in Theorem 2.3 and the last paragraph of its proof. Suppose that λI − T is invertible for all λ ∈ ℝ. Pick any α + βi ∈ σ(T), and put S = β^{2}I +(T −αI)^{2}. It can be easily verified that S is non-zero and not invertible.

Hence, there exists A ∈ A such that F := ker(I − AS) is proper and non-trivial. Clearly, F is T -invariant.

3. Minimal vectors of quasi-localizing algebras

In this section we consider another generalization of Theorem 1.3.

Definition 3.1. Let X be a Banach space, and A a subalgebra of L(X). We say that A is quasi-localizing if there exists a ball B in X not containing the origin, such that given two sequences (x_{n}) in B and (z_{n}) in X \ {0}, there exist subsequences (x_{n_{i}}) and (z_{n_{i}}), a sequence (S_{i}) in A, and a real C > 0 such that ∥S_{i}z_{n_{i}}∥ ⩽ C∥z_{n_{i}}∥ for all i and (S_{i}x_{n_{i}}) converges to a non-zero vector w.

Note that we do not require that S_{i}’s are uniformly bounded. It is easy to see that every localizing algebra is quasi-localizing.

Recall some notation and terminology from the method of minimal vectors that will be needed in the proof of Theorem 3.2 (we refer the reader to [AE98] and [Tro04] for details). Suppose that x_{0} ∈ X and r < ∥x_{0}∥, so that the closed ball B = B(x_{0}, r) doesn’t contain the origin. Suppose that Q is a one-to-one operator with dense range, and ε > 0. Let d = dist(0, Q^{−1}B). Choose y ∈ Q^{−1}B such that ∥y∥ ⩽ (1 + ε)d. Such a y is called a (1 + ε)-minimal vector. Using the Hahn-Banach Theorem, find f ∈ X^{∗} of norm one, such that f_{|B} ⩾ c and f_{|QB(0,d)} ⩽ c for some c > 0. We call f a minimal functional. It is easy to see that f(x_{0}) ⩾ r and that the hyperplane Q^{∗}f = c separates (non-strictly) B(0, d) and QB. It follows easily that

(1)  (Q^{∗}f)(y) ⩾ (1 + ε)^{−1}∥Q^{∗}f∥∥y∥.

Repeating the preceding procedure with Q replaces with Q^{n} for every n ∈ ℕ, we produce y_{n} and f_{n}. Thus, we end up with sequences (y_{n}) and (f_{n}) such that y_{n} is a (1 + ε)-minimal vector and f_{n} is a minimal functional for Q^{n} and B.

Since every localizing algebra is quasi-localizing, the following theorem is a generalization of Theorem 1.3. The proof is similar to that of the main theorem of [Tro04].

Theorem 3.2. Suppose that X is a Banach space and A is a transitive quasi-localizing subalgebra of L(X). Then A′ contains no non-zero quasinilpotent operators.

Proof. We present a proof for the case of a real Banach space. The complex case can be obtained by straightforward modifications. Suppose that Q is a non-zero quasinilpotent operator in A′. Without loss of generality, A is unital. Since A is transitive, Q is oneto-one and has dense range. Let B = B(x_{0}, r) be the ball as in Definition 3.1. Fix

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 7

ε > 0. Let (y_{n}) and (f_{n}) be the sequences of (1 + ε)-minimal vectors and minimal functionals for B and (Q^{n}). Then there is a subsequence (n_{i}) such that \frac{∥y_{n_{i}−1}∥}{∥y_{n_{i}}∥} → 0. Indeed, otherwise there would exist δ > 0 such that \frac{∥y_{n−1}∥}{∥y_{n}∥} > δ for all n, so that ∥y_{1}∥ ⩾ δ∥y_{2}∥ ⩾ … ⩾ δ^{n}∥y_{n+1}∥. Since Q^{n}y_{n+1} ∈ Q^{−1}B, it follows from the definition of y_{1} that

∥Q^{n}y_{n+1}∥ ⩾ \frac{∥y_{1}∥}{1 + ε} ⩾ \frac{δ^{n}}{1 + ε}∥y_{n+1}∥.

It follows that ∥Q^{n}∥ ⩾ \frac{δ^{n}}{1+ε}, which contradicts the quasinilpotence of Q.

Since ∥f_{n_{i}}∥ = 1 for all i, we can assume (by passing to a further subsequence) that (f_{n_{i}}) weak*-converges to some g ∈ X^{∗}. Since f_{n}(x_{0}) ⩾ r for all n, it follows that g(x_{0}) ⩾ r, hence g ≠ 0.

Observe that the sequence (Q^{n_{i}−1}y_{n_{i}−1})_{i=1}^{∞} is contained in B. Since A is quasilocalizing, by passing to yet a further subsequence if necessary, we find a sequence (S_{i}) in A such that ∥S_{i}y_{n_{i}−1}∥ ⩽ C∥y_{n_{i}−1}∥ and S_{i}Q^{n_{i}−1}y_{n_{i}−1} → w ≠ 0. Put

Y = AQw = {T Qw | T ∈ A}.

One can easily verify that Y is a linear subspace of X invariant under A. Since Q is one-to-one, we have 0 ≠ Qw. Since A is transitive, Y is dense in X. On the other hand, we will show that Y ⊆ ker g, which would lead to a contradiction.

Let T ∈ A, we will show that g(T Qw) = 0. It follows from (1) that

\frac{|f_{n_{i}}(Q^{n_{i}}T S_{i}y_{n_{i}−1})|}{f_{n_{i}}(Q^{n_{i}}y_{n_{i}})} = \frac{|(Q^{∗n_{i}}f_{n_{i}})(T S_{i}y_{n_{i}−1})|}{(Q^{∗n_{i}}f_{n_{i}})(y_{n_{i}})}

⩽ \frac{∥Q^{∗n_{i}}f_{n_{i}}∥∥T S_{i}y_{n_{i}−1}∥}{(1 + ε)^{−1}∥Q^{∗n_{i}}f_{n_{i}}∥∥y_{n_{i}}∥} ⩽ \frac{∥T ∥ · C∥y_{n_{i}−1}∥}{(1 + ε)^{−1}∥y_{n_{i}}∥} → 0

Since ∥f_{n_{i}}∥ = 1 and Q^{n_{i}}y_{n_{i}} ∈ B, we have

f_{n_{i}}(Q^{n_{i}}y_{n_{i}}) ⩽ ∥Q^{n_{i}}y_{n_{i}}∥ ⩽ ∥x_{0}∥ + r,

it follows that f_{n_{i}}(Q^{n_{i}}T S_{i}y_{n_{i}−1}) → 0. On the other hand,

f_{n_{i}} ⟶^{w^{∗}} g and Q^{n_{i}}T S_{i}y_{n_{i}−1} = T QS_{i}Q^{n_{i}−1}y_{n_{i}−1} → T Qw,

therefore g(T Qw) = 0. □

Recall that an operator S on a Banach space X is strictly singular if the restriction of S to any infinite-dimensional subspace of X fails to be an isomorphism. It is easy to see that every compact operator is strictly singular, and that strictly singular operators form a norm closed two-sided algebraic ideal in L(X). There is an example of a strictly

8 V. I. LOMONOSOV, H. RADJAVI, AND V. G. TROITSKY

singular operator without invariant subspaces [Read99]. See [LT77] for further details on strictly singular operators.

It is easy to see that the spectrum of a strictly singular operator consists of eigenvalues and zero. If it has eigenvalues, then every eigenspace is a hyperinvariant subspace. Otherwise, it is quasinilpotent. Therefore, from Theorem 3.2 we can immediately deduce the following result.

Corollary 3.3. If T is strictly singular and {T }′ is quasi-localizing, then T has a hyperinvariant subspace.

Recall that a Banach space is said to be hereditarily indecomposable if no closed subspace of it can be written as a direct sum of two infinite-dimensional closed subspaces, [GM93]. Every operator on a hereditarily indecomposable Banach spaces is of the form λI + S, where S is strictly singular.

Corollary 3.4. If A is a transitive quasi-localizing subalgebra on a hereditarily indecomposable Banach space then A′ is trivial.

Proof. Let T ∈ A′, then T = μI + S for some strictly singular operator S. It follows that S ∈ A′. Furthermore, given any scalar λ, then λI − S ∈ A′, so that ker(λI − S) is invariant under A, hence trivial. Therefore, S has no eigenvalues. It follows that S is quasinilpotent, and Theorem 3.2 yields S = 0. □

Suppose now that S is a collection of one-to-one operators with dense range. Again, fix a ball B = B(x_{0}, r) with 0 ∉ B, fix ε > 0 and for each A ∈ S choose a (1 + ε)- minimal vector y_{A} for A and B.

Proposition 3.5. Suppose that S, B, ε, and (y_{A})_{A∈S} are as above. If there are nets (z_{α}) in X and (A_{α}) in S such that \frac{∥z_{α}∥}{∥y_{Aα}∥} → 0 while A_{α}z_{α} converges to some w ≠ 0, then S′ is non-transitive.

Proof. For A ∈ S let f_{A} be a minimal functional for A and B. By passing to a sub-net if necessary, we can assume that f_{A_{α}} ⟶^{w^{∗}} g for some g ∈ X^{∗}. Again, g ≠ 0 because g(x_{0}) ⩾ r.

Put Y = S′w. Then, clearly, Y is invariant under S′ and non-trivial as w ∈ Y. We will show that Y ⊆ ker g, this will imply that Y is not dense in X. Let T ∈ S′, then it follows from (1) that

\frac{|f_{A_{α}}(A_{α}T z_{α})|}{f_{A_{α}}(A_{α}y_{A_{α}})} = \frac{|(A_{α}^{∗}f_{A_{α}})(T z_{α})|}{(A_{α}^{∗}f_{A_{α}})(y_{A_{α}})} ⩽ (1 + ε)\frac{∥A_{α}^{∗}f_{A_{α}}∥∥T z_{α}∥}{∥A_{α}^{∗}f_{A_{α}}∥∥y_{A_{α}}∥} ⩽ (1 + ε)\frac{∥T ∥∥z_{α}∥}{∥y_{A_{α}}∥} → 0.

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 9

Since

0 ⩽ f_{A_{α}}(A_{α}y_{A_{α}}) ⩽ ∥f_{A_{α}}∥∥A_{α}y_{A_{α}}∥ ⩽ ∥x_{0}∥ + r,

it follows that f_{A_{α}}(A_{α}T z_{α}) → 0. On the other hand, since

f_{A_{α}} ⟶^{w^{∗}} g and A_{α}T z_{α} = T A_{α}z_{α} → T w,

it follows that g(T w) = 0. □

Consider the condition in Proposition 3.5. We can assume without loss of generality (by scaling) that ∥A_{α}∥ = 1 for all α. Then (z_{α}) cannot converge to zero, as this would imply A_{α}z_{α} → 0. Thus, it is necessary that ∥y_{A_{α}}∥ → ∞. This leads to the following question.

Question. Under what conditions on S is the set {y_{A} | A ∈ S, ∥A∥ = 1} unbounded?

4. Sesquitransitivity

Recall that a set S ⊆ L(X) is said to be n-transitive for n ∈ ℕ if for every linearly independent n-tuple x_{1}, …, x_{n} in X, for every n-tuple y_{1}, …, y_{n} in X, and for every ε > 0 there exists A ∈ S such that ∥Ax_{i} − y_{i}∥ < ε, i = 1, …, n. Motivated by the notion of quasi-localizing algebras, we introduce sesquitransitive sets of operators.

Definition 4.1. We say that a set S in L(X) is uniformly sesquitransitive if there exists a constant C > 0 such that for every linearly independent x and z in X, for every y ∈ X, and for every ε > 0 there exists A ∈ S such that ∥Ax − y∥ < ε and ∥Az∥ ⩽ C∥z∥. We say that S is sesquitransitive if for every non-zero z ∈ X there is a positive real C = C(z) such that for every x linearly independent of z, for every y ∈ X, and every ε > 0 there exists A ∈ S such that ∥Ax − y∥ < ε and ∥Az∥ ⩽ C∥z∥.

Clearly, the following implications hold.

2-transitivity ⇒ uniform sesquitransitivity ⇒ sesquitransitivity ⇒ transitivity.

Remark 4.2. It can be easily verified that a uniformly sesquitransitive algebra is quasi-localizing for any ball B not containing the origin and for every non-zero w in Definition 3.1. Indeed, suppose A is uniformly sesquitransitive with constant C. Let B be any ball centered at x_{0} of radius r with r > ∥x_{0}∥, and let w be any non-zero vector in B. We claim that A is quasi-localizing for this ball B and w with constant C̃ = C ∧ \frac{∥w∥+1}{∥x_{0}∥+r}. Indeed, given a sequence (x_{n}) in B, and a sequence (z_{n}) in X \ \{0\}. Fix n ∈ ℕ. If x_{n} and z_{n} are linearly independent, then we can find A_{n} ∈ A such that ∥A_{n}x_{n} − w∥ < \frac{1}{n} and ∥A_{n}z_{n}∥ ⩽ C∥z_{n}∥. On the other hand, if z_{n} = λx_{n} then transitivity of A implies that there is A_{n} ∈ A such that ∥A_{n}x_{n} − w∥ < \frac{1}{n}, so that

∥A_{n}z_{n}∥ = |λ|∥A_{n}x_{n}∥ ⩽ |λ|(∥w∥ + \frac{1}{n}) ⩽ |λ|(∥w∥ + 1)\frac{∥x_{n}∥}{∥x_{0}∥ + r} ⩽ C̃∥z_{n}∥.

It is known (see, e.g., [RT05]) that the commutant of a 2-transitive algebra is trivial. The following theorem extends this fact to sesquitransitive algebras. It can be viewed as a counterpart of Theorem 3.2.

Proposition 4.3. If X is a Banach space and A is a sesquitransitive subalgebra of L(X), then A′ is trivial.

Proof. Suppose that A is sesquitransitive, but there exists S ∈ A′ such that S is not a multiple of the identity. Then we can find a non-zero z ∈ X such that Sz is not a multiple of z. Put x = Sz. Let C = C(z) in the definition of sesquitransitivity. Choose y ∉ Range S such that ∥y∥ > C∥S∥∥z∥. Then sesquitransitivity of A implies that for every n ∈ ℕ there exists A_{n} ∈ A such that ∥A_{n}x − y∥ ⩽ \frac{1}{n} and ∥A_{n}z∥ ⩽ C∥z∥. It follows that A_{n}x → y, so that ∥A_{n}x∥ → ∥y∥. However,

∥A_{n}x∥ = ∥A_{n}Sz∥ = ∥SA_{n}z∥ ⩽ ∥S∥ · C∥z∥,

so that ∥y∥ ⩽ C∥S∥∥z∥; a contradiction. □

Next, we consider the algebraic version of sesquitransitivity. Recall that a set S of linear maps on a vector space is called strictly transitive if for every two non-zero vectors x and y there exists A ∈ S such that Ax = y. One says that S is strictly n-transitive for n ∈ ℕ if for every n linearly independent vectors x_{1}, …, x_{n}, for n vectors y_{1}, …, y_{n} there exists A ∈ S such that Ax_{i} = y_{i}, i = 1, …, n.

Definition 4.4. We will say that S is algebraically sesquitransitive if for any two non-zero linearly independent vectors x_{1} and x_{2} there exists a non-zero vector z such that for every non-zero y there exists A ∈ S such that Ax_{1} = y and Ax_{2} ≠ z.

It should be immediately clear that

strict 2-transitivity ⇒ algebraic sesquitransitivity ⇒ strict transitivity. Algebraic sesquitransitivity is similar to sesquitransitivity in the sense that we can send x_{1} to any prescribed destination, while keeping some control over the image of x_{2}. At the first glance it might seem that algebraic sesquitransitivity is just slightly stronger than strict transitivity. However, we will see that for rings it actually implies strict 2-transitivity (hence, we, in fact, have complete control over x_{2}).

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 11

Recall that a set S of operators is strictly dense if it is strictly n-transitive for every n ∈ ℕ. Jacobson’s Density Theorem asserts that every strictly 2-transitive ring of linear maps on a vector space over any field is strictly dense. The following is a generalization of Jacobson’s Density Theorem.

Theorem 4.5. Suppose that X is a vector space over an arbitrary field, and R is a sub-ring of L(X). If R is algebraically sesquitransitive then it is strictly dense in L(X).

Proof. It suffices to show that R is strictly 2-transitive, then the result would follows from Jacobson’s Density Theorem. Suppose that R is not strictly 2-transitive.

Using a standard argument we will show that there exist linearly independent vectors x_{1}, x_{2} ∈ X such that if Ax_{1} = 0 for some A ∈ R, then Ax_{2} = 0. Indeed, otherwise, for every two linearly independent vectors x_{1} and x_{2} we would find operators A, B ∈ R such that Ax_{1} = 0 and Ax_{2} ≠ 0, and Bx_{1} ≠ 0 and Bx_{2} = 0. Furthermore, since R is strictly transitive, for every y_{1} and y_{2} in X we would find C, D ∈ R such that C(Ax_{2}) = y_{2} and D(Bx_{1}) = y_{1}. Let S = CA + DB ∈ R, then Sx_{1} = y_{1} and Sx_{2} = y_{2}, so that R is strictly 2-transitive, contradiction.

By Definition 4.4 there exists z ∈ X such that for every non-zero y there exists A ∈ R such that Ax_{1} = y and Ax_{2} ≠ z.

Define a linear operator T ∈ L(X) by T (Ax_{1}) = Ax_{2} for every A ∈ R. It can be easily verified that T is well defined and commutes with every operator in R. Together with strict transitivity of R this yields that T is a bijection. Therefore, one can find a non-zero y ∈ X such that T y = z. Then there exists A ∈ R such that Ax_{1} = y and Ax_{2} ≠ z. However, Ax_{2} = T Ax_{1} = T y = z, contradiction. □

We would like to mention that strict n-semitransitivity introduced in [RT] is another generalization of strict n-transitivity. In a result similar to Theorem 4.5, [RT] shows that every strictly 2-semitransitive ring is strictly dense.

5. Transitive and sesquitransitive algebras in real spaces

It is easy to see that if dim X < ∞ then a subalgebra of L(X) is transitive iff it is strictly transitive. Recall that a classical theorem of Burnside asserts that M_{n}(ℂ) has no proper transitive subalgebras (see, e.g., [RR00]). Clearly, this is false in the real case: the algebra generated by the rotation through π/2 in L(ℝ^{2}) is transitive but proper. In this section we establish several analogues of the Burnside Theorem as well as of [Lom73] for algebras on real Banach spaces.

12 V. I. LOMONOSOV, H. RADJAVI, AND V. G. TROITSKY

Recall that a unital algebra is a division algebra if every non-zero element in it is invertible. It was proved by Rickart [Ric60, Theorem 1.7.6.] that every real normed division algebra is algebraically isomorphic to either ℝ, ℂ, or ℍ. Here ℍ stands for the quaternion algebra.

Suppose that A is a transitive subalgebra of M_{n}(ℝ). It follows from Schur’s Lemma and from Wedderburn-Artin Theorem that A′ is algebraically isomorphic to either ℝ, ℂ, or ℍ. Furthermore, A′′ = A, and A is (algebraically isomorphic to) M_{n}(ℝ), M\frac{n}{2} (ℂ), or M\frac{n}{4} (ℍ), respectively.

Next, we consider some consequences of these facts for infinite dimensional real Banach spaces. In particular, we consider a version of [Lom73] for algebras of operators. Namely, it follows from [Lom73] that if X is a complex Banach space then every transitive subalgebra of L(X) containing a compact operator is WOT-dense in L(X), see [RR03, Theorem 8.23]. This statement fails in real Banach spaces. We will prove a version of this statement for transitive algebras in real Banach spaces.

Proposition 5.1. Suppose that A is a transitive algebra of operators on a real Banach space X, and A′ has a finite-dimensional invariant subspace. Then A′ is algebraically isomorphic to ℝ, ℂ, or ℍ.

Proof. Suppose that M is a finite-dimensional subspace invariant under A′. Consider the restriction map Φ: A′ → L(M) given by Φ(T) = T_{|M}. For every non-zero T ∈ A′, transitivity of A implies that ker T and, therefore, ker T_{|M} are trivial. It follows that Φ is one-to-one. Moreover, Φ(T) is invertible. Hence, Φ(A′) is a division algebra. Now Rickart’s Theorem implies that Φ(A′) and, therefore, A′ is isomorphic to ℝ, ℂ, or ℍ. □

Theorem 5.2. Suppose that X is a real Banach space and A is a transitive subalgebra of L(X) containing a compact operator. Then A′ is algebraically isomorphic to either ℝ, ℂ, or ℍ.

Proof. By Lemma 2.2, A contains a compact operator K with eigenvalue 1. Then the corresponding eigenspace M := ker(K − I) is a finite-dimensional subspace invariant under A′. Now the conclusion follows from Proposition 5.1. □

Proposition 5.3. Suppose that V is a finite-dimensional vector space over ℝ, A is a transitive subalgebra of L(V), and T ∈ A′ such that T is not a multiple of the identity. Then {T }′ is least possible, that is, {T }′ is the algebra generated by T and A.

SESQUITRANSITIVE AND LOCALIZING OPERATOR ALGEBRAS 13

Proof. We reduce the real case to the complex case as in [Sir05]. It follows from Proposition 5.1 that by replacing T with λI + μT for some μ, λ ∈ ℝ we can assume that T^{2} = −I. Define a complex structure on V by putting ix = T x for x ∈ V. One can easily check that with this complex scalar multiplication V becomes a vector space over ℂ; denote it by V_{ℂ}. Observe that V and V_{ℂ} coincide as sets. Note that an operator S ∈ L(V) belongs to L(V_{ℂ}) iff it is complex-linear, or, equivalently, if ST = T S. Hence, L(V_{ℂ}) = {T }′. It follows from T ∈ A′ that we can view A as a subset of L(V_{ℂ}). Note that A is still transitive (as the definition of transitivity doesn’t involve scalar multiplication). However, A need not be closed under complex multiplication in L(V_{ℂ}). Let Ã = {A + iB | A, B ∈ A} in L(V_{ℂ}). Then Ã is a transitive subalgebra of L(V_{ℂ}). Hence, by the Burnside Theorem, A is all of L(V_{ℂ}), so that {T }′ = Ã = {A + T B | A, B ∈ A}. □

Finally, we will prove that the Burnside Theorem and the algebraic version of [Lom73] remain valid over real scalars if transitivity is replaced with sesquitransitivity.

Proposition 5.4. Suppose that A is a subalgebra of M_{n}(ℝ). If A is sesquitransitive or algebraically sesquitransitive then A = M_{n}(ℝ).

Proof. If A is algebraically sesquitransitive then it is strictly n-transitive by Theorem 4.5, hence A = M_{n}(ℝ). Now, if A is sesquitransitive then A′ is trivial by Proposition 4.3, so that A = A′′ = M_{n}(ℝ). □

The following is well known for complex Banach spaces.

Lemma 5.5. Suppose that K is a compact operator on a real Banach space X such that K has a non-zero fixed vector. Then the uniformly closed subalgebra of L(X) generated by K contains an idempotent of finite rank.

Proof. It follows immediately from the hypotheses that 1 is an eigenvalue of K. Since K is compact, so is its complexification K_{c} on X_{c}. Let Z be the spectral subspace of K corresponding to {1}. It follows from σ(K_{c}_{|Z}) = {1} that K_{c}_{|Z} is invertible. Since K_{c}_{|Z} is compact, it follows that Z is finite-dimensional. Using the usual Functional Calculus, we can find the canonical spectral projection onto Z. Recall that we can write this projection as f(K), where f is the characteristic function of an open subset U of ℂ such that U ∩ σ(K) = {1}.

Let A be the uniformly closed algebra generated by K_{c} in L(X_{c}). It follows from Theorem 5.4(a) of [Con90] that σ_{A}(K) = σ(K), so that f(K) ∈ A. It is left to show that f(K) is actually a real operator, that is, that f(K) leaves X invariant.

14 V. I. LOMONOSOV, H. RADJAVI, AND V. G. TROITSKY

Again, by Function Calculus we can write f(K) = \frac{1}{2πi} ∫_{Γ} R(λ; K) dλ, where R(λ; K) is the resolvent of K at λ and the integration is done over a circle centered at 1 and contained in U. Observe that for x ∈ X we have the following relation for the complex conjugates in X_{c}: R(\overline{λ}; K)x = \overline{R(λ; K)x}. Indeed, direct verification shows that if x = (λ − K)(y + iz) then x = (\overline{λ} − K)(y − iz) for y, z ∈ X. It follows that f(K)x = \frac{1}{2πi} ∫_{Γ} R(λ; K)x dλ belongs to X. □

Theorem 5.6. If X is a real Banach space then every sesquitransitive subalgebra A of L(X) containing a compact operator is WOT-dense in L(X).

Proof. Without loss of generality we assume that A is uniformly closed. Lemma 2.2 yields that there is a compact operator K in A such that K has a non-zero fixed vector. It follows from Lemma 5.5 that A contains an idempotent operator P of finite rank. Let Y = Range P, then dim Y < ∞.

We will show that the restriction algebra P AP is still sesquitransitive on Y. Indeed, let z ∈ Y, then there exists C such that for all x, y in X such that x and z are linearly independent and for every ε > 0 there is A ∈ A such that ∥Ax − y∥ < ε and ∥Az∥ ⩽ C∥z∥. In particular, when x, y ∈ Y we have

∥P AP x − y∥ = ∥P (Ax − y)∥ ⩽ ∥P ∥ε

and ∥P AP z∥ ⩽ ∥P ∥ · C∥z∥.

Proposition 5.4 implies that the restriction of P AP is all of L(Y). It follows that P AP and, therefore, A contains an operator of rank one. Now a standard argument (see, e.g., the proof of Lemma 7.4.5 in [RR00]) shows that A contains all operators of finite rank, hence is WOT-dense in L(X). □

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Department of Mathematical Sciences, Kent State University, Kent, OH, 44242. USA.

  E-mail address: lomonoso@math.kent.edu

Department of Pure Mathematics, University of Waterloo, Waterloo, ON, N2L 3G1. Canada.

  E-mail address: hradjavi@math.uwaterloo.ca

Department of Mathematical and Statistical Sciences, University of Alberta, Edmonton, AB, T6G 2G1. Canada.

  E-mail address: vtroitsky@math.ualberta.ca