ON SEMITRANSITIVE COLLECTIONS OF OPERATORS

J. BERNIK, L. GRUNENFELDER, M. MASTNAK, H. RADJAVI, V.G. TROITSKY

Abstract. A collection F of operators on a vector space V is said to be semitransitive if for every pair of nonzero vectors x and y in V there exists a member T of F such that either T x = y or T y = x (or both). We study semitransitive algebras and semigroups of operators. One of the main results is that if the underlying field is algebraically closed, then every semitransitive algebra of operators on a space of dimension n contains a nilpotent element of index n. Among other results on semitransitive semigroups, we show that if the rank of nonzero members of such a semigroup acting on an n-dimensional space is a constant k, then k divides n.

1. Introduction

As is well known, a collection F of linear transformations on a vector space V is said to be transitive if for every pair of nonzero vectors x, y ∈ V there exists a transformation T ∈ F such that T x = y.

We shall consider a weaker version of this property, which was recently introduced in [6] and proved not only interesting in its own right, but also fruitful in connection with the properties of lattices of invariant subspaces of algebras of operators. Although our primary interest in this paper lies in algebras and semigroups, we state the definition for an arbitrary collection. We say that F is semitransitive if for every pair of nonzero vectors x, y ∈ V there exists T ∈ F such that either T x = y or T y = x. Note, that for groups the notions of transitivity and semitransitivity coincide.

It should be noted that if V is a topological vector space, then these two properties are sometimes referred to as strict transitivity and strict semitransitivity respectively, while transitivity and semitransitivity are understood in the topological sense, that is, for every pair of nonzero vectors x, y ∈ V and for every neighborhood U of 0 there exists a transformation T ∈ F such that T x ∈ y + U (or T y ∈ x + U, for semitransitivity). We will refer to the latter properties as topological transitivity and topological semitransitivity.

Semigroup forum, 70(3), 2005, 436–450.

1 2 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

Date: October 30, 2004.

2000 Mathematics Subject Classification. 15A30, 47D03, 47L10.

Key words and phrases. Semitransitive operator algebra, semitransitive operator semigroup.

Semitransitive algebras of bounded operators on a Banach space were introduced and investigated in [6]. There it is observed that a unital algebra of bounded operators on a Banach space is semitransitive if and only if its invariant linear subspaces are totally ordered by inclusion, and it is topologically semitransitive if and only if all its closed invariant subspaces are totally ordered by inclusion. This implies that in the finite-dimensional case the terms topological semitransitivity and semitransitivity are equivalent for unital algebras of real or complex matrices.

If S is a uniformly bounded semitransitive semigroup of operators on a Banach space B, then S has no invariant subspaces. Indeed, suppose that ∥T ∥ < M for all T ∈ S. Consider a subspace Z of B and choose a vector x ∈ Z with ∥x∥ > M and a vector y ∉ Z with ∥y∥ = 1. Since no operator in S can map y to x, there must exist an operator T ∈ S that maps x to y and hence Z is not invariant under T.

In this paper we study semitransitive algebras and semigroups of matrices, or equivalently, operators on a finite-dimensional space. It turns out that even the case of dimension one is not totally trivial.

2. Semitransitivity and preorders for abelian groups

We start with some algebraic considerations. For this section, let G be an abelian group, and S a subsemigroup of G. We say that G is preordered (ordered) if it is equipped with a reflexive, transitive (and antisymmetric for ordered) binary relation “⪯” such that x ⪯ y implies xz ⪯ yz for all x, y, z ∈ G. We write G_{+} for the positive cone {x ∈ G | x ⪰ e}. Note that G_{+} is a subsemigroup of G. Clearly, a preorder is total if and only if G_{+} ∪ G_{+}^{−1} = G. A preorder is an order if and only if G_{+} ∩ G_{+}^{−1} = {e}.

We say that S is semitransitive, or acts semitransitively on G, if for any x, y ∈ G there exists s ∈ S such that sx = y or sy = x. One can easily verify the following characterization.

Proposition 1. The following are equivalent.

(i) S is semitransitive; (ii) S^{−1} is semitransitive; (iii) S ∪ S^{−1} = G; (iv) S = G_{+} for some total preorder on G.

Proposition 2. Every semitransitive semigroup S in G contains the torsion group t(G) of G.

Proof. Let a ∈ t(G), then a^{m} = e for some m. By Proposition 1(iii) either a ∈ S or a ∈ S^{−1}. In the latter case, a^{−1} ∈ S, so that a = (a^{−1})^{m−1} ∈ S. □

We say that S is minimal semitransitive if it is semitransitive, and no proper subsemigroup of S acts semitransitively on G.

Let Ĝ = G/ t(G), then Ĝ is a torsion-free group. Let π: G → Ĝ be the canonical epimorphism.

Theorem 3. The following are equivalent:

(i) S is minimal semitransitive; (ii) S ∪ S^{−1} = G and S ∩ S^{−1} = t(G); (iii) S = π^{−1}(Ĝ_{+}) for some total order on Ĝ.

Furthermore, every semitransitive subsemigroup of G contains a minimal semitransitive subsemigroup.

Proof. The implication (ii)⇒(i) follows from Propositions 1 and 2. To show that (iii)⇒(ii), let S = π^{−1}(Ĝ_{+}) for some total order on Ĝ, then

\begin{aligned}S ∪ S^{−1} & = π^{−1}(Ĝ_{+} ∪ Ĝ_{+}^{−1}) = π^{−1}(Ĝ) = G, and \\ S ∩ S^{−1} & = π^{−1}(Ĝ_{+} ∩ Ĝ_{+}^{−1}) = π^{−1}(\{e\}) = t(G). \end{aligned} ; (1)

Suppose S is a semitransitive subsemigroup of G. Consider the group G = S ∩ S^{−1}. Propositions 1 and 2 yield t(G) ⊆ G. Let B = G/ t(G), then B is a torsion-free subgroup of Ĝ. It is well known (see [2] or [3, p.5]) that every torsion-free group admits a total order. Let C = π^{−1}(B_{+}) for some total order on B. Note that C ⊆ G is a semigroup. Also, as in (1), C ∪ C^{−1} = G, and C ∩ C^{−1} = t(G). Put S_{0} = C ∪ (S \ G). Then S_{0} ⊆ S. Show that S_{0} is a semigroup: if x, y ∈ S_{0} then xy ∈ S_{0}. Suppose not, since xy ∈ S then xy ∈ G ⊆ S^{−1} so that x ∈ y^{−1}S^{−1} ⊆ S^{−1}. Hence, x, y ∈ S^{−1}, so that x, y ∈ G and, therefore, x, y ∈ C. It follows that xy ∈ C ⊆ S_{0}. Thus, S_{0} is a semigroup. Furthermore,

S_{0} ∪ S_{0}^{−1} = C ∪ (S \ G) ∪ C^{−1} ∪ (S \ G)^{−1} = (C ∪ C^{−1}) ∪ ((S ∪ S^{−1}) \ G) = G ∪ (G \ G) = G, and

S_{0} ∩ S_{0}^{−1} = (C ∪ (S \ G)) ∩ (C^{−1} ∪ (S \ G)^{−1})

= (C ∩ C^{−1}) ∪ (C ∩ (S \ G)^{−1}) ∪ ((S \ G) ∩ C^{−1}) ∪ ((S \ G) ∩ (S \ G)^{−1})

= t(G) ∪ ∅ ∪ ∅ ∪ ∅ = t(G),

4 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

so that S_{0} is minimal semitransitive. Hence, every semitransitive semigroup in G contains a minimal one.

Finally, show that (i)⇒(iii). Suppose that S is minimal semitransitive and construct S_{0} as in the previous paragraph. The minimality of S implies that S_{0} = S, hence G = C, and, therefore,

G = G ∩ G^{−1} = C ∩ C^{−1} = t(G).

Let K = π(S), then it is easy to verify that S = π^{−1}(K). Observe that K is a semigroup, K ∪ K^{−1} = Ĝ, and K ∩ K^{−1} = {e}. Then K is the positive cone of a total order on Ĝ as claimed. □

Corollary 4. If S is minimal semitransitive then T = S \ t(G) is a semigroup ideal in S.

Proof. Suppose that x ∈ S \ t(G), y ∈ S. Theorem 3(ii) implies that x ∉ S^{−1}. But xy ∈ t(G) would imply x = (xy)y^{−1} ∈ S^{−1}, a contradiction. □

3. Examples: one-dimensional case

Consider a semigroup of operators acting on a vector space V. In case dim V = 1, the field 𝔽 acts on itself by multiplication. Then 𝔽^{•} = 𝔽 \ {0} is an abelian group, and it follows from Theorem 3 that every semitransitive semigroup contains a minimal such semigroup and that minimal semitransitive semigroups are the preimages of positive cones of total preorders on 𝔽̂^{•}.

We also study those semitransitive semigroups of 𝔽 ∈ {ℝ, ℂ} that are either bounded or compact.

3.1. Real line. The semigroups [−1, 1] and [−1, 1]\{0} are the only bounded semitransitive semigroups in ℝ.

3.2. Complex plane. It is easy to see that the closed unit disk D = {|z| ⩽ 1} is the only compact semitransitive semigroup in ℂ and that any bounded semitransitive subsemigroup of ℂ is contained in D.

Proposition 5. A bounded semigroup S in ℂ^{•} is semitransitive if and only if it is the union of the punctured open disk {0 < |z| < 1} and a semigroup S′ ⊆ ∂D, which acts semitransitively on the circle ∂D = {|z| = 1}.

Proof. It is clear that any z outside D can be mapped to any nonzero element in D by multiplication with a nonzero number in the interior of D. It is also easy to see,

ON SEMITRANSITIVE COLLECTIONS OF OPERATORS 5

that the entire punctured interior of D is needed for that purpose. Hence S ⊆ D is semitransitive if and only if it contains {0 < |z| < 1} and S ∩ ∂D acts semitransitively on ∂D. □

Remark 6. It follows from Theorem 3 that a semigroup S′ ⊆ ∂D acts semitransitively on ∂D if an only if it contains the preimage of the positive cone of some total order on ˆ∂D.

Theorem 3 and Proposition 5 also yield the following.

Corollary 7. There exists a bounded minimal semitransitive semigroup S in ℂ.

4. Examples: n-dimensional case

In this section we are going to consider some examples of semitransitive algebras and semigroups of matrices. Let M_{n} = M_{n}(𝔽). For the following examples let S_{𝔽} = S′_{𝔽}∪{0}, where S′_{𝔽} is a fixed minimal semitransitive semigroup in 𝔽^{•}.

4.1. Upper-triangular Toeplitz algebra and semigroup. Let A be the algebra of all upper-triangular Toeplitz operators. More precisely, A is the algebra generated by the shift S, that is, the algebra of all the operators of the form a_{1}I + a_{2}S + a_{3}S^{2} + ⋯ + a_{n}S^{n−1} or, in matrix form,

(\begin{matrix}a_{1} & & a_{2} & a_{3} & & … & & & a_{n} \\ 0 & & a_{1} & a_{2} & & … & & & a_{n−1} \\ . & . & . & . & . & . & . & . & . & . & . & . & . & . & . \\ 0 & & 0 & 0 & & … & & & a_{1} \end{matrix})

where a_{1}, …, a_{n} are arbitrary scalars. Let S be the set of all matrices in A with a_{1} ∈ S_{𝔽}. We claim that A is a semitransitive algebra and S is a semitransitive semigroup.

The following shows that S is semitransitive. We claim, that given x, y ∈ 𝔽^{n}, there is a T ∈ S such that either T x = y or T y = x. Let x_{k} and y_{m} be the last non-zero components of x and y respectively. Without loss of generality we may assume that k ⩾ m and that \frac{y_{k}}{x_{k}} ∈ S_{𝔽}. Finding T ∈ A satisfying T x = y is equivalent to solving the system

(2)  { a_{1}x_{1} + a_{2}x_{2} + … + a_{k−1}x_{k−1} + a_{k}x_{k} = y_{1} ; a_{1}x_{2} + a_{2}x_{3} + … + a_{k−1}x_{k}  = y_{2} ; ⋮ ; a_{1}x_{k}  = y_{k}

for a_{1}, a_{2}, …, a_{k}.

6 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

The system (2) is consistent, so that there exists T ∈ A such that T x = y. Since \frac{y_{k}}{x_{k}} ∈ S_{𝔽}, it follows from the last equation of (2) that a_{1} ∈ S_{𝔽}, hence T ∈ S.

Any semitransitive subset of A must contain the identity matrix (I is the only element of A that maps e_{n} to itself) and also any matrix of the form a_{2}S +…+a_{n}S^{n−1} (we denote the semigroup of all the matrices of this form by R), since such a matrix is the only matrix in A mapping e_{n} to x = a_{n}e_{1} + … + a_{2}e_{n−1} and no matrix in A can map x to e_{n}. Therefore A is a minimal semitransitive algebra and, as we shall see later, is unique up to conjugation if the field 𝔽 is algebraically closed.

We will show that S contains a minimal semitransitive semigroup. To make notation less cumbersome we identify the field 𝔽 with the scalars in M_{n}. As above, let R denote the semigroup of all the matrices of the form a_{2}S + … + a_{n}S^{n−1}. Observe that R coincides with the set of matrices in S of determinant zero. Thus, S is a disjoint union of R and the semigroup of the form S′_{𝔽} + R and R is a semigroup ideal in S. As we noted, any minimal subsemigroup of S will contain R. So we study the semigroup S′_{𝔽} + R. It is easy to see that any semitransitive subsemigroup of S must also contain all matrices of the form a_{1}I +R, where a_{1} ∈ S′_{𝔽} is not a root of unity. Now, S′_{𝔽}+R is a disjoint union of t(𝔽^{•})+R, which is a group, and (S′_{𝔽}\ t(𝔽^{•}))+R, which is a semigroup ideal in S′_{𝔽} + R by Corollary 4. So, finally, we consider the abelian group t(𝔽^{•}) + R. Observe that it acts transitively and freely on the set of vectors with x_{n} ∈ t(𝔽^{•}). So by Theorem 3 there exists a minimal semitransitive semigroup for this action, which we denote by P. It is easy to see that P ∪ ((S′_{𝔽}\ t(𝔽^{•})) + R) ∪ R is minimal semitransitive. Now, the structure of this semigroup depends on the characteristic of the field 𝔽. If char(𝔽) = p > 0, then every element of t(𝔽^{•}) + R is torsion, so P = t(𝔽^{•}) + R in this case. If, in addition, 𝔽 is algebraic over the prime field, then 𝔽^{•} = t(𝔽^{•}) and S was minimal to begin with.

If char(𝔽) = 0, then t(𝔽^{•}) + R, as an abelian group, is a product of its torsion part t(𝔽^{•}) and the group I +R, where the latter is torsion free. So P, and thus any minimal subsemigroup in S, corresponds to a certain total order on I + R by Theorem 3.

4.2. Upper triangular one-column matrices. Consider the set

S = {xe_{i}^{T} | 1 ⩽ i ⩽ n, x ∈ 𝔽^{n}, x_{i} ∈ S_{𝔽} and x_{j} = 0 for j > i}

in M_{n}(𝔽). It can be easily verified that S is a semigroup. We claim that it is minimal semitransitive.

First we show that S is semitransitive. Let x and y be arbitrary nonzero vectors in 𝔽^{n}. Define i and j to be the smallest integers such that x_{k} = 0 = y_{l} for all k > i

ON SEMITRANSITIVE COLLECTIONS OF OPERATORS 7

and l > j. Without any loss of generality we can assume that i ⩾ j and that \frac{y_{i}}{x_{i}} ∈ S_{𝔽}. Now note that \frac{1}{x_{i}} ye_{i}^{T} ∈ S maps x to y.

Now suppose that S′ ⊆ S is a semitransitive semigroup. Choose A = xe_{i}^{T} ∈ S. We will show that A ∈ S′, thus proving minimality of S. We consider two cases.

Case: x_{i} is not a root of unity. Then A is the only element of S that maps e_{i} to x and there is no element in S that would map x to e_{i}. Hence A ∈ S′.

Case: x_{i}^{k} = 1 for some positive integer k. Then A is the only element of S that maps x to x_{i}x and \frac{1}{x_{i}^{2}} A is the only element of S that maps x_{i}x to x. Hence A ∈ S′, since (\frac{1}{x_{i}^{2}} A)^{k−1} = A.

4.3. One-column contractions on ℓ_{∞}^{n}. Consider the set

S = {xe_{i}^{T} | 1 ⩽ i ⩽ n, ∥x∥_{∞} ⩽ 1, |x_{j}| < 1 for all j > i, and x_{i} ∈ S_{𝔽}}. Here we assume that 𝔽 ∈ {ℝ, ℂ} and that S_{𝔽} is bounded. Clearly, S is a subsemigroup of contractions on ℓ_{∞}^{n}. We claim that S is a minimal semitransitive semigroup. First, show that S is semitransitive. Fix x, y ∈ 𝔽^{n}.

Case ∥x∥_{∞} > ∥y∥_{∞}. If k is such that ∥x∥_{∞} = |x_{k}| then the operator T = \frac{1}{x_{k}} ye_{k}^{T} ∈ S, since ∥\frac{1}{x_{k}} y∥_{∞} < 1, and maps x to y.

Case ∥x∥_{∞} = ∥y∥_{∞}. Let k and l be largest integers, such that ∥x∥_{∞} = |x_{k}| and ∥y∥_{∞} = |y_{l}|. Without any loss of generality assume that k ⩾ l and that \frac{y_{k}}{x_{k}} ∈ S_{𝔽} and note that the matrix T = \frac{1}{x_{k}} ye_{k}^{T} ∈ S maps x to y.

Now assume that S′ is a semitransitive subsemigroup of S. We show that any operator T = xe_{j}^{T} ∈ S, must also belong to S′. Note that T is the only matrix in S mapping e_{j} to x.

If x_{j} is not a root of unity, then we show that no matrix (ye_{i}^{T}) ∈ S maps x to e_{j} and therefore T ∈ S′. Indeed, if (ye_{i}^{T})x = e_{j} for some ye_{i}^{T} ∈ S, then y = \frac{1}{x_{i}} e_{j}, thus |x_{i}| = |y_{j}| = 1 (since |x_{i}| ⩽ 1 and |y_{j}| = \frac{1}{|x_{i}|} ⩽ 1) and hence also i = j. But then y_{j} ∉ S_{𝔽} (the only elements of S_{𝔽} whose inverses are in S_{𝔽} as well, are roots of unity), contradicting ye_{j}^{T} ∈ S.

If x_{j} is a root of unity, say x_{j}^{k} = 1, then the only element of S mapping x to x_{j}x is T. Indeed, if (ye_{i}^{T})x = x_{j}x, then x_{i}y = x_{j}x, hence |x_{i}| = 1 = |y_{j}|, thus i = j and therefore ye_{i}^{T} = T. Also, \frac{1}{x_{j}^{2}} T is the unique matrix in S mapping x_{j}x to x. Hence T ∈ S′, since (\frac{1}{x_{j}^{2}} T)^{k−1} = T.

4.4. Rank-one Euclidean contractions. Let S be the set of all rank-one complex matrices with Euclidean norm at most one together with the zero matrix. This is clearly a closed semigroup. It is easy to see that it is semitransitive. Indeed, given

8 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

x, y ∈ ℂ^{n}, without loss of generality ∥x∥_{2} ⩾ ∥y∥_{2}. Let T = \frac{1}{∥x∥_{2}^{2}} yx^{∗}. Then T is clearly a contraction, and T x = y. It will follow from Corollary 21 that S contains a minimal closed semitransitive subsemigroup.

We claim that S also contains a minimal semitransitive subsemigroup T . Note that any semitransitive semigroup contained in S must necessarily contain all self-adjoint rank-one projections. Indeed, if ∥x∥ = 1 then xx^{∗} is the only rank-one matrix that takes x to x. As a consequence, it must contain all rank-one proper contractions (operators of norm strictly less than one) since every rank-one proper contractions is a product of rank-one projections; see [4]. Thus, we only have to care about the normone rank-one elements. These are uniquely determined by pairs of norm-one vectors: if ∥x∥ = 1 = ∥y∥, then yx^{∗} is the only element of S that maps x to y.

For ∥x∥ = ∥y∥ = 1 we write x ∼ y if λx = y for some |λ| = 1, and [x] = {z | ∥z∥ = 1 and z ∼ x}. Fix a total order on the set of all the equivalence classes of norm-one vectors. Let L be a set of norm one vectors such that L contains exactly one vector from each equivalence class. As before, let S_{ℂ} denotes a fixed bounded minimal semitransitive subsemigroup of the complex plane. Put T = T_{1} ∪ T_{2} ∪ T_{3}, where

T_{1} = all rank-one proper contractions,
T_{2} = { λxx^{∗} | λ ∈ S_{ℂ}, x ∈ L }, and
T_{3} = { yx^{∗} | ∥x∥ = ∥y∥ = 1 and [y] < [x] }.

We claim that T is a minimal semitransitive semigroup. First, show that T is a semigroup. Let S, T ∈ T . Clearly, if either T or S belongs to T_{1}, then the product is also in T_{1}. Suppose that S, T ∈ T_{2} ∪ T_{3}. Then S = yx^{∗} and T = vu^{∗} where ∥x∥ = ∥y∥ = ∥u∥ = ∥v∥ = 1, [y] ⩽ [x], and [v] ⩽ [u]. Note that T attains its norm only on the elements of [u], so that ∥ST ∥ = 1 only when ∥ST u∥ = 1, but ∥ST u∥ = ∥yx^{∗}vu^{∗}u∥ ⩽ ∥x^{∗}v∥. It follows that if [x] ≠ [v] then ST ∈ T_{1}. On the other hand, if [x] = [v], then v = λx for some |λ| = 1, and then ST = yx^{∗} · λxu^{∗} = λyu^{∗} ∈ T_{2} ∪ T_{3} because [y] ⩽ [x] = [v] ⩽ [u].

In order to show that T is semitransitive, fix nonzero x and y, and observe that if ∥x∥ > ∥y∥ then T x = y where T = \frac{1}{∥x∥_{2}^{2}} yx^{∗} ∈ T_{1}. Suppose that ∥x∥ = ∥y∥, we can assume that ∥x∥ = ∥y∥ = 1. If x ∼ y then there is T ∈ T_{2} such that T x = y or T y = x. Finally, if x ≁ y then there is T ∈ T_{3} such that T x = y when [y] < [x] or T y = x when [x] < [y].

Finally, show that T is minimal. Suppose T_{0} is a semitransitive subsemigroup of T . Then T_{0} contains all self-adjoint rank-one projections, hence T_{1} ⊂ T_{0}. Since the action

ON SEMITRANSITIVE COLLECTIONS OF OPERATORS 9

of T_{0} on [x] is semitransitive for every x with ∥x∥ = 1, it follows that T_{2} ⊂ T_{0}. Finally, if [y] < [x] for some ∥x∥ = ∥y∥ = 1 then yx^{∗} is the only element of T that maps x to y, while no element of T maps y to x, so that yx^{∗} ∈ T_{0}. Therefore, T_{3} ⊂ T_{0}. It follows that T_{0} = T .

4.5. Multiples of unitaries. Let S be the set of all matrices of the form λU, where λ ∈ [0, 1] and U is a unitary matrix. This is clearly a closed semitransitive semigroup. As usual, let SU_{n} be the group of all special unitary matrices A in M_{n}(ℂ), i.e., A^{∗}A = I and det A = 1.

Proposition 8. The group SU_{2} is a minimal semigroup acting semitransitively on norm-one vectors in ℂ^{2}.

Proof. Since the columns of every matrix in SU_{2} form an orthonormal basis, it follows that SU_{2} consists of all the matrices of the form ( \begin{matrix}x & −ȳ \\ y & x̄ \end{matrix} ) with |x|^{2} + |y|^{2} = 1. A direct calculation shows that whenever ∥u∥ = ∥v∥ = 1 in ℂ^{2} then there exists a unique matrix A ∈ SU_{2} such that Au = v. It follows that SU_{2} acts transitively on the norm one vectors.

Let S ⊆ SU_{2} be a semigroup acting semitransitively on the unit sphere in ℂ^{2}. Then S contains all the elements of SU_{2} of finite order. Indeed, suppose that A ∈ SU_{2} and A^{n} = I. Fix any u with ∥u∥ = 1, then A is the only matrix in SU_{2} that takes u to Au, while A^{−1} is the only matrix in SU_{2} that takes Au to u. Hence, either A ∈ S or A^{−1} ∈ S. In the later case, S ∋ (A^{−1})^{n−1} = A.

Since ( \begin{matrix}0 & −ᾱ \\ α & 0 \end{matrix} ) is of order 4 whenever |α| = 1, it belongs to S. In particular, ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) ∈ S. Since

(\begin{matrix}α & 0 \\ 0 & ᾱ \end{matrix}) = (\begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}) · (\begin{matrix}0 & −ᾱ \\ α & 0 \end{matrix}) ,

it follows that ( \begin{matrix}α & 0 \\ 0 & ᾱ \end{matrix}) ∈ S whenever |α| = 1. Since every conjugate of ( \begin{matrix}α & 0 \\ 0 & ᾱ \end{matrix}) ∈ S can still be written as a product of matrices of order 4, it also has to be in S.

It is left to show that every matrix in SU_{2} is a conjugate of ( \begin{matrix}α & 0 \\ 0 & ᾱ \end{matrix}) for some |α| = 1. Indeed, every matrix A in SU_{2} has an orthonormal basis of eigenvectors. Denote the first eigenvector by ( \begin{matrix}x \\ y \end{matrix}), then, without loss of generality, we can assume that ( −ȳ ; x̄) is the second eigenvector. Let α and β be the eigenvalues of A. Since A is unitary, then |α| = |β| = 1. Furthermore, αβ = det A = 1, so that β = ᾱ. It follows that

A = (\begin{matrix}x & −ȳ \\ y & x̄ \end{matrix}) (\begin{matrix}α & 0 \\ 0 & ᾱ \end{matrix}) (\begin{matrix}x & −ȳ \\ y & x̄ \end{matrix})^{−1} .

□

Corollary 9. The semigroup (0, 1]SU_{2} is minimal semitransitive.

5. Main results

Theorem 10. If 𝔽 is an algebraically closed field, then every semitransitive subalgebra of M_{n}(𝔽) contains a nilpotent element of index n (i.e., T^{n} = 0, but T^{n−1} ≠ 0).

Proof. Let A be a semitransitive subalgebra of M_{n}(𝔽). It is easy to see that the invariant subspaces of A are totally ordered. First, we consider the case when A is unital. Furthermore, we can also assume that A has the block upper triangular form given in [5, p. 13]. Denoting the block form of A ∈ A by (A_{ij})_{i,j=1}^{k}, observe that the Jacobson radical R of A consists precisely of those R for which all diagonal blocks R_{ii} are zero. We shall show that for fixed i, 1 ⩽ i ⩽ k − 1, the set {R_{i,i+1} | R ∈ R} is the full space of matrices of the given size. First of all, if this set were just {0}, then the 2 × 2 algebra

L = \{(\begin{matrix}A_{ii} & A_{i,i+1} \\ 0 & A_{i+1,i+1} \end{matrix})| A ∈ A\}

would be semisimple; we can assume after a similarity that A_{i,i+1} = 0 for all A ∈ A. This is a contradiction to the total order of the invariant subspaces. Indeed, if the invariant subspaces of A given by the above triangularization are denoted by

{0} ⊂ M_{1} ⊂ … ⊂ M_{k} = 𝔽^{n},

then the identity A_{i,i+1} = 0 would give an invariant subspace

M_{i−1} ⊕ (M_{i+1} ⊖ M_{i}),

not comparable to M_{i}.

Now fix an R with R_{i,i+1} ≠ 0, and note that the (i, i + 1)-block of AR + RB, with A, B ∈ A is

A_{ii}R_{i,i+1} + R_{i,i+1}B_{i+1,i+1}.

Since A and B here can be chosen in A independently of each other, and since {A_{jj} | A ∈ A} is the full matrix algebra of its size for each j, we conclude that the (i, i + 1) block can be arbitrarily chosen in R as desired.

By construction, since 𝔽 is algebraically closed, A has a member N whose diagonal is N_{1} ⊕ … ⊕ N_{k}, where each N_{i} is the m_{i} × m_{i} Jordan cell

(\begin{matrix}0 & 1 \\ & 0 & 1 \\ & & ... & ... \\ & & & 0 & 1 \\ & & & & 0 \end{matrix}) ,

ON SEMITRANSITIVE COLLECTIONS OF OPERATORS 11

with m_{i} the size of the i^{th} block. If k = 1, then there is nothing to prove. Otherwise, we view each member of A as an operator expressed in the ordered basis

{e_{1,1}, …, e_{1,m_{1}}; e_{2,1}, …, e_{2,m_{2}}; …; e_{k,1}, …, e_{k,m_{k}}}

of the n-dimensional column space. By the preceeding paragraph, for each j ⩽ k − 1, R has a member R_{j} with

R_{j}e_{j+1,1} = e_{j,m_{j}}.

One can inductively choose the scalars α_{1}, …, α_{k−1} such that the k − 1 entries of α_{1}R_{1} + … + α_{k−1}R_{k−1} at (m_{1} + … + m_{j}, m_{1} + … + m_{j} + 1) are all nonzero. Put

A = N + α_{1}R_{1} + … + α_{k−1}R_{k−1},

then all the (i, i + 1)-entries of the nilpotent matrix A are nonzero, so that we have A^{n−1} ≠ 0 and A^{n} = 0 as desired.

In the case when A is not unital, by applying the previous argument to the unitalization A′ of A we know that there is a nilpotent A of index n in A′. If A ∈ A, we are done. Otherwise, we may assume, after multiplying A with a scalar, that there exists T ∈ A such that A = I + T. Then T^{2} + T is a nilpotent element of index n in A. □

Every semitransitive algebra of matrices contains a minimal semitransitive subalgebra because it is finite-dimensional.

Corollary 11. The algebra of upper triangular Toepliz matrices, described in Section 4.1 is the unique (up to simultaneous similarity) minimal semitransitive algebra of matrices over an algebraically closed field. In particular, every minimal semitransitive algebra of matrices over an algebraically closed field is commutative.

Proof. Let A be a semitransitive algebra of n × n matrices over an algebraically closed field. By Theorem 10, A must contain an matrix S such that S^{n} = 0, but S^{n−1} ≠ 0 (i.e. S is the backward shift in some basis). The subalgebra of A generated by S is similar to the algebra of upper triangular Toepliz matrices and is semitransitive (see Section 4.1).

On the other hand, if A is minimal semitransitive, then it must be equal to the subalgebra generated by S. □

We now turn our attention to semitransitive semigroups of matrices. The lemma below is needed to show that semitransitive semigroups in M_{n} of constant rank k exist only for those k that divide n.

12 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

Lemma 12. Let S be a semigroup of linear transformations of constant rank k (or 0) acting on 𝔽^{n} and assume that U is a subspace of 𝔽^{n} of dimension m > k containing the range of S (i.e. T x ∈ U for all x ∈ 𝔽^{n} and T ∈ S). If S acts semitransitively on U, then there exists a subspace W of U of dimension m − k and a subsemigroup S′ of S whose range is contained in W and acts semitransitively on W.

Proof. It is easy to see that S must contain an element T for which Range T^{2} = Range T (any nonnilpotent operator will suffice). We will show that S′ = {S ∈ S | T S = 0} and W = (ker T) ∩ U = ker T |_{U} satisfy the required criteria. It is clear that Range S′ ⊆ W and that dim(W) = m − k (rank T |_{U} = k, since rank T = k and Range T^{2} = Range T). We complete the proof by showing that S′ acts semitransitively on W. Let x and y be nonzero vectors in W. Since S acts semitransitively on U (hence also on W) there is S ∈ S that maps one of the vectors into the other; suppose Sx = y. Note that T Sx = T y = 0. If T S ≠ 0, then rank(T S) = k and therefore S has the same kernel as T S (clearly ker T S ⊆ ker S, we must have equality since dim ker T S = n − k = dim ker S); but this would mean that Sx = 0. Thus T S = 0 and hence S ∈ S′. □

The following result extends a known theorem for transitivity of semigroups [1, Theorem 3.13].

Theorem 13. If S is a semitransitive semigroup of matrices in M_{n}(𝔽) of rank k or 0, then k divides n.

Proof. Suppose, if possible, that k does not divide n. Then we can write n = sk + r, where 0 < r < k. Using Lemma 12 (s times) we obtain a subspace W of 𝔽^{n} of dimension r and a subsemigroup S′ of S whose image is contained in W and acts semitransitively on W. But then S′ would be a nonzero semigroup of rank at most r; a contradiction to the assumption that all ranks in S and hence in S′ are either 0 or k. □

Example 14. If n = kl and V = V_{1} ⊕ V_{2} ⊕ … ⊕ V_{l}, is a decomposition of an n-di-mensional vector space V into k-dimensional subspaces, then all linear transformations A: V → V of the form A = A_{1,i} + … + A_{l,i}, where each A_{j,i}: V_{i} → V_{j} is either 0 or invertible, form a semitransitive semigroup of transformations of rank k or 0.

The lemma below will be used to construct examples of semitransitive semigroups that contain no minimal semitransitive semigroups.

Lemma 15. If a semigroup S is either transitive or semitransitive, then so is {AB | A, B ∈ S}, the subsemigroup of S consisting of products of two elements of S.

Proof. If x, y is a pair of nonzero vectors, then note that ABx = y, provided that Bx = y and Ay = y. □

Corollary 16. If 𝔽 is a field of characteristic zero and n ⩾ 2, then

S = {A ∈ M_{n}(𝔽) | det(A) = 2^{k} for some positive integer k}

is a transitive semigroup that contains no minimal transitive (semitransitive) subsemigroups. In particular, if 𝔽 ∈ {ℝ, ℂ}, then S is closed and contains no minimal closed transitive (semitransitive) subsemigroups.

Proof. Note that if S′ is a subsemigroup of S, then S′ · S′ is a proper subsemigroup S′. The exercise of showing that S is transitive is left to the reader. □

Corollary 17. Let S_{α} be the subsemigroup of M_{n}(ℂ), generated by unitary matrices of determinant α with |α| = 1. If n ⩾ 2 and α is not a root of unity, then

S′_{α} = (0, 1] · S_{α}

is a bounded transitive semigroup that contains no minimal transitive (semitransitive) subsemigroups.

Proof. Note that T · T ⊊ T ,whenever T is a subsemigroup of S′_{α}. It remains to show that the semigroup S′_{α} is transitive. For that, it is sufficient to prove that unitary matrices of determinant α act transitively on the unit sphere of vectors in ℂ^{n}. If x, y is a pair of unit vectors in ℂ^{n}, then choose an orthonormal basis V = {v_{1}, …, v_{n}} for ℂ^{n}, so that x = v_{1} and y = y_{1}v_{1} + y_{2}v_{2}. Now note that (in basis V),

A = (\begin{matrix}y_{1} & −αȳ_{2} \\ y_{2} & αȳ_{1} \end{matrix}) ⊕ I_{n−2}

is a unitary transformation of determinant α that maps x to y. □

The examples above show that in general semitransitive semigroups need not contain any minimal semitransitive semigroups. In the remainder of our paper, we give some conditions which guarantee the existence of such a subsemigroup.

Theorem 18. If S_{0} is a semitransitive semigroup in M_{n}(𝔽) consisting of matrices with at most one non-zero column then it contains a minimal semitransitive subsemigroup.

Proof. Let G be the set of all semitransitive subsemigroups of S_{0}, ordered by the reverse inclusion. We use Zorn’s lemma to prove that G contains a maximal element. Suppose that (S_{α})_{α∈Λ} is a chain in G. We claim that the semigroup S = ⋂_{α∈Λ} S_{α} is

14 BERNIK, GRUNENFELDER, MASTNAK, RADJAVI, TROITSKY

semitransitive. We can assume that Λ is infinite. Pick a pair of nonzero vectors x, y and define

Λ′ = { α ∈ Λ | Ax = y for some A ∈ S_{α}}
Λ′′ = { α ∈ Λ | Ay = x for some A ∈ S_{α}}.

Since Λ′ ∪ Λ′′ = Λ we can, without any loss of generality, assume that ⋂_{Λ′} S_{α} = S. For each α ∈ Λ′ choose a matrix A_{α} ∈ S_{α} so that A_{α}x = y and note that A_{α} = z_{α}e_{j_{α}}^{T} for some integer j_{α} and nonzero vector z_{α}. If Λ_{j} = {α ∈ Λ′ | j_{α} = j} then clearly Λ′ = ⋃_{j=1}^{n} Λ_{j} and hence ⋂_{Λ_{j_{0}}} S_{α} = S for some 1 ⩽ j_{0} ⩽ n, since at least one of the Λ_{j} has no upper bound.

For every α ∈ Λ_{j_{0}} we have y = A_{α}x = z_{α}e_{j_{0}}^{T}x = x_{j_{0}}z_{α}, thus z_{α} = \frac{1}{x_{j_{0}}} y, hence A_{α} = \frac{1}{x_{j_{0}}} ye_{j_{0}} for all α ∈ Λ_{j_{0}} and therefore \frac{1}{x_{j_{0}}} ye_{j_{0}} ∈ S. □

The proof of the theorem above can be modified to show the following.

Theorem 19. Suppose that X is a Banach space and S is a semitransitive semigroup of bounded operators on X. If S is compact in operator norm, then it contains a minimal compact semitransitive semigroup.

Proof. Let G be the set of all compact semitransitive subsemigroups of S_{0}, ordered by the reverse inclusion. We use Zorn’s lemma to show that G has a maximal element. If (S_{α})_{α∈Λ} is a chain in G then define S = ⋂_{Λ} S_{α}. We claim, that S is semitransitive (it is clearly a compact semigroup). Pick a pair of nonzero vectors x, y and define

Λ′ = { α ∈ Λ | Ax = y for some A ∈ S_{α}}
Λ′′ = { α ∈ Λ | Ay = x for some A ∈ S_{α}}.

Since Λ′ ∪ Λ′′ = Λ we can, without any loss of generality, assume that ⋂_{Λ′} S_{α} = S. If α ∈ Λ′ then choose an operator A_{α} ∈ S_{α}, so that A_{α}x = y. Since (A_{α})_{α∈Λ′} is a chain in S_{0}, it contains a convergent subnet (A_{α})_{α∈Λ′_{0}} → A. Note that A ∈ S_{α} for all α, since every S_{α} is closed and contains a tail of the net (A_{α})_{α∈Λ′_{0}}. Hence A ∈ S and Ax = lim_{α∈Λ′_{0}} A_{α}x = y. □

Remark 20. Observe that operator norm compactness in Theorem 19 can be replaced with compactness in any Hausdorff operator topology τ that satisfies the following condition: if A_{α} ⟶^{τ} A and A_{α}x = y for some net (A_{α}) and some nonzero vectors x and y, then Ax = y. In particular, Theorem 19 remains valid for the weak operator topology and, in case when X is a dual space, for the weak* operator topology.

Corollary 21. A closed bounded semitransitive semigroup in M_{n}(𝔽), where 𝔽 ∈ {ℝ, ℂ}, contains a minimal closed semitransitive subsemigroup.

Remark 22. The only semigroup property that we used in the proof of Theorem 19 is that the intersection of a chain of semigroups is a semigroup. Thus Theorem 19, Remark 20 and Corollary 21 apply also to semitransitive sets or groups of operators.

References

[1] R. Drnovšek, L. Livshits, G. MacDonald, B. Mathes, H. Radjavi, and P. Šemrl, On transitive linear semigroups, Lin. Alg. & Appl. 305 (2000), 67–86.

[2] F.W. Levi, Arithmetische Gesetze im Gebiet diskreter Gruppen, Rend. Palermo 35 (1913), 225– 236.

[3] A.I. Kokorin and V.M. Kopytov, Fully ordered groups, Translated from the Russian by D. Louvish, Halsted Press [John Wiley & Sons], New York, 1974.

[4] K.H. Kuo, P.Y. Wu, Factorization of matrices into partial isometries, Proc. Amer. Math. Soc. 105 (1989), 263–272.

[5] H. Radjavi, P. Rosenthal, Simultaneous triangularization, Springer 2000.

[6] H. Rosenthal and V.G. Troitsky, Strictly semi-transitive operator algebras, to appear in J. of Operator Theory.

Department of Mathematics, University of Ljubljana, Jadranska 19, 1000 Ljubljana. Slovenia.

Department of Mathematics, Dalhousie University, Halifax, NS, B3H 3J5. Canada.

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

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

  E-mail address: janez.bernik@fmf.uni-lj.si, luzius@mathstat.dal.ca,
mastnak@mathstat.dal.ca, hradjavi@uwaterloo.ca, vtroitsky@math.ualberta.ca