# IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction

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1 IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible semigroup of compact linear operators on a Banach space of dimension at least 2 with the property that the products AB and BA are proportional for each pair of elements A and B in S. We show that S is necessarily a nilpotent matrix group of nilpotency class 2 with possible addition of the zero matrix. We also study generalizations of the property. 1. Introduction Several authors have studied semigroups of linear operators satisfying polynomial identities and considered the problem of irreducibility of these semigroups (see e.g. [1, 2, 3]). Then one assumes that every sequence of elements satisfies the given identities. Here we consider a weaker problem when the coefficients of the identities can depend on the chosen elements. We first study the reducibility of an operator semigroup S such that for each pair of elements A and B in S the products AB and BA are proportional. It is easy to find examples of irreducible matrix groups having this property. Indeed, let U be the permutation n n (n 2) matrix defined on the standard basis vectors {e i } n i=1 by Ue i = e i+1 for 1 i < n and Ue n = e 1, and let D be the diagonal matrix diag(1, w, w 2,..., w n 1 ), where w is the primitive n-th root of unity. Then the group G generated by U and D is irreducible and for each pair A, B G there exists j {0, 1, 2,..., n 1} such that AB = w j BA. These matrix groups are often minimal counterexamples for reducibility when one studies various reducibility questions (see e.g. Sections 3.3, 4.2 and 4.3 in [3]). Most of these reducibility questions were considered in the infinite-dimensional setting as well. One may also ask whether there are infinitedimensional examples of irreducible operator semigroups S such that AB and BA are proportional for all A, B S. Date: November 10, Mathematics Subject Classification. 15A30, 47A15, 47D03. Key words and phrases. Groups and semigroups of operators, irreducibility, proportional products. 1

2 2 R. DRNOVŠEK, T. KOŠIR In this note we show that the only such irreducible semigroups of compact linear operators on a Banach space of dimension at least 2 are irreducible nilpotent matrix groups of class 2 with possible addition of the zero matrix. In these groups for each pair A, B the commutator ABA 1 B 1 is a scalar matrix, say γi, and so AB = γba. We also obtain weaker results in the case when the pair {AB, BA} in the condition is replaced by pairs {ABA k, BA k+1 } or {A k+1 B, A k BA} for some k 1 depending on A, B. In our results the compactness assumption can not be omitted: there exists a bounded operator on l 1 without nontrivial invariant subspace (see e.g. [4]) so that the semigroup it generates is irreducible and commutative. 2. Results Throughout the paper we assume that the dimension of the underlying Banach space is at least 2. We denote by R + the set of all nonnegative real numbers. The closure of a set S in a topological space is denoted by S. In the proof of our first result we use several times the well-known fact that r(ab) = r(ba) for all bounded operators A and B, where r denotes the spectral radius. Theorem 2.1. Let S be an irreducible semigroup of compact operators on a complex Banach space X with the property that for each pair of elements A and B in S the products AB and BA are proportional. Then (a) the dimension n of X is finite. (b) the set G = R + S \ {0} is a nilpotent group of class 2, and its subgroup G 0 = {G G : det (G) = 1} is finite. (c) there exists a basis of X in which operators of the group G 0 are represented by monomial unitary matrices. (d) for each pair A, B G there is a (unique) n-th root γ A,B of unity such that AB = γ A,B BA. (e) for every A G the operator A n is a multiple of the identity. Proof. With no loss of generality we may assume that R + S = S. Denote by I the set of all quasinilpotents in S. We claim that I is an ideal of S. Assume on the contrary that r(as) > 0 for some A I and S S. By the assumption, there exist complex numbers α and β, not both equal to zero, such that αas = βsa. Since r(as) = r(sa) > 0, we

3 IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL 3 conclude that α = β > 0. It follows that (AS) m = A m S m A m S m for all positive integers m, and so r(as) r(a) r(s) = 0. This contradiction shows that I is an ideal of S. By the famous Turovskii s theorem (see [6] or [3]), the ideal I is reducible. Since every non-zero ideal of an irreducible semigroup is also irreducible we obtain that I = {0}. This implies that for A, B S we have AB = 0 if and only if BA = 0. Consequently, we may assume for each pair A, B S there exists a nonzero complex number γ A,B such that AB = γ A,B BA. Furthermore, we may assume that γ A,B is on the unit circle. Indeed, for each pair A, B S with AB 0 we have r(ab) > 0, and so it follows from r(ab) = γ A,B r(ba) that γ A,B = 1. Now, it is easy to verify that there is no loss of generality in assuming that S is closed. By [3, Theorem 7.4.5], S contains non-zero finite rank operators. Denote by n the minimal rank of non-zero members of S and pick E S of rank n. Since AE = γ A,E EA for all A S, the range im E of E is invariant under every member of S. In view of irreducibility of S we obtain that im E = X so that (a) holds and every member of G is invertible. Moreover, G is a group by [3, Lemma 3.1.6]. The property AB = γ A,B BA implies that ABA 1 B 1 belongs to the center of G for all A, B G. Since G is not commutative (otherwise it was reducible), G is a nilpotent group of class 2. Since its subgroup G 0 = {G G : det (G) = 1} is finite by [5, Theorem 1, p. 208], the proof of (b) is finished. By [5, Lemma 6, p. 207] there exists a basis {e 1, e 2,..., e n } of X in which operators of G are represented by monomial matrices. We recall a well known argument to prove (c). Denote by (, ) the standard inner product on X with respect to the basis {e 1, e 2,..., e n }. If G 1, G 2,..., G m are all members of G 0, then we define a new inner product on X as follows x, y = 1 m m (G i x, G i y). i=1 It is easily verified that every member G G 0 has a monomial unitary matrix with respect to the basis {f 1,..., f n } defined by f i = e i / e i, e i, i = 1,..., n. This completes the proof of (c). For every pair A, B G we have det (AB) = γa,b n det (BA) which yields that γn A,B = 1 proving (d). To show (e), take A G. It follows from (d) that the operator A n commutes

4 4 R. DRNOVŠEK, T. KOŠIR with any B G. Since G is irreducible, we conclude that A n must be a multiple of the identity. Corollary 2.2. Let S be an irreducible semigroup of n n complex matrices. Then for all pairs A, B S there exists a complex number γ A,B such that AB = γ A,B BA if and only if R + S \ {0} is a nilpotent group of class 2. Proof. Only the converse implication needs a proof. If G = R + S \{0} is a nilpotent group of class 2, then for any pair A, B G the group commutator ABA 1 B 1 belongs to the center of G. Since G is irreducible, the center consists of scalar matrices only, and so there is a complex number γ A,B such that AB = γ A,B BA. We denote by G(k) the irreducible subgroup of GL k (C) generated by the cyclic permutation matrix U defined on the standard basis vectors {e i } k i=1 by Ue i = e i+1 for 1 i < k and Ue k = e 1, and by the diagonal matrix diag(1, w, w 2,..., w k 1 ), where w is a primitive k-th root of unity. Corollary 2.3. A semigroup S of n n complex matrices is a maximal irreducible semigroup such that for all pairs A, B S there exists a complex number γ A,B AB = γ A,B BA if and only is there exists a factorization n = k 1 k 2 k t (all k j 2 and k j k i for j > i) such that S is conjugate to the semigroup C G(k 1 ) G(k 2 ) G(k t ). Proof. By Corollary 2.2 we know that R + S \ {0} is an irreducible nilpotent matrix group of class 2. It is contained in a maximal nilpotent matrix group of class 2 which we denote by G. with Another application of Corollary 2.2 implies that for all pairs A, B G there exists a complex number γ A,B such that AB = γ A,B BA. By maximality it follows that S = R + S = G {0}. By [5, Theorem 7, pp ] we know that G is conjugate to the matrix group C G(k 1 ) G(k 2 ) G(k t ) for a factorization n = k 1 k 2 k t such that all k j 2 and k j k i for j > i. Here C = C \ {0}. Then it follows that S is of the required form. Theorem 2.4. Let Γ be a non-void compact subset of C \ {0}, and let S be an irreducible semigroup of compact operators on a complex Banach space X with the property that for each pair A, B S \ {0} there exist γ = γ A,B Γ and k = k A,B N such that (AB γba)a k = 0. Then the conclusions of Theorem 2.1 hold. Proof. With no loss of generality we may assume again that R + S = S. By Turovskii s theorem the semigroup S contains an operator that is not quasinilpotent. Since R + S = S,

5 IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL 5 there is an operator K S of spectral radius 1. As in the proof of [3, Lemma 7.4.5] we obtain an unbounded sequence {m i } in N and a sequence {t i } in (0, ) such that the sequence {t i K m i } converges to a non-zero finite-rank operator F that is either idempotent or nilpotent with F 2 = 0. (If F is idempotent, we can even take t i = 1.) By the assumption for any B S \ {0} there exist γ i = γ K m i,b Γ and k i = k K m i,b N such that (K m i B γ i BK m i )(K m i ) k i = 0. We can find a subsequence {m ij } of {m i } such that m ij+1 m ij k ij for all j, and so (K m i j B γij BK m i j )K m ij+1 = 0. Since Γ is a compact set, by passing to a subsequence if necessary we may assume that the sequence {γ ij } converges to γ Γ. Multiplying the last equation by t ij t ij+1 and taking the limit we obtain (F B γbf )F = 0. If F is nilpotent of index 2, we have F BF = 0 so that B(im F ) ker F for all B S. Then for any non-zero vector x im F the closed linear span of the set {x} {Sx : S S} is contained in ker F X and is invariant under each member of S which is a contradiction with the irreducibility of S. So, F has to be idempotent, and it follows from F BF = γbf that the range im F is invariant under every member of the irreducible semigroup S, so that im F = X. Now note that with n = dim X it holds that im (A k ) = im (A n ) for all A S and all k > n, and so the semigroup S has the property that for each pair A, B S \ {0} there exists γ A,B Γ such that (AB γ A,B BA)A n = 0. By a simple compactness argument we may assume that S is closed. Then by [3, Lemma 3.1.6] S contains an idempotent E of minimal rank in S \ {0}. Since EBE = γ E,B BE for all B S, the range im E is invariant under every member of S, so that E is the identity operator. It follows that every non-zero member of S is invertible, and so S has the property that for each pair A, B S \ {0} there exists γ A,B Γ such that AB = γ A,B BA. Now Theorem 2.1 can be applied to complete the proof. The following example shows that the condition 0 / Γ in Theorem 2.4 is necessary. Example 2.5. Let n 2. We denote by E ij, i, j = 1, 2,..., n the standard linear basis for M n (C), i.e., E ij is the matrix with the only nonzero entry (i, j) equal to 1. Then S = {E ij, i, j = 1, 2,..., n} {0} is a matrix semigroup. It is easy to check that for each

6 6 R. DRNOVŠEK, T. KOŠIR pair A, B S, A B we have ABA 2 = 0, and so for each pair A, B S there exists a complex number γ such that (AB γba)a 2 = 0. The following result is a dual version of Theorem 2.4. We omit its proof, since it is a slight modification of the proof of Theorem 2.4. Theorem 2.6. Let Γ be a non-void compact set of C \ {0}, and let S be an irreducible semigroup of compact operators on a complex Banach space X with the property that for each pair A, B S \ {0} there exist γ = γ A,B Γ and k = k A,B N such that A k (AB γba) = 0. Then the conclusions of Theorem 2.1 hold. The semigroup S in Example 2.5 satisfies the condition that A 2 (AB BA)B 2 = 0 for each pair A, B S. Therefore further weakening of conditions in Theorems 2.4 and 2.6 to the condition that for each pair A, B S \ {0} there exist γ A,B 0 and k = k A,B N such that A k (AB γba)b k = 0 will no more yield the conclusions of Theorem 2.1. Acknowledgement. The authors were supported in part by the Ministry of Education, Science, and Sport of Slovenia. References [1] J. Okniński. Semigroups of matrices. World Scientific Publishing Co., Inc., River Edge, N.J., [2] H. Radjavi. Polynomial conditions on operator semigroups, submitted. [3] H. Radjavi and P. Rosenthal. Simultaneous Triangularization. Springer-Verlag, Berlin, Heidelberg, New York, [4] C. J. Read. Quasinilpotent operators and the invariant subspace problem, J. London Math. Soc. (2) 56 (1997), [5] D. A. Suprunenko. Matrix groups. Translations of Mathematical Monographs, Vol. 45. American Mathematical Society, Providence, R.I., [6] Yu.V. Turovskii, Volterra semigroups have invariant subspaces, J. Funct. Anal. 162 (1999), R. Drnovšek, T. Košir: Department of Mathematics, University of Ljubljana, Jadranska 19, 1000 Ljubljana, Slovenia address:

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