LOMONOSOV’S THEOREM CANNOT BE EXTENDED TO CHAINS OF FOUR OPERATORS

VLADIMIR G. TROITSKY

Abstract. We show that the celebrated Lomonosov theorem cannot be improved by increasing the number of commuting operators. Specifically, we prove that if T: ℓ_{1} → ℓ_{1} is the operator without a non-trivial closed invariant subspace constructed by C. J. Read, then there are three operators S_{1}, S_{2} and K (non-multiples of the identity) such that T commutes with S_{1}, S_{1} commutes with S_{2}, S_{2} commutes with K, and K is compact. It is also shown that the commutant of T contains only series of T.

All Banach spaces in this note are assumed to be infinite dimensional and separable; all operators are linear and bounded. We say that an operator is a non-scalar operator if it is not a multiple of the identity operator.

One of the major results in the history of the Invariant Subspace Problem was obtained by V. Lomonosov in [L] who proved that if an operator T on a Banach space commutes with another non-scalar operator S and S commutes with a non-zero compact operator K, then T has an invariant subspace. Motivated by their study of the Invariant Subspace Problem for positive operators on Banach lattices, Y. A. Abramovich and C. D. Aliprantis have asked recently whether or not Lomonosov’s theorem can be extended to chains of four or more operators. The purpose of this note is to answer this question in the negative. For our initial operator T we will take an operator without an invariant subspace on ℓ_{1} coming from the famous construction of C. J. Read (see [R1]). Then we will produce three non-scalar operators S_{1}, S_{2}, and K with K compact (as a matter of fact K has rank one) such that T S_{1} = S_{1}T, S_{1}S_{2} = S_{2}S_{1}, and S_{2}K = KS_{2}. This will be done in Section 1.

After that, in Section 2, we will consider a related question of describing the commutant of the C. J. Read operator.

1. A chain from C. J. Read’s operator to a rank-one operator

We begin with reminding the reader of the construction in [R1] that will be central for us. As in [R1], we denote the standard unit vectors of ℓ_{1} by (f_{i})_{i=0}^{∞}. The symbol F denotes the linear subspace of ℓ_{1}, spanned by f_{i}’s, and thus, F consists of eventually vanishing sequences.

Proc. Amer. Math. Soc. 128 (2000), 521–525.

1 2 V. G. TROITSKY

1991 Mathematics Subject Classification. 47A15.

Key words and phrases. Invariant subspaces, commutant.

Supported in part by NSF Grant DMS 96-22454.

Let d = (a_{1}, b_{1}, a_{2}, b_{2}, …) be a strictly increasing sequence of positive integers. Also let a_{0} = 1, v_{0} = 0, and v_{n} = n(a_{n} + b_{n}) for n ⩾ 1. Then there is a unique sequence (e_{i})_{i=0}^{∞} ⊂ F with the following properties:

0) f_{0} = e_{0};

A) if integers r, n, and i satisfy 0 < r ⩽ n, i ∈ [0, v_{n−r}] + ra_{n}, we have f_{i} =

a_{n−r}(e_{i} − e_{i−ra_{n}});

B) if integers r, n, and i satisfy 1 ⩽ r < n, i ∈ (ra_{n} + v_{n−r}, (r + 1)a_{n}), (respectively, 1 ⩽ n, i ∈ (v_{n−1}, a_{n})), then f_{i} = 2^{(h−i)/\sqrt{a_{n}}}e_{i}, where h = (r + \frac{1}{2})a_{n} (respectively,

h = \frac{1}{2}a_{n});

C) if integers r, n, and i satisfy 1 ⩽ r ⩽ n, i ∈ [r(a_{n} + b_{n}), na_{n} + rb_{n}], then

f_{i} = e_{i} − b_{n}e_{i−b_{n}};

D) if integers r, n, and i satisfy 0 ⩽ r < n, i ∈ (na_{n} + rb_{n}, (r + 1)(a_{n} + b_{n})), then

f_{i} = 2^{(h−i)/\sqrt{b_{n}}}e_{i}, where h = (r + \frac{1}{2})b_{n}.

Indeed, since f_{i} = ∑_{j=0}^{i} λ_{ij}e_{j} for each i ⩾ 0 and λ_{ii} is always nonzero, this linear relation is invertible. Further,

(∗)  lin{e_{i} : i = 1, …, n} = lin{f_{i} : i = 1, …, n} for every n ⩾ 0.

In particular all e_{i} are linearly independent and also span F. Then C. J. Read defines T: F → F to be the unique linear map such that T e_{i} = e_{i+1}, and in Lemma 5.1 he proves that ∥T f_{i}∥ ⩽ 1 for every i ⩾ 0 provided d increases sufficiently rapidly, i. e., satisfies several conditions of the form

a_{n} ⩾ G(n, a_{0}, b_{0}, a_{1}, b_{1}, …, a_{n−1}, b_{n−1}), and
b_{n} ⩾ H(n, a_{0}, b_{0}, a_{1}, b_{1}, …, a_{n−1}, b_{n−1}, a_{n}),

where the G and H are some real-valued functions. It follows that T can be extended to a bounded operator on ℓ_{1}. Finally, C. .J. Read proves that this extension, which is still denoted by T, has no invariant subspaces provided d increases sufficiently rapidly.

Throughout this section we will assume, without loss of generality, that all integers a_{i} and b_{i} are even. We are going to construct non-scalar operators S_{1}, S_{2}, and K such that K has rank one and commutes with S_{2}, S_{2} commutes with S_{1}, and S_{1} commutes with T. In fact, we take S_{1} = T^{2}, so that the equality T S_{1} = S_{1}T is automatic. Define S_{2} on F via

S_{2}e_{i} = \{ \begin{matrix}e_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

We claim that

S_{2}f_{i} = \{ \begin{matrix}f_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

To prove this we consider all possible cases:

0) In this case S_{2}f_{0} = S_{2}e_{0} = e_{0} = f_{0}; A) Since a_{n} is even then

S_{2}f_{i} = a_{n−r}(S_{2}e_{i} − S_{2}e_{i−ra_{n}}) = \{ \begin{matrix}a_{n−r}(e_{i} − e_{i−ra_{n}}) = f_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

B) In this case

S_{2}f_{i} = 2^{(h−i)/\sqrt{a_{n}}}S_{2}e_{i} = \{ \begin{matrix}2^{(h−i)/\sqrt{a_{n}}}e_{i} = f_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

C) Since b_{n} is even, we have

S_{2}f_{i} = S_{2}e_{i} − b_{n}S_{2}e_{i−b_{n}} = \{ \begin{matrix}e_{i} − b_{n}e_{i−b_{n}} = f_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

D) Finally, in this case

S_{2}f_{i} = 2^{(h−i)/\sqrt{b_{n}}}S_{2}e_{i} = \{ \begin{matrix}2^{(h−i)/\sqrt{b_{n}}}e_{i} = f_{i} & if i is even; \\ 0 & otherwise. \end{matrix}

In particular, S_{2} is bounded on F and can be extended to ℓ_{1}. For every i ⩾ 0 we have

T^{2}S_{2}e_{i} = \{ \begin{matrix}T^{2}e_{i} = e_{i+2} & if i is even; \\ 0 & otherwise. \end{matrix}

On the other hand

S_{2}T^{2}e_{i} = S_{2}e_{i+2} = \{ \begin{matrix}e_{i+2} & if i is even; \\ 0 & otherwise, \end{matrix}

so that T^{2}S_{2}x = S_{2}T^{2}x for every x ∈ F. Since F is dense in ℓ_{1}, it follows that T^{2} and S_{2} commute on ℓ_{1}.

Finally, define K on ℓ_{1} via Kf_{0} = f_{0} and Kf_{i} = 0 for all i > 0. Then K is a bounded rank one operator on ℓ_{1}, and K commutes with S_{2}.

Note that if m divides a_{n} and b_{n} for every n, then, similarly to the previous construction, we could take for S_{1} the operator T^{m} instead of T^{2}. It follows from Lomonosov’s theorem that T^{m} has an invariant subspace (confer [R1, Lemma 6.4]).

In [R2] C. J. Read presents as a modification of his original example a quasinilpotent operator on ℓ_{1} without closed nontrivial invariant subspaces. The same argument as above provides a chain of four commuting operators connecting this operator to a compact operator.

2. Commutants of C. J. Read’s operators

Let 𝔐 denote the collection of all real infinite matrices A = (a_{ij})_{i,j=0}^{∞}. Elements of 𝔐 can be added entry-wise, the zero and identity matrices are defined in the natural way. Let F denote the subfamily of 𝔐 consisting of the matrices with finite number of nonzero entries in every column and every row. For A, B, and C in 𝔐 we say that AB = C if for all i, j ∈ ℕ we have a_{ij} = ∑_{k=0}^{∞} b_{ik}c_{kj} and the series converge absolutely. Though AB may not exist in general, AB exists if A or B belongs to F.

4 V. G. TROITSKY

We can also define the action of a matrix on a sequence: we say that Ax = y for A ∈ 𝔐 and x, y ∈ ℝ^{ℕ} if y_{i} = ∑_{j=0}^{∞} a_{ij}x_{j} for every i ⩾ 0. Again, if A ∈ F, then Ax is defined for every x ∈ ℝ^{ℕ}. Let A^{(j)} denote the j-th column of A ∈ 𝔐, then A^{(j)} = Af_{j}. Finally, (A)_{ij} or a_{ij} will denote the (i, j)-th entry of A ∈ 𝔐.

The space B(ℓ_{1}) of all (bounded) operators on ℓ_{1} can be naturally embedded in 𝔐: if R ∈ B(ℓ_{1}), then r_{ij} = (Rf_{j})_{i}. Obviously, the sum of operators in B(ℓ_{1}) corresponds to the sum of matrices in 𝔐. Moreover, the action of R on an element of ℓ_{1} is in accord with the definition of the action of a matrix on a sequence: if x = (x_{1}, x_{2}, …) ∈ ℓ_{1} then x = ∑_{j=0}^{∞} x_{j}f_{j}, so that Rx = ∑_{j=0}^{∞} x_{j}Rf_{j}, and (Rx)_{i} = ∑_{j=0}^{∞} x_{j}(Rf_{j})_{i} = ∑_{j=0}^{∞} r_{ij}x_{j}. Also, the product of two operators in B(ℓ_{1}) corresponds to the product of two matrices: if R, P ∈ B(ℓ_{1}), then RP f_{j} = RP^{(j)} = R ∑_{k=0}^{∞} p_{kj}f_{k} = ∑_{k=0}^{∞} p_{kj}Rf_{k}, so that (RP)_{ij} = (RP f_{j})_{i} = ∑_{k=0}^{∞} p_{kj}r_{ik}. Notice that two operators in B(ℓ_{1}) commute if and only if they commute as matrices in 𝔐. Finally, the identity and zero operators correspond to the identity and zero matrices respectively.

Let S be the right shift operator, i.e., Sf_{j} = f_{j+1}. Given a formal power series p(t) = ∑_{n=0}^{∞} p_{n}t^{n}, the matrix p(S) = p_{0}I + p_{1}S + p_{2}S^{2} + … belongs to 𝔐 and is of the form

(1)  ( \begin{matrix}p_{0} & 0 & 0 & 0 & … \\ p_{1} & p_{0} & 0 & 0 & … \\ p_{2} & p_{1} & p_{0} & 0 & … \\ p_{3} & p_{2} & p_{1} & p_{0} & … \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} ) .

On the other hand, every matrix of this form is obviously a series of S.

Lemma 1. If A ∈ 𝔐 commutes with S, then A = p(S) for some series p.

Proof. Since

(AS)_{ij} = ∑_{k=0}^{∞} a_{ik}s_{kj} = a_{i,j+1}

and

(SA)_{ij} = ∑_{k=0}^{∞} s_{ik}a_{kj} = \{ \begin{matrix}a_{i−1,j} & if i ⩾ 1; \\ 0 & if i = 0 \end{matrix}

for every pair (i, j), it follows that A is of the form (1). □

Consider Q ∈ 𝔐 such that Qf_{j} = Q^{(j)} = e_{j} for every j ⩾ 0. It follows from (∗) that Q ∈ F. It also follows from (∗) that Q is invertible, and Q^{−1} ∈ F. Further, we can define “change of basis” map A ∈ 𝔐 ↦ Ã = Q^{−1}AQ. Since Q, Q^{−1} ∈ F, this map is defined for every A ∈ 𝔐, one-to-one, onto, and A = QÃQ^{−1}. Clearly, Ã describes the action of A in terms of the e_{i}’s. Finally, AB = BA if and only if ÃB̃ = B̃Ã for A, B ∈ 𝔐.

Recall that T denotes the Read operator introduced in the previous section. Since T e_{j} = e_{j+1} for every j ⩾ 0, then T Qf_{j} = Qf_{j+1}, so that Q^{−1}T Qf_{j} = f_{j+1} which implies T̃ = S. Suppose RT = T R for some R ∈ B(ℓ_{1}), then R̃S = SR̃, so that R̃ = p(S) for some series p(t) = ∑_{n=0}^{∞} p_{n}t^{n} by Lemma 1. Therefore, R = QR̃Q^{−1} = Qp(S)Q^{−1} = p(QSQ^{−1}) = p(T). Since every bounded operator of the form ∑_{n=0}^{∞} p_{n}T^{n} commutes with T, we have proved the following proposition:

Proposition 2. The commutant of T is the set of all bounded operators of the form ∑_{n=0}^{∞} p_{n}T^{n}.

I would like to thank Prof. Y. A. Abramovich for suggesting this line of investigation to me and for our discussions. I am thankful to Professors C. D. Aliprantis, V. J. Lomonosov, and C. J. Read for their interest in this work.

We remark in conclusion that in [TV] we study the modulus |T | of the Read operator of [R2], and we prove that, unlike T, the modulus |T | does have an invariant subspace.

References

[L] V. I. Lomonosov, Invariant subspaces of the family of operators that commute with a completely continuous operator, Funktsional. Anal. i Prilozhen. 7 (1973), No. 3, 55–56. (Russian)

[R1] C. J. Read, A short proof concerning the invariant subspace problem, J. Lond. Math. Soc., (2) 33 (1986), 335–348.

[R2] C. J. Read, Quasinilpotent Operators and the Invariant Subspace Problem, J.Lond.Math.Soc., (2) 56 (1997), No. 3, 595–606.

[TV] V. G. Troitsky, On the modulus of C. J. Read’s operator, Positivity, 2 (1998), No 3, 257–264.

Mathematics Department, University of Illinois at Urbana-Champaign, 1409 West Green St., Urbana, Il 61801, USA

  E-mail address: vladimir@math.uiuc.edu