IDEALS IN A TRIDIAGONAL ALGEBRA ALGL_∞

Abstract

We find examples of Ideals in a tridiagonal algebra ALGL_∞ and study some properties of Ideals in ALGL_∞. We prove the following theorems: Let k and j be fixed natural numbers. Let A be a subalgebra of ALGL_∞ and let A_2,{k} ⊂ A ⊂ {T ∈ ALGL_∞ | T_(2k-1,2k) = 0}. Then A is an ideal of ALGL_∞ if and only if A = A_2,{k} where A_2,{k} = {T ∈ ALGL_∞ | T_(2k-1,2k) = 0, T_(2k-1,2k-1) = T_(2k,2k) = 0}. Let B be a subalgebra of ALGL_∞ such that B_2,{j} ⊂ B ⊂ {T ∈ ALGL_∞ | T_(2j+1,2j) = 0}. Then B is an ideal of ALGL_∞ if and only if B = B_2,{j}, where B_2,{j} = {T ∈ ALGL_∞ | T_(2j+1,2j) = 0, T_(2j,2j) = T_(2j+1,2j+1) = 0}.

1. Introduction

Let H be an infinite-dimensional separable Hilbert space with a fixed orthonormal base {e1, e2, · · · } and let B(H) be the algebra of all bounded operators acting on H. If x1, x2, · · · , xk are vectors in H, we denote by [x1, x2, · · · , xk] the closed subspace spanned by the vectors x1, x2, · · · , xk. A subspace lattice L is a strongly closed lattice of orthogonal projections acting on H. We denote by L∞ the subspace lattice generated by the subspaces [e1], [e3], · · · , [e2n−1], · · · , [e1, e2, e3], [e3, e4, e5], · · · , [e2n−3, e2n−2, e2n−1], · · · .

By AlgL∞, we mean the algebra of bounded operators which leave invariant all of the subspaces in L∞. It is easy to see that all such operators have the matrix form

where all non-starred entries are zero.

Let A be a subalgebra of AlgL∞. We say that A is a left ideal of AlgL∞ if AT ∈ A for all A in AlgL∞ and T in A. A is called a right ideal of AlgL∞ if TA ∈ A for all A in AlgL∞ and T in A. A is said to be an ideal of AlgL∞ if A is a left ideal of AlgL∞ and a right ideal of AlgL∞.

Let R be a ring and let A be an ideal in R. A is called a maximal ideal if there exists no proper ideal B such that Let R be a (commutative)ring and let P be an ideal in R. P is prime if P ≠ R and if ab ∈ P for a, b ∈ R implies either a ∈ P or b ∈ P.

In this paper, let I be the identity operator on H. Let ℂ be the set of all complex numbers and ℕ = {1, 2, · · · }

 

2. Examples of ideals in AlgL∞

We can easily prove the following examples by simple calculation. We denote T(i,j) or tij by the (i, j)-component of an operator T in AlgL∞.

Example 1. Let A0 = {T ∈ AlgL∞ | T(k,k) = 0, k ∈ ℕ}. Then A0 is a left ideal in AlgL∞ and a right ideal in AlgL∞. Hence A0 is an ideal in AlgL∞.

Example 2. Let J be a nonempty subset of ℕ. Let AJ = {T ∈ AlgL∞ | T(i,i) = 0, i ∈ J}. Then AJ is an ideal of AlgL∞.

Example 3.

Example 4. Let Ω be a nonempty subset of ℕ and let AΩ = {T ∈ AlgL∞ | T(i,i) = T(j,j) for i, j ∈ Ω and for all T ∈ A}. Then AΩ is a Lie ideal but not an ideal in AlgL∞.

Example 5.

Example 6.

Example 7.

Example 8.

Example 9.

Example 10.

 

3. Properties of ideals in AlgL∞

In this section we investigate some properties of ideals of AlgL∞. If J = { k } for a fixed natural number k, we denote AJ by A{k}.

Theorem 1. Let k be a fixed natural number. Then

(1) A{k} is prime and (2) A{k} is maximal.

Proof. (1) Let A = (aij) and let T = (tij) be operators in AlgL∞. Let AT be in A{k}. Since (AT)kk = akktkk = 0, akk = 0 or tkk = 0. So A is in A{k} or T is in A{k}.

(2) It is sufficient to show that the case k=1. Let A be an ideal of AlgL∞ such that A{1} ⊂ A ⊂ AlgA∞. Suppose that there exists T = (tij) in A such that t11 ≠ 0. Let S = (sij) be an element of AlgA∞. If s11 = 0, then S ∈ A{1}. Let s11 ≠ 0. Let A = (aij) be an operator defined by

Then A + T is an operator in A. Put A1 = A + T. Then A1(1,1) = t11 and A1(i,j) = 0 if (i, j) ≠ (1, 1). Let B = (bij) be an operator defined by

Then B is an element of A{1}. Put Then B+xA1 = S is an operator of A. Hence A = AlgL∞. □

Theorem 2. Let A be a subalgebra of AlgL∞. Let k be a fixed natural number and let A2,{k} ⊂ A ⊂ {T ∈ AlgL∞ | T(2k−1,2k) = 0 }. Then A is an ideal of AlgA∞ if and only if A = A2,{k}.

Proof. Let A = (aij) be an operator in AlgL∞ and let T = (tij) be an operator in A. Suppose that A is an ideal of AlgA∞. Then the (2k − 1, 2k)-component of AT is

Since (*) holds for all A in AlgL∞, t2k 2k = 0. Also the (2k − 1, 2k)-component of TA is

Since (**) holds for all A in AlgL∞, t2k−1 2k−1 = 0. Hence A ⊂ A2,{k} and A = A2,{k}. Suppose that A = A2,{k}. Then A is an ideal of AlgL∞ by Example 8-(2). □

Theorem 3. Let j be a fixed natural number and let B be a subalgebra of AlgL∞ such that B2,{j} ⊂ B ⊂ {T ∈ AlgL∞ | T(2j+1,2j) = 0 }. Then B is an ideal of AlgL∞ if and only if B = B2,{j}.

Proof. Let A = (aij) be an operator in AlgL∞ and let T = (tij) be an operator in B. Suppose that B is an ideal of AlgA∞. Then the (2j +1, 2j)-component of AT is

Since (∗) holds for all A in AlgL∞, t2j 2j = 0. Also the (2j + 1, 2j)-component of TA is

Since (∗∗) holds for all A in AlgL∞, t2j+1 2j+1 = 0. Hence T ∈ B2,{j} and so B ⊂ B2,{j}. Thus B = B2,{j}. Assume that B = B2,{j}. Then B is an ideal of AlgL∞ by Example 9-(2). □

Theorem 4. Let k be a fixed natural number and let A be a subalgebra of AlgL∞ such that A{k,k+1} ⊂ A ⊂ A{k}. Then A is an ideal of AlgL∞ if and only if A = A{k,k+1} or A = A{k}.

Proof. It is sufficient to show that the case k = 1. Let A be an ideal of AlgL∞. Suppose that A ≠ A{1,2}. Then there exists an element T = (tij) in A and T ∉ A{1,2}. Then t22 ≠ 0. Let A = (aij) be an element of A{1}. If a22 = 0, then A ∈ A{1,2} and so A ∈ A. Let a22 ≠ 0. Define an operator B = (bij) by

Then B ∈ A. Put B + T = D. Then D ∈ A. Put Then xD ∈ A. Let S = (sij) be an operator defined by

Then S ∈ A{1,2} and hence A = S + xD ∈ A. □

Theorem 5. Let k be a fixed natural number and let A be a subalgebra of AlgL∞ such that { S ∈ AlgL∞ | S(2k−1,2k) = 0 }⊂ A ⊂ AlgL∞. Then A is an ideal of AlgL∞ if and only if A = AlgL∞.

Proof. It is sufficient to show that the case k = 2. Let A is an ideal of AlgL∞. Since { S ∈ AlgL∞ | S(3,4) = 0 } is not an ideal of AlgL∞, there exists T = (tij) ∈ A such that t34 ≠ 0. Let A = (aij) be an operator in AlgL∞. If a34 = 0, then A ∈ A. Let a34 ≠ 0. Let S = (sij) be an operator defined by

Then S ∈ A. So S + T ∈ A. Put D = S + T. Define an operator B = (bij) by

Then B ∈ A. Put Then B + xD = A ∈ A. □

Theorem 6. Let j be a fixed natural number and let B be a subalgebra of AlgL∞ such that {T ∈ AlgL∞ | T(2j+1,2j) = 0} ⊂ B ⊂ AlgL∞. Then B is an ideal of AlgL∞ if and only if B = AlgL∞.

Proof. It is sufficient to show that j = 1. Let B be an ideal of AlgL∞. Since {T ∈ AlgL∞ | T(3,2) = 0 } is not an ideal of AlgL∞, there exists T = (tij) in B such that t32 ≠ 0. Let A = (aij) be an operator in AlgL∞. If a32 = 0, then A ∈ B. Let a32 ≠ 0. Let S = (sij) be an operator defined by

Then S ∈ B. So S + T ∈ B. Put S + T = D. Define an operator B = (bij) by

Then B ∈ B. Put Then B + xD = A ∈ B. □

Theorem 7. Let Γ = {k1, k2} be a subset of ℕ such that Let A be a subalgebra of AlgL∞ such that A2,Γ ⊂ A ⊂ {T ∈ AlgL∞ | T (2ki−1,2ki) = 0, i = 1, 2 }. Then A is an ideal of AlgL∞ if and only if A = A2,Γ.

Proof. Let A is an ideal of AlgL∞ and let T = (tij) be an operator in A. Then T(2ki-1,2ki) = 0, i = 1, 2. Since A is an ideal of AlgA∞, T(2ki-1,2ki-1) = 0 = T(2ki,2ki)(i = 1, 2) by Theorem 2. So A ⊂ A2,Γ. Hence A = A2,Γ. □

Theorem 8. Let Γ = {k1, k2} be a subset of ℕ such that Let A be a subalgebra of AlgL∞ such that A2,Γ ⊂ A ⊂ B, where B = A2,{k1} ∩ A{2k2-1,2k2}. Then A is an ideal of AlgL∞ if and only if A = A2,Γ or A = B.

Proof. It is sufficient to show that Γ = {k1, k2} = {2, 3}. Let A be an ideal of AlgL∞ and A ≠ A2,Γ. Then there exists an element T ∈ A such that T ∉ A2,Γ, i.e. T(5,6) ≠ 0. Let A = (aij) ∈ B. If a56 = 0, then A ∈ A2,Γ and so A ∈ A. Let a56 ≠ 0. Let S = (sij) be an operator defined by

Then S ∈ A2,Γ and so S ∈ A. Put S + T = A1. Then A1 is an operator of A. Let B = (bij) be an operator defined by

Then B ∈ A2,Γ. Put Then A = B + xA1 ∈ A. Hence A = B. □

Theorem 9. Let n be a fixed natural number(n > 1) and let Γ = {j1, j2} be a subset of ℕ such that Let B be a subalgebra of AlgL∞ such that B2,Γ ⊂ B ⊂ {T ∈ AlgL∞ | T(2ji+1,2ji) = 0, i = 1, 2 }. Then B is an ideal of AlgL∞ if and only if B = B2,Γ.

Proof. Let B be an ideal of AlgL∞ and let A = (aij) ∈ B. Then a2ji+1 2ji = 0(i = 1, 2). Since B is an ideal, a2ji 2ji = 0 = a2ji+1 2ji+1 for all ji ∈ Γ by Theorem 2. Hence A ∈ B2,Γ. □

Theorem 10. Let Γ = {j1, j2} be a subset of natural numbers. Let B = B2,{j1} ∩ A{2j2,2j2+1}. Let A be a subalgebra of AlgL∞ such that B2,Γ ⊂ A ⊂ B. Then A is an ideal of AlgL∞ if and only if A = B2,Γ or A = B.

Proof. Let A be an ideal of AlgL∞ and let A ≠ B2,Γ. Then there exists an operator T = (tij) such that T ∈ A and T ∉ B2,Γ, i.e. t(2j2+1 2j2) ≠ 0. Let A = (aij) ∈ B. If a2j2+1 2j2 = 0, then A ∈ A. Let a2j2+1 2j2 ≠ 0. Define an operator B = (bij) by

Then B ∈ A and B+T ∈ A. Put D = B+T and Then αD ∈ A. Define an operator T1 by

Then T1 ∈ A. So αD + T1 = A ∈ A. Hence A = B. □

Theorem 11. Let k and j be fixed natural numbers. Let C be a subalgebra of AlgL∞ such that C2,{k,j} ⊂ C ⊂ B = {T ∈ AlgL∞ | T(2k−1,2k) = 0, T(2j+1,2j) = 0 }. Then C is an ideal of AlgL∞ if and only if C = C2,{k,j}.

Proof. Let C be an ideal of AlgL∞ and let A ∈ C. Since C ⊂ B, A ∈ B, i.e. A(2k−1,2k) = 0, A(2j+1,2j) = 0. Since C is an ideal, A(2k−1,k-1) = 0, A(2k,2k) = 0, A(2j,2j) = 0, A(2j+1,2j+1) = 0 by Theorem 3. So A ∈ C2,{k,j}. Hence C = C2,{k,j}. □

Theorem 12. Let k be a fixed natural number and let A be a subalgebra of AlgL∞ such that A0,{k} ⊂ A ⊂ A0. Then A is an ideal of AlgA∞ if and only if A = A0,{k} or A = A0.

Proof. Let A be an ideal of AlgL∞. It is sufficient to show that the case k = 1, i.e. if A0,{1} ⊂ A ⊂ A0, then A = A0,{1} or A = A0. Assume that A ≠ A0,{1}. Then there exists T = (tij) in A such that T ∉ A0,{1}. Then t12 ≠ 0 and tii = 0 for all i ∈ ℕ. Let A = (aij) be an element of A0. If a12 = 0, A ∈ A0,{1} ⊂ A. If a12 ≠ 0, let A1 be an operator defined by

Then A1 ∈ A0,{1} ⊂ A. Let T1 be an operator defined by

Then T1 ∈ A0,{1} ⊂ A. Let T2 = T + T1 ∈ A. Then T2(1,2) = t12 and T2(i,j) = 0 for (i, j) ≠ (1, 2). Let Then xT2 + A1 = A ∈ A. Hence A = A0. □

Theorem 13. Let Γ = {k1, k2} and let A be a subalgebra of AlgL∞ such that A0,Γ ⊂ A ⊂ A0,{ki} i = 1 or i = 2. Then A is an ideal of AlgA∞ if and only if A = A0,Γ or A = A0,{ki}.

Proof. Let A be an ideal of AlgL∞. It is sufficient to show that the case k1 = 1, k2 = 2, ki = 1. Suppose that A ≠ A0,{1,2}. Then there exists an element T = (tij) in A and T ∉ A0,{1,2}. Then t34 ≠ 0. Let A = (aij) be an element of A0,{1}. If a34 = 0, then A ∈ A0,{1,2} and so A ∈ A. Let a34 ≠ 0. Define an operator S = (sij) by

Then S ∈ A0,{1,2}. Put S + T = A1. Then A1 ∈ A. Put Then xA1 ∈ A. Define an operator B = (bij) by

Then B ∈ A0,{1,2} and hence B + xA1 = A ∈ A. So A = A0,{1}. □

Theorem 14. Let k be a fixed natural number and A be a subalgebra of AlgL∞ such that A2,{k} ⊂ A ⊂ A2,{k-1,2k}. Then A is an ideal of AlgL∞ if and only if A = A2,{k} or A = A{2k−1,2k}.

Proof. Let A be an ideal of AlgL∞. It is sufficient to show that the case k = 1. i.e. if A2,{1} ⊂ A ⊂ A{1,2}, then A = A2,{1} or A = A{1,2}. Assume that A ≠ A2,{1}. Then there exists T = (tij) in A such that T ∉ A2,{1}. Then t12 ≠ 0 and t11 = 0 and t22 = 0. Let A = (aij) be an element of A{1,2}. If a12 = 0, A ∈ A2,{1} ⊂ A. If a12 ≠ 0, let A1 be an operator defined by

Then A1 ∈ A2,{1} ⊂ A. Let T1 be an operator defined by

Then T1 ∈ A2,{1} ⊂ A. Let T2 = T + T1 ∈ A. Then T2(1,2) = t12 and T2(i,j) = 0 for (i, j) ≠ (1, 2). Let Then xT2+A1 = A ∈ A. Hence A = A{1,2}. □

If we modify the proof of Theorem 14, then we can prove the following Theorem.

Theorem 15. Let j be a fixed natural number and let A be a subalgebra of AlgL∞ such that B2,{j} ⊂ A ⊂ A{2j,2j+1}. Then A is an ideal of AlgL∞ if and only if A = B2,{j} or A = A{2j,2j+1}.

Theorem 16. Let Γ = {j1, j2} and let B be a subalgebra of AlgL∞ such that B0,Γ ⊂ B ⊂ B0,{ji} i = 1 or i = 2. Then B is an ideal of AlgA∞ if and only if B = B0,Γ or B = B0,{ji}.

Proof. Let B be an ideal of AlgL∞. It is sufficient to show that the case j1 = 1, j2 = 2 and ji = 2. Let B0,Γ ⊂ B ⊂ B0,{2}. Suppose that B ≠ B0,Γ. Then there exists an element T = (tij) in B and T ∉ B0,Γ. Then t32 ≠ 0. Let A = (aij) be an element of B0,{2}. If a32 = 0, then A ∈ B0,Γ and so A ∈ B. Let a32 ≠ 0. Define an operator B = (bij) by

Then B ∈ B0,Γ. Put B + T = A1. Then A1 ∈ B. Put Then xA1 ∈ B. Let S = (sij) be an operator defined by

Then S ∈ B0,Γ and hence S + xA1 = A ∈ B. So B = B0,{2}. □

If we repeat the proof of the Theorem 12 and the Theorem 13, then we can get the following theorem.

Theorem 17. Let k, j be natural numbers and Ω1,1 = { k, j }.

i) Let C be a subalgebra of AlgL∞ such that C0,Ω1,1 ⊂ C ⊂ C0,Ω1,, where C0,Ω1,1 = {T ∈ A0 | T(2k−1,2k) = 0 = T(2j+1,2j)} and C0,Ω1, = {T ∈ A0 | T(2k−1,2k) = 0 }. Then C is an ideal of AlgL∞ if and only if C = C0,Ω1,1 or C = C0,Ω1,.

ii) Let C be a subalgebra of AlgL∞ such that C0,Ω1,1 ⊂ C ⊂ C0,Ω,1. Then C is an ideal of AlgL∞ if and only if C = C0,Ω1,1 or C = C0,Ω,1.

Let Λ and Γ be nonempty subsets of N and A2,Γ ∩ AΛ will be denoted by A2,Γ,Λ. And we will prove only one case of relationships between ideals of A2,Γ,Λ. The other relations will be proved by the same way.

Theorem 18. Let Γ = {k1, k2} = {1, 2} and Λ = {3, 4}. Let A be a subalgebra of AlgL∞ such that A2,Γ ⊂ A ⊂ A2,{1},{3,4}. Then A is an ideal of AlgL∞ if and only if A = A2,Γ or A = A2,{1},{3,4}.

Proof. Let A be an ideal of AlgL∞ and A ≠ A2,Γ. Then there exists an element T = (tij) ∈ A and T = (tij) ∉ A2,Γ, i.e. t34 ≠ 0. Let A = (aij) be an element of A2,{1},{3,4}. If a34 = 0, then A ∈ A2,Γ and so A ∈ A. If a34 ≠ 0, we let define an operator S = (sij) by

Then S ∈ A2,Γ and so S ∈ A. We define an operator B = (bij) by

Then B ∈ A. Put D = S + T and Then A = B + xD ∈ A. Hence A = A2,{1},{1,4}. □

If we denote B2,Γ ∩ BΛ by B2,Γ,Λ and C2,Γ ∩ CΛ by C2,Γ,Λ, then we will prove relationships between ideals B2,Γ,Λ and C2,Γ,Λ by modifying the method of the proof of Theorem 15.

Theorem 19. Let ki be natural numbers such that Let Γ1 = {k1}, Γ2 = {k1, k2}, · · · , Γn = {k1, k2, · · · , kn} and Γ = {k1, k2, · · · }. Then

A0,Γ ⊂ · · · ⊂ A0,Γn ⊂ A0,Γn-1 ⊂ · · · ⊂ A0,Γ2 ⊂ A0,Γ1 = A0,{k1}

A2,Γ ⊂ · · · ⊂ A2,Γn ⊂ A2,Γn-1 ⊂ · · · ⊂ A2,Γ2 ⊂ A2,Γ1 = A2,{k1}

B0,Γ ⊂ · · · ⊂ B0,Γn ⊂ B0,Γn-1 ⊂ · · · ⊂ B0,Γ2 ⊂ B0,Γ1 = B0,{k1}

B2,Γ ⊂ · · · ⊂ B2,Γn ⊂ B2,Γn-1 ⊂ · · · ⊂ B2,Γ2 ⊂ B2,Γ1 = B2,{k1}