Abstract. We consider general bounded derivations on the Banach algebra of Hilbert-Schmidt operators on an underlying complex inﬁnite dimensional separable Hilbert space

ℋ

. Their structure is described by means of unique inﬁnite matrices. Certain classes of derivations are identiﬁed together in such a way that they correspond to a unique matrix derivation. In particular, Hadamard derivations, the action of general derivations on Hilbert-Schmidt and nuclear operators and questions about innerness are considered.

1. Introduction

Throughout this article

ℋ

will be a separable inﬁnitely dimensional complex Hilbert space. Let

ℬ (ℋ)

𝒦 (ℋ)

𝒮 (ℋ)

and

𝒩 (ℋ)

be the classes of bounded, compact, Hilbert-Schmidt and nuclear operators on

ℋ

respectively. As it is well known,

𝒦 (ℋ)

is the only non-trivial closed self-adjoint two-sided ideal in

ℬ (ℋ)

(cf. [5]). Furthermore, by the polar decomposition theorem any

A \in ℬ (ℋ)

can be written uniquely as

A = U \circ |A|

, where

U

is a partial isometry and

|A|

is a positive operator (see [8], Vol. 2, Th. 6.1.2, p. 401). Remember that if

A \in 𝒦 (ℋ)

and

\{ρ_{n}\}

is the sequence of eigenvalues of

|A|

then

A

is said to be a Hilbert-Schmidt operator (resp. a nuclear operator) if

\sum ρ_{n}^{2} < \infty

(resp. if

\sum ρ_{n} < \infty

). Moreover, an operator

A

is of Hilbert-Schmidt type if and only if the series

\sum {↑A f_{n}↑}^{2}

converges for at least one orthonormal basis

\{f_{n}\}

ℋ

. In that case it is readily seen that

\sum ρ_{n}^{2} = \sum {↑A g_{n}↑}^{2}

\{g_{n}\}

is any orthonormal basis of

ℋ

. So, if

\{e_{n}\}

is an orthonormal basis of

ℋ

and

A, B \in 𝒮 (ℋ)

then

deﬁnes an inner product on

𝒮 (ℋ)

. If

{↑A↑}_{2} = {〈A, A〉}_{2}^{1 ∕ 2}

then

(𝒮 (ℋ), {〈\circ, \circ〉}_{2})

becomes a Hilbert space. Indeed,

(𝒮 (ℋ), {↑\circ↑}_{2})

is a Banach

*

- algebra without unit. Analogously, if

A \in 𝒩 (ℋ)

and

\{e_{n}\}

is an orthonormal basis of

ℋ

then

\sum 〈A e_{n}, e_{n}〉 = \sum ρ_{n},

i.e. the sum of the former series does not depend on the choice of

\{e_{n}\} .

This value is known as the trace of

A

and it is denote as

tr (A) .

Further, if we let

{↑A↑}_{1} = tr (|A|)

then

(𝒩 (ℋ), {↑\circ↑}_{1})

is a Banach algebra. If

A \in ℬ (ℋ)

then

A \in 𝒩 (ℋ)

if and only if

{|A|}^{1 ∕ 2}

\in 𝒮 (ℋ)

. For more details on this matter the reader can see [7], Ch. I.

§

2. In this article, by a derivation on a Banach algebra

𝔘

we will mean any linear operator

δ : 𝔘 \to 𝔘

so that the usual Leibnitz rule

δ (a \cdot b) = δ (a) \cdot b + a \cdot δ (b)

holds for all

a, b \in 𝔘

. In particular, given

a \in 𝔘

the maps

[a, b] = a \cdot b - b \cdot a

deﬁned for all

b \in 𝔘

are the so called inner derivations. We will say that a derivation is outer if it is not inner.

The authors previously studied the existence and structure of general derivations in the frame of weight sequence Banach algebras (cf. [1]). Our matter in this article is to consider questions concerning to innerness of bounded derivations on

𝒮 (ℋ)

. This problem has been solved in other contexts; for instance in the frame of von Neumann algebras every bounded derivation is inner (cf. [10], [11]). In Section 2 we consider the structure of general (bounded) derivations on

𝒮 (ℋ) .

We develop, in Th. 2 and Prop. 5, the intrinsic relationship between bounded derivations on

𝒮 (ℋ)

and derivations on a Hilbert space of inﬁnite complex matrices (cf. [2], [3]). The precise structure of derivations on

𝒮 (ℋ)

is given in Prop. 7 and Prop. 8 allows us to deﬁne an equivalence relation on them. So, in Corollary 11 we see how each inﬁnite matrix derivation determines a unique equivalence class of bounded derivations on

𝒮 (ℋ) .

In particular, from this development it follows the simple structure of the so called Hadamard derivations. Finally, in Section 3 we describe the action of bounded derivations on self-adjoint Hilbert-Schmidt operators. Hadamard derivations and their restrictions to nuclear operators on

ℋ

are considered in Prop. 15. In Prop. 16 we realize certain derivations on

𝒮 (ℋ)

, in general not inner, as certain series of inner derivations on

𝒮 (ℋ)

Notation 1. Throughout this article $ω$ will be an inﬁnite countable set and, if $𝔸$ is a Banach algebra, $𝒟 (𝔸)$ will denote the class of bounded derivations on $𝔸$ . Let $l^{2} (ω \times ω)$ be the Hilbert space of inﬁnite matrices $a = {(a_{n, m})}_{n, m \in ω}$ endowed with the norm ${↑a↑}_{2} = {(\sum_{n, m \in ω} {|a_{n, m}|}^{2})}^{1 ∕ 2}$ . We will write by means of $\bar{a}$ , $a^{t}$ and $a^{*}$ the conjugate, the transpose and the adjoint of $a$ respectively. If $m, n \in ω$ it is easy to see that $a_{n, m}^{*} = \bar{a_{m, n}}$ , as usual $a_{n, m}^{t} = a_{m, n}$ and $\bar{a} = a^{* t} .$

2. On the structure of general derivations on

𝒮 (ℋ)

Theorem 2. (cf. [2], [4]) A bounded linear endomorphism $Δ$ of $l^{2} (ω \times ω)$ is a derivation if and only if there are matrices $α = {\{α_{n, m}\}}_{n, m \in ω}$ and $β = {\{β_{n, m}\}}_{n, m \in ω}$ of complex numbers uniquely determined so that

For any $n \in ω,$ $α_{n, n} = 0 .$
$sup_{n, m \in ω} |β_{n, m}| < \infty .$
The matrix $α$ is nearly-inner, i.e. the formal mapping $ℒ_{α} : z \to α \cdot z - z \cdot α$ deﬁnes a bounded linear operator on $l^{2} (ω \times ω) .$
For any $n, m, p \in ℕ$ the identities $β_{n, m} + β_{m, p} = β_{n, p}$ hold.

Then

Δ (z) = \sum_{n, m \in ω} (\sum_{p \in ω} (z_{p, m} α_{n, p} - α_{p, m} z_{n, p}) + z_{n, m} β_{n, m}) e_{n, m}

(1)

for $z \in l^{2} (ω \times ω)$ , where for all $n, m \in ω$ we are writing $e_{n, m} = {(δ_{n, m}^{p, q})}_{p, q \in ω}$ and $δ_{n, m}^{p, q}$ denotes the usual Kroneckerˊs function. In particular, we can denote $Δ = Δ_{α, β} .$

Remark 3. Let $Δ \in 𝒟 (l^{2} (ω \times ω))$ . So, the corresponding entries of the matrices $α$ and $β$ related to $Δ$ according to Th.2 are deﬁned by the relations

Δ (e_{n, m}) = \sum_{p \in ω} (α_{p, n} e_{p, m} - α_{m, p} e_{n, p}) + β_{n, m} e_{n, m}, n, m \in ω .

In particular, observe that

{sup}_{n, m \in ω} \sum_{p \in ω} ({|α_{p, n}|}^{2} + {|α_{m, p}|}^{2}) + {|β_{n, m}|}^{2} \leq {↑Δ↑}^{2} < \infty,

i.e. all rows and columns of a nearly-inner matrix are bounded and square summable.

Proposition 4. Let $α, α_{1}, α_{2}$ be nearly-inner matrices with null diagonals and let $β, β_{1}, β_{2}$ be bounded matrices that verify condition (iv) of Th.2. So, the following formulae hold:

$[Δ_{α_{1}, 0}, Δ_{α_{2}, 0}] = Δ_{α (α_{1}, α_{2}), β (α_{1}, α_{2})}$ , where for each $n, m \in ω$
$\begin{matrix} \begin{aligned} α {(α_{1}, α_{2})}_{n, m} & = (1 - δ_{n, m}) \cdot {[α_{1}, α_{2}]}_{n, m}, \\ β {(α_{1}, α_{2})}_{n, m} & = {[α_{1}, α_{2}]}_{n, n} + {[α_{2}, α_{1}]}_{m, m} . \end{aligned} \end{matrix}$ (2)
$[Δ_{α, 0}, Δ_{0, β}] = - Δ_{α ⊙ β, 0},$ where $α ⊙ β = {\{α_{n, m} \cdot β_{n, m}\}}_{n, m \in ω}$ denotes the Hadamard product of the matrices $α$ and $β .$
$[Δ_{0, β_{1}}, Δ_{0, β_{2}}] = 0 .$
$[Δ_{α_{1}, β_{1}}, Δ_{α_{2}, β_{2}}] = Δ_{α (α_{1}, α_{2}) - α_{1} ⊙ β_{2} + α_{2} ⊙ β_{1}, β (α_{1}, α_{2})} .$

Proof. Given $n, m \in ω$ we obtain that

\begin{matrix} [Δ_{α_{1}, 0}, Δ_{α_{2}, 0}] (e_{n, m}) \\ = \sum_{p \in ω} (δ_{n}^{p} {[α_{1}, α_{2}]}_{p, n} \cdot e_{p, m} + δ_{m}^{p} {[α_{2}, α_{1}]}_{m, p} \cdot e_{n, p}) \\ + ({[α_{1}, α_{2}]}_{n, n} + {[α_{2}, α_{1}]}_{m, m}) \cdot e_{n, m} . & (3) \end{matrix}

By Remark 3 the inﬁnite matrix ${\{{[α_{1}, α_{2}]}_{n, m}\}}_{n, m \in ω}$ is deﬁned. Hence, relations (2) will be established if we show that $α (α_{1}, α_{2})$ and $β (α_{1}, α_{2})$ satisfy the conditions of Th.2. For, by Remark 3 and the deﬁnitions in (2) they do for $β (α_{1}, α_{2}) .$ By deﬁnition $α (α_{1}, α_{2})$ has null diagonal. In order to see that $α (α_{1}, α_{2})$ is nearly-inner let $z \in l^{2} (ω \times ω)$ be a matrix with only a ﬁnite number of non zero entries. If $n, m \in ω$ we have

\begin{matrix} (ℒ_{α_{1}} \circ ℒ_{α_{2}} - ℒ_{α_{2}} \circ ℒ_{α_{1}}) {(z)}_{n, m} \\ = ℒ_{α (α_{1}, α_{2})} {(z)}_{n, m} + z_{n, m} \cdot β {(α_{1}, α_{2})}_{n, m} . & (4) \end{matrix}

In consequence, the formal operator $ℒ_{α (α_{1}, α_{2})}$ is deﬁned and obviously linear on the dense subspace of matrices with ﬁnite support of $l^{2} (ω \times ω) .$ Since $α_{1}$ and $α_{2}$ are nearly-inner matrices by (4) the restriction of $ℒ_{α (α_{1}, α_{2})}$ to this subspace is continuous. Thus by completeness it extends to a unique bounded operator on $l^{2} (ω \times ω),$ i.e. the matrix $α (α_{1}, α_{2})$ is nearly-inner and (i) follows. Since $ℒ_{α ⊙ β} = - [ℒ_{α}, Δ_{0, β}]$ and $ℒ_{α}, Δ_{0, β} \in ℬ (l^{2} (ω \times ω))$ then $ℒ_{α ⊙ β}$ is also bounded. So $α ⊙ β$ is a nearly-inner matrix and as it has null diagonal (ii) holds. Now, the proofs of (iii) and (iv) are straightforward. __

Proposition 5. Let $e = {\{e_{n}\}}_{n \in ω}$ , $f = {\{f_{m}\}}_{m \in ω}$ be orthonormal basis of $ℋ$ , let $U \in ℬ (ℋ)$ be the unitary operator so that $U e_{n} = f_{n}$ if $n \in ω$ and let $A \in 𝒮 (ℋ) .$

Proof. Since

\sum_{n, m \in ω} {|〈A e_{n}, f_{m}〉|}^{2} = \sum_{n \in ω} {↑A e_{n}↑}^{2} = {↑A↑}_{2}^{2} < \infty

then $𝔍_{e, f}$ is well deﬁned and it is clearly an $*$ - isometry from $𝒮 (ℋ)$ into $l^{2} (ω \times ω) .$ If $a = {(a_{n, m})}_{n, m \in ω}$ belongs to $l^{2} (ω \times ω)$ it is easy to see that

h \to A h = \sum_{m} f_{m} \sum_{n} a_{n, m} 〈h, e_{n}〉

deﬁnes a Hilbert - Schmidt operator on $ℋ$ so that $𝔍_{e, f} (A) = a .$ Hence by the open mapping theorem (i) holds. For (ii) it suﬃces to observe that for all $n, m \in ω$ is

〈A e_{n}, f_{m}〉 = 〈A e_{n}, U e_{m}〉 = 〈A U^{*} f_{n}, f_{m}〉 .

Moreover,

\begin{array}{l} 𝔖_{e, f} {(A)}_{n, m}^{*} & = \bar{𝔖_{e, f} {(A)}_{m, n}} = 〈U A^{*} e_{m}, f_{n}〉 \\ = 〈e_{m}, A e_{n}〉 = \bar{〈U A e_{n}, f_{m}〉} = \bar{𝔍_{e, f} {(U A)}_{n, m}} \\ = 𝔖_{e, f} {(A^{*})}_{n, m} \end{array}

It is clear that $𝔖_{e, f}$ is linear and as ${↑U A^{*}↑}_{2} = {↑A↑}_{2}$ if $A \in 𝒮 (ℋ)$ by (i) $𝔖_{e, f}$ becomes an isometry. On the other hand, if $a \in l^{2} (ω \times ω)$ it is easily seeing that $𝔖_{e, f}^{- 1} (a) = 𝔍_{e, f}^{- 1} {(\bar{a})}^{*} \circ U$ and $𝔖_{e, f}^{- 1}$ becomes also isometric. Finally, if $A, B \in 𝒮 (ℋ)$ and $m, n \in ω$ we have

\begin{matrix} {(𝔖_{e, f} (A) \cdot 𝔖_{e, f} (B))}_{n, m} = \sum_{p \in ω} 𝔖_{e, f} {(A)}_{n, p} \cdot 𝔖_{e, f} {(B)}_{p, m} \\ = \sum_{p \in ω} 〈f_{p}, U A^{*} e_{n}〉 \cdot 〈f_{m}, U B^{*} e_{p}〉 = \sum_{p \in ω} 〈A e_{p}, e_{n}〉 \cdot 〈B e_{m}, e_{p}〉 \\ = 〈\sum_{p \in ω} 〈B e_{m}, e_{p}〉 A e_{p}, e_{n}〉 = 〈A B e_{m}, e_{n}〉 = 〈U^{*} f_{m}, {(A B)}^{*} e_{n}〉 \\ = \bar{〈U {(A B)}^{*} e_{n}, f_{m}〉} = 𝔖_{e, f} {(A B)}_{n, m} \end{matrix}

and (iii) follows. __

Remark 6. In what follows, if $a, b \in ℋ$ we will write $a^{*} \otimes b$ to denote the vector map

(a^{*} \otimes b) (h) = 〈h, a〉 \cdot b, (h \in ℋ) .

It is easy to see that

$a^{*} \otimes b$ is a Hilbert-Schmidt if $a, b \in ℋ$ since it is a ﬁnite rank operator.
$〈A a, b〉 = {〈A, a^{*} \otimes b〉}_{2}$ if $a, b \in ℋ$ and $A \in 𝒮 (ℋ) .$
${(a_{1} + a_{2})}^{*} \otimes b = a_{1}^{*} \otimes b + a_{2}^{*} \otimes b$ if $a_{1}, a_{2}, b \in ℋ$ $.$
$a^{*} \otimes (b_{1} + b_{2}) = a^{*} \otimes b_{1} + a^{*} \otimes b_{2}$ if $a, b_{1}, b_{2} \in ℋ .$
${(λ a)}^{*} \otimes b = \bar{λ} (a^{*} \otimes b) = a^{*} \otimes (\bar{λ} b)$ if $a, b \in ℋ$ and $λ \in ℂ .$
${(a^{*} \otimes b)}^{*} = b^{*} \otimes a$ if $a, b \in ℋ$ .
If $e = {\{e_{n}\}}_{n \in ω}$ and $f = {\{f_{m}\}}_{m \in ω}$ are orthonormal basis of $ℋ$ the set ${\{e_{n}^{*} \otimes f_{m}\}}_{n, m \in ω}$ becomes orthonormal basis of $𝒮 (ℋ) .$ In fact, the class of ﬁnite rank operators is dense in $𝒮 (ℋ)$ , (cf. [7], p. 36).

Proposition 7. Let $δ \in 𝒟 (𝒮 (ℋ))$ , $A \in 𝒮 (ℋ)$ and let $e = {\{e_{n}\}}_{n \in ω}$ be an orthonormal basis of $ℋ$ $.$ There exist a unique set of bounded linear forms ${\{γ_{e}^{n, m}\}}_{n, m \in ω}$ on $𝒮 (ℋ)$ so that $δ (A)$ can be written in $𝒮 (ℋ)$ as

δ (A) = \sum_{m, n \in ω} γ_{e}^{n, m} (A) e_{n}^{*} \otimes e_{m} .

(5)

Further, there exist unique matrices $α$ and $β$ as in Th. 2 so that for each $n, m \in ω$ is $γ_{e}^{n, m} (A) = {〈A, B_{e}^{n, m}〉}_{2}$ ,

B_{e}^{n, m} = e_{n}^{*} \otimes [e ({\bar{α}}_{m}) + {\bar{β}}_{m, n} \cdot e_{m}] - e {(α^{n})}^{*} \otimes e_{m},

$e ({\bar{α}}_{m}) = \sum_{p \in ω} \bar{α_{m, p}} \cdot e_{p}$ and $α^{n} = \sum_{p \in ω} α_{p, n} \cdot e_{p} .$

Proof. Given two orthonormal basis $e = {\{e_{n}\}}_{n \in ω}$ , $f = {\{f_{m}\}}_{m \in ω}$ of $ℋ$ , by (iii) of Prop. 5 there is a 1-1 correspondence

Ψ_{e, f} : 𝒟 (l^{2} (ω \times ω)) \to 𝒟 (𝒮 (ℋ)), Ψ_{e, f} (Δ) = 𝔖_{e, f}^{- 1} \circ Δ \circ 𝔖_{e, f} .

(6)

So, if $δ \in 𝒟 (𝒮 (ℋ))$ there are unique matrices $α = {(α_{m, n})}_{m, n \in ω}, β = {(β_{m, n})}_{m, n \in ω}$ in the conditions of Th. 2 so that $𝔖_{e, f} \circ δ \circ 𝔖_{e, f}^{- 1} = Δ_{α, β} .$ Now, with the above notation if $A \in 𝒮 (ℋ)$ then

\begin{matrix} δ (A) = (𝔖_{e, f}^{- 1} \circ Δ) (\bar{𝔍_{e, f} (U A^{*})}) \\ = \sum_{m, n \in ω} (\sum_{p \in ω} (〈A e_{n}, e_{p}〉 α_{m, p} - α_{p, n} 〈A e_{p}, e_{m}〉) + 〈A e_{n}, e_{m}〉 β_{m, n}) \\ \cdot 𝔖_{e, f}^{- 1} (e_{m, n}) & (7) \end{matrix}

\begin{matrix} = \sum_{m, n \in ω} (〈A e_{n}, e ({\bar{α}}_{m})〉 - 〈A e (α^{n}), e_{m}〉 + 〈A e_{n}, e_{m}〉 β_{m, n}) \cdot e_{n}^{*} \otimes e_{m} \\ = \sum_{m, n \in ω} {〈A, B_{e}^{n, m}〉}_{2} \cdot e_{n}^{*} \otimes e_{m} \end{matrix}

and so (5) follows. __

Proposition 8. Let $e = {\{e_{n}\}}_{n \in ω}, f = {\{f_{m}\}}_{m \in ω}, g = {\{g_{n}\}}_{n \in ω}, h = {\{h_{m}\}}_{m \in ω}$ be orthonormal basis of $ℋ$ , $Δ \in 𝒟 (l^{2} (ω \times ω))$ , $A \in 𝒮 (ℋ) .$ Then

Ψ_{e, f} (Δ) (V^{*} A U) = V^{*} Ψ_{g, h} (Δ) (A) U,

(8)

where $U, V \in ℬ (ℋ)$ are the unitary operators so that $U e_{n} = g_{n}$ and $V f_{m} = h_{m}$ if $n, m \in ω .$

Notation 9. Let $δ_{1}, δ_{2} \in 𝒟 (𝒮 (ℋ)) .$ We will write $δ_{1} \sim δ_{2}$ if and only if there are unitary operators $U, V$ on $ℋ$ so that $δ_{1} (V^{*} A U) = V^{*} δ_{2} (A) U$ if $A \in 𝒮 (ℋ) .$ It is readily seeing that $\sim$ deﬁnes an equivalence relation on $𝒟 (𝒮 (ℋ)) .$

Corollary 10. Given $δ_{1}, δ_{2} \in 𝒟 (𝒮 (ℋ))$ , $δ_{1} \sim δ_{2}$ if and only if for all orthonormal basis $e = {\{e_{n}\}}_{n \in ω}, f = {\{f_{m}\}}_{m \in ω}$ of $ℋ$ there are orthonormal basis $g = {\{g_{n}\}}_{n \in ω}, h = {\{h_{m}\}}_{m \in ω}$ of $ℋ$ so that $Ψ_{e, f}^{- 1} (δ_{1}) = Ψ_{g, h}^{- 1} (δ_{2}) .$

Proof.

$(\Rightarrow)$ :: Let $U, V$ unitary operators on $ℋ$ so that $δ_{1} (V^{*} A U) = V^{*} δ_{2} (A) U$ for all $A \in 𝒮 (ℋ) .$ Given orthonormal basis $e = {\{e_{n}\}}_{n \in ω}, f = {\{f_{m}\}}_{m \in ω}$ of $ℋ$ there exists $Δ \in 𝒟 (l^{2} (ω \times ω))$ so that $Ψ_{e, f} (Δ) = δ_{1}$ . If we write $U e_{n} = g_{n}$ and $V f_{m} = h_{m}$ if $n, m \in ω$ then $g = {\{g_{n}\}}_{n \in ω}, h = {\{h_{m}\}}_{m \in ω}$ are orthonormal basis of $ℋ$ . So, by Prop. 8 we have $Ψ_{g, h} (Δ) = δ_{2}$ and the condition is necessary.
$(\Leftarrow)$ :: Let $e = {\{e_{n}\}}_{n \in ω},$ $g = {\{g_{n}\}}_{n \in ω}, h = {\{h_{m}\}}_{m \in ω}$ be ﬁxed orthonormal basis of $ℋ$ so that $Ψ_{e, e}^{- 1} (δ_{1})$ and $Ψ_{g, h}^{- 1} (δ_{2})$ determine a same element $Δ$ in $𝒟 (l^{2} (ω \times ω)) .$ Hence $δ_{1} \sim δ_{2}$ since by Prop. 8 there are unitary operators $U, V \in ℬ (ℋ)$ so that for all $A \in 𝒮 (ℋ)$ is $δ_{1} (V^{*} A U) = Ψ_{e, e} (Δ) (V^{*} A U) = V^{*} Ψ_{g, h} (Δ) (A) U = V^{*} Ψ_{g, h} (Δ) (A) U .$

Corollary 11. There is a 1-1 correspondence between $𝒟 (𝒮 (ℋ)) ∕ \sim$ onto $𝒟 (l^{2} (ω \times ω)),$ i.e. $𝒟 (𝒮 (ℋ)) ∕ \sim$ $\equiv 𝒟 (l^{2} (ω \times ω)) .$

Proof. Let us write

Ψ : 𝒟 (l^{2} (ω \times ω)) \to 𝒟 (𝒮 (ℋ)) ∕ \sim, Ψ (Δ) = {[Ψ_{e, f} (Δ)]}^{\sim},

where $Δ \in 𝒟 (l^{2} (ω \times ω)),$ $e$ and $f$ are orthonormal basis of $ℋ$ and ${[Ψ_{e, f} (Δ)]}^{\sim}$ denotes the $\sim$ equivalence class of $Ψ_{e, f} (Δ)$ in $𝒟 (𝒮 (ℋ)) ∕ \sim .$ By Prop. 8 the function $Ψ$ is well deﬁned, i.e. $Ψ (Δ)$ does not depend on the choice of the orthonormal basis $e$ nor $f$ . If $Δ_{1}, Δ_{2} \in 𝒟 (l^{2} (ω \times ω))$ and $Ψ (Δ_{1}) = Ψ (Δ_{2})$ then $Ψ_{e, e} (Δ_{1}) = Ψ_{e, e} (Δ_{2}) .$ By Prop. 8 we have $Δ_{1} = Δ_{2},$ i.e. $Ψ$ is injective. We have already seen that for any orthonormal basis $e$ and $f$ of $ℋ$ the map $Ψ_{e, f},$ introduced in (6), deﬁnes a bijection between $𝒟 (l^{2} (ω \times ω))$ and $𝒟 (𝒮 (ℋ)) .$ Therefore $Ψ$ is also surjective. __

Notation 12. From now on we’ll denote any bounded derivation $δ$ on $𝒟 (𝒮 (ℋ))$ as $δ = δ_{α, β},$ where $α, β$ are the unique inﬁnite matrices determined by $δ$ as we have pointed out in Prop. 7. This matrices are uniquely by means of Th. 2 and they identify the corresponding $\sim$ equivalence class of $δ$ as in Corollary 11. So, $δ$ is intrinsically determined by these matrices while its coordinate representation changes according to the rules of Prop. 4 and (8) of Prop. 8. In particular, we shall say that any bounded derivation $δ_{0, β}$ on $𝒮 (ℋ)$ is a Hadamard derivation.

Remark 13. Any Hadamard derivation is induced by bounded sequences of complex numbers. For, if ${\{b_{n}\}}_{n \in ω}$ is such a sequence and

b = {\{b_{n} - b_{m}\}}_{n, m \in ω},

then $δ_{o, b}$ is a Hadamard derivation. On the other hand, given $δ_{0, β}$ we already know that $β_{n, m} = β_{n, l} + β_{l, m}$ for all $n, l, m \in ω .$ In consequence $β$ has null diagonal and $β_{n, m} = β_{n, 1} + β_{1, m} = - β_{1, n} + β_{1, m}$ , i.e. the ﬁrst row determines the whole $β .$ Of course, although $β$ deﬁnes $δ_{0, β}$ uniquely it does not give rise to a unique bounded sequence.

3. On certain particular derivations

Proposition 14. Let $δ_{α, β} \in 𝒟 (𝒮 (ℋ))$ and $A$ be a self-adjoint Hilbert-Schmidt operator on $ℋ$ . If $e = {\{e_{n}\}}_{n \in ω}$ is the orthonormal basis of $ℋ$ induced by the sequence ${\{ρ_{n}\}}_{n \in ω}$ of eigenvalues of $A$ then

\begin{matrix} δ_{α, β} (A) = \sum_{n \in ω} e_{n}^{*} \otimes (A - ρ_{n} \cdot I d_{ℋ}) e (α_{n}) \\ = - \sum_{n \in ω} {((A - ρ_{n} \cdot I d_{ℋ}) e ({\bar{α}}^{n}))}^{*} \otimes e_{n} \end{matrix}

Proof. If $n \in ω$ then

\begin{align} δ_{α, β} (e_{n}^{*} \otimes e_{n}) & = Ψ_{e, e} (Δ_{α, β}) (e_{n}^{*} \otimes e_{n}) & (9) \\ = 𝔖_{e, e}^{- 1} (Δ_{α, β} (e_{n, n})) = \sum_{m \in ω} (α_{m, n} e_{n}^{*} \otimes e_{m} - α_{n, m} e_{m}^{*} \otimes e_{n}) \end{align}

Since $A = \sum_{n \in ω} ρ_{n} e_{n}^{*} \otimes e_{n}$ in $𝒮 (ℋ)$ by (9) we obtain that

δ_{α, β} (A) = \sum_{n, m \in ω} α_{n, m} \cdot (ρ_{m} - ρ_{n}) e_{n}^{*} \otimes e_{m} .

(10)

Now from (10) our claim follows easily. __

Proposition 15.

Any Hadamard derivation on $𝒮 (ℋ)$ is the restriction of an inner one on $ℬ (ℋ) .$
The restriction of any Hadamard derivation to $𝒩 (ℋ)$ belongs to $𝒟 (𝒩 (ℋ))$ .

Proof.

Let $δ_{0, β}$ be a Hadamard derivation and let $e = {\{e_{n}\}}_{n \in ω}$ be an orthonormal basis in $ℋ$ . By Remark (13) we can assume that $β = {\{β_{m} - β_{n}\}}_{m, n \in ω}$ , where ${\{β_{n}\}}_{n \in ω}$ is a bounded sequence in $ℂ$ . Let us prove that if $A \in 𝒮 (ℋ)$ then
$δ_{0, β} (A) = [\sum_{p \in ω} {({\bar{β}}_{p} \cdot e_{p})}^{*} \otimes e_{p}, A] .$ (11)

For, if $n, m \in ω$ then
$\begin{array}{l} [\sum_{p \in ω} {({\bar{β}}_{p} \cdot e_{p})}^{*} \otimes e_{p}, e_{n}^{*} \otimes e_{m}] & = (β_{m} - β_{n}) \cdot (e_{n}^{*} \otimes e_{m}) \\ = \sum_{p, q \in ω} γ_{e}^{p, q} (e_{n}^{*} \otimes e_{m}) e_{p}^{*} \otimes e_{q} \\ = δ_{0, β} (e_{n}^{*} \otimes e_{m}), \end{array}$
i.e. (11) holds by Prop.7. Indeed, $δ_{0, β} (\circ) = [𝔥 (β), \circ],$ where $𝔥 (β) \in ℬ (ℋ)$ is the diagonal operator $𝔥 (β) = \sum_{p \in ω} β_{p} \cdot (e_{p}^{*} \otimes e_{p}) .$
With the notation of (i), since $𝒩 (ℋ)$ is an ideal our claim is clear if $𝔥 (β) \in 𝒩 (ℋ)$ (for instance, if ${\{β_{n}\}}_{n \in ω} \in l^{1} (ω)$ ). For the general case, let $A \in 𝒩 (ℋ) .$ Actually, let $A = U \circ |A|$ be the polar decomposition of $A$ and ${\{ρ_{n}\}}_{n \in ω}$ the sequence of eigenvalues of $|A|$ associated to an orthonormal basis $e = {\{e_{n}\}}_{n \in ω} .$ Since $|A| = \sum_{n \in ω} ρ_{n} \cdot (e_{n}^{*} \otimes e_{n})$ in $𝒮 (ℋ)$ , we get $δ_{0, β} (|A|) = \sum_{n \in ω} ρ_{n} \cdot δ_{0, β} (e_{n}^{*} \otimes e_{n}) = 0$
as follows by Prop. 14. Consequently we see that
$\begin{matrix} δ_{0, β} (A) = [𝔥 (β), U \circ |A|] \\ = [𝔥 (β), U] \circ |A| + U \circ [𝔥 (β), |A|] = [𝔥 (β), U] \circ |A| . & (12) \end{matrix}$
But for all $S, T \in ℬ (ℋ)$ we have ${↑S A T↑}_{1} \leq ↑S↑ {↑A↑}_{1} ↑T↑$ (cf. [9], $§$ 3.34 (xi), p. 133). Accordingly we obtain
${↑δ_{0, β} (A)↑}_{1} \leq ↑[𝔥 (β), U]↑ \cdot {↑A↑}_{1} \leq 2 ↑𝔥 (β)↑ \cdot {↑A↑}_{1}$
and $↑𝔥 (β)↑ =$ ${sup}_{n \in ω} |β_{n}| < \infty$ , i.e. $δ_{0, β} ∣_{𝒩 (ℋ)}$ $\in 𝒟 (𝒩 (ℋ))$ .

Proposition 16. Let $α$ be a nearly-inner matrix with null diagonal. Given an orthonormal basis $e = {\{e_{n}\}}_{n \in ω}$ of $ℋ$ and $A \in 𝒮 (ℋ)$ then

δ_{α, 0} (A) = \sum_{p \in ω} [e_{p}^{*} \otimes e (α^{p}), A] .

(13)

Proof. By Prop. 7, (7) we have

δ_{α, 0} (A) = \sum_{n, m \in ω} \sum_{p \in ω} (〈A e_{n}, e_{p}〉 α_{m, p} - α_{p, n} 〈A e_{p}, e_{m}〉) \cdot e_{n}^{*} \otimes e_{m} .

(14)

Hence, if $A$ belongs to the ﬁnite linear span of ${\{e_{n}^{*} \otimes e_{m}\}}_{n, m \in ω}$ by (14) we obtain that

δ_{α, 0} (A) = \sum_{p \in ω} \{{(A^{*} e_{p})}^{*} \otimes e (α^{p}) - e_{p}^{*} \otimes A e (α^{p})\} .

(15)

Since ${\{e_{n}^{*} \otimes e_{m}\}}_{n, m \in ω}$ is a basis of the complete space $𝒮 (ℋ)$ and $δ_{α, 0} \in ℬ (𝒮 (ℋ))$ then (15) holds for all $A \in 𝒮 (ℋ)$ . Now, it is straightforward to see that each summand in (15) deﬁnes a bounded derivation on $𝒮 (ℋ)$ . Indeed, those derivations are all inner, if $p \in ω$ is

{(A^{*} e_{p})}^{*} \otimes e (α^{p}) - e_{p}^{*} \otimes A e (α^{p}) = [e_{p}^{*} \otimes e (α^{p}), A]

and we get (13). __

Example 17. Let $ℋ = l^{2} (ℕ)$ and let $α$ be the nearly-inner matrix so that $α_{n, m}$ is $0$ or $1$ , according as $m \neq n + 1$ or $m = n + 1$ respectively. On identifying $e_{0}^{*}$ with the zero form on $ℋ$ then $α$ induces the bounded derivation

δ_{α, 0} (A) = \sum_{n \geq 1, m \geq 1} {〈A, e_{n}^{*} \otimes e_{m + 1} - e_{n - 1}^{*} \otimes e_{m}〉}_{2} \cdot e_{n}^{*} \otimes e_{m}

deﬁned for $A \in 𝒮 (ℋ)$ . Let us assume that $δ_{α, 0}$ is inner, say $δ_{α, 0} (\circ) = [C, \circ]$ for same $C \in 𝒮 (ℋ)$ . So, if $r, s \in ℕ$ we get

[C, e_{r}^{*} \otimes e_{s}] = e_{r}^{*} \otimes C e_{s} - {(C^{*} e_{r})}^{*} \otimes e_{s}

(16)

and

δ_{α, 0} (e_{r}^{*} \otimes e_{s}) = \{\begin{matrix} - e_{r + 1}^{*} \otimes e_{1} & i f & s = 1, \\ e_{r}^{*} \otimes e_{s - 1} - e_{r + 1}^{*} \otimes e_{s} & i f & s > 1 . \end{matrix}

(17)

From (16) and (17), if $r \geq 1$ and $s > 1$ we deduce that

[C, e_{r}^{*} \otimes e_{s}] (e_{r}) = C e_{s} - 〈C e_{r}, e_{r}〉 \cdot e_{s} = e_{s - 1} .

Hence if $r \neq s$ is $〈C e_{s}, e_{r}〉 = δ_{r}^{s - 1}$ . Thus ${\{〈C e_{s}, e_{r}〉\}}_{s, r \in ℕ}$ becomes not square summable, which contradicts Prop. 5. So $δ_{α, 0}$ is not inner, although it is realized as a generalized series of inner derivations as stated in Prop. 16.

References

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[2] Barrenechea, A. L. & Peña, C. C.: Examples of bounded derivations on some none algebras. Actas del VII Congreso Dr. A. Monteiro, 23 – 25, (2004).

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[8] Kadison, R. V. & Ringrose, J. R.: Fundamentals of the theory of operator algebras. Volumes 1 and 2. Graduate Studies in Math., 15/16, AMS, (1997).

[9] Peña, C.: Sobre análisis funcional, variable compleja, topología, álgebra y teoría de la medida. Publicaciones Electrónicas de la Sociedad Matemática Mejicana. Serie Textos, Vol. 3, 1 – 337, Méjico. Copyright (C), (2003). Free Software Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA02111-1307, USA, (http://www.smm.org.mx/SMMP).

[10] Sakai, S: Derivations on simple C $^{*}$ algebras. J. Funct. Anal., 2, 202 - 206, (1968).

[11] Sakai, S.: Derivations on simple C $^{*}$ algebras, II. Bull. Soc. Math. France, 99, 259 - 263, (1971).

UNCentro – FCExactas – NuCoMPA – Dpto. de MatemÁticas. Tandil, Pcia, de Buenos Aires, Argentina

1. Introduction

2. On the structure of general derivations on 𝒮(ℋ)

3. On certain particular derivations

References

2. On the structure of general derivations on $𝒮 (ℋ)$