Nilpotent elements of operator ideals as single commutators

TL;DR

This paper proves that all nilpotent elements within any operator ideal can be expressed as single commutators of operators from a related ideal, with the exponent depending on the nilpotency degree.

Contribution

It establishes a general result linking nilpotent elements of operator ideals to single commutators, extending previous understanding in operator theory.

Findings

Nilpotent elements are single commutators within operator ideals.

The exponent t depends on the nilpotency degree.

The result applies to arbitrary operator ideals.

Abstract

For an arbitrary operator ideal I, every nilpotent element of I is a single commutator of operators from I^t, for an exponent t that depends on the degree of nilpotency.

Equations

$$\Vert y\xi {\Vert }^2=⟨y^{\ast }y\xi ,\xi ⟩\le ⟨x^{\ast }x\xi ,\xi ⟩=\Vert x\xi {\Vert }^2.$$

$$x\xi \mapsto y\xi .$$

$$B({\mathcal{H}}^{\oplus n})=M_n(B(\mathcal{H}))=B(\mathcal{H})\otimes M_n(\mathbf{C})$$

$${\mathcal{V}}_1$$

$${\mathcal{V}}_k$$

$${\mathcal{W}}_1\subseteq {\mathcal{W}}_2\subseteq \cdots \subseteq {\mathcal{W}}_n=\mathcal{H}$$

$${\mathcal{W}}_k\subseteq \ker A^k$$

$$\dim ({\mathcal{W}}_k\ominus {\mathcal{W}}_{k-1})=\dim \mathcal{H}$$

$$\dim ((\ker A^{k+1})\ominus {\mathcal{W}}_k)$$

$$A({\mathcal{V}}_{k+1})$$

$$A({\mathcal{V}}_2)\subseteq {\mathcal{W}}_1\subseteq \ker A$$

$$\dim {\mathcal{W}}_1=\dim \mathcal{H}=\dim (\ker A\ominus {\mathcal{W}}_1).$$

$$\dim \mathcal{H}\ge \dim ((\ker A^2)\ominus {\mathcal{W}}_1)\ge \dim ((\ker A)\ominus {\mathcal{W}}_1)=\dim \mathcal{H},$$

$$A({\mathcal{V}}_{k+1})+{\mathcal{W}}_{k-1}\subseteq {\mathcal{W}}_k\subseteq \ker A^k$$

$$\dim ({\mathcal{W}}_k\ominus {\mathcal{W}}_{k-1})=\dim \mathcal{H}=\dim ((\ker A^k)\ominus {\mathcal{W}}_k).$$

$$\dim ((\ker A^k\ominus {\mathcal{W}}_{k-1})=\dim \mathcal{H}$$

$$\dim \mathcal{H}\ge \dim ((\ker A^{k+1})\ominus {\mathcal{W}}_k)\ge \dim ((\ker A^k)\ominus {\mathcal{W}}_k)=\dim \mathcal{H}.$$

$$A({\mathcal{W}}_k)\subseteq A(\ker A^k)=A({\mathcal{V}}_1)+\cdots +A({\mathcal{V}}_k)\subseteq {\mathcal{W}}_{k-1}.$$

$$A({\mathcal{H}}_1)$$

$$A({\mathcal{H}}_k)$$

$$A=\sum _{1\le i<j\le n}{a}_{i,j}\otimes e_{i,j}$$

$$B=\sum _{i=1}^{n-1}{b}_i\otimes e_{i,i+1},C=\sum _{2\le i\le j\le n}{c}_{i,j}\otimes e_{i,j}$$

$$a_{1,j}$$

$$a_{i,j}$$

$$b_1{c}_{2,j}$$

$$b_i{c}_{i+1,j}$$

$$A=\sum _{1\le i<j\le n}{a}_{i,j}\otimes e_{i,j}$$

$$c_{2,j}$$

$$c_{i+1,j}$$

$$A=\sum _{1\le i<j\le n}{a}_{i,j}\otimes e_{i,j}$$

$$b_1$$

$$b_i$$

$$b_{n-2}$$

$$b_i^2{r}_{i,j}=a_{i,j}(1\le i\le n-2,i<j\le n).$$

Full text

1. Introduction

By operator ideal we mean a proper, nonzero, two-sided ideal of the algebra $B({\mathcal{H}})$ of bounded operators on a separable, infinite Hilbert space ${\mathcal{H}}$. These ideals consist of compact operators. For a compact operator, $A$ on ${\mathcal{H}}$, let $s(A)=(s_{1}(A),s_{2}(A),\ldots)$ be the sequence of singular numbers of $A$. This is the non-increasing sequence of nonzero eigenvalues of $|A|:=(A^{*}A)^{1/2}$, listed in order of multiplicity, with a tail of zeros in case $A$ has finite rank. As Calkin showed [C41], an operator ideal ${\mathcal{I}}$ is characterized by $s({\mathcal{I}})=\{s(A)\mid A\in{\mathcal{I}}\}$. (See also, e.g., [GK69] or [DFWW04] for expositions). For a positive real number $t$ and an operator ideal ${\mathcal{I}}$, we let ${\mathcal{I}}^{\,t}$ denote the operator ideal generated by $\{|A|^{t}\mid A\in{\mathcal{I}}\}$.

Questions about additive commutators $[B,C]:=BC-CB$ involving elements of operator ideals have been much studied. One of the questions asked in [PT71], by Pearcy and Topping, is whether every compact operator $A$ is a single commutator $A=[B,C]$ of compact operators $B$ and $C$. This question is still open. Important results about single commutators in operator ideals were obtained in [PT71] and by Anderson [A77]. Further results are found in Section 7 of [DFWW04]. More recently, Beltiţă, Patnaik and Weiss [BPW14] have made progress on the above mentioned question.

Our purpose in this note is to show that every nilpotent compact operator is a single commutator of compact operators. In fact, we show (Theorem 3.2) that for a general operator ideal ${\mathcal{I}}$, every nilpotent element $A\in{\mathcal{I}}$ is a single commutator $A=[B,C]$ of $B,C\in{\mathcal{I}}^{\,t}$, where the value of $t>0$ depends on the value of $n$ for which $A^{n}=0$. Except in the case $n\leq 4$, we don’t know if we have found the optimal value of $t$.

2. Preliminaries

Let ${\mathcal{H}}$ be an infinite dimensional Hilbert space. Nothing in this section is new, but we include proofs for convenience.

Lemma 2.1**.**

Suppose $x,y\in B({\mathcal{H}})$ and $t\in{\mathbf{R}}$, $t>0$.

1. (i)

If $t(x^{*}x)\geq y^{*}y$, then there exists $r\in B({\mathcal{H}})$ such that $\|r\|\leq\sqrt{t}$ and $y=rx$. 2. (ii)

If $t(xx^{*})\geq yy^{*}$, then there exists $r\in B({\mathcal{H}})$ such that $\|r\|\leq\sqrt{t}$ and $y=xr$.

Proof.

The assertion (ii) follows from (i) by taking adjoints. If we prove the assertion (i) when $t=1$, then the case of arbitrary $t$ follows, by replacing $x$ with $\sqrt{t}x$. So it will prove (i) in the case $t=1$.

Suppose $x^{*}x\geq y^{*}y$. Given $\xi\in{\mathcal{H}}$, we have

[TABLE]

Thus, we may define a contractive linear operator from $\operatorname{ran}(x)$ into ${\mathcal{H}}$ by

[TABLE]

This extends uniquely to a contractive linear operator, which we call $r_{0}$, from $\overline{\operatorname{ran}(x)}$ into ${\mathcal{H}}$. We have $r_{0}x=y$. Letting $p$ be the orthogonal projection from ${\mathcal{H}}$ onto $\overline{\operatorname{ran}(x)}$, we set $r=r_{0}p$. Thus, $r\in B({\mathcal{H}})$ is a contraction and $rx=y$. ∎

For $n\geq 1$, we make the natural identifications

[TABLE]

and we let $(e_{i,j})_{1\leq i,j,\leq n}$ be the usual system of matrix units in $M_{n}({\mathbf{C}})$.

Recall that ${\mathcal{H}}$ is assumed to be infinite dimensional (and here we do not need to assume it is separable.)

Lemma 2.2**.**

Let $A\in B({\mathcal{H}})$ satisfy $A^{n}=0$. Then there exists a unitary $U:{\mathcal{H}}\to{\mathcal{H}}^{\oplus n}$ such that $UAU^{*}$ is a strictly upper triangular element of $M_{n}(B({\mathcal{H}}))$.

Proof.

We will first show that $\dim\ker A=\dim{\mathcal{H}}$. Consider $B=A|_{\ker A^{2}}$. Note that $\ker A=\ker B$. If $\dim\ker A^{2}=\dim{\mathcal{H}}$, then either $\dim\ker B=\dim{\mathcal{H}}$, or $\dim\operatorname{ran}B=\dim{\mathcal{H}}$. But $\operatorname{ran}B\subset\ker B$, so $\dim\ker A^{2}=\dim{\mathcal{H}}$ implies $\dim\ker A=\dim{\mathcal{H}}$. Since $A$ is nilpotent, we have $\dim\ker A^{2^{k}}=\dim{\mathcal{H}}$, for some $k$. Thus, (arguing by induction on $k$), we must have $\dim\ker A=\dim{\mathcal{H}}$.

Let

[TABLE]

We will construct subspaces

[TABLE]

with

[TABLE]

such that, letting ${\mathcal{W}}_{0}=\{0\}$, we have, for every $1\leq k\leq n$,

[TABLE]

and for every $k\leq n-1$,

[TABLE]

Fixing $k=1$, if $\dim{\mathcal{V}}_{2}=\dim{\mathcal{H}}$, then let ${\mathcal{W}}_{1}=\ker A$. We know $\dim\ker A=\dim{\mathcal{H}}$, so (2) holds. Moreover, $\ker A^{2}\ominus{\mathcal{W}}_{1}={\mathcal{V}}_{2}$, so (3) holds and $A({\mathcal{V}}_{2})\subseteq A(\ker A^{2})\subseteq\ker A$, so (4) holds. Otherwise, if $\dim{\mathcal{V}}_{2}<\dim{\mathcal{H}}$, then choose ${\mathcal{W}}_{1}$ so that

[TABLE]

and

[TABLE]

This choice is possible because we know $\dim\ker A=\dim{\mathcal{H}}$ and by hypothesis $\dim A({\mathcal{V}}_{2})\leq\dim{\mathcal{V}}_{2}<\dim{\mathcal{H}}$. Then (2) and (4) (for $k=1$) hold by construction. We have

[TABLE]

so (3) holds.

Now suppose $2\leq k\leq n-1$ and ${\mathcal{W}}_{1},\ldots,{\mathcal{W}}_{k-1}$ have been constructed with the required properties. If $\dim{\mathcal{V}}_{k+1}=\dim{\mathcal{H}}$, then let ${\mathcal{W}}_{k}=\ker A^{k}$. Then (2) for $k$ is just (3) for $k-1$ while (3) for $k$ is just the hypothesis $\dim({\mathcal{V}}_{k+1})=\dim{\mathcal{H}}$. Moreover, $A({\mathcal{V}}_{k+1})\subseteq A(\ker A^{k+1})\subseteq\ker A^{k}$, so (4) holds for this $k$ as well.

Othewrwise, if $\dim{\mathcal{V}}_{k+1}<\dim{\mathcal{H}}$, then choose ${\mathcal{W}}_{k}$ so that

[TABLE]

and

[TABLE]

This is possible because, by hypothesis (namely, (3) for $k-1$),

[TABLE]

and $\dim(A({\mathcal{V}}_{k+1}))\leq\dim{\mathcal{V}}_{k+1}<\dim{\mathcal{H}}$. Then (2) and (4) hold by construction, while for (3), we use

[TABLE]

Finally, set ${\mathcal{W}}_{n}={\mathcal{H}}=\ker A^{n}$. Then (2) for $k=n$ follows from (3) for $k=n-1$.

Using (4), we get

[TABLE]

Let ${\mathcal{H}}_{1}={\mathcal{W}}_{1}$, and ${\mathcal{H}}_{k}={\mathcal{W}}_{k}\ominus{\mathcal{W}}_{k-1}$, $2\leq k\leq n$. Then $\dim{\mathcal{H}}_{k}=\dim{\mathcal{H}}$ for all $k$ and

[TABLE]

Choosing unitaries $U_{k}:{\mathcal{H}}_{k}\to{\mathcal{H}}$ yields a unitary $U=\oplus_{k=1}^{n}U_{j}:{\mathcal{H}}\to{\mathcal{H}}^{\oplus n}$ so that $UAU^{*}$ is a strictly upper triangular matrix. ∎

Remark 2.3*.*

We work in $B({\mathcal{H}})\otimes M_{n}({\mathbf{C}})$ and suppose

[TABLE]

for $a_{i,j}\in B({\mathcal{H}})$. If

[TABLE]

with $b_{i},c_{i,j}\in B({\mathcal{H}})$, then the condition $A=BC-CB$, is equivalent to

[TABLE]

or, equivalently,

[TABLE]

3. Nilpotents in operator ideals

Let ${\mathcal{I}}$ be an operator ideal. It is well known and easy to see that, under any identification of $B({\mathcal{H}})$ with $M_{n}(B({\mathcal{H}}))$ as in (1), the ideal ${\mathcal{I}}$ is identified with $M_{n}({\mathcal{I}})$.

We first prove the following easy result, whose proof is similar to that of Proposition 3.2 of [DS12].

Proposition 3.1**.**

Let ${\mathcal{I}}$ be an operator ideal and suppose $A\in{\mathcal{I}}$ is nilpotent. Then there exist $B\in B({\mathcal{H}})$ and $C\in{\mathcal{I}}$ such that $A=BC-CB$.

Proof.

Let $n\geq 2$ be such that $A^{n}=0$. By Lemma 2.2, we may work in $B({\mathcal{H}})\otimes M_{n}({\mathbf{C}})$ and suppose

[TABLE]

for $a_{i,j}\in{\mathcal{I}}$. We need only find elements $b_{i}\in B({\mathcal{H}})$ and $c_{i,j}\in{\mathcal{I}}$, as in Remark 2.3, so that (5) and (6) hold. This is easily done by setting $b_{i}=1$ for all $i$ and recursively assigning

[TABLE]

∎

Theorem 3.2**.**

Let ${\mathcal{I}}$ be an operator ideal and suppose $A\in{\mathcal{I}}$ satisfies $A^{n}=0$, for some integer $n\geq 4$. Then there exist $B,C\in{\mathcal{I}}^{1/2^{n-3}}$ such that $A=BC-CB$.

Proof.

By Lemma 2.2, we may work in $B({\mathcal{H}})\otimes M_{n}({\mathbf{C}})$ and suppose

[TABLE]

for $a_{i,j}\in{\mathcal{I}}$. We will find elements $b_{i}$ and $c_{i,j}$ of ${\mathcal{I}}^{1/2^{n-3}}$, as in Remark 2.3, so that (5) and (6) hold.

Step 1: assign values to $b_{1},\ldots,b_{n-2}$. Let

[TABLE]

Since for every $1\leq i\leq n-2$ and every $i<j\leq n$, we have $b_{i}^{4}\geq|a_{i,j}^{*}|^{2}$, by Lemma 2.1 there exists $r_{i,j}\in B({\mathcal{H}})$ such that

[TABLE]

Moreover, for every $2\leq i\leq n-3$, since $b_{i}^{4}\geq b_{i-1}^{2}$, by the same lemma there exists $x_{i}\in B({\mathcal{H}})$ such that

[TABLE]

Furthermore, since $b_{n-2}^{4}\geq b_{n-3}^{4}$ and the square root function is operator monotone, we have $b_{n-2}^{2}\geq b_{n-3}^{2}$. Thus, by Lemma 2.1 there exists $z\in B({\mathcal{H}})$ so that

[TABLE]

Step 2: assign values to $y_{2,j}$ and $c_{2,j}$ for $2\leq j\leq n$ and verify (5). Let

[TABLE]

Thus, $c_{2,j}\in{\mathcal{I}}^{1/2}$. Then we have

[TABLE]

namely, (5) holds.

Step 3: assign values to $y_{p,j}$ and $c_{p,j}$ for $3\leq p\leq n-2$ and $p\leq j\leq n-1$ and verify the equality in (6) for $2\leq i\leq n-3$ and $i<j\leq n-1$. We let $p$ increase from $3$ to $n-2$ and for each such $p$ we define (recursively in $p$) for every $j\in\{p,p+1,\ldots,n-1\}$,

[TABLE]

Thus, $c_{p,j}\in{\mathcal{I}}^{1/2^{p-1}}$ and we have

[TABLE]

and the equality in (6) holds for these values of $i$ and $j$.

Step 4: assign a value to $c_{n-1,n-1}$ and verify the equality in (6) for $i=n-2$ and $j=n-1$. Let

[TABLE]

Then $c_{n-1,n-1}\in{\mathcal{I}}^{1/2^{n-3}}$ and

[TABLE]

Thus, the equality in (6) holds for $i=n-2$ and $j=n-1$.

Step 5: assign a value to $b_{n-1}$. Let

[TABLE]

Then $b_{n-1}\in{\mathcal{I}}^{1/2^{n-3}}$. Since $b_{n-1}^{4}\geq|a_{n-1,n}^{*}|^{2}$, by Lemma 2.1 there is $r_{n-1,n}\in B({\mathcal{H}})$ so that

[TABLE]

Since $b_{n-1}^{4}\geq|c_{n-1,n-1}^{*}|^{4}$, we have $b_{n-1}^{2}\geq|c_{n-1,n-1}^{*}|^{2}$ and, from Lemma 2.1, we have $s\in B({\mathcal{H}})$ so that

[TABLE]

Step 6: assign values to $c_{p,n}$ for all $3\leq p\leq n-2$ and verify the equality in (6) for all $2\leq i\leq n-3$ and $j=n$. Let

[TABLE]

Then $c_{p,n}\in{\mathcal{I}}^{1/2^{p-1}}$ and

[TABLE]

namely, the equality in (6) holds for these values of $i$ and for $j=n$.

Step 7: assign a value to $c_{n-1,n}$ and verify the equality in (6) for $i=n-2$ and $j=n$. Let

[TABLE]

Then $c_{n-1,n}\in{\mathcal{I}}^{1/2^{n-3}}$ and

[TABLE]

namely, the equality in (6) holds for $i=n-2$ and for $j=n$.

Step 8: assign a value to $c_{n,n}$ and verify the equality in (6) for $i=n-1$ and $j=n$. Let

[TABLE]

Then $c_{n,n}\in{\mathcal{I}}^{1/2^{n-3}}$ and

[TABLE]

as required. ∎

Corollary 3.3**.**

Let ${\mathcal{I}}$ by any operator ideal such that ${\mathcal{I}}^{\,t}\subseteq{\mathcal{I}}$ for every $t>0$. Then for every nilpotent element $A$ of ${\mathcal{I}}$, there exist $B,C\in{\mathcal{I}}$ such that $A=BC-CB$.

Examples of operator ideals ${\mathcal{I}}$ satisfying the conditions of Corollary 3.3 include

1. (a)

the ideal ${\mathcal{K}}$ of all compact operators; 2. (b)

the ideal of all operators $A$ whose singular numbers have polynomial decay: $s_{n}(A)=O(n^{-t})$ for some $t>0$; note that this ideal is equal to the union of all Schatten $p$-class ideals, $p\geq 1$; 3. (c)

the ideal of all operators $A$ whose singular numbers have exponential decay: $s_{n}(A)=O(r^{n})$ for some $0<r<1$; 4. (d)

the ideal of all finite rank operators.

Question 3.4**.**

Is $1/2^{n-3}$ the optimal exponent of ${\mathcal{I}}$ in Theorem 3.2? Clearly, the answer is yes when $n=4$. But as far as we know, it is possible that the best exponent is $1/2$ for arbitrary $n$.