1
$\begingroup$

Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Definition : Let $(e_{n})_{n \in \mathbb{N}}$ be an orthonormal basis.
$T \in B(H)$ is banded if $\exists r \in \mathbb{N}$ such that $ (Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$.

Definition : A $C^{*}$-algebra is exact if it preserves exact sequences under the minimum tensor product.

Property : A $C^{*}$-algebra is exact if and only if :

  • it's nuclearly embeddable into $B(H)$.
  • it's isomorphic to a subalgebra of the Cuntz algebra $\mathcal{O}_2$.

Definition : An operator $T \in B(H)$ is exact if it generates an exact $C^{*}$-algebra $C^{*}(T)$.

The following is known by N. Ozawa (see here) : if $T \in B(H)$ unitary equivalent to a banded operator, then $T$ is exact. What about the converse ?

Is an exact operator, unitary equivalent to a banded operator ?

Improve this question
$\endgroup$
13
  • 2
    $\begingroup$ What happens for the operator $T$ given by multiplication on the coordinate on $L^2[0,1]$? Is there an orthonormal basis making this banded? $\endgroup$
    – Matthew Daws
    Commented Aug 7, 2013 at 10:46
  • 1
    $\begingroup$ No: let $H=L^2[0,1]$ and for $\xi\in L^2[0,1]$ define $T$ to be the multiplication operator $T(\xi)(t) = t\xi(t)$. Then $C^*(T) \cong C[0,1]$ is commutative... $\endgroup$
    – Matthew Daws
    Commented Aug 7, 2013 at 11:27
  • 2
    $\begingroup$ Taking one of the two standard generators of $\mathcal O_2$, is this unitarily equivalent to a banded operator? I suspect not, and this would show that whether or not an operator $T$ is unitarily equivalent to a banded operator does not just depend on the abstract C*-algebra $C^*(\{T\})$ together with the placement of $T$ in this C*-algebra. (Since the C*-algebra of the one-sided shift generates the same C*-algebra as one of the Cuntz algebra generators.) $\endgroup$
    – Aaron Tikuisis
    Commented Aug 8, 2013 at 9:36
  • 1
    $\begingroup$ @MatthewDaws's example is surely the first thing that should have been tried... $\endgroup$
    – Yemon Choi
    Commented Aug 9, 2013 at 16:25
  • 1
    $\begingroup$ Also, surely it is more natural to look not just at banded operators, but at operators which are limits (in norm topology) of banded operators. This would correspond to the Roe algebra of Z if I remember correctly $\endgroup$
    – Yemon Choi
    Commented Aug 9, 2013 at 16:29

0

Reset to default

You must log in to answer this question.